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Solve $\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $
Solve $$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$
Preface; I think there should be an algebraic method of solving this equation for $x$ since graphing these two graphs
We get whole number solutions such as $x=0,2,3$
So I think there is someway of manipulating this equation into a disguised quadratic somehow!
So my attempt is this:
$$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$
Let $u=x^2-5x+8$ and $u-2=x^2-5x+6$ which means we can rewrite our equation as
$$\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x + \left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x = 2^{\frac{x+4}{4}} $$
Squaring both sides we get
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x + 2\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x\left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x+ \left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+4}{2}}$$
$$\left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x+ 2(\sqrt{2})^x+ = 2^{\frac{x+4}{2}}$$
Now some little algebra
$2^{\frac{x+4}{2}}-2(\sqrt{2})^x=2^{\frac{x}{2}} \cdot 2^2 - 2 \cdot 2^{\frac{x}{2}}=2^{\frac{x}{2}} \cdot 2=2^{\frac{x+2}{2}}$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+2}{2}} $$
Square both sides again
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2\left( \sqrt{u} + \sqrt{u-2}\right)^x\left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{x+2} $$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2(2^x) = 2^{x+2} $$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x} = 2^{x+1} $$
Now I've hit a roadblock.. :(
| HINT:
Replace $\sqrt u-\sqrt{u-2}=\dfrac2{\sqrt u+\sqrt{u-2}}$ to form a Quadratic equation in $$\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}$$ which is
$$\left(\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}-1\right)^2=0$$
| {
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"url": "https://math.stackexchange.com/questions/2068147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$ I have tried factoring it already, but it doesn't seem to evolve much:
First I multiply each side by $2$:
$ 2(a^2+b^2)>2ab$
Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:
$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$
and then:
$3(a^2+b^2)>(a+b)^2$
And that's pretty much it, I'm stuck.
| We have that
$$0\le (a-b)^2=a^2+b^2-2ab.$$ Thus we have
$$2ab\le a^2+b^2.$$
Now, if $ab$ is positive then
$$ab< 2ab\le a^2+b^2.$$ And if $ab$ is negative then
$$ab< 0 <a^2+b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Special Orthogonal Group $SO(2)$ The special orthogonal group for $n=2$ is defined as:
$$SO(2)=\big\{A\in O(2):\det A=1\big\}$$
I am trying to prove that if $A\in SO(2)$ then:
$$A=\left(\begin{array}{cc}
\cos\theta& -\sin\theta\\
\sin\theta&\cos\theta
\end{array}\right)$$
My idea is show that $\Phi:S^1\to SO(2)$ defined as:
$$z=e^{\theta i}\mapsto \Phi(z)=\left(\begin{array}{cc}
\cos\theta& -\sin\theta\\
\sin\theta&\cos\theta
\end{array}\right)$$
is an isomorphism of Lie groups. It is easy prove that is an monomorphism of Lie groups. How can I prove that is also surjective?
| Let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in \mathrm{SO}(2)$. Then,
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
and
$$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc=1.$$
Thus, $\mathrm{SO}(2)$ is the subset of $\mathbb{R}^4$ satisfying the following four equations:
$$
\begin{align*}
a^2+c^2 &= 1 \\
b^2+d^2 &= 1\\
ad-bc &= 1\\
ab+cd &= 0.
\end{align*}
$$
The first two equations imply that $(a,c)$ and $(b,d)$ lie on a circle, so
$$a=\cos\alpha,\quad c=\sin\alpha,\quad b=\cos\beta,\quad d=\sin\beta$$
for some angles $\alpha,\beta\in\Bbb R$. Inserting in the last two equations, we get
$$
\begin{align*}
\cos\alpha\sin\beta-\cos\beta\sin\alpha &= 1 \\
\cos\alpha\cos\beta+\sin\alpha\sin\beta &= 0.
\end{align*}
$$
Using the angle sum trigonometric identities, these equations are
$$
\begin{align*}
\sin(\beta-\alpha) &= 1 \\
\cos(\beta-\alpha) &= 0.
\end{align*}
$$
Hence, $\beta-\alpha\in \pi/2+2\pi\Bbb Z$ and we get
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Which of these quantity systems corresponds to sigma algebra? Given is the set $Ω = \left\{ 3,4,5,6,7\right\} $.
Determine which of the given quantity systems corresponds to an sigma algebra and justify it.
$1. \left\{\left\{\right\},\left\{3\right\},\left\{5\right\},\left\{3,4\right\},\left\{3,4,6,7\right\},\left\{3,4,5,6,7\right\}\right\}$
$2. \left\{\left\{\right\},\left\{5\right\},\left\{3,4,6,7\right\},\left\{3,4,5,6,7\right\}\right\}$
$3. \left\{\left\{\right\},\left\{3\right\},\left\{4\right\},\left\{5\right\},\left\{6\right\},\left\{7\right\},\left\{3,4\right\},\left\{3,5\right\},\left\{3,6\right\},\left\{3,7\right\},\left\{4,5\right\},\left\{4,6\right\},\left\{4,7\right\},\left\{5,6\right\},\left\{5,7\right\},\left\{6,7\right\},\left\{3,4,5\right\},\left\{3,4,6\right\},\left\{3,4,7\right\},\left\{3,5,6\right\},\left\{3,5,7\right\},\left\{3,6,7\right\},\left\{4,5,6\right\},\left\{4,5,7\right\},\left\{4,6,7\right\},\left\{5,6,7\right\},\left\{3,4,5,6\right\},\left\{3,4,5,7\right\},\left\{3,4,6,7\right\},\left\{3,5,6,7\right\},\left\{4,5,6,7\right\},\left\{3,4,5,6,7\right\}\right\}$
$4. \left\{\left\{\right\},\left\{5\right\}\right\}$
I think that 1. and 3. are corresponded. Is that correct and if yes how to justify that?
| *
*is not a $\sigma$-algebra because $\{3\}\cup \{5\}=\{3,5\}$ is not in 1.
*is a $\sigma$-algebra because $\{\}$, $\Omega$, and all the possible intersections and unions of the elements of 2. are in 2.
*works because all the $2^{\Omega}=32$ subsets (of $\Omega $) are in 3.
*would not work because $\Omega$ is not in 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's Taylor series is close to this.
However, I'm not sure how to approach this. I tried to take several derivates of second term to see its value on $x_0 = 0$, but it becomes complicated and I don't see general formula for $n$-th derivative at point zero to find $a$ and $b$.
| Your function is
$$\cos(x)-\frac{a}{b}-\frac{1-\frac{a}{b} }{ bx^2+1 }$$
$$=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{a}{b}-1+\frac{a}{b}+bx^2-ax^2-b(b-a)x^4+x^4\epsilon(x).$$
thus, we need
$b-a=-\frac{1}{2}$ and $b(b-a)=-\frac{1}{24}$.
which gives
$b=\frac{1}{12}$ and $a=\frac{7}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
How to calculate expected value for piecewise constant distribution function? The distribution function of a discrete random variable X is given
$F_X(x)=\begin{cases} 0, &x<1\\ \frac{5}{13},& 1\leq x< 2 \\ \frac{10}{13}, & 2\leq x<3 \\ \frac{11}{13}, & 3\leq x<4 \\ 1, & 4\leq x \end{cases} $
$A=(X=2)\cup (X=4)$
Calculate: $P(A)$ and $E(X)$
I was thinking to solve $P(A)$ with formula: $P(a)=\begin{pmatrix} n \\ a \end{pmatrix} p^a (1-p)^{n-a} $, but I dont $p$ and $n$. Which formula I should use?
| the random variable $X$ can take four values, which are exactly the points of discontinuity of $F_X$:
$$
\mathbb P (X=1)= \frac 5 {13}, \quad \mathbb P (X=2)=\frac {10} {13}- \frac 5 {13}, \quad
\mathbb P (X=3)=\frac {11} {13} -\frac {10} {13}, \quad \mathbb P (X=4)=1- \frac {11} {13}. \quad
$$
Therefore
$$
\mathbb P(A)=\mathbb P(X=2)+\mathbb P(X=4) = \frac 7 {13},
$$
and
$$
\mathbb E [X]= 1 \cdot \frac {5}{13}+2 \cdot \frac {5}{13}+3 \cdot \frac {1}{13}+4 \cdot \frac {2}{13}=2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Matrix determinant lemma derivation While reading this wikipedia article on the determinant lemma, I stumbled upon this expression (in a proof section):
\begin{equation}
\begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix} =
\begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
\end{equation}
Although I see that this equation "works", I'm interested in HOW this thing was invented. For example, why we have $u$ term in a central block matrix of the left side?
