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A simple conditional expectation problem $X, Y$ iid uniform random variables on $[0,1]$
$$Z =
\left\{
\begin{aligned}
X+Y \quad&\text{ if } X>\frac{1}{2}
\\
\frac{1}{2} + Y \quad & \text{ if } X\leq\frac{1}{2}
\end{aligned}
\right.$$
The question is $E\{Z|Z\leq 1\}= ?$
I tried $\displaystyle \int_0^1 E\{Z|Z = z\} P\{Z = z\}dz$ and got $5/8$, but I am not so sure about the result since I haven't touched probability for years.
| For $x \in (1/2,1]$, let
$$
F_{Z|Z \leq 1}(x) = {\rm P}(Z \leq x | Z \leq 1) = \frac{{{\rm P}(Z \le x)}}{{{\rm P}(Z \le 1)}},
$$
and let
$$
f_{Z|Z \leq 1}(x) = \frac{{\rm d}}{{{\rm d}x}}F_{Z|Z \le 1} (x) = \frac{1}{{{\rm P}(Z \le 1)}}\frac{{\rm d}}{{{\rm d}x}}{\rm P}(Z \le x).
$$
Then,
$$
{\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} .
$$
So now the problem reduces to calculating ${\rm P}(Z \leq x)$, for $x \in (1/2,1]$.
This can be done using the law of total probability, conditioning on $X$, leading to
$$
{\rm E}[Z|Z \le 1] = 7/9.
$$
EDIT:
Fix $x \in (1/2,1]$. Then, by the law of total probability,
$$
{\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}(Z \le x|X = s)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Z \le x|X = s)\,{\rm d}s} .
$$
It thus follows from the definition of $Z$ (and the independence of $X$ and $Y$) that
$$
{\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} .
$$
Now,
$$
\int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} = \frac{1}{2}{\rm P}(Y \le x - 1/2) = \frac{{x - 1/2}}{2} = \frac{x}{2} - \frac{1}{4}
$$
and
$$
\int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {(x - s)\,{\rm d}s} = \frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}.
$$
Hence
$$
{\rm P}(Z \leq x) = \bigg(\frac{x}{2} - \frac{1}{4}\bigg) + \bigg(\frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}\bigg) = \frac{{x^2 }}{2} - \frac{1}{8}.
$$
In particular,
$$
{\rm P}(Z \leq 1) = \frac{3}{8}.
$$
Thus,
$$
F_{Z|Z \leq 1}(x) = \frac{8}{3}\bigg(\frac{{x^2 }}{2} - \frac{1}{8}\bigg) = \frac{{4x^2 - 1}}{3},
$$
and in turn
$$
f_{Z|Z \leq 1}(x) = \frac{{8x}}{3}.
$$
Finally,
$$
{\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} = \frac{8}{3}\int_{1/2}^1 {x^2 \,{\rm d}x} = \frac{8}{3}\bigg(\frac{{1 - 1/8}}{3}\bigg) = \frac{7}{9}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/52317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and
$6 \mid ab(a^2+b^2)(a^2-b^2)$?
Indeed this method implies studying numerous congruences and is quite long.
| Hint: First write $f(a,b)=ab(a^2+b^2)(a^2-b^2)=a^5b-b^5a$. Recall Fermat's Little Theorem which states $x^{p-1}\equiv 1 \pmod{p}$ for $x$ not divisible by $p$.
Assume $5$ does not divide either of $a,b$. (otherwise it follows automatically that $5|f(a,b)$) Then
$$a^5b-b^5a\equiv ab-ba\equiv 0\pmod{5}$$ and we see that $5|f(a,b)$.
You can prove that $2$ and $3$ divide $f(a,b)$ in the same way.
(An alternative way to see that it is divisible by $2$ is to note that one of $a$, $b$, $a^2+b^2$ must be divisible by $2$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Why does this expression equal $\pi$? I noticed that the following expression equals $\pi$ and I was curious as to why. Is it just a coincidence, or is there a meaningful explanation?
$$\int_{-\infty}^\infty\frac{1}{x^2+1}~dx=\pi$$
| Here is another approach, which avoids taking $\arctan(\infty)$, etc.: let $u = \frac{1}{x}$, so that $\frac{du}{dx} = \frac{-1}{x^2}.$ Then we obtain $\int_{1}^{\infty} \frac{1}{1+u^{2}} du = \int_{0}^{1} \frac{1}{1+x^2}dx$. Hence $\int_{- \infty}^{\infty} \frac{1}{1+x^2} dx = 4\int_{0}^{1} \frac{1}{1+x^2}dx = 4 \arctan(1) = 4\frac{\pi}{4} = \pi$.
As is well-known, the last integral can also be evaluated by integrating $\frac{1}{1+x^2} = (1 - x^2 + x^4 -\ldots )$ term by term (this needs a little justification but is permissible here), to obtain the famous formula $\frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Formula for $1^2+2^2+3^2+...+n^2$ In example to get formula for $1^2+2^2+3^2+...+n^2$ they express $f(n)$ as:
$$f(n)=an^3+bn^2+cn+d$$ also known that $f(0)=0$, $f(1)=1$, $f(2)=5$ and $f(3)=14$
Then this values are inserted into function, we get system of equations solve them and get a,b,c,d coefficients and we get that $$f(n)=\frac{n}{6}(2n+1)(n+1)$$
Then it's proven with mathematical induction that it's true for any n.
And question is, why they take 4 coefficients at the beginning, why not $f(n)=an^2+bn+c$ or even more? How they know that 4 will be enough to get correct formula?
| It is a consequence of the following algebraic identity
$$1+2^{2}+3^{2}+\ldots +n^{2}=\frac{1}{3}\left( n^{3}+3n^{2}+3n+1\right) -
\frac{1}{3}n-\frac{1}{2}(n^2+n)-\frac{1}{3}.$$
$$\tag{1}$$
The RHS is a cubic function of $n$: $\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n$.
Proof. From the algebraic identity
$$\left( 1+k\right) ^{3}=1+3k+3k^{2}+k^{3}$$
we get the following $n$ identities:
$$\begin{eqnarray*}
\left( 1+1\right) ^{3} &=&1+3+3+1 \\
\left( 1+2\right) ^{3} &=&1+3\cdot 2+3\cdot 2^{2}+2^{3} \\
\left( 1+3\right) ^{3} &=&1+3\cdot 3+3\cdot 3^{2}+3^{3} \\
&&\ldots \\
\left( 1+n\right) ^{3} &=&1+3n+3n^{2}+n^{3}.
\end{eqnarray*}$$
Now if we sum these equalities and cancel the common terms, $\left(
1+1\right) ^{3}$ on the LHS of the first and $2^{3}$ on the RHS of the
second, $\left( 1+2\right) ^{3}$ on the LHS of the second and $3^{3}$ on the
RHS of the third, etc., and $(1+n-1)^3$ on the LHS of the second last and $n^3$ on the RHS of the last, we get:
$$\left( 1+n\right) ^{3}=n+3(1+2+3+\ldots +n)+3(1+2^{2}+3^{2}+\ldots +n^{2})+1,\tag{2}$$
and $(1)$ follows from $(2)$ and the sum formula for the arithmetic progression $$1+2+3+\ldots +n=\frac{\left( n+1\right) n}{2}.$$
(Adapted from Proof 1 in this answer to the question Prove that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? ; see the above Theo Buehler's comment.)
Note: This is a particular case of the sum $1+2^{p}+3^{p}+\ldots +n^{p}$,
which is a polynomial of degree $p+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/59175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ show that $x=-c/b$ when $a=0$ OK, this one has me stumped. Given that the solution for $ax^2+bx+c =0$ $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad(*)$$
How would you show using $(*)$ that $x=-c/b$ when $a=0$
(Please dont use $a=0$ hence, $bx+c=0$.)
| You don't. The quadratic formula only works if $a\neq 0$ (remember, if you try dividing by $0$, the universe explodes).
If $a=0$, then the equation is
$$0x^2 + bx + c = 0$$
or equivalently,
$$bx + c = 0.$$
So, solve that equation for $x$. If $b=0$, then it's always false if $c\neq 0$, and always true if $c=0$.
And if $b\neq 0$, then...
Given your comment... you could look at limiting behavior. Then
$$\begin{align*}
\lim_{a\to 0}\frac{-b\pm\sqrt{b^2-4ac}}{2a} &= \lim_{a\to 0}\frac{b^2-(b^2-4ac)}{2a\left(-b\mp\sqrt{b^2-4ac}\right)}\\
&= \lim_{a\to 0}\frac{2c}{-b\mp\sqrt{b^2-4ac}}.
\end{align*}$$
As $a\to 0$, $\sqrt{b^2-4ac}\to |b|$. If $b\gt 0$, only the minus sign in the denominator makes sense, giving you $\frac{2c}{-2b} = -\frac{c}{b}$. If $b\lt 0$, only the plus sign in the denominator makes sense, giving you again $\frac{2c}{-2b} = -\frac{c}{b}$. (The "other root" goes to $\infty$ or $-\infty$, and so is "out of sight, out of mind..."; unless $c=0$; but if $c=0$, then the original function has one value equal to $0$ and the other to $-\frac{b}{a}$; the latter goes to $\pm\infty$ as $a\to 0$, and the former has limit $0=-\frac{c}{b}$).
And if $b=0$, then of course the whole thing doesn't makes sense unless $c=0$, in which case $x$ can be anything.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/60792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How to prove that the Binet formula gives the terms of the Fibonacci Sequence? This formula provides the $n$th term in the Fibonacci Sequence, and is defined using the recurrence formula: $u_n = u_{n − 1} + u_{n − 2}$, for $n > 1$, where $u_0 = 0$ and $u_1 = 1$.
Show that
$$u_n = \frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n \sqrt{5}}.$$
Please help me with its proof. Thank you.
| Let
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
$$M_n =
\begin{pmatrix}
F_{n+1} & F_n \\ F_n & F_{n-1}
\end{pmatrix}$$
By induction, you can show that $M_n=A^n$. Now, the eigenvalues of $A$ satisfy $\lambda^2-\lambda-1=0$. Let them be $\lambda_1$ and $\lambda_2$.
$A$ can be diagonalized and written as $A=SBS^{-1}$, where $B=\begin{pmatrix}
\lambda_1 & 0\\ 0 & \lambda_2
\end{pmatrix}$, $S=\begin{pmatrix}
\lambda_1 & \lambda_2 \\ 1 & 1
\end{pmatrix}$
Thus,
$$A^n = SBS^{-1}SBS^{-1}SBS^{-1}SBS^{-1}.... = SB^nS^{-1}$$
$$=\begin{pmatrix}
\lambda_1 & \lambda_2 \\ 1 & 1
\end{pmatrix}\begin{pmatrix}
\lambda_1^n & 0\\ 0 & \lambda_2^n
\end{pmatrix}(\frac{1}{\lambda_1-\lambda_2}\begin{pmatrix}
1 & -\lambda_2 \\ -1 & \lambda_1
\end{pmatrix})$$
$$=\frac{1}{\lambda_1-\lambda_2}\begin{pmatrix}
\lambda_1 & \lambda_2 \\ 1 & 1
\end{pmatrix}\begin{pmatrix}
\lambda_1^n & -\lambda_1^n\lambda_2\\ -\lambda_2^n & \lambda_2^n\lambda_1
\end{pmatrix}$$
$$=\frac{1}{\lambda_1-\lambda_2}\begin{pmatrix}
\lambda_1^{n+1}-\lambda_2^{n+1} & -\lambda_1\lambda_2(\lambda_1^n-\lambda_2^n)\\ \lambda_1^n-\lambda_2^n &-\lambda_1\lambda_2(\lambda_1^{n-1}-\lambda_2^{n-1})
\end{pmatrix}$$
Using $\lambda_1\lambda_2=-1$, $A^n =
\begin{pmatrix}
F_{n+1} & F_n \\ F_n & F_{n-1}
\end{pmatrix}$, $\lambda_1=\frac{1+\sqrt{5}}{2}$ and $\lambda_2=\frac{1-\sqrt{5}}{2}$, we get
$$\begin{pmatrix}
F_{n+1} & F_n \\ F_n & F_{n-1}
\end{pmatrix} = \frac{1}{\sqrt{5}}\begin{pmatrix}
\lambda_1^{n+1}-\lambda_2^{n+1} &\lambda_1^n-\lambda_2^n\\ \lambda_1^n-\lambda_2^n &\lambda_1^{n-1}-\lambda_2^{n-1}
\end{pmatrix}$$
Which gives us $F_n=\frac{\lambda_1^n-\lambda_2^n}{\sqrt{5}}$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/65011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 1
} |
Just checking: sine series for $x^2$ Is the Fourier sine series for $x^2$ equal to $\sum {2\pi\over 2m+1}-{8\over (2m+1)^3\pi} \sin ((2m+1)x)$? (just want to check that those multiple steps of intergation by parts did not slip me up). Thanks.
| The Fourier sine series is $x^2 = \sum_{n=1}^\infty a_n \sin( n x)$, where $a_n = \frac{2}{\pi} \int_0^\pi x^2 \sin( n x) \mathrm{d} x$. Now
$$
\begin{eqnarray}
\frac{\pi}{2} a_n &=& \left. (-\frac{x^2}{n} \cos(n x)) \right\vert_{0}^{\pi} + \int_0^\pi ( \frac{2}{n} x) \cos(n x) \mathrm{d} x \\
&=& \left. (-\frac{x^2}{n} \cos(n x)) \right\vert_{0}^{\pi} +
\left. (\frac{2 x}{n^2} \sin(n x)) \right\vert_{0}^{\pi} - \int_0^\pi \frac{2}{n^2} \sin( n x) \mathrm{d} x \\
&=&
\left. \left( (-\frac{x^2}{n} \cos(n x)) + (\frac{2 x}{n^2} \sin(n x)) + \frac{2}{n^3} \cos(n x) \right) \right\vert_{0}^{\pi} \\
&=& (-1)^n \left(\frac{2}{n^3} -\frac{\pi^2}{n} \right) - \frac{2}{n^3}
\end{eqnarray}
$$
Thus
$$
\begin{eqnarray}
x^2 &=& \frac{2}{\pi} \sum_{n=1}^\infty \left( (-1)^n \left(\frac{2}{n^2} - \pi^2 \right) - \frac{2}{n^2} \right) \frac{\sin(n x)}{n} \\
&=& \frac{2}{\pi} \sum_{n=1}^\infty \left( \pi^2 - \frac{4}{(2n-1)^2}\right) \frac{\sin((2n-1)x}{2n-1} -\pi \sum_{n=1}^\infty \frac{\sin(2 n x)}{n}
\end{eqnarray}
$$
Where the last expression was arrived at by splitting summation over even and odd integers.
Added: As robjohn pointed out in comments, the resulting series approximates odd function $\operatorname{sign}(x) x^2$ on interval $(-\pi, \pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\log_{12} 2=m$ what's $\log_6 16$ in function of $m$? Given $\log_{12} 2=m$ what's $\log_6 16$ in function of $m$?
$\log_6 16 = \dfrac{\log_{12} 16}{\log_{12} 6}$
$\dfrac{\log_{12} 2^4}{\log_{12} 6}$
$\dfrac{4\log_{12} 2}{\log_{12} 6}$
$\dfrac{4\log_{12} 2}{\log_{12} 2+\log_{12} 3}$
$\dfrac{4m}{m+\log_{12} 3}$
And this looks like a dead end for me.
| Writing everything without logarithms:
$$
\begin{array}{ccc}
12^m=2&\therefore&3=2^{1/m-2}\\
6^x=16&\therefore&3=2^{4/x-1}
\end{array}
$$
Thus we get
$$
\begin{array}{ccc}
1/m-2=4/x-1&\therefore&x=\frac{4m}{1-m}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine a point $$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$
Find a D point so this equality is true:
$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$
| $$\text{The given vectors } \overrightarrow{AB}=B-A\text{ and }\overrightarrow{AC}=C-A \text{ and the solution }D=A+\overrightarrow{AD}$$
Let $(x,y)$ be the coordinates of $D$. The equation
$$5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}\tag{0}$$
means that
$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2),\tag{1}$$
because the vectors $\overrightarrow{AD}=D-A$, $\overrightarrow{AB}=B-A$ and
$\overrightarrow{AC}=C-A$.
The vectors $5\overrightarrow{AD}=5\left( D-A\right) =\left( 5D-5A\right) $,
$2\overrightarrow{AB}=\left( 2B-2A\right) $, etc.
A possible way of solving the equation $(1)$ is as follows.
$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2)$$
$$\begin{eqnarray*}
&\Leftrightarrow &(5x-20,5y-10)=2(-6,-1)-3(-1,-4) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)-(-3,-12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)+(3,12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12+3,-2+12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-9,10) \\
&\Leftrightarrow &\left\{
\begin{array}{c}
5x-20=-9 \\
5y-10=10
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
x=\frac{11}{5} \\
y=4
\end{array}
\right.
\end{eqnarray*}\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Diverging sequence I can't understand diverging sequences. How can I prove that $a_n=1/n^2-\sqrt{n}$ is divering? Where to start? What picture should I have in my mind? I tried to use $\exists z \forall n^* \exists n\ge n^*: |a_n-A|\ge z$, but how should I see this? And how can I solve the question with this property?
| A convergent sequence is also a Cauchy sequence. If you can find a constant $\epsilon>0$ so that for all $N>0$, there are $n>N$ and $m>N$ so that $|a_n-a_m|\ge\epsilon$, then $\{a_n\}$ is a divergent sequence.
In the case of $a_n=\frac{1}{n^2}-\sqrt{n}$, we can let $\epsilon=1$ and for a given $N>0$, let $n=N+1$ and $m=N+2+\lceil2\sqrt{N+1}\;\rceil$. Then
$$
\begin{align}
&a_n-a_m\\
&=\left(\frac{1}{(N+1)^2}-\sqrt{N+1}\right)-\left(\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}-\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}\right)\\
&=\left(\frac{1}{(N+1)^2}-\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}\right)+\left(\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1}\right)\\
&\ge1
\end{align}
$$
since
$$
0\le\frac{1}{(N+1)^2}-\frac{1}{(N+2+\lceil2\sqrt{N+1}\;\rceil)^2}
$$
and
$$
\begin{align}
1&=\sqrt{N+2+2\sqrt{N+1}}-\sqrt{N+1}\\
&\le\sqrt{N+2+\lceil2\sqrt{N+1}\;\rceil}-\sqrt{N+1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/67476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding sum form for a particular recursive function Consider a finite sequence of zeros and ones of length $3n$, with $n$ an integer. We write an element of this sequence as $a_i$. How many sequences are there such that there exists an integer $k$, $0<k\le n$, such that $\sum^{3k}_{j=1}a_j=2k$? Here is what I have as of now: let $x_n$ be this number. We notice that $x_1=\binom{3}{2}=3$, and $x_n=\binom{3n}{2n}-\binom{3}{2}x_{n-1}+2^{3}x_{n-1}=\binom{3n}{2n}+5x_{n-1}$. How would I find a sum form solution to this? Also, does this seem correct? I got this because $\binom{3n}{2n}$ counts the total number of sequences satisfying the condition for $k=n$, $\binom{3}{2}x_{n-1}$ is the number of sequences satisfying it for both $k=n$ and $k=n-1$, and the number of sequences satisfying $k=n-1$ should be $x_{n-1}$, and we can choose the last three elements at random, so we multiply by $2^3$.
| If $x_n$ was $\binom{3n}{2n}+5x_{n-1}$, which it isn’t, we would have
$$
x_n=\sum_{k=1}^{n} 5^{(n-k)} \binom{3k}{2k}
$$
Mathematica gives the following “closed form”:
$$
x_n=-\frac{1}{5} \binom{3 n+3}{2 n+2} \, _3F_2\left(1,n+\frac{4}{3},n+\frac{5}{3};n+\frac{3}{2},n+2;\frac{27}{20}\right)
-\alpha \;5^n $$
where
$
\alpha=1+2 i \sqrt{\frac{5}{7}} \cos \left(\frac{1}{6} \cos ^{-1}\left(-\frac{17}{10}\right)\right)
$ is a (complex) root of $7 z^3-21 z^2+36 z-27$, and $_3F_2$ is a generalized hypergeometric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Complex solutions for Fermat-Catalan conjecture The Fermat-Catalan conjecture is that $a^m + b^n = c^k$ has only a finite number of solutions when $a, b, c$ are positive coprime integers, and $m,n,k$ are positive integers satisfying $\frac{1}{m} + \frac{1}{n} +\frac{1}{k} <1$. There are currently only 10 solutions known (listed at the end of this post). My question concerns the case where $a, b, c$ are positive coprime Gaussian integers. I've found two solutions. Is there a clever method for finding more? I've used brute force techniques.
*
*$(8+5i)^2+(5+3i)^3=(1+2i)^7$
*$(20+9i)^2+(1+8i)^3=(1+i)^{15}$
*$(1+2i)^7+(49+306i)^2=(27+37i)^3$ (Zander)
*$(44+83i)^2+(31+39i)^3=(5+2i)^7$ (Zander)
*$(238+72i)^3+(7+6i)^8=(7347−1240i)^2$ (Oleg567)
Here are the known solutions over integers.
