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Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdot i.$
We now need to find the sum $S=x_1+x_2$ and the product $P=x_1\cdot x_2$ to form an equation $x^2-Sx+P=0$ which has the solutions $x_1$ and $x_2$.
\begin{align*}
sum = x_1+x_2&=6m \tag{1}\\
\iff(r+n\cdot i)+(r-n\cdot i)=2r&=6m \tag{2}\\
\implies r&=3m \tag{3}
\end{align*}
The $product = x_1\cdot x_2=(r+n\cdot i)(r-n\cdot i)=r^2+n^2 = (\cdots).\quad $
From here I don't know what relation to find between $\Delta$ and $x_1\cdot x_2.$
I tried: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$
| Alternative approach:
Without loss of generality, the equation is
$x^2 + Bx + C = 0.$
Since $x_1 + x_2 = 6m$,
and since you must have that
$(x - x_1) \times (x - x_2) = x^2 + Bx + C$,
you must have that
$B = -6m.$
Here, there is some ambiguity involved. Taking the constraint of (in effect)
$B^2 - 4C = -36$
at face value, you have that
$\displaystyle B^2 + 36 = 4C \implies C = \frac{36m^2 + 36}{4} = 9m^2 + 9.$
Therefore, the quadratic equation is
$\displaystyle x^2 + [-6m]x + \left[9m^2 + 9\right] = 0.$
I mentioned a possible ambiguity.
Generally, for an equation of the form $Ax^2 + Bx + C = 0$, whose solutions are given by
$\displaystyle \frac{1}{2A} \left[-B \pm \sqrt{B^2 - 4AC}\right]$
it is unclear whether the discriminant should be considered to be
$$ B^2 - 4AC ~~~\text{or}~~~ \frac{B^2 - 4AC}{4A^2}. \tag1 $$
In (1) above, I went with the LHS, on a guess.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Rolling a die - Conditional Probability A die is thrown repeatedly.
Let $X$ ~ First 5 is thrown and $Y$ ~ First 6 is thrown
Calculate $\mathbb{E}(X|Y=3)$
You may use the identity: $\sum_{n=k}^\infty nz^{n-k} = \frac{1}{(1-z)^2}+\frac{k-1}{1-z}$
I know from the definition of expectation, we have:
$\mathbb{E}(X|Y=3) = (1*\frac{1}{5})+(2*\frac{4}{5} * \frac{1}{5}) + (3* \frac{4}{5}* \frac{4}{5} * 0) + (5* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac{1}{6}) + (6* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac {5}{6} * \frac{1}{6}) + ...$, where every following term, has an extra '$*\frac{5}{6}$' term and constant increases by 1.
However I am unsure of how to apply this to the identity given to find the value of the infinite sum?
| While it is not necessary, if the question expects you to use the given identity then here is how you would go about it -
$ \small \displaystyle E(X|Y = 3) = 1 \cdot \frac{1}{5} + 2 \cdot \frac{4 \cdot 1}{5^2} + 4 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{1}{6} + 5 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{5 \cdot 1}{6^2} + ...$
$ \small \displaystyle = \frac{13}{25} + \frac{4 \cdot 4}{5^2} \cdot \frac{1}{6} \left (4 \left(\frac{5}{6}\right)^0 + 5 \cdot \left(\frac{5}{6}\right)^1 + ...\right)$
Now note that comparing the below against the given identity,
$ \small \displaystyle 4 \left(\frac{5}{6}\right)^0 + 5 \cdot \left(\frac{5}{6}\right)^1 + ...$
$ \displaystyle \small z = \frac{5}{6}, k = 4$ and $\sum \limits_{n=k}^\infty nz^{n-k} = 54$
and we get, $ \displaystyle \small E(X|Y = 3) = \frac{157}{25}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Express an integer $m$ as sum of 1, 0 and -1 for fixed number of summands $j$ Let $j \in \mathbb{N}_0$. I'm looking for the number of all possible combinations, such that for a given $|m| \leq j, m \in \mathbb{Z}$
$m = \sum \limits_{k=1}^{j} a_k\quad$ where $a_k \in {-1, 0, 1}$
Examples:
$j=3;m=3 \Rightarrow a \in A=\left\{\begin{pmatrix}1\\1\\1\end{pmatrix}\right\}\quad \#A = 1$
$j=3;m=2 \Rightarrow a \in A=\left\{\begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix}\right\}\quad\#A=3$
$j=3;m=1 \Rightarrow a \in A=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix},\begin{pmatrix}1\\1\\-1\end{pmatrix},\begin{pmatrix}1\\-1\\1\end{pmatrix},\begin{pmatrix}-1\\1\\1\end{pmatrix}\right\}\quad \#A = 6$
$j=3;m=0 \Rightarrow a \in A=\left\{\begin{pmatrix}0\\0\\0\end{pmatrix},\begin{pmatrix}1\\-1\\0\end{pmatrix},\begin{pmatrix}1\\0\\-1\end{pmatrix},\begin{pmatrix}0\\1\\-1\end{pmatrix},\begin{pmatrix}0\\-1\\1\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix},\begin{pmatrix}-1\\1\\0\end{pmatrix}\right\}\quad \#A = 7$
Is there some formula that relates $\#A$ to $m$ and $j$?
| I'll use $n$ instead of $j$. We can assume $m$ is nonnegative, since replacing $m$ with $-m$ gives the same answer (by negating all the solutions).
We can count the solutions in groups according to how many $1$s there are. In order for the sum $m$ to be possible, we need at least $m$ $1s$ and at most $\left\lfloor (m+n)/2\right\rfloor$ of them. Once we've chosen to use $k$ of them, we can then choose where they go ($k$ out of $n$ positions) and then where the $-1$s go ($k-m$ out of $n-k$ positions). The total number of solutions is therefore
$$\sum_{k=m}^{\left\lfloor (m+n)/2\right\rfloor}\binom nk\binom{n-k}{k-m}.$$
We could also just let the sum run from $k=0$ to $n$ and let $\binom ab=0$ whenever $b\notin[0,a]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both sides of the equation and continue from there, I tried by solving the equation:
$(x^3+x)(ax^3+bx^2+cx+d) = 1$
But the solution I get is that $a=0,b=1,c=1,d=0$ which doesn't produce the inverse element.
What is the inverse element of $(1010)$ in this field and how can I arrive at that solution?
Any help is greatly appreciated.
| Here are some methods to compute this using a computer. The software is sagemath.
First, if the program supports defining finite fields with a given polynomial, you can just use that:
K.<a> = GF(2^4, modulus=x^4 + x^3 + 1)
(a^3 + a)^-1
=> a^3 + a + 1
You can also use the extended GCD which gives polynomials $u, v$ such that $uf + vg = 1$ so, modulo $f$, that gives $v = g^{-1}$.
K = GF(2)[x]
xgcd(K(x^4 + x^3 + 1), K(x^3 + x)) # K(...) means "as an element of K"
=> (1, x^2 + x + 1, x^3 + x + 1)
So $(x^3 + x + 1)(x^4 + x^3 + 1) + (x^3 + x + 1)(x^3 + x) = 1$.
You can also do it "by hand" using companion matrices. If $x^4 + ax^3 + bx^2 + cx + d$ is the minimal polynomial of your generator $\alpha$, then form the matrix
$$A = \begin{pmatrix}
0 & 0 & 0 & -d \\
1 & 0 & 0 & -c \\
0 & 1 & 0 & -b \\
0 & 0 & 1 & -a
\end{pmatrix}$$
In this way, $A$ also has that minimal polynomial. More specifically, each column represents $\alpha \cdot t$ for $t \in \{1, \alpha, \alpha^2, \alpha^3\}$. Now you can compute $(A^3 + A)^{-1}$, and again the columns represent $(\alpha^3 + \alpha)^{-1} \cdot t$ for $t$ in that ordered basis. Of course, you want $t = 1$ to find $(\alpha^3 + \alpha)^{-1}$.
Here $(a,b,c,d) = (1,0,0,1)$ and
$$(A^3 + A)^{-1} = \begin{pmatrix}
1&1&1&0\\
1&1&1&1\\
0&1&1&1\\
1&1&0&1
\end{pmatrix}.$$
Thus $(\alpha^3 + \alpha)^{-1} = 1 + \alpha + \alpha^3$ by reading the first column.
You can also just solve the linear system $(A^3 + A)u = (1,0,0,0)$ rather than the full inverse $(A^3 + A)U = I$. This works out to be the same linear system as given in GEdgar's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the sum of geometric progression Evaluate the sum:
$$\sum_{x=0}^\infty x(x-1) {2+x \choose x}(0.008)(0.8)^x $$
I was able to make this into:
$$0.004\sum_{x=0}^\infty x(x-1) (x+1)(x+2)(0.8)^x $$
Let $x=n-2$ then $n=x+2$:
$$0.004\sum_{n=2}^\infty n(n-1)(n-2)(n-3)(0.8)^{n-2} $$
$$0.004\sum_{n=4}^\infty (n)_3~(0.8)^{n-2} $$
Let $j=n-4$ then $n=j+4$:
$$0.004\sum_{j=0}^\infty (j+4)_3~(0.8)^{j+2} $$
$$0.00256\sum_{j=0}^\infty (j+4)_3~(0.8)^{j} $$
I thought I could just use the formula for geometric sum on this but I know the answer's $192 $ and right now my solution's pretty far off.
| There are a couple ways to go about this. An important identity is that $\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{n+k-1}{k-1}x^n$ for $|x|<1$. This can be proven with just binomial theorem and the negative binomial coefficients. It can also be proven by noting that $\frac{1}{(1-x)^k}=(1+x+x^2+\ldots)^k$. The coefficient of $n$ is $\binom{n+k-1}{k-1}$ by stars-and-bars.
Method 1
Let's say we wanted to find the value of
$$g(x)=\sum_{n=0}^\infty n(n-1)\binom{2+n}{n} x^n$$
Consider the function $f(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{n}x^n$. We know that
$$f''(x)=\frac{(3)(4)}{(1-x)^5}=\sum_{n=0}^\infty n(n-1)\binom{n+2}{n}x^{n-2}$$
So
$$g(x)=x^2f''(x)=\frac{12x^2}{(1-x)^5}$$
Our desired expression is the value of $.008g(.8)=\frac{12(.008)(.8^2)}{(.2)^5}=\frac{12(8^3)}{2^5}=\boxed{192}$
Method 2
This aligns a little more closer with your original method. You already showed that the desired expression is equivalent to
$$.004\sum_{n=0}^\infty (n+2)(n+1)(n)(n-1)(.8)^n$$
$$=.004(4!)\sum_{n=0}^\infty \binom{n+2}{4}(.8)^n$$
$$=.004(4!)\sum_{n=0}^\infty \binom{n+2}{4}(.8)^n$$
Note that $\binom{n+2}{4}=0$ for $n=0,1$, so this sum is equivalent to
$$=.004(4!)\sum_{n=2}^\infty \binom{n+2}{4}(.8)^n$$
$$=.004(4!)\sum_{n=2}^\infty \binom{(n-2)+4}{4}(.8)^n$$
We can shift the indices down by $2$ to see that this is equivalent to
$$=.004(4!)\sum_{n=0}^\infty \binom{n+4}{4}(.8)^{n+2}$$
$$=.004(.8)^2(4!)\sum_{n=0}^\infty \binom{n+4}{4}(.8)^n$$
From our initial identity, we know that $\sum_{n=0}^\infty \binom{n+4}{4}(.8)^n=\frac{1}{(.2)^5}$, so our expression is equivalent to
$$=\frac{.004(.8)^2(4!)}{(.2)^5}$$
$$=\frac{4(8)^2(4!)}{2^5}$$
$$=\boxed{192}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
The statement tells us that $x\in[5,9]$. Also,
$$\sqrt{x-5}\gt1+\sqrt{9-x}$$
Since both sides are positive, we can square
$$x-5>1+9-x+2\sqrt{9-x}\\2x-15\gt2\sqrt{9-x}$$
$\implies 2x-15\gt0\implies x\gt7.5$
Since $x\in\mathbb Z\implies x=8,9$
But on back substitution, $x=8$ doesn't satisfy. Is there a way we could get the final answer without back sustitution?
| Hint:
Let $\sqrt{x-5}=a\ge0$ and $\sqrt{9-x}=b\ge0$
$$\implies a^2+b^2=4$$
and $a-b>1\iff a> b+1$
$$4=a^2+b^2>(b+1)^2+b^2\iff 2b^2+2b-3<0$$
Now for $(x-a)(x-b)<0, a<b;$ $$a<x<b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to? Two circles whose centres lie on the x axis, whose radii are $\sqrt2cm$ and $1cm$ and whose centres are 2 cm apart intersect at a point A.The chord AC of the larger circle cuts the smaller circle at a point B and is bisected by that point. What is the length of chord AC?
.
My attempt :
.
.
As shown in the diagram above I started by assuming the centre of smaller circle S1 (Of radius 1) to be origin and the centre of the larger circle S2 (Of radius $\sqrt2$ ) to be $(2,0)$ as the centres are separated by 2 units.
I then solved :
S1 : $ x^2 + y^2 = 1 $ ; and
S2 : $ (x-2)^2 + y^2 = 2 $ to obtain $A(\frac{3}4 , \frac{\sqrt7}4)$
I noted the following equations:
*
*Since B is given to be mid point of chord AC:
XB = $\frac{X_A + X_C}2$ ; $Y_B = \frac{Y_A +Y_C}2$
*Since C lies on $S_2$ :
$ (X_C-2)^2 + Y_C^2 = 2 $
*The distance of B from origin is 1 unit :
$(X_B-0)^2 + (Y_B-0)^2 = 1$
$(\frac{X_A + X_C}2 - 0)^2$ +$(\frac{Y_A +Y_C}2 - 0)^2 = 1$
$(\frac{\frac{3}4 + X_C}2 - 0)^2$ +$(\frac{\frac{\sqrt7}4 +Y_C}2 - 0)^2 = 1$
This equation and equation generated in point 2 together are two equations in two variables and i should be able to solve them to get the co-ordinate of C. This however is proving to be cumbersome.
.
Is there a better way to avoid this approach.
| Drop a perp from $C_1$ to $AB$ and extend. Then $D$ is the midpoint of $AB$. Drop a perp from $C_2$ to $C_1D$ extend.
If $AC = 4a$, $C_2E = BD = a$ and $C_1E = C_1D + C_2B$
$C_1D = \sqrt{1-a^2}, C_2B = \sqrt{2 - 4a^2}$
Using Pythagoras in $\triangle C_1EC_2$,
$ (\sqrt{1-a^2} + \sqrt{2 - 4a^2})^2 + a^2 = 4$
$ 3 - 4a^2 + 2 \sqrt{1-a^2} \sqrt{2 - 4a^2} = 4$
$ 4 (1-a^2) (2 - 4a^2) = (1 + 4a^2)^2$
$ 8 - 24a^2 + 16a^4 = 1 + 16 a^4 + 8 a^2$
$ \displaystyle 32a^2 = 7 \implies AC = 4a = \sqrt{\frac{7}{2}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
relation between roots and coefficient in a cubic polynomial If $\alpha,\beta,\gamma$ are roots of the cubic equation
$$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.
I have $\alpha+\beta+\gamma$= - $\frac{3}{2}$.
Therefore roots become $\frac{\alpha}{-\frac{3}{2} -\alpha}$,$\frac{\beta}{-\frac{3}{2} -\beta}$, $\frac{\gamma}{-\frac{3}{2} -\gamma}$.
To find the equation we have to find their sums and products, but it looks like complicated.
Is their any easy trick ?
| If
$$
2x^3+3x^2-x-1=0
$$
has roots $\alpha,\beta,\gamma$, then substituting $x\mapsto\frac1x$ (and multiplying by $-x^3$ to clear denominators)
$$
\begin{align}
&-x^3\left(\frac2{x^3}+\frac3{x^2}-\frac1x-1\right)\\
&=x^3+x^2-3x-2=0
\end{align}
$$
has roots $\frac1\alpha,\frac1\beta,\frac1\gamma$. Then substituting $x\mapsto-\frac23x$ (and multiplying by $-\frac{27}8$ to clear denominators)
$$
\begin{align}
&-\frac{27}8\left(\left(-\frac23x\right)^3+\left(-\frac23x\right)^2-3\left(-\frac23x\right)-2\right)\\
&x^3-\frac32x^2-\frac{27}4x+\frac{27}4=0
\end{align}
$$
has roots $\frac{-3/2}\alpha=\frac{\alpha+\beta+\gamma}\alpha,\frac{-3/2}\beta=\frac{\alpha+\beta+\gamma}\beta,\frac{-3/2}\gamma=\frac{\alpha+\beta+\gamma}\gamma$. Next, substituting $x\mapsto x+1$
$$
\begin{align}
&(x+1)^3-\frac32(x+1)^2-\frac{27}4(x+1)+\frac{27}4\\
&=x^3+\frac32x^2-\frac{27}4x-\frac12=0
\end{align}
$$
has roots $\frac{\beta+\gamma}\alpha,\frac{\alpha+\gamma}\beta,\frac{\alpha+\beta}\gamma$. Finally, substituting $x\mapsto\frac1x$ (and multiplying by $-4x^3$ to clear denominators)
$$
\begin{align}
&-4x^3\left(\frac1{x^3}+\frac32\frac1{x^2}-\frac{27}4\frac1x-\frac12\right)\\
&=2x^3+27x^2-6x-4=0
\end{align}
$$
has roots $\frac\alpha{\beta+\gamma},\frac{\beta}{\alpha+\gamma},\frac{\gamma}{\alpha+\beta}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the ODE $y'=1-\frac{y}{x}$ I substituted $u=\frac{y}{x}$ then tried to solve the ODE $$\frac{u'}{2u-1}= -\frac{1}{x}$$ and I came this far $$\frac{1}{2}\ln |{2u-1}|=- \ln |{x}| + c_1$$ but then in the solution there was the step $$c_1=\ln c_2 \in \mathbb{R}, c_2 > 0$$ to get $$\ln |2u-1|=\ln{(\frac{c_2}{x})^2}$$ but why do we have this additional step? Couldn't we just calculate the solution without this step?
| $y' = 1 - \dfrac{y}{x}; \tag 1$
$y' + \dfrac{y}{x} = 1; \tag 2$
$xy' + y = x; \tag 3$
$(xy)' = y + xy'; \tag 4$
$(xy)' = x; \tag 5$
$xy = \dfrac{x^2}{2} + C, \; C \; \text{the arbitrary constant}; \tag 6$
$y = \dfrac{x}{2} + \dfrac{C}{x}; \tag 7$
We Check:
$y' = \dfrac{1}{2} - \dfrac{C}{x^2}; \tag 8$
$\dfrac{y}{x} = \dfrac{1}{2} + \dfrac{C}{x^2}; \tag 9$
$1 - \dfrac{y}{x} = \dfrac{1}{2} - \dfrac{C}{x^2} = y'! \; \checkmark \tag{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ Solve the equation $$(2x^2-3x+1)(2x^2+5x+1)=9x^2$$
The given equation is equivalent to $$4x^4+4x^3-11x^2+2x+1=9x^2\\4x^4+4x^3-20x^2+2x+1=0$$ which, unfortunately, has no rational roots. What else can I try?
| Hint:
As $x\ne0$ we can divide both sides by $x^2$ to find
$$(a-3)(a+5)=9\iff a^2+2a-24=0$$ where $2x+\dfrac1x=a$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $x^4-8x^3+23x^2-30x+15=0$ Solve the equation $$x^4-8x^3+23x^2-30x+15=0$$
As $x=0$ is obviously not a solution, we can consider $x\ne0$, so I have tried to divide both sides by $x^2$ to get $$x^2-8x+23-\dfrac{30}{x}+\dfrac{15}{x^2}=0$$ Clearly this does not help. I have also tried to find the rational roots using Horner's method. It seems that the polynomial in the left-hand side does not have rational roots.
| Since you tried the rational root theorem and didn't find any rational roots, then next thing to try is factoring as a product of two quadratics. Set your polynomial equal to
$$(x^2+ax+b)(x^2+cx+d)$$ $$=x^4 + (a+c)x^3+(d+ac+b)x^2+(ad+bc)x+bd$$
and equate coefficients. Since $bd=15$ you can try $b=3$, $d=5$ and see what $a$ and $c$ turn out to be. If that doesn't work, try $b=1$, $d=15$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Closed-form for $\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm looking for a closed-form for the following sum:
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $|m|>1.$
In a previous question of mine, the following similar sum was determined:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)=\frac{1}{2}+\frac{1}{2} \log \left|\cot\left(\frac{\pi}{2m}\right)\right|+\frac{m}{2\pi} \left(\frac{1}{2}\text{Cl}_2\left(\frac{2\pi}{m}\right)-2\text{Cl}_2\left(\frac{\pi}{m}\right)\right)$$
for $|m|>1,$ where $\text{Cl}_2$ is the Clausen function of order 2.
I determined the following closed-forms for example as mentioned in that question:
$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
I suspect a similar method to that used by @skbmoore involving the Barnes G function might be applicable. Any help would be much appreciated.
Here is my attempt:
$$S(m) := \sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right) = \frac{m}{2} \log \left( \prod_{n=1}^{\infty} \frac{1}{e} \left(\frac{1+1/(m n)}{1-1/(m n)}\right)^{n}\right)\\ = \frac{1}{2} + \frac{m}{2} \zeta' \left(-1,1-\frac{1}{m}\right)-\frac{m}{2}\zeta' \left(-1,1+\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1-\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1+\frac{1}{m}\right) \\ =\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) \\+ \frac{1}{2} \ln\left( \Gamma \left(1-\frac{1}{m}\right)\Gamma \left(1+\frac{1}{m}\right)\right) - \frac{1}{2}\ln (2\pi) \\=\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) + \frac{1}{2} \ln\left( \frac{\pi}{m} \csc \left(\frac{\pi}{m}\right)\right) - \frac{1}{2}\ln (2\pi)$$
However, I would like to write the solution in terms of the Clausen function if possible, like in the alternating sum, but I cannot see how to do that.
| EDIT: There was a question asked a long time ago about the related infinite series $$\sum_{n=1}^{\infty} \left( n \operatorname{arccot}(n)-1 \right). $$
The approach used in the accepted answer is similar to the approach I used for this question.
For $|x| >1 $, the inverse hyperbolic cotangent function has the Laurent series representation $$\operatorname{arcoth}(x) = \frac{1}{2} \log \left(\frac{1+ \frac{1}{x}}{1- \frac{1}{x}} \right) = \frac{1}{2} \log \left[ \left(1+ \frac{1}{x} \right) - \log\left(1- \frac{1}{x} \right) \right]= \sum_{n=0}^{\infty}\frac{1}{(2n+1)x^{2n+1}}. $$ And for $0 < |x| < 1$, $\cot (\pi x)$ has the Laurent series representation $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}. $$
Therefore, for $m>1$, we have $$ \begin{align} \sum_{n=1}^{\infty} \left(mn \operatorname{arcoth} (mn)-1 \right) &= \sum_{n=1}^{\infty} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1) (mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{(2k+1)(mn)^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{1}{(2k+1)m^{2k}} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{\zeta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \, \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x \cot (\pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{m x \log(\sin \pi x)}{2}\Bigg|_{0}^{1/m} + \frac{m}{2} \int_{0}^{1/m} \log(\sin \pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi} \int_{0}^{2 \pi /m} \log \left(\sin \frac{ u}{2}\right) \, \mathrm du \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi}\int_{0}^{2 \pi /m} \log \left(2 \sin \frac{u}{2}\right) \, \mathrm du - \frac{\log 2}{2} \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} - \frac{m \operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{4 \pi } - \frac{\log 2}{2}. \end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\underbrace{\int\frac{\sqrt{1-x}}{x} d x}_{K}\right]
$$
To deal with $J$, we use rationalization instead of substitution. $$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=2 \int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=2 \int \frac{d(\sqrt{1+x})}{x}+2 \sqrt{1+x} \\
&=2 \int \frac{d(\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+2 \sqrt{1+x} \\
&=\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| +2 \sqrt{1+x}+C_{1}
\end{aligned}
$$
$\text {Replacing } x \text { by } -x \text { yields }$
$$
\begin{array}{l} \\
\displaystyle K=\int \frac{\sqrt{1-x}}{-x}(-d x)=\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+2 \sqrt{1-x}+C_{2}
\end{array}
$$
Now we can conclude that $$
I=\sqrt{1+x}+\sqrt{1-x}+\frac{1}{2}\left(\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|\right)+C
$$
My question is whether there are any simpler methods such as integration by parts , trigonometric substitution, etc…
Please help if you have. Thank you for your attention.
| The answer is no.
There are only two different representations of the result at all.
$ 1/2 (2 \sqrt{1-x}+2 \sqrt{x+1}+ln(\sqrt{1-x}-1)-ln(\sqrt{1-x}]+1)+ln(\sqrt{x+1}-1)-ln(\sqrt{x+1}+1))+constant$
or alternatively:
$ \sqrt{1-x}+\sqrt{x+1}+1/2 ln(((\sqrt{1-x}-1) (\sqrt{x+1}-1))/((\sqrt{1-x}+1) (\sqrt{x+1}+1)))+constant$
This can be rewritten as:
$\sqrt{1-x}+\sqrt{x+1}-tanh^{-1}(\sqrt{1-x})-tanh^{-1}(\sqrt{x+1})+constant$
All valid on the domains where the given function is valid and the individual terms are valid for example real.
There is no other way to integrate the given function. Like the example by Etemon and J.G. shows it is really important to obey for each of the substitutions the domains on/in which they are valid. There are several branches that do not lead to a proper integration.
The integration path is:
(1) $u=\sqrt{x+1}$
(2) $s=u+1$
(3) $p=u-1$
(4) $w=\sqrt{1-x}$
(5) $v=w+1$
(6) $z=w-1$
And the long divisions in between. So written this in length is much longer if done correctly than Your path. Look for example in the step-by-step solution of Wolfram Alpha or Maplesoft Online. Which are both free or the given one by Mikasa. These are standardized digital step-by-step solution conform with suggestions of the NIST. NIST digests are targeted for efficiency and accurateness. These are usually unique integration solution in step-by-step format.
There is only the identity between $ln$ and $arctanh$ that can be used. Six steps and backsubstitution is really long and tedious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$ Source: https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
Let $x,y,z>0,xyz=1$ then $$
\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1 \ \ \ \ (1)
$$
and$$
\sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1\ \ \ (2)
$$
For this type of inequalities, I usually use local inequalities$$
x,y,z>0,xyz=1,\sum_{cyc}{\frac{1}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x^2}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x+1}{x^2+x+1}}\leqslant 2
$$
For (1) it's smooth,because$$
\frac{x^4+1}{x^8+16 x^4+1}-\left(\frac{1}{x^2+x+1}\right)^2=\frac{(x-1)^2 x \left(2 x^4+7 x^3+14 x^2+7 x+2\right)}{\left(x^2+x+1\right)^2 \left(x^8+16 x^4+1\right)} \geqslant 0$$
Then it's easy to prove.
But for (2) I get
$$\frac{x^2+1}{4 x^4+10 x^2+4}-\left(\frac{x+1}{2 \left(x^2+x+1\right)}\right)^2=\frac{(x-1)^2 x^2}{4 \left(x^2+2\right) \left(x^2+x+1\right)^2 \left(2 x^2+1\right)} \geqslant 0$$
Then $$
\sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\geqslant \sum_{cyc}{\frac{1}{2}\frac{\sqrt{x}+1}{x+\sqrt{x}+1}}\leqslant 1
$$
Failed. How can I prove (2)? Any solution is welcome.
| Hint :
For $x\leq 0$ we have :
$$\frac{d}{dx}\left(\frac{d}{dx}f\left(e^{x}\right)\right)<0$$
Where :
$$f(x)=\sqrt{\frac{x+1}{4x^{2}+4+10x}}$$
Next we show that for $x>0$ and $n= 3$ a natural number :
$$f\left(x\right)+\left(n-1\right)f\left(\frac{1}{x^{\frac{1}{n-1}}}\right)-nf\left(1\right)\leq 0$$
In the case $n=3$ we can use the bound given in the link https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
We have for $x>0$:
$$\frac{1}{1+\sqrt{1+3x}}-\sqrt{\frac{x+1}{4x^{2}+4+10x}}\geq 0$$
To show this inequality we use the substitution $y^2=3x+1$ then square both side then clearing the denominator and use a factorization .