UPD
A little clarification of the question above.
Let
\begin{equation}L = \begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix} \end{equation}
I see that
\begin{equation}
L^{-1}=
\begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\end{equation}
and hence the first equation looks like
\begin{equation}
L\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{u} \\ 0 & 1\end{pmatrix}L^{-1} =
\begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
\end{equation}
I see that $\det(L) = \det(L^{-1}) = 1 $. Hence determinants of RHS and LHS are equal as well.
What I do not understand is how we jumped from simple $\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{0} \\ 0 & 1\end{pmatrix}$ or $\begin{pmatrix}\mathbf{I} & \mathbf{u} \\ \mathbf{-v}^T & 1\end{pmatrix}$ to $\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{u} \\ 0 & 1\end{pmatrix}$ for a central part of LHS.
Thank you.
| I have no idea how this was invented and what was the original motivation, but let me outline a different proof for the formula $\det(I + uv^T) = 1 + v^T u$ which I think is much less magical and more natural.
Set
$$ B = uv^T = \begin{pmatrix} u_1 v_1 & \dots & u_1 v_n \\
u_2 v_1 & \dots & u_2 v_n \\
\vdots & \ddots & \vdots \\
u_n v_1 & \dots & u_n v_n \end{pmatrix} \in M_n(\mathbb{F}). $$
Let us start by finding the characteristic polynomial $p_B(x) = \det(B - xI)$ of $B$. Since $B$ has $\operatorname{rank}(B) \leq 1$, we know that it has at least $n - 1$ eigenvectors associated to the eigenvalue $0$. Since the sum of all eigenvalues must be $\operatorname{tr}(B) = v^T u$, we see that
$$ p_B(x) = \det(B - xI) = (-1)^{n} x^{n-1}(x - v^T u). $$
Now plug in $x = (-1)$ and deduce the required formula:
$$ p_B(1) = \det(B - (-1)I) = \det(B + I) = \det(I + uv^T) = (-1)^n (-1)^{n-1} (-1 - v^T u) = 1 + v^T u.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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The valid interval of the maclaurin series for $\frac{1}{1+x^2}$ The Maclaurin series for $\frac{1}{1-x}$ is $1 + x + x^2 + \ldots$ for $-1 < x < 1$.
To find the Maclaurin series for $\frac{1}{1+x^2}$, I replace $x$ by $-x^2$.
The Maclaurin series for $\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots$.
This is valid for $-1 < -x^2 < 1$ if I replace $x$ by $-x^2$. So if I multiply each side by $-1$, I get $-1 < x^2 < 1$. If I take the square roots, I get $i < |x| < 1$. And I am stuck.
And my book says that this Maclaurin series is valid for $-1 < x < 1$ anyway. There is no further explanation.
How can I derive $-1 < x < 1$ from $i < |x| < 1$? Please help.
| Taking square roots in $a<x^2<b$ to get $\sqrt a < |x| < \sqrt b$ is valid if $a\ge0$ and $b\ge0$.
But notice that the solution of $-4< x^2$ is $-\infty<x<\infty$ since every square of a real number is $>-4.$
Since the square of a real number is never negative, the inequality $-1<x^2<1$ is equivalent to $x^2<1.$
That is equivalent to $x^2-1<0$, which is $(x-1)(x+1)<0$. A product of two numbers is negative only if one of them is negative and the other is positive. Since the second factor is bigger than the first, it needs to be the one that is positive. So we have $x+1>0$ and $x-1<0$. Thus $x>-1$ and $x<1.$
| {
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"url": "https://math.stackexchange.com/questions/2074149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x)^2}$. (Ignoring terms which have powers of x greater than $x^n$)
So one can say that
coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$
=coefficient of $x^n$ in $(1-x+x^2)^2(1-x)^{-4}$
Is there a shorter way.
| Such coefficient is clearly
$$ 2n+\sum_{k=1}^{n-1} k(n-k) = \frac{n(n^2+11)}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$ proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$
for $t\leq 1,$ left side expression is $>0$
for $t\geq 1,$ left side expression $t^5(t-1)+t^3(t-1)+t(t-1)+0.4$ is $>0$
i wan,t be able to prove for $0<t<1,$ could some help me with this
| If $0< t < 1$
$\frac {t^7 + 1}{t+1} \ge .6 \iff t^7 + 1 \ge .6t + .6 \iff t^7 - .6t \ge -.4$
$\frac {d(t^7 - .6t)}{dt} = 7t^6 - .6 = 0$ if $t = \sqrt[6] \frac 6{70}$
$\frac{d^2(t^7 - .6t)}{d^2t} = 42t^5 > 0 $ if $t > 0$ so $t = \sqrt[6] \frac 6{70}$ is a minimum value of $t^7 - .6t$. And so $t^7 - .6t \ge \sqrt[6] \frac 6{70}^7 -.6*\sqrt[6] \frac 6{70} = \sqrt[6] \frac 6{70}(\frac 6{70} - .6)\approx -.341 > -.4$
so $t^7 - .6t \ge -.4$ for $0 < t < 1$.
So $\frac {t^7 + 1}{t+1}= t^6 - t^5 + t^4 - t^3 +t^2 -t + 1 \ge .6$
and $t^6 - t^5 + t^4 - t^3 +t^2 -t + .4 > 0$ for $0 < t < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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proving $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{n-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}$ proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdots (-1)^{n-1}\frac{n.(n-1)\cdot (n-2)\cdots(n-m+2)}{(m-1)!}$
Added
writting LHS as $\displaystyle \binom{n}{0}-\left(\binom{n-1}{0}+\binom{n-1}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\cdots \cdots +(-1)^{n-1}\left(\binom{n-1}{m-2}+\binom{n-1}{m-1}\right)=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \binom{n}{0}-\binom{n-1}{0}+\binom{n-1}{1}-\cdots +(-1)^{n-1}\binom{n-1}{m-2}-\left(\binom{n-1}{1}-\binom{n-1}{2}+\cdots +(-1)^n\binom{m-1}{m-1}\right)$
wan,t be able to solve after that, help me to solve it
| Extended HINT: The result is incorrect as originally stated; it should read
$$\binom{n}0-\binom{n}1+\binom{n}2-\ldots+(-1)^{\color{crimson}m-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}\;,$$
or, more compactly,
$$\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^{m-1}\binom{n-1}{m-1}\;.\tag{1}$$
Fix $n\in\Bbb N$. For $m=1$ the desired result is
$$\binom{n}0=(-1)^0\binom{n-1}0\;,$$
which is indeed true, since both sides are equal to $1$. Suppose as an induction hypothesis that $(1)$ holds for some $m$; for the induction step we want to prove that
$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n-1}m\;.\tag{2}$$
Using the induction hypothesis we can rewrite the lefthand side of $(2)$:
$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n}m+\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}\;,$$
so to complete the induction step we need only show that
$$(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}=(-1)^m\binom{n-1}m\;.$$
This is easily done using one of the most basic identities involving binomial coefficients.
Added: After some thought I realize that $(1)$ can be proved by direct calculation:
$$\begin{align*}
\sum_{k=0}^{m-1}(-1)^k\binom{n}k&=\sum_{k=0}^{m-1}(-1)^k\left(\binom{n-1}k+\binom{n-1}{k-1}\right)\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-1}(-1)^k\binom{n-1}{k-1}\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=1}^{m-1}(-1)^k\binom{n-1}{k-1}\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-2}(-1)^{k+1}\binom{n-1}k\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\
&=(-1)^{m-1}\binom{n-1}{m-1}+\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\
&=(-1)^{m-1}\binom{n-1}{m-1}\;.
\end{align*}$$
| {
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How to calculate Limit of $(1-\sin x)^{(\tan \frac{x}{2} -1)}$ when $x\to \frac{\pi}{2}$.
How to calculate Limit of $(1-\sin x)^{(\tan \frac{x}{2} -1)}$ when $x\to \frac{\pi}{2}$.
We can write our limit as $\lim_{x\to \frac{\pi}{2}}e^{(\tan \frac{x}{2} -1) \log(1-\sin x)}~ $ but I can not use L'Hopital rule.