*
*$1^m+2^3=3^2$
*$2^5+7^2=3^4$
*$13^2+7^3=2^9$
*$2^7+17^3=71^2$
*$3^5+11^4=122^2$
*$33^8+1549034^2=15613^3$
*$1414^3+2213459^2=65^7$
*$9262^3+15312283^2=113^7$
*$17^7+76271^3=21063928^2$
*$43^8+96222^3=30042907^2$
| Here is a small portion of what I have found:
$i^{4 m}+2^3=3^2$ where integer $m \ge 2$, i.e. the smallest is $i^8+2^3=3^2$
$(1+i)^2=i^{4 m+1}+i^{4 n+1}$ where integer $m \ge 1$, $n \ge 1$, i.e. the smallest is $(1+i)^2=i^5+i^5$
$(78-78 i)^2=(23 i)^3+i^{4 m+3}$ where integer $m \ge 1$, i.e. the smallest is $(78-78 i)^2=(23 i)^3+i^7$
$(11+11 i)^2+i^{4 m+1}=(3 i)^5$ where integer $m \ge 1$, i.e. the smallest is $(11+11 i)^2+i^5=(3 i)^5$
$(34-34 i)^2=(5 i)^3+(3 i)^7$
$(36+19 i)^2+(9-8 i)^3=(1+i)^{13}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$ Given the geometric series:
$1 + x^2 + x^4 + x^6 + x^8 + \cdots$
We can recast it as:
$S = 1 + x^2 \, (1 + x^2 + x^4 + x^6 + x^8 + \cdots)$, where $S = 1 + x^2 + x^4 + x^6 + x^8 + \cdots$.
This recasting is possible only because there is an infinite number of terms in $S$.
Exactly how is this mathematically possible?
(Related, but not identical, question: General question on relation between infinite series and complex numbers).
| There is a finite version of which the expression you have is the limit.
Suppose $S=1+x^2+x^4+x^6+x^8$, then we can put
$S+x^{10}=1+x^2(1+x^2+x^4+x^6+x^8)=1+x^2S$
And obviously this can be taken as far as you like, so you can replace 10 with 10,000 if you choose. If the absolute value of $x$ is less than 1, this extra term approaches zero as the exponent increases.
There is also a theory of formal power series, which does not depend on notions of convergence.
| {
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Solving a recurrence using substitutions I have to solve this recurrence using substitutions:
$(n+1)(n-2)a_n=n(n^2-n-1)a_{n-1}-(n-1)^3a_{n-2}$ with $a_2=a_3=1$.
The only useful substitution that I see is $b_n=(n+1)a_n$, but I don't know how to go on, could you help me please?
| it seems that I solved. $$(n-2)b_n=(n^2-n-1)b_{n-1}-(n-1)^2b_{n-2}$$
So divide it by $n-1$ then we get $$\begin{align*}
\left(1-\frac{1}{n-1}\right)b_n&=\left(n-\frac{1}{n-1}\right)b_{n-1}-(n-1)b_{n-2}\\
&=\left(n-1+1-\frac{1}{n-1}\right)b_{n-1}-(n-1)b_{n-2}\\
&=(n-1)(b_{n-1}-b_{n-2})+b_{n-1}\left(1-\frac{1}{n-1}\right)
\end{align*}$$
then $$\left(1-\frac{1}{n-1}\right)(b_n-b_{n-1})=(n-1)(b_{n-1}-b_{n-2})$$
then we make subtitution $p_n=b_n-b_{n-1}$ as $b_2=3,b_3=4$ we get that $p_3=1$.
we have that $$\begin{align*}
&\left(1-\frac{1}{n-1}\right)p_n =(n-1)p_{n-1} \implies\\ &p_n=\frac{(n-1)^2}{n-2}p_{n-1}\implies\\
&p_n=\prod\limits_{k=4}^n{\frac{(k-1)^2}{k-2}}=\frac{n-1}{2}\frac{(n-1)!}{2} \implies\\
&b_n-b_{n-1}=\frac{n!-(n-1)!}{4} \implies\\
&b_n=4+\sum\limits_{k=4}^n{\frac{k!-(k-1)!}{4}}=4+\frac{1}{4}(n!-6)=\frac{10+n!}{4} \implies\\
&a_n=\frac{10+n!}{4(n+1)}
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to sum this series for $\pi/2$ directly? The sum of the series
$$
\frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1}
$$
can be derived by accelerating the Gregory Series
$$
\frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2}
$$
using Euler's Series Transformation. Mathematica is able to sum $(1)$, so I assume there must be some method to sum the series in $(1)$ directly; what might that method be?
| Notice that for $c_k = \frac{k!}{(2k+1)!!}$ the ratio of successive terms $\frac{c_{k+1}}{c_k} = \frac{k+1}{2k +3} = \frac{1}{2} \frac{k+1}{k+3/2}$.
This means that the series is hypergeometric with the value ${}_2 F_1(1, 1, \frac{3}{2}, \frac{1}{2})$.
This particular Gaussian hypergeometric is elementary:
$$
{}_2 F_1(1, 1, \frac{3}{2}, x) = \frac{\arcsin\left(\sqrt{x}\right)}{\sqrt{1-x} \sqrt{x}}
$$
Upon substitution of $x=\frac{1}{2}$ we recover the result $ 2 \arcsin(\frac{1}{\sqrt{2}}) = \frac{\pi}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series.
| Ok, here is an approach with generating functions. Let
$$
g_1(z) = \sum_{s=0}^\infty \binom{p+s}{s} z^s = \frac{1}{\left(1-z\right)^{p+1}}
$$
$$
g_2(z) = \sum_{s=0}^\infty \binom{2p+m}{s} z^s = \left(1+z\right)^{m+2p}
$$
Now
$$ \begin{eqnarray}
\sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} &=& \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{m-2s} = [z]^m g_1(z^2) g_2(z) = [z]^m \frac{\left(1+z\right)^{m+2p}}{(1-z^2)^{p+1}} \\
&=& [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}}
\end{eqnarray}
$$
Here is a verification:
In[27]:= With[{p = 5,
m = 7}, {SeriesCoefficient[(1 + z)^(m + 2 p)/(1 - z^2)^(
p + 1), {z, 0, m}],
Sum[Binomial[p + s, s] Binomial[2 p + m, 2 p + 2 s], {s,
0, \[Infinity]}]}]
Out[27]= {71808, 71808}
Let's continue:
$$ \begin{eqnarray}
[z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}} &=& \sum_{s=0}^\infty \binom{p+m-1}{m-s} \binom{p+s}{s} = \sum_{s=0}^\infty \binom{p+m-1}{p+s-1} \binom{p+s}{s}\\ &=& \sum_{s=0}^\infty \frac{(p+s) (m+p-1)!}{p! s! (m-s)!} =
\sum_{s=0}^\infty \frac{p (m+p-1)!}{p! s! (m-s)!} + \sum_{s=0}^\infty \frac{s (m+p-1)!}{p! s! (m-s)!} \\
&=& \binom{m+p-1}{m} \left( \sum_{s=0}^\infty \binom{m}{s} + \sum_{s=0}^\infty \frac{s}{p} \binom{m}{s} \right) \\
&=& \binom{m+p-1}{m} \left( 2^m + 2^{m-1} \frac{m}{p} \right)
\end{eqnarray}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $\lim_{n\to\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac1{\sqrt k}\right)$ How to find $\lim\limits_{n\to\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac1{\sqrt k}\right)$ ?
And generally does the limit of the integral of $f(x)$ minus the sum of $f(x)$ exist?
How to prove that and find the limit?
| Use $\sqrt{n} = \sum_{k=1}^n \left( \sqrt{k} - \sqrt{k-1} \right)$, then
$$
\begin{eqnarray}
2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} &=& \sum_{k=1}^n \left( 2 \sqrt{k} - 2 \sqrt{k-1} - \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \sqrt{k}-\sqrt{k-1} \right)^2\\
&=& \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \frac{(\sqrt{k}-\sqrt{k-1})(\sqrt{k}+\sqrt{k-1})}{(\sqrt{k}+\sqrt{k-1})} \right)^2 \\
&=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}
\end{eqnarray}
$$
This shows the limit does exist and $\lim_{n \to \infty} \left( 2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^\infty \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}$.
The value of this sums equals $-\zeta\left(\frac{1}{2} \right) \approx 1.4603545$. This value is found by other means, though:
$$
2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2 \sqrt{n} - \left( \zeta\left(\frac{1}{2}\right) - \zeta\left(\frac{1}{2}, n+1\right)\right) \sim -\zeta\left(\frac{1}{2}\right) - \frac{1}{2\sqrt{n}} + o\left( \frac{1}{n} \right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Converting multiplying fractions to sum of fractions I have the next fraction: $$\frac{1}{x^3-1}.$$
I want to convert it to sum of fractions (meaning $1/(a+b)$).
So I changed it to:
$$\frac{1}{(x-1)(x^2+x+1)}.$$
but now I dont know the next step. Any idea?
Thanks.
| $x^2+x+1=(x-a)(x-\bar{a})$ where $a=\exp(\frac{2\pi i}{3})=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, so
$$
\frac{1}{x^3-1}=\frac{1}{(x-1)(x-a)(x-\bar{a})}\tag{1}
$$
and then you can use partial fractions on $(1)$ to get
$$
\frac{1}{x^3-1}=\frac{1}{3}\left(\frac{1}{x-1}+\frac{a}{x-a}+\frac{\bar{a}}{x-\bar{a}}\right)\tag{2}
$$
Partial Fractions (Heaviside Method):
Suppose we wish to write
$$
\frac{A}{x-1}+\frac{B}{x-a}+\frac{C}{x-\bar{a}}+\frac{1}{(x-1)(x-a)(x-\bar{a})}\tag{3}
$$
To compute $A$, multiply both sides by $x-1$ and set $x=1$:
$$
\begin{align}
A
&=\frac{1}{(1-a)(1-\bar{a})}\\
&=\frac{1}{3}\tag{3a}
\end{align}
$$
To compute $B$, multiply both sides by $x-a$ and set $x=a$:
$$
\begin{align}
B
&=\frac{1}{(a-1)(a-\bar{a})}\\
&=\frac{a}{3}\tag{3b}
\end{align}
$$
To compute $C$, multiply both sides by $x-\bar{a}$ and set $x=\bar{a}$:
$$
\begin{align}
C
&=\frac{1}{(\bar{a}-1)(\bar{a}-a)}\\
&=\frac{\bar{a}}{3}\tag{3c}
\end{align}
$$
Collecting equations $(3)$, yields $(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}$? This is the inverse function of sin. Why is $\cos(\sin^{-1}x)=\sqrt{1-x^2}$?
Thanks a lot.
| Draw a right angled triangle a marked angle $\theta $. We can scale the triangle to set the hypotenuse equal to 1. Label the side opposite $\theta $ by some length $x$, then the adjacent side has length $\sqrt{1-x^2}.$
$\phi = \sin^{-1} x $ denotes the angle such that $\sin \phi = x $. So in the triangle, $ \sin^{-1} (x) = \theta. $
Then $ \displaystyle \cos \theta = \frac{ \text{Adjacent} }{ \text{Hypotenuse} } = \frac{ \sqrt{1-x^2} }{1} = \sqrt{1-x^2} $ so $$ \cos ( \sin^{-1} (x) ) = \sqrt{1-x^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine if the equation is valid/true The equation is:
$$\log_b \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 2\log_b(\sqrt{3}+\sqrt{2}).$$
I can get as far as:
$$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$
Which looks almost too simple, but I can't get the signs to match up right to solve the problem. Do I need to further break out the logarithmic functions that are there?
| $$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$
is equivalent to
$$\log_b(\sqrt{3}+\sqrt{2}) = \log_b(\sqrt{3}-\sqrt{2}) + 2\log_b(\sqrt{3}+\sqrt{2})$$
which is equivalent to
$$\log_b(\sqrt{3}+\sqrt{2}) = \log_b((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2)$$
which you can multiply out to check, or you might spot $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove equations in modular arithmetic Prove or disprove the following statement in modular arithmetic.
*
*If $a\equiv b \mod m$, then $ a^2\equiv b^2 \mod m$
*If $a\equiv b \mod m$, then $a^2\equiv b^2 \mod m^2$
*If $a^2\equiv b^2\mod m^2$, then $a\equiv b\mod m$
My proofs.
*
*$$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$$
$$ a^2-b^2 = (a+b)(a-b) = (a+b)mr = ms \text{ where } s = (a+b)\cdot r$$
So the first statement is true
*$$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$$
$$a^2-b^2 = (a+b)(a-b) = (a+b)mr$$
but $(a+b)\neq ms$ $\forall s\in\mathbb{Z}$ in general. So the second one is false.
*$$a^2-b^2 = m^2r, \exists r\in\mathbb{N}$$
$$a^2-b^2= (a+b)(a-b) $$
Then I kind of got stuck here. I'm not sure how to continue it. Am I missing some properties I don't know? Or there is a algebra trick that could be applied here?
| For Question 3, you are told that $m^2$ divides $a^2-b^2$, and are asked whether (necessarily) $m$ divides $a-b$.
Note that $a^2-b^2=(a-b)(a+b)$. Maybe the "factor" $m^2$ comes in whole or in large part from $a+b$, not from $a-b$. Let's see whether we can find an example. Let $m=3$, so $m^2=9$. Make $a+b$ divisible by $9$, for example by taking $a=8$, $b=1$. Then $a-b=7$, very much not divisible by $3$.
There is no example with $m=2$, but you can use the same idea to find an example with $m=4$. We want $(a-b)(a+b)$ to be divisible by $16$, with $a-b$ not divisible by $4$. We can for example take $a=15$, $b=1$, or $a=13$, $b=3$, or $a=7$, $b=1$. For all these examples, $a^2\equiv b^2\pmod{4^2}$ but $a \not\equiv b\pmod{4}$.
Now make up your own counterexample!
| {
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Maxima of bivariate function [1] Is there an easy way to formally prove that,
$$
2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$
$${0<x,y<1}$$
without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima?
[2] Similarly, is there an easy way for finding $$\max_{0<x,y<1} [f(x,y)]$$
where,
$$f(x,y)=2x(1+x)+2y(1+y)-8xy-4(2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y)^{2}$$
| Certainly, there is no need for taking the quotient, since $a \ge b \Leftrightarrow \min \{a-b\} \ge 0$.
Here's a cool trick called the S.O.S. (sum of squares) method. The idea is to try and factor out $(x-y)^2$:
$$\begin{align}
LHS-RHS &=(x^4+y^4-2x^2y^2)-2(x^3+y^3-x^2y-xy^2)+2(x^2+y^2-2xy)\\
&=(x^2-y^2)^2-2(x^2-y^2)(x-y)+2(x-y)^2\\
&=(x-y)^2(x+y)^2-2(x-y)^2(x+y)+2(x-y)^2\\
&=(x-y)^2((x+y)^2-2(x+y)+2)\\
&=(x-y)^2((x+y-1)^2+1)\\
&\ge 0
\end{align}$$
Note that this holds for all $x, y \in \mathbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my trig result unique? I recently determined that for all integers $a$ and $b$ such that $a\neq b$ and $b\neq 0$,
$$
\arctan\left(\frac{a}{b}\right) + \frac{\pi}{4} = \arctan\left(\frac{b+a}{b-a}\right)
$$
This implies that 45 degrees away from any angle with a rational value for tangent lies another angle with a rational value for tangent. The tangent values are related.
If anyone can let me know if this has been done/shown/proven before, please let me know. Thanks!
| As written, the formula is not true: the values of $\arctan(x)$ are always between $-\frac{\pi}2$ and $\frac{\pi}{2}$. Pick a rational number $\frac{a}{b}$ with $\frac{\pi}{4}\lt \frac{a}{b}\lt \frac{\pi}{2}$. For example, $a=11$, $b=10$. Then the left hand side,
$$\arctan\left(\frac{11}{10}\right)+\frac{\pi}{4}\approx 1.6184$$
whereas the right hand side is negative:
$$\arctan\left(\frac{11+10}{10-11}\right) = \arctan(-21) \approx -1.5232.$$
I think that what you mean is that if $\alpha$ is an angle such that $\tan(\alpha)$ is rational, different from $1$,
$$\tan(\alpha)=\frac{a}{b}\neq 1,\qquad a,b\text{ integers},$$
then
$$\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{b+a}{b-a}.$$
Certainly, well done if you discovered it by yourself! However, it is not new. In fact, the result is true even if $a$ and $b$ are not integers; all you need is for $a$ to be different from $b$, that is, for $\alpha\neq\frac{\pi}{4}$.
There are well-known formulas that express the sine, cosine, and tangent of a sum of angles in terms of the sines, cosines, and tangents of the summands:
$$\begin{align*}
\sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\
\cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\
\tan(\alpha+\beta) &= \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}.
\end{align*}$$
Taking $\beta=\frac{\pi}{4}$, since $\tan(\frac{\pi}4) = 1$, we get
$$\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{\frac{a}{b}+1}{1-\frac{a}{b}} = \frac{\quad\frac{a+b}{b}\quad}{\frac{b-a}{b}} = \frac{a+b}{b-a},$$
giving your formula.
| {
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Adding a different constant to numerator and denominator Suppose that $a$ is less than $b$ , $c$ is less than $d$.
What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?
| Note that if $b$ and $d$ have the same sign, then
$$
\frac{a}{b}-\frac{a+c}{b+d}=\frac{ad-bc}{b(b+d)}
$$
and
$$
\frac{a+c}{b+d}-\frac{c}{d}=\frac{ad-bc}{d(b+d)}
$$
also have the same sign.
Therefore, if $b$ and $d$ have the same sign, then $\dfrac{a+c}{b+d}$ is between $\dfrac{a}{b}$ and $\dfrac{c}{d}$.
Comment: As Srivatsan points out, if $b$ and $d$ are both positive,
$$
\frac{a}{b}\lesseqqgtr\frac{a+c}{b+d}\text{ if }\frac{a}{b}\lesseqqgtr\frac{c}{d}
$$
| {
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} |
What is the meaning of $\mathbf{Q}(\sqrt{2},\sqrt{3})$ I know that : $\mathbf{Q}(\sqrt{2}) = \mathbf{Q}+ \sqrt{2} \mathbf{Q}$ , but then what is $\mathbf{Q}(\sqrt{2},\sqrt{3})$?
| $\mathbf{Q}(\sqrt{2},\sqrt{3})$ means $\mathbf{Q}+\sqrt{2}\mathbf{Q}+\sqrt{3}\mathbf{Q}+\sqrt{6}\mathbf{Q}$, or in other words
$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}.$$
Be careful though. For example, $\mathbf{Q}(\sqrt{2},\sqrt[4]{2})=\mathbf{Q}(\sqrt[4]{2})=\mathbf{Q}+\sqrt[4]{2}\mathbf{Q}+(\sqrt[4]{2})^2\mathbf{Q}+(\sqrt[4]{2})^3\mathbf{Q}$, because adding in the $\sqrt{2}$ is redundant: we already have $\sqrt{2}=(\sqrt[4]{2})^2$ inside $\mathbf{Q}(\sqrt[4]{2})$.
In general, the field $\mathbf{Q}(a_1,\ldots,a_n)$ is the smallest field containing $\mathbf{Q}$ and the elements $a_1,\ldots,a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is every Mersenne prime of the form : $x^2+3 \cdot y^2$? How to prove or disprove following statement :
Conjecture :
Every Mersenne prime number can be uniquely written in the form : $x^2+3 \cdot y^2$ ,
where $\gcd(x,y)=1$ and $x,y \geq 0$
Since $M_p$ is an odd number it follows that : $M_p \equiv 1 \pmod 2$
According to Fermat little theorem we can write :
$2^p \equiv 2 \pmod p \Rightarrow 2^p-1 \equiv 1\pmod p \Rightarrow M_p \equiv 1 \pmod p$
We also know that :
$2 \equiv -1 \pmod 3 \Rightarrow 2^p \equiv (-1)^p \pmod 3 \Rightarrow 2^p-1 \equiv -1-1 \pmod 3 \Rightarrow$
$\Rightarrow M_p \equiv -2 \pmod 3 \Rightarrow M_p \equiv 1 \pmod 3$
So , we have following equivalences :
$M_p \equiv 1 \pmod 2$ , $M_p \equiv 1 \pmod 3$ and $M_p \equiv 1 \pmod p$ , therefore for $p>3$
we can conclude that : $ M_p \equiv 1 \pmod {6 \cdot p}$
On the other hand : If $x^2+3\cdot y^2$ is a prime number greater than $5$ then :
$x^2+3\cdot y^2 \equiv 1 \pmod 6$
Proof :
Since $x^2+3\cdot y^2$ is a prime number greater than $3$ it must be of the form $6k+1$ or $6k-1$ .
Let's suppose that $x^2+3\cdot y^2$ is of the form $6k-1$:
$x^2+3\cdot y^2=6k-1 \Rightarrow x^2+3 \cdot y^2+1 =6k \Rightarrow 6 | x^2+3 \cdot y^2+1 \Rightarrow$
$\Rightarrow 6 | x^2+1$ , and $ 6 | 3 \cdot y^2$
If $6 | x^2+1 $ then : $2 | x^2+1$ , and $3 | x^2+1$ , but :
$x^2 \not\equiv -1 \pmod 3 \Rightarrow 3 \nmid x^2+1 \Rightarrow 6 \nmid x^2+1 \Rightarrow 6 \nmid x^2+3 \cdot y^2+1$ , therefore :
$x^2+3\cdot y^2$ is of the form $6k+1$ , so : $x^2+3\cdot y^2 \equiv 1 \pmod 6$
We have shown that : $M_p \equiv 1 \pmod {6 \cdot p}$, for $p>3$ and $x^2+3\cdot y^2 \equiv 1 \pmod 6$ if $x^2+3\cdot y^2$ is a prime number greater than $5$ .