So we need to show :
$$\frac{1}{1+\sqrt{1+3x}}+2\frac{1}{1+\sqrt{1+3x^{-0.5}}}-3\frac{1}{1+\sqrt{1+3}}\leq 0$$
Wich is easy !
So now we can apply the LCF corollary (first reference) and conclude .
Last remark : We have also the case $n=4$ for $a_i\in(e^{-1},e^{1})$ such that $\prod_{i=1}^{4}a_i=1$.
Reference :
*
*Cirtoaje and Baiesu: An extension of Jensen’s discrete inequality to half convex functions.
Journal of Inequalities and Applications 2011 2011:101.
2)https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
Edit 11/02/2022 :
For $0<x\leq 1$ we have :
$$\frac{\left(x+2\right)\left(x+1\right)}{\frac{11}{2}x^{2}+4+\frac{17}{2}x}\geq f(x)$$
On the other hand we have $x\geq 1$ :
$$h\left(x\right)=\frac{2}{3}-\frac{1}{1.5x^{-0.5}+1.5}\geq f(x)$$
Edit 12/02/2022 :
We can do better on $x\in (0,1]$ we have :
$$r(x)=\frac{\frac{1}{2}x^{2}+\frac{7}{2}x+2}{4x^{2}+4+10x}\geq f(x)$$
Using the same approach as above we have on $(-\infty,0]$ :
$$\frac{d}{dx}\left(\frac{d}{dx}r\left(e^{x}\right)\right)<0$$
So we have for $0<a\leq 1$ and $0<b\leq 1$ and $c\geq 1$ :
$$f(a)+f(b)+f(c)\leq 2r(\sqrt{ab})+h(c)=2r(\sqrt{ab})+h(1/ab)\leq 1$$
Wich is smooth .
For the other case I use Buffalo's ways see Wolfram alpha
*I made a mistake I solve another problem ...
Here you can find the solution for the other case wich use Buffalo's way
See Wolfram alpha (2).
| {
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"url": "https://math.stackexchange.com/questions/4346877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Let $X$ be a random variable with Cauchy distribution, compute the density function of $Y=\frac{1}{1+X^2}$
Let $X$ be a random variable with Cauchy distribution, if $$Y=\frac{1}{1+X^2}$$ then compute the density function of $Y$
I have tried to use some random variable transformation theorem using the functions $\frac{1}{1+X}$ and $X^2$ but I have not been able to get very far, any suggestion or help would be greatly appreciated.
| The problem can also be solved via the change of the variable in PDF.
$$X:\,f(x)=\frac{1}{\pi}\frac{1}{1+x^2}; \,\,Y=\frac{1}{1+X^2}; \,\,y=f(x);\,\,g(y)-?$$
$$1+x^2=\frac{1}{f(x)}\,\Rightarrow x=\sqrt{\frac{1-f(x)}{f(x)}}$$
$$ dx=-\frac{1}{2}\bigg(\frac{1}{\sqrt{1-f(x)}\sqrt f(x)}+\frac{\sqrt{1-f(x)}}{(\sqrt f(x))^3}\bigg)df=-\frac{1}{2}\frac{df}{\sqrt{f(x)}\sqrt{1-f(x)}f(x)}$$
Therefore, after changing the variable, the PDF looks ($y=f(x)$)
$$\frac{dx}{\pi}\frac{1}{1+x^2}=-\frac{1}{2\pi}\frac{f(x)\,df}{\sqrt{f(x)}\sqrt{1-f(x)}f(x)}=-\frac{1}{2\pi}\frac{dy}{\sqrt{y}\sqrt{1-y}}$$
We also have to note that $y>0$ at any $x$, and $y\in(1;0)$ for $x\in(0;\infty)$; taking the usual order of integration (from $0$ to$1$) and multiplying PDF by $2$ (taking into consideration the part of $x\in(-\infty;0)\,$), we get $$g(y)=\frac{1}{\pi}\frac{1}{\sqrt{y}\sqrt{1-y}}$$
$$\frac{1}{\pi}\int_0^1\frac{dy}{\sqrt{y}\sqrt{1-y}}=\frac{1}{\pi}B\Big(\frac{1}{2};\frac{1}{2}\Big)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove $1If $a$, $b$, $c$ are three positive real numbers such that:
$a+b>c$
$b+c>a$
$c+a>b$
$a+b+c=2$
$a$, $b$ and $c$ might be or might not be equal in value
Show that:
$$1<ab+bc+ca-abc<\frac{28}{27}$$
From the first half of the question, I realised that $a$, $b$ and $c$ are the sides of a triangle whose perimeter is $2$. However, I'm unable to solve this problem, even after using some standard inequalities.
Can someone help me with this problem?
| I'll write up my comment as an answer. Consider the polynomial
$$
\begin{split}
f(x)&=(x-a)(x-b)(x-c)\\
&=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc\\
&=x^3-2x^2+(ab+bc+ca)x-abc.
\end{split}
$$
Then $f(1)=1-2+ab+bc+ca-abc=ab+bc+ca-abc-1$, so we need to show that
$0<f(1)\le 1/27$.
Note that $1=(a+b+c)/2=p$, the semiperimeter of the triangle with sides $a,b,c$. This triangle is not degenerate since all the inequalities in the problem statement are strict. Therefore, its area is positive. But the area of this triangle is
$$
A=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{1\cdot f(1)}=\sqrt{f(1)},
$$
so $f(1)=A^2>0$. On the other hand, by the AM-GM inequality,
$$
\begin{split}
\sqrt[3]{f(1)}&=\sqrt[3]{(p-a)(p-b)(p-c)}\\
&\le \frac{(p-a)+(p-b)+(p-c)}{3}\\
&=\frac{3p-(a+b+c)}{3}=\frac{3-2}{3}=\frac{1}{3},
\end{split}
$$
i.e. $f(1)\le (1/3)^3=1/27$.
In fact, we can use AM-GM inequality on any triangle to show that its area $A\le\dfrac{p^2}{3\sqrt{3}}$, where $p$ is its semiperimeter.
| {
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"url": "https://math.stackexchange.com/questions/4347633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a pattern to the roots of continued-fraction equations with depth $n$? I've found what looks like the start of a pattern, but I don't have the tools or background to know if one really exists here. Does anyone know how to analyze this mathematical behavior and continue the pattern?
Consider the continued fraction $f_n(x) = [0; x, x, x, ...]$, a zero followed by $n$ $x$s. This represents the continued fraction
$$f_n(x) = 0 + \frac{1}{x + \frac{1}{x + \frac{1}{x + ...}}}$$
with $n$ layers of depth. Thus
*
*for $n = 0$ we have $f_0(x) = 0$,
*for $n = 1$ we have $f_1(x) = 0 + \frac{1}{x} = \frac{1}{x}$,
*for $n = 2$ we have $f_2(x) = 0 + \frac{1}{x + \frac{1}{x}} = \frac{x}{x^2 + 1}$,
*for $n = 3$ we have $f_3(x) = 0 + \frac{1}{x + \frac{1}{x + \frac{1}{x}}} = \frac{x^2 + 1}{x^3 + 2 x}$,
and so on.
If we include a single complex point at infinity and claim that $\frac{1}{∞}$ is zero, $f_n(x)$ has $n$ roots, all of which lie on the imaginary axis. All roots are symmetrical about the real axis. These roots seem to follow patterns related to the Beraha Constants, $B(n) = 2 + 2 \cos\left(\frac{2 π}{n}\right)$, at least for the first several $n$s.
The sets of roots for $n$ from 1 to 10, in terms of $B(n)$ where I can figure out the connection, are as follows:
*
*$n$ = 1: ∞
*$n$ = 2: ∞, $ i \sqrt{B( 2)}$
*$n$ = 3: ∞, $±i \sqrt{B( 3)}$
*$n$ = 4: ∞, $±i \sqrt{B( 4)}$, $ i (B( 4) - 2)$
*$n$ = 5: ∞, $±i \sqrt{B( 5)}$, $±i (B( 5) - 2)$
*$n$ = 6: ∞, $±i \sqrt{B( 6)}$, $±i (B( 6) - 2)$, 0
*$n$ = 7: ∞, $±i \sqrt{B( 7)}$, $±i (B( 7) - 2)$, $±i \frac{1}{1 + 2 \cos\left(\frac{2 π}{7}\right)}$
*$n$ = 8: ∞, $±i \sqrt{B( 8)}$, $±i (B( 8) - 2)$, $±i \sqrt{2 - \sqrt{2}}$, 0
*$n$ = 9: ∞, $±i \sqrt{B( 9)}$, $±i (B( 9) - 2)$, $±i, ±2 i \sin\left(\frac{π}{18}\right)$
*$n$ = 10: ∞, $±i \sqrt{B(10)}$, $±i (B(10) - 2)$, $±i \sqrt{\frac{5 - \sqrt{5}}{2}}$, $±i \sqrt{\frac{3 - \sqrt{5}}{2}}$, 0
The first set of roots for each value of $n$ greater than 1 has a magnitude of $\sqrt{B(n)}$. The second set for $n$ greater than 3 has a magnitude of $B(n) - 2$. The third set for $n$ greater than 5 and the fourth set for $n$ greater than 7 seem to be following a similar structure, starting at zero and then rising towards an upper limit with a shallower slope each time, but I haven't been able to figure out a relation to the Beraha Constants for them, and I haven't been able to see any meta-pattern in the relations with only two instances of such a a relation to build off of.
So my question is this: is there a pattern to the roots beyond $\sqrt{B(n)}$ and $B(n) - 2$? Some way to know the roots of $f_n(x)$ for arbitrary values of $n$ without needing to calculate a continued fraction dozens of layers deep (equivalent to a function of degree $n$, which quickly becomes unmanageable)?
At minimum I'd like an equation $g(B(n))$ that gives the values $0$, $\frac{1}{1 + 2 \cos\left(\frac{2 π}{7}\right)}$, $\sqrt{2 - \sqrt{2}}$, $1$, and $\sqrt{\frac{5 - \sqrt{5}}{2}}$ for $n$ from 6 to 10, and an equation $h(B(n))$ that gives the values $0$, $2 \sin\left(\frac{π}{18}\right)$, and $\sqrt{\frac{3 - \sqrt{5}}{2}}$ for $n$ from 8 to 10, plus an explanation of how they can be derived, since I haven't been able to figure that out myself. Ideally, the procedure for finding those equations would extend to additional roots for higher values of $n$, allowing arbitrary roots to be found.
[ EDIT WITH ANSWER ]
Based on the answer from @dxiv, the magnitude of the roots can more correctly be written in the form $R(n, k) = 2 \cos\left(\frac{k π}{n}\right)$, where $n$ is the depth of the continued fraction function and $k$ is the index of the root counting from largest (one) to smallest ($\lfloor\frac{n}{2}\rfloor$). This just happens to be equal to $\sqrt{B(n)}$ for $k = 1$ and equal to $B(n) - 2$ for $k = 2$. Thus the roots for all values of $n$ up to ten, ordered from largest to smallest, are as follows:
*
*$n$ = 1: ∞
*$n$ = 2: ∞, 0
*$n$ = 3: ∞, $±2 i \cos\left(\frac{π}{ 3}\right)$
*$n$ = 4: ∞, $±2 i \cos\left(\frac{π}{ 4}\right)$, 0
*$n$ = 5: ∞, $±2 i \cos\left(\frac{π}{ 5}\right)$, $±2 i \cos\left(\frac{2 π}{ 5}\right)$
*$n$ = 6: ∞, $±2 i \cos\left(\frac{π}{ 6}\right)$, $±2 i \cos\left(\frac{2 π}{ 6}\right)$, 0
*$n$ = 7: ∞, $±2 i \cos\left(\frac{π}{ 7}\right)$, $±2 i \cos\left(\frac{2 π}{ 7}\right)$, $±2 i \cos\left(\frac{3 π}{ 7}\right)$
*$n$ = 8: ∞, $±2 i \cos\left(\frac{π}{ 8}\right)$, $±2 i \cos\left(\frac{2 π}{ 8}\right)$, $±2 i \cos\left(\frac{3 π}{ 8}\right)$, 0
*$n$ = 9: ∞, $±2 i \cos\left(\frac{π}{ 9}\right)$, $±2 i \cos\left(\frac{2 π}{ 9}\right)$, $±2 i \cos\left(\frac{3 π}{ 9}\right)$, $±2 i \cos\left(\frac{4 π}{ 9}\right)$
*$n$ = 10: ∞, $±2 i \cos\left(\frac{π}{10}\right)$, $±2 i \cos\left(\frac{2 π}{10}\right)$, $±2 i \cos\left(\frac{3 π}{10}\right)$, $±2 i \cos\left(\frac{4 π}{10}\right)$, 0
| Let $\,f_n = \frac{p_n}{q_n}\,$ then for $\,n \ge 1\,$:
$$
\frac{p_{n+1}}{q_{n+1}}= f_{n+1} = \frac{1}{x+f_n}=\frac{1}{x+\frac{p_n}{q_n}}=\frac{q_n}{xq_n+p_n} \;\;\implies\;\;\\ \begin{cases}p_{n+1}=q_n \\ q_{n+1}=xq_n+p_n=xq_n+q_{n-1}\end{cases}
$$
It follows that $\,p_{n+1}\,$ satisfies $\,p_{n+1}=xp_n+p_{n-1}\,$ with $\,p_0(x)=0, \,p_1(x)=1, \,p_2(x)=x\,$.
Let $\,p_n(x)=i^n \,r_n\left(\frac{x}{2i}\right)\,$, then after substituting and canceling a factor of $\,i^{n+1}\,$:
$$
r_{n+1}\left(\frac{x}{2i}\right) = 2\,\frac{x}{2i} r_n\left(\frac{x}{2i}\right) - r_{n-1}\left(\frac{x}{2i}\right) \;\;\implies\;\; \\r_{n+1}(z) = 2z\, r_n(z)-r_{n-1}(z)
$$
This is the same recurrence satisfied by the Chebyshev polynomials. Moreover:
$$
\begin{align}
r_1(z) &= i^{-1}\,p_1(2iz) = -i \cdot 1
\\ r_2(z) &= i^{-2}\,p_2(2iz) = -i \cdot 2z
\end{align}
$$
Aside from an index shift of $\,1\,$, these match the initial conditions for the Chebyshev polynomials of the second kind $\,U_0(z)=1\,$, $\,U_1(z)=2z\,$ scaled by a factor of $\,-i\,$, so $\,r_{n+1}(z)=-i\,U_n(z)\,$.
The roots of $\,r_{n}(z)\,$ are therefore the roots of $\,U_{n-1}(z)\,$, known to be $\,z_k=\cos \frac{k\pi}{n} \;\big|_{k = 1,2,\dots,n-1}\,$ and the roots of $\,p_{n}(x)\,$ are $\,x_k = 2i\,z_k\,$, which are also the zeros of $\,f_{n}\,$.
[ EDIT ] $\;$ The relation with Beraha constants follows from trig identities between their definition $B(n) = 2\left(1 + \cos \frac{2 \pi}{n}\right)$ and the expression for the roots $\,x_k = 2i \, \cos \frac{k\pi}{n} \;\big|_{k = 1,2,\dots,n-1}\,$.
*
*$x_1^2 = x_{n-1}^2 = -4 \cos^2 \frac{\pi}{n} = -2\left(\cos \frac{2\pi}{n}+1\right) = -B(n)$
*$x_2 = -x_{n-2} = 2i \,\cos\frac{2 \pi}{n} = i \,\big(B(n) - 2\big)$
| {
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how this simplification was done in the integral? I'm stuck at simplification
Here is the Integral ,
$$I=\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}} \ dx$$
To solve this ,
First let's Substitute, we got
$$x^2+2x+3=t^2$$
$$\implies {x}=\sqrt{(t^2-2)}-1$$
$$\implies dx=\frac{t}{\sqrt{t^2-2}}dt$$
Putting this back into the Integral followed by some simplification we have,
$$I=\int\sqrt{t^{2}-2}+\frac{1}{\sqrt{t^{2}-2}}-1 dt$$
Here, how they did it or how they "put back integral " and simplified it ? please, can anybody explain it step by step ?
| Substitution is not the way to go here,
Write the integral as
$$=\int\sqrt{x^2+2x+3}dx-\int\frac{x+1}{\sqrt{x^2+2x+3}}dx-\int\frac{1}{\sqrt{x^2+2x+3}}dx$$
We have three types of integrals. The middle is a straightforward substitution.
The first can be done by parts,
$$\int\sqrt{x^2+2x+3}dx=x\sqrt{x^2+2x+3}-\int\frac{x^2+x}{\sqrt{x^2+2x+3}}dx$$
which gives,
$$2\int\sqrt{x^2+2x+3}dx=x\sqrt{x^2+2x+3}+\int\frac{x+3}{\sqrt{x^2+2x+3}}dx$$
and the final type,
$$\int\frac{1}{\sqrt{x^2+2x+3}}dx=\int\frac{1}{\sqrt{(x+1)^2+2}}dx$$ use a $$x+1=\sqrt{2}\sinh y$$ substitution.
| {
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Transformation Matrix with respect to a basis and the General Linear Group I need some help going over this problem because I'm not entirely sure if my solution is sound.
Let $$\phi:\mathbb{R^2} \rightarrow \mathbb{R^2},\begin{pmatrix} x\\y
\end{pmatrix} \mapsto \begin{pmatrix} \frac{3}{2}x- \frac{1}{2}y \\ -\frac{1}{2}x+\frac{3}{2}y
\end{pmatrix} $$
a) Show that there exists a basis $B$ of $\mathbb{R^2}$ such that the transformation matrix is equal to ${^B}A_\phi^B= \begin{pmatrix} 1&0 \\ 0&2
\end{pmatrix}$.
b) Show that the set $Z= \bigg\{T\in GL_2(\mathbb{R})| \ T \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2}
\end {pmatrix}T^{-1} = \begin{pmatrix} 1&0 \\ 0&2 \end{pmatrix} \bigg\} $ is not empty.
c) Show that $\forall \ T\in Z$ and $\forall \lambda\in \mathbb{R}- \{0 \}$ also $ \lambda T \in Z$ is.
d) For any $T_1,T_2 \in Z$ does a $\lambda \in \mathbb{R}-\{0\}$ exist such that $T_2=\lambda T_1 $?
My work:
a) Let $B=\{B_1,B_2\}$ be a basis for $\mathbb{R^2}$ with $B_1= \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}, B_2= \begin{pmatrix} b_3 \\ b_4 \end{pmatrix}$. We find the image of $B_1$ and $B_2$ and then express them in terms of the basis with real coefficients $p_i,q_i$.
$$ \phi(B_1)= \begin{pmatrix} \frac{3}{2}b_1-\frac{1}{2}b_2 \\ -\frac{1}{2}b_1+\frac{3}{2}b_2 \end{pmatrix}= p_1\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}+ p_2\begin{pmatrix} b_3 \\ b_4 \end{pmatrix}$$
$$\phi(B_2)= \begin{pmatrix} \frac{3}{2}b_3-\frac{1}{2}b_4 \\ -\frac{1}{2}b_3+\frac{3}{2}b_4 \end{pmatrix}= p_1\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}+ p_2\begin{pmatrix} b_3 \\ b_4 \end{pmatrix} $$
Since the transformation matrix has been given we find the coefficients as follows: $p_1=1,p_2=0,q_1=0,q_2=2$. Which simply means that $\phi(B_1)=B_1, \phi(B_2)=2B_2$. I then simply chose the vectors $\begin{pmatrix} 2\\2 \end{pmatrix}$, $\begin{pmatrix} 2\\-2 \end{pmatrix}$.
b) This question was much easier and I guessed, correctly, that the matrix $\begin{pmatrix} 2 &2 \\2&-2 \end{pmatrix}$ was an element of the set.
The last two parts of this problem is where I need help. I tried simply computing the results of $\lambda T \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2}
\end {pmatrix}T^{-1} $ but it got quickly very complicated and I was wondering if there was an easier way to show it?
| c) If $T\in Z$ and $\lambda\in\Bbb R\setminus\{0\}$, then\begin{align}(\lambda T)\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}(\lambda T)^{-1}&=\lambda T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}\lambda^{-1}T^{-1}\\&=\lambda\lambda^{-1}T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}T^{-1}\\&=\begin{bmatrix}1&0\\0&2\end{bmatrix}.\end{align}
d) Not necessarily, since$$\begin{bmatrix}2&2\\2&-2\end{bmatrix},\begin{bmatrix}2&1\\2&-1\end{bmatrix}\in T.$$
| {
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Prove $\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2\left(x+y\right)\sqrt{xy}}$, with $x,\:y\in \mathbb{R}$ and $x\ne 0$ My thinking:
Starting with the basic fact $\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0$ which is trivial.
$\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0\:$
$\rightarrow \:x^2-4\sqrt{yx}\sqrt{x}+6xy-4\sqrt{xy}\sqrt{y}+y^2\ge \:0\:$
$\rightarrow x^2+4\sqrt{yx}\sqrt{x}+6xy+4\sqrt{xy}\sqrt{y}+y^2\ge \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow $ $\left(\sqrt{x}+\sqrt{y}\right)^4\ge \: \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow \frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge \sqrt{yx}\sqrt{x}+\sqrt{xy}\sqrt{y}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge x\sqrt{xy}+y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{4}\ge 2x\sqrt{xy}+2y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2}\ge \sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$
$\rightarrow \:\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$ as needed. QED
I'm not sure if this is correct. If someone can provide some feedback that would be great!
| Your proof works, but there is a much more direct proof.
First, just write $a=\sqrt x,b=\sqrt y$ and rewrite it as $$(a+b)^2\geq 2\sqrt{(a^2+b^2)(2ab)}$$
Letting $u=a^2+b^2, v=2ab,$ use AM/GM:
$$\frac{u+v}{2}\geq\sqrt{uv}$$
That first step is important from an aesthetic point of view, as much as anything - all those square roots are hard on the eyes, and it is difficult to see what is going on.
If we were to prove AM/GM ourselves, we would, of course, square $u-v=(a-b)^2,$ and you will see quite a lot of your proof is just proving AM/GM for a special case.
| {
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Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.
| The following fills-in the remaining step in OP's approach.
Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$?
*
*$3^{20} = 9^{10} \gt 8^{10}=2^{30}\,$ which excludes $\,\sqrt{2} = \sqrt[4]{4}\,$ as possible maximums;
*$3^{20} = 9^{10} \gt 6^{10}$ which excludes $\,\sqrt[6]{6}\,$ as a possible maximum;
*$3^{20} \gt 3^{18} = 27^6 \gt 25^{6} = 5^{12}$ which excludes $\,\sqrt[5]{5}\,$ as a possible maximum.
| {
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Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| There's not realy a need to use De Moivre's Theorem to solve this.
Let a=3-2i:
\begin{align}
(z-a)^4=z^4 &\Leftrightarrow (z-a)^4 - z^4= 0 \\
& \Leftrightarrow \big((z-a)^2-z^2 \big)\big( (z-a)^2 + z^2\big) =0 \\
& \Leftrightarrow z^2 -2az +a^2 - z^2=0 \quad \vee \quad z^2-2az + a^2 +z^2=0\\
& \Leftrightarrow z=\frac{a}{2} \quad \vee \quad 2z^2 -2az+a^2=0 \\
\end{align}
Solving the quadratic equation
\begin{align}
2z^2 -2az+a^2=0 & \Leftrightarrow z= \frac{2a \pm \sqrt{4a^2-8a^2}}{4}\\
& \Leftrightarrow z = \frac{2a \pm 2i\sqrt{a^2}}{4} \\
& \Leftrightarrow z=\frac{a \pm ai}{2}
\end{align}
Substituting $a$, we get the following solutions:
$$z=\frac{3}{2}-i \vee z=\frac{5}{2}+\frac{i}{2} \vee z=\frac{1}{2}-\frac{5i}{2}$$
I hope I have been of some assistance.
| {
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Problem of finding values of $ a $ for which two matrices are similar Problem: Let $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 1 & 2 & 7-a^{2} \\ 2 & 2+a & 6\end{array}\right) $ , $ B=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 9\end{array}\right)$ where $ a \in \mathbb{R} $. Find all the values of $ a $ for which $ A $ is similar to $ B $
Attempt:
We know the necessary conditions for two matrices to be similar are:
Their traces are equal
They have the same determinant
They have the same characteristic polynomial
They have the same eigenvalues
Their ranks are equal
Thus, $ |A| = [ 12 - (7-a^2)(2+a) ] - 2[ 6-2(7-a^2)] + 3[ 2+a - 4 ] = (2-a)^2\cdot (2+a) $,
$ |B| = 0 $, hence $ |A| = 0 $ and therefore either $ a= 2 $ or $ a = -2 $
The characteristic polynomial of $ B $ is $ \Delta_B(x) = x^2(9-x) $
How do I continue from here? I'm stuck. ( I thought to jordanize $ A $ and $ B $ but that seems to really complicate things [ jordanizing $ A $ seems really hard here ], I also thought I'd find the eigenspaces of the eigenvalues of $ B $ but I still wasen't sure how I'd continue from there )
Thanks in advance for help!
| You have $\det(A)=a^3-2a^2-4a+8$. On the other hand, if $a=\pm2$, then the first two line of $A$ will be equal, and therefore $\det(A)=0$. And if you divide $\det(A)$ by $(a+2)(a-2)$, then you will get $a-2$, and therefore $\det(A)=0\iff a=\pm2$. So, since $\det(B)=0$, we only have to consider the cases $a=2$ and $a=-2$. But, if $a=-2$, then the characteristic polynomial of $A$ is $-\lambda ^3+9 \lambda ^2-12 \lambda$, whereas the characteristic polynomial of $B$ is $-\lambda^3+9\lambda^2$. So, the only possibility that remains is that $a=2$ (in which case the characteristic polynomial is $-\lambda^3+9\lambda^2$. The roots of the characteristic polynomial are then $0$ and $9$. Besides, the eigenvectors corresponding to the eigenvalue $9$ are the multiplies of $(1,1,2)$, whereas $(-3,0,1)$ and $(-2,1,0)$ are linearly independent eigenvectors corresponding to the eigenvalue $0$. So, $A$ is similar to $B$ in the case $a=2$.
| {
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Prove that $\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}$ Prove that $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}, a>0, x>0,x\ne a.$$ My try $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=\dfrac{\sqrt{a}^2-\sqrt{x}^2}{\sqrt{a}-\sqrt{x}}-\dfrac{a+\sqrt[4]{ax^3}-\sqrt[4]{a^3x}-\sqrt[4]{a^2x^2}}{\sqrt{a}+\sqrt[4]{ax}}$$ Can we do something more directly to arrive at the solution? Thank you!
| $$
\begin{aligned}
\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right) &= \frac{\left(\sqrt{a}\right)^2-\left(\sqrt{x}\right)^2}{\sqrt{a}-\sqrt{x}}-\frac{\left(\sqrt[4]{a}\right)^4+\sqrt[4]{a}\left(\sqrt[4]{x}\right)^3}{\left(\sqrt[4]{a}\right)^2+\sqrt[4]{a}\sqrt[4]{x}}+\sqrt[4]{ax} = \\
&=\frac{\left(\sqrt{a}-\sqrt{x}\right)\left(\sqrt{a}+\sqrt{x}\right)}{\sqrt{a}-\sqrt{x}}-\frac{\sqrt[4]{a}\left((\sqrt[4]{a})^3+(\sqrt[4]{x})^3\right)}{\sqrt[4]{a}\left(\sqrt[4]{a}+\sqrt[4]{x}\right)}+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-\frac{(\sqrt[4]{a}+\sqrt[4]{x})((\sqrt[4]{a})^2-\sqrt[4]{a}\sqrt[4]{x}+(\sqrt[4]{x})^2)}{\sqrt[4]{a}+\sqrt[4]{x}}+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-(\sqrt{a}-\sqrt[4]{ax}+\sqrt{x})+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-\sqrt{a}+\sqrt[4]{ax}-\sqrt{x}+\sqrt[4]{ax} = 2\sqrt[4]{x}
\end{aligned}
$$
| {
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Indefinite integral with geometric series I was trying to calculate the following integral
$$\int \frac{\text{d}x}{(1 + x^n)^n}$$
for $n > 0$.