Is there another way?
| Making the substitution $
x = \dfrac{\pi}{2} + y$ the required limit is
$\lim_{y \to 0} \exp h(y)$ where $h(y)= \ln(1-\cos y) \left( \tan(\pi/4 + y/2) - 1 \right) = \ln(1-\cos y) \times \dfrac{2\tan(y/2)}{1-\tan(y/2)}$.
Since $1-\cos y = 2 \sin^2(y/2)$ we have $$h(y) = (\sqrt{2}\sin(y/2)) \ln(2\sin^2(y/2)) \times \dfrac{2}{\sqrt{2}} \times \dfrac{\dfrac{\tan (y/2)}{y/2}}{\dfrac{\sin(y/2)}{y/2}} \times \dfrac{1}{1-\tan(y/2)} $$.
Since $\lim_{x\to0}x\ln(x^2) = 2\lim_{x \to 0} x \ln |x| = 0$ and so we have $\lim_{y\to 0}(\sqrt{2}\sin(y/2)) \ln(2\sin^2(y/2)) = 0$ and $\lim_{y\to 0}h(y) = 0 \times \dfrac{2}{\sqrt{2}} \times \dfrac{1}{1} \times 1 = 0.$
So the required limit is 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$ Is it possible to know if $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$ without using decimal numbers?
| It is not hard to verify following inequalities (just power both sides and it should result into simple inequalities in natural numbers only):
\begin{align}
\frac{4}{3} &< \sqrt{2} < \frac{5}{3}\\
\frac{4}{3} &< \sqrt[3]{3} < \frac{5}{3}\\
\frac{4}{3} &< \sqrt[5]{5} < \frac{5}{3}\\
\end{align}
Summing these up will give you
$$
4 < \sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5} < 5\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Consider a function $f(x, y) = Ax^5 + Bx^4y + Cx^3y^2 + Dx^2y^3 + xy^4 − y^5$
Consider a function $f(x, y) = Ax^5 + Bx^4y + Cx^3y^2 + Dx^2y^3 + xy^4 − y^5$
where $A$, $B$, $C$, $D$ are unspecified real numbers. Determine the values of
$A$, $B$, $C$, $D$ such that $f(x,y)$ satisfies $f_{xx}(x,y) + f_{yy}(x,y) = 0$
What I did so far, I took double derivative of x and y:
$f_{xx}(x,y) = 20Ax^3+12Bx^2y+6Cxy^2+2Dy^3$
$f_{yy}(x,y) = 2Cx^3+6Dx^2y+12xy^2-20y^3$
How can I satisfies $f_{xx}(x,y) + f_{yy}(x,y) = 0$? There are no terms that cancels, subtracts or add.
| Add up and you get
$[20A+2C]x^3 + [12B+6D]x^2y + [6C+12]xy^2 + [2D-20]y^3=0$
So
$D = 10$
$C = -2$
$B = -5$
$A = \frac{1}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.
| \begin{align}
I:&=\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dz\left(x^2-2xy\right)\exp\left(-3x^2 - 2y^2 - z^2 - 2xy\right)\\
&=\sqrt{\pi}\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\left(x^2-2xy\right)\exp\left(-3x^2-2y^2-2xy\right) \tag1\\
\end{align}
We split the integral in $(1)$ into two parts:
\begin{align}
I_1&=\sqrt{\pi}\int_{-\infty}^{\infty}dxx^2\exp\left(-3x^2\right)\int_{-\infty}^{\infty}dy\exp\left(-2y^2-2xy\right) \tag2
\end{align}
We complete the square:
\begin{align}
-2y^2-2xy&=-2\left(y^2+xy\right)\\
&=-2\left(\left(y+\frac{x}{2}\right)^2-\frac{x^2}{4}\right)\\
&=-2\left(y+\frac{x}{2}\right)^2+\frac{x^2}{2}
\end{align}
Thus, $(2)$ becomes
\begin{align}
I_1&=\sqrt{\pi}\int_{-\infty}^{\infty}dxx^2\exp\left(-\frac{5}{2}x^2\right)\int_{-\infty}^{\infty}dy\exp\left(-2\left(y+\frac{x}{2}\right)^2\right)\\
&=\frac{\pi}{\sqrt{2}}\int_{-\infty}^\infty dxx^2\exp\left(-\frac{5}{2}x^2\right)\\
&=\frac{\pi^{\frac{3}{2}}}{5\sqrt{5}} \tag3
\end{align}
Then, we look at the second integral:
\begin{align}
I_2&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-3x^2\right)\int_{-\infty}^\infty dyy\exp\left(-2y^2-2xy\right)\\
&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty dyy\exp\left(-2\left(y+\frac{x}{2}\right)^2\right)\\
&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty du\left(u-\frac{x}{2}\right)\exp\left(-2u^2\right) \tag4
\end{align}
Splitting $(4)$ again, we have
\begin{align}
I_{2,1}&=2\sqrt{\pi}\int_{-\infty}^\infty dx x\exp\left(-2x^2\right)\int_{-\infty}^\infty duu\exp\left(-2u^2\right)\\
&= 0
\end{align}
\begin{align}
I_{2,2}&=\sqrt{\pi}\int_{-\infty}^\infty dx x^2\exp\left(-\frac{5}{2}x^2\right)\int_{-\infty}^\infty du\exp\left(-2u^2\right)\\
&=\frac{\pi}{\sqrt{2}}\int_{-\infty}^\infty dx x^2\exp\left(-\frac{5}{2}x^2\right)\\
&=\frac{\pi^{\frac{3}{2}}}{5\sqrt{5}} \tag5
\end{align}
Adding up $(3)$ and $(5)$, we have
\begin{align}
\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dz\left(x^2-2xy\right)\exp\left(-3x^2 - 2y^2 - z^2 - 2xy\right)=\frac{2\pi^{\frac{3}{2}}}{5\sqrt{5}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$
Does this series converge? ($p$ is finite number greater than $1$)
| For $p>1$ the series converges.
Lets take the general geometric series;
$$|x|<1\\a_1(1+x+x^2+x^3+...)=\frac{a_1}{1-x}\\a_1(1+2x+3x^2+4x^3+...)=a_1(1+x+x^2+x^3+...)'=(\frac{a_1}{1-x})'=\frac{a_1}{(1-x)^2}$$
So in our case;
$$a_1=\frac{1}{p}\ \ ,\ \ x=\frac{1}{p}\\\frac{1}{p}+\frac{2}{p^2}+...=\frac{1}{p}\cdot\frac{1}{(1-\frac{1}{p})^2}=\frac{1}{p(1-\frac{2}{p}+\frac{1}{p^2})}=\frac{1}{\frac{p^2-2p+1}{p}}=\frac{p}{(p-1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Compute the $n$-th power of triangular $3\times3$ matrix I have the following matrix
$$
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}
$$
and I am asked to compute its $n$-th power (to express each element as a function of $n$). I don't know at all what to do. I tried to compute some values manually to see some pattern and deduce a general expression but that didn't gave anything (especially for the top right). Thank you.
| Here is another variation based upon walks in graphs.
We interpret the matrix $A=(a_{i,j})_{1\leq i,j\leq 3}$ with
\begin{align*}
A=
\begin{pmatrix}
1 & 2 & 3\\
\color{grey}{0} & 1 & 2\\
\color{grey}{0} & \color{grey}{0} & 1
\end{pmatrix}
\end{align*}
as adjacency matrix of a graph with three nodes $P_1,P_2$ and $P_3$ and for each entry $a_{i,j}\neq 0$ with a directed edge from $P_i$ to $P_j$ weighted with $a_{i,j}$.
Note: When calculating the $n$-th power $A^n=\left(a_{i,j}^{(n)}\right)_{1\leq i,j\leq 3}$ we can interpret the element $a_{i,j}^{(n)}$ of $A^n$ as the number of (weighted) paths of length $n$ from $P_i$ to $P_j$. The entries of $A=(a_{i,j})_{1\leq i,j\leq 3}$ are the weighted paths of length $1$ from $P_i$ to $P_j$.
See e.g. chapter 1 of Topics in Algebraic Combinatorics by Richard P. Stanley.
Let's look at the corresponding graph and check for walks of length $n$.