This result is a necessary condition but it seems that I am not much closer to the solution of the conjecture than in the begining of my reasoning ...
| It is a theorem that may have been first proved by Euler (but was known to Fermat) that every prime $p$ of the form $6k+1$ can be expressed in the form $p=x^2+3y^2$, where $x$ and $y$ are integers.
This representability question has been discussed on StackExchange. Please see here for a compact complete proof by Ewan Delanoy.
| {
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Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?
| de Moivre's formula says
$$
\begin{align}
\cos(3x)+i\sin(3x)
&=(\cos(x)+i\sin(x))^3\\
&=\left(\cos^3(x)-3\cos(x)\sin^2(x)\right)+i\left(3\cos^2(x)\sin(x)-\sin^3(x)\right)\\
&=\left(4\cos^3(x)-3\cos(x)\right)+i\left(3\sin(x)-4\sin^3(x)\right)\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\cos(3x)&=4\cos^3(x)-3\cos(x)\tag{2}\\
\sin(3x)&=3\sin(x)-4\sin^3(x)\tag{3}
\end{align}
$$
Solving $(3)$ for $\sin^3(x)$ yields
$$
\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)\tag{4}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
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} |
Convergence of the next series I'm trying to determine the convergence of this series:
$$\sum \limits_{n=1}^\infty\left(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}\right)^a$$
I've tried using D'Alambert criteria for solving it.
$$\lim_{n->\infty}\frac{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}\frac{2n}{2n+1})^a}{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n})^a} =
\lim_{n->\infty}\left(\frac{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}·\frac{2n}{2n+1})}{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n})}\right)^a$$
Which becomes:
$$\lim_{n->\infty}\left(\frac{2n}{2n+1}\right)^a$$
But after that, the limit is 1, so its convergence is unknown.
Any idea?
| Let $$a_n={1\over2}\cdot{3\over4}\cdot{5\over6}\cdot\ \cdots\ \cdot {{2n-3}\over{2n-2}}\cdot{{2n-1}\over{2n}}.$$
Note that $a_{n+1}=a_n\cdot {2n +1\over2(n+1)}$.
We will show that for all $n$:
$$
{1\over\sqrt{ 4n}} \le a_n\le {1\over\sqrt{ 2n+1}}.
$$
Having established this, it will follow by comparison with $p$-series that
$\sum\limits_{n=1}^\infty a_n^a$ converges if and only if $a>2$.
We establish the inequalities by induction.
Towards showing that $ a_n\le {1\over\sqrt{ 2n+1}}$, first
note that $a_1\le {1\over \sqrt{2\cdot 1+1}}$.
Now
suppose $a_n \le {1\over\sqrt{ 2n+1}}$.
Then $$ a_{n+1}^2\le \biggl[\,{2n +1\over 2(n+1)}\biggr]^2\cdot{1\over 2n+1}
={2n+1\over 4(n+1)^2}={2n+1\over 4n^2+8n+4}\le{2n+1\over 4n^2+8n+1 }={1\over 2n+1}.$$
So, $a_n\le {1\over\sqrt {2n+1}}$ for all $n$.
Towards showing that $ {1\over\sqrt{ 4n}} \le a_n$, first
note that $a_1\ge {1\over \sqrt{4\cdot 1 }}$.
Now
suppose $a_n \ge {1\over\sqrt{ 4n}}$.
Then $a_{n+1}^2\ge \Bigl[\,{2n +1\over 2(n+1)}\,\Bigr]^2\cdot{1\over 4n}$.
Now
$$
\biggl[\,{2n +1\over 2(n+1)}\,\biggr]^2\cdot{1\over 4n}-{1\over 4(n+1)}
={1\over 16 n(n+1)^2}>0;$$
thus
$$
a_{n+1}^2\ge{1\over 4(n+1)}.
$$
So, $a_n\ge {1\over\sqrt {4n}}$ for all $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \equiv 1 \pmod n $ then we can say $b^2 \equiv -1 \pmod n $ then it is not possible since solution exists only for $n=4k_2+1 $ so $a \equiv b\equiv 0 \pmod n $
So from the fact that $a^{2^{2k+1}} \equiv 1 \pmod n$ I have to conclude something.
| Hint: If $p \equiv 3 \pmod{4}$, then $x$ is a quadratic residue $\bmod{p}$ if and only if $-x$ is a quadratic nonresidue $\bmod{p}$.
Edit, to be more direct:
Lemma:
Suppose that $p$ is an odd prime, and assume that $x$ is a nonzero quadratic residue $\bmod{p}$. Then, $-x$ is a quadratic residue $\bmod{p}$ if and only if $p \equiv 1 \pmod{4}$.
Proof:
First suppose that $p \equiv 1 \pmod{4}$. By assumption, as $x$ is a nonzero quadratic residue, there exists an integer $a \neq 0$ such that $a^2 = x$. Since $-1$ is also a quadratic residue, there also exists an integer $b$ such that $b^2 = -1$. Thus,
$$
-x = -1 x = b^2 a^2 = (ba)^2,
$$
and hence $-x$ is a quadratic residue.
On the other hand, suppose that $p \equiv 3 \pmod{4}$ and $x = a^2$ is a quadratic residue. Since $x \neq 0$, it follows that $-x \neq 0$ as well. Then, suppose for a contradiction that $-x$ is also a nonzero quadratic residue, and hence $-x = c^2$ for some $c$. If this were the case, it would follow that
$$
-1 = -x/x = c^2/a^2 = (c/a)^2
$$
which would imply that $-1$ is a square $\bmod{p}$. This gives a contradiction, and we conclude that if $p \equiv 3\pmod{4}$ and $x$ is a nonzero quadratic residue, then $-x$ must be a quadratic nonresidue.
| {
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proof that if $AB=BA$ matrix $A$ must be $\lambda E$ Let $A \in Mat(2\times 2, \mathbb{Q})$ be a matrix with $AB = BA$ for all matrices $B \in Mat(2\times 2, \mathbb{Q})$.
Show that there exists a $\lambda \in \mathbb{Q}$ so that $A = \lambda E_2$.
Let $E_{ij}$ be the matrix with all entries $0$ except $e_{ij} = 1$.
$$
\begin{align}
AE_{11} &= E_{11}A \\
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\\
\left( \begin{array}{cc}
a_{11} & 0 \\
a_{21} & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
a_{11} & a_{12} \\
0 & 0 \\
\end{array} \right) \\
\end{align}
$$
$\implies a_{12} = a_{21} = 0$
And then the same for the other three matrices $E_{12}, E_{21}, E_{22}$ …
I guess it's not the most efficient way of argueing or writing it down … ? Where's the trick to simplify this thingy?
| Take $B$ to be each of the basis matrices $E_{ij}$ that is all zeros except that it is 1 at position $ij$.
| {
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Finding the n-th root of a complex number I am trying to solve $z^6 = 1$ where $z\in\mathbb{C}$.
So What I have so far is :
$$z^6 = 1 \rightarrow r^6\operatorname{cis}(6\theta) = 1\operatorname{cis}(0 + \pi k)$$
$$r = 1,\ \theta = \frac{\pi k}{6}$$
$$k=0: z=\operatorname{cis}(0)=1$$
$$k=1: z=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3} + i}{2}$$
$$k=2: z=\operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{1 + \sqrt{3}i}{2}$$
$$k=3: z=\operatorname{cis}\left(\frac{\pi}{2}\right)=i$$
According to my book I have a mistake since non of the roots starts with $\frac{\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots.
Can anyone please tell me where my mistake is? Thanks!
| I will solve $z^6=1$ where $z∈\mathbb{C}$. First I will take off all data: $r=1$, $\theta = 0$, $n=6$ and $k=0,1,2,3,4,5$ then
For $k=0: w_1=\sqrt[6]{1}[\cos{\frac{2\pi (0)}{6}}+ i\sin{\frac{2\pi (0)}{6}}]=1[\cos{0}+i \sin{0}]$ therefore $z_1=\cos{0}+i\sin{0}$ finally $z_1=1$
For $k=1: w_2=\sqrt[6]{1}[\cos{\frac{2\pi (1)}{6}}+ i\sin{\frac{2\pi (1)}{6}}]=1[\cos{\frac{\pi}{3}}+i \sin{\frac{\pi}{3}}]$ therefore $z_2=\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$ finally $z_2=\frac{1}{2}+\frac{\sqrt{3}}{2}i$
For $k=2: w_3=\sqrt[6]{1}[\cos{\frac{2\pi (2)}{6}}+ i\sin{\frac{2\pi (2)}{6}}]=1[\cos{\frac{2\pi}{3}}+i \sin{\frac{2\pi}{3}}]$ therefore $z_3=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}$ finally $z_3=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
For $k=3: w_4=\sqrt[6]{1}[\cos{\frac{2\pi (3)}{6}}+ i\sin{\frac{2\pi (3)}{6}}]=1[\cos{\pi}+i \sin{\pi}]$ therefore $z_4=\cos{\pi}+i\sin{\pi}$ finally $z_4=-1$
For $k=4: w_5=\sqrt[6]{1}[\cos{\frac{2\pi (4)}{6}}+ i\sin{\frac{2\pi (4)}{6}}]=1[\cos{\frac{4\pi}{3}}+i \sin{\frac{4\pi}{3}}]$ therefore $z_5=\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}$ finally $z_5=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$
For $k=5: w_6=\sqrt[6]{1}[\cos{\frac{2\pi (5)}{6}}+ i\sin{\frac{2\pi (5)}{6}}]=1[\cos{\frac{5\pi}{3}}+i \sin{\frac{5\pi}{3}}]$ therefore $z_6=\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}}$ finally $z_6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$
I hope this help you!!!
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving Triangles (finding missing sides/angles given 3 sides/angles) What is a general procedure for "solving" a triangle—that is, for finding the unknown side lengths and angle measures given three side lengths and/or angle measures?
| First, some notation: let $A$, $B$, and $C$ be the measures of the three angles and let $a$, $b$, and $c$ be the lengths of the sides opposite those angles, respectively. Now, let's look case-by-case at the possible sets of information we could have.
SSS
Let's start with the case where we know the three sides, $a$, $b$, and $c$. We can use the Law of Cosines, in the form where it is solved for the cosine of an angle, to find the measures of two of the angles, then use the fact that the sum of the measures of the angles is 180° to find the third:
$$A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)$$
$$B=\arccos\left(\frac{a^2+c^2-b^2}{2ac}\right)$$
$$C=180°-A-B$$
SAS
If we know two sides and the measure of the angle they include, say $a$, $b$, and $C$, we can use the Law of Cosines to find the unknown side length, then pick up with the SSS process to find the unknown angles:
$$c=\sqrt{a^2+b^2-2ab\cos C}$$
$$A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)$$
$$B=180°-A-C$$
ASA or AAS
If we know the measures of two angles, we can find the measure of the third angle using $A+B+C=180°$, so let's assume we know all three angle measures $A$, $B$, and $C$, and the side length $a$. We can use the Law of Sines to find each of the unknown side lengths:
$$b=\frac{a\sin B}{\sin A}$$
$$c=\frac{a\sin C}{\sin A}$$
SSA
If we know two side lengths and the measure of an angle that isn't included between the two known sides, say $a$, $b$, and $A$, we can start by using the Law of Sines to find $B$, but this may give two solutions:
$$\sin B=\frac{b\sin A}{a}$$
If $\sin B=\frac{b\sin A}{a}>1$, then there is no solution and the given information does not determine a triangle (it is impossible for the given information to describe a triangle).
If $\sin B=\frac{b\sin A}{a}=1$, then $B$ is a right angle and the given information determines a single triangle.
If $\sin B=\frac{b\sin A}{a}<1$, then there are two solutions for $B$:
$$B_1=\arcsin\left(\frac{b\sin A}{a}\right)\text{ or }B_2=180°-B_1$$
In each case, we can use $A+B+C=180°$ to determine $C$:
$$C_1=180°-A-B_1\text{ or }C_2=180°-A-B_2$$
At this point, $B_1$ and $C_1$ will definitely describe a triangle, but if $C_2\le0$ then $B_2$ and $C_2$ do not describe a triangle. Whether we only have the $B_1$ case or both the $B_1$ and $B_2$ cases, we can use the Law of Cosines to find the unknown side:
$$c=\sqrt{a^2+b^2-2ab\cos C}$$
AAA
Knowing three angles only determines the triangle up to similarity—that is, if we don't know at least one length, we're not going to be able to find any lengths.
Reference
Law of Cosines
Wikipedia MathWorld
$$c^2=a^2+b^2-2ab\cos C$$
$$\cos C=\frac{a^2+b^2-c^2}{2ab}$$
Law of Sines
Wikipedia MathWorld
$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$
| {
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Solving modular equations Is there a procedure to solve this or is it strictly by trial and error?
$5^x \equiv 5^y \pmod {39}$ where $y > x$.
Thanks.
| We can rewrite your problem as
$$5^{y-x}=1\bmod 39.$$
So we are trying to find the positive integers $k$ such that $5^k=1\bmod{39}$.
Since we can find the prime power factorization of $39$ easily, our problem reduces to finding the positive $n$ such that (simultaneously) $5^n=1\bmod 3$ and $5^n=1 \bmod{13}$.
If $p$ is a prime which does not divide $a$, then the smallest positive $d$ such that $a^d=1 \bmod{p}$ is a divisor of $p-1$.
Let $a=5$ and $p=3$. Then $p-1$ has only the divisors $1$ and $2$. Obviously $5^1 \neq 1\bmod{3}$. So the smallest $d$ that works for $p=3$ is $d=2$.
Let $a=5$ and $p=13$. Then $p-1=12$, so our candidates for $d$ are the divisors of $12$. It is easy to check that $1$, $2$, and $3$ don't work. But $4$ does.
Now we want the smallest $d$ that works for both $3$ and $13$. This is the least common multiple of the $d$ that works for $3$ and the $d$ that works for $13$. The least common multiple of $2$ and $4$ is $4$.
We conclude that the least positive $d$ such that $5^d=1\bmod{39}$ is $4$. By a general theorem, the $n$ such that $5^n=1\bmod{39}$ are the multiples of $4$. So we can write $n=4k$, where $n$ ranges over the positive integers.
Thus in the original problem, the pairs $(x,y)$ that satisfy your condition are all pairs $(x,x+4k)$. You say that you want $x\ge 1$, so the pairs are all pairs $(x,y)$ where $1\le x$ and $y-x$ is a positive multiple of $4$.
Remark: If instead of working with the small number $39$, we work with a huge number $m$, then the prime power factorization of $m$ can be computationally very difficult.
Even after we have factored $m$, things can be computationally very unpleasant, even in the simplest case where $m$ is prime, or a product of two primes.
There are algorithms that are much better than the crude try everything approach, but still, they are not fast enough to deal with $m$ if it is $150$ digits long.
For small $m$, like our $39$, or even modestly large $m$, say of size less than $10^6$, there are programs that give virtually instantaneous answers.
| {
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The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:
In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
coefficients of $x^3$?
| The multinomial expansion gives $\mathbf{S}= (1 - \frac{x}{2} + \frac{1}{\sqrt{x}})^8 = \sum_{k_1+k_2 +k_3=8} \binom{8}{k_1,k_2,k_3} (1)^{k_1}(\frac{-x}{2})^{k_2} (\frac{1}{\sqrt{x}})^{k_3}.$ Simplifying we get $\mathbf{S}=\sum_{k_1+k_2 +k_3=8} \binom{8}{k_1,k_2,k_3} (\frac{-1}{2})^{k_2} (x)^{k_2 -k_3/2}.$ Now identify the tuples for which $(k_1,k_2,k_3) \in \mathbb{Z}_{+}^{3}$ such that $k_1+k_2+k_3=8$ and $k_2-k_3/2=3.$ Then add the coefficients for these tuples to get your answer.
| {
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere.
Can anybody provide a hint as to how I should be looking at this problem?
| I'm carrying out a congruence procedure, so that you have different approaches.
If $p \, | \, n^3 + 1$ and $p \, | \, n^2 + 2$, then $2n \equiv 1 \pmod p$, which means
$$
-8 \equiv 8n^3 = (2n)^3 \equiv 1^3 \equiv 1 \pmod p,
$$
hence $9 \equiv 0 \pmod p$ and assuming $p$ is a prime means $p = 3$. This means the $\gcd$ is a power of $3$. Now I'm checking powers of $3$ by hand using congruences.
EDIT : As miracle's comment says, I got far too carried away by liking primes so much : a good way to say that this proof is done is that $9 \equiv 0 \mod p$ means $p \, | \, 9$, hence getting examples is enough to get our answer.
If $n \equiv 0 \pmod 3$, this $\gcd$ is clearly $1$.
If $n \equiv 1 \pmod 3$, $n^3 + 1 \equiv 1 \pmod 3$ so that the $\gcd$ is again $1$.
If $n \equiv 2 \pmod 3$, then $9 \, | \, n^3 + 1$ and $3 \, | \, n^2 + 2$. But letting $n = 3k+2$ we notice that
\begin{align*}
(3k+2)^3 + 1 & = (3k)^3 + 3 \cdot (3k)^2 \cdot 2 + 3 \cdot 3k \cdot 2^2 + 2^3 +1 \\& \equiv 9(k+1)\pmod{27}
\end{align*}
which is $0$ if and only if $k \equiv 2 \pmod 3$. Carry on! We get
$$
(3k+2)^2 + 2 \equiv (3k)^2 + 2 \cdot (3k) \cdot 2 + 2^2 + 2 = 9k^2 + 12k + 6 \equiv 0 \pmod{27}
$$
if and only if
$$
3k^2 + 4k + 2 \equiv 0 \pmod 9
$$
but reading this $\bmod 3$, we get $k \equiv 1 \pmod 3$, a contradiction.
I must say it is a little longer than the $\gcd$ proof : I expected people to put up to $\gcd$ proof, so I've shown this one instead. I like those proofs because they're mechanical ; I didn't have to think much to write it, I just chose to go this way and things always go as expected (assuming the question asks something that's true, obviously)... Now I've only proven that the $\gcd$ divides $9$ : again, to show that the $3$ possibilities actually happen, pull off examples like Andre Nicolas.
Hope that helps,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Area of a trapezoid from given the two bases and diagonals Find the area of trapezoid with bases $7$ cm and $20$ cm and diagonals $13$ cm and $5\sqrt{10} $ cm.
My approach:
Assuming that the bases of the trapezoid are the parallel sides, the solution I can think of is a bit ugly,
*
*Find the other two non-parallel sides of the trapezoid by using this formula.
*Find the height using this $$ h= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}{2(b-a)}$$
Now, we can use $\frac12 \times$ sum of the parallel sides $\times$ height.
But, this is really messy and I am not sure if this is correct or feasible without electronic aid, so I was just wondering how else we could solve this problem?
| First Solution: Let our trapezoid be $ABCD$ as in the diagram supplied by pedja. Let the diagonals meet at $O$.
Note that $\triangle OAB$ and $\triangle OCD$ are similar. Indeed we know the scaling factor. Since $AB=20$ and $CD=7$, the sides of $\triangle OCD$ are $\frac{7}{20}$ times the corresponding sides of $\triangle OAB$.
That is very useful. We have $AC=13=AO+\frac{7}{20}AO$. It follows that
$$AO=\frac{(20)(13)}{27}, \quad\text{and similarly,}\quad BO=\frac{(20)(5\sqrt{10})}{27}.$$
If we want to use the usual formula for the area of a trapezoid, all we need is the height of the trapezoid. That is $1+\frac{7}{20}$ times the height of $\triangle OAB$.
The height of $\triangle OAB$ can be found in various ways. For example, we can use the Heron Formula to find the area of $\triangle OAB$, since we know all three sides. Or else we can use trigonometry. The Cosine Law can be used to compute the cosine of $\angle OAB$. Then we can find an exact (or approximate) expression for the sine of that angle. From this we can find the height of $\triangle OAB$.
Second Solution: This is a variant of the first solution that uses somewhat more geometry. Let $\alpha$ be the area of $\triangle OAB$.
We first compute the area of $\triangle COB$. Triangles $OAB$ and $COB$ can be viewed as having bases $OA$ and $CO$ respectively, and the same height. But the ratio of $CO$ to $OA$ is $\frac{7}{20}$, so the area of $\triangle COB$ is $\frac{7}{20}\alpha$.
Since triangles $ABC$ and $ABD$ have the same area, by subtraction so do $\triangle COB$ and $\triangle DOA$. And since $\triangle OCD$ is $\triangle OAB$ scaled by the linear factor $\frac{7}{20}$, the area of $\triangle OCD$ is $\left(\frac{7}{20}\right)^2\alpha$. Putting things together, we find that the area of our trapezoid is
$$\alpha +2\frac{7}{20}\alpha +\left(\frac{7}{20}\right)^2\alpha,\quad\text{that is,}\quad \left(\frac{27}{20}\right)^2\alpha.$$
Pretty! Finally, by the similarity argument of the first solution, we know the sides of $\triangle OAB$, so we can find $\alpha$ by using Heron's Formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What does $\sin^{2k}\theta+\cos^{2k}\theta=$?
What is the sum $\sin^{2k}\theta+\cos^{2k}\theta$ equal to?