I tried this road:
$$\int \frac{\text{d}x}{\left(x^n\left(1 + \frac{1}{x^n}\right)\right)^n} = \int \frac{\text{d}x}{x^{n^2}} \sum_{k = 0}^{+\infty} (-1)^k \left(\frac{1}{x^n}\right)^k = \sum_{k = 0}^{+\infty} (-1)^k \int \frac{\text{d}x}{x^{n^2 + kn}} = \sum_{k = 0}^{+\infty} (-1)^k \frac{x^{1-n (k+n)}}{1-n (k+n)}$$
The $x$ term can be taken outside the sum. With the help of Mathematica, I found out the sum is a Hypergeometric function:
$$x \sum_{k = 0}^{+\infty} (-1)^k \frac{x^{-n (k+n)}}{1-n (k+n)} = -\frac{x^{-n^2} \, _2F_1\left(1,n-\frac{1}{n};n+1-\frac{1}{n};-x^{-n}\right)}{n^2-1}$$
Yet this result doesn't match with the integration that Mathematica performs. Indeed it "should" be:
$$\int \frac{\text{d}x}{(1 + x^n)^n} = x \, _2F_1\left(\frac{1}{n},n;1+\frac{1}{n};-x^n\right)$$
which also seems to hold for every $n$ except for $n = 0$ and $n = -1$.
Any help in understanding the passages and the result?
| There are mistakes in the proposed derivation, since the geometric series should have been taken with an exponent $n$.
To obtain the quoted result, one can use the generalized binomial expansion
\begin{equation}
\frac{1}{(1-z)^{s}}=\sum _{k=0}^{\infty }\binom{s+k-1}{k}z^{k}
\end{equation}
to express
\begin{align}
f(x)&=\int \frac{dx}{(1 + x^n)^n}\\
&=\int \sum_{k=0}^\infty\binom{n+k-1}{k}(-1)^kx^{nk}\,dx\\
&=\sum_{k=0}^\infty\binom{n+k-1}{k}(-1)^k\frac{x^{nk+1}}{nk+1}
% &=\frac{y^{\frac1n}}n\sum_{k=0}^\infty(-1)^k\frac{(n+k-1)_k}{k!}\frac{y^k}{\frac{1}{n}+k}
\end{align}
which can be rearranged as
\begin{align}
f(x)&=x\sum_{k=0}^\infty\frac{\Gamma(n+k)}{k!\Gamma(n)}\frac{\left( \frac{1}{n} \right)_k}{\left( \frac{1}{n} +1\right)_k}(-1)^kx^{nk}\\
&=x\sum_{k=0}^\infty\frac{(n)_k\left( \frac{1}{n} \right)_k}{\left( \frac{1}{n} +1\right)_k}\frac{(-x^n)^k}{k!}\\
&=x\,_2F_1\left(n,\frac{1}{n};1+\frac{1}{n};-x^n \right)
\end{align}
as expected. The case $n=1$ is correctly found by simplifying the hypergeometric function, while for $n=0$ the result is to be taken as the limit for $n\to0$.
| {
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Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is not that brought me in a right way.
Do you have any idea how to solve this one?
| $$\cos x \sin y=\frac{\sqrt2}{2}$$
$$\cos x \sin y=\frac{1}{\sqrt2}$$
$\frac{1}{\sqrt2}$ can be written as $1*\frac{1}{\sqrt2}$.
This gives rise to two cases-->
*
*$\cos x=1$ and $\sin y=\frac{1}{\sqrt2}$. So, $x=0$ and $y=\frac{\pi}{4}$.
*$\cos x=\frac{1}{\sqrt2}$ and $\sin y=1$. So, $x=\frac{\pi}{4}$ and $y=\frac{\pi}{2}$.
Solving Equation 1 for both cases-->
*
*$$\tan x+\tan y$$
$$\tan 0+\tan\frac{\pi}{4}$$
$$0+1=1$$
2.$$\tan x+\tan y$$
$$\tan\frac{\pi}{4}+\tan\frac{\pi}{2}$$
$$1+\infty=\infty$$
We see that only Case 1 satisfies the given equation. So, $x=0$ and $y=\frac{\pi}{4}$.
| {
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Finding bounds for a function I would like to show that
$$\frac{1}{\pi^2}<\int_{\pi/2}^\pi\frac{\sin x}{x^3}<\frac{3}{2\pi^2}.$$
We know that $\frac{\sin x}{x^3}\leq\frac{1}{x^3}$ and integrating gives $\int_{\pi/2}^\pi\sin x/x^3\leq3/(2\pi^2)$. I don't know how to make $\leq$ into $<$. On the other hand, $\frac{1}{\pi^3}\leq\frac{\sin x}{x^3}$ and integrating gives $1/\pi^3\leq\int_{\pi/2}^\pi\sin x/x^3$. But $1/\pi^3<1/\pi^2$ so this clearly is not what I want.
Any suggestions in how to fix these issues?
| Just for the fun.
Without any special function, we can have a more than decent approximation of the integral using, as an approximation,
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed, more than $\large 1,400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
This gives
$$\int_{\frac \pi 2}^\pi \frac {sin(x)}{x^3}\,dx \sim \int_{\frac \pi 2}^\pi \frac{16 (\pi -x)}{x^2 \left(5 \pi ^2-4 (\pi -x) x\right)}\,dx$$
$$5 \pi ^2-4 (\pi -x) x=4 (x-a)(x-b)$$ where
$$a=\left(\frac{1}{2}-i\right) \pi \qquad \text{and} \qquad b=\left(\frac{1}{2}+i\right) \pi$$
$$\frac{16 (\pi -x)}{4(x-a)(x-b)}=-\frac{4 (a b-\pi ( a+ b))}{a^2 b^2 x}+\frac{4 (\pi -a)}{a^2 (a-b) (x-a)}-\frac{4
(\pi -b)}{b^2 (a-b) (x-b)}+\frac{4 \pi }{a b x^2}$$ Simple integrals leading to
$$\int_{\frac \pi 2}^\pi \frac{16 (\pi -x)}{x^2 \left(5 \pi ^2-4 (\pi -x) x\right)}\,dx=\frac{4 \pi ^2}{25}\left(20-11 \pi -8 \log (2)+2 \log (5)+22 \tan ^{-1}(2)\right)$$ which is $0.121155$ while the exact value is $0.121210$ (so a relative error of $0.045$%).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trying to understand the sigma expansion in an answer Iteration 2: Reassign cluster labels according to the smallest squared distances:
$\begin{array}{ccccc}\text { Data point } X_{i} & \left\|X_{i}-(0,1)\right\|^{2} & \left\|X_{i}-(4,4)\right\|^{2} & \left\|X_{i}-(6,1)\right\|^{2} & \text { Cluster Label } \\ (6,-1) & 40 & 29 & 4 & 3 \\ (4,3) & 20 & 1 & 8 & 2 \\ (6,3) & 40 & 5 & 4 & 3 \\ (1,1) & 1 & 18 & 25 & 1 \\ (4,5) & 32 & 1 & 20 & 2 \\ (-1,1) & 1 & 34 & 49 & 1\end{array}$
$$
\begin{aligned}
&\left\{C_{l_{1}}, C_{l_{2}}, C_{l_{3}}\right\}=\{\{(1,1),(-1,1)\},\{(4,3),(4,5)\},\{(6,-1),(6,3)\}\} \\
&\left\{\mu_{l_{1}}, \mu_{l_{2}}, \mu_{l_{3}}\right\}=\{(0,1),(4,4),(6,1)\}
\end{aligned}
$$
The cost function which is the sum of the distances of the data points from each cluster to their centroids is given by
$$
\begin{aligned}
C\left(l_{1}, l_{2}, l_{3}, \mu_{l_{1}}, \mu_{l_{2}}, \mu_{l_{3}}\right)=& \frac{1}{6} \sum_{l_{j}=1}^{K=3} \sum_{i=1}^{N=6}\left\|X_{i}-\mu_{l_{j}}\right\|^{2} \\
=& \frac{1}{6}\left\{\left\|X_{1}-\mu_{l_{3}}\right\|^{2}+\left\|X_{2}-\mu_{l_{2}}\right\|^{2}+\left\|X_{3}-\mu_{l_{3}}\right\|^{2}\right.\\
&\left.+\left\|X_{4}-\mu_{l_{1}}\right\|^{2}+\left\|X_{5}-\mu_{l_{2}}\right\|^{2}+\left\|X_{6}-\mu_{l_{1}}\right\|^{2}\right\} \\
=& \frac{1}{6}\{4+1+4+1+1+1\}=\frac{12}{6}=2
\end{aligned}
$$
Hi, I was reading a solution. But I didn't understand how the author find this sigma expansion, that is I didn't understand the following:How did get the author these, can you explain?
$\begin{aligned}=& \frac{1}{6}\left\{\left\|X_{1}-\mu_{l_{3}}\right\|^{2}+\left\|X_{2}-\mu_{l_{2}}\right\|^{2}+\left\|X_{3}-\mu_{l_{3}}\right\|^{2}\right.\\ &\left.+\left\|X_{4}-\mu_{l_{1}}\right\|^{2}+\left\|X_{5}-\mu_{l_{2}}\right\|^{2}+\left\|X_{6}-\mu_{l_{1}}\right\|^{2}\right\} \end{aligned}$
| You have three cluster centroids $\left\{\mu_{l_{1}}, \mu_{l_{2}}, \mu_{l_{3}}\right\}=\{(0,1),(4,4),(6,1)\}$. Now you (re-)assign 6 points to one of the 3 clusters with the regard of the centroid. You choose the smallest (squared) distance between the point and the three centroids and assign the corresponding point to the corresponding cluster.
In total you have 6 distances due the 6 points. The other combination/distances are not regarded.
Finally you calculate the arithmetic mean. The squared distance between $X_1$ and the centroid of cluster $3$ is $(6-6)^2+(-1-1)^2=0+4=4$
| {
"language": "en",
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Plane passing through a given point so that the tetrahedron is of the minimum possible positive volume
Find the plane passing through the point $(1,1,2)$ s. t. the volume of the tetrahedron enclosed by the plane and the coordinate axes is of the minimum possible positive volume.
$$A(x-1)+B(y-1)+C(z-2)=0\\Ax+By+Cz=A+B+2C\tag 1$$ the plane shouldn't be parallel to any of the axes so $A,B,C\ne 0.$ Plugging $(x_0,0,0),(0,y_0,0),(0,0,z_0)$ into $(1),$ I got: $$\begin{aligned}x_0&=1+\frac{B}A+\frac{2C}A\\y_0&=\frac{A}B+1+\frac{2C}B\\z_0&=\frac{A}C+\frac{B}C+2\end{aligned}$$
Then, I tried to find the minimum of the function $\begin{aligned}f(x,y,z)=\left(1+\frac{y}x+\frac{2z}x\right)\left(\frac{x}y+1+\frac{2z}y\right)\left(\frac{x}z+\frac{y}z+2\right)\\=\frac{(x+y+2z)^3}{xyz}\end{aligned}$
$x,y,z$ should be of the same sign.
$$\frac{(x+y+2z)^3}{xyz}\ge\frac{(3\sqrt[3]{2xyz})^3}{xyz}=27\cdot 2=54,$$ which is attained when $x=y=2z.$
So, my answer is $$\pi\ldots 2(x-1)+2(y-1)+z-2=0$$
Can somebody verify my answer?
| Let the normal to the plane be $N = (1, A, B) $, then the equation of the plane is
$N \cdot (r - (1, 1, 2) ) = 0 $
which simplifies to
$ (x - 1) + A (y - 1) + B (z - 2) = 0 $
Setting $y=z=0$, gives us
$ a = 1 + A + 2 B $
Setting $x = z = 0 $ gives us
$ b = 1 + \dfrac{1}{A} ( 1 + 2 B ) $
Setting $ x = y = 0$ gives us
$ c = 2 + \dfrac{1}{B} (1 + A) $
Then we want the product $abc$ to be minimum, i.e.
$abc = (1 + A + 2 B) (1 + \dfrac{1 + 2 B}{A} ) (2 + \dfrac{1+A}{B} )$
This simplifies to
$ abc = f(A, B) = \dfrac{ (1 + A + 2 B)^3}{AB} $
$ f_A = \dfrac{ 3(1 + A + 2 B)^2 (A B) - B (1 + A + 2 B)^3 }{A^2 B^2} $
$f_B = \dfrac{ 3 (1 + A + 2 B)^2 (2 A B) - A (1 + A + 2 B)^3 }{A^2 B^2} $
Therefore, at the critical point, $A , B$ satisfy
$ 3 A - (1 + A + 2 B) = 0 $
$ 6 B - (1 + A + 2 B) = 0 $
whose solution is $ A = 1 , B = \dfrac{1}{2} $
Hence,
$ a = 3 , b = 3, c = 6 $
Which gives the minimum volume of the tetrahedron as $ \dfrac{1}{6} (3)(3)(6) = 9 $
And the equation of the plane is
$ (x - 1) + (1) (y - 1) + \dfrac{1}{2} ( z - 2) = 0 $
which simplifies to
$ 2 (x - 1) + 2 (y - 1) + z - 2 = 0 $
So your answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of
$\triangle ABC$.
I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.
|
Let us give name $a$ for the area of triangles with common area and $x,y,z$ for the other areas. We have (using the following result: the areas of triangles sharing a same altitude with collinear bases are in the ratio of the lengths of these bases):
$$\begin{cases}\dfrac{EA}{EC}&=&\dfrac{x}{a}&=&\dfrac{a+z}{a+y}\\ \dfrac{DC}{DB}&=&\dfrac{y}{a}&=&\dfrac{a+x}{a+z}\\
\dfrac{FB}{FA}&=&\dfrac{z}{a}&=&\dfrac{a+y}{a+x}\end{cases}\tag{1}$$
By taking new variables
$$X=\dfrac{x}{a}, \ \ Y=\dfrac{y}{a}, \ \ Z=\dfrac{z}{a}$$
(1) becomes :
$$\begin{cases}X&=&\dfrac{1+Z}{1+Y}\\ Y&=&\dfrac{1+X}{1+Z}\\
Z&=&\dfrac{1+Y}{1+X}\end{cases}\tag{2}$$
*
*The product of these 3 equations (2) gives
$$\underbrace{XYZ}_P=1 \tag{3}$$
*Expanding the equations of (2) gives
$$\begin{cases}
1+Z&=&X+XY\\
1+X&=&Y+YZ\\
1+Y&=&Z+ZX
\end{cases} \tag{4}$$
Adding equations (4) gives:
$$\underbrace{XY+YZ+ZX}_R=3 \tag{5}$$
Multiplying equations in (4), by $Z$, $X$ and $Y$ resp., we get:
$$\begin{cases}
Z+Z^2&=&XZ+XYZ\\
X+X^2&=&YX+XYZ\\
Y+Y^2&=&ZY+XYZ
\end{cases} \tag{6}$$
adding them and setting $S:=X+Y+Z$, we get:
$$S+(S^2-2R)=R+P$$
Taking into account (3) and (5), we get:
$$S+S^2-12=0, \tag{7}$$
a quadratic whose unique positive root is $S=3$.
Therefore, due to Vieta's relationships, $X,Y,Z$ are roots of the third degree equation in $T$ :
$$T^3-ST^2+RT-P=0 \ \iff \ T^3-3T^2+3T-1=0 \ \iff \ (T-1)^3=0\tag{8}$$
giving the triple root $X=Y=Z=1$. Otherwise said:
$$x=y=z=a\tag{9}$$
As a consequence, the areas of triangles $MAB, AMC, ABM$ are identical with common value $2a$.
Therefore, the barycentric coordinates of $M$ are $([MAB]/[ABC], [AMC]/[ABC], [ABM]/[ABC])=(1/3,1/3,1/3)$ characterizing the centroid of $ABC$.
Edit: I just discovered this similar question If three cevians are concurrent at a point and form triangles of equal area, the point is the centroid with an interesting answer by @almagest whose reasoning is partly the same as mine, with a final clever use of the Arithmetic/Geometric means inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}}$ Trying to prove $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$
I tried using ratio-root criteria with $a_n= \frac{1}{n^n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n}}$ and end up with $\frac{n^n}{(n+1)^{n+1}}\cdot 2(n+1)^2 \Pi_{k=1}^n\frac{\left(k^2\right)^{\frac{k}{n+1}} \left(1+\left(\frac{n+1}{k} \right)^2 \right)^{\frac{k}{n+1}}}{\left(k^2\right)^{\frac{k}{n}} \left(1+\left(\frac{n}{k} \right)^2 \right)^{\frac{k}{n}}}$ and can’t see how to simplify this to get the limit as the numerator and denominator seem to end up going to 1 and some kind of $e^{\Sigma \frac{1}{n}}$ but the $(n+1)^2$ seems to persist and send everything to infinity.
So I tried taking e^ natural log(I’ll denote them as log here, I don’t mind ln either) and got this sum in the exponent $-\log (n) + \Sigma_{k=1}^n \frac{k}{n^2}\log(n^2+k^2)$ comparison showed I could get a $\log 2$ and a $\log n$ But where is the $-1/2$ I need? My guess is that this approach won’t simplify further.
Am doing this for practice/review so would be interested in seeing solutions or hints as I already spent too much time on this. My hunch is that I’m either overlooking something or there’s some typo in the problem.
This is also equivalent to $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)^{\frac{1}{n^2}}(n^2+2^2)^{\frac{2}{n^2}} \cdots (n^2 + n^2)^{\frac{1}{n}} \right] = 2e^{-1/2}$
Try explaining this before calling this duplicates. I get something different:
$\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = n^{\frac{1}{n}}\left[(1 +(\frac{1}{n})^2)^{\frac{1}{n^2}}(1+(\frac{2}{n})^2)^{\frac{2}{n^2}} \cdots (1 + (\frac{n}{n})^2)^{\frac{1}{n}} \right]$.
How do you get $\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = (1 +(\frac{1}{n})^2)(1+(\frac{2}{n})^2)^{2} \cdots (1 + (\frac{n}{n})^2)^{n}$?
| We have
\begin{align*}
& \frac{1}{n}\left( {\prod\limits_{k = 1}^n {(n^2 + k^2 )^k } } \right)^{1/n^2 } = \frac{1}{n}\exp \left( {\frac{1}{{n^2 }}\sum\limits_{k = 1}^n {k\log (n^2 + k^2 )} } \right) \\ & = \frac{1}{n}\exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{2\log n}}{{n^2 }}\sum\limits_{k = 1}^n k } \right)
\\ &
= \exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{n(n + 1)}}{{n^2 }}\log n - \log n} \right) \\ &= \exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{\log n}}{{n }}} \right)
\\ &
\to \exp \left( {\int_0^1 {x\log (1 + x^2 )dx} +0} \right) = \exp \left( {\log 2 - \frac{1}{2}} \right) = 2 e^{-1/2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin{equation}\frac{1}{(-n)^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3} \end{equation}
so that $\Delta n = dn = -n-(n+1) = -2n-1$.
For $f(n) = 1/n^2$, $f'(n) = -2/n^3$, and
\begin{equation}
df = d(\frac{1}{n^2}) = f'(n)dn
\end{equation}
\begin{equation}
df = \frac{(2)(2n+1)}{n^3}
\end{equation}
Now, for very large $n$, $2n+1 \approx 2n$. Thus, \begin{equation}
df = \frac{4}{n^2}
\end{equation}
But, $4/n^2$ is not approximately equal to $2/n^3$ (required ans.) even if $n$ is very large.
So, please point out my mistake(s).
Thanks in advance (;
| It's simpler than that:
$$\frac{1}{n^2}-\frac{1}{(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac{2n+1}{n^2(n+1)^2}=\frac{2+1/n}{n(n+1)^2}\approx \frac{2}{n^3}$$
as for large $n$ one has $1/n\approx 0$ and $n+1\approx n$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve tangent of $y=\left(\log_{a}{x}\right)^2$ and $y=-ax+2$ How can I solve the tangent point and $a$ when $f(x)=\left(\log_{a}{x}\right)^2$ is tangent to $g(x)=-ax+2$?
Although this can be solved by substituting $a=e^2$ and $x=e^{-2}$, then
$f\left(e^{-2}\right)=g\left(e^{-2}\right)$ and $f^\prime\left(e^{-2}\right)=g^\prime\left(e^{-2}\right)$ can be proved,
is there any general solution for this, rather than just substituting random numbers? Maybe using Lambert-W could help?
| for real $a,x$
$$\frac{d}{dx}\left(\log_a(x)^2\right)=\frac{d}{dx}\left(-ax+2\right)$$
$$\frac{2\ln(x)}{\ln(a)^2x}=-a$$
$a\to e^t$:
$$t^2e^t=-\frac{2\ln(x)}{x}$$
$$\sqrt{t^2e^t}=\sqrt{-2\frac{\ln(x)}{x}}$$
$$te^{\frac{1}{2}t}=\pm\sqrt{-2\frac{\ln(x)}{x}}$$
$$\frac{1}{2}te^{\frac{1}{2}t}=\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}$$
$k\in\{-1,0\}$:
$$\frac{1}{2}t=W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)$$
$$t=2W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)$$
$$a=e^{2W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)}$$
$$a=-\frac{1}{2}\frac{\ln(x)}{x\ W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)^2}$$
$$a=-\frac{1}{2}\frac{\ln(x)}{x\ W_k\left(\pm\frac{1}{\sqrt{-\frac{2x}{\ln(x)}}}\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380862",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
"answer_id": 1
} |
From homogeneous to non-homogeneous linear recurrence relation I'm trying to do the following exercise:
Find a non-homogeneous recurrence relation for the sequence whose general term is
$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$
From this expression we can obtain the roots of the characteristic polynomial $P(x)$, which are $3$ and $7$, so $P(x) = x^2 - 10x + 21$ and $a_n = 10a_{n-1} - 21a_{n-2} \; \forall \; n \ge 2, \; a_0 = \frac {1}{10}, \; a_1 = -\frac{13}{10}$.
Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation.
| Write the relations for two consecutive terms:
$$
\begin{align}
\begin{cases}
a_n &=\, \dfrac{1}{2}\,3^n - \dfrac{2}{5}\, 7^n
\\ a_{n+1} &=\, \dfrac{1}{2}\,3^{n+1} - \dfrac{2}{5}\, 7^{n+1} \,=\, \dfrac{3}{2}\,3^n-\dfrac{14}{5}\,7^n
\end{cases}
\end{align}
$$
Eliminate (for example) $\,7^n\,$ between the two:
$$
\require{cancel}
a_{n+1} - 7 a_n = \left(\dfrac{3}{2}\,3^n-\cancel{\dfrac{14}{5}\,7^n}\right) - \left(\dfrac{7}{2}\,3^n - \cancel{\dfrac{14}{5}\, 7^n}\right) \;\iff\; a_{n+1} = 7a_n-2\cdot 3^n
$$
Note that the non-homogeneous recurrence is not unique. If you chose to eliminate the other power $\,3^n\,$, for example, you would get $\,a_{n+1} = 3a_n-\dfrac{8}{5}\,7^n\,$, which is equally valid, as are many others.
| {
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=} -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle 1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$
| Inspired by metamorphy, I do want to try his wonderful trick
$$
I_{n}=\frac{1}{4} F^{\prime} _n\left(\frac{1}{2}\right) \text { where } F_{n}(\alpha)=2 \int_{0}^{\infty} \frac{x^{2 \alpha-1}}{\left(x^{2}+1\right)^{2}} d x
$$
In my post, I had found $$ \displaystyle \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+a\right)^{n}}=\frac{\pi}{m(n-1) !} \csc \frac{(r+1) \pi}{m} \prod_{j=1}^{n-1}\left(j-\frac{r+1}{m}\right) a^{-\left(n-\frac{r+1}{m}\right)}\tag*{} $$
Replacing $m=2, r=2 \alpha-1 $ yields
$$
\begin{aligned}
F_{n}(\alpha) &=\frac{2 \pi}{2(n-1) !} \csc \left(\frac{2 \alpha \pi}{2}\right) \prod_{j=1}^{n-1}\left(j-\frac{2 \alpha}{2}\right) \\
&=\frac{\pi}{(n-1) !} \csc (\alpha \pi) \prod_{j=1}^{n-1}(j-\alpha)
\end{aligned}
$$
By logarithmic differentiontin, we get
$$
\frac{F_{n}^{\prime}(\alpha)}{F_{n}(\alpha)}=-\pi \cot (\alpha \pi)-\sum_{j=1}^{n-1}\left(\frac{1}{j-\alpha}\right)
$$
Putting $\alpha=\frac{1}{2}$ yields
$$
\begin{aligned}
F_{n}^{\prime}\left(\frac{1}{2}\right) &=-F_{n}\left(\frac{1}{2}\right) \sum_{j=1}^{n-1} \frac{1}{j-\frac{1}{2}} \\
&=-\frac{\pi}{(n-1) !} \prod_{j=1}^{n-1}\left(j-\frac{1}{2}\right) \sum_{j=1}^{n-1} \frac{1}{j-\frac{1}{2}}\\& =-\frac{\pi(2 n-3) ! !}{2^{n-2}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}
\end{aligned}
$$
Now we can conclude that
$$\boxed{I_n=-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$3$-digit numbers are formed by arranging three of the digits $1$, $2$, $3$, $4$ and $5$
Three-digit numbers are formed by arranging three of the digits $1$, $2$, $3$, $4$ and $5$. No digit may be used more than once. Find the number of three-digit numbers if there exists at least $1$ prime number in the $3$ digits.
I have found the answer to be $3C1 \cdot 4P2 = 36$, but the correct answer is $60$. Can someone help with this?
Thought process: one of the digits must be $2, 3$ or $5$ so $3C1 = 3$ ways and remaining $2$ digits to choose from the $4$ digits remaining, hence $4P2 = 12$.
| Three of the five numbers in the set $S = \{1, 2, 3, 4, 5\}$ are prime numbers. Therefore, any three-digit number formed using three distinct digits of the set will contain at least one prime number. Thus, the number of three-digit numbers which contain at least one prime number in the three digits is $$5 \cdot 4 \cdot 3 = 60$$
Note: Be careful of choosing a representative of a set, then selecting additional elements from the remaining numbers of the set. This leads to multiple counting.
You chose one of the three prime numbers as a representative of the set. As Henry pointed out in the comments, you did not choose its position. There are three ways to do so. If you then multiply by the $P(4, 2) = 4 \cdot 3$ ways of filling in the remaining numbers, you would get $$3 \cdot 3 \cdot 4 \cdot 3 = 108$$ which is more than the total number of three-digit numbers with distinct digits which can be formed using the numbers in set $S$.
As pointed out above, it is not possible to form a number which does not contain at least one prime.
Exactly one prime appears among the three distinct digits: There are three ways to select one of the three prime numbers in the set $S$. There is one way to select both of the nonprime numbers in the set $S$. There are $3!$ ways to permute the three distinct digits. Hence, there are
$$\binom{3}{1}\binom{2}{2}3!$$
such numbers.
Exactly two primes appear among the three distinct digits: There are $\binom{3}{2}$ ways to select two of the three primes in the set $S$, two ways to select one of the two nonprimes in the set $S$, and $3!$ ways to permute the three distinct digits. Hence, there are
$$\binom{3}{2}\binom{2}{1}3!$$
such numbers.
Exactly three primes appear among the three distinct digits: There is one way to select all three primes and $3!$ ways to permute the digits. Hence, there are
$$\binom{3}{3}\binom{2}{0}3!$$
such numbers.
Total: Since the three cases are mutually exclusive and exhaustive, there are
$$\binom{3}{1}\binom{2}{2}3! + \binom{3}{2}\binom{2}{1}3! + \binom{3}{3}\binom{2}{0}3! = 60$$
three-digit numbers which can be formed by using three distinct digits of the set $S = \{1, 2, 3, 4, 5\}$ which contain at least one prime.