*
*We see there are no directed edges from $P_2$ to $P_1$ and no directed edges from $P_3$ to $P_2$ and from $P_3$ to $P_1$ which implies there are no walks of length $n$ either. So, $A^n$ has due to the specific triangle structure of $A$ necessarily zeroes at the same locations as $A$.
\begin{align*}
A^n=
\begin{pmatrix}
. & . & .\\
\color{grey}{0} & . & .\\
\color{grey}{0} & \color{grey}{0} & .
\end{pmatrix}
\end{align*}
*It is also easy to consider the walks of length $n$ from $P_i$ to $P_i$. There is only one possibility to loop along the vertex weighted with $1$ from $P_i$ to $P_i$ and so the entries $a_{i,i}^{(n)}$ are
\begin{align*}
1\cdot 1\cdot 1\cdots 1 = 1^n=1
\end{align*}
and we obtain
\begin{align*}
A^n=
\begin{pmatrix}
1& . & .\\
\color{grey}{0} & 1 & .\\
\color{grey}{0} & \color{grey}{0} & 1
\end{pmatrix}
\end{align*}
and now the more interesting part
*
*$P_1$ to $P_2$:
The walks of length $n$ from $P_1$ to $P_2$ can start with zero or more loops at $P_1$ followed by a step (weigthed with $2$) from $P_1$ to $P_2$ and finally zero or more loops at $P_2$. All the loops are weighted with $1$. There are $n$ possibilities to walk this way
\begin{align*}
a_{1,2}^{(n)}=2\cdot 1^{n-1}+1\cdot 2\cdot 1^{n-2}+\cdots +1^{n-2}\cdot 2\cdot 1+1^{n-1}\cdot 2=2n
\end{align*}
*$P_2$ to $P_3$:
Symmetry is trump. When looking at the graph we observe the same situation as before from $P_1$ to $P_2$ and conclude
\begin{align*}
a_{2,3}^{(n)}=2n
\end{align*}
*
*$P_1$ to $P_3$:
Here are two different types of walks of length $n$ possible. The first walk uses the weight $3$ edge from $P_1$ to $P_3$ as we did when walking from $P_1$ to $P_2$ along the weight $2$ edge. This part gives therefore
\begin{align*}
3\cdot 1^{n-1}+1\cdot 3\cdot 1^{n-2}+\cdots +1^{n-2}\cdot 3\cdot 1+1^{n-1}\cdot 3=3n\tag{1}
\end{align*}
The other type of walk of length $n$ uses the hop via $P_2$. We observe it is some kind of concatenation of walks as considered before from $P_1$ to $P_2$ and from $P_2$ to $P_3$. In fact there are $\binom{n}{2}$ possibilities to place two $2$'s in a walk of length $n$. All other steps are loops at $P_1,P_2$ and $P_3$ and we obtain
\begin{align*}
\binom{n}{2}\cdot 2\cdot 2=2n(n-1)\tag{2}
\end{align*}
Summing up (1) and (2) gives
\begin{align*}
a_{1,3}^{(n)}=3n+2n(n-1)=n(2n+1)
\end{align*}
and we finally obtain
\begin{align*}
A^n=\left(a_{i,j}^{(n)}\right)_{1\leq i,j\leq 3}=\begin{pmatrix}
1& 2n & n(2n+1)\\
\color{grey}{0} & 1 & 2n\\
\color{grey}{0} & \color{grey}{0} & 1
\end{pmatrix}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 6,
"answer_id": 5
} |
Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large number of steps I achieved the correct answer:
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx=\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}+C$$
I was able to check my answer using Mathematica.
expr = D[1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(2 x^2), x];
Assuming[x >= 3, FullSimplify[expr]]
Which returned the correct response:
Sqrt[-9 + x^2]/x^3
Mathematica returns the following answer:
Integrate[Sqrt[x^2 - 9]/x^3, x, Assumptions -> x >= 3]
-(Sqrt[-9 + x^2]/(2 x^2)) - 1/6 ArcTan[3/Sqrt[-9 + x^2]]
Which I can write to make more clear.
$$-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}+D$$
Now, you can see that part of my answer is there, but here is my question. How can I show that
$$\frac16\sec^{-1}\frac{x}{3}\qquad\text{is equal to}\qquad -\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}$$
plus some arbitrary constant? What identities can I use? Also, can anyone share the best web page for inverse trig identities?
Update: I'd like to thank everyone for their help. The Trivial Solution's suggestion gave me:
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$
Then the following identity came to mind:
$$\tan^{-1}x+\tan^{-1}\frac1x=\frac{\pi}{2}$$
So I could write:
\begin{align*}
\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}
&=\frac16\tan^{-1}\frac{\sqrt{x^2-9}}{3}-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac16\left(\frac{\pi}{2}-\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac{\pi}{12}-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}
\end{align*}
Using Olivier's and Miko's thoughts, I produced this plot in Mathematica.
Plot[{1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(
2 x^2), -(1/6) ArcTan[3/Sqrt[x^2 - 9]] - Sqrt[x^2 - 9]/(
2 x^2)}, {x, -6, 6},
Ticks -> {Automatic, {-\[Pi]/12, \[Pi]/12}}]
Which shows that the two answers differ by $\pi/12$, but only for $x>3$.
| What you are asking to prove is incorrect, I believe.
By the substitution, we have that
$$\frac{x}{3}=\sec(\theta)\Leftrightarrow\frac{3}{x}=\cos(\theta).$$
By the Pythagorean identity,
$$\sin(\theta)=\sqrt{1-\frac{9}{x^2}}=\sqrt{\frac{x^2-9}{x^2}}.$$
Therefore,
$$ \tan(\theta)=\sqrt{\frac{x^2-9}{x^2}}\frac{x}{3}=\frac{1}{3}\sqrt{x^2-9}$$
Hence,
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{1}{3}\sqrt{x^2-9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.
claim 1 : If $a_n>0$ and
${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime.
The table below illustrates what I am seeing:
\begin{array}{| l | l | l | l }
\hline
n & a_n & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline
0 & 1 & .333333.. & 2.2360679.. \\
1 & 1 & .333333.. & 3 \\
2 & 2 & .666666.. & 3 \\
3 & 6 & 2 & 5 \\
4 & 12 & 4 & 7 \\
6 & 30 & 10 & 11 \\
7 & 42 & 14 & 13 \\
9 & 72 & 24 & 17 & \\
10 & 90 & 30 & 19 \\
12 & 132 & 44 & 23 \\
15 & 210 & 70 & 29 \\
16 & 240 & 80 & 31 \\
19 & 342 & 114 & 37 \\
21 & 420 & 140 & 41 \\
22 & 462 & 154 & 43 \\
\hline
\end{array}
claim 2: If $a_n<0$ then $2a_n+5$ is composite
| Claim 1 is vaccuously true, since given any $n$, $a_{n}\in\mathbb{Z}$.
It also looks like n=14 is a counterexample to claim 2.
Proof:
$3-14=-11$ and $2(-11)+5=-17$ has positive divisors $1$ and $17$ only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Using the Arithmetic Mean-Geometric Mean Inequality Let a, b, c be positive real numbers. Prove that
$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq \sqrt{3(a^{2}+b^{2}+c^{2})}$
| Rewrite as $a^2b^2 + b^2c^2 + c^2a^2 \ge \sqrt{3(a^4b^2c^2 + a^2b^4c^2 + a^2b^2c^4)}$
Let $x = a^2b^2, y = b^2c^2, z = c^2a^2$.
Therefore, now we have to prove : $x + y + z \ge \sqrt{3(xy + yz + zx)}$
Squaring both sides, $(x + y + z)^2 \ge 3(xy + yz + zx)$
$x^2 + y^2 + z^2 \ge xy + yz + zx$.
Now, use A.M. $\ge$ G.M. $(\frac{x^2 + y^2}2 \ge xy)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using the Arithmetic Mean-Geometric Mean Inequality. Let a, b, c be real numbers. Prove that:
$(a+b-c)^{2}(b+c-a)^{2}(c+a-b)^{2}\geq(a^{2}+b^{2}-c^{2})(b^{2}+c^{2}-a^{2})(c^{2}+a^{2}-b^{2}) $
| We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$.
If $a^2+b^2-c^2<0$ and $a^2+c^2-b^2<0$ so $2a^2<0$, which is contradiction.
Thus, we can assume $a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $b^2+c^2-a^2\geq0$.