Besides Mathematical Induction,more solutions are desired.
| I do not think there is a closed form for all values of $k$, but one can play around with trigonometric identities to simplify the expression for certain values of $k$. For instance:
*
*If $k=2$, then:
$$\sin^4 x + \cos^4 x = (1-\cos^2 x)^2 + \cos^4 x\\
= 1-2\cos^2x + 2\cos^4 x \\
= 1-2\cos^2x(1-\cos^2x)\\
= 1-2\sin^2x\cos^2x\\
= 1 - \frac{\sin^2(2x)}{2}.$$
*
*If $k=3$, then:
$$\sin^6 x + \cos^6 x = (1-\cos^2 x)^3 + \cos^6 x\\
= 1-3\cos^2x + 3\cos^4 x - \cos^6 x + \cos^6 x \\
= 1-3\cos^2x + 3\cos^4x\\
= 1-3\cos^2x(1-\cos^2x)\\
= 1-3\sin^2x\cos^2x\\
= 1 - \frac{3\sin^2(2x)}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/111265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing $ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$ I would like to prove the existence and the exact value of the following series:
$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$
| If $ n+1 $ is a square, $\lfloor \sqrt{n+1} \rfloor=\sqrt{n+1} $
$$ 0<\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<1$$
So: $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$
So: $$ \lfloor \sqrt{n} \rfloor= \sqrt{n+1}-1$$
$$ \lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor=1 $$
I have proved:
$n+1$ is a square $\Longrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$
Now I must show that
$ \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor \Longrightarrow $ $n+1$ is a square
If $ n+1$ is not a square:
If $ n $ is a square, $\lfloor \sqrt{n} \rfloor=\sqrt{n}$.
As
$$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$
$ \lfloor \sqrt{n+1} \rfloor= \sqrt{n} $
So: $ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $
If $ n $ is not a square, there exists $a\in \mathbb{N} $ such that
$$ a^2<n<n+1<(a+1)^2$$
So:
$$ a<\sqrt{n}<\sqrt{n+1}<a+1$$
So :
$$ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $$
Finally:
$n+1$ is a square $\Longleftrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$
So: $$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}=\sum_{n=2}^{\infty} \frac{1}{n^2-1}=\cdots=\frac{3}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding a simple expression for this series expansion without a piecewise definition I am doing some practice Calculus questions and I ran into the following problem which ended up having a reduction formula with a neat expansion that I was wondering how to express in terms of a series. Here it is: consider
$$
I_{n} = \int_{0}^{\pi /2} x^n \sin(x) dx
$$
I obtained the reduction formula
$$
I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - n I_{n-1}.
$$
I started incorrectly computing up to $I_{6}$ with the reduction formula
$$
I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - I_{n-1}
$$
by accident which ended up having a way more interesting pattern than the correct reduction formula. So, after computing $I_{0} = 1$, the incorrect reduction expansion was,
$$
I_{1} = 0 \\
I_{2} = \pi \\
I_{3} = \frac{3\pi^2}{2^2} - \pi \\
I_{4} = \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\
I_{5} = \frac{5\pi^4}{2^4} - \frac{4\pi^3}{2^3} + \frac{3\pi^2}{2^2} - \pi \\
I_{6} = \frac{6\pi^5}{2^5} - \frac{5\pi^4}{2^4} + \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\
$$
Note that $\pi = \frac{2\pi}{2^1}$, of course, which stays in the spirit of the pattern. How could I give a general expression for this series without defining a piecewise function for the odd and even cases? I was thinking of having a term in the summand with $(-1)^{2i+1}$ or $(-1)^{2i}$ depending on it was a term with an even or odd power for $n$, but that led to a piecewise defined function. I think that it will look something like the following, where $f(x)$ is some function that handles which term gets a negative or positive sign depending on whether $n$ is an even or odd power in that term: $$\sum\limits_{i=1}^{n} n \left(\frac{\pi}{2} \right)^{n-1} f(x)$$
Any ideas on how to come up with a general expression for this series?
| How about
$$\sum_{k=1}^{n} k \left(\frac{\pi}{2}\right)^{k-1} \cdot (-1)^{n+k+1}$$
| {
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Divide inside a Radical It has been so long since I have done division inside of radicals that I totally forget the "special rule" for for doing it. -_-
For example, say I wanted to divide the 4 out of this expression:
$\sqrt{1 - 4x^2}$
Is this the right way to go about it?
$\frac{16}{16} \cdot \sqrt{1 - 4x^2}$
$16 \cdot \frac{\sqrt{1 - 4x^2}}{16}$
$16 \cdot \sqrt{\frac{1 - 4x^2}{4}} \Longleftarrow \text{Took the square root of 16 to get it in the radicand as the divisor}$
I know that this really a simple, question. Can't believe that I forgot how to do it. :(
| Square roots obey the rule $\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$ . You wanted to take the $4$ out of $\sqrt{1-4x^2}$. $1-4x^2=4\cdot\frac{1-4x^2}{4}$. So $\sqrt{1-4x^2}=\sqrt{4\cdot\frac{1-4x^2}{4}}=\sqrt{4}\cdot\sqrt{\frac{1-4x^2}{4}}=2\cdot\sqrt{\frac{1-4x^2}{4}}$. I think that's the best explanation I can give.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $ x^2+4=y^d$ in integers with $d\ge 3$
Find all triples of integers $(x,y,d)$ with $d\ge 3$ such that $x^2+4=y^d$.
I did some advance in the problem with Gaussian integers but still can't finish it. The problem is similar to Catalan's conjecture.
NOTE: You can suppose that $d$ is a prime.
Source: My head
| Here's a proof sketch. It's missing two major steps Propositions 1 and 2, but I'll try to fill them in and come back with an update.
First consider $x$ and $y$ even. Then we can only have $x\equiv 4\pmod{16}$, $y\equiv 8\pmod{16}$ and hence that $d=3$, for which @André Nicolas showed that $x=y=2$ is the only solution.
Now consider $y$ odd.
$d$ must be odd, so we can rewrite the equation as
$$
yz^2-x^2=4
$$
When $y=a^2+4$ then this has solutions
$$
\begin{array}{l}
z_0=z_1=1,z_{k+2}=(y-2)z_{k+1}-z_k \\
x_0=-a,x_1=a,x_{k+2}=(y-2)x_{k+1}-x_k
\end{array}
$$
Proposition 1: These enumerate all solutions, and there are no solutions in integers when $\sqrt{y-4}$ is not an integer.
Now to find a solution to the original problem we would need $z_i$ to be a power of $y$ for some $i$.
We can look at $z_k$ modulo $y$ and modulo $y^2$:
$$
\begin{align}
z_k & \equiv (-1)^{k+1}(2k-1) \pmod{y}\\
z_k & \equiv (-1)^{k}(2k-1)\left(\frac{k(k-1)}{6}y-1\right) \pmod{y^2}\\
\end{align}
$$
(These can be shown by induction.)
Thus $y\mid z_k$ iff $y\mid 2k-1$ and the first $z_k$ divisible by $y$ is when $k=(y+1)/2$.
Proposition 2: If $y\mid z_k$ then $z_{(y+1)/2} \mid z_k$.
Taking $z_{(y+1)/2}$ mod $y^2$:
$z_{(y+1)/2} \equiv -y\left(\frac{y^2-1}{24}y-1\right) \equiv -yw \pmod{y^2} $
where $\gcd(y,w)=1$. Hence $z_{(y+1)/2}$ is never divisible by $y^2$ and is either equal to $y$ or has a factor greater than 1 that does not divide $y$.
Since $z_k$ is increasing in $k$ and $z_3 = y^2-5y+5$ we find that $z_3=5$ when $y=5$ and $y<z_3<z_{(y+1)/2}$ when $y>5$. Thus $z_{(y+1)/2}$ is not a power of $y$ for $y>5$, and assuming Prop 2 nor can any other $z_k$ be, so a solution can only exist if $y=5$.
If $y=5$, $z_3=5,d=3,x=11$ is one solution.
The case $y=5$ is special in that $z_k=F_{2k-1}$ where $F_n$ is the $n$th Fibonacci number. We can use the divisibility properties of $F_n$ to show that if $25\mid z_k$ then $3001 \mid F_{25} \mid z_k$ so $z_k$ is not a power of 5. (This also shows Prop 2 is true when $y=5$ and hints how it might be proved in general.)
Thus the only solutions are $(x,y,d)=(\pm 2,2,3)$ or $(x,y,d)=(\pm 11,5,3)$.
Edit: Proposition 1 is incorrect, there may be solutions when $\sqrt{y-4}$ is not an integer, e.g. $39^2 = 61\cdot5^2-4$. This looks to be fatal to the argument.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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finding variance from a piecewise function How do you calculate the variance of a piecewise function? For example, what would be the variance of the probability density function $f_x(x)= \frac{3}{4}, 0\leq x\leq 1; \frac{1}{4}, 2\leq x \leq 3; 0$ otherwise?
| Let $U$ be a continuous uniform random variable on a unit interval. It's mean and variance are
$$\mathbb{E}(U) = \frac{1}{2} \qquad \qquad
\mathbb{Var}(U) = \mathbb{E}(U^2) - \left(\mathbb{E}(U)\right)^2 = \frac{1}{12}
$$
Now, your continuous random variable $X$ is a component mixture of a uniform $U$ and shifted uniform $2+U$ with weights $w_1 = \frac{3}{4}$ and $w_2 = \frac{1}{4}$.
Then
$$
\mathbb{Var}(X) = \mathbb{E}(X^2) - \left(\mathbb{E}(X)\right)^2 =
\left( w_1 \mathbb{E}(U^2) + w_2 \mathbb{E}((2+U)^2) \right) - \left( w_1 \mathbb{E}(U) + w_2 \mathbb{E}(2+U) \right)^2
$$
Since $\mathbb{E}(U^2) = \mathbb{Var}(U) + \left(\mathbb{E}(U)\right)^2 = \frac{1}{3}$, $\mathbb{E}((2+U)^2) = 4 + 4 \mathbb{E}(U) + \mathbb{E}(U^2) = \frac{19}{3}$. Therefore
$$
\mathbb{Var}(X) = \left( \frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{19}{3} \right) - \left( \frac{3}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{5}{2} \right)^2 = \frac{11}{6} - 1^2 = \frac{5}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute $\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$ Compute
$$\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$$
I did
$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x = \lim_{x\to\infty} \exp(x\cdot \ln(\frac{x-2}{x+2})) = \exp( \lim_{x\to\infty} x\cdot \ln(\frac{x-2}{x+2}))$$
But how do I continue? The hint is to use L Hopital's Rule. I tried changing to
$$\exp(\lim_{x\to\infty} \frac{\ln(x-2)-\ln(x+2)}{1/x})$$
This is
$$(\infty - \infty )/0 = 0/0$$
But I find that I can keep differentiating?
| $$\lim_{x\to\infty} (\frac{x-2}{x+2})^x$$
$$\lim_{x\to\infty} (1-\frac{4}{x+2})^x = y$$
taking log on both sides we get
$$ln(y) = x ln (1- \frac{4}{x+2})$$
the expansion for $ln (1+r) $ is $ r- \frac{r^2}{2} +\frac{r^3}{3}$ ....
where r tends to zero
$$ln(y) = x ( \frac{-4}{x+2} - \frac{\frac{-4}{x+2}^2}{2} +\frac{\frac{-4}{x+2}^3}{3} ....)$$
$ln (y) = \frac{-4x}{x+2}$ {rest all terms will terminate to zero}
$$ln (y) =\lim_{x\to\infty} \frac{-4x}{x+2} = -4$$
$$y = \frac{1}{e^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the following integral is divergent $$\int_0^\infty \frac{7x^7}{1+x^7}$$
Im really not sure how to even start this. Does anyone care to explain how this can be done?
| The answer by Davide Giraudo has all the right elements, but I think the OP may appreciate to see all of the details spelled out. First of all, an improper integral is defined as a limit, as follows:
$$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx.$$
Next, we use the fact that $1+x^7\leq 2x^7$ when $x\geq 1$, and the properties of the definite integral to bound the integral inside the limit:
$$\begin{align*} \int_0^N \frac{7x^7}{1+x^7} dx &=\int_0^1 \frac{7x^7}{1+x^7} dx + \int_1^N \frac{7x^7}{1+x^7} dx \\
&\geq \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7x^7}{2x^7} dx \\
&= \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7}{2} dx \\
& = \int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2}.\end{align*}$$
Hence:
$$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx\geq \lim_{N\to \infty} \left(\int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2} \right)= \infty.$$
Therefore, the improper integral diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
| Any proof will need to make use of induction at some point.
*
*Note that $10^n \equiv 1 \bmod 9$ and $4^n \equiv (1,4,7) \bmod 9$. (You need induction to prove previous statements.) Hence, $3 \times 4^n \equiv 3 \bmod 9$. Assimilating these together, we get that $$\left(10^n + 3 \times 4^n + 5 \right) \equiv (1 + 3 + 5) \bmod 9 = 0 \bmod 9.$$ Hence, $\displaystyle \frac{10^n + 3 \times 4^n + 5}{9}$ is an integer for all $n \in \mathbb{N}$.
*For a proof where induction is easily aparent, note that $$10^{n+1} + 3 \times 4^{n+1} + 5 = 10 \times \left( 10^{n} + 3 \times 4^{n} + 5 \right) - 9 \times \left( 4^n + 5 \right).$$
Hence,
$$\frac{10^{n+1} + 3 \times 4^{n+1} + 5}{9} = 10 \times \left( \frac{10^{n} + 3 \times 4^{n} + 5}{9} \right) - \left( 4^n + 5 \right).$$ Note that by induction hypothesis $\displaystyle \frac{10^{n} + 3 \times 4^{n} + 5}{9}$ is an integer and hence, $\displaystyle \frac{10^{n+1} + 3 \times 4^{n+1} + 5}{9}$ is also an integer since $4^n + 5$ is an integer for all $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the integral $S_k = (-1)^k \int_0^1 (\log(\sin \pi x))^k dx$ My nephew asked me this, so I suggested him to sign up here. But anyways this question I was
trying to solve myself. I got part of the solution. Let me know the rest.
$(1)$
Solve the integral defined as
$\displaystyle{S_k = (-1)^k \int_0^1 (\log(\sin \pi x))^k dx}$ and show
$(2)$
$\displaystyle{S_k = \frac{(-1)^k}{\sqrt\pi 2^k} \frac{d^k}{d\alpha^k} \frac{\Gamma(\alpha+\frac{1}{2})}{\Gamma(\alpha+1)}}$
with $\alpha=0$.
$(3)$
Show that
$\displaystyle{S_4 = \frac{19 \pi^4}{240}+\frac{1}{2} \pi^2 \log^2 2 + \log^4 2 + 6 \log 2 \, \zeta(3)}$
$(4)$
Show the following:
$\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$
For $(3)$, if I substitute $y=\pi x$, I can transform this to a well known log-sine function as described
here at Wolfram, and so showing
$\displaystyle{S_4 = \frac{19 \pi^4}{240}+\frac{1}{2} \pi^2 \log^2 2 + \log^4 2 + 6 \log 2 \, \zeta(3)}$
is not hard.
I would like suggestions to read or partial solutions (not complete solutions).
(NOTE: I also noted that someone have asked a similar question for $k=2$ here)
| As for the the problem (2) we have
$$
\begin{align}
S_k&=(-1)^k \int\limits_0^1\log^k(\sin\pi x)dx\\
&=\frac{(-1)^k}{\pi} \int\limits_0^\pi\log^k(\sin y)dy\\
&=\frac{(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy+
\frac{(-1)^k}{\pi} \int\limits_{\pi/2}^\pi\log^k(\sin y)dy\\
&=\frac{(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy+
\frac{(-1)^k}{\pi} \int\limits_{0}^{\pi/2}\log^k(\sin y)dy\\
&=\frac{2(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy\\
&=\frac{2(-1)^k}{\pi} \int\limits_0^{1}\frac{\log^k(t)}{\sqrt{1-t^2}}dt
\end{align}
$$
Now following Sasha's idea consider integral
$$
I(\alpha)=
\int\limits_0^1\frac{t^{2\alpha}}{\sqrt{1-t^2}}dt=
\frac{1}{2}B\left(\frac{1}{2},\frac{2\alpha+1}{2}\right)=
\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac{2\alpha+1}{2}\right)}{\Gamma\left(\frac{2\alpha+2}{2}\right)}
$$
Note that
$$
I^{(k)}(\alpha)=\int\limits_0^1\frac{t^{2\alpha}2^k\log^k(t)}{\sqrt{1-t^2}}dt
$$
so
$$
S_k=
\frac{2(-1)^k}{\pi} \int\limits_0^{1}\frac{\log^k(t)}{\sqrt{1-t^2}}dt=
\frac{2(-1)^k}{\pi} \frac{1}{2^k}I^{(k)}(0)=
\frac{(-1)^k}{2^k\sqrt{\pi}}\frac{d^k}{d\alpha^k}\left(\frac{\Gamma\left(\alpha+\frac{1}{2}\right)}{\Gamma\left(\alpha+1\right)}\right)_{\alpha=0}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Closed form for coefficient of $x^N$ of $\frac{1}{(1-x^2)(1-x^3)(1-x^5)}$ I'm trying to get a closed form for the generating function which counts the number of non-negative integer solutions to
$$2e_1+3e_2+5e_3 = N$$
I note that we have something like:
$$\frac{1}{(1-x^2)(1-x^3)(1-x^5)}$$
but can't get something to translate nicely. I've thought of using $e^{2x} e^{3x} e^{5x}$, but is it valid?
| According to Maple:
convert(1/(1-x^2)/(1-x^3)/(1-x^5),parfrac,x);
$$-1/30\, \left( x-1 \right) ^{-3}+1/8\, \left( x+1 \right) ^{-1}-{
\frac {77}{360}}\, \left( x-1 \right) ^{-1}+1/5\,{\frac {2+x+{x}^{2}+{
x}^{3}}{{x}^{4}+{x}^{3}+{x}^{2}+x+1}}+{\frac {7}{60}}\, \left( x-1
\right) ^{-2}+1/9\,{\frac {-x+1}{{x}^{2}+x+1}}$$
map(convert, %, FPS, x);
$$\sum _{k=0}^{\infty } \left( \sum _{{\it \alpha}={\it RootOf} \left(
{{\it \_Z}}^{4}+{{\it \_Z}}^{3}+{{\it \_Z}}^{2}+{\it \_Z}+1 \right) }-
1/25\,{\frac {{x}^{k} \left( -2\,{\it \alpha}+1+{{\it \alpha}}^{2}
\right) }{{{\it \alpha}}^{k+1}}} \right) +\sum _{k=0}^{\infty }
\left( -{\frac {1}{54}}\, \left( -\sqrt {3}+3\,i \right) \sqrt {3}
\left( -1/2-1/2\,i\sqrt {3} \right) ^{k}-{\frac {1}{54}}\,i\sqrt {3}
\left( -3+i\sqrt {3} \right) \left( -1/2+1/2\,i\sqrt {3} \right) ^{k
} \right) {x}^{k}+\sum _{k=0}^{\infty }{\frac {77}{360}}\,{x}^{k}+
\sum _{k=0}^{\infty } \left( {\frac {7}{60}}+{\frac {7}{60}}\,k
\right) {x}^{k}+\sum _{k=0}^{\infty }1/8\, \left( -1 \right) ^{k}{x}^
{k}+\sum _{k=0}^{\infty } \left( 1/30+1/20\,k+{\frac {1}{60}}\,{k}^{2}
\right) {x}^{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Why is the differentiability of a piecewise defined function studied using the definition of derivative? Why is the differentiability of a piecewise defined function often studied using the definition of derivative?
For example, let:
$$\begin{align*}
f(x)
&= x\cdot |x-1|\\
&= \left\{ \begin{array}{lcl}x (-x+1)& \text{if} & x <
1 \\ x (x-1) & \text{if} & x \geq 1 \end{array} \right.\\
&= \left\{ \begin{array}{lcl}-x^2+x& \text{if} & x <
1 \\ x^2-x & \text{if} & x \geq 1 \end{array} \right.
\end{align*}$$
$f$ is continuous in $\mathbb{R} $ except, maybe, when $x=1$:
$$\begin{align*}
\lim_{x \to 1^{-}} f(x) &= \lim_{x \to 1} (-x^2+x) = 0\\
\lim_{x \to
1^{+}} f(x)&= \lim_{x \to 1} (x^2 -x) = 0 \end{align*}$$
$f$ is continuous when $x=1$ because:
$$ \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1) $$
As to the differentiability at that point (by derivative's definition):
$$\begin{align*}
\lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0}
\frac{\left[-(1+h)^2 + (1+h)\right] - \left[-1^2 + 1\right]}{h}\\
&= \lim_{h\to 0}
\frac{- h(h+1)}{h} = -1\\
\strut\\
\lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0} \frac{\left[(1+h)^2
- (1+h)\right] - \left[1^2 - 1\right]}{h}\\
&= \lim_{h\to 0} \frac{h(h+1)}{h} = 1
\end{align*}$$
$$ \lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} \neq \lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} $$
so $f$ is not differentiable when $x=0$.
The "second" method is:
$$ f'(x) = \left\{ \begin{array}{lcl} -2x + 1& \text{if} & x <
1 \\ 2x-1 & \text{if} & x > 1 \end{array} \right. $$
$$ f'(1^{-}) = -1 \neq f'(1^{+}) = 1 $$
Is there any case in which the differentiability study by using the derivative (the second method) would be wrong?
| Consider the function
$$f(x) = \left\{\begin{array}{ll}
x^2\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\
0 &\text{if }x=0.
\end{array}\right.$$
The function is differentiable at $0$, with derivative equal to $0$:
$$\begin{align*}
\lim_{h\to 0}\frac{f(h)-f(0)}{h} &= \lim_{h\to 0}\frac{h^2\sin(\frac{1}{h})}{h} \\
&=\lim_{h\to 0}h\sin(\frac{1}{h}) = 0.