Had you remembered to take the position of your designated prime into account, you would have counted each number with two primes twice, once for each way of designating one of the two primes as the prime number among its digits, and each number with three primes three times, once for each way of designating one of the three digits as the prime digit. Note that
$$\binom{3}{1}\binom{2}{2}3! + \color{red}{\binom{2}{1}}\binom{3}{2}\binom{2}{1}3! + \color{red}{\binom{3}{1}}\binom{3}{3}\binom{2}{0}3! = \color{red}{108}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$ Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
$$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$$for all $x,$ $y.$ Find the sum of all possible values of $f(1).$
I tried plugging in combinations of $0$'s and $1$'s, but it didn't really get me anywhere. I would appreciate any help!!
Thanks!!
| Hint: try plugging $x=y$ in the formula.
Full answer: This yields $f(f(0)) = f(x)^2-x^2$. Let $c=f(f(0))$. Then $$f(x)^2 = x^2+c.$$ Therefore
$$c^2=f(f(0))^2 = f(0)^2+c = 0^2+2c=2c.$$
Thus $c$ is either $0$ or $2$.
If $c=0$, then $f(1)^2=1$ and therefore $f(1)$ is either $1$ or $-1$. As the function $f(x)=-x$ is a solution of the original equation with $f(1)=-1$, $-1$ is indeed a possible value for $f(1)$.
However, by looking at $x=1$ and $y=0$, we see that $f(f(1)) = -f(1)$, which implies that $f(1)\ne1$, and $1$ is therefore not a possible value for $f(1)$.
If $c=2$, we have that $f(1)^2=3$. Thus $f(f(1))^2 = 5$. But with $x=1$ and $y=0$ again, we find that $\pm\sqrt{5}=f(f(1))=\pm\sqrt{6}\pm\sqrt{2}\pm\sqrt{3}$, which is never true (see this question).
Therefore the only acceptable value for $f(1)$ is $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How is this differential equation solved? I have a differential equation as follows:
$$ y \cdot y'' +(y')^2 +1=0.$$
I'm interested in how to solve it. So far I have found a few solutions like $y=\sqrt{r^2-(x-k)^2}$ and $y=x+k.$ [In these, $r$ and $k$ are real constants.]
It came up while looking at a property of surface area of revolutions which held for spheres and certain cones. I am wondering if a general soution can be found, and in particular whether only circles and certain lines are solutions.
Note: As user RadialArmSaw noted, $y=x+k$ is not a solution.
| You can observe that
$(y^2)’’=(2yy’)’=2(yy’’+(y’)^2)$
Thus you get
$(y^2)’’=-2 $
$(y^2)’-(y^2)’(0)=-2x$
$y^2=y^2(0)+ (y^2)’(0)x-x^2$
Set $a= y^2(0)$, $b= (y^2)’(0)$, then
$y^2=a+bx-x^2$ so that all the possible solutions are of kind
$y=+/-\sqrt{a+bx-x^2}$ for every $a,b\neq 0$
The function it makes sense only for $ \frac{b-\sqrt{b^2+4a}}{2}\leq x\leq \frac{b+\sqrt{b^2+4a}}{2}$
and it’s the solution of your problem (I.e. the maximal definition domain of your solution is) for
$\frac{b-\sqrt{b^2+4a}}{2}< x< \frac{b+\sqrt{b^2+4a}}{2}$
Another way to see the solution is the following one:
The graph of $y$ is the circle (minus the two points $(\frac{b+/-\sqrt{b^2+4a}}{2} ,0)$ ) of center $(\frac{b}{2}, 0)$ and radius $R=\frac{1}{2}\sqrt{b^2+4a}$
$x^2+y^2-bx-a=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Joint PDF of $U=XY$ and $V=\frac{X}{Y}$ Given that $X$ and $Y$ are jointly distributed with pdf:
$$f(x, y)=\left\{\begin{array}{cc}
\frac{1}{x^{2} y^{2}}, & x \geqslant 1, y \geqslant 1 \\
0 & \text { else }
\end{array}\right\}$$
Find the joint pdf of $U=XY, V=\frac{X}{Y}$
My try:
First i found the Jacobian of the transformation:
$$\begin{aligned}
&\Rightarrow \quad x=\sqrt{u v} \quad y=\sqrt{\frac{u}{v}} \\
&J=\left|\begin{array}{ll}
\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\
\frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}
\end{array}\right|=\left|\begin{array}{ccc}
\frac{\sqrt{v}}{2 \sqrt{u}} & \frac{1}{2 \sqrt{u v}} \\
\frac{\sqrt{u}}{2 \sqrt{v}} & \frac{-1}{2} \sqrt{u} v^{\frac{-3}{2}}
\end{array}\right|=\frac{-1}{2v}
\end{aligned}$$
Now we have:
$$f(u, v)=|J| f\left(x\left(u, v\right), y(u, v)\right)$$
$$\begin{aligned}
&f(u,v)=\frac{1}{2 v} \frac{1}{(\sqrt{u v})^{2}} \frac{1}{\left(\sqrt{\frac{u}{v}}\right)^{2}} \\
&f(u,v)=\frac{1}{2 v u^{2}}
\end{aligned}$$
We have $u=xy \geq 1 $
But i am not able to find limits of $v$?
| $x \geq 1$ and $y \geq 1$ become $\sqrt {uv} \geq 1$ and $\sqrt {\frac u v} \geq 1$. So the limits are $\frac 1 u \leq v \leq u$ and $u \geq 1$.
| {
"language": "en",
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Using diagonalisation of a symmetric on quadratic form Let $f (x, y) = 7x_2 + 4y_2 + 4xy$. Use the diagonalisation of a symmetric matrix to write this quadratic form in the form $λ_1u_2^2 +λ_2v_2^2$, with u and v linear combinations of x and y.
Here's what I have tried:
step 1. get the symmetric matrix
$$f (x, y) = 7x_2 + 4y_2 + 4xy=\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)$$
step 2. find the eigenvalues
$$det(A-\lambda I) \\
\left(\begin{matrix}7-\lambda&2\\2&4-\lambda\\\end{matrix}\right)=(7-\lambda)(4-\lambda)-4=(\lambda-3)(\lambda-8)$$
step 3. find the eigenvectors
case for $\lambda =3$
$$\left(\begin{matrix}7-3&2\\2&4-3\\\end{matrix}\right)=\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right)
$$
Row reduced
$$\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right)=\left(\begin{matrix}1&\frac{1}{2}\\0&0\\\end{matrix}\right) \implies \left(\begin{matrix}x\\y\\\end{matrix}\right)=y_1\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right)$$
Doing the same for $\lambda =8$ (i'll skip the calculation), I get $\left(\begin{matrix}x\\y\\\end{matrix}\right)=y_2\left(\begin{matrix}2\\1\\\end{matrix}\right)$
Finding the orthogonal eigenvectors
$$u_1 = \frac{y_1}{||y_1||} = \frac{2}{\sqrt{5}}\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right) \\
u_2 = \frac{y_1}{||y_1||} = \frac{1}{\sqrt{5}}\left(\begin{matrix}2\\1\\\end{matrix}\right) \\
P=(u_1,u_2)=\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) \\
D = P^TAP = \left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right)\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) = \left(\begin{matrix}3&0\\0&8\\\end{matrix}\right) \\
x^TAx = \lambda^TD\lambda = \left(\begin{matrix}\lambda_1&\lambda_2\\\end{matrix}\right)\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)\left(\begin{matrix}\lambda_1\\\lambda_2\\\end{matrix}\right) = 3\lambda_1^2+8\lambda_2^2$$
However, my result does not look like it's in the same form as the question:
$λ_1u_2^2 +λ_2v_2^2 \ne 3\lambda_1^2+8\lambda_2^2$, how do I proceed from here?
| You start with the quadratic form
$$
7x^2 + 4y^2 + 4xy=
\left(\begin{matrix}x&y\end{matrix}\right)A
\left(\begin{matrix}x\\y\\\end{matrix}\right)=
\left(\begin{matrix}x&y\end{matrix}\right)
\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)
\left(\begin{matrix}x\\y\\\end{matrix}\right)
$$
Write the eigenvalues decomposition of $A$ as
$$
A=PDP^T=\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)
\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
$$
the quadratic form becomes
$$
\left[\left(\begin{matrix}x&y\end{matrix}\right)
\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)
\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)
\left(\begin{matrix}x\\y\\\end{matrix}\right)\right]$$
can you see what the two new variables $u,v$ are?
Note that your notation is a bit inaccurate and confusing.
| {
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"url": "https://math.stackexchange.com/questions/4397322",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Closed form for $T_n$ if $T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$ Consider the recurrence relation $$T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$$ with first three terms as:
$T_{1}=20, T_{2}=12, T_{3}=70$
Find the closed form expression for $T_n$.
My try:
Since it is a constant coefficient difference equation, the auxiliary equation is:
$$\lambda^3-2\lambda^2-2\lambda+1=0$$ whose roots are:
$-1,\frac{3\pm\sqrt{5}}{2}$
Thus we have:
$$T_{n}=A(-1)^{n}+B\left(\frac{3+\sqrt{5}}{2}\right)^{n}+C\left(\frac{3-\sqrt{5}}{2}\right)^{n}$$
where the constants $A,B,C$ to be determined by values of $T_1,T_2,T_3$ but its becoming too tedious to solve. Any alternate approach?
|
$$T_{n}=A(-1)^{n}+B\left(\frac{3+\sqrt{5}}{2}\right)^{n}+C\left(\frac{3-\sqrt{5}}{2}\right)^{n}$$
Where the constants $A,B,C$ to be determined by values of $T_1,T_2,T_3$ but its becoming too tedious to solve. Any alternate approach?
The approach is correct, and can be made less tedious by noting that:
*
*It is easier to shift the powers $\,T_n=A'(-1)^{n-1}+B'\left(\frac{3+\sqrt{5}}{2}\right)^{n-1}+C'\left(\frac{3-\sqrt{5}}{2}\right)^{n-1}\,$ since the sequence starts with index $\,1\,$.
*All terms in the sequence are integers, so $\,B',C'\,$ must be rational conjugates, therefore $\,A'=a\,$ and $\,B',C'=b \pm c\sqrt{5}\,$ where $\,a,b,c \in \mathbb Q\,$.
Then:
*
*$\require{cancel} \color{red}{20} = T_1 = a+b+\cancel{c\sqrt{5}}+b-\cancel{c\sqrt{5}} \color{red}{= a + 2 b}$
*$\color{red}{24} = 2 \cdot T_2 = -2a + (b+c\sqrt{5})(3+\sqrt{5}) + (b-c\sqrt{5})(3-\sqrt{5})$ $= -2a+3b+5c+\cancel{(b+3c)\sqrt{5}}+3b+5c-\cancel{(b+3c)\sqrt{5}}$ $\color{red}{= -2a + 6b + 10c}$
*$\color{red}{280} = 4 \cdot T_3 = 4a + (b+c\sqrt{5})(3+\sqrt{5})^2 + (b-c\sqrt{5})(3-\sqrt{5})^2$ $= 4a + (b+c\sqrt{5})(14 + 6 \sqrt{5}) + (b-c\sqrt{5})(14-6\sqrt{5})$ $= 4a+14b+30c+\cancel{(6b+14c)\sqrt{5}}+14b+30c-\cancel{(6b+14c)\sqrt{5}}$ $\color{red}{= 4a + 28 b + 60 c}$
Solving the system for $\,a,b,c\,$ gives $\,a = \dfrac{54}{5}, b = \dfrac{23}{5}, c = \dfrac{9}{5}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the minimum of $a^4+2b^4+3c^4$ Suppose that $a, b,$ and $c$
are positive real numbers satisfying
$a+b+c=3$. Find the minimum of $$a^4+2b^4+3c^4$$
We know $f(x)=x^4$ is convex on the positive reals so by Jensen's Inequality, we have $$a^4+b^4+c^4\ge \frac{1}{27}$$
Hence $$a^4+2b^4+3c^4\ge \frac{1}{27}+b^4+2c^4$$
What now?
| Using Holder's inequality:
$$\left(a^4+2b^4+3c^4\right)\cdot\left(1+\frac1{\sqrt[3]2}+\frac1{\sqrt[3]3}\right)^3 \geqslant (a+b+c)^4=3^4$$
Equality is possible when $a^3:b^3:c^3 =1:\frac12:\frac13$, so this gives the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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inverse trigonometric differentiation - 2 approaches, 2 different answers I have the following question:
$\arctan x^2+\arctan y^2=\frac{\pi}{4}$
Find $\frac{dy}{dx}$ in terms of $x$ and $y$.
I am aware we can differentiate implicitly straight away to get $\frac{2x}{1+x^4}+\frac{2y}{1+y^4}\cdot\frac{dy}{dx}=0$
then $\frac{dy}{dx}=\frac{-x\left(1+y^4\right)}{y\left(1+x^4\right)}$
However, I wasn't sure what was incorrect about the following approach and wondered if someone could highlight?
$\tan\left(\arctan x^2+\arctan y^2\right)=\tan\left(\frac{\pi}{4}\right)$
$\frac{x^2+y^2}{1-x^2y^2}=1$
$\frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(1-x^2y^2\right)$
$2x+2y\cdot\frac{dy}{dx}=-2xy^2-2x^2y\cdot\frac{dy}{dx}$
$\left(2+2x^2\right)y\cdot\frac{dy}{dx}=-2xy^2-2x$
$\frac{dy}{dx}=\frac{-2xy^2-2x}{\left(2+2x^2\right)\cdot y}=\frac{-x\left(1+y^2\right)}{y\left(1+x^2\right)}$
| From $x^2 + y^2 = 1 - x^2y^2$, we get:
$$x^2 = \frac{1-y^2}{1+y^2} $$
Then,
$$\frac{1+y^2}{1+x^2} = \frac{1+y^2}{1+\frac{1-y^2}{1+y^2}} = \frac12(1+y^2)^2$$
And,
$$\frac{1+y^4}{1+x^4} = \frac{1+y^4}{1+ \left( \frac{1-y^2}{1+y^2} \right)^2} = \frac12 (1+y^2)^2$$
The two expressions are equal and there's nothing wrong with any of the two methods you used.
| {
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Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}$, we have $xyz=1$ and using $abc=1$, the inequality can be written as:
$$\dfrac{x}{\sqrt{x+y}}+\dfrac{y}{\sqrt{y+z}}+\dfrac{z}{\sqrt{z+x}}\le \dfrac{xy+yz+zx}{\sqrt{2}}$$
I'm trying to use Cauchy-Schwarz:
$$LHS\le\sqrt{(x+y+z)(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x})}$$ but now I have to prove $$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$$ because $ab+bc+ca\ge\sqrt{3(a+b+c)}$, but I can't prove it. Can anyone give me a hint? Not necessarily a complete solution.
By the way, I also relized a problem that seems quite similar to the above problem $\sqrt{\frac{2 x}{x+y}}+\sqrt{\frac{2 y}{y+z}}+\sqrt{\frac{2 z}{z+x}} \leq 3$ if $x,y,z>0$ (Vasile Cirtoaje) (and then we can use $3\le xy+yz+zx$ ?Hope it helps)
| Use this inequality:
If a, b and c are all positive, show that:
$$\frac a{b+c}+\frac b{a+c}+\frac c{b+a}\geq\frac 32$$
Solution:
Since a, b and c are positive we have a+b, b+c, and c+a are also positive. Let b+c<c+a<a+b, so that $\frac 1{b+c}\geq \frac 1{c+a} \geq \frac 1{a+b}$
Using:
$(a_1+a_2+a_3+...+a_n)(b_1+b_2+b_3+...b_n)\geq n(a_1b_1+a_2b_2+a_3b_3+ ...a_nb_n)$
we have:
$[(a+b)+(b+c)+(a+c)]\cdot \big[\frac 1{ a+b} +\frac 1{b+c}+\frac 1{a+c}\big]\geq3\big[ (a+b)\cdot \frac 1 {a+b}+(b+c)\cdot \frac 1 {b+c}+(a+c)\cdot \frac 1 {a+c}\big]$
or:
$2(a+b+c)\big[\frac 1{a+b}+\frac 1{b+c} +\frac 1 {c+a}\big]\geq 9$
finally:
$\big(\frac a{b+c}+1+\frac b{c+a}+1+\frac c{a+b}+1\big)\geq \frac 9 2$
or:
$\big(\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\big)\geq \frac 3 2$
Now since you let $a=\frac 1x$, $b=\frac 1y$ and $c=\frac 1z$, the conditions changes and you have to get what you wanted to prove, i. e:
$\big(\frac x{x+y}+\frac y{y+z}+\frac z{z+x}\big)\leq \frac 3 2$
| {
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Is it possible to prove algebraically that these formulas for calculating a rotated point produce the same result? The following shows 2 formulas to calculate a point $(x',y')$, which is $(x,y)$ rotated $\theta$ degrees.
$$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}h\cos(\theta')\\h\sin(\theta')\end{bmatrix}$$
where
$$h=\sqrt{x^2+y^2}\\\theta'=\theta+\operatorname{arctan2}(x, y)$$
where
$$\operatorname{arctan2}(x, y) =
\begin{cases}
\arctan(\frac y x) &\text{if } x > 0 \\
\arctan(\frac y x) + \pi &\text{if } x < 0 \text{ and } y \ge 0 \\
\arctan(\frac y x) - \pi &\text{if } x < 0 \text{ and } y < 0 \\
+\frac{\pi}{2} &\text{if } x = 0 \text{ and } y > 0 \\
-\frac{\pi}{2} &\text{if } x = 0 \text{ and } y < 0 \\
0 &\text{if } x = 0 \text{ and } y = 0
\end{cases}$$
The first one is applying a 2D rotation matrix, and the second one is from a computer program I wrote some time ago, which directly handles the process of rotating a point, but computationally much slower.
I was wondering why these two produce the same result. Of course, they do the same operation geometrically, but is it possible to prove algebraically that the equality always hold for any real number $x$, $y$, and $\theta$?
| Following about the study of the geometric transformations of the plane it's known that the counterclockwise rotations of center $O$ are described by the linear equations:
$$
\begin{cases}
x' = x\,\cos\theta - y\,\sin\theta \\
y' = x\,\sin\theta + y\,\cos\theta \\
\end{cases}\,.
$$
At this point we can decline two cases:
*
*if $x^2+y^2=0$, i.e. $x=y=0$, we have $x'=y'=0$;
*if $x^2+y^2\ne0$, instead, we can write:
$$
\begin{cases}
x' = \sqrt{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\,\cos\theta - \frac{y}{\sqrt{x^2+y^2}}\,\sin\theta\right) \\
y' = \sqrt{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\,\sin\theta + \frac{y}{\sqrt{x^2+y^2}}\,\cos\theta\right) \\
\end{cases}
$$
and given that:
$$
\small
-1 \le \frac{x}{\sqrt{x^2+y^2}} \le 1\,,
\quad \quad
-1 \le \frac{y}{\sqrt{x^2+y^2}} \le 1\,,
\quad \quad
\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2+\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2=1
$$
it's possible to write:
$$
\begin{cases}
x' = \sqrt{x^2+y^2}\left(\cos\varphi\,\cos\theta - \sin\varphi\,\sin\theta\right) \\
y' = \sqrt{x^2+y^2}\left(\cos\varphi\,\sin\theta + \sin\varphi\,\cos\theta\right) \\
\end{cases}
$$
which thanks to the cosine and sine addition formulas are equivalent to:
$$
\begin{cases}
x' = \sqrt{x^2+y^2}\,\cos(\varphi+\theta) \\
y' = \sqrt{x^2+y^2}\,\sin(\varphi+\theta) \\
\end{cases}\,.
$$
This is exactly what you wrote, where:
$$
\cos\varphi := \frac{x}{\sqrt{x^2+y^2}}\,,
\quad \quad \quad
\sin\varphi := \frac{y}{\sqrt{x^2+y^2}}
$$
and from trivial graphical considerations it's possible to retrace all the cases with which 2-argument arctangent function was built:
$$
\varphi = \text{atan2}(y,\,x)\,.
$$
The fact that you, artificially, also defined the case $x=y=0$ means that all possible cases are covered and this proves the equality between the two formulations, even if the first is cleaner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Pythagorean triples and primes.
Determine whether there are any right-angled triangles with integer lengths such that the
lengths of both of the sides adjacent to the right angle are primes.
This was the question I was asked by my teacher. My first answer is no, which I still believe is correct, but then she asked me "why?" and I got stumped. My first thought was that Pythagorean triples can be generalized as:
$$2k,\quad k^2-1,\quad k^2+1$$
Is this generalization correct? Because, if so, $2k$, for $k > 1$, cannot be a prime as it is divisible by $2$. Even if $k= 1$, there are no Pythagorean integer triples that have $2$.
Also, $k^2-1$ cannot be a prime, for $k > 2$, because it can be factored like this:
$(k+1)(k-1)$
Even if $\quad k = 2,\quad 2k = 4,\space k^2-1=3, \space k^2+1 = 5.\quad$ Thus, the two sides adjacent to the right angle would be 3 and 4, one of which (4) is not prime.
Only $k^2+1$ could be prime, proving that either a or b can be prime, but both can't.
Is this proof correct? If the generalization is wrong, the proof is also flawed.
Thanks.
| If $a=2$ and $b\ge 2$ is prime, we need an integer $c$ with $c^2=b^2+2^2$. As $(b+1)^2=b^2+2b+1>b^2+4$, we’d need $b^2<c^2<(b+1)^2$, which is impossible.
Hence we must have odd sides $a=2r+1$ and $b=2s+1$, so
$$\tag1c^2=(2r+1)^2+(2s+1)^2
=4(r^2+r+s^2+s)+2,$$
a number that is even, but not a multiple of $4$. If $c$ is odd, then $c^2$ is odd; and if $c$ is even, $c^2$ is a multiple of $4$. Neither of these options is compatible with $(1)$, hence no solution exists.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Fair price to play a dice game We play a game with a pot and a single die. The pot starts off empty.
*
*if the die roll is 1, 2 or 3, I put 1 pound in the pot, and the die is thrown again
*if its 4 or 5, the game finishes, and you win whatever is in the pot
*if its 6, you leave with nothing
What is the fair price for you to play this game?
*
*1/2
*1/3
*3/2
*2/3
| This question is equivalent to asking the expected value of the price in the pot when the game ends.
There’s a $\frac{1}{2}\cdot \frac{1}{3}$ chance the total is 1 (1, 2, or 3 and then 4 or 5), $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{3}$ chance the total is 2 (1, 2, or 3 and then 1, 2, or 3 and then 4 or 5), and so on, such that we want to find $\frac{1}{3}(1\cdot (\frac{1}{2})^1+2\cdot (\frac{1}{2})^2+3\cdot (\frac{1}{2})^3+\dots).$
Let $S = \frac{1}{3}(1\cdot (\frac{1}{2})^1+2\cdot (\frac{1}{2})^2+3\cdot (\frac{1}{2})^3+\dots)$
Then $\frac{S}{2} = \frac{1}{3} (1\cdot (\frac{1}{2})^2+2\cdot (\frac{1}{2})^3 + \dots)$
Subtracting the two equations gets $S-\frac{S}{2} = \frac{1}{3}(1\cdot (\frac{1}{2})^1 + 1\cdot (\frac{1}{2})^2+1\cdot (\frac{1}{2})^3+\dots).$ The parentheses is just equal to $1,$ so $\frac{S}{2} = \frac{1}{3}\cdot 1,$ so $S = \boxed{\frac{2}{3}.}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$
I want to prove that
$$\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$$
for all $n\ge 0$, where $1<\alpha\le\beta$ and $K$ a constant.
Everything I tried to do, I always arrived at the constant depending on $n$, which cannot happen. And I didn't want to use induction, is it possible?
| Complement to Diger's proof:
According to Diger's idea, we have
\begin{align*}
&\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\beta}}\\
\le\,& \sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
\le\,& \sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}
+ \sum_{n/2\le k\le n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
=\,& \sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}
+ \sum_{0\le m\le n/2}{\frac{1}{(n-m+1)^\alpha}\frac{(n + 1)^\alpha}{(m + 1)^\alpha}}\\
=\,& 2\sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
\le\,& 2\sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-n/2)^\alpha}}\\
\le\,& 2^{\alpha + 1}
\sum_{k=0}^ {\infty}\frac{1}{(k+1)^\alpha}.
\end{align*}
Since $\sum_{k=0}^ {\infty}\frac{1}{(k+1)^\alpha}$ is convergent (p-series test), the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.
| We may also work from the disposition of the points of interest toward obtaining the slope of the line. We then need to treat this question as two separate cases.