Since $(a+b-c)^2(a+c-b)^2-(a^2+b^2-c^2)(a^2+c^2-b^2)=2(b-c)^2(b^2+c^2-a^2)\geq0$,
we obtain:
$$\prod\limits_{cyc}(a+b-c)^2(a+c-b)^2\geq\prod\limits_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2),$$
which is
$$\prod\limits_{cyc}(a+b-c)^2\geq\prod\limits_{cyc}(a^2+b^2-c^2)$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality with a+b+c=1 Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$
Progress: This is equivalent to show that
$$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$
I'm not sure how to proceed further. Also equality doesn't occur when $a=b=c$, which makes me more confused.
Edit: The original inequality is as follows:
Let $a_1,a_2,\cdots{},a_n$ be $n>2$ positive reals such that $a_1+a_2+\cdots{}+a_n=1$. Prove that,
$$\sum_{k=1}^n \dfrac{\prod\limits_{j\ne k}a_j}{a_k+n+2}\le \dfrac{1}{(n-1)^2}.$$
| Notice that the conditions imply that $0\leq a,b,c\leq 1$; in particular, we have
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{bc}{5}+\dfrac{ca}{5}+\dfrac{ab}{5} \leq \frac{b+c+a}{5} < \dfrac{1}{4}.$$
I don't think this works for the general case though, because for $n\geq 4$, we have $(n-1)^2\geq n+2$.
Now for the general case, by AM-GM inequality
$$\prod_{j\neq k} a_j \leq \left(\frac{\sum_{j\neq k}a_j}{n-1}\right)^{n-1}=\left(\frac{1-a_k}{n-1}\right)^{n-1}\leq \frac{1}{(n-1)^{n-1}},$$
and thus
$$\sum_{k=1}^n \dfrac{\prod_{j\ne k}a_j}{a_k+n+2}\leq \frac{n}{(n+2)(n-1)^{n-1}} \leq \frac{1}{(n-1)^{n-1}}\leq \dfrac{1}{(n-1)^2}$$since $n-1\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{\dfrac{(4-x)(4-x)}{(4+x)(4-x)}}dx\\
&=\int\sqrt{\dfrac{(4-x)^2}{16-x^2}}dx\\
&=\int\dfrac{4-x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16-x^2}}dx-\int\dfrac{x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16(1-x^2/16)}}dx+\int\dfrac{-2x}{2\sqrt{16-x^2}}dx\\
&=\underbrace{\int\dfrac{1}{\sqrt{1-(x/4)^2}}dx}_{\text{set $t=x/4$}}+\sqrt{16-x^2}+C\\
&=\underbrace{\int\dfrac{4}{\sqrt{1-t^2}}dt}_{\text{set $t=\sin \theta$}}+\sqrt{16-x^2}+C\\
&=\int\dfrac{4\cos\theta}{\sqrt{\cos^2\theta}}d\theta+\sqrt{16-x^2}+C\\
\end{align}
So finally, I get
$$G(x)=\pm\theta+\sqrt{16-x^2}+C'.$$
With wolframalpha I found some different answer. Could you provide any suggestions?
Also, multiplying by $4-x$ is it correct at the beginning? because I should say then that $x\neq 4$.
| First of all, you are right that there is trouble in multiplying by $\frac{4-x}{4-x}$ when $x=4$. But why bother with the domain $\langle-4,4]$ in the first place? You can change the integrand at one point without changing integral, so that one point is irrelevant. Thus, choose $D_g = \langle-4,4\rangle$.
The only other issue is the one that I already mentioned in the comments. If you substitute $t = \sin\theta \in\langle -1,1\rangle$, just choose $\theta$ to be in $\langle-\frac\pi 2,\frac\pi 2\rangle$ (substitution is natural bijection that way). Then you have that $\cos\theta>0$, and thus $\sqrt{\cos^2\theta}=\cos\theta$.
So, your final result should be $$4\theta + \sqrt{16-x^2} + C = 4\arcsin\frac x4 + \sqrt{16-x^2} + C$$ and if you differentiate it, you can see that the result is just fine.
Although your procedure is fine, in this case you might want to make trigonometric substitution much sooner:
$$\int\sqrt{\frac{4-x}{4+x}}\,dx = \int\frac{4-x}{\sqrt{16-x^2}}\, dx = [x = 4\sin t] =\int \frac{4-4\sin t}{4\cos t}\cdot 4\cos t\, dt=\\ = 4(t+\cos t)+C = 4\arcsin\frac x4 +4\cos(\arcsin\frac x4) + C = 4\arcsin\frac x4 +4\sqrt{1-\frac{x^2}4} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find all such $n$ that $a+b+c+d=0\implies a^7+b^7+c^7+d^7=0$ in $\mathbb{Z}/n\mathbb{Z}$
The problem is to specify all such $n>1$ that for any $a,b,c,d\in\mathbb{Z}/n\mathbb{Z}$ the following implication stands:
$$a+b+c+d=0\implies a^7+b^7+c^7+d^7=0.$$
One can note that when $n=7$ we have $(a+b+c+d)^7=a^7+b^7+c^7+d^7$ so the above implication stands.
If $n=2,3$ then $x^7=x$, so it's also true.
For $n=4$ it's false. Counterexample: $(a,b,c,d)=(2,3,3,0)$.
How to find other such $n$ and prove that those are the only ones?
| The examples $(2,-1,-1,0)$ and $(3,-1,-1,-1)$ show that you need $2^7=2$ and $3^7=3$. This means that $n$ must divide both $2^7-2$ and $3^7-3$, and their greatest common divisor is $42$.
It turns out that $k^7-k$ is divisible by $42$ for every natural $k$ (divisible by $7$ by Fermat's little theorem, by $2$ and $3$ by checking), so the $n$ that work are $1,2,3,6,7,14,21$ and $42$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Power series expansion of $\frac{z}{(z^3+1)^2}$ around $z=1$ I want to expand $f(z)=\frac{z}{(z^3+1)^2}$ around $z=1$. That is, I want to find the coefficients $c_n$ such that $f(z) = \sum_{n=0}^\infty c_n (z-1)^n$.
So far, my first strategy was using long division after I expanded both the numerator and the denominator about $z=1$. To this end, we first apply a substitution $u:= z-1$, after we'll expand around $u=0$.
For the denominator we simply get $$1 + u$$.
For the numerator we have
$$
\frac{1}{\big((u+1)^3+1\big)^2} = \frac{1}{2^2} \frac{1}{\big(1 + \frac{u^3+3u(u+1)}{2}\big)^2} \quad \mbox{put $x:= \frac{u^3+3u(u+1)}{2}$}\\
=-\frac{1}{2^2}\frac{\partial d}{\partial x}\frac{1}{x+1} \\
= -\frac{1}{2^2}(-1+2x-3x^2 +\ \ldots\ ) \\
=-\frac{1}{2^2}\big(1+6u-21u^2-52u^3+\ \ldots \ \big)
$$
After using long division, we could try to see a pattern in the coefficients of $u^n$ but this doesn't seem to bear any quick results.
My second attempt was using the binomial theorem to expand the power series of $$f(z) = \sum_{n=0}^\infty a_n z^{3n+1} \quad \mbox{with} \quad a_n = (n+1)(-1)^n$$ about $z=1$.
To this end, we need to reorganize the $u^n$ in the following sum
$$
\sum_{n=0}^\infty (u+1)^n = \sum_{n=0}^\infty \sum_{k=0}^n \binom nk u^k .
$$
This seems fairly laborious. Below we have the terms for increasing $n$.
$$
1 \\
1 + u \\
1 + 2u + u^2 \\
1 + 3u + 3u^2 + u^3 \\
\vdots \\
1 + mu + \binom m2 + \ldots + mu^{m-1} + u^m \\
1 + (m+1)u + \binom{m+1}{2}u^2 + \ldots + (m+1)u^m + u^{m+1} \\
1 + (m+2)u + \binom{m+2}{2}u^2 + \ldots + (m+2)u^{m+1} + u^{m+2} \\
\vdots
$$
If we look at the terms of $1$, $u$ and $u^2$, their coefficients $b_n$ in the sum over $n$ suggest the following coefficients for $1$, $u$ and $u^2$ respectively.
$$
b_0 = \sum_{n=0}^\infty 1 \\
b_1 =\sum_{n=0}^\infty n \\
b_2 = \sum_{n=0}^\infty \binom n2 \ .
$$
Expanding on this, we arrive at
$$
\sum_{i=0}^\infty b_i u^i = \sum_{i=0}^\infty \bigg( \sum_{k=0}^\infty \binom ki \bigg)u^i
$$
, which seems very incorrect, since we have a double infinite sum.
| Notice that
$$\frac z{(1+z^3)^2}=-z\frac{\partial}{\partial z^3}\frac1{1+z^3}=-z\frac\partial{\partial z^3}\sum_{n=0}^\infty(-1)^n(z^3)^n=\sum_{n=0}^\infty n(-1)^{n+1}z^{3n-2}$$
This expansion works for $|z|<1$, and since we know the expansion exists at $z=1$, we may apply the method described in this answer to get the expansion at $z=1$.
$$=\sum_{j=0}^\infty \left(\sum_{k=j}^\infty \binom kj a_k \right)(z-1)^j$$
where $a_{3n+1}=(-1)^n(n+1)$ for $n\in\mathbb N$, else $a_k=0$.