\end{align*}$$
However, using the "second method", we look at the derivative of $x^2\sin(\frac{1}{x})$:
$$\left(x^2\sin\left(\frac{1}{x}\right)\right)' = 2x\sin\frac{1}{x} - \cos\frac{1}{x}.$$
And
$$\lim_{x\to 0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)$$
does not exist.
Using the method, we would incorrectly conclude that the derivative does not exist at $0$.
In general, a derivative does not have to be continuous; so the second method may yield "false negatives", functions that are differentiable but which the technique suggests are not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to expand $\cos nx$ with $\cos x$? Multiple Angle Identities:
How to expand $\cos nx$ with $\cos x$, such as
$$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$
See a list of trigonometric identities in english/ chinese
| You can always repeatedly use
$$\begin{align*}
\cos(a\pm b) &= \cos a\cos b \mp \sin a\sin b\\
\sin(a\pm b) &= \sin a\cos b \pm \cos a\sin b\\
\sin^2(r) &= 1-\cos^2(r).
\end{align*}$$
For example,
$$\begin{align*}
\cos(4x) &= \cos(2x+2x)\\
&= \cos(2x)^2 - \sin^2(2x)\\
&= \cos(2x)^2 - (1-\cos^2(2x))\\
&= 2\cos(2x)^2 - 1\\
&= 2(\cos(x+x))^2 - 1\\
&= 2(\cos x\cos x - \sin x\sin x)^2 - 1\\
&= 2(\cos^2 x - \sin^2 x)^2 - 1\\
&= 2(\cos^2x - (1-\cos^2 x))^2 - 1\\
&= 2(2\cos^2x - 1)^2 - 1\\
&= 2(4\cos^4 x - 4\cos^2 x + 1) - 1\\
&= 8\cos^4 x - 8\cos^2 x + 1.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
The $n^{th}$ root of the geometric mean of binomial coefficients. $\{{C_k^n}\}_{k=0}^n$ are binomial coefficients. $G_n$ is their geometrical mean.
Prove
$$\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}$$
| $G_n$ is the geometric mean of $n+1$ numbers:
$$
G_n=\left[\prod_{k=0}^n{n\choose k}\right]^{\frac1{n+1}}
$$
or with $\log$ representing the natural logarithm (to the base $e$),
$$
\log G_n
= \frac1{n+1} \sum_{k=0}^n \log {n\choose k}
= \log n! - \frac2{n+1} \sum_{k=0}^n \log k!
\,.
$$
Stirling's approximation is
$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$
or
$$
\log n!
\approx
\frac12\log{(2\pi n)}+n\log\left(\frac{n}{e}\right)
= \left(n+\frac12\right)\log n+\frac12\log 2\pi-n
$$
so
$$
\eqalign{
\log \left(G_n\right)^\frac1n
&
= \frac1n \log G_n
= \frac1n \log n! - \frac2{n(n+1)} \log \prod_{k=0}^n k!
\\
&
= \frac1n \log n! - \frac2{n(n+1)} \sum_{k=0}^n \log k!
\\
&
\approx \left(1+\frac1{2n}\right) \log n
- \frac2{n(n+1)} \sum_{k=1}^n \left(k+\frac12\right)\log k
- \frac1{2n}\log 2\pi
\\
&
\approx \left(1+\frac1{2n}\right) \log n
- \frac2{n(n+1)}
\left[
\frac{n(n+1)}{2}\log n -
\frac{n(n+2)}{4}
\right]
- \frac1{2n}\log 2\pi
\\
&
= \frac{\log n-\log 2\pi}{2n}
+ \frac{n+2}{2(n+1)}
\\
&
\rightarrow \frac12
\,,
}
$$
where the sum of logarithms was approximated
using the definite integrals
$$
\sum_{k=1}^n \log k \approx
\int_1^n \log x\,dx =
\Big[x\log x-x\Big]_1^n \approx
\Big[x\log x-x\Big]_0^n
$$
and
$$
\sum_{k=1}^n k \log k \approx
\int_0^n x\log x\,dx=\left[\frac{x^2}{2}\log x - \frac{x^2}{4}\right]_0^n
$$
(using integration by parts as shown in a comment), so that
$$
\eqalign{
\sum_{k=1}^n \left(k+\frac12\right)\log k
&=
\sum_{k=1}^n k \log k + \frac12
\sum_{k=1}^n \log k
\\
&\approx
\left( \frac{n^2}{2}\log n - \frac{n^2}{4} \right) + \frac12
\Big( n \log n - n \Big)
\\
&=
\frac{n^2+n}{2}\log n - \frac{n^2+2n}{4}
\,.
}
$$
Thus
$$
G_n=e^{\log G_n}\rightarrow e^{\frac12}=\sqrt{e}
\,.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Evaluation of $ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$ I am trying to calculate $$ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$$
Whenever I try it all I can seem to reach is the conclusion that it goes to infinity, however when I try it in wolframalpha it gives me the answer -1/4! How is that value reached?! I can't see any way of cancelling x's or using binomial expansion or anything!
| $$\displaystyle \lim_{x \to \infty}x^{3/2}(\sqrt {x+2}+\sqrt x-2\sqrt{x+1})=\displaystyle \lim_{x \to \infty} \frac{x^2(-2x-2+2\sqrt{x^2+2x})}{\sqrt x(\sqrt {x+2}+\sqrt x+2\sqrt{x+1})}=$$
$$=\displaystyle \lim_{x \to \infty}\frac{2x^2(\sqrt{x^2+2x}-(x+1))(\sqrt{x^2+2x}+(x+1))}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$$
$$=\displaystyle \lim_{x \to \infty} \frac{-2x^2}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$$
$$=\displaystyle \lim_{x \to \infty} \frac{-2}{\left(\sqrt{1+\frac{2}{x}}+\sqrt 1+2\sqrt{1+\frac{1}{x}}\right)\left(\sqrt{1+\frac{2}{x}}+1+\frac{1}{x}\right)}=$$
$$=\frac{-2}{ 4 \cdot 2} =\frac{-1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate the integral: $\int_{0}^{1}\frac{dx}{1+x^{3}}$ I am a bit lost on how to evaluate the integral: $$\int_{0}^{1}\frac{dx}{1+x^{3}}$$
I tried the substitution: $y=x^{3}$, but I got a more complicated integrand. Any ideas?
| There's a factorization of a sum of two cubes:
$$
1+x^3 = (1+x)(1-x+x^2)
$$
This quadratic polynomial is irreducible unless you allow imaginary numbers since if you write it as $ax^2+bx+c$, it turns out that $b^2-4ac<0$.
So
$$
\frac{1}{1+x^3} = \frac{1}{(1+x)(1-x+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1-x+x^2}
$$
As usual with partial fractions, you then find $A$, $B$, and $C$. The integral of $A/(1+x)$ is easy. The other one you can start with a substitution: $u=x^2-x+1$, $du = (2x-1)\,dx$. So
$$
(Bx+C)\,dx = \frac B 2 \left(2x + \frac{-2}{B} \right)\,dx + (\text{some constant (find it!)})\,dx
$$
Then you need
$$
\frac B 2 \int \frac{2x-1}{x^2-x+1}\,dx
$$
and you can do that by using the substitution.
Finally, you need
$$
\int \frac{\text{constant}}{x^2-x+1}\,dx.
$$
Complete the square:
$$
x^2 - x + 1 = \left(x^2 - x + \frac 1 4\right) + \frac 3 4 = \left(x - \frac 1 2\right)^2 + \frac 3 4
$$
We'd like $1$ where we see $3/4$, so that it will look like the derivative of the arctangent. So
$$
\left(x - \frac 1 2\right)^2 + \frac 3 4 = \frac 3 4 \left( \frac 4 3\left(x - \frac 1 2\right)^2 + 1 \right) = \frac 4 3 \left(\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1\right) = \frac 4 3 (w^2 + 1)
$$
and then $dw = \dfrac{2}{\sqrt{3}}\,dx$
Finally you have
$$
(\text{constant})\cdot \int \frac{dw}{w^2 + 1} = \text{constant}\cdot\arctan(w) + c
$$
and then you convert it back to a function of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is this proof that if $a_{n+1} = \sqrt{2 + \sqrt{a_n}}$ and $a_1 = \sqrt{2}$, then $\sqrt{2} \leq a_n \leq 2$ correct? Let $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + \sqrt{a_n}}$. Now I want to show by induction that $\sqrt{2} \leq a_n \leq 2$ for all $n$.
The base case is $n=1$ and it is clear $\sqrt{2} \leq a_1 \leq 2$. Then I assume that $\sqrt{2} \leq a_n \leq 2$ holds and I want to show $\sqrt{2} \leq a_{n+1} \leq 2$.
Then I note that $\sqrt{2} \leq a_{n+1} \leq 2 \implies \sqrt{2} \leq \sqrt{2 + \sqrt{a_n}} \leq 2$. By squaring both sides I get $2 \leq 2 + \sqrt{a_n} \leq 4$. Then by subtracting $2$, I get $0 \leq \sqrt{a_n} \leq 2$. This means that $0 \leq a_n \leq 4$. This is ok because I assumed $0 \leq \sqrt{2} \leq a_n \leq 2 \leq 4$.
Edit
How about this: Since I know $\sqrt{2} \leq a_n \leq 2$. Then it is clear that $0 \leq a_n \leq 4$. By taking square roots I get $ 0 \leq \sqrt{a_n} \leq 2$. Now if I add 2, $2 \leq \sqrt{a_n} + 2 \leq 4$. Taking another square root I get $\sqrt{2} \leq \sqrt{\sqrt{a_n} + 2} \leq 2$. So $\sqrt{2} \leq a_{n+1} \leq 2$.
| Note that you want to show the case $n$ implies $n+1$, but you're going the other way.
The good thing about this sequence is the inductive step is indeed easy.
The base case is $\sqrt 2 \leq a_1 \leq 2$
We now assume $\sqrt 2 \leq a_n \leq 2$
Take a square root, then add two, then take a square root.
$$\sqrt{\root 4 \of 2+2} \leq \sqrt{2+\sqrt a_n}\leq \sqrt{\sqrt2+2}$$
$$\sqrt{\root 4 \of 2+2} \leq a_{n+1}\leq \sqrt{\sqrt2+2}$$
Note that since $\sqrt 2 + 2 < 4 \Rightarrow \sqrt {\sqrt 2 + 2} < 2$ and similarily $2 < \root 4 \of 2 + 2 \Rightarrow \sqrt 2 < \sqrt {\root 4 \of 2 + 2} $
Add: Your edit makes perfect sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Showing a series is a solution to a differential equation I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$
So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$
$$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$
then substituting these into the differential equation I get:
$$(1-2x+x^2)\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
$$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
relabeling the indexes:
$$\sum_{n=-2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=-1}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
and then cancelling the $n=-2$ and $n=-1$ terms:
$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need
$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$
I'm not sure where I have gone wrong?
Thanks very much for any help
| Everything seems correct. Before you expand out powers of $n$, notice that equating like powers of $x^n$ in your last sum gives:
$(n+2)(n+1)a_{n+2}-2n(n+1)a_{n+1}+[n(n-1)-2]a_n=0$
Notice that $[n(n-1)-2]=n^2-n-2=(n+1)(n-2)$, so since $n\geq 0$, you can divide the recurrence equation by $(n+1)$ to get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determinant of a 3x3 matrix with 6 unknowns given the determinants of two 3x3 matrices with same unknowns? Given:
$$
det(A) = 3 \\ det(B) = -4
$$
$$
A =
\begin{pmatrix}
a & b & c \\
1 & 1 & 1\\
d & e & f
\end{pmatrix} \\
B =
\begin{pmatrix}
a & b & c \\
1 & 2 & 3 \\
d & e & f
\end{pmatrix} \\
C =
\begin{pmatrix}
a & b & c \\
4 & 6 & 8 \\
d & e & f
\end{pmatrix}
$$
Find $det(C)$.
$$
det(A) = (af-cd)+(bd-ae)+(ce-bf) = 3 \\
det(B) = 2(af-cd)+3(bd-ae)+(ce-bf) = -4 \\
det(C) = 6(af-cd)+8(bd-ae)+4(ce-bf) = x
$$
I've written this as an augmented matrix with $(af-cd), (bd-ae), (ce-bf)$ as the unknowns and found the reduced row echelon form to be:
$$
\begin{pmatrix}
1 & 0 & 2 & 3 \\
0 & 1 & -1 & -10 \\
0 & 0 & 0 & x+2
\end{pmatrix}
$$
Can I then conclude that $det(C) = -2$?
| The determinant is a multilinear function of the rows (or columns). Since $(4,6,8) = 2(1,1,1)+2(1,2,3)$, we have $\det C = 2 \det A + 2 \det B$. Hence the answer is $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/140144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?
| For what is worth:
$$S_n = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{{{2^n}}}$$
Then
$$\frac{{{S_n}}}{2} = {S_n} + \frac{1}{{{2^{n + 1}}}} - \frac{1}{2} \Rightarrow {S_n} = 2{S_n} + \frac{1}{{{2^n}}} - 1 \Rightarrow 1 - \frac{1}{{{2^n}}} = {S_n}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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laplace form and matrix exponential Given the matrix $(I-A)^{-1}$ and $B$, can we compute $e^{A+B}$, where $e^X$ is defined to be $\sum_{i=0}^{\infty} \frac{X^i}{i!}$.
(Note that $A$ and $B$ do not commute, and hence $e^A \cdot e^B \neq e^{A+B}$).
Now I've observed that Laplace transformation might be a useful tool. I've obtained that
$$\mathcal{L}[e^{tA+B}](s) ={(sI-A)}^{-1}e^{B}.$$
So is the above (inverse) laplace transformation really useful to compute $e^{A+B}$ from $(I-A)^{-1}$ and $B$? How can I get the resultant $e^{A+B}$ from the Laplace transformation?
Hope anyone who is familiar with linear algebra and Laplace transformation could give me a hand. Thanks!
| $e^{A+B}$ is not uniquely determined by $e^A$ and $e^B$.
First take $A = \left[ \begin{array}{cc} 0 & -\pi \\\ \pi & 0 \end{array} \right]$ and $B = \left[ \begin{array}{cc} \pi & 0 \\\ 0 & -\pi \end{array} \right]$. Then $A + B$ squares to zero, so we have
$$e^A = \left[ \begin{array}{cc} \cos \pi & - \sin \pi \\\ \sin \pi & \cos \pi \end{array} \right] = \left[ \begin{array}{cc} -1 & 0 \\\ 0 & -1 \end{array} \right], e^B = \left[ \begin{array}{cc} e^{\pi} & 0 \\\ 0 & e^{-\pi} \end{array} \right], e^{A+B} = \left[ \begin{array}{cc} 1 + \pi & -\pi \\\ \pi & 1 - \pi \end{array} \right].$$
Now replace $A$ with $\left[ \begin{array}{cc} 0 & - 3\pi \\\ 3\pi & 0 \end{array} \right]$. Then $e^A$ is the same, but now
$$A + B = \left[ \begin{array}{cc} \pi & - 3\pi \\\ 3 \pi & - \pi \end{array} \right]$$
has eigenvalues $\pm \pi i \sqrt{7}$, so the eigenvalues of $e^{A+B}$ are different from what they were before.
| {
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"source": "stackexchange",
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What is the $x(t)$ function of $\dot{v} = a v² + bv + c$ to obtain $x(t)$ How to solve $$\frac{dv}{dt} = av^2 + bv + c$$ to obtain $x(t)$, where $a$, $b$ and $c$ are constants, $v$ is velocity, $t$ is time and $x$ is position. Boundaries for the first integral are $v_0$, $v_t$ and $0$, $t$ and boundaries for the second integral are $0$, $x_{max}$ and $0$, $t$.
| As suggested by Arturo:
$$ \begin{eqnarray}
\int_{0}^{t} \frac{\dot{v}(t)}{a v(t)^2+ b v(t) + c} \mathrm{d} t &=& \int_0^t \mathrm{d} t \\
\int_{v_0}^{v_t} \frac{\mathrm{d} u}{a u^2 + b u + c} &=& t \\
\int_{v_0}^{v_t} \frac{4 a }{\left(2 a u + b\right)^2 + 4 a c - b^2} \mathrm{d} u &=& t \\
\left.\frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a u + b}{\sqrt{4 c a - b^2}}\right)\right|_{v_0}^{v_t} &=& t \\
\frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_t + b}{\sqrt{4 c a - b^2}}\right) &=& \frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + t \\
\frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \tan\left( \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + \frac{\sqrt{4 c a - b^2}}{2} t \right) \\
\frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \frac{ \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} + \tan\left(\frac{\sqrt{4 c a - b^2}}{2} t\right)}{1 - \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} \tan\left(\frac{\sqrt{4 c a - b^2}}{2} t\right)}
\end{eqnarray}
$$
Now one solves for $v_t$, and integrates over time:
$$
x(t) = x(0) + \int_0^t v_t \mathrm{d} t
$$
The integral is not likely to admit evaluation in closed form. Typically trajectories for such system are defined parametrically.
Added
As pointed out in @ChristianBlatter answer, the integral is indeed elementary:
$$
x(t) = x(0) - \frac{b}{2a} t - \frac{1}{2a} \log\left(\frac{4 a \left(a v_0^2 + b v_0 + c\right)}{4 a c-b^2} \cos^2\left(\frac{t}{2} \sqrt{4 a c - b^2} + \arctan\left(\frac{2 a v_0 + b }{\sqrt{4 a c - b^2}}\right)\right)\right)
$$
Here is a comparison of the exact solution to the numerical solution of the differential equation in Mathematica:
Several physical conclusions:
*
*if $4 a c - b^2 < 0$, trigonometric cosine becomes hyperbolic cosine, and position $x(t)$ increases with time linearly, approximately linearly.
*if $4 a c - b^2 > 0$, the position $x(t)$ plunges to $-\infty$ in finite time, determined by zero of the cosine function.
*if $4 a c = b^2$, the solution is easy to re-derived. It also goes to $-\infty$ in finite time (if $2a v_0 + b > 0$), or goes $\pm\infty$ (if $2 a v_0 + b < 0$), depending on initial conditions:
$$
x(t) = x(0) - \frac{ b t}{2a} - \frac{1}{a} \log\left( 1 - \frac{b t}{2} - a t v_0\right)
$$
| {
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Number of partitions contained within Young shape $\lambda$ It is well known that the number of partitions contained within an $m\times n$ rectangle is $\binom{m+n}{n}$.
Furthermore, it is not difficult to calculate the number of partitions contained within a Young shape $\lambda$, where $\lambda $ is also a partition, for "small" $\lambda$ by recursively counting lattice paths with steps up and to the right.
For example, the number of partitions contained within the shape $\lambda = (3,2,1,1)$ is 19.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
Is there a simpler way to determine the number of partitions contained within the shape $\lambda=(\lambda_1,\dots,\lambda_n$)?
| Ira Gessel has very kindly explained how this can be solved by counting nonintersecting lattice paths. Again, working through the example of $(3,2,1,1)$ will explain the general approach.
First, convert the problem to counting distinct part partitions contained in $(3+3,2+2,1+1,1+0) = (6,4,2,1)$. Now a collection of four nonintersecting paths from $(0,0), (1,0), (2,0), (3,0)$ to $(1,6), (2,4), (3,2), (4,1)$ gives a subpartition by looking at the height of each column of boxes. E.g.,
corresponds to $(5,3,2,0)$ in $(6,4,2,1)$ and the partition $(5-3,3-2,2-1,0-0) = (2,1,1)$ in $(3,2,1,1)$. Using the Gessel-Viennot Lemma, the number of such quartets of nonintersecting paths is given by
$$ \begin{vmatrix} \binom{7}{1} & \binom{6}{2} & \binom{5}{3} & \binom{5}{4} \\
\binom{6}{0} & \binom{5}{1} & \binom{4}{2} & \binom{4}{3} \\
\binom{7}{-1} & \binom{4}{0} & \binom{3}{1} & \binom{3}{2} \\
\binom{8}{-2} & \binom{5}{-1} & \binom{2}{0} & \binom{2}{1} \end{vmatrix}
= \begin{vmatrix} 7 & 15 & 10 & 5 \\
1 & 5 & 6 & 4 \\
0 & 1 & 3 & 3 \\
0 & 0 & 1 & 2 \end{vmatrix}
= 19.$$
Note that moving to distinct part partitions is necessary for this approach since, for instance, the empty partition $(0,0,0,0)$ in $(3,2,1,1)$ would correspond to paths that intersect other dots along the bottom. In the equivalent distinct part subpartition problem, the empty partition corresponds to $(3,2,1,0)$.
Let me close with a historical note. In the context of Young diagrams, this determinant result is usually attributed to Kreweras 1965. In the context of partitions, MacMahon's solution is in his 1915 collection and may date to even earlier. (The MacMahon and Kreweras solutions are very similar, and Gessel and Viennot connect to Kreweras as an application of their results.)
| {
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"timestamp": "2023-03-29T00:00:00",
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$3^2 \ 5^2 \ldots (p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \ (\mathrm{mod} \ p)$
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
Why is $3^2 \ 5^2 \ldots (p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \ (\mathrm{mod} \ p)$, where $p$ is an odd prime?