The "easier" of the two to set up geometrically occurs for $ \ c \ < \ 0 \ \ , \ $ in which $ \ P \ , \ $ the $ \ y-$intercept of the line $ \ y \ = \ mx + c \ $ lies "below" the vertex of the parabola $ \ y \ = \ x^2 \ \ . \ $ Both intersection points $ \ A \ $ and $ \ B \ $ of the line with the parabola lie on the same side of the $ \ y-$axis (since we want $ \ m \ > \ 0 \ ) \ , \ $ so determining the circumstance for which $ \ AP - BP \ = \ 1 \ $ involves requiring the distance between $ \ A \ $ and $ B \ $ to equal $ \ 1 \ \ . \ $ Since the slope $ \ m \ $ of the line is the tangent of the angle $ \ \theta \ $ that the line makes to the "direction" of the positive $ \ x-$axis, then if we call $ \ \Delta \ $ the difference in the $ \ x-$coordinates of the intersection points, we have $ \ B \ (X \ , \ X^2) \ $ and $ \ A \ (X + \Delta \ , \ [X + \Delta]^2) \ \ , \ $ where we may also write
the $ \ y-$coordinate of $ \ A \ $ as $ \ X^2 + \Delta·\tan \theta \ = \ X^2 + m·\Delta \ \ . \ $ The requirement for the separation between the intersection points then gives us
$$ \Delta^2 \ + \ (m·\Delta)^2 \ \ = \ \ 1 \ \ \Rightarrow \ \ \Delta \ \ = \ \ \frac{1}{\sqrt{1 \ + \ m^2}} \ \ . \ $$
If we solve directly for the intersection points (if they exist), they must satisfy $ \ x^2 \ = \ m·x + c \ \Rightarrow \ x^2 - m·x - c \ = \ 0 \ \ , \ $ which has the discriminant $ \ \mathcal{D} \ = \ m^2 + 4c \ \ . \ $ The line will be just tangent to the parabola when this discriminant is zero; since $ \ c \ < \ 0 \ \ , \ $ the slope of the tangent line $ \ m \ = \ 2·\sqrt{|c|} \ \ . \ $
Thus, we must have $ \ m \ > \ 2·\sqrt{|c|} \ $ in order to have two intersection points on the parabola, these lying at $ \ x \ = \ \frac{m}{2} \ \pm \ \frac{\sqrt{\mathcal{D}}}{2} \ \ . \ $ This leads to
$$ \sqrt{\mathcal{D}} \ \ = \ \ \Delta \ \ \Rightarrow \ \ \sqrt{m^2 - 4·|c|} \ \ = \ \ \frac{1}{\sqrt{1 \ + \ m^2}} $$
and the biquadratic equation $ \ m^4 \ + \ (1 - 4·|c|)·m^2 \ - \ (1 + 4·|c|) \ = \ 0 \ \ . $ (It becomes clear why the problem asks for $ \ m^2 \ ) \ . \ $ The slope solution that guarantees intersection points between the line and the parabola for this case is then
$$ m^2 \ \ = \ \ \frac{4·|c| \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 - 4·|c|)^2 \ + \ 16·|c| + 4} }{2} $$ $$ = \ \ \frac{4·|c| \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 + 4·|c|)^2 \ + \ 4}}{2} \ \ > \ \ 4·|c| \ \ . $$
Using this for $ \ c \ = \ -4 \ \ , \ $ we find that $ \ m^2 \ = \ \frac{15}{2} + \frac{\sqrt{293}}{2} \ \approx \ 16.0586 \ \Rightarrow \ m \ \approx \ 4.0073 \ \ . \ $ With $ \ \mathcal{D} \ \approx \ 0.0586 \ , \ $ the intersection points are then $ \ B \ (1.883 \ , \ 3.544) \ $ and $ \ A \ (2.125 \ , \ 4.514) \ \ ; \ $ the distance separating them is indeed $ \ 1 \ $ . [Alternatively, with $ \ P \ (0 \ , \ -4) \ \ , \ $ we can verify that $ \ AP \ \approx \ 8.775 \ $ and $ \ BP \ \approx \ 7.775 \ \ . \ ] $
For the case with $ \ c \ > \ 0 \ \ , \ $ the intersection points are on opposite sides of the $ \ y-$axis, with $ \ x_A \ > \ 0 \ $ and $ \ x_B \ < \ 0 \ \ ; \ $ because $ \ m \ > \ 0 \ \ , \ $ we will have $ \ AP \ > \ BP \ \ . \ $ We can form two similar triangles with vertices $ \ A \ , \ P \ , \ (x_A \ , \ c) \ $ and $ \ B \ , \ P \ , \ (x_B \ , \ c) \ \ ; \ $ if we arrange the smaller triangle "within" the larger one, the requirement that $ \ AP - BP \ = \ 1 \ $ again is equivalent to having the separation between $ \ A \ (x_A \ , \ x_A^2 - c) \ $ and $ \ B' \ (-x_B \ , \ 2c - x_B^2) \ $ be equal to $ \ 1 \ \ . $ Constructing the small triangle $ \ \triangle AB'C \ $ with $ \ C \ (x_A \ , \ 2c - x_B^2) \ \ , \ $ we have $ \ AC \ = \ m·B'C \ \ , \ $ and so $ \ \Delta^2 \ + \ (m·\Delta)^2 \ \ = \ \ 1 \ \ . \ $ In this way, we can again use the relation $ \ \Delta \ = \ \frac{1}{\sqrt{1 \ + \ m^2}} \ \ , \ $ with this quantity now being $ \ x_A - (-x_B) \ = \ x_A + x_B \ \ . $
The intersection points are also found in the same way as before, except there is no "tangency" restriction for $ \ m \ \ . \ $ We now obtain
$$ \Delta \ \ = \ \ \left( \ \frac{m}{2} \ + \ \frac{\sqrt{\mathcal{D}}}{2} \ \right) \ + \ \left( \ \frac{m}{2} \ - \ \frac{\sqrt{\mathcal{D}}}{2} \ \right) \ \ = \ \ m $$
$$ \Rightarrow \ \ \Delta \ \ = \ \frac{1}{\sqrt{1 \ + \ m^2}} \ \ = \ \ m \ \ \Rightarrow \ \ m^2 \ · \ (1 \ + \ m^2) \ \ = \ \ 1 \ \ \Rightarrow \ \ m^4 \ + \ m^2 \ - \ 1 \ \ = \ \ 0 \ \ , $$
for which the permissible solution is $ \ m^2 \ = \ -\frac12 + \frac{\sqrt5}{2} \ \ . \ $ Perhaps surprisingly, the condition $ \ AP - BP \ = \ 1 \ \ $ is satisfied by only one slope for the line, regardless (!) of the value of $ \ c \ > \ 0 \ \ . \ $ [The "Golden Ratio" emerges stealthily, in that we have $ \ m^2 \ = \ \frac{1}{\phi} \ \Rightarrow \ m \ = \ \frac{1}{\sqrt{\phi}} \ \ . \ ] $
We may also note that the two cases meld seamlessly at $ \ c \ = \ 0 \ \ : \ $ applying our slope formula for $ \ c \ < \ 0 \ $ produces
$$ m^2 \ \ = \ \ \frac{4·0 \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 + 4·0)^2 \ + \ 4}}{2} \ \ = \ \ -\frac{1}{2} \ + \ \frac{\sqrt{ 1 \ + \ 4}}{2} \ \ = \ \ -\frac{1}{2} \ + \ \frac{\sqrt{5}}{2} \ \ . $$ The point $ \ B \ $ becomes "identified" with $ \ P \ \ , $ now at the origin, and the intersection point $ \ A \ $ of the line $ \ y \ = \ \frac{1}{\sqrt{\phi}}·x \ $ with the parabola is located at $ \ \left(\frac{1}{\sqrt{\phi}} \ , \ \frac{1}{\phi} \right) \ \ ; \ $ its distance from $ \ P \ $ is indeed $$ \ \left(\frac{1}{\sqrt{\phi}} \right)^2 \ + \ \left(\frac{1}{\phi} \right)^2 \ \ = \ \ \frac{1}{\phi} \ + \ \frac{1}{\phi^2} \ \ = \ \ \frac{\phi \ + \ 1}{\phi^2} \ \ = \ \ \frac{\phi^2}{\phi^2} \ \ = \ \ 1 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find equation of the plane that this $x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}$ curve lies
Prove that all points of the given curve lie in one plane, and find
the equation of that plane:
$$x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}.$$
If the given curve lies in one plane, then
$$a\left(\frac{1+t}{1-t}\right)+b\left(\frac{1}{1-t^2}\right)+c\left(\frac{1}{1+t}\right)+d=0.$$
Solving this I get $2a=c,a=d,a=\frac{-b}{4}.$
How from this find equation of plane? Or maybe I did something wrong?
When putting values back into equation of plane I get
$$ax-4ay+2az+a=0.$$ Now, problem is I can't cancel $a$ here because first of all, I need to prove that such plane exists.
| Observe that (for $t\neq\pm1$)
$$\begin{align}
x&=\frac{2}{1-t}-1\iff\frac{1/2}{1-t}=\frac{x+1}{4},\\
y&=\frac{1/2}{1-t}-\frac{1/2}{1+t},\\
z&=\frac{1}{1+t}\iff\frac{1/2}{1+t}=\frac{z}{2}.
\end{align}$$
From here
$$y=\frac14(x+1)-\frac12z\iff x-4y+2z=-1.$$
Alternatively, the curves equation is
$$\begin{pmatrix}
-1\\ 0\\ 0
\end{pmatrix}+
\frac{1}{1-t}
\begin{pmatrix}2\\ 1/2 \\0
\end{pmatrix}+
\frac{1}{1+t}
\begin{pmatrix}
0\\-1/2\\1
\end{pmatrix},
$$
clearly an equation of a part of a plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Quadrilateral ABCD satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $Quadrilateral ABCD is inscribed in a circle satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $<BAC=60$
I've never been good at Euclidean geometry questions like this... Really, what strategies could I employ to begin an analysis of the situation? I'm working through a text on Euclidean geometry and this is a question. The specific questions being asked about this scenario are the following:
1). The radius of the circumscribed circle
2). $\overline{AC}$
3). $<BDC$
4). The area of $\Delta BDC$
5). $\overrightarrow{CA} \cdot \overrightarrow{DC}$
Now, The fact that $\overline{BD}=\overline{DC}$ struck me as odd. The length of a diagonal is equal to the length of one of the sides? I want to say that this implies that the quadrilateral we are dealing with must not be convex.
Since $<BAC=60$, we can use the law of cosines to deduce that $\overline{AC}=2$ and so $\overline{AB}=1$.
My analysis sort of hits a road block here. I have not used the fact that the quadrilateral is inscribed in a circle. For a quadrilateral inscribed in a circle, it is well known that:
*
*The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.
*
*The opposite angles of quadrilateral inscribed in a circle are supplementary. i.e., the sum of the opposite angles is equal to 180˚.
However, I have not been able to make use of these facts.
Can anyone help me out here?? Thanks in advance!
| Let $x = \overline{AB}$ , then $\overline{AC} = 2 x$, and we are given that $\overline{BC} = \sqrt{3} $ and that $\angle BAC = 60^\circ$
Applying the law of cosines to $\triangle ABC$, we get
$ (\sqrt{3})^2 = x^2 + (2 x)^2 - 2 x (2 x) (\frac{1}{2}) $
Hence,
$ 3 = x^2 + 4 x^2 - 2 x^2 = 3 x^2 $
From which, $x = 1$. Therefore, $\overline{AB} = 1 $ and $\overline{AC} = 2 $
Now using coordinate geometry, we can set vertex $A = (0,0)$ and $B = (1, 0)$, then $C = (1, \sqrt{3}) $ (Assuming $C$ is in the upper half plane).
Next, we find the circle on which $A,B,C$ lies. The center is the intersection of the perpendicular bisector of $AB$, which is the line $x = \dfrac{1}{2}$, and the perpendicular bisector of $BC$, which is the line $ y = \dfrac{\sqrt{3}}{2} $
Hence the center of the circumcircle of $\triangle$ ABC is $O = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) $
The radius of this circumcircle is $R = \overline{OA} = 1 $
Since point $D$ lies on the cicrcumcircle and on perpendicular bisector of $BC$ (because it is equidistant from $B$ and $C$), then we $D$ can be
$D_1 = O + (-R, 0) = (-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) \hspace{25pt} (1)$
or
$D_2 = O + (R, 0) = ( \dfrac{3}{2} , \dfrac{\sqrt{3}}{2} ) \hspace{25pt} (2)$
However, from the cyclic order of the vertices, $D_2$ is rejected. Hence $D = D_1 $.
Now we can compute the area.
The area of $ABCD = [ABC] + [ACD_1] = \frac{1}{2} \left( \sqrt{3} + \dfrac{\sqrt{3}}{2} - \sqrt{3} (- \dfrac{1}{2} ) \right) = \sqrt{3} $
Also, we have $\vec{CA} = A - C = (-1, -\sqrt{3} ) $, and $\vec{DC} = C - D = ( \dfrac{3}{2} , \dfrac{\sqrt{3}}{2} ) $
So $\vec{CA} \cdot \vec{DC} = - \dfrac{3}{2} - \dfrac{3}{2} = -3 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413992",
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"source": "stackexchange",
"question_score": "2",
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Rewriting $\cos^4 x \sin^2 x $ with exponent no higher than $1$ I'm having some trouble finishing this one off.
Rewrite with exponent no higher than $1$:
$$\cos^4 x \sin^2 x$$
The answer is:
$$\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
So I started like this:
$$\cos^4 x \sin^2 x = \frac{1+\cos(2x)}{2}\frac{1+\cos(2x)}{2}\frac{1-\cos(2x)}{2}$$
$$= \frac{1}{8}\left(\{1+\cos(2x)\}\{1^2 - \cos^2(2x)\}\right)$$
$$\frac{1}{8}\left(\{1 + \cos(2x)\}\sin^2(2x)\right)$$
$$\frac{1}{16}\left(1 + \cos(2x)\{1-\cos(4x)\}\right)$$
Now this is where I start to get lost:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \cos(2x)\cos(4x) \right)$$
I really can't find a way from here - I try this, but not sure if this is the right path.
$$\require{cancel} \cancel{\frac{1}{16}\left(1 - \cos(4x) + \cos(2x)\{1 - \cos(4x)\} \right)}$$
Completing thanks to help below:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \left(\frac{\cos(6x) + \cos(-2x)}{2}\right)\right)$$
$$\frac{1}{32}\left(2 - 2\cos(4x) + 2\cos(2x) - \cos(6x) - \cos(2x)\right)$$
$$=\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
| Since $\cos(3\theta) = 4\cos^{3}(\theta) - 3\cos(\theta)$, you can also proceed backwards:
\begin{align*}
\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32} & = \frac{2 + \cos(2x) - 2(2\cos^{2}(2x) - 1) - (4\cos^{3}(2x) - 3\cos(2x))}{32}\\\\
& = \frac{-4\cos^{3}(2x) - 4\cos^{2}(2x) + 4\cos(2x) + 4}{32}\\\\
& = \frac{-\cos^{3}(2x) - \cos^{2}(2x) + \cos(2x) + 1}{8}\\\\
& = \frac{-\cos^{2}(2x)(\cos(2x) + 1) + \cos(2x) + 1}{8}\\\\
& = \frac{(1 - \cos^{2}(2x))(\cos(2x) + 1)}{8}\\\\
& = \frac{2\sin^{2}(2x)\cos^{2}(x)}{8}\\\\
& = \frac{\sin^{2}(2x)\cos^{2}(x)}{4}\\\\
& = \frac{4\sin^{2}(x)\cos^{2}(x)\cos^{2}(x)}{4}\\\\
& = \cos^{4}(x)\sin^{2}(x)
\end{align*}
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand... like just adding up the fractions until I get to $\frac{7}{8}$ and got $n=7$. Surely there must be a better way. Help appreciated here! Thanks, I really appreciate it.
| This is a simple telescoping series trick
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1} \\ $$
By choosing values of k for $k = 1, \cdots, n$ then we have
$$\begin{align}k&=1 \implies 1-\frac{1}{2} \\ k&=2 \implies \frac{1}{2} - \frac{1}{3} \\\vdots \\ k&= n-1 \implies \frac{1}{n-1}- \frac{1}{n} \\ k&=n \implies \frac{1}{n} - \frac{1}{n+1} \end{align}$$
Everything cancels except for $1 -\frac{1}{n+1}$, therefore we have that
$$1-\frac{1}{n+1} = \frac{7}{8} \implies n = 7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to prove by induction $\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $ I'm trying to solve this problem about mathematical induction but every time I try to solve it I end up with an incorrect answer.
The problem is as follows:
Prove by induction : $$\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $$
my work:
first I prove that it holds for n=1: $\frac 12 = 1- \frac {1}{2^1} = \frac 12$
Now assume it holds for n=k, prove for n=k+1 the sum is equal to $1 - \frac{1}{2^{k+1}}$
then $\frac 12 + \frac 14 +...+ \frac {1}{2^k} + \frac {1}{2^{k+1}} = 1-\frac {1}{2^k} + \frac {1}{2^{k+1}} = 1 - \frac {2^{k+1}+2^k}{2^{k+1}2^k} = 1- \frac {2(2^k)+2^k}{2(2^k)2^k} = 1 - \frac {3(2^k)}{2(2^k)2^k} = 1- \frac {3}{2(2^k)} = 1- \frac {3}{2^{k+1}} $
I would very much appreciate if at least someone can tell me where I'm wrong so I could try to solve it by myself. I think the mistake is at factoring $2^{k+1}+2^k$
| Be careful with the minus sign.
$$-\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{-2^{k+1} + 2^k}{2^k 2^{k+1}}
= \frac{-2 + 1}{2^{k+1}} = -\frac{1}{2^{k+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that $\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {a + b + c} \right)}}$
Let $a,b,c>0,\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) =10.$ Prove that
$$\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {a + b + c} \right)}}.$$
Side Note: The familiar inequality is:
$\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{9}}{{2\left( {a + b + c} \right)}}.$
| It suffices to prove that
$$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \ge \frac{19}{12}.$$
Since the condition and the inequality are both homogeneous and symmetric, assume that $c = \max(a, b, c) = 1$.
Let $p = a + b, q = ab$. We have $0 < p \le 2$.
It suffices to prove that
$$\frac{p^2 - 2q + p}{1 + p + q} + \frac{1}{p} \ge \frac{19}{12}.$$
From $(a + b + c)(1/a + 1/b + 1/c) = 10$, we have $(p + 1)(p/q + 1) = 10$
and $q = p(1 + p)/(9 - p)$.
It suffices to prove that
$$\frac{p^2 - 2\cdot p(1 + p)/(9 - p) + p}{1 + p + p(1 + p)/(9 - p)} + \frac{1}{p} \ge \frac{19}{12}$$
or
$$\frac{(4 - p)(3 - 2p)^2}{36p} \ge 0 $$
which is true.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/4422755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find: $\lim\limits _{n \rightarrow \infty} n^{3} \int\limits_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$ Problem:
What is the following limit?
$\lim _{n \rightarrow \infty} n^{3} \int_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$
Attempt:
Denote $ f(x) = \frac{1}{1+x^8} $ for $ x \in [ -\pi , \pi] $ ( Note that this function is infinitely differentiable. )
$ I_n = n^{3} \int_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x = \frac{n^{3}}{2} \int_{-\pi}^{\pi} \frac{ e^{2inx} + e^{-2inx} }{1+x^{8}} d x = \frac{2\pi \cdot n^{3}}{2} ( \hat{f}(2n) + \hat{f}(-2n) ) $.
Now we use $ \hat{f^{(k)}}(n) = (in)^{k}f(n) $ ( since $ f$ is indeed differentiable on its interval ), thus
$ I_n = \frac{2\pi \cdot n^{3}}{2} ( \hat{f}(2n) + \hat{f}(-2n) ) = \pi \cdot n^3 ( \frac{\hat{f^{(3)}}(2n)}{-2^3n^3}
+ \frac{\hat{f^{(3)}}(-2n)}{-2^3n^3} ) = \frac{-\pi}{8} ( \hat{f^{(3)}}(2n) + \hat{f^{(3)}}(-2n) ) $
And by Riemann-Lebesgue we get $ I_n \to 0 $.
I'm not really sure whether the limit I calculated is justified or the proof's specious ( Which, I think it is but I don't know why, probably where I said that $ \hat{f^{(k)}}(n) = (in)^{k}f(n) $ ). then I had doubts whether the limit is $ 0, \infty, -\infty $ or none of them, I think it's none of them but don't have a gist of justification as to why. Can you please help as to how to find the limit? thanks very much for the help in advance!
| Too long for a comment
Using the recommendation of @Gribouillis, integrating by part several times, using the fact that the integrand is even and denoting $\,\Big[\frac{1}{1+x^8}\Big]^{(k)}=\frac{d^k}{dx^k}\frac{1}{1+x^8}$
$$I(n)=2\int_0^\pi \frac{\cos (2 n x)}{1+x^{8}} d x=\frac{1}{n}\frac{\sin 2nx}{1+x^8}\bigg|_{x=0}^{x=\pi}+\frac{1}{n}\int_0^\pi \sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(1)}d x$$
$$=-\frac{1}{2n^2}\cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(1)}\bigg|_{x=0}^{x=\pi}+\frac{1}{2n^2}\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(2)}d x$$
$$=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+\frac{1}{4n^3}\sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(2)}\bigg|_{x=0}^{x=\pi}-\frac{1}{4n^3}\int_0^\pi \sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(3)}d x$$
$$=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+\frac{1}{8n^4}\cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(3)}\bigg|_{x=0}^{x=\pi}-\frac{1}{8n^4}\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(4)}d x$$
Evaluating the third term:
$$\frac{1}{8n^4}\bigg|\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(4)}d x\bigg|<\frac{1}{8n^4}\bigg|\int_0^\pi \Big[\frac{1}{1+x^8}\Big]^{(4)}d x\bigg|<\frac{\operatorname{Const}}{8n^4}$$
Therefore,
$$I(n)=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+O\Big(\frac{1}{n^4}\Big)$$
and
$$\lim_{n\to\infty}n^3I(n)=\lim_{n\to\infty}\Big(-n\frac{4\pi^7}{(1+\pi^8)^2}\Big)=-\infty$$
Numeric check at WolframAlpha confirms the result. At $n=10 \,000$
$$10^{12}\int_{-\pi}^\pi\frac{\cos(20\,000\, x)}{1+x^8}dx=-1.341589316...$$
while
$$-\frac{4\pi^7}{(1+\pi^8)^2}\cdot 10\,000=-1.341589346...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$
If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$
My Attempt:
Replacing $x$ by $-x$ in the given expression, $$(1-x+x^2)^n=a_0-a_1x+a_2x^2-\dots+a_{2n}x^{2n}$$
Therefore, $$(1+x+x^2)^n(1-x+x^2)^n=(a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n})(a_0-a_1x+a_2x^2-\dots+a_{2n}x^{2n})$$
LHS $=((1+x^2)^2-x^2)^n=(1+x^2+x^4)^n$
And if from the right side, I try to find the coefficient of $x^{2n}$, I get $a_0a_{2n}-a_1a_{2n-1}+\dots$
It seems like I need to do $a_0=a_{2n}, a_1=a_{2n-1}, \dots$
$a_0=^nC_0$ and $a_{2n}=^nC_n$, so, they are equal.
I am confused with $a_{n-1}$ and $a_{n+1}$.
$a_{n+1}=^nC_{\frac{n+1}2}$. But for this, $n$ should be odd, right? What if $n$ is even?
For $a_{n-1}$, would we obtain coefficients both from $x$ and $x^2?$
| Your approach is absolutely the way to go. Just a tiny step is missing.
So, you let $f(x)=(1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k$ and compute the coefficient of $x^{2n}$ in $$f(x)f(-x)=f(x^2)$$ in two ways: obviously $[x^{2n}]f(x^2)=a_n$; on the other hand, $$[x^{2n}]f(x)f(-x)=\sum_{k=0}^{2n}(-1)^k a_k a_{2n-k}=\sum_{k=0}^{2n}(-1)^k a_k^2=(-1)^na_n^2+2\sum_{k=0}^{n-1}(-1)^ka_k^2;$$ the last two equalities follow from $a_k=a_{2n-k}$ which is a consequence of $f(x)=x^{2n}f(1/x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Straight line $y = x + 2$ cuts circle at $x^{2} + y^{2} = 20$ at $A$ and $B$ determine the values of $A$ and $B$ Im currently studying circles so now that I have two equations I'm confused. I'm asked to determine the values of $A$ and $B$ given straight line $y = x+2$ that cuts circle at $x^{2} + y^{2} = 20$.
| If $(x,y)$ is in the intersection of the line $y=x+2$ and the circle $x^2+y^2=20$, then\begin{align*}x^2+(x+2)^2 & =20 \\
x^2+x^2+4x+4 & =20 \\
2x^2+4x-16 & =0 \\
x^2+2x-8 & =0 \\
(x-2)(x+4) & =0
\end{align*}hence $x=2$ or $x=-4$. This gives the points $(2,4)$ and $(-4,-2)$, because $y=x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4433181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\iint_D\frac{\sin y}{y}dxdy$, where $D$ is the region enclosed by $y=2$, $y=1$, $y=x$, $y=2x$ I have been given this exercise: Calculate the double integral:
$$\iint_D\frac{\sin(y)}{y}dxdy$$
Where $D$ is the area enclosed by the lines: $y=2$, $y=1$, $y=x$, $2y=x$ (not $y = 2x$).
Visualising $D$ is easy. You can split D in two sub areas and get the bounds for the integrals. The problem I face is:
Let's split D in two sub areas, $D_1$ and $D_2$. $D_1$ is the left, upright triangle of $D$ and $D_2$ is the right, upside down one.
Then $D_1$ is defined by the lines $y=1$, $y=x$, and $x=2$.
You can express the area in a $y$-normal form as:
$$\begin{align}
1 \le y \le 2\\
y \le x \le 2
\end{align}$$
then the integral can be written as
$$ \begin{align}
&\int_1^2\int_y^2\frac{\sin(y)}{y}dxdy \\
&=\int_1^2\frac{\sin(y)}{y}[x]^2_y \space dxdy \\
&=\int_1^2\frac{2\sin(y)}{y} - \sin(y)dy \\
&=2\int_1^2\frac{\sin(y)}{y}dy -\int_1^2 \sin(y)dy \\
\end{align}$$
The second integral is trivial, but in the first one is not. I have tried substituting, integrating by parts but to no avail. What am I doing wrong?
Any answer is really appreciated.
| Note that the integrand does not depend on $x$:
$$
\int\int_D \frac{\sin y}{y} dx dy
= \int_1^2 \left( \int_{y/2}^y dx\right) \frac{\sin y}{y} dy
= \int_1^2 \frac{y}{2}\frac{\sin y}{y} dy
$$
$$
= \int_1^2 \frac{\sin y}{2} dy = \frac{\cos 1 - \cos 2}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute the following integral: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$ Question: Compute: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$. This is from Folland's Real Analysis book.
If we can find an integrable majorant, then the integral will equal $0$, since $\lim_{n\rightarrow\infty}\sin(x/n)=0$. We can use the Binominal Theorem to show $(1+\frac{x}{n})^n\geq 1+x+(\frac{1}{2}-\frac{1}{2n})x^2$, by noticing that all the terms will be positive since $x$ only takes nonnegative values, so we just "cut it off" after those couple of terms. Next, since $|\sin x|\leq 1$, we can say $|\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^n}|\leq \frac{1}{1+x+(\frac{1}{2}-\frac{1}{2n})x^2}$, and $\int_0^\infty\frac{1}{1+x+(\frac{1}{2}-\frac{1}{2n})x^2}<\infty$, so we have our integrable majorant, and by Dominated Convergence Theorem we can pass the limit inside the integral giving us a value of $0$.
I am curious if there are any (unseen by me) holes in this argument, or, maybe, are there any other ways of using DCT to solve this? Thank you!
EDIT:
Just a quick note, I wanted to use Bernouli's inequality to get $\frac{1}{(1+\frac{x}{n})^n}\leq\frac{1}{1+n\frac{x}{n}}\leq\frac{1}{1+x}$... but $\int_0^\infty\frac{1}{1+x}dx$ doesn't converge.
| Let
$$I_1(n) := \int_0^n (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x$$
and
$$I_2(n) := \int_n^\infty (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x.$$
Clearly, $I_1(n) \ge 0$.
Using $\sin u \le u$ for all $u\ge 0$, we have
$$I_1(n) \le \int_0^n (1 + x/n)^{-n} \frac{x}{n} \mathrm{d} x = \frac{n(1 - n 2^{1 - n})}{(n - 1)(n - 2)}.$$
(Note: $\int (1 + x/n)^r \mathrm{d} x = \frac{n}{r + 1}(1 + x/n)^{r + 1} + C$ for $r \ne -1$.)
Thus, we have $\lim_{n\to \infty} I_1(n) = 0$.
Also, we have
$$|I_2(n)| \le \int_n^\infty (1 + x/n)^{-n} \left|\sin \frac{x}{n}\right| \mathrm{d} x
\le \int_n^\infty (1 + x/n)^{-n} \mathrm{d} x = \frac{n 2^{1 - n}}{n - 1}.$$
Thus, we have $\lim_{n\to \infty} I_2(n) = 0$.
The desired result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$ for $a,b\in\mathbb{Z}^+$ Evaluate $$\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
where $a$ and $b$ are positive integers and $\{x\}:=x-\lfloor x\rfloor$
WLOG, we may assume that $a\le b$. Via the substitution $x \mapsto x/a$, we obtain
$$\int_0^a \left(\{x\}-\frac{1}{2}\right)\left(\left\{\frac{bx}{a}\right\}-\frac{1}{2}\right) dx$$
I know that we should partition the interval $(0,1)$ in such a way that the values of $\{x\}$ and $\{bx/a\}$ are constant in each subinterval. How do I think of such a partition? I also tried to express $a=gu$ and $b=gv$ where $u:=\gcd(a,b)$. But then again I face the same issue of partitioning $(0,1)$
| This is a brute force approach:
$$\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
Break the interval up into $ab$ parts to avoid the switch:
$$\sum_{k=1}^{ab}\int_{(k-1)/(ab)}^{k/(ab)} \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
Now $\{ax\}$ is just a straight line segment with a slope of $a$ starting at a height of $(k \bmod b)/b$.
$$\sum_{k=1}^{ab}\int_{0}^{1/(ab)} \left( \frac{k \bmod b}{b} + xa -\frac{1}{2}\right)\left(\frac{k \bmod a}{a} + xb -\frac{1}{2}\right) dx$$
$$\frac{1}{12a^2b^2} \sum_{k=1}^{ab} \left(\begin{array} {rl}
&(k \bmod a)(k \bmod b)\cdot 12
\\ + &(k \bmod a)(6 - 6b)
\\ + &(k \bmod b)(6 - 6a)
\\ + &3ab-3b-3a+4
\end{array}\right)$$
And now just a number theory problem. Using $d=\gcd(a,b)$
$$\sum_{k=1}^{ab} (k \bmod a)(k \bmod b)
= d\sum_{0 \le p < a \\ 0 \le q < b \\ p \equiv q \pmod d } pq
= \frac{ab(d^2+3ab-3b-3a+2)}{12}
$$
$$
\sum_{k=1}^{ab} (k \bmod a)
= b\sum_{k=0}^{a-1} k
= \frac{ab(a-1)}{2}
$$
So altogether:
$$\frac{1}{12a^2b^2} \left(\begin{array} {l}
\frac{ab(d^2+3ab-3b-3a+2)}{12}\cdot 12
\\ + \frac{ab(a-1)}{2}(6 - 6b)
\\ + \frac{ab(b-1)}{2}(6 - 6a)
\\ + ab(3ab-3b-3a+4)
\end{array}\right)$$
Which is:
$$\frac{d^2}{12ab}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\bar{c})z^2+(b+\bar{b})z+(\bar{a}+c)=0$ Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\overline{c})z^2+(b+\overline{b})z+(\overline{a}+c)=0$. If $|z_1|<1,|z_2|<1$, then prove that $|w_1|=|w_2|=1$
My Progress:
We can see that whenever $z$ is a root of $(a+\overline{c})z^2+(b+\overline{b})z+(\overline{a}+c)=0,\,\, \frac{1}{\overline{z}}$ is also a root.