However, note that this method is usually only viable when the center for the radius of convergence was within the radius of convergence of the original series. For example, plugging $z=1$ into our "expansion" yields
$$\frac14\stackrel?=1-2+3-4+5-6+\dots$$
However, an interesting case happens that we may regularize our divergent series to equal our original function by applying an Euler sum to it:
$$f(z)=\sum_{j=0}^\infty\sum_{k=\lfloor j/3\rfloor}^\infty \binom{3k+1}j (-1)^k(k+1)(z-1)^j\\=\sum_{j=0}^\infty\sum_{i=\lfloor j/3\rfloor}^\infty\frac1{2^i}\sum_{k=\lfloor j/3\rfloor}^i\binom ik\binom{3k+1}j (-1)^k(k+1)(z-1)^j$$
which now converges as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the numbers in A.P. those sum is $24$ and product is $440$
If the sum of three numbers in A.P. is $24$ and their product is $440$, find the numbers.
My Approach: Let the numbers be $a,a+d,a+2d$
So, according to question $$3a+3d=24$$ $$a+d=8$$ and $$a(a+d)(a+2d)=440$$ $$8a+ad=55$$ I can’t proceed from here.
Please help.
| Another way to view it is let $m$ equal the middle term.
So they are $m -d, m, m+d$
$(m-d) + m + (m+d) = 3m = 24$ so $m = 8$.
So the numbers are $8-d, 8, 8+d$ and
$(8-d)8(8-d) = 8(64 -d^2)=440$
$64 -d^2 = 55$
$d^2 = 64 - 55 = 9$
$d = \pm 3$
so the numbers are $5,8,11$. Or $11, 8, 5$
Even if we did it your way.
$a + d = 8$ and $8a + ad = 55$ we'd have.
$d = 8 -a$
$8a + a(8-a) = 8a +8a - a^2 = 55$
$a^2 - 16a + 55 = 0$
so
$a = \frac {16 \pm \sqrt{16^2 - 4*55}}{2} = $
$\frac {16 \pm \sqrt {256 - 220}}{2} = 8 \pm \frac{\sqrt{36}}2 = 8 \pm 3 = 5;11$
so $d = 8 -5 = 3$ or $d = 8 -11 = -3$
And the numbers are $5, 8, 11$ or $11, 8, 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Wrong solution - but why ?
Find all solutions to the ODE $$y'=\begin{pmatrix}0 & 1 \\ \frac{-2}{1-x^2} & \frac{2x}{1-x^2}\end{pmatrix}y$$
What I did :
Guessing $y_1=\begin{pmatrix}x \\1\end{pmatrix}$ and reduce the order:
complete $y_1$ to an invertible matrix such that $$H^{-1}=\begin{pmatrix}0 & 1 \\ 1 & -x\end{pmatrix}$$
Calculate $$B=\begin{pmatrix}0 & 1 \\ 1 & -x\end{pmatrix} \begin{pmatrix}0 & 1 \\ \frac{-2}{1-x^2} & \frac{2x}{1-x^2}\end{pmatrix}\begin{pmatrix}1 \\0\end{pmatrix}-\begin{pmatrix}0 \\0\end{pmatrix}$$
So $$B=\frac{2}{x^2-1}\begin{pmatrix}1 \\-x\end{pmatrix}$$ and
$B_1=\frac{2}{x^2-1}$ and $B_2=\frac{-2x}{x^2-1}$
solve $z'=B_2z$
$\Rightarrow C_2=x^2-1$ and
$C_1=\int B_1 C_2dx = \int 2dx=2x$
Calculate $HC$
$$HC=\begin{pmatrix}x & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}2x \\ x^2-1\end{pmatrix}=\begin{pmatrix}3x^2-1 \\ 2x\end{pmatrix}$$
This should be an soultion but it doesn't work and I don't know why :(
| In the wording of the question some symbols are undefined such as $B$, $C_1$ , $C_2$ , $z\quad$ This is confusing and makes difficult to answer with any certainty.
Nevertheless, the end of calculus might be :
$$z'=-\frac{2x}{x^2-1}z \quad\to\quad z=\frac{c_2}{x^2-1}$$
$$\int \frac{1}{(x^2-1)^2}dx=\frac{1}{2}\ln\left|\frac{x+1}{x-1}\right|-\frac{x}{x^2-1}$$
$$y=c_1\left(\begin{matrix}x\\1 \end{matrix}\right)+c_2\left(\begin{matrix}\frac{1}{2}x\ln\left|\frac{1+x}{1-x}\right|-1\\ \frac{1}{2}\ln\left|\frac{1+x}{x-1}\right|-\frac{x}{x^2-1} \end{matrix}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity
$$\frac{1}{n(n+1)} > \frac{1}{(n+1)^2} > \frac{1}{(n+1)(n+2)}.$$
But I am not able to implement it
Please help me.
| Another way:
$$\begin{align} \sum_{k=1}^{2^5} \frac{1}{k^2}
&\leq 1 + \sum_{k=2}^{2^5}\int_{k-1}^{k}\frac{1}{t^2} dt\\
&=1+\int_1^{2^5}\frac{1}{t^2} dt \\
&=2-\frac{1}{2^5}\\
&<2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
It's $(X+Y)^3- (X^2+Y^2)\in \mathbb{C}[X,Y]$ irreducible? It's $(X+Y)^3- (X^2+Y^2)\in \mathbb{C}[X,Y]$ irreducible? I can't apply Eisenstein Criterion.
| The brute force method :
See it as a polynomial of degree $3$ of $K[X]$ with coefficients in $K = \mathbb{C}(y)$. If it is not irreducible then $$X^3+y^3+3X^2y+3Xy^2-X^2-y^2=( X +a)(X^2+bX+c)=X^3+bX^2 +cX+aX^2+abX+ac$$
$c = \frac{y^3-y^2}{a}$, $b = 3y-1-a$
$$= X^3+(3y-1-a)X^2 +\frac{y^3-y^2}{a}X+aX^2+a(3y-1-a)X+1$$
I made a mistake, you have to solve $\frac{y^3-y^2}{a}+a(3y-1-a) =3y^2 $ and show the solution $a \not\in \mathbb{C}(y)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.
Thanks for the attention.
|
\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}
What is the value of $xyz$?
Surprisingly, the Ravi substitution
works in this case, despite that
not all the numbers $x,y,z$ are positive,
and hence, the corresponding triangle is "unreal".
So, let
\begin{align}
a &= y + z
,\quad
b = z + x
,\quad
c = x + y
\tag{4}\label{4}
,\\
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{5}\label{5}
,
\end{align}
where the triplet $a, b, c$ represents the sides of a triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$.
Then
\begin{align}
x + z + y &= \rho
\tag{6}\label{6}
,\\
x^2 + z^2 + y^2 &= \rho^2-2(r^2+4rR)
\tag{7}\label{7}
,\\
x^3 + z^3 + y^3 &= \rho(\rho^2-12rR)
\tag{8}\label{8}
,\\
xyz&=\rho\,r^2
\tag{9}\label{9}
.
\end{align}
Excluding $rR$ from \eqref{7}-\eqref{8},
we get
\begin{align}
r^2&=
\tfrac16\,\rho^2-\tfrac12\,v+\tfrac13\,\frac w{\rho}
\tag{10}\label{10}
,
\end{align}
and from \eqref{9} we have the answer
\begin{align}
xyz&=
\tfrac16\,\rho^3-\tfrac12\,v\rho+\tfrac13\,w
=
\tfrac16\,5^3-\tfrac12\,21\cdot5+\tfrac13\,80
=-5
\tag{11}\label{11}
.