I can't seem to figure it out. Any help would be appreciated. Thanks!
| Since $a\equiv -(p-a)\pmod{p}$, we can write
\begin{align}
3^2 \cdot 5^2 \cdots (p-2)^2&\equiv \left(3\cdot 5\cdots (p-2)\right)\times\left((p-3)\cdot (p-5)\cdots 2\right)\times (-1)^{(p-3)/2}\\
&\equiv (-1)^{(p-3)/2}(2\cdot 3\cdots (p-2))\cdot (p-1)\cdot (p-1)\\
&\equiv (-1)^{(p-1)/2}(p-1)!\\
&\equiv (-1)^{(p+1)/2}\pmod{p},
\end{align}
where in the third step we introduced a factor of $(p-1)(p-1)\equiv -1\cdot -1\equiv 1$, and in the last step we used Wilson's Theorem.
| {
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If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$. In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that:
If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$?
I'm not familiar with these kind of calculations, so I'd like to see, if my answer is correct:
*
*Let $n=3k+2$ so then
$2^{3k+2}+3\equiv 2^{3k+2}-4 \equiv 4\left(2^{3k}-1\right)\equiv \phantom{4\cdot } 0 \pmod 7$.
*Reading a bit about Modular arithmetics, I found that
$
a_1 \equiv b_1 \pmod N $ and $ a_2 \equiv b_2 \pmod N
$
then
$
a_1 a_2 \equiv b_1 b_2 \pmod n$, if $a_i$ and $b_i$ are integers.
*Since we have $4\equiv 4 \pmod 7$, I conclude that $2^{3k}-1 \equiv 0 \pmod 7$.
*Finally we use that
$
2^{3n}-1=(2^3-1)\cdot \left(1+2^3+2^{2\cdot 3}+2^{3\cdot 3}+\cdots+2^{(k-1)3}\right)
$
and are done.
Are there alternative ways to prove it?
| It doesn't require factorization or any other such ingenuity, just simple modular arithmetic:
$$\rm mod\ 7\!:\ \ 3 + 2^{\:\!2+3\:\!K} = 3 + 4\cdot 8^K \equiv 3 + 4\cdot 1^K\equiv 3+4 \equiv 0$$
| {
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Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
| another method, using Stolz–Cesàro theorem:
let ${ x }_{ n }=\left( p+1 \right) \left( { 1 }^{ p }+{ 2 }^{ p }+...+{ n }^{ p } \right) -{ n }^{ p+1 },{ y }_{ n }=\left( p+1 \right) { n }^{ p }$
$$\lim _{ x\rightarrow \infty }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty }{ \frac { \left( p+1 \right) { \left( n+1 \right) }^{ p }-{ \left( n+1 \right) }^{ p+1 }+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { \left( n+1 \right) }^{ p }-{ n }^{ p } \right) } = } \\ =\lim _{ x\rightarrow \infty }{ \left( \frac { \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1 \right) }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right) } \right) + } \\ +\frac { -{ n }^{ p+1 }-\left( p+1 \right) { n }^{ p }-\frac { p\left( p+1 \right) }{ 2 } { n }^{ p-1 }-...-1+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right) } $$
let's cobmine all coefficients of n,then divide numerator and denominator by $n^{ p-1 }$ and define sum of the all terms no more -1 power with $o\left( \frac { 1 }{ n } \right) $
$$\\ \\ \lim _{ x\rightarrow \infty }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty }{ \frac { \frac { p\left( p+1 \right) }{ 2 } +o\left( \frac { 1 }{ n } \right) }{ p\left( p+1 \right) +\left( \frac { 1 }{ n } \right) } =\frac { 1 }{ 2 } } \\ $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Long division in integration by partial fractions I am trying to figure out what my book did, I can't make sense of the example.
"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write
$$\int \frac{x^3 + x}{x -1} dx = \int \left(x^2 + x + 2 + \frac{2}{x-1}\right)dx = \frac{x^3}{3} + \frac{x^2}{2} + 2x + 2\ln|x-1| + C$$
I am mostly concerned with the transformation of the problem by long division I think.
I attempt to do this on my own.
$(x+1)$ and $(x^3 + x)$ inside the long division bracket
I am left with $x^2 - 1$ on top and a leftover -1
This is not in their answer, I do not know how they did that.
| Your first term is indeed $x^2$. Then $$x^3+x-x^2(x-1)=x^3+x-(x^3-x^2)=x^2+x,$$ and so your next term is $x$. Then $$x^2+x-x(x-1)=x^2+x-(x^2-x)=2x,$$ so your next term is $2$. Finally, $2x-2(x-1)=2$, so your remainder is $2$. I suspect you simply made an arithmetic error. It always helps to check your answer, and indeed, $$(x^2-1)(x-1)+-1=x^3-x^2-x+1-1=x^3-x^2-x\not\equiv x^3+x.$$
| {
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Sum with binomial coefficients: $\sum_{k=1}^m \frac{1}{k}{m \choose k} $ I got this sum, in some work related to another question:
$$S_m=\sum_{k=1}^m \frac{1}{k}{m \choose k} $$
Are there any known results about this (bounds, asymptotics)?
| We will make use of Euler-Maclaurin to get the asymptotic. The final asymptotic is $$\sum_{k=1}^{m} \dfrac1k\dbinom{m}{k} = \dfrac{2^{m+1}}{m} \left(\sum_{n=0}^{N} \dfrac{2^n \Gamma(n+1/2)}{m^n \sqrt{\pi}} \right) + \mathcal{O} \left( \dfrac{2^{m+1}}{m^{N+2}}\right)$$
Let us denote $\displaystyle \sum_{k=1}^{m} \dfrac1k\dbinom{m}{k}$ as $S$ i.e $$S = \sum_{k=1}^{m} \dfrac1k\dbinom{m}{k}$$
Let $f(k) = \dfrac1k$ and $A_m(k) = \displaystyle\sum_{n=1}^{k} \dbinom{m}{n}$. Hence, $$S = \sum_{k=1}^{m} f(n) \left( A_m(k) - A_m(k-1)\right) = \int_1^m f(t) dA_m(t) = \left. f(t)A_m(t) \right \rvert_{1}^m - \int_1^m A_m(t) df(t)\\
=\dfrac{2^m}{m} - \dfrac{m}{1} + \int_1^m \dfrac{A_m(t)}{t^2} dt$$
Now for large $m$, by central limit theorem, $$A_m(t) \sim \dfrac{2^m}{\sqrt{2 \pi \dfrac{m}4}} \int_1^t \exp \left( - \dfrac{(x-m/2)^2}{m/2}\right) dx = \dfrac{2^{m+1}}{\sqrt{2 \pi m}} \int_1^t \exp \left( - \dfrac{(x-m/2)^2}{m/2}\right) dx$$
Hence, we want an estimate for the integral
$$I = \int_1^m \dfrac1{t^2} \int_1^t \exp \left( - \dfrac{(x-m/2)^2}{m/2}\right) dx \, dt$$
$$\int_1^m \int_x^m \dfrac1{t^2} \exp \left( - \dfrac{(x-m/2)^2}{m/2}\right) dt \, dx= \int_1^m \left( \dfrac1x - \dfrac1m \right) \exp \left( - \dfrac{(x-m/2)^2}{m/2}\right) dx$$
If we let $\dfrac{x-m/2}{\sqrt{m/2}} = y$, and let $s = \sqrt{m/2}$, then
$$I = \displaystyle\int_{-s + 1/s}^{s} \left( \dfrac1{s^2 + ys} - \dfrac1{2s^2}\right) \exp(-y^2) s dy$$
$$I = \displaystyle\int_{-s + 1/s}^{s} \left( \dfrac1{s+y} - \dfrac1{2s}\right) \exp(-y^2) dy$$
$$I = \dfrac1s \displaystyle\int_{-s + 1/s}^{s} \left( \dfrac1{1+y/s} - \dfrac12\right) \exp(-y^2) dy$$
$$I = \dfrac1s \displaystyle\int_{-s + 1/s}^{s} \left( \dfrac12 - \dfrac{y}{s} + \dfrac{y^2}{s^2} - \dfrac{y^3}{s^3} + \dfrac{y^4}{s^4} \pm \cdots \right) \exp(-y^2) dy$$
For large $s$, $$\displaystyle \int_{-s+1/s}^s \exp(-y^2) dy = \sqrt{\pi} + \text{ some exponentially decaying error term}$$
$$\displaystyle \int_{-s+1/s}^s y \exp(-y^2) dy = 0 + \text{ some exponentially decaying error term}$$
$$\displaystyle \int_{-s+1/s}^s y^2 \exp(-y^2) dy = \dfrac{\sqrt{\pi}}{2} + \text{ some exponentially decaying error term}$$
$$\displaystyle \int_{-s+1/s}^s y^3 \exp(-y^2) dy = 0 + \text{ some exponentially decaying error term}$$
$$\displaystyle \int_{-s+1/s}^s y^4 \exp(-y^2) dy = \dfrac{3\sqrt{\pi}}{4} + \text{ some exponentially decaying error term}$$
Hence, $$I = \dfrac1s \left( \dfrac{\sqrt{\pi}}{2} + \dfrac{\sqrt{\pi}}{2s^2} + \dfrac{3 \sqrt{\pi}}{4s^4} + \mathcal{O} \left( \dfrac1{s^6} \right)\right)$$
Putting these together we get that $$S = \dfrac{2^m}{m} - m + \dfrac{2^{m+1}}{\sqrt{2 \pi m}} \times \left( \dfrac{\sqrt{2}}{\sqrt{m}} \left(\dfrac{\sqrt{\pi}}{2} + \dfrac{\sqrt{\pi}}{m} + \dfrac{3 \sqrt{\pi}}{m^2} + \mathcal{O} \left( \dfrac1{m^3} \right) \right)\right)$$
Hence, we get that $$S = \dfrac{2^m}{m} - m + \dfrac{2^{m+1}}{m} \times \left( \dfrac12 + \dfrac1m + \dfrac3{m^2} + \mathcal{O} \left( \dfrac1{m^3} \right) \right)$$
Hence, we get that $$S = \dfrac{2^{m+1}}{m} \left( 1 + \dfrac1m + \dfrac3{m^2} + \mathcal{O} \left(\dfrac1{m^3} \right) \right)$$
Extending this we see that $$S = \dfrac{2^{m+1}}{m} \left(\sum_{n=0}^{N} \dfrac{2^n \Gamma(n+1/2)}{m^n \sqrt{\pi}} \right) + \mathcal{O} \left( \dfrac{2^{m+1}}{m^{N+2}}\right)$$
i.e. writing out the first few terms, we get that
$$S = \dfrac{2^{m+1}}{m} \left( 1 + \dfrac1m + \dfrac3{m^2} + \dfrac{15}{m^3} + \dfrac{105}{m^4} + \cdots \right)$$
One of the advantages is that it is easy to compute better asymptotic. The saddle method by oenamen can also be extended to get better asymptotic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finite Sum of Power? Can someone tell me how to get a closed form for
$$\sum_{k=1}^n k^p$$
For $p = 1$, it's just the classic $\frac{n(n+1)}2$.
What is it for $p > 1$?
| By Binomial Series
$(n+1)^x=1 + {x \choose 1}\sum_{k=1}^n{k^{x-1}} + {x \choose 2}\sum_{k=1}^n{k^{x-2}}+{x \choose 3}\sum_{k=1}^n{{k^{x-3}}}
...+{x \choose x-1}\sum_{k=1}^n{{k^{x-x+1}}}+{x \choose x}\sum_{k=1}^n{{k^{x-x}}}$
which becomes
$(n+1)^x = 1 + {x \choose 1}\sum_{k=1}^n{k^{x-1}} + {x \choose 2}\sum_{k=1}^n{k^{x-2}}+{x \choose 3}\sum_{k=1}^n{{k^{x-3}}}
....+{x \choose x-1}\sum_{k=1}^n{{k}}+{x \choose x}\sum_{k=1}^n{{1}}$
for example consider $x=3$
$(n+1)^3 = 1 + {3 \choose 1}\sum_{k=1}^n{k^{2}} + {3 \choose 2}\sum_{k=1}^n{k^{1}}+{3 \choose 3}\sum_{k=1}^n{{1}}$
$(n+1)^3 = 1 + 3\sum_{k=1}^n{k^{2}} + 3*\frac{n*(n+1)}{2}+n$
which gives
$\sum_{k=1}^n{k^2} =(1/6)*n*(n+1)*(2n+1)$
| {
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Simplify these expressions with radical sign 2 My question is
1) Rationalize the denominator:
$$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$
My answer is:
$$\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{18}$$
My question is
2) $$\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$$
My answer is: $$\frac{1}{\sqrt{2}}$$
I would also like to know whether my solutions are right.
Thank you,
| \begin{align}
\dfrac1{\sqrt{2} + \sqrt{3} + \sqrt{5}} & = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(\sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3} - \sqrt{5})}\\
& = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(\sqrt{2} + \sqrt{3})^2 - 5}\\
& = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(2+3+2 \sqrt{6}) - 5}\\
& = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{2 \sqrt{6}}\\
& = \dfrac{(\sqrt{2} + \sqrt3 - \sqrt5) \sqrt{6}}{2 \times 6}\\
& = \dfrac{(\sqrt{12} + \sqrt{18} - \sqrt{30} )}{12}\\
\end{align}
Hence, your denominator should be $12$ and not $18$ for the first problem.
For the second problem,
\begin{align}
\dfrac1{\sqrt{2} + \sqrt{3} - \sqrt{5}} + \dfrac1{\sqrt{2} - \sqrt{3} - \sqrt{5}} & = \dfrac1{(\sqrt{2} - \sqrt{5}) + \sqrt{3}} + \dfrac1{(\sqrt{2} - \sqrt{5}) - \sqrt{3}} \\
& = \dfrac{2(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})^2 - 3}\\
& = \dfrac{2(\sqrt{2} - \sqrt{5})}{7 - 2 \sqrt{10} - 3}\\
& = \dfrac{\sqrt{2}(2 - \sqrt{10})}{4 - 2 \sqrt{10}}\\
& = \dfrac{\sqrt{2}(2 - \sqrt{10})}{2(2 - \sqrt{10})}\\
& = \dfrac{\sqrt{2}}{2}
\end{align}
Hence, your second answer is indeed correct. You may want to rationalize the denominator by multiplying by $\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the square root of the polynomial My question is:
Find the square root of the polynomial-
$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
| The polynomial is
$ \frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left( \frac{x}{y} + \frac{y}{x} \right) + 3$
Denote $ z = \frac{x}{y}$ and have
$ z^2 + z^{-2} - 2(z+z^{-1}) + 3 $
Assume $y \ne 0$ and multiply by $z^2$ and have
$ z^4 + 1 - 2z^3 - 2z + 3z^2 = 0$
Instructions how to solve a general quartic equation can be found in Wikipedia.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
determine whether this series converges for this value of z Does $$f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$$ converge for $z=\frac{-3}{2}$?
| $$ \frac{2^n + n^2}{3^n + n^3} z^n = \frac{2^n}{3^n + n^3} z^n + \frac{n^2}{3^n + n^3} z^n $$
$$ \frac{n^2}{3^n + n^3} z^n < n^2 \left ( \frac{z}{3} \right )^n = \frac{n^2 }{(-2)^n } \text{ Which converges from Ratio test }$$
$$ \frac{2^n}{3^n + n^3} z^n = \left ( \frac{2}{3} z\right )^n \frac{1}{1 + \frac{n^3}{3^n}} \text{ which is } (-1)^n \text{ for } n \rightarrow \infty \text{ (It does not converge) } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
| Encouraged by Hans Lundmark's answer, I'm posting my own solution without derivatives and integrals.
The triple-angle formula for $\tan$ is $$\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.$$
Suppose $\lim_{x\to0}(\tan x-x)/x^3 = c$. Letting $x = 3\theta$, we then have
$$\begin{align}
c &= \lim_{x\to0} \frac{\tan x-x}{x^3} \\
&= \lim_{\theta\to0} \frac{\tan 3\theta-3\theta}{27\theta^3} \\
&= \lim_{\theta\to0} \frac{3\tan\theta - \tan^3\theta-3\theta+9\theta\tan^2\theta}{27\theta^3(1-3\tan^2\theta)}
\end{align}$$
We can get rid of $1/(1-3\tan^2\theta)$ because its limit is $1$. Next we start pulling out terms and find
$$\begin{align}
c &= \lim_{\theta\to0}\frac{3\tan\theta-3\theta}{27\theta^3} - \lim_{\theta\to0}\frac{\tan^3\theta}{27\theta^3} + \lim_{\theta\to0}\frac{9\theta\tan^2\theta}{27\theta^3} \\
&= \frac19c - \frac1{27} + \frac{1}{3},
\end{align}$$
because $\lim_{\theta\to0}(\tan\theta)/\theta = 1$. So $8c/9 = 8/27$, or $c = 1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Evaluting: $\int\frac{1}{(1+\tan x)^2} dx$ I can solve this integral
$$
\int\frac{1}{(1+\tan x)^2} dx
$$
using the substitution $t=\tan x$ i.e $x=\arctan t$. Does anyone know another way to solve this integral?
| Using J.M.'s suggestion
$$ \int \frac{1}{( 1 + \tan x)^2} dx = \int \frac{\frac 12 (1 + \cos 2x)}{ \sin ^2x +2 \sin x \cos 2 + \cos ^2 x} dx = \frac 12 \int \frac{1 + \cos 2x}{1 + \sin 2x}dx$$
$$ = \frac 12 \left [ \int \frac 1 {1 + \sin 2x} dx + \int \frac{\cos 2x}{1 + \sin 2x}dx\right ]$$
$$ = \frac 1 2 \frac{\sin x}{\sin x + \cos x} + \frac 1 4 \ln (1 + \sin 2x) + C$$
$$ = \frac 12 \left [ \frac{-1}{1 + \tan x} + \ln(\sin x + \cos x)\right ] + C $$
EDIT:: first half of split up integral
$$ \int \frac{1}{1 + \sin 2x} dx = \frac 1 2 \int \frac{1}{1 + \sin u} du $$
and Weierstrass substitution $$ \frac 12 \int \frac{1}{1 + \frac{2t}{1 + t^2}}\frac{2dt}{1 + t^2} = \int \frac{1}{(1 + t)^2} dt $$
$$ \implies \frac{-1}{1 + t} = \frac{-1}{1 + \tan x} = \frac{-\cos x}{\sin x + \cos x}$$
From wolfram I got $ \int \frac{\sin x}{ \sin x + \cos x} $
Using Weierstrass substitution I got $ \frac{-\cos x}{\sin x + \cos x} $ And it seems $$ \frac{\sin x}{ \sin x + \cos x} - \frac{-\cos x}{\sin x + \cos x} = 1 $$
Hence both are vaid :D :D
And by clicking show steps at Wolframalpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recursive method.)
| How much is a direct summation worth?
$$\begin{align*}
1 + \sum_{i=0}^n 2^i &= 1 + (2^0 + 2^1 + 2^2 + \cdots + 2^n)\\
&= (2^0 + 2^0) + (2^1 + 2^2 + \cdots + 2^n)\\
&= 2^1 + (2^1 + 2^2 + \cdots + 2^n)\\
&= (2^1 + 2^1) + (2^2 + \cdots + 2^n)\\
&= 2^2 + (2^2 + \cdots + 2^n)\\
\vdots &= \ddots\\
&= 2^n + (2^n)\\
&= 2^{n+1}.
\end{align*}$$
Hence, $\displaystyle \sum_{i=0}^n 2^i = 2^{n+1} - 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
Question on the unsolvability a group Let $G$ be a finite group. Let $\pi(G)=\{2,3,5\}$ be the set of prime
divisors of its order. If 6 divide the number of Sylow 5-subgroups of G
and 10 divide the number of Sylow $3$-subgroups of $G$, then whether the
group $G$ group with those properties is unsolvable?
In particular if the number of Sylow $5$-subgroups of $G$ is 6 or
the number of Sylow $3$-subgroups of $G$ is 10, then by the Hall's
theorem $G$ is unsolvable group. For example if the number of Sylow $5$-subgroups of $G$ is 6 and $G$ is solvable, then $2\equiv 1$ (mod $5$), a contradiction.
Hall's theorem: Let $G$ be a finite soluble group and $|G|=m.n$, where $m=p_{1}^{\alpha
_{1}}...p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $\pi =\{p_{1},...,p_{r}\}$ and $
h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $
h_{m}=q_{1}^{\beta _{1}}...q_{s}^{\beta _{s}}$, satisfies the following
conditions for all $i\in \{1,2,...,s\}$:
1) $q_{i}^{\beta_{i}} \equiv 1$ (mod $p_{j}$), for some $p_{j}$.
2) The order of some chief factor of $G$ is divisible by $
q_{i}^{\beta_{i}}$.
Thank you so much.
| No. There is a solvable group of order $2^{22} 3^5 5^3$ with $n_2=1$, $n_3=2^{18} 5^2$, and $n_5 = 2^{20} 3^4$.$\newcommand{\GF}{\operatorname{GF}}\newcommand{\AGL}{\operatorname{AGL}}$
$$G =
\left(\left(3\ltimes\GF(5^2)\right) \ltimes \left(\GF(2^4)^3\right)\right) ~ \times ~ \left(\left(5\ltimes\GF(3^4)\right) \ltimes \left(\GF(2^2)^5\right)\right)$$
I'll describe the pieces:
$H = 3\ltimes\GF(5^2) \leq \AGL(1,5^2)$ is the collection of affine maps $x\mapsto ax+b$ where $a,x,b$ are elements of the finite field $\GF(5^2)$ of order 25 and $a^3=1$.