Case 1: $w_1=\frac{1}{\overline{w_1}}$ and $w_2=\frac{1}{\overline{w_2}}$
$\implies |w_1|=|w_2|=1$
Case 2: $w_1=\frac{1}{\overline{w_2}}$ and $w_2=\frac{1}{\overline{w_1}}$
$\implies w_1\overline {w_2}=1$
| Let $a+\overline{c} = t$ and $b+\overline{b} = 2s$ with $s$ real.
Claim: $t\neq 0\text{ and }s^2-|t|^2 <0$
proof:
(i) $t=0\implies a=-\bar{c}\implies |z_1z_2|=1$ contradiction!!
(ii) let $\frac{\overline{a}}{a}=\alpha.$
\begin{align*}
4(s^2-|t|^2)&= |a(z_1+z_2)+\overline{a(z_1+z_2)}|^2-4|a+\overline{az_1z_2}|^2\\
&=-|a|^2[4+4|z_1|^2|z_2|^2-2|z_1+z_2|^2-2\Re{(\overline\alpha(z_1-z_2)^2)}]\\
&\le-|a|^2[4+4|z_1|^2|z_2|^2-2|z_1+z_2|^2-2|\overline\alpha||z_1-z_2|^2]\\
&=-4|a|^2(1-|z_1|^2)(1-|z_2|^2)\\
&\le 0
\end{align*}
Now
\begin{align*}
tw^2+2sw+\overline{t} &= 0\\
\implies w_{1,2} &= \dfrac{-s\pm\sqrt{s^2-|t|^2}}{t}\\
\implies w_{1,2} &= \dfrac{-s\pm\iota\sqrt{|t|^2-s^2}}{t}\\
\implies |w_{1,2}| &= \dfrac{\sqrt{|-s|^2+|t|^2-s^2}}{|t|}\\
&=1
\end{align*}
| {
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Find all solutions to the equation:$3^{\sin x} \cos x-3^{\cos x} \sin x=0$ Find all solutions to the equation:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0$
My attempt:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0\implies 3^{\sin x} \cos x=3^{\cos x} \sin x\implies3^{\sin x-\cos x}=\tan x.$
I have found that $x=45^{\circ}$ or $225^{\circ}$ are satisfying the equation.
But,I don't know if these are the only solutions.
If these are the only solutions,then how to prove it?
Any guidance is welcome!
Thank you!
| Assume $x \in [0,2\pi]$. Rewrite it : $3^{\cos x}\sin x = 3^{\sin x}\cos x$. If $\sin x > 0 \implies \cos x > 0$ and the same for the case negative sign. Thus you might consider $x \in (0,\pi/2)\cup(\pi, 3\pi/2)$. For the first case, rewrite further: $\dfrac{3^{\sin x}}{\sin x}= \dfrac{3^{\cos x}}{\cos x}$. Consider $f(t) = \dfrac{3^t}{t}$. We have: $f'(t) = \dfrac{t3^t\ln 3 - 3^t}{t^2}= \dfrac{3^t(t\ln 3 - 1)}{t^2}= 0 \implies t = \dfrac{1}{\ln 3}$. Observe that $\dfrac{1}{\ln 3} = 0.91024 > 0.78539 =\dfrac{\pi}{4} > 0$. So if $0 < t < \pi/4 \implies f'(t) < 0$, and in this interval $0 <\sin x < \cos x\implies f(\sin x) > f(\cos x)\implies \dfrac{3^{\sin x}}{\sin x} > \dfrac{3^{\cos x}}{\cos x}$. Thus the equation has no solution. If $\pi/4 < t < \dfrac{1}{\ln 3}\implies f'(t) < 0$ and on this interval $0 < \cos x < \sin x\implies f(\cos x) > f(\sin x) \implies \dfrac{3^{\cos x}}{\cos x} > \dfrac{3^{\sin x}}{\sin x}$. Thus the equation also has no solution on this interval. At $x=\pi/4$, the equation holds and therefore $x = \pi/4$ is a solution. If $\dfrac{1}{\ln 3} < t \le \pi/2\implies f'(t) > 0$ and on this interval $\sin x > \cos x \implies f(\sin x) > f(\cos x)\implies \dfrac{3^{\sin x}}{\sin x} > \dfrac{3^{\cos x}}{\cos x}$. So the equation has no solution on this interval. At $t = \dfrac{1}{\ln 3} = 0.91024$, $\dfrac{3^{\sin(0.91024)}}{\sin(0.91024)}= 3.01525 < 3.19806 = \dfrac{3^{\cos(0.91024)}}{\cos(0.91024)}$. Thus $x = \dfrac{1}{\ln 3}$ is not a solution.Thus there is only one solution on the interval $(0,\pi/2)$ and that solution is $x = \pi/4 = 45^{\circ}$. Similarly you can show that the other solution is $x = 5\pi/4 = 225^{\circ}$, and this completes the proof that there are only two named solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
| First notice that
$${\rm Dom}(f)=\left]-\infty,-3\right]\cup \left[1,+\infty\right[$$
Now, setting
$$y=f(x)\implies y=\frac{\sqrt{(x-1)(x+3)}}{x+2}\implies y^{2}=\frac{(x-1)(x+3)}{(x+2)^{2}}$$
Then,
$$y^{2}(x+2)^{2}=(x-1)(x+3)$$
Expanding and collecting all in terms of $x$, we have
$$(1-y^{2})x^{2}+(2-4y^{2})x+(-3-4y^{2})=0,\quad (*)$$
Using the quadratic formula we get
$$x=\frac{1-2y^{2}\pm \sqrt{4-3y^{2}}}{y^{2}-1},\quad (**)$$
Then,
*
*If $y=-1$, then $x=-\frac{7}{2}\in {\rm Dom}(f)$ using $(*)$ so not problem.
*If $y\not=\pm 1$ so only we need $4-3y^{2}\geqslant 0$ so $-\frac{2}{\sqrt{3}}\leqslant y\leqslant \frac{2}{\sqrt{3}}$. $[◇]$
Joint all, we have
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},1\right[$$
as desired.
Fixed and more details:
Notice with $y=1$, we have $1=\frac{\sqrt{(x-1)(x+3)}}{x+2}$ which don't have solutions over $\mathbb{R}$. Then exclude the $1$ of the interval.
Well, by $(**)$ we get
$$y\in \left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[\cup \left]-1,\frac{2}{\sqrt{3}}\right[$$
then we have the conclusion give in $[◇]$. However the interval $\left]-1,\frac{2}{\sqrt{3}}\right[$ is a strange solution (just study $f$ using the domain and the conclusion as follows), so we remove it. In my previous post I included the value $1$ in the range, however it was a mistake because as said above. Therefore the range of $f$ over the natural domain is given by
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[=\left[-\frac{2}{\sqrt{3}},1\right[.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$? I stumble upon the expression
$$
\sum_{k=0}^\infty
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}
$$
and it seems to converge to
$$
\frac{3-\sqrt{5}}{2}
$$
Do they equate ? How to prove that ?
Not sure if it's helpful :
$\frac{1}{k+1}\binom{3k+1}{k} , \; k\ge0$ is the OEIS sequence $A006013$.
I only have fundamental knowledge in combinatorics so I could only check it numerically that the convergence seems to be true . However I wouldn't mind learning new theories .
| Let
$$S(z)=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}(1-z)^{2k+1}.$$
One has
\begin{align}S(z)&=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}\sum_{l=0}^{2k+1}(-1)^l{{2k+1}\choose{l}}z^l \\
&=\sum_{n=0}^\infty\sum_{k+l=n}(-1)^l{{3k+1}\choose{k}}{{2k+1}\choose{l}}z^n\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}{{3k+1}\choose{k}}{{2k+1}\choose{n-k}}z^n
\end{align}
On the other hand,
$${{3k+1}\choose{k}}{{2k+1}\choose{n-k}}=\frac{(3k+1)!}{k!(2k+1)!} \frac{(2k+1)!}{(n-k)!(3k+1-n)!}={n\choose k}{3k+1\choose n}.$$
Moreover, we have the following combinatorial identity:
$$\sum_{k=0}^n(-1)^{n-k}{n\choose k}{3k+1\choose n}=3^n.$$
One way to verify this identity is to write
$$f(x)=x(1+x^3)^n=\sum_{k=0}^n {n\choose k}x^{3k+1}.$$
and notice that the $n$th derivative of $x(1+x^3)^n$ at $x=-1$ is $-3^nn!$, while the $n$th derivative of $\sum_{k=0}^n {n\choose k}x^{3k+1}$ is $\sum_{k=0}^n n!{3k+1\choose n}{n\choose k}x^{3k+1-n}$.
Combining these, we deduce that
$$S(z)=\sum_{n=0}^\infty 3^nz^n=\frac{1}{1-3z}.$$
Next, let
$$G(u)=\sum_{k=0}^\infty \frac{1}{k+1}{3k+1\choose k}u^{k+1},$$
and $u=z(1-z)^2$. Then
$$\frac{dG}{dz}=\frac{dG}{du}\frac{du}{dz}.$$
Since $dG/du=\sum_{k=0}^\infty {3k+1 \choose k}(z(1-z)^2)^k=\frac{S(z)}{1-z}=\frac{1}{(1-z)(1-3z)}$ and $du/dz=1-4z+3z^2$, we have $\frac{dG}{dz}=1$, and so $G(z)=z$ (note that $G(0)=0$).
We have proved that
$$\sum_{k=0}^\infty \frac{1}{k+1}{3k+1 \choose k}(z(1-z)^2)^{k+1}=z.$$
Finally, let $z=(3-\sqrt{5})/2$, for which $z(1-z)^2=1/8$, to conclude that
$$\sum_{k=0}^\infty \frac{1}{k+1}{3k+1 \choose k}\frac{1}{2^{3k+3}}=\frac{3-\sqrt{5}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Asymptotic for $\sum_{k=1}^n k^n$ Consider the OEIS sequence A031971, which is defined as:
$$a_n=\sum\limits_{k=1}^n k^n\quad\color{gray}{(1,\,5,\,36,\,354,\,4425,\,67171,\,1200304,\,.\!.\!.\!)}\tag{1}$$ I'm interested in the asymptotic behavior of $a_n$ for $n\to\infty$.
Empirically, it appears that
$$a_n\stackrel{\color{gray}?}\sim\frac{e}{e-1}\,n^n\cdot\left(1-\frac{e+1}{2\,(e-1)^2}\,n^{-1}+c\,n^{-2}+\mathcal O\!\left(n^{-3}\right)\right),\tag{2}$$
where $c\approx0.6310116...$ (I haven't found a plausible closed form it). The leading term $\frac{e}{e-1}\,n^n$ is given in the OEIS.
How can we prove this formula and find higher terms in it?
| Let us write
$$
\sum\limits_{k = 1}^n {k^n } = n^n \sum\limits_{k = 0}^{n - 1} {\left( {1 - \frac{k}{n}} \right)^n } .
$$
Now by power series expansions
\begin{align*}
\left( {1 - \frac{k}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{k}{n}} \right)} \right) = \exp \left( { - n\sum\limits_{j = 1}^\infty {\frac{{k^j }}{j}\frac{1}{{n^j }}} } \right)
\\ & = e^{ - k} - e^{ - k} \frac{{k^2 }}{2}\frac{1}{n} + e^{ - k} \left( {\frac{{k^4 }}{8} - \frac{{k^3 }}{3}} \right)\frac{1}{{n^2 }} - \ldots \,.
\end{align*}
Substituting back to the original sum, extending that sum to $n=\infty$, and using closed forms of $\sum_{k=0}^\infty k^px^k$ (with $x=e^{-1}$), we deduce
\begin{align*}
\sum\limits_{k = 1}^n {k^n } & \sim n^n \left( {\frac{e}{{e - 1}} - \frac{{e^2 + e}}{{2(e - 1)^3 }}\frac{1}{n} - \frac{{e(5e^3 - 9e^2 - 57e - 11)}}{{24(e - 1)^5 }}\frac{1}{{n^2 }} + \ldots } \right)
\\ & =\frac{e}{{e - 1}}n^n \left( {1 - \frac{{e + 1}}{{2(e - 1)^2 }}\frac{1}{n} - \frac{{5e^3 - 9e^2 - 57e - 11}}{{24(e - 1)^4 }}\frac{1}{{n^2 }} + \ldots } \right)
\end{align*}
as $n\to +\infty$. You can obtain more terms of the expansion if you like, the process should be clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solutions of PDE I am looking for solutions $g \colon \mathbb R^2 \times \mathbb R^2 \to \mathbb C$ of the PDE
$$\partial_{x_1} g(x,y) + \partial_{y_1} g(x,y) = a(y_1+x_1+i(x_2-y_2))g(x,y), \\ \partial_{x_2} g(x,y) + \partial_{y_2} g(x,y) = a(y_2+x_2+i(y_1-x_1))g(x,y),$$
where $a>0$ and $(x,y) = (x_1,x_2,y_1,y_2) \in \mathbb R^2 \times \mathbb R^2$. I already know that $f(x,y) := \exp(a(x\cdot y - i x \wedge y))$ with $x\wedge y := x_1 y_2 - x_2 y_1$ is a solution but I was so far not able to find any other (smooth) solution. Does anyone have an answer to this? General approaches to solve this kind of PDE are also appreciated!
| As the system of PDEs strongly decouples, with one equation only dependent on the $(x_{1}, y_{1})$ coordinates while the other is only dependent on the $(x_{2}, y_{2})$ coordinates, it makes sense to solve the problems separately.
For the first PDE, the method of characteristics implies (labelling $g \to g_{1}$)
$$\frac{d x_{1}}{1} = \frac{d y_{1}}{1} = \frac{d g_{1}}{a (y_{1} + x_{1} + i (x_{2} - y_{2})) g_{1}}$$
The first equality gives
$$x_{1} - y_{1} = C_{1}$$
Using componendo-divindendo on the first equality and setting it equal to the last ratio gives
\begin{align}
\frac{d g_{1}}{a (y_{1} + x_{1} + i (x_{2} - y_{2})) g_{1}} &= \frac{d (x_{1} + y_{1})}{2} \\
\implies \frac{dg_{1}}{g_{1}} &= \frac{a (y_{1} + x_{1} + i (x_{2} - y_{2})) d(x_{1} + y_{1})}{2} \\
\implies \ln g_{1} &= \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} + C_{2} \\
\implies g_{1} &= C_{2} \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right) \\
&= f_{1}(x_{1} - y_{1}) \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right)
\end{align}
which you can check satisfies the first equation. A similar calculation for the second PDE (by symmetry, let $x_{1} \leftrightarrow x_{2}, y_{1} \leftrightarrow y_{2}$ and relabel $g \to g_{2}$) then shows
\begin{align}
\implies g_{2} &= f_{2}(x_{2} - y_{2}) \exp \left( \frac{a}{4} (x_{2} + y_{2})^{2} + \frac{i a (x_{1} - y_{1})(x_{2} + y_{2})}{2} \right)
\end{align}
The final solution is then a superposition of the two
\begin{align}
g = \ &g_{1} + g_{2} \\
= \ &f_{1}(x_{1} - y_{1}) \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right) \\
+ \ &f_{2}(x_{2} - y_{2}) \exp \left( \frac{a}{4} (x_{2} + y_{2})^{2} + \frac{i a (x_{1} - y_{1})(x_{2} + y_{2})}{2} \right)
\end{align}
Note that we get the result you found by choosing $f_{1}, f_{2}$ appropriately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$ If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$
My attempt:
$f(x) = \sqrt{1+{\sqrt{x}} f(x)} \implies f(4) =\sqrt{1+2f(4)} \implies
f(4)(f(4)-2)=1.$
I don't know how to proceed from here.
The answer should be $2.$ Where am I getting wrong?
Kindly, explain it.Thank you for any guidance!
| For $y>0$,
\begin{eqnarray}
1+y & = & \sqrt{y^2+(2y+1)}\\
& = & \sqrt{y^2+\sqrt{(2y+1)^2}}\\
& = & \sqrt{y^2+\sqrt{4y^2+(4y+1)}}\\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{(4y+1)^2}}}\\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{16y^2+(8y+1)}}}\\
& \vdots & \\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{4^2y^2+\sqrt{4^3y^2+\sqrt{4^4y^2+\cdots}}}}}
\end{eqnarray}
If you set $y=1$ you get the value of your function for the special case $x=4$, i.e. $f(4)=1+1=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality involving sums with binomial coefficient I am trying to show upper- and lower-bounds on
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$
(where $n\geq 1$) in order to show that it basically grows as $\Theta(n)$.
The upper-bound is easy to get since $\min(i, n-i)\leq i$ for $i\in\{0, \dots n\}$ so that
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)\leq \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}i = \frac{n}{2}.$$
Thanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $f(n)=\frac{n-1}{3}$ does provide a lower-bound. One can in fact rewrite
$$\frac{n-1}{3}=\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\frac{2i-1}{3}.$$
I was thus hoping to show that for each term we have $\frac{2i-1}{3}\leq \min(i, n-i)$, but this is only true if $i\leq \frac{3n+1}{5}$ and not generally for $i\leq n$. I imagine there is a clever trick to use at some point but for some reason I am stuck here.
Any help would be appreciated, thank you!
EDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the "accepted" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $\Theta(n)$ growth was sufficient here.
| We start with
$$
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k}
=\left\{\begin{array}{}
2^{n-1}&\text{if $n$ is odd}\\
2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if $n$ is even}
\end{array}\right.\tag1
$$
Substitute $n\mapsto n-1$:
$$
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k}
=\left\{\begin{array}{}
2^{n-2}&\text{if $n$ is even}\\
2^{n-2}+\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag2
$$
Subtract the $k=\left\lfloor\frac{n-1}2\right\rfloor$ term:
$$
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor-1}\binom{n-1}{k}
=\left\{\begin{array}{}
2^{n-2}-\left(\raise{2pt}{n-1}\atop{\frac{n}2-1}\right)&\text{if $n$ is even}\\
2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag3
$$
Substitute $k\mapsto k-1$:
$$
\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}
=\left\{\begin{array}{}
2^{n-2}-\frac12\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag4
$$
Then
$$
\begin{align}
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n}{k}k
&=n\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}\tag{5a}\\
&=\left\{\begin{array}{}
n2^{n-2}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
n2^{n-2}-\frac{n}2\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag{5b}
\end{align}
$$
Explanation:
$\text{(5a)}$: $\binom{n}{k}k=\binom{n-1}{k-1}n$ and the $k=0$ term vanishes
$\text{(5b)}$: multiply $(4)$ by $n$
Double and add the middle term in the even $n$ case:
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}\min(k,n-k)
&=\left\{\begin{array}{}
n2^{n-1}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
n2^{n-1}-n\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag6
\end{align}
$$
The estimates for the central binomial coefficients from this answer give
$$
\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac13\right)}}\le\binom{n}{\frac{n}2}\le\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac14\right)}}\tag7
$$
which gives upper and lower bounds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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} |
How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) .$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d x}{x^{6}+1} \\
&=\frac{1}{2} \oint_{\gamma} \frac{d z}{z^{6}+1} \\
&=\frac{1}{2} \cdot 2 \pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^{6}+1} , z_k \right)
\end{aligned}
$$
with its simple poles at $ z_{k}=e^{\frac{(2 k+i) \pi}{6} i}$, where $k=0,1,2$.
Now lets evaluate the residues of $\frac{1}{z^{6}+1} $ at $ z_{k}$.
$$
\operatorname{Res}\left( \frac{1}{z^{6}+1} , z_{k}\right)=\frac{1}{6 z _k^{5}}=\frac{z_{k}}{6 z_{k}^{6}}=-\frac{z_{k}}{6}
$$
Putting the residues back yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\pi i\left(-\frac{1}{6}\left(z_{0}+z_{1}+z_{2}\right)\right) \\
&=-\frac{\pi i}{6}\left(e^{\frac{\pi}{6}}+e^{\frac{\pi}{2} i}+e^{\frac{5 \pi}{6} i}\right) \\
&=-\frac{\pi i}{6}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i+i-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
Similarly, we can deal with the integral in general
$$I_{2n}=\int_{0}^{\infty} \frac{1}{x^{2n}+1} d x$$
with the same contour $\gamma$ and $n $ simple poles whose residues are
$$
\operatorname{Res}\left( \frac{1}{z^{2n}+1} , z_{k}\right)=\frac{1}{2n z _k^{2n-1}}=\frac{z_{k}}{2n z_{k}^{2n}}=-\frac{z_{k}}{2n}
$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{2 n}+1} d x&=\pi i \sum_{k=0}^{n-1} \operatorname{Res}\left(\frac{1}{z^{2 n}+1}, z_k\right) \\
&=\pi i \sum_{k=0}^{n-1}\left(-\frac{z _k}{2 n}\right)\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} z_{k}\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} e^{\frac{2 k+1}{2 n} \pi i}\\ &=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{1-e^{\frac{\pi i}{n}(n)}}{1-e^{\frac{\pi i}{\pi}}}\\
&=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{2}{1-e^{\frac{\pi}{n} i}}\\
&=-\frac{\pi i}{n} \frac{e^{\frac{\pi i}{2 n}}}{1-e^{\frac{\pi}{n} i}}\\&= -\frac{\pi i}{n} \cdot \frac{1}{e^{-\frac{\pi i}{2 n}}-e^{\frac{\pi i}{2 n}}}\\&= \frac{\pi}{2 n} \csc \frac{\pi}{2 n}
\end{aligned}
$$
However, I can’t use the same contour to deal with the odd one
$$\int_{0}^{\infty} \frac{1}{x^{2 n-1}+1} d x$$
as the simple pole $-1$ is on $\gamma_1$.
My Question: How can we evaluate the odd one?
| Let $f(z) = \frac{1}{z^n + 1}$.
The trick is to pick a different contour. We choose the ray parametrised by $z(t) = e^{2 \pi i /n} t$ for $t \in [0, \infty)$.
We then have, after taking a bit of care with limits, $\int\limits_0^\infty f(t) dt - e^{2 \pi i / n} \int\limits_0^\infty f(t) dt = 2 \pi i Res(f(z), z = e^{\pi i / n}) = \frac{1}{ne^{\pi i (1 - 1/n))}}$.
Let $I = \int\limits_0^\infty f(t) dt$. Then we have $I (1 - e^{2 \pi i / n}) = \frac{2 \pi i}{n e^{\pi i (1 - 1/n)}}$.
Taking the modulus of both sides gives us $2I\sin(\pi / n) = \frac{2 \pi}{n}$. That is, $I = \frac{\pi}{n}\csc\frac{\pi}{n}$.
| {
"language": "en",
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"answer_count": 3,
"answer_id": 1
} |
$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ I found this problem in a high school text book.
Let $ \displaystyle s = \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001}$. Find $s$.
How I tried:
Observe that $T_n = \frac{n^2}{(2n-1)(2n+1)}$. Here, $T_n$ is the $n$th term of the sequence. So, we need to find the value of $\sum_{n=1}^{500}\frac{n^2}{(2n -1)(2n+1)}$. We can find its value with Telescope Cancellation Method, but it requires breaking the expression we got into simpler terms.
How to simplify $T_n = \frac{n^2}{(2n-1)(2n+1)}$ ?
| I don't think your $T_n$ is correct. Here's how I broke it down:
$$\frac{1^2}{1.3} + \frac{2^2}{3.5} + \frac{3^2}{5.7}+ ... = \frac{1^2}{1 + 0.1 \cdot 3} + \frac{2^2}{3 +0.1 \cdot 5} + \frac{3^2}{5+ 0.1 \cdot 7} + ...
\\ = \sum_{n=1}^N \frac{n^2}{(2n-1)+0.1(2n+1)}
\\ = \sum_{n=1}^N \frac{10n^2}{10(2n-1)+(2n+1)}
\\ = \sum_{n=1}^N \frac{10n^2}{22n-9}
$$
From here, it becomes a matter of polynomial long division, which yields:
$$\sum_{n=1}^N \left(\frac{5n}{11} + \frac{45}{242} +\frac{\frac{405}{242}}{22n-9} \right)
\\= \frac{5N(N+1)}{22}+\frac{45N}{242}+\frac{405}{242} \cdot \sum_{n=1}^N \frac{1}{22n-9}
$$
Not the prettiest expression, I know. What's more, there exists no closed-form expression for that final sum. But if you plug this in for $N = 500$ into a calculator, it will yield $\approx 57025.3858937$
| {
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} |
Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfied.
Alternatively, if $\ell_1\leq\ell_2\leq\ell_3$ are the values of $a,b$ and $c$ ordered in ascending order, then the triangle exists iff $\ell_1+\ell_2\geq\ell_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in\mathbb R^2$ be the three vertices of the triangle, with $\|x\|=a,\,\|y\|=b$ and $\|x-y\|=c$. Then $c^2=\|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2=a^2+b^2-2\langle x,y\rangle$. Therefore $x^Ty=\langle x,y\rangle=\frac{1}{2}(a^2+b^2-c^2)$ and
$$\pmatrix{x^T\\ y^T}\pmatrix{x&y}=\frac{1}{2}\pmatrix{2a^2&a^2+b^2-c^2\\ a^2+b^2-c^2&2b^2}.\tag{1}$$
The RHS of $(1)$ must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain $x$ and $y$ and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of $(1)$ is positive semidefinite if and only if $(2a^2)(2b^2)-(a^2+b^2-c^2)^2\geq0$, by Sylvester's criterion. That is, the triangle exists if and only if
$$-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\geq0.\tag{2}$$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of $(2)$ is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both $(0)$ and $(2)$ are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why $(0)$ and $(2)$ are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?
| For non-negative $a,b,c$:
$$a+b\geq c; a+c\geq b;b+c\geq a \Leftrightarrow |b-a|\leq c \leq b+a\Leftrightarrow\\
(b-a)^2\leq c^2 \leq (b+a)^2 \Leftrightarrow (c^2-(b+a)^2)(c^2-(b-a)^2)\leq 0 \Leftrightarrow\\
c^4-((b+a)^2+(b-a)^2)c^2+(b^2-a^2)^2\leq 0\Leftrightarrow
c^4+b^4+a^4\leq 2(a^2b^2+a^2c^2+b^2c^2)$$
| {
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} |
Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
This is related to the following problem:
Problem 2: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}.$$
Problem 2 is in "Problems From the Book", 2008, Ch. 2, which was proposed by Vasile Cartoaje.
See: Prove that $\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)} \le \frac{n(n-1)}{2(2n-1)^2}$
Background:
I proposed Problem 1 when I tried to find my 2nd proof for Problem 2.
It is not difficult to prove that
$$\frac{1}{(2n-1)^4} + \frac{16n^2(n-1)^2}{(2n-1)^4}\cdot \frac{x_ix_j}{1-x_i-x_j}
\ge \frac{x_ix_j}{(1-x_i)(1-x_j)}.$$
(Hint: Use $\frac{x_ix_j}{(1-x_i)(1-x_j)}= 1 - \frac{1}{1 + x_ix_j/(1-x_i-x_j)}$
and $\frac{1}{1+u} \ge \frac{1}{1+v} - \frac{1}{(1+v)^2}(u-v)$
for $u = x_ix_j/(1-x_i-x_j)$ and $v=\frac{1}{4n(n-1)}$. Or simply
$\mathrm{LHS} - \mathrm{RHS} = \frac{(4x_ix_jn^2 - 4x_ix_j n + x_i + x_j - 1)^2}{(2n-1)^4(1-x_i-x_j)(1-x_i)(1-x_j)}\ge 0$.)