\end{align}
As a bonus, we can find that
\begin{align}
rR &= \tfrac1{12}\,\frac{\rho^3-w}{\rho}
\tag{12}\label{12}
\end{align}
and $x=\rho-a,\ y=\rho-b,\ z=\rho-c$ are the roots of cubic equation
\begin{align}
x^3-\rho\,x^2+(r^2+4rR)\,x-\rho r^2&=0
\tag{13}\label{13}
,\\
\text{or }\quad
x^3-\rho\,x^2+\tfrac12\,(\rho^2-v)\,x-\tfrac13\,w-\tfrac16\,\rho\,(\rho^2-3v)&=0
\tag{14}\label{14}
,\\
x^3-5\,x^2+2\,x+5&=0
\tag{15}\label{15}
.
\end{align}
One of the roots of \eqref{15} is
\begin{align}
x &= \tfrac53+
\tfrac23\,\sqrt{19}\,\cos\Big(\tfrac13\,\arctan(\tfrac9{25}\,\sqrt{331})\Big)
\approx 4.253418
\tag{16}\label{16}
,\\
\text{the other two are }\quad
y&\approx -0.773387
,\quad
z\approx 1.519969
\tag{17}\label{17}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Linear equation with different variable on the denominator? I have this given problem. That asking me to solve for $x$. Although this example has answered. I've had troubles on a certain part.
Here's the equation with answer
\begin{align*}
\frac{2x-a}b &= \frac{4x-b}a\\
a(2x-a) &= b(4x-b) \\
2ax-a^2 &= 4bx - b^2 \\
2ax-4bx &= a^2 - b^2 \\
x(2a-4b) &= a^2 - b^2 \\
x&=\frac{a^2-b^2}{2a-4b}
\end{align*}
How did we arrive to
$$a(2x -a) = b(4x-b)?$$
| Maybe we need to write again the equation:
$$\frac{2x-a}{b}=\frac{4x-b}{a}.$$
Using the Multiplication Property of Equality, we get
$$ab\cdot\left(\frac{2x-a}{b}\right)=ab\cdot\left(\frac{4x-b}{a} \right).$$
Simplifying, we get
$$\frac{a\cdot b\cdot (2x-a)}{b}=\frac{a\cdot b\cdot (4x-b)}{a}.$$
Apply Cancellation Law in Multiplication (meaning we can cancel $b$ at the left hand side and same to $a$ at the right hand side), we get
$$a\cdot(2x-a)=b\cdot(4x-b).$$
Hope this help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Summation of $\arcsin $ series. What is $a $ if
$$\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?$$
Attempt: What I tried is to convert the series to $\arctan$ and then convert it telescoping series. So in terms of $\arctan $ it becomes
$$\arctan \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n}+\sqrt {n-1}}\right) $$
but now if I divide by $n$ it simplifies as $n\frac {\pi}4-\sum _1^{\infty} \arctan \left(\frac {\sqrt {n-1}}{\sqrt {n}}\right) $ but as $n$ is tending towards infinity it will lead to infinity which seems wrong. Also note that $a$ is an integer . Thanks!
| Taking the principal branch of $\arcsin$ (with values in $\bigl[-\frac{\pi}{2}, \frac{\pi}{2}\bigr]$), we have
$$\tan\bigl(\arcsin s\bigr) = \frac{\sin \bigl(\arcsin s\bigr)}{\cos \bigl(\arcsin s\bigr)} = \frac{s}{\sqrt{1 - s^2}}.$$
With $s = \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}$, we get
\begin{align}
1 - s^2 &= 1 - \frac{(\sqrt{n} - \sqrt{n-1})^2}{n(n+1)}\\
&= \frac{n^2 + n - (n - 2\sqrt{n(n-1)} + n-1)}{n(n+1)}\\
&= \frac{n(n-1) + 2\sqrt{n(n-1)} + 1}{n(n+1)}\\
&= \frac{(1 + \sqrt{n(n-1)})^2}{n(n+1)},
\end{align}
and so
\begin{align}
\tan \biggl(\arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\biggr)
&= \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\cdot \frac{\sqrt{n(n+1)}}{1 + \sqrt{n(n-1)}} \\
&= \frac{\sqrt{n} - \sqrt{n-1}}{1 + \sqrt{n} \sqrt{n-1}} \\
&= \tan\bigl(\arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr),
\end{align}
whence we obtain
$$\sum_{n = 1}^{\infty} \arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}} = \sum_{n = 1}^\infty \bigl( \arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr) = \frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How prove $\left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{c+a}\right)\left(1+\frac{4c}{a+b}\right)\ge 25$ let $a,b,c>0$ show that
$$\left(1+\dfrac{4a}{b+c}\right)\left(1+\dfrac{4b}{c+a}\right)\left(1+\dfrac{4c}{a+b}\right)\ge 25$$
It seem hard to prove AM-GM.Cauchy-Schwarz
| $$\Leftrightarrow a^{3}+b^3+c^3+7abc\geq ab(a+b)+bc(b+c)+ca(c+a) > 0$$
Right by Schur:
$$a^{3}+b^3+c^3+7abc> a^{3}+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099548",
"timestamp": "2023-03-29T00:00:00",
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
| Note that
\begin{align*}
(a+b+c)^2 \ge 3ab+3bc+3ac.
\end{align*}
Therefore,
\begin{align*}
&\ \frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b}\\
=&\ \frac{a^2}{a^2+3ab+3ac}+\frac{b^2}{b^2+3ab+3bc}+\frac{c^2}{c^2+3ac+3bc}\\
\ge&\ \frac{(a+b+c)^2}{a^2+b^2+c^2 + 6ab + 6ac+6bc}\\
=&\ \frac{(a+b+c)^2}{(a+b+c)^2 + 4ab + 4ac+4bc}\\
\ge&\ \frac{(a+b+c)^2}{(a+b+c)^2 + \frac{4}{3}(a+b+c)^2}\\
=&\ \frac{3}{7}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Identifying left- and right-Riemann sums of $\int_9^{14}e^{-x^4}\ dx$
My attempt:
Relooking at it, I think $L_{20}$ would be the highest, so like
$R_{1200} < L_{1200} < L_{20}$, but I have no way to justify it, any help is appreciated.
| You're last thought is correct. One rigorous justification goes as follows:
$\displaystyle L_{20} = \sum_{i=0}^{19}\frac{5}{20}f\left(9+\frac{5i}{20}\right)$, and
$\displaystyle L_{1200} = \sum_{i=0}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.
Now, we are going to split up the $L_{1200}$ sum into groups of 60. In particular,
$\displaystyle L_{1200} = \sum_{i=0}^{59}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.
Now, by monotonicity of the function $f$ we have that this sum is less than
$\displaystyle\sum_{i=0}^{59}\frac{5}{1200}f\left(9\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5\cdot 60}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5\cdot 1140}{1200}\right) \\
=60\cdot\frac{5}{1200}f\left(9\right) + 60\cdot\frac{5}{1200}f\left(9 + \frac{5}{20}\right) + \ldots + 60\cdot\frac{5}{1200}f\left(9 + \frac{5\cdot 19}{20}\right) =: L_{20}$.
Therefore, $L_{1200} \leq L_{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible projective duality between two determinantal formulas for triangle area The shoelace formula for the area of a polygon in terms of consecutive vertices is well-known. In the particular case of a triangle, this may be written using a 3-by-3 determinant as
$$\text{area of triangle}=\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}$$
where $(x_k,y_k)_{k=1,2,3}$ are the three vertices. Several proofs have appeared on this site already.
What may be surprising is that there's another determinantal formula for the area in terms of the three lines. Specifically, a triangle with lines $a_k x+b_k y+c_k=0$ for $k=1,2,3$ satisfies
$$\text{area of triangle}=\frac{1}{2C_1C_2C_3}\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}^2$$ where $C_1,C_2,C_3$ are the cofactors of the third column. (An older question has multiple proofs.)
The parallels between these formulas intrigue me: Both express the area of a triangle in terms of a determinant, but one in terms of points and the other in terms of lines. This is reminiscent of the duality of lines and points in projective geometry. Hence my question: Can these the two formulas indeed be understood through projective duality?
| I believe the connection between these two formulas is just a consequence of Cramer's rule.