$H$ has an irreducible $\GF(2^4)$ module $V$ of dimension 3 formed by inducing a non-principal one-dimensional module from $\GF(5^2) \leq H$.
$H \ltimes V =\left(3\ltimes\GF(5^2)\right) \ltimes \left(\GF(2^4)^3\right) \leq \AGL(3,2^4)$ consists of all maps $x\mapsto ax+b$ where $a \in H$ and $x,b \in V$.
$K = 5 \ltimes \GF(3^4) \leq \AGL(1,3^4)$ is the collection of affine maps $x\mapsto ax+b$ where $a,x,b$ are elements of the finite field $\GF(3^4)$ of order 81 and $a^5=1$.
$K$ has an irreducible $\GF(2^2)$ module $W$ of dimension 5 formed by inducing a non-principal one-dimensional module from $\GF(3^4) \leq K$.
$K \ltimes W = \left(5\ltimes\GF(3^4)\right) \ltimes \left(\GF(2^2)^5\right) \leq \AGL(5,2^2)$ consists of all maps $x\mapsto ax+b$ where $a \in K$ and $x,b \in W$.
$G = \left( H \ltimes V \right) \times \left( K \ltimes W \right)$ is the direct product of these two groups.
$\begin{array}{c|cccc}
& o & n_2 & n_3 & n_5 \\ \hline
H & 2^{~0} 3^1 5^2 & 3^0 5^0 & 2^{~0} 5^2 & 2^{~0} 3^0 \\
H \ltimes V & 2^{12} 3^1 5^2 & 3^0 5^0 & 2^{~8} 5^2 & 2^{12} 3^0 \\
K & 2^{~0} 3^4 5^1 & 3^0 5^0 & 2^{~0} 5^0 & 2^{~0} 3^4 \\
K \ltimes W & 2^{10} 3^4 5^1 & 3^0 5^0 & 2^{10} 5^0 & 2^{~8} 3^4 \\
G & 2^{22} 3^5 5^3 & 3^0 5^0 & 2^{18} 5^2 & 2^{20} 3^4 \\
\end{array}$
I think this is approximately minimal order. I think I misunderstand Hall's theorem if there is any example of order less than $2^{22} 3^4 5^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
| If you expand the two sides, you will get
\begin{equation}
\frac{c^2}{4}-\frac{cn}{2}-c+\frac{n^2}{4}+2n\leq\frac{n^2}{4}-\frac{n}{2}+\frac{5}{4}
\end{equation}
\begin{equation}
\frac{c^2}{4}-\frac{5}{4}-c\leq n\frac{c-5}{2}
\end{equation}
If c=5 than the inequality is satisfied, assuming $5\leq c$ we have
\begin{equation}
\frac{c-5}{2} -\frac{2c}{c-5}\leq n
\end{equation}
I think you can use induction to prove this inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$ I try to solve the following sum:
$$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$
I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.
| Let's write:
$$f(x) = \sum_{k = 1}^{+\infty}\sum_{n = 1}^{+\infty} \frac{x^{n+k+2}}{nk (n+k+2)}$$
then:
$$f'(x) = \sum_{k = 1}^{+\infty}\sum_{n = 1}^{+\infty} \frac{x^{n+k+1}}{nk} = - \sum_{k = 1}^{+\infty}\frac{x^{k+1} \ln (1-x)}{k}$$
We want to know the value of $f(1)$, so we integrate:
$$f(1) = - \sum_{k = 1}^{+\infty}\frac{1}{k} \int_0^1 x^{k+1} \ln (1-x) \, dx = \sum_{k = 1}^{+\infty} \frac{H_{k+2}}{k(k+2)}\\
= \frac{1}{2} \sum_{k = 1}^{+\infty} \left( \frac{H_{k+2}}{k} - \frac{H_{k+2}}{k+2} \right)=\frac{1}{2} \sum_{k = 1}^{+\infty} \left( \frac{H_{k} + \frac{1}{k+1} + \frac{1}{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\
= \frac{1}{2} \left( H_1 + \frac{1}{2}H_2 + \sum_{k = 1}^{+\infty} \left( \frac{3}{2k} - \frac{1}{1+k} - \frac{1}{2(k+2)} \right) \right) = \frac{7}{4}$$
The very last sum telescopes:
$$\sum_{k = 1}^{+\infty} \left( \frac{3}{2k} - \frac{1}{1+k} - \frac{1}{2(k+2)} \right) = \sum_{k = 1}^{+\infty} \left( \left( \frac{1}{k} - \frac{1}{1+k} \right) + \left(\frac{1}{2k} - \frac{1}{2(k+2)} \right) \right) = \frac{7}{4}$$
And the integral I've used is obtained by integrating by parts:
$$\int_0^1 x^{k+1} \ln(1-x)\,dx = \frac{x^{k+2} \ln(1-x)}{k+2} \Big|_0^1 + \frac{1}{k+2} \int_0^1 \frac{x^{k+2} - 1 + 1}{1-x} \,dx \\
= -\frac{H_{k+2}}{k+2} + \lim_{\epsilon \to 1^-} \left( x^{k+2} \ln(1-x) + \int_0^\epsilon \frac{dx}{1-x} \right) = -\frac{H_{k+2}}{k+2}$$
I admit it is a little bit long solution but the trick of making power series from regular series is very useful :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Tangent line of parametric curve I have not seen a problem like this so I have no idea what to do.
Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.
$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$
I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$
But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.
| *
*Eliminating the parameter $t$. The given system of two parametric equations $$\begin{eqnarray*}\left\{
\begin{array}{c}
x=1+\ln t \\
y=t^{2}+2
\end{array}
\right. \end{eqnarray*} \tag{A}$$ is equivalent successively to
$$\begin{eqnarray*}
\left\{
\begin{array}{c}
x-1=\ln t \\
y=t^{2}+2
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
t=e^{x-1} \\
y=t^{2}+2
\end{array}
\right.
\Leftrightarrow \left\{
\begin{array}{c}
t=e^{x-1} \\
y=\left( e^{x-1}\right) ^{2}+2
\end{array}
\right. \end{eqnarray*}$$ and finally to $$\left\{
\begin{array}{c}
t=e^{x-1} \\
y=e^{2\left( x-1\right) }+2.
\end{array}
\right. \tag{B}
$$
We have thus eliminated the parameter $t$.
Here is a plot of $y=e^{2\left( x-1\right) }+2$
Differentiating the equation of the curve $$y=e^{2\left( x-1\right) }+2\tag{C}$$ gives by the chain rule applied to $e^u$, with $u=2(x-1)$
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{d}{dx}\left( e^{2\left( x-1\right) }+2\right) =\frac{d
}{dx}e^{2\left( x-1\right) }=\left( \frac{
d}{du}e^{u}\right) \frac{du}{dx}
=e^{u}\times 2=2e^{2\left( x-1\right) }.
\end{eqnarray*}$$
At $x=1$ this derivative has the value
$$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2e^{2\left( x-1\right)
}\right\vert _{x=1}=2e^{2(1-1)}=2.\tag{1}
\end{equation*}$$
Since at $x=1$, $$y=e^{2(1-1)}+2=3,$$ we have confirmed that the point $(x,y)=(1,3)$ belongs to the graph of the curve given by equation $(C)$. Hence the equation of the tangent line to the graph of the curve at $(1,3)$ is
$$\begin{equation*}
y-3=2(x-1)\Leftrightarrow y=2x+1\tag{2}
\end{equation*}$$
*Without eliminating the parameter $t$. (Reformulated in view of OP's comment.) To compute the derivative we use now the parametric equations $(A)$ and the formula $$\frac{dy}{dx} =\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}.\tag{D}$$ We have
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{dy}{dt}/\frac{dx}{dt}=
\left( \frac{d}{dt}\left( t^{2}+2\right) \right) /\frac{d}{dt}\left( 1+\ln
t\right) \\
&=&2t/\frac{1}{t}=2t^2,\tag{3}
\end{eqnarray*}$$
which confirms your result. Since for $x=1$ the equation $$x=1+\ln t $$ gives $$1=1+\ln t\Leftrightarrow 0=\ln t \Leftrightarrow t=1,$$ we get the same value as in $(1)$ for the derivative $$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2t^2\right\vert
_{t=1}=2\cdot 1^2=2.\tag{3a}
\end{equation*}$$
The equation of the tangent line is as above
$$\begin{equation*}
y=2x+1.\tag{4}
\end{equation*}$$
In terms of the parameter $t$ the tangent line at $t=1$, i.e. at $(x,y)=(1,3)$ is given by the parametric equations
$$\begin{equation*}
\left\{
\begin{array}{c}
x=t \\
y=2t+1,
\end{array}
\right.\tag{4a}
\end{equation*}$$
because
$$\begin{equation*}
\left. \frac{dx}{dt}\right\vert _{t=1}=\left. \frac{1}{t}\right\vert _{t=1}=1
\end{equation*}$$
and
$$\begin{equation*}
\left. \frac{dy}{dt}\right\vert _{t=1}=\left. 2t\right\vert _{t=1}=2.
\end{equation*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Triples of positive real numbers $(a,b,c)$ such that $\lfloor a\rfloor bc=3,\; a\lfloor b\rfloor c=4,\;ab\lfloor c\rfloor=5$ Find the all ordered triplets of positive real numbers $(a,b,c)$ such that: $$\lfloor a\rfloor bc=3,\quad a\lfloor b\rfloor c=4,\quad ab\lfloor c\rfloor=5,$$
where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
| Now we have by dividing equations and putting together the ratios we get the ratio $$\frac a{\lfloor a\rfloor}:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$
This shows that $\frac a{\lfloor a\rfloor}\ge\frac53$ since $\frac c{\lfloor c\rfloor}\ge1$. It is checkable that $\frac a{\lfloor a\rfloor}\ge\frac53$ forces $\lfloor a\rfloor=1$, since $a,b,c\ge1$
Similarly, we get $\frac b{\lfloor b\rfloor}\ge\frac54$ This forces $\lfloor b\rfloor\le3$
Now,$$a=\frac a{\lfloor a\rfloor}\ge\frac53$$
we already have $ab\ge\frac{25}{16}$. But $ab\lfloor c\rfloor=5$ so that $\lfloor c\rfloor\le\frac{80}{25}$ whence $\lfloor c\rfloor\le 3$ But if one of $b,c$ is $3$ or more then it forces the other two to be $1$ or less which contradicts either the ratios above or the fact that all of a,b,c are at least 1.
So, $\lfloor b\rfloor=\lfloor c\rfloor\le2$ and $\lfloor a\rfloor=1$
The ratios become
$$a:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$
Case 1: $\lfloor b\rfloor=1$
Then $a:b=4:3$, so $b=\frac{3a}4$. The third equation in data then gives$$\frac{3a^2}4\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=20$. But $a<2$ implies $\lfloor c\rfloor=2$ Now we get $a=\sqrt\frac{10}3$ $b=\frac{\sqrt{30}}4$ Since $\lfloor a\rfloor=1$, we have by the first equation in data that $\frac{\sqrt{30}c}4=3$ That is, $c=\frac25{\sqrt{30}}$
Case 2: $\lfloor b\rfloor=2$
Then $a:b=2:3$, so $b=\frac{3a}2$. The third equation in data then gives$$\frac{3a^2}2\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=10$.
Suppose $\lfloor c\rfloor=1$, we get $3a^2=10$ again and as above using ratios we get the same $a$ and $b$, which contradicts that $\lfloor b\rfloor\ne1$ here! This contradiction gives $\lfloor c\rfloor=2$ and we get $6a^2=10$ and $a=\sqrt\frac53$ But we saw earlier that $a\ge\frac53$. So case 2 is contradictory and we have only 1 solution!
$$(a,b,c)=\left(\sqrt\frac{10}3,\frac{\sqrt{30}}4,\frac25{\sqrt{30}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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How can I calculate this limit: $\lim\limits_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$? I was given this limit to solve, without using L'Hospital rule. It's killing me !! Can I have the solution please ?
$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$
| $$ \lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} $$
$$ = \lim_{x\rightarrow 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} \times \frac{2+\sqrt{2+x}}{2+\sqrt{2+x}} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3} }{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}$$
$$ = \lim_{x\rightarrow 2} \frac{2 - x}{-(2-x)} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}{2 + \sqrt{2+x}}$$
$$= -\frac{3 (2) ^{2/3}}{4} = -\frac{3}{2(2)^{1/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Linear algebra: power of diagonal matrix? Let A = $\begin{pmatrix} 3 & -5 \\ 1 & -3 \end{pmatrix}$. Compute $A^{9}$. (Hint: Find a matrix P such that $P^{-1}AP$ is a diagonal matrix D and show that $A^{9}$= $PD^{9}P^{-1}$
Answer: $\begin{pmatrix} 768 & -1280 \\ 256 & -768 \end{pmatrix}$
I keep getting $\begin{pmatrix} -768 & 1280 \\ -256 & 768 \end{pmatrix}$ but could it be still right? I have D=$\begin{pmatrix} -2 & 0\\ 0& 2\end{pmatrix}$ and P =$\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ with $P^{-1}$= $\begin{pmatrix} \frac{1}{4} & \frac{1}{-4} \\ \frac{1}{-4} & \frac{5}{4} \end{pmatrix}$
| With $P=\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ we have $P^{-1}AP = \left(\begin{array}{rr}2 & 0\\0 & -2\end{array}\right)$ then you can use your formula $PA^{9}P^{-1}$ and calculate $PA^{9}P^{-1}$.
$ \left(\begin{array}{rr}2 & 0\\0 & -2\end{array}\right)^9=\left(\begin{array}{rr}\,\,2^9 & 0\,\,\,\,\,\,\,\\0 & \,(-2)^9\end{array}\right)$
Now you can calculate $A^9$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate $\lim\limits_{ x\rightarrow 100 }{ \frac { 10-\sqrt { x } }{ x+5 } }$ $$\lim_{ x\rightarrow 100 }{ \frac { 10-\sqrt { x } }{ x+5 } } $$
Could you explain how to do this without using a calculator and using basic rules of finding limits?
Thanks
| Recall that $\lim_{x \to a} f(x) = f(a)$ if $f(x)$ is continuous at $a$. The function $f(x) = \dfrac{10-\sqrt{x}}{x+5}$ is continuous at $100$ since the numerator $10-\sqrt{x}$ is continuous for all $x > 0$ and the denominator $5+x$ is continuous for all $x$. Hence, the function $f(x) = \dfrac{10-\sqrt{x}}{x+5}$ is continuous for all $x > 0$. Now you should be able to finish it off.
Move your mouse below the gray area for the answer.
Hence, $$\lim_{x \to 100} \dfrac{10-\sqrt{x}}{x+5} = \dfrac{10 - \sqrt{100}}{100 + 5} = \dfrac{10 - 10}{105} = 0$$
| {
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Power (Laurent) Series of $\coth(x)$ I need some help to prove that the power series of $\coth x$ is:
$$\frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + O(x^5) \ \ \ \ \ $$
I don't know how to derive this, should I divide the expansion of $\cosh(x)$ by the expansion of $\sinh(x)$? (I've tried but without good results :( )
Or I have to use residue calculus?
Anyone can suggest me a link where I can find a detailed explanation of this expansion?
Thanks.
| Long division of series with $\cosh(x) = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{24} + \ldots$
and $\sinh(x) = x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \ldots$. Unfortunately I don't know how to typeset this nicely in LaTeX.
First term is $1/x$,
$$1 + \dfrac{x^2}{2} + \dfrac{x^4}{24} + \ldots - \dfrac{1}{x} \left(x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \ldots\right) = \frac{x^2}{3} + \frac{x^4}{30} + \ldots$$
Next term is $x/3$,
...
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $x=\dfrac 1y$, show that $∫\frac {dx}{x \sqrt{(x^2-1)}} = -∫\frac {dy}{\sqrt{1-y^2}}$ Given that $x=\dfrac 1y$, show that $\displaystyle \int \frac 1{x\sqrt{x^2-1}}\,dx = -\int \frac 1{\sqrt{1-y^2}}\,dy$
Have no idea how to prove it.
here is a link to wolframalpha showing how to integrate the left side.
| Given that $x=1/y$
Put $x = \dfrac{1}{y},dx = \dfrac{-1}{y^2}.dy$:
$$\int \dfrac{1}{(x\sqrt{(x^2-1)})}dx$$
$$=\int \dfrac{1}{(\frac{1}{y} \sqrt{((\frac{1}{y})^2-1)})}\cdot\dfrac{-1}{y^2}dy$$
$$\int \dfrac{1}{\frac{1}{y} \sqrt{\frac{1-y^2}{y^2}}}\cdot \dfrac{-1}{y^2}dy$$
Simplify and you get your answer.
| {
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A problem dealing with even perfect numbers. Question: Show that all even perfect numbers end in 6 or 8.
This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.
What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions.
Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$.
I would appreciate any comments and or alternate approaches to arrive at a good proof.
| In this proof, n is any integer(not just primes)
The number 2^(n-1) last digit repeats in the four number cycle(2,4,8,6)
The number 2^(n) -1 last digit repeats in the four number cycle (3,7,5,1)
The product's last digit repeats in the four number cycle (6,8,0,6)
If n is odd, the pattern repeats (8,6)
You will find the number end in 6 or 8 for all odd numbers(not just primes)
| {
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how to solve system of linear equations of XOR operation? how can i solve this set of equations ? to get values of $x,y,z,w$ ?
$$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$
$$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$
$$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$
$$\begin{aligned}1=w \oplus y \oplus z \end{aligned}$$
this is not a real example, the variables don't have to make sense, i just want to know the method.
| As others have noted, all the usual methods of solving systems of linear equations (such as Gaussian elimination) in the field of real numbers work just as well in the finite field of integers modulo 2, also known as $GF(2)$.
In this field, addition corresponds to the XOR operation, while multiplication corresponds to AND (as it does in the reals, if the operands are restricted to $0$ and $1$). As both $0$ and $1$ are their own additive inverses in $GF(2)$ (since $0 \oplus 0 = 1 \oplus 1 = 0$), subtraction is also equivalent to XOR, while division is trivial (dividing by $1$ does nothing, dividing by $0$ is undefined).
So, let's try solving your example equations. Since martini already did Gaussian elimination, let me do Gauss–Jordan:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
1x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
The first coefficient of the first equation is already $1$, so we just subtract (XOR) that equation from the second and the third:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
Now the second coefficient of the second equation is $0$, so we need to choose some later row that has a $1$ for that coefficient and swap those rows:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
... and then subtract that row from all the others with a non-zero second coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\
\end{array}$$
Then we subtract the third row from those with a non-zero third coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\
\end{array}$$
... and finally the fourth row from those with a non-zero fourth coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 0w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\
0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\
\end{array}$$
And now we can read out the result: $x = 0$, $y = 1$, $z = 0$, $w = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.
| We know that $-3/2$ is a solution. Then we divide $x^3+27$ by $x+3/2$. Hence we have
$x^3+27=(x+3/2)(8x^2-12x+18)=0$. But, 8x^2-12x+18 don't have real solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$
| Of course, $x$ cannot be an integer multiple of $\pi$ plus $\frac{\pi}{2}$. We have here a quadratic equation after a little manipulation, which you, yourself, started.
$$\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$$
After multiplying by $\cos x$, using the identity $1=\sin^2 x+\cos^2 x$, and bringing everything to the left hand side:
$$2\sin^2 x+\sin x-1=0$$
If we replace $y=\sin x$, then you will be able find a familiar expression for $y$:
$$2y^2+y-1 = 0$$
Which gives, $y = \frac{1}{2}$ or $-1$. So, $\sin x$ is equal to either of those two values, which yields $x= 2n\pi+\frac{\pi}{6}$, $x=2n\pi+\frac{5\pi}{6}$ or $x=2n\pi+\frac{\pi}{2}$. However, we ruled out the latter solution initially.
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $x^2+y^2=z^5+z$ Has infinitely many relatively prime integral solutions How to show that this equation:
$$x^2+y^2=z^5+z$$
Has infinitely many relatively prime integral solutions
| It is necessary to write down the formula!
In the equation:
$X^2+Y^2=Z^5+Z$
I think this formula should be written in a more general form:
$Z=a^2+b^2$
$X=a(a^2+b^2)^2+b$
$Y=b(a^2+b^2)^2-a$
And yet another formula:
$Z=\frac{a^2+b^2}{2}$
$X=\frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$
$Y=\frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$Z=\frac{(a^2+b^2)^2}{2}$
$X=\frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$
$Y=\frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$
where $a,b$ - any integers asked us.
Well, a simple solution:
$Z=(a^2+b^2)^2$
$X=a^2+2(a^2+b^2)^4ab-b^2$
$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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The number $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ is always an integer For each $n$ consider the expression $$\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$$
I am trying to prove by induction that this is an integer for all $n$.
In the base case $n=1$, it ends up being $1$.
I am trying to prove the induction step:
*
*if $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ is an integer, then so is
$\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.
I have tried expanding it, but didn't get anywhere.
| Try writing
$$
\left(\frac{1+\sqrt{5}}{2}\right)^n = \frac{a_n+b_n\sqrt{5}}{2}
$$
with $a_1=b_1=1$ and
$$
\frac{a_{n+1}+b_{n+1}\sqrt{5}}{2} = \left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{a_n+b_n\sqrt{5}}{2}\right)
$$
see what you get, then repeat as much as you need to with $\left(\frac{1-\sqrt{5}}{2}\right)^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
| \begin{align*}
ax^2+bx+c &= 0 \\
ax^2+bx &= -c \\
x^2+\frac{b}{a}x &= -\frac{c}{a} \\
x^2+2x\Bigl(\frac{b}{2a}\Bigr)+\Bigl(\frac{b}{2a}\Bigr)^2-\Bigl(\frac{b}{2a}\Bigr)^2 &= -\frac{c}{a} \\
\Bigl(x+\frac{b}{2a}\Bigr)^2-\frac{b^2}{4a^2} &= -\frac{c}{a} \\
\Bigl(x+\frac{b}{2a}\Bigr)^2 &= \frac{b^2-4ac}{4a^2} \\
x+\frac{b}{2a} &= \frac{\pm\sqrt{b^2-4ac}}{2a} \\
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$
I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox.