To prove Problem 2, it suffices to prove that
$$\frac{1}{(2n-1)^4}\cdot \frac{n(n-1)}{2} + \frac{16n^2(n-1)^2}{(2n-1)^4}\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac{n(n-1)}{2(2n-1)^2} $$
or
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
For $n=2, 3, 4$, the inequality is true.
For $n=5, 6$, numerical evidence supports the statement.
Any comments and solutions are welcome and appreciated.
| Partial proof with hint :
Let us consider the inequality for $x,y>0$ and $x,y\leq 0.5$ :
$$l\left(x,y\right)=g\left(2xy+\frac{4}{3}xy\left(x-y\right)^{2}\right)-\frac{x^2y^2}{1-x^2-y^2}\ge 0$$
Where $g\left(x\right)=\frac{\frac{x^{2}}{4}}{1-x}$
As $g$ is convex on $[0,0.5]$ we use Jensen-Mercer inequality
We need to show :
$$\frac{n\left(n-1\right)}{2}\left(h\left(x,y\right)+h\left(a,b\right)-g\left(f\left(x,y\right)+f\left(a,b\right)-d\right)\right)\le\frac{1}{8}\tag{I}$$
Where $f\left(x,y\right)=2xy+\frac{4}{3}xy\left(x-y\right)^{2},h\left(x,y\right)=g\left(f\left(x,y\right)\right)$ and $f(a,b)\leq d \leq f\left(x,y\right)$
Now we use the inverse function of $f(x)$ with a positive value or :
$$r\left(x\right)=2\sqrt{x^{2}+x}-2x$$
So $(I)$ is true with the constraint (where we have used Karamata's inequality)
$$f\left(x,y\right)+f\left(a,b\right)\le r\left(\frac{\frac{1}{8}2}{n\left(n-1\right)}+g\left(f\left(x,y\right)+f\left(a,b\right)-d\right)\right)+r(0)$$
| {
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"source": "stackexchange",
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What is $\sum_{k = 1}^n (k \log k)\binom{n}{k}$? If the exact answer is difficult to find, what is the tightest asymptotic upper bound? While trying to solve the complexity of my program I came across the the following summation:
$$\sum_{k = 1}^n (k \log k)\binom{n}{k}$$
Could you please provide a solution to this sum. If it is difficult to obtain the exact solution, could you please provide an asymptotic upper bound that is as close as possible?
I was able to obtain the following asymptotic upperbound:
\begin{align*}
\sum_{k = 1}^n (k \log k)\binom{n}{k}
&= \mathop{O}\left(\sum k(k-1) \binom{n}{k} \right) \\
&= \mathop{O}\left(\sum n(n-1) \binom{n-2}{k-2} \right) \\
&= \mathop{O}\left(n^2 \sum \binom{n-2}{k-2} \right) \\
&= \mathop{O}(n^2 2^n)
\end{align*}
Is it possible to get smaller upper bound, for example $O(2^n n \log n)$.
| a) rewriting the sum
First of all we have better to rewrite the sum as
$$
\eqalign{
& S(n) = \sum\limits_{k = 1}^n {k\ln k\left( \matrix{
n \cr k \cr} \right)} = \sum\limits_{k = 1}^n {\ln \left( {k^k } \right)\left( \matrix{
n \cr k \cr} \right)} = \sum\limits_{k = 0}^n {\ln \left( {k^k } \right)\left( \matrix{
n \cr k \cr} \right)} \cr
& = \sum\limits_{0 \le k} {k\ln \left( k \right)\left( \matrix{
n \cr k \cr} \right)} = n\sum\limits_{1 \le k} {\ln \left( k \right)\left( \matrix{
n - 1 \cr k - 1 \cr} \right)} = n\sum\limits_{0 \le k} {\ln \left( {1 + k} \right)\left( \matrix{
n - 1 \cr k \cr} \right)} = \cr
& = n\,R(n - 1) \cr}
$$
where $R(n)$ is more manageable.
b) properties of R
(in the attempt to find a closed form or interesting identities)
One interesting property of $R$ is that the binomial inversion theorem tells
$$
R(n) = \sum\limits_{0 \le k} {\ln \left( {1 + k} \right)
\left( \matrix{ n \cr k \cr} \right)} \quad \Leftrightarrow \quad \ln \left( {1 + n} \right)
= \sum\limits_{0 \le k} {\left( { - 1} \right)^{n - k}
\left( \matrix{ n \cr k \cr} \right)R(k)} = \left. {\Delta ^n R(n)\;} \right|_{n = 0}
$$
Another interesting fact is that we can rewrite $R$ as
$$
\eqalign{
& R(n) = \sum\limits_{0 \le k} {\ln \left( {1 + k} \right)\left( \matrix{ n \cr
k \cr} \right)} = n!\sum\limits_{0 \le k} {{{\ln \left( {1 + k} \right)} \over {k!}}{1 \over {n - k}}} \cr
& {{R(n)} \over {n!}} = \sum\limits_{0 \le k} {{{\ln \left( {1 + k} \right)} \over {k!}}{1 \over {n - k}}} \cr}
$$
and thus obtain the e.g.f.
$$
\eqalign{
& F(z) = \sum\limits_{0 \le n} {{{R(n)} \over {n!}}z^n }
= \sum\limits_{0 \le k} {{{\ln \left( {1 + k} \right)z^k } \over {k!}}{{z^{n - k} } \over {\left( {n - k} \right)!}}} = \cr
& = \left( {\sum\limits_{0 \le k} {{{\ln \left( {1 + k} \right)} \over {k!}}z^k } } \right)
\left( {\sum\limits_{0 \le j} {{{z^j } \over {j!}}} } \right)
= e^z \left( {\sum\limits_{0 \le k} {{{\ln \left( {1 + k} \right)} \over {k!}}z^k } } \right) \cr}
$$
c) approximation
Since the binomial is symmetric wrt $n/2$ and the log
gets quite "flat" for large values of $k$, we have better
to center the approximation at $n/2$.
$$
\eqalign{
& R(n) = \sum\limits_{0 \le k} {\ln \left( {1 + k} \right)
\left( \matrix{ n \cr k \cr} \right)} = \sum\limits_{\left( { - \left\lfloor {n/2} \right\rfloor \le } \right)
j\left( { \le \left\lfloor {n/2} \right\rfloor } \right)} {\ln \left( {1 + n/2 + j} \right)
\left( \matrix{ n \cr n/2 + j \cr} \right)} = \cr
& = \sum\limits_{\left( { - \left\lfloor {n/2} \right\rfloor \le } \right)j
\left( { \le \left\lfloor {n/2} \right\rfloor } \right)} {\left( {\ln \left( {1 + n/2} \right)
+ \ln \left( {1 + {j \over {\left( {1 + n/2} \right)}}} \right)} \right)
\left( \matrix{ n \cr n/2 + j \cr} \right)} = \cr
& = \ln \left( {1 + n/2} \right)2^n + \sum\limits_{\left( { - \left\lfloor {n/2} \right\rfloor \le } \right)
j\left( { \le \left\lfloor {n/2} \right\rfloor } \right)} {\ln \left( {1 + {j \over {\left( {1 + n/2} \right)}}} \right)
\left( \matrix{ n \cr n/2 + j \cr} \right)} \cr
& \le \ln \left( {1 + n/2} \right)2^n + {1 \over {\left( {1 + n/2} \right)}}
\sum\limits_{\left( { - \left\lfloor {n/2} \right\rfloor \le } \right)j\left( { \le \left\lfloor {n/2} \right\rfloor } \right)}
{j\left( \matrix{ n \cr n/2 + j \cr} \right)} = \cr
& = \ln \left( {1 + n/2} \right)2^n + {1 \over {\left( {1 + n/2} \right)}}
\sum\limits_{\left( { - \left\lfloor {n/2} \right\rfloor \le } \right)j\left( { \le \left\lfloor {n/2} \right\rfloor } \right)}
{\left( {j + n/2 - n/2} \right)\left( \matrix{n \cr n/2 + j \cr} \right)} = \cr
& = \ln \left( {1 + n/2} \right)2^n - {{n/2} \over {\left( {1 + n/2} \right)}}2^n
+ {1 \over {\left( {1 + n/2} \right)}}\sum\limits_{0 \le k}
{k\left( \matrix{ n \cr k \cr} \right)} = \cr
& = \ln \left( {1 + n/2} \right)2^n \cr}
$$
i.e.
$$
R(n) \le \ln \left( {1 + n/2} \right)2^n ,\quad
S(n) \le n\ln \left( {1 + \left( {n - 1} \right)/2} \right)2^{n - 1}
$$
and
$$
{{\ln \left( {1 + n/2} \right)2^n } \over {R(n)}}\;\mathop \to \limits_{n \to \infty } \;1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464680",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular points $v = \pm 1, \pm i$ all lie inside the contour $G$. Thus, from this theorem (*), we have
\begin{align*}
\int_{G}{\dfrac{1}{v^4-1} \text{d}v} &= \dfrac{1}{4} \left( \int_{G}{\dfrac{1}{v-1}\text{d}v} - \int_{G}{\dfrac{1}{v+1}\text{d}v} + \int_{G}{\dfrac{i}{v-i}\text{d}v} - \int_{G}{\dfrac{i}{v+i}\text{d}v} \right) \\
&= \dfrac{1}{4}\left( 2\pi i - 2\pi i + i\left( 2\pi i \right) - i \left( 2\pi i \right) \right) \\
&= \dfrac{1}{4} \left( 0 \right) \\
&= 0
\end{align*}
(*) Theorem: Let $C$ be a simple closed contour with a positive orientation such that $v_0$ lies interior to $C$, then $\displaystyle{\int_{C} {\dfrac{dv}{(v-v_0)^n}} = 2\pi i}$ for $n =1$ and $0$ when $n \neq 1$ is an integer.
Is that proof correct? If so, could you also point out if there are still theorems I have to mention to make it more accurate?
I'm trying to solve (perhaps overthink) this with the other approach:
We see that it is analytic except at $\pm 1$ and $ \pm i$.
Also, we can apply deformation of the contour $G$ by forming a leaf-like contour and forming the respective circles $C_1, C_2, C_3,$ and $C_4$. As shown here:
The integration can then be evaluated as
$$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} $$
And,
$$\int_{C_n}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4} \left( \int_{C_n}{\dfrac{1}{v-1}\text{d}v} - \int_{C_n}{\dfrac{1}{v+1}\text{d}v} + \int_{C_n}{\dfrac{i}{v-i}\text{d}v} - \int_{C_n}{\dfrac{i}{v+i}\text{d}v} \right) $$
Note that when $v_n$ lies exterior to $C_n$, then by Cauchy-Goursat theorem, $\displaystyle{\int_{C_n}{\dfrac{dv}{v-v_n}} = 0}$.
Thus, for $n = 1,$,
$$\int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0-0 + i(2\pi i)- 0) = \dfrac{- \pi }{2} $$
for $ n = 2,$
$$\int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}( 2\pi i - 0 + 0-0) = \dfrac{ \pi i}{2}$$
for $ n = 3,$
$$\int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 - 0 + 0 - i (2\pi i) ) = \dfrac{\pi }{2}$$
for $ n = 4,$
$$\int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 -(2\pi i) + 0-0 ) = \dfrac{ - \pi i }{2}$$
Therefore, $$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{- \pi }{2} + \dfrac{ \pi i}{2} + \dfrac{\pi }{2} + \dfrac{ - \pi i }{2} = 0$$
Did I just overcomplicate it? Is my first proof already enough? If any of these proofs are correct, could you also point out if there are still theorems I have to mention for them to make it more accurate?
| You could do it easier by using the special symmetry your function and your path have: They both behave nicely under rotations of $e^{i\frac{\pi}{2}}$. For the path it is evident that $C_k = e^{k\cdot i\frac{\pi}{2}}C_1$, and for the function it is the same:
\begin{align}f(e^{i\frac{\pi}{2}}z) &= \frac{1}{(e^{i\frac{\pi}{2}}z-1)(e^{i\frac{\pi}{2}}z-i)(e^{i\frac{\pi}{2}}+i)(e^{i\frac{\pi}{2}}+1)(e^{i\frac{\pi}{2}}+i)} \\ &= e^{-i\frac{\pi}{2}}\frac{1}{(z-e^{i\frac{\pi}{2}})(z-ie^{i\frac{\pi}{2}})(z+e^{i\frac{\pi}{2}})(z+ie^{i\frac{\pi}{2}})}\\ &= e^{-i\frac{\pi}{2}}\frac{1}{(z-1)(z-i)(z+1)(z+i)}\\ &= e^{-i\frac{\pi}{2}}f(z)
\end{align}
So with the same notation as in your own post, you can write
\begin{align}\int\limits_{C_k} f(z) dz = \int\limits_{C_1}f(e^{k\cdot i\frac{\pi}{2}}z) dz = e^{-k\cdot i\frac{\pi}{2}} \int\limits_{C_1}f(z) dz,\end{align}
so by adding up the different terms you end up with
$$\int\limits_G f(z) dz = \int\limits_{C_1}f(z) dz\cdot\sum\limits_{k=1}^4 e^{-k\cdot i\frac{\pi}{2}} = \int\limits_{C_1}f(z) dz\cdot(1+i-1-i) = 0$$
| {
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"question_score": "4",
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Distribution of random variable $Y = (X + 1)^2$. Could you please help to solve the following problem:
The random variable $X$ has a uniform distribution on the segment $[0,2]$. Find the density of the distribution of a random variable $Y = (X + 1)^2$.
My attempt (but it is obviously wrong):
$$ f(x) = \begin{cases}
(x+1)^2, & x \notin [0,2], \\
0, & x \notin [0,2].
\end{cases} $$
$$ \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
= \int_{0}^{2} (x+1)^2 \, \mathrm{d}x
= \left[ (x+1)^2 \right]_{0}^{2}
= (x+1)^2 (2 - 0)
= 1. $$
\begin{align*}
&\implies \quad 2(x+1)^2 = 1 \\[0.5em]
&\implies \quad (x+1)^2 = \frac{1}{2} \\[0.5em]
&\implies \quad x^2 + 2x + \frac{1}{2} = 0 \\[0.5em]
&\implies \quad x_1 = \frac{-2 + 2^{1/2}}{2}, \quad x_2 = \frac{-2 - 2^{1/2}}{2}
\end{align*}
| $$ P (Y \leq u) = P((X+1)^2 \leq u) = P(X \leq \sqrt{u}-1)$$
Thus, using $F$ for the cummulative distribution function we have
$$F_Y (u) = F_x (\sqrt{u}-1)$$
The density $f$ is a derivative of $F$:
$$f_Y(u) = \frac{d}{du} F_X(\sqrt{u}-1) = F_X^\prime (\sqrt{u}-1) \frac{1}{2\sqrt{u}}$$
$$ = \frac{f_x\left(\sqrt{u}-1 \right)}{2\sqrt{u}}
=
\left\{
\begin{array}{ccc}
\frac{1}{4\sqrt{u}} & \;\;if\;\; & 1 \leq u \leq 9
\\
0 & \;\; elsewhere
\end{array}
\right.
$$
| {
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
~
EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
| A not very strict approach using a rather standard geometric reuslt is as follows.
$5x+3y=100$ depicts a line on the plane, specifically the $y=-\frac{5}{3}x-\frac{100}{3},$ which intersects the $y-$axis on $A=(0,33,333)$ and the $x-$axis on $B=(20,0)$.
Thus the right triangle $\triangle AOB$ is formed, where $O=(0,0)$.
Now, the desired greatest possible value of $xy$ is equivalent to the area of a rectangle inside $\triangle AOB$.
But we know that the maximum area of a rectangle inscribed in a triangle is 1/2 the area of the triangle.
By a direct computation, $E_{\triangle AOB}=333,33$. Moreover $\triangle AOB$'s perpendicular sides are of a similar length. Thus the maximum of $xy$ is expected to be near the value of $\frac{1}{2}E_{\triangle AOB}=166,67$ with $x,y$ also of approximately similar lengths.
By experimenting with pairs of $x,y$ such that $xy\approx167$, we readily observe that $(11,15)$ and $(15,11)$ are the optimal. But the second pair doesn't belong to $y=-\frac{5}{3}x-\frac{100}{3}$ and thus $(x,y)=(11,15)$ is the answer.
| {
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Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$ While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ terms, that is, a single formula for the product $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)(x^{a^4}+1)\ldots(x^{a^n}+1)$$
After multiplying the first four terms I could see a pattern develop in the form $x^{a+a^2+a^3+a^4...a^n}+x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}+x^{a+a^2+a^3}+x^{a+a^2+a^4}+x^{a^2+a^3+a^4}+x^{a+a^3+a^4} ...+x^{a^{n-2}+a^{n-1}+a^n} + x^{a+a^2}
+x^{a+a^3}+x^{a+a^4}+x^{a^2+a^3}+x^{a^2+a^4}+x^{a^3+a^4}...x^{a^{n-1}+n}...+x^{a+a^2+a^3+....a^{n-2}+a^{n-1}+1}$
Now I can easily find the summation of powers in the first term but cant find a formula for the summation of $x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}$ and also can't figure out how to account for the other terms.
| Using $[n]=\{1,2,\ldots,n\}$ we can write the product as
\begin{align*}
\prod_{j=1}^n\left(x^{a^j}+1\right)=\sum_{S\subseteq [n]}x^{\sum_{j\in S}a^j}
\end{align*}
where $S$ runs over all subsets of $[n]$. A special case is $S=\emptyset$, which gives the empty sum. The empty sum is zero per definition, resulting in $x^0=1$.
| {
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How to show that $\forall n \geq 1$, $\frac{n^{2n}}{n!^2} \geq (\frac{n+1}{n})^{n^2-n}$? I tried using the fact that $(n+1)/n \leq 2$ and thus $\frac{n^{2n}}{n!^2} \geq 2^{n^2-n}$ but this does not seem to be true.
| We can proceed by induction on $n$. The case $n=1$ is easy to check. Suppose that the inequality holds for $1,2,\ldots,n$. Then
\begin{align*}
\frac{{(n + 1)^{2n + 2} }}{{(n + 1)!^2 }} = \frac{{(n + 1)^{2n} }}{{n!^2 }} &= \frac{{n^{2n} }}{{n!^2 }}\left( {\frac{{n + 1}}{n}} \right)^{2n} \ge \left( {\frac{{n + 1}}{n}} \right)^{n^2 - n} \left( {\frac{{n + 1}}{n}} \right)^{2n} \\& = \left( {\frac{{n + 1}}{n}} \right)^{(n + 1)^2 - (n + 1)} \ge \left( {\frac{{n + 2}}{{n + 1}}} \right)^{(n + 1)^2 - (n + 1)} ,
\end{align*}
which completes the proof.
| {
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Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers
Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers.
I think a useful idea for tackling this question is considering congruences. If we consider the equation modulo 4, we get that the RHS is congruent to 2, so $x=1.$ If $y > 1,$ the LHS is congruent to 0 modulo 9. The residues of $5^z$ modulo 9 for $z=1,2,\cdots, 6$ are $5,7,8,4,2,1$. So 5 is a primitive root modulo 9. Thus in order to have $1+5^z\equiv 0\mod 9,$ we must have $z \equiv 3\mod 6$. But then considering the equation modulo 6 or 3 doesn't seem to yield any contradiction.
| If $z=6k+3,$ then $5^z+1$ is divisible by $7,$ so it cannot yield a solution.
That finishes your solution, since it means that $x=y=1$ are the only solutions.
| {
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Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ integer. Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$
Clearly if $n=0$ then we will have $20$ but I couldn't decide that how can I find the other integers. Any hint?
If I say that $x=\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ then we have $200+2\sqrt{10^4-n}=x^2$
| Alternative approach:
You want
$$M^2 = \left(\sqrt{100 + \sqrt{n}} + \sqrt{100 - \sqrt{n}}\right)^2 = 200 + 2\sqrt{10000 - n}.$$
To minimize $n$, you need $M^2$ as large as possible,
and still less than $200 + 2\sqrt{10000} = 400$.
The largest such square is
$19^2 = 361 \implies 2\sqrt{10000 - n} = 161 \implies $
$\sqrt{10000 - n} = 80.5 \implies (10000 - n)$ is not an integer.
Therefore, the largest feasible square is
$\displaystyle 18^2 = 324 \implies 2\sqrt{10000 - n} = 124 = 2\sqrt{(62)^2}.$
Therefore, $10000 - n = (62)^2 \implies n = 10000 - (62)^2 = 6156.$
| {
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.
| Here is another method to solve. Let
\begin{eqnarray}
I&=&\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\frac{2}{D^3}\int_{0}^\infty \frac{1}{\bigg[\big(\frac{x}{D}\big)^2 + 1)\bigg]^2}d(\frac xD)\\
&\stackrel{x/D\to x}{=}&\frac{2}{D^3}\int_{0}^\infty \frac{1}{(x^2 + 1)^2}dx\tag1\\
&\stackrel{1/x\to x}{=}&\frac{2}{D^3}\int_{0}^\infty \frac{x^2}{(x^2 + 1)^2}dx\\\tag2.
\end{eqnarray}
Adding (1) to (2) gives
\begin{eqnarray}
2I&=&\frac{2}{D^3}\int_{0}^\infty \frac{x^2+1}{(x^2 + 1)^2}dx\\
&=&\frac{2}{D^3}\int_{0}^\infty \frac{1}{x^2+1}dx\\
&=&\frac{2}{D^3}\cdot\frac{\pi}{2}
\end{eqnarray}
and hence $I=\frac{\pi}{2D^3}$.
| {
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Where is my mistake in the computation of $\int \frac{1}{\sin x} \text{d}x$ Here is my attempt:
$$\quad \ \sin x = \frac{2 \tan(x/2)}{1 + \tan^2 (x/2)} \\
\implies \int \frac{\text{d}x}{\sin x}=\int \frac{1+\tan^2(x/2)}{2 \tan(x/2)} \ \text{d}x$$
I do the following variable change: $u = \tan (x/2)$
$$\quad \ \frac{\text{d}u}{\text{d}x} = \frac{1}{2} \left(1 + \tan^2(x/2) \right) \\
\implies \text{d}u = \frac{1}{2} \left(1 + \tan^2 (x/2) \right) \text{d}x \\
\implies \int \frac{\text{d}x}{\sin x} = \int \frac{\text{d}u}{u} = \ln u = \ln ( \tan(x/2))$$
But if I try to compute the derivation of $\ln ( \tan(x/2))$ I will not find $1/\sin x$. Where is the mistake ?
| The derivative is precisely $$\frac{d}{dx}\bigg(\ln\big(\tan(\frac{x}{2})\big)\bigg)=\frac{1}{2}\cdot \frac{\sec^{2}(\frac{x}{2})}{\tan(\frac{x}{2})}=\frac{1}{2\cdot\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1}{\sin(x)}$$
| {
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About a Conjecture: $-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$ Hi it's a conjecture wich refine for $0< x\leq 1$ the inequality Refinement of a famous inequality :
Problem/Conjecture
Let $0<x\leq 1$ then for $n\geq 3$ a natural number it seems we have the inequality :
$$f(x)=-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$$
Some material :
Here New bound for Am-Gm of 2 variables I have proved using Gerber's theorem , the inequality :
$$\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}\geq x^n$$
And here https://mathoverflow.net/questions/337457/prove-that-left-fracxn1xn-11-rightn-left-fracx12-rightn the inequality in the first link is shown greatly .
Idea for a proof (weaker in fact):
It seems we have under the constraint above :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\geq 0$$
And :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)-\frac{\left(x+1\right)}{2}\geq 0$$
And for $x\in(0,0.5]$:
$$x^{\frac{x}{x+1}}+\sqrt{x}-1-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\le 0$$
And finally for $x\in(0,0.5]$ :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)+\frac{\left(x+1\right)}{2}-\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\ge0$$
So applying the Karamata's inequality on $f(x)=x^n$ with $x\in[0,0.5]$ gives a weaker result if i'm not wrong .
Edit :
Using the bounds above we have a bound for the derivative for $n\geq 6$ :
$$j\left(x\right)=-n\left(\frac{x^{2n}+1}{x^{2\left(n-1\right)}+1}\right)^{\left(n-1\right)}\cdot\frac{2x^{2n+1}\left(x^{2n}+n\left(x^{2}-1\right)+1\right)}{\left(x^{2n}+x^{2}\right)^{2}}-2^{-n}\cdot2n\cdot x\left(x^{2}+1\right)^{\left(n-1\right)}+n\left(x^{2}\right)^{\left(n-1\right)}\cdot\left(\left(2x^{2}\right)x\cdot\frac{\left(x^{2}+2\ln\left(x\right)+1\right)}{\left(x^{2}+1\right)^{2}}+1\right)\leq \frac{d}{dx}f\left(x^{2}\right)$$
How to (dis)prove it ?
| Not an answer just a useful remark :
We can split the problem because it seems we have :
Let $\exists x\in(0,1]$ and $\exists n\geq 3$ a natural number such that
$$\frac{\left(1+\frac{1}{2^{n}}+\frac{1}{n^{2}}-nx\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}+nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}-1\geq 0$$
And :
$$\frac{\left(x+1\right)^{n}}{2^{n}}+\frac{\left(nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}-nx+\frac{1}{2^{n}}+\frac{1}{n^{2}}-\left(x^{\frac{x}
{x+1}}+\sqrt{x}-1\right)^{n}\ge0$$
I go a little bit further :
Define :
$$f\left(x\right)=\frac{\left(nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}-\left(n-\frac{1}{2^{n}}\right)x+1/n^{2}+\frac{1}{2^{n}},g(x)=\frac{\left(x+1\right)^{n}}{2^{n}}-\frac{1+2x}{2^{n}},h\left(x\right)=\frac{x}{2^{n}}-\left(\sqrt{x}+x^{\frac{x}{x+1}}-1\right)^{n}$$
Let $a,b\in(0,1)$ $\exists x\in(a,b]$ and $\exists n\geq 3$ a natural number such that :
$$h(x)>0,f(x)>0,g(x)>0$$
We can also study the inequality for $x,y,z\in[0.5,1]$ and $n\geq 10$:
$$0\le y^{n}+z^{n}-1-\left(x-1\right)^{n}$$
This inequality seems true for :
Let :
$r(x)=\frac{x^{n}+1}{x^{n-1}+1}$
Denotes by $d=x_{min}\in[0.5,1]$ the value such that :
$$r'(d)=0$$
Then taking $z\in[d-1/2^{n},1]$ and $x\in[0.5,1]$ and $y=r(x)$ the inequality generalized seems true .
I haven't a proof yet but with all this stuff we can show the inequality .
| {
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Why can't we solve $\lim\limits_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}$ through dividing by xy in both numerator and denominator? When solving this limit, we use methods like $\epsilon$ - $\delta$ definition and polar coordinate convertion. And I wonder if we can solve it by dividing by $xy$ in both numerator and denominator like this
$\lim\limits_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} = \lim\limits_{(x, y) \to (0, 0)} \frac{1}{\sqrt{\frac{1}{y^2} + \frac{1}{x^2}}} = 0$
Thanks in advance!
| You can actually divide by $x$ and $y$, with a grain of salt:$\def\sgn{\operatorname{sgn}}$
$$\begin{align}
f(x,y) = \frac{xy}{\sqrt{x^2 + y^2}}
&= \frac{xy}{|x||y|\sqrt{\frac1{x^2} + \frac1{y^2}}} \\
&= \frac{\sgn(x)\sgn(y)}{\sqrt{\frac{1}{y^2} + \frac{1}{x^2}}} \tag 1
\end{align}$$
which is due to $\sqrt {x^2}=|x|$. The function $\sgn x$ denotes the sign of $x$:
$$\sgn(x) = \begin{cases}
+1,& x > 0 \\
-1,& x < 0 \\
0,& x = 0 \\
\end{cases}$$
Now the denominator of $(1)$ goes to $\infty$ as $(x,y)\to(0,0)$ and the numerator is bounded irrespective of $x$ and $y$, thus $f(x,y)\to0$.
However there is a loophole: $f(x,y)$ is defined for any $(x,y)\neq(0,0)$, for example $f(0,y)$ is perfectly fine provided $y\neq0$. But when you factor out $|x|$ and $|y|$ in the first step, this requires that neither $x$ nor $y$ may be 0.
A way around that loophole could be to use polar coordinates $(x,y) = r(\cos \phi,\sin\phi)$ and using $\sin^2\phi + \cos^2\phi = 1$:
$$\begin{align}
\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2 + y^2}}
&= \lim_{r\to0^+} \frac{r\cos\phi\cdot r\sin\phi}{\sqrt{r^2\cos^2\phi +r^2\sin^2\phi}} \\
&= \lim_{r\to0^+} \frac{r^2 \cos\phi \sin\phi}{\sqrt{r^2}} \\
&= \lim_{r\to0^+} r \cos\phi \sin\phi \\
&= 0\\
\end{align}$$
| {
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Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be appreciated.
| $$\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\sum_{i=1}^n \frac{1}{\sqrt{i}}\right) &= \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n^2}\right) \\
&< \frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n(n-1)}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n - 1} - \frac{1}{n}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{1}{i - 1} \\
&<\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{2}{i}
\end{align*}$$
which converges.
For those interested, the bound I gave above is $\frac{\pi^2}{6} + \zeta\left(\frac{3}{2}\right) - 1 \approx 6.86968476421920$, while the correct sum seems to be about $3.44\ldots$ by summing the first $10^6$ terms. I wonder what the correct sum is.
If you don't understand the first line, try drawing $16$ points in a $4$ by $4$ square, label them $(n, i) = (1, 1)$ to $(n, i) = (4, 4)$ from bottom left to top right, then colour/circle the points that will be included in the sum. You should colour the points $i \leq n$ only.
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Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.
| Simpler Solution.
\begin{align} & (ac+bd)^2+(ad-bc)^2 \\ = \; & (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2 \\ = \; & a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\ = \; & (a^2+b^2)(c^2+d^2) \end{align}
| {
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"answer_count": 6,
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} |
Integrate $\int\frac{3x}{x^5+x^4+1}dx$ I have an integral which I solved. But, I am not sure whether my answer is right or not. The integral is $$\int\frac{3x}{x^5+x^4+1}dx$$ My answer $$3\left(-\dfrac{\displaystyle\sum_{\left\{Z:\>Z^3-Z+1=0\right\}}\frac{\left(2Z^2-3Z-1\right)\ln\left(\left|x-Z\right|\right)}{3Z^2-1}}{7}\\+\dfrac{\ln\left(x^2+x+1\right)}{7}-\dfrac{4\arctan\left(\frac{2x+1}{\sqrt{3}}\right)}{7\sqrt{3}}\right)$$
| $x^3-x+1= (x-a)(x^2+ax-a^{-1}) $ has a single real root $a=-1.3247$, or
$$a =-\sqrt[3]{\frac{1}2+\frac16\sqrt{\frac{23}3}}-\sqrt[3]{\frac{1}2-\frac16\sqrt{\frac{23}3}}
$$
Then
\begin{align}
&\int\frac{3x}{x^5+x^4+1}dx\\
=& \ \frac17\int
\frac{3(2x-1)}{x^2+x+1}-\frac{2(3x^2-1)}{x^3-x+1} + \frac{9x+1}{(x-a)(x^2+ax-a^{-1})}\ dx\\
=& \ \frac17\bigg(\ln\frac{(x^2+x+1)^3}{(x^3-x+1)^2}
-\frac {9a^2+a}{2a-3}\ln\frac{x-a}{\sqrt{x^2+ax-a^{-1}}}\\
& \ \>\>\>\>\>-{4\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt{3}}
+\frac{3(a^2-3a+9)}{(2a-3)\sqrt{3a^2-4}}\tan^{-1}\frac{2x+a}{\sqrt{3a^2-4}}\bigg)+ C
\end{align}
| {
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What is wrong with my proof that $\int 2x dx= 2x^2$ by writing $2x=\underbrace{2+2+\cdots+2}_{x\;\text{times}}$? I know $\int 2x \,dx = x^2 + C$ (by the power rule) but why does the following proof not give the same answer?
\begin{align*}
\int 2x \,dx &= \int \underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}} \, dx \\
&= \underbrace{\int{2} \, dx + \int{2} \, dx + \dots \ + \int{2}_ \, dx}_{x \text{ times}}\\
&= 2x + 2x + \dots + 2x + C \\
&= 2x \times x + C \\
&= 2x^2 + C
\end{align*}
(And I have the same question for this false proof that $\int{2^x} \, dx = 2^{x}x+ C$)
\begin{align*}
\int{2^x} \,dx &= \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{x \text{ times}} \cdot 1 \, dx \\
&= 2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-1) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\
&= 2^2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-2) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\
&= 2^x \cdot \int{1} \, dx \\
&= 2^{x}x+ C \\
\end{align*}
I suspect that it has something to do with not being able to:
*
*Change integral of sums to sums of integrals for an arbitrary $x$, and
*Remove a constant out of an integral if there are variable numbers of those constants.
But I'm not sure why these do not hold. If this is the reason, is there a theorem stating it?
Thanks in advance!
| An integral might run from $x=0$ to $x=X$.
In $\int2xdx$, the function is increasing as you go along. So the first $2$ is there all the way as x goes from 1 to X, but the second $2$ doesn't start until x is 2, and the final $2$ doesn't appear until x has already reached X.
In $\int2dx+\int2dx+...$, all the $2$s are there all the way as x goes from 0 to X.
| {
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Limit of sequence given by $x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$? Let $x_1,x_2,x_3\in\mathbb{R}$ be three distinct real numbers. I am interested in the convergence of the sequence
$$
x_{n} = \frac{x_{n-2}+x_{n-3}}{2},\quad n = 4,5,\ldots
$$
ie,
$$
x_{4} = \frac{x_{2}+x_{1}}{2},\quad x_{5} = \frac{x_{3}+x_{2}}{2},\quad x_{6} = \frac{x_{4}+x_{3}}{2},\quad\cdots
$$
Please note that, although related to, this is not a recursive averaging sequence with two initial values, as described and studied elsewhere --- see, for instance, Smith, Scott G. "Recursive Averaging". The Mathematics Teacher, Vol. 108, No. 7 (March 2015), pp. 553-557.
In the present case, convergence is easily verified by noting that
$$
\big|\,x_{n}-x_{n-1}\,\big| = \big|\frac{x_{n-2}+x_{n-3}}{2}-\frac{x_{n-3}+x_{n-4}}{2}\big| = \big|\,\frac{1}{2} (x_{n-2}-x_{n-4})\,\big| = \big|\,\frac{1}{2^{\,n-4}}\,\big|\cdot\big|\, x_{3}-x_{1}\,\big|
$$
Therefore the sequence is Cauchy and hence it converges in $\mathbb{R}$.
However, I was not able to find the limit to where the sequence converges. Computer experiments carried out with several different sets of initial values suggest that this limit should be a linear combination of the those initial values, as occurs in the case mentioned above. Also, the results of these experiments point to the existence of (how many?) sub-sequences, all converging to the very same limit.
I would appreciate some help in the analysis of this problem.
| The characteristic equation of the linear recurrence is
$$\lambda^3 - \frac{1}{2}\lambda-\frac{1}{2} = 0$$ with roots $\lambda_1=1$, $\lambda_{2,3} = \frac{1}{\sqrt{2}}\cdot(-\frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}})$. There exist unique $a$, $b$, $c$ such that for all $n \ge 0$ we have
$$x_n = a \cdot 1^n + b \cdot \lambda_2^n + c \cdot \lambda_3^n$$
Since $|\lambda_{2,3}|= \frac{1}{\sqrt{2}}< 1$, we have
$$\lambda_2^n \to 0, \ \ \lambda_3^n \to 0$$
for $n \to \infty$. We conclude that $x_n \to a$. Now we only need to obtain $a$ from the equalities
$$x_0 = a + b + c \\
x_1 = a + b \lambda_2 + c \lambda_3\\
x_2 = a + b \lambda_2^2 + c \lambda_3^3$$
We get with Cramer's rule
$$ a = \frac{\left| \begin{matrix} x_0 & 1 & 1 \\ x_1 & -\frac{1}{2} + \frac{i}{2} & -\frac{1}{2} - \frac{i}{2}\\ x_2 & (-\frac{1}{2} + \frac{i}{2})^2 & (-\frac{1}{2} - \frac{i}{2})^2 \end{matrix} \right |} {\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -\frac{1}{2} + \frac{i}{2} & -\frac{1}{2} - \frac{i}{2}\\ 1 & (-\frac{1}{2} + \frac{i}{2})^2 & (-\frac{1}{2} - \frac{i}{2})^2 \end{matrix} \right |} = \frac{1}{5}(x_0 + 2 x_1 + 2 x_2)$$
$\bf{Added:}$ Since we are dealing with linear recurrences it's worth getting comfortable with determinants of the form
$$D(x_0, x_1, \ldots x_{n-1}) \colon = \left| \begin{matrix} x_0 & 1 & \ldots &1 \\
x_1 & \mu_2& \ldots& \mu_{n} \\
x_2 & \mu_2^2& \ldots & \mu_{n}^2 \\
\ldots& & & \ldots \\
x_{n-1} & \mu_2^{n-1} & \ldots & \mu_{n}^{n-1} \end{matrix} \right |$$
Note that when $x_k = x^k$ for $0\le k \le n-1$ we have a Vandermonde determinant, which equals
$$V(x, \mu_2, \ldots, \mu_{n}) = V( \mu_2, \ldots, \mu_{n}) \cdot \prod_{i=2}^{n}( \mu_i - x)= \\ =(-1)^{n-1} V( \mu_2, \ldots, \mu_{n})\cdot (x-\mu_2)\cdot (x-\mu_{n}) = (-1)^{n-1} V( \mu_2, \ldots, \mu_{n})\cdot ( x^{n-1} - s_1 x^{n-2} + s_2 x^{n-3} - \cdots)$$
It follows that the determinant with first column $(x_0, x_1, \ldots x_{n-1})$ equals
$$(-1)^{n-1} V(\mu_2, \ldots, \mu_{n+1}) \cdot ( x_{n-1} - s_1 x_{n-2} + s_2 x_{n-3} - \cdots)$$
Now when solving the system we have a quotient of such determinants $D$, the extra factor $(-1)^{n-1} \cdot V$ can be omitted. We get: the coefficient in front of a power $\mu_1^n$ equals
$$\frac{P(x)/(x-\mu_1)}{P'(\mu_1)}$$
where the numerator is considered "in the umbral sense".
Example with our problem at hand:
The characteristic polynomial equals $P(x)=x^3 - \frac{1}{2} x - \frac{1}{2}$.
*
*For the numerator : divide $P(x)$ by $x-\mu_1= x-1$ and get
$\frac{1}{2}(2 x^2 + 2 x + 1)$. This umbrally is $\frac{1}{2}(2 x_2 + 2x_1 + x_0)$.
*For the denominator we have $P'(x) = 3 x^2 - \frac{1}{2}$ so $P'(1) = \frac{5}{2}$
*Set up the fraction
$$a_1 = a =\frac{\frac{1}{2}(2 x_2 + 2 x_1 + x_0)}{\frac{5}{2}} = \frac{2 x_2 + 2 x_1 + x_0}{5}$$
| {
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"url": "https://math.stackexchange.com/questions/4485083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $\int_0^\infty \frac{x^k}{(x^2 + b^2)^l}\,dx=\frac1{2b^{2l-k-1}}\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})/\Gamma(l)$ . My quantum mechanics textbook states the following integral without proof:
$$
\int_0^\infty \frac{x^k}{(x^2 + b^2)^l} \mathrm dx = \frac1{2b^{2l-k-1}}\frac{\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})}{\Gamma(l)}
$$
What is this class of integrals called? Why are the gamma functions involved, and how do we prove this identity?
| Letting $x=b \tan \theta, \quad$ then $d x=b \sec ^{2} \theta d \theta$ and
$$
I=\int_{0}^{\frac{\pi}{2}} \frac{b^{k} \tan ^{k} \theta}{b^{2 l} \sec ^{2 l} \theta} \cdot b \sec ^{2} \theta d \theta =\frac{1}{b^{2 l-k-1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta \cos ^{2 l-k-2} \theta d \theta
$$
Using $$
B(x, y)=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2 x-1} \theta \cos^{2y-1} \theta d \theta
$$
yields
$$
I=\frac{1}{2 b^{2 l-k-1}} B\left(\frac{k+1}{2}, l-\frac{k+1}{2}\right)
$$
By the property of Beta function
$$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
we have
$$
I=\frac{1}{2 b^{2 l-k-1}} \frac{\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{2 l-k-1}{2}\right)}{\Gamma(l)}
$$
| {
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How to find the sum of constants given the following systems of equations? Given that $q,r,s$, and $t$ are different constant values in the following systems of equations containing $a,b,c$, and $d$. Find the sum $q+r+s+t$.
$\frac{1}{qa+1} + \frac{1}{qb+1} + \frac{1}{qc+1}+ \frac{1}{qd+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} + \frac{1}{rc+1}+ \frac{1}{rd+1} = 1$
$\frac{1}{sa+1} + \frac{1}{sb+1} + \frac{1}{sc+1}+ \frac{1}{sd+1} = 1$
$\frac{1}{ta+1} + \frac{1}{tb+1} + \frac{1}{tc+1}+ \frac{1}{td+1} = 1$
I tried using a simpler version of the problem but I can't find the pattern.
I am thinking that there might be a pattern that I can get from finding the sum $q+r$ from this simpler version of the problem.
$\frac{1}{qa+1} + \frac{1}{qb+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} = 1$
I also tried adding the four equations that leads me to factor out the sum $q+r+s+t$ but it becomes more complicated. But I noticed that the sum of the each set of denominators per equation is equal to the product of all those denominators. However, I am stuck at looking for the pattern where I can use that fact.
Thanks in advance for your comments and suggestions on how to solve this particular algebra problem.
| Hint: $\,$ the equations are of the form:
$$
\frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c}+ \frac{1}{x+d} = \frac{1}{x} \quad\quad\text{for}\;\; x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t} \tag{1}
$$
Let $P(x) = (x+a)(x+b)(x+c)(x+d)$ then $(1)$ can be written as:
$$
P(x) - x P'(x) = 0 \quad\quad\text{for}\;\; x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t} \tag{2}
$$
The LHS of $(2)$ is a polynomial of degree $4$ which is zero for the distinct values $x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t}$, so those are its $4$ roots, and the rest follows from Vieta's relations.
| {
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Which interval is correct here$?$ The equation $$2\textrm{sin}^2\theta x^2-3\textrm{sin}\theta x+1=0$$ where $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$ has one root lying in the interval
$(0,1)$
$(1,2)$
$(2,3)$
$(-1,0)$
I know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in general odd number of roots of the equation $f(x)=0$ lie between $a$ and $b$. But I am not able to use this piece of information here, maybe because of two variables. I also tried to assume $\textrm{sin}\theta x$ as $y$ but nothing good came out.
Any help is greatly appreciated.
EDIT Answer given is $(1,2)$
| $\begin{align}
2\sin^2 \theta x^2 − 3 \sin \theta x +1 & = 0 \\
x & = \frac{3 \sin \theta \pm \sqrt{9 \sin^2 \theta - 8 \sin^2 \theta}}{4 \sin^2 \theta} \\
x & = \frac{3 \pm 1}{4 \sin \theta} \\
x & = \frac{1}{\sin\theta} \; \; \text{or} \; \; x = \frac{1}{2\sin\theta}
\end{align}$
$$ \therefore \, \theta = (2n+1)\pi - \sin^{-1} \frac{1}{x}, \; 2n\pi + \sin^{-1} \frac{1}{x}, \; \theta = 2n\pi + \sin^{-1} \frac{1}{2x}, \; \theta = (2n+1)\pi - \sin^{-1} \frac{1}{2x} $$
Now, you can finish it.
| {
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Computing the value of $(x+y)^4$ if $x^4+y^4=5$ and $x^2+xy+y^2=10$
Let $x^4+y^4=5$ and $x^2+xy+y^2=10.$ Find $(x+y)^4.$
First, I tried expanding $(x+y)^4$ using the binomial theorem to get $5+4x^3y+6x^2y^2+4xy^3,$ so simplifying I got $5+4xy(x^2+y^2)+6(xy)^2.$ Then I rearranged the given equation to get $x^2+y^2=10-xy,$ so the expansion becomes $5+4xy(10-xy)+6(xy)^2.$ I further simplified to get $2x^2y^2+40xy+5,$ but I'm not sure how to continue off here. Factoring this expression doesn't seem to help. May I have some help? Thanks in advance.
| An alternative approach is to rearrange the second equality and square:
$$
x^2 + y^2 = 10 - xy \implies \underbrace{x^4 + y^4}_{=\, 5} + 2 x^2y^2 = 100 - 20 xy +x^2 y^2 \implies x^2y^2 + 20 xy = 95
$$
The last quadratic could be solved for $\,xy\,$, then $\,x,y\,$ determined and $\,(x+y)^4\,$ calculated by brute force. A shortcut, instead, is to again rearrange the second equality in a different way and square:
$$
(x+y)^2 = 10 + xy \implies (x+y)^4 = 100 + 20 xy + x^2y^2 = \dots
$$
| {
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"source": "stackexchange",
"question_score": "3",
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Computing the integral of trig function under square root How can we solve this integral
$$\int \sqrt{\csc^2x -2}
\mathrm{d}x$$
My idea was substituting $\csc^2x=2\csc^2\theta$. Then the integral became $$\sqrt{2}\int \frac{\csc^2\theta-1}{\sqrt{2\csc^2\theta-1}} \mathrm{d}\theta$$ after a few simplifications. I don't know how to go about at this point. If i try breaking the numerator,still i would be left with an equivalent question of $$\int \sqrt{2\csc^2\theta-1}\mathrm{d}\theta$$ and $$\int \frac{\mathrm{d}\theta}{\sqrt{2\csc^2\theta-1}}$$
Could someone provide a cleaner approach to this problem or give new insights on how to further simply the above integrals?
| Letting $u=\sqrt{\csc ^{2} x-2}$ transforms the integral into
$$\begin{aligned}
I&=-\frac{1}{2} \underbrace{\int \frac{u^{2} d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}}_{J}\end{aligned}
$$
For the integral $J$, letting $t=\frac{u}{\sqrt{u^{2}+1}} $, we have $$ u^{2}=\frac{t^{2}}{1-t^{2}} \Rightarrow u d u=\frac{t d t}{\left(1-t^{2}\right)^{2}}\Rightarrow \frac{d u}{\sqrt{u^{2}+1}}=\frac{t^{2}}{u^{2}} \cdot \frac{d t}{\left(1-t^{2}\right)^{2}}=\frac{d t}{1-t^{2}} $$
Plugging them yields
$$
\begin{aligned}
J&= \int \frac{1}{\frac{2-t^{2}}{1-t^{2}}} \cdot \frac{t^2}{1-t^2} \frac{d t}{1-t^{2}}\\&= \int\left(\frac{1}{1-t^{2}}-\frac{1}{2-t^{2}}\right) d t\\& =\frac{1}{2} \ln \left|\frac{1-t}{1+t}\right|+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2}-t}{\sqrt{2}+t}\right|+C \\&= \frac{1}{2} \ln \left|\frac{u-\sqrt{u^{2}+1}}{u+\sqrt{u^{2}+1}}\right|+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2}-\sqrt{u^{2}+1}}{\sqrt{2}+\sqrt{u^{2}+1}}\right|+C\end{aligned}$$
Now we can conclude that
$$I= \frac{1}{4} \ln \left|\frac{u+\sqrt{u^{2}+1}}{u-\sqrt{u^{2}+1}}\right|+\frac{1}{4\sqrt{2}} \ln \left|\frac{\sqrt{2}+\sqrt{u^{2}+1}}{\sqrt{2}-\sqrt{u^{2}+1}}\right|+C $$
| {
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"question_score": "1",
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Solve $\sin{(70^{\circ}-x)}=\frac{1}2 + \frac{1}{4\cos{(20^{\circ})}}$ $$\text{Solve} \ \ \sin{(70^{\circ}-x)}=\frac{1}2 + \frac{1}{4\cos{(20^{\circ})}}$$
I came across this equation while solving a problem. The computer provides $x=20^{\circ}$, and I tried to draw a shape with that in mind:
Then I got the equation $\sin{(50^{\circ})}=2\sin{(20^{\circ})}\tan{(30^{\circ})}+\sin{(40^{\circ})}+\tan{(30^{\circ})}$.
Can this be somehow converted to $\frac{1}2 + \frac{1}{4\cos{(20^{\circ})}}$? Any help will be appreciated.
| $$\sin{(70^{\circ}-x)}=1/2 + \frac{1}{4\cos{(20^{\circ})}}$$
Start from RHS:
$$\begin{align}
\frac{1}{2} + \frac{1}{4\cos{(20^{\circ})}}&=\frac{2\cos(20^{\circ})+1}{4\cos(20^{\circ})}=\frac{2\sin(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})}{4\sin(20^{\circ})\cos(20^{\circ})}\\
\\
&=\frac{\sin(40^{\circ})+\sin(20^{\circ})}{2\sin(40^{\circ})}=\frac{2\sin(30^{\circ})\cos(10^{\circ})}{2\sin(40^{\circ})}\tag{1}\\
\\
&=\frac{\cos(10^{\circ})}{2\sin(40^{\circ})}
=\frac{\sin(80^{\circ})}{2\sin(40^{\circ})}\\
\\
&=\frac{2\sin(40^{\circ})\cos(40^{\circ})}{2\sin(40^{\circ})}=\cos(40^{\circ})=\sin(50^{\circ})
\end{align}$$
Note: Eq.$(1)$ comes from $\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$. The equation goes to:
$$\sin(70^{\circ}-x)=\sin(50^{\circ})$$
So you can solve for $x$
| {
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$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
| $\begin{align}p(x)&=x^{20}+x^{19}+\dots+x^{10}\\&=x^{17}(x^3+x)+x^{16}(x^3+x)+x^{13}(x^3+x)+x^{12}(x^3+x)+x^9(x^3+x)+\color{red}{x^{11}}\end{align}$
Again $x^{11}=x^8(x^3+x) -x^6(x^3+x) +x^4(x^3+x) -x^2(x^3+x) +(x^3+x) -x$
Hence remainder is $=-x$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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The least possible number of tokens on a $n×n$ board Tokens are placed on the squares of a
$2021
×
2021$
board in such a way that each square contains at most one token. The token set of a square of the board is the collection of all tokens which are in the same row or column as this square. (A token belongs to the token set of the square in which it is placed.) What is the least possible number of tokens on the board if no two squares have the same token set?.
It's a problem from https://gonitzoggo.com/archive/problem/441/english Recently, I've been taking preparation for junior Math Olympiad Contest and found this problem.
I have calculated least possible number of tokens for $ 2×2, 3×3, 4×4, 5×5$ but finding no pattern between them.
| On an $N\times N$ board, we can do this with $\frac32N$ tokens if $N$ is even: fill the main diagonal with tokens, and a parallel sub-diagonal of length $\frac12N$. And if $N$ is odd, we only need $\frac12(3N-1)$ tokens, because the parallel sub-diagonal can be of length $\frac12(N-1)$. For example ($N=10$ and $N=11$):
$$
\begin{bmatrix}
\bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot \\
\bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\
\cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\
\cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \bullet \\
\end{bmatrix}
\,\,\,\,\,\,\,\,\,\,\,\,
\begin{bmatrix}
\bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot \\
\bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot \\
\cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot \\
\cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot \\
\cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \bullet & \cdot \\
\cdot & \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot & \cdot & \bullet \\
\end{bmatrix}
$$
So if $N=2021$, there exists an arrangment of $3031$ tokens in which no two squares have the same token set.
I have checked on my laptop that these arrangements are the best possible for $N\le 7$, but for $N\ge 8$ I don't have any kind of proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}
Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}
Is it correct? How to compute $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
| This is just a supplement to the nice answer of @Cathedral. Here we close a gap and show
\begin{align*}
\color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r+1}=\frac{(2r)!!}{(2r+1)!!}}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1}{2r+1}}\\
&=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\frac{1}{r+\frac{1}{2}}
\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q}\tag{2.1}\\
&=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\underbrace{\operatorname{res}_{z=r}\left(\frac{1}{z+\frac{1}{2}}
\prod_{q=0}^{n}\frac{1}{z-q}\right)}_{f(z)}\tag{2.2}\\
&=\frac{1}{2}(-1)^nn!\left(-\operatorname{res}_{z=-\frac{1}{2}}f(z)
-\operatorname{res}_{z=-\infty}f(z)\right)\tag{2.3}\\
&=\frac{1}{2}(-1)^nn!\left(-\lim_{z=-\frac{1}{2}}\prod_{q=0}^n\frac{1}{z-q}
+\underbrace{\operatorname{res}_{z=0}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)}_{=0}\right)\tag{2.4}\\
&=\frac{1}{2}(-1)^{n+1}n!\prod_{q=0}^n\frac{1}{-\frac{1}{2}-q}\\
&=2^nn!\prod_{q=0}^n\frac{1}{2q+1}\\
&=\frac{2^nn!}{(2n+1)!!}\\
&\,\,\color{blue}{=\frac{(2n)!!}{(2n+1)!!}}
\end{align*}
and the claim (1) follows. Similarly we can show
\begin{align*}
\color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r-1}=-\frac{(2n)!!}{(2n-1)!!}}
\end{align*}
Comment:
*
*In (2.1) we use
\begin{align*}
\binom{n}{r}&=\frac{n!}{r!(n-r)!}\\
&=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{n-r-q}\\
&=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{q+1}\tag{$q\to\ n-r-1-q$}\\
&=n!(-1)^{n-r}\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q}
\end{align*}
*In (2.2) we write the summands as residue of a meromorphic function at the pole $z=q$.
*In (2.3) we use the sum of the residues of a meromorphic function at the poles $z=q, 0\leq q\leq r$ and $z=-\frac{1}{2}$ plus the residue at $\infty$ sum up to zero. This way we get rid of the sum and what is left are just two residues, the one at $z=-\frac{1}{2}$ and the one at $z=\infty$.
*In (2.4) we use the identity
\begin{align*}
\operatorname{res}_{z=\infty}f(z)=\operatorname{res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)
\end{align*}
which transforms a residue at infinity to a residue at zero. We then find by inspection the residue of $f(z)$ at $z=\infty$ is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2$ I have given a partial answer to a very interesting recent post Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$, where it remains to prove that
$$\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2.$$
Wolfarm Mathematica supports this result, I want your help in proving this.
| We seek to prove that
$$\sum_{j=1}^{n+1} {2n+2\choose 2j-1} {j-1/2\choose n}
= 2(n+1)^2.$$
The LHS is
$$\sum_{j=0}^{n} {2n+2\choose 2j+1} {j+1/2\choose n}
= \sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
\frac{1-(-1)^j}{2}.$$
The first piece $A$ is
$$\frac{1}{2}
\sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
= \frac{1}{2} [w^n]
\sum_{j=0}^{2n+2} {2n+2\choose j} (1+w)^{j/2}
\\ = \frac{1}{2} [w^n] (1+\sqrt{1+w})^{2n+2}
= \frac{1}{2}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1+\sqrt{1+w})^{2n+2}.$$
Now put $1-\sqrt{1+w} = v$ so that $w=v(v-2)$ and $dw = 2(v-1) \; dv$
to obtain
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1} (v-2)^{n+1}} (2-v)^{2n+2} (v-1)
\\ = \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1}} (v-2)^{n+1} (v-1)
\\ = {n+1\choose n-1} ((-1)\times 2)^2
- {n+1\choose n} ((-1)\times 2)
\\ = \frac{1}{2} (n+1) n \times 4
+ (n+1) \times 2 = 2(n+1)^2.$$
This is the claim, piece $B$ must be zero. And indeed we find
$$\frac{1}{2}
\sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n} (-1)^j
= \frac{1}{2}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1-\sqrt{1+w})^{2n+2}.$$
Apply the same substitution to get
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1} (v-2)^{n+1}} v^{2n+2} (v-1)
= \; \underset{v}{\mathrm{res}} \;
\frac{1}{(v-2)^{n+1}} v^{n+1} (v-1) = 0$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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