If the series continues infinitely then each term is just going to get smaller and smaller.
Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$?
| You can prove it with partial sums:
$$
S_n=\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\left(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\right)=\frac{1}{2}\left(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=2}^{n+1}\frac{1}{2k-1}\right)
$$ $$
=\frac{1}{2}\left(\frac{1}{2(1)-1}-\frac{1}{2(n+1)-1}\right)=\frac{1}{2}-\frac{1}{2(2n+1)}
$$
Hence,
$$
\sum_{k=1}^\infty\frac{1}{(2k-1)(2k+1)}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{2(2n+1)}\right)=\frac{1}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for any $x \in \mathbb N$ such that $xProve that every positive integer $x$ with $x<n!$ is the sum of at most $n$ distinct divisors of $n!$.
| People do not seem to be going along with my comment. So this is CW, and directly from the answer by Sanchez.
For $n=2,$ we need only 1 divisor of $2!,$ as $1=1.$
For $n=3,$ we need only 2 divisors of $3!=6,$ as $1=1, 2=2,3=3,4=3+1,5=3+2.$
Induction hypothesis: for some $n \geq 2,$ we need at most $(n-1)$ distinct divisors of $n!$ to write any $1 \leq x < n!$ as a sum.
Induction step (Sanchez, above). Let $N = n+1.$ Let $1 \leq x < N! = (n+1)!$ Write
$$ x = N q + r, \; \; \mbox{with} \; \; 0 \leq r < N. $$
Because $q < (N-1)! = n!,$ we need at most $(n-1) = (N-2)$ divisors of $n!$ to write $q$ as a sum. So
$$ q = \sum_{i=1}^{n-1} d_i, $$
where each $d_i | n!$ Therefore each $Nd_i | N!$
At this stage, we have at most $N-2$ divisors of $N!$ What about $r?$ Well, $r < N,$ so it is automatically a factor of $N!$ So we have finished the decomposition as a sum with at most $(N-1)$ divisors of $N!,$ where $N=n+1.$
CONCLUSION: For all $N \geq 2,$ every integer $1 \leq x < N!$ can be written as the sum of (at most) $N-1$ distinct divisors of $N!$
SUGGESTION: try it for $N=4, \; \; N! = 24.$
NEVER MIND, do it myself. Aliquot divisors 1,2,3,4,6,8,12.
$$1=1,2=2,3=3,4=4,5=4+1,6=4+2,7=4+3, 8=8,9=8+1,10=8+2, $$
$$11=8+3, 12= 12, 13 = 12+1, 14 = 12+2, 15 = 12+3, 16 = 12+4,$$
$$17=12+4+1, 18=12+6,19=12+4+3,20=12+8,$$
$$21=12+8+1,22=12+8+2,23 = 12+8+3. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Image of $\partial B_r $ under an holomorphic function Let
\begin{equation*}
\begin{split}
f \colon & \mathbb C \setminus \left\{0\right\} \to \mathbb C \\
& z \mapsto \frac{1}{2}\left( z+ \frac{1}{z}\right)
\end{split}
\end{equation*}
I am asked to find the image of $\partial B_r := \{z \in \mathbb C : \vert z \vert = r\}$ (where $r >0$) under $f$.
Let me show you what I've done: hope it's correct.
Let $\vert z \vert = r$. Then
\begin{equation*}
\begin{split}
\vert f(z) \vert & = \sqrt{f(z) \overline{f(z)}} = \sqrt{\frac{1}{2}\left( z+ \frac{1}{z}\right)\frac{1}{2}\left( \overline{z}+ \frac{1}{\overline{z}}\right)} = \\
& = \frac{1}{2}\sqrt{\left( z\overline{z} + \frac{z}{\overline{z}} + \frac{\overline{z}}{z} + \frac{1}{z\overline{z}}\right)} = \\
& = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)}
\end{split}
\end{equation*}
Now we calculate
$$
\Re{\frac{z}{\overline{z}}} = \Re{\frac{x+iy}{x-iy}} = \Re{\frac{x^2+y^2 + 2ixy}{x^2+y^2}} = 1
$$
hence we get
\begin{equation*}
\begin{split}
\vert f(z) \vert & = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)} = \\
& = \frac{1}{2}\sqrt{\left( r^2 + 2 + \frac{1}{r^2} \right)} = \frac{r^2+1}{2r} =: k.
\end{split}
\end{equation*}
So we can conclude that
$$
f(\partial B_{r}) = \partial B_{k}
$$
What do you think? Is it correct? Thanks for your help.
| We have, writing $a+ib:=re^{it}$ that
\begin{align}
f(re^{it})&=\frac 12\left(a+ib+\frac 1{a+ib}\frac{a-ib}{a-ib}\right)\\
&=\frac 12\left(a+ib+\frac{a-ib}{r^2}\right)\\
&=\frac 12\left(a\left(1+\frac 1{r^2}\right)+ib\left(1-\frac 1{r^2}\right)\right)\\
&=\frac 12\left(r\cos \theta\frac{r^2+1}{r^2}+ir\sin\theta\frac{r^2-1}{r^2}\right)\\
&=\frac 1{2r}(\cos\theta (r^2+1)+i\sin\theta(r^2-1)).
\end{align}
Do you recognize curve in the plane of the form $\gamma(t)=(A\cos t,B\sin t)$, where $A$ and $B$ are fixed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Two sums with Fibonacci numbers
*
*Find closed form formula for sum: $\displaystyle\sum_{n=0}^{+\infty}\sum_{k=0}^{n} \frac{F_{2k}F_{n-k}}{10^n}$
*Find closed form formula for sum: $\displaystyle\sum_{k=0}^{n}\frac{F_k}{2^k}$ and its limit with $n\to +\infty$.
First association with both problems: generating functions and convolution. But I have been thinking about solution for over a week and still can't manage. Can you help me?
| The first one can be solved using the fact that the generating
function of the Fibonacci numbers is
$$\frac{z}{1-z-z^2}.$$
Introduce the function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n \frac{F_{2k} F_{n-k}}{10^n}$$
so that we are interested in $f(1).$
Re-write $f(z)$ as follows:
$$f(z) = \sum_{k\ge 0} F_{2k}
\sum_{n\ge k} \frac{z^n}{10^n} F_{n-k}
= \sum_{k\ge 0} F_{2k} \frac{z^k}{10^k}
\sum_{n\ge 0} \frac{z^n}{10^n} F_n.$$
Now we have
$$ \sum_{k\ge 0} F_{2k} z^{2k} =
\frac{1}{2} \frac{z}{1-z-z^2}
- \frac{1}{2} \frac{z}{1+z-z^2}$$
and therefore $f(1)$ is
$$\left(\frac{1}{2} \frac{1/\sqrt{10}}{1-1/\sqrt{10}-1/10}
- \frac{1}{2} \frac{1/\sqrt{10}}{1+1/\sqrt{10}-1/10}\right)
\times
\frac{1/10}{1-1/10-1/100}$$
which simplifies to
$$\frac{1}{2\sqrt{10}}
\frac{2/\sqrt{10}}{81/100-1/10} \times \frac{10}{89}
= \frac{1}{10} \times \frac{1}{71/100} \times \frac{10}{89} =
\frac{100}{89\times 71}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove $\cos^2x\sin^4x = \frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))$ I need help with a problem some may consider odd but here it is.
I have the following trig identity I been working on and I managed to get it to.
$$\cos^2x\sin^4x=\frac{3}{16}-\frac{\cos(2x)}{4}+\frac{3\cos(2x)}{16}+\frac{1}{8}(1+\cos(4x))+\frac{1}{32}(\cos(6x)+\cos(2x))$$
However I am not sure how to make this simplify to $\frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))=\cos^2x\sin^4x$
and thus I am stuck.
| The equation you have written is not correct in general.
What you can do is note that there exists some $\theta \in [0, \pi)$ such that $\sin^2 x = \sin \theta$.
Then $1 - \sin^2 x = 1 - \sin \theta \Rightarrow \cos^2 x = 1 - \sin \theta$.
Thus we know that $\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta$.
Let $1 - 2 \sin \theta = T$. Then we know:
$\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta = \frac{1}{8} (1 + T)(1 - T)^2$.
Similarly,
$\cos(2x) = T,$
$\cos(4x) = 2T^2 - 1,$
$\cos(8x) = 8T^4 - 8T^2 + 1$.
The sum-to-product formula tells us:
$\cos (2x) + \cos (6x) = 2 \cos(4x) \cos (8x) = 32T^6 - 48T^4 + 20T^2 - 2$.
Perhaps you can go somewhere from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing sign table for $2^k$ factorial experiment design all!
I need to compute the sign table for a generic $2^k$ factorial design. For $k$ factors we compute $2^k$ experiments and need to compute a $2^k \times 2^k$ matrix, as the following example for $k=3$:
\begin{matrix}
& I & A & B & C & AB & AC & BC & ABC\\
1 & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
2 & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
3 & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
4 & +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\
5 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
6 & +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\
7 & +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\
8 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
\end{matrix}
It is easy to compute the symbol's ($A$, $B$, $C$) columns seeing the experiment number as a bit array, where a bit zero maps to $-1$ and a bit one to $+1$. The following columns are calculated as the product of the combined symbols.
I want to compute the matrix directly, looping through $i$ and $j$, for any generic $k$. How can I find to which combination (and thus, which symbol's signs to multiply) a given $j$ corresponds to?
I hope it is clear enough; if not, please ask. Thanks for any attention!
| It's hard to see the bit structure in the table the way you've ordered it. If you write it like this:
$$
\begin{matrix}
& I & A & B & AB & C & AC & BC & ABC\\
8 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
7 & +1 & -1 & +1 & -1 & +1 & -1 & +1 & -1\\
6 & +1 & +1 & -1 & -1 & +1 & +1 & -1 & -1\\
5 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
4 & +1 & +1 & +1 & +1 & -1 & -1 & -1 & -1\\
3 & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
2 & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
1 & +1 & -1 & -1 & +1 & -1 & +1 & +1 & -1\\
\end{matrix}
$$
then the entries are just the parity of the AND of the row and column indices from $0$ to $2^k-1$, which you can most efficiently get by precomputing a lookup table of size $2^k$ for the parities and indexing it with the AND.
If you want the table in the order you showed it in, the reversal of the rows is trivial (the index is $2^k$ minus your row label), but the rearrangement of the columns isn't – you've got the columns sorted by first subset size and then lexicographical order; you can get that either by sorting the columns in the end, or by making another lookup table of size $2^k$ beforehand that translates between the two column indexing schemes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y). $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
| Let, $y=2n-1$
given, $2n-1\lt61$, $2n \lt 62$
$\frac{1}{x}+\frac{4}{y}=\frac{1}{12}$
$\frac{1}{x}+\frac{4}{2n-1}=\frac{1}{12}$
$\frac{1}{x}=\frac{1}{12}-\frac{4}{2n-1}$
$\frac{1}{x}=\frac{2n-1-48}{(12)(2n-1)}$
$\frac{1}{x}=\frac{2n-49}{(12)(2n-1)}$
$x=\frac{(12)(2n-1)}{2n-49}$
$x=\frac{(1)(3)(4)(2n-1)}{2n-49}$
By careful observation, we see that $2n-49$ needs to be odd as $2n$ is even
i.e., $2n-49=1$, $2n-49=3$, $2n-49=(3)(k)$ where, k is an odd integer
i.e., $n=25$, $n=26$, $2n=49+3k$
i.e., $n=25$, $n=26$, $2n=58$
i.e., $n=25$, $n=26$, $n=29$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/182882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a basis for the solution space of a system of Diophantine equations Let $m$, $n$, and $q$ be positive integers, with $m \ge n$.
Let $\mathbf{A} \in \mathbb{Z}^{n \times m}_q$ be a matrix.
Consider the following set:
$S = \big\{ \mathbf{y} \in \mathbb{Z}^m \mid \mathbf{Ay} \equiv \mathbf{0} \pmod q \big\}$.
It can be easily shown that $S$ constitutes a lattice, because it is a discrete additive subgroup of $\mathbb{R}^m$.
I want to find the basis of this lattice. In other words, I want to find a matrix $\mathbf{B} \in \mathbb{Z}^{m \times m}$, such that the following holds:
$S = \{\mathbf{Bx} \mid \mathbf{x} \in \mathbb{Z}^m \}$.
Let me give some examples:
*
*$q=2$, and $\mathbf{A} = [1,1]$ $\quad \xrightarrow{\qquad}\quad$
$\mathbf{B} = \left[ \begin{array}{cc} 2&1 \\ 0&1 \end{array} \right]$
*$q=3$, and $\mathbf{A} = \left[ \begin{array}{ccc} 0&1&2 \\ 2&0&1 \end{array} \right]$
$\quad \xrightarrow{\qquad}\quad$
$\mathbf{B} = \left[ \begin{array}{ccc} 3&0&1 \\ 0&3&1 \\ 0&0&1 \end{array} \right]$
*$q=4$, and $\mathbf{A} = \left[ \begin{array}{ccc} 0&2&3 \\ 3&1&2 \end{array} \right]$
$\quad \xrightarrow{\qquad}\quad$
$\mathbf{B} = \left[ \begin{array}{ccc} 4&2&1 \\ 0&2&1 \\ 0&0&2 \end{array} \right]$
Note that in all cases, $\mathbf{AB} =\mathbf{0} \pmod q$. However, $\mathbf{B}$ is not an arbitrary solution to this equivalence, since it must span $S$. For instance, in the example 1 above, matrix $\mathbf{\hat B} = \left[ \begin{array}{cc} 2&0\\ 0&2 \end{array} \right]$ satisfies $\mathbf{A \hat B} =\mathbf{0} \pmod 2$, but generates a sub-lattice of $S$.
Also note that if $\mathbf{B}$ is a basis of $S$, any other basis $\mathbf{\bar B}$ is also a basis of $S$, provided that there exists a unimodular matrix $\mathbf{U}$, for which $\mathbf{\bar B} = \mathbf{BU}$.
My questions:
*
*How to obtain $\mathbf{B}$ from $\mathbf{A}$ and $q$?
*Is it possible that $\mathbf{B}$ is not full rank, i.e. $\text{Rank}(\mathbf{B})< m$?
*Is there any difference between the case where $q$ is prime and the case where it is composite?
Side note: As far as I understood, $S$ is the solution space of a system of linear Diophantine equations. The solution has something to do with Hermite normal forms, but I can't figure out how.
| Well, here is how it works over a field. We take $q=5.$ We will start with $A$ as a 2 by 4,
$$ A \; = \; \left( \begin{array}{cccc}
2 & 3 & 4 & 1 \\
3 & 4 & 0 & 1
\end{array} \right) $$
We begin a sequence of elementary row operations, first multiply the first row times 3 and the second by 2,
$$ \left( \begin{array}{cccc}
1 & 4 & 2 & 3 \\
1 & 3 & 0 & 2
\end{array} \right) $$
Next subtract row 1 from row 2.
$$ \left( \begin{array}{cccc}
1 & 4 & 2 & 3 \\
0 & 4 & 3 & 4
\end{array} \right) $$
then multiply the second row by 4,
$$ \left( \begin{array}{cccc}
1 & 4 & 2 & 3 \\
0 & 1 & 2 & 1
\end{array} \right) $$
Finally add row 2 to row 1,
$$ \left( \begin{array}{cccc}
1 & 0 & 4 & 4 \\
0 & 1 & 2 & 1
\end{array} \right). $$
This is the most favorable case. It allows us to place a little 2 by 2 identity matrix at the bottom when writing the null space we need
$$ \left( \begin{array}{cc}
? & ? \\
? & ? \\
1 & 0 \\
0 & 1
\end{array} \right) $$
which is then forced to become
$$ \left( \begin{array}{cc}
1 & 1 \\
3 & 4 \\
1 & 0 \\
0 & 1
\end{array} \right) $$
This can be readily filled in the way Buchmann, Lindner, Ruckert, Schneider demand at the bottom of page 2,
$$ B \; = \; \left( \begin{array}{cccc}
5 & 0 & 1 & 1 \\
0 & 5 & 3 & 4 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array} \right). $$
You can check that $AB \equiv 0 \pmod 5$ as the matrix of appropriate size.
What happens instead if the row echelon form comes out with staggered nontrivial 1's? Let us begin again with
$$ A \; = \; \left( \begin{array}{cccc}
1 & 2 & 0 & 3 \\
0 & 0 & 1 & 1
\end{array} \right) $$
It is first necessary to stagger the little 2 by 2 identity matrix in the same way, as in
$$ \left( \begin{array}{cc}
? & ? \\
1 & 0 \\
? & ? \\
0 & 1
\end{array} \right) $$
and forces us to the penultimate
$$ \left( \begin{array}{cc}
3 & 2 \\
1 & 0 \\
0 & 4 \\
0 & 1
\end{array} \right). $$
Note that it is impossible to just place this 4 by 2 as the final two columns of a 4 by 4, BLRS demand nonzero entries on the main diagonal. So what we do is simply stagger the columns with 5's in the same way, giving
$$ B \; = \; \left( \begin{array}{cccc}
5 & 3 & 0 & 2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 5 & 4 \\
0 & 0 & 0 & 1
\end{array} \right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving a literal equation containing fractions. I know this might seem very simple, but I can't seem to isolate x.
$$\frac{1}{x} = \frac{1}{a} + \frac{1}{b} $$
Please show me the steps to solving it.
| $\frac{1}{x} = \frac{b}{ab} + \frac{a}{ab}$
$\frac{1}{x} = \frac{a + b}{ab}$
$x = \frac{ab}{a + b}$
note that $\frac{1}{x} = \frac{1}{a} + \frac{1}{b}$ is possible if and only if $\frac{1}{a} + \frac{1}{b} \neq 0$. This implies that $a \neq -b$; and, hence $a + b \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplify $\tan^{-1}[(\cos x - \sin x)/(\cos x + \sin x)]$
Write the following functions in simplest form:
$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right), \quad 0<x<\pi$$
Please help me to solve this problem. I have been trying to solve this from last 3 hours. I can solve simple inverse trigonometric functions
| It is $\tan ^{-1}\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}$
We will divide in bracket with cosx to get it in tan form which will be easy for us to simplify
$\implies \tan ^{-1}\dfrac{\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}{\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}$
$\implies \tan ^{-1}\dfrac{1-\tan \left(x\right)}{1+\cos \left(x\right)}$
Now we know that $\tan ^{-1}x + \tan ^{-1}y= \dfrac{x-y}{1-xy}$ when $xy<1$
and we have the bracket in the same form as $\dfrac{\tan \left(1\right)-\tan \left(x\right)}{1-\tan \left(1\right)\tan \left(x\right)}$
So we get $\tan ^{-1} \left(\tan \left(\dfrac{π}{4}\right) + \tan \left(X\right)\right)$
i.e $\dfrac{π}{4-x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding the sum of this alternating series with factorial denominator. What is the sum of this series?
$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$
| Maybe one can do it without using power series:
$$
\begin{align}
\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{n!}
&=\sum_{n=0}^{\infty}(-1)^n\frac{n}{n!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\
&=\sum_{n=1}^{\infty}(-1)^n\frac{1}{(n-1)!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\
&=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{1}{k!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\&=0.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?
| Because $x \mapsto e^x$ is convex, $\displaystyle \left( 1- \frac{u}{n} \right)^n \leq e^{-u} \leq \left( 1+ \frac{u}{n} \right)^{-n}$, so $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx \leq \int_0^{\sqrt{n}} e^{-x^2} dx \leq \int_0^{\sqrt{n}} \left( 1+ \frac{x^2}{n} \right)^{-n} dx$.
We use the change of variables $x= \sqrt{n} \sin(\theta)$: $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx = \int_0^{\pi/2} \cos^{2n+1} (\theta) d\theta= \sqrt{n} W_{2n+1}$, where $W_{2n+1}$ is a Wallis integral.
If $p \geq \sqrt{n}$, $\displaystyle \int_0^{\sqrt{n}} \left(1+ \frac{x^2}{n} \right)^{-n} dx \geq \int_0^p \left(1+ \frac{x^2}{n} \right)^{-n} dx$. We use the change of variables $x=\sqrt{n} \tan (\theta)$ to find: $\displaystyle \int_0^{\arctan(p)} \sqrt{n} \cos^{2n+2} (\theta) d \theta \underset{p \to + \infty}{\longrightarrow} \sqrt{n} W_{2n+2}$.
Finally, $\displaystyle \sqrt{n} W_{2n+1} \leq \int_0^{\sqrt{n}} e^{-x^2}dx \leq \sqrt{n} W_{2n+2}$. But $\displaystyle W_n \underset{n \to + \infty}{\sim} \sqrt{\frac{\pi}{2n}}$ so $\displaystyle \int_0^{+ \infty} e^{-x^2}dx= \frac{\sqrt{\pi}}{2}$.
With this method, an asymptotic development of $W_n$ gives an asymptotic development of the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 0
} |
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