Let $A=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{pmatrix}$ and $B=\begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{pmatrix}$.
Let us assume that $a_jx_i+b_jy_i+c_j=0$, whenever $i\neq j$, and $d_i=a_ix_i+b_iy_i+c_i$. Since this is a triangle then $d_i\neq 0$, for every $i$.
Hence, $AB=\begin{pmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3\end{pmatrix}$ and $\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{pmatrix}\begin{pmatrix} \frac{x_1}{d_1} & \frac{x_2}{d_2} & \frac{x_3}{d_3} \\ \frac{y_1}{d_1} & \frac{y_2}{d_2} & \frac{y_3}{d_3} \\ \frac{1}{d_1} & \frac{1}{d_2} & \frac{1}{d_3}\end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.
By Cramer's rule, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0\end{vmatrix}=\frac{1}{d_1}$, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 1 \\ a_3 & b_3 & 0\end{vmatrix}=\frac{1}{d_2}$, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 1\end{vmatrix}=\frac{1}{d_3}$.
So $\det(AB)=d_1d_2d_3=\dfrac{\det(A)^3}{C_1C_2C_3}$. Thus, $\det(B)=\dfrac{\det(A)^2}{C_1C_2C_3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find a polynomial of which this field is a splitting field? Consider the field $Q(\sqrt{2} + i \sqrt{5})$. I've proven that $\mathbb{Q}(\sqrt{2} + i \sqrt{5}) = \mathbb{Q}(\sqrt{2}, i \sqrt{5})$. Also, by the degree product rule for extension degrees, I have $$ [\mathbb{Q}(\sqrt{2}, i \sqrt{5}) : \mathbb{Q}] = 4. $$
I'm now being asked to give a polynomial $f \in \mathbb{Q}[X]$ such that $Q(\sqrt{2} + i \sqrt{5})$ is the splitting field of $f$ over $\mathbb{Q}$.
I don't know how to handle this problem. I tried setting $x = \sqrt{2} + i \sqrt{5}$ and squaring etc. but I cannot get a polynomial over $\mathbb{Q}$. The problem gives me a hint, saying I should look for a suitable field $E$ such that $$ \mathbb{Q} \subset E \subset Q(\sqrt{2} + i \sqrt{5})$$ but I'm not sure how this will help me. I know that $\mathbb{Q}(\sqrt{2}, i \sqrt{5}) = (\mathbb{Q}(\sqrt{2})(i\sqrt{5})$. So then I would maybe let $E = \mathbb{Q}(\sqrt{2})$. Then $x^2 + 5$ is the minimal polynomial of $i \sqrt{5}$ over $\mathbb{Q}(\sqrt{2})$. But how to find a polynomial from this over $\mathbb{Q}$?
| Let $\alpha = \sqrt 2 + \mathrm i \sqrt 5$ and consider successive powers:
\begin{eqnarray*}
\alpha &=& 0+1\sqrt 2 + \mathrm i \sqrt 5 + 0\sqrt{10}\\ \\
\alpha^2 &=& -3+0\sqrt 2 + 0\sqrt 5 + \mathrm i \sqrt{10} \\ \\
\alpha^3 &=& 0-13\sqrt 2+\mathrm i \sqrt 5 + 0\sqrt{10}\\ \\
\alpha^4 &=& -31+0\sqrt 2 + 0\sqrt 5 -12\mathrm i \sqrt{10}
\end{eqnarray*}
Putting this into a matrix equation gives
$$\left[\begin{array}{c} \alpha \\ \alpha^2 \\ \alpha^3 \\ \alpha^4 \end{array}\right] =
\left[\begin{array}{cccc} 0 & 1 & \mathrm i & 0 \\
-3 & 0 & 0 & 2\mathrm i \\
0 & -13 & \mathrm i & 0 \\
-31 & 0 & 0 & -12\mathrm i \end{array}\right]
\left[\begin{array}{c} 1 \\ \sqrt 2 \\ \sqrt 5 \\ \sqrt{10} \end{array}\right]$$
The four-by-four matrix has non-zero determinant, and hence:
$$\frac{1}{98}\left[\begin{array}{cccc}
0 & -12 & 0 & -2 \\
7 & 0 & -7 & 0 \\
-91\mathrm i & 0 & -7 \mathrm i & 0 \\
0 & -31\mathrm i & 0 & 3\mathrm i \end{array}\right]
\left[\begin{array}{c} \alpha \\ \alpha^2 \\ \alpha^3 \\ \alpha^4 \end{array}\right] = \left[\begin{array}{c} 1 \\ \sqrt 2 \\ \sqrt 5 \\ \sqrt{10} \end{array}\right]$$
Expanding the first row gives $-\frac{12}{98}\alpha^2-\frac{2}{98}\alpha^4=1$, i.e.
$$2(\alpha^4 + 6\alpha^2 + 49) = 0$$
The polynomial in question is then $x^4+6x^2+49$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove $\sin2x+\sin4x+\sin6x=4\cos x\cos2x\sin3x$ Prove $\sin{2x}+\sin{4x}+\sin{6x}=4\cos{x}\cos{2x}\sin{3x}$
I have reached the point where the LHS equation has turned into $2\cos{x}\cos2x\sin{x}(2\sin2x+1)$
But I have no idea how to turn $\sin{x}(2\sin2x+1)$ into $2\sin3x$
A quicker method if it exists would be greatly appreciated
Thanks in advance
| To solve this problem, we can use the identities:
$$
\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A - B}{2},
$$
$$
\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A - B}{2},
$$
and
$$
\sin 2\phi = 2\sin \phi \cos \phi.
$$
Going back to the question,
$
\text{LHS} = \sin 2x + \sin 4x + \sin 6x \\
= 2\sin 3x \cos x + \sin 6x \\
= 2\sin 3x \cos x + 2\sin 3x \cos 3x \\
= 2\sin 3x (\cos x + \cos 3x) \\
= 2\sin 3x \times 2\cos 2x \cos x \\
= 4\cos x \cos 2x \sin 3x \\
= \text{RHS}.
$
Hence, proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove $n^5+n^4+1$ is not a prime I have to prove that for any $n>1$, the number $n^5+n^4+1$ is not a prime.With induction I have been able to show that it is true for base case $n=2$, since $n>1$.However, I cannot break down the given expression involving fifth and fourth power into simpler terms. Any help?
| $$n^5+n^4+1=n^5-n^2+n^4+n^2+1=n^2(n-1)(n^2+n+1)+(n^2+n+1)(n^2-n+1)=$$
$$=(n^2+n+1)(n^3-n^2+n^2-n+1)=(n^2+n+1)(n^3-n+1)$$
I think, the best way is the following:
$$n^5+n^4+1=n^5+n^4+n^3-(n^3-1)=(n^2+n+1)(n^3-n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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The notion of equality when considering composition of functions When going through some (very introductory) calculus homework I was given the function $f(x) = \frac{x}{1+x}$ and asked to find the composition $(f\circ f)(x)$ and then find its domain. Substituting, we find that $$(f\circ f)(x)= \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}}$$ The domain is then found by solving $\frac{x}{1+x} = -1$ and to find that the composition is undefined at $-\frac{1}{2}$. It is also, of course, undefined at $-1$. Thus our domain is $\{x \in \mathbb{R} \mid x \neq -\frac{1}{2} $ and $x \neq -1$}. My question comes from noticing that if we take the algebra further we find that $$(f\circ f)(x)= \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}} = \frac{x}{2x+1}$$ The domain is still surely unchanged. However, suppose I never did this problem and for some reason I simply desired to write down the function $\frac{x}{2x+1}$ on a sheet of paper and find its domain. I would find that it is defined for all reals except $-\frac{1}{2}$. (Wolfram alpha also verifies this). This would then imply that $$(f\circ f)(x)= \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}} \neq \frac{x}{2x+1}$$ since the domain of the two functions is unequal. Couldn't we also work backwards from $\frac{x}{2x+1}$ in the following manner? $$\frac{x}{2x+1} = \frac{\frac{x}{1+x}}{\frac{1+2x}{1+x}} = \frac{\frac{x}{1+x}}{\frac{1+x}{1+x} + \frac{x}{1+x}} = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$$ This is the function we orignally found the domain for. Did I some how just remove a point from the domain by just doing algebraic manipulations?
My guess is perhaps there are two (or more?) notions of equality going on here. One notion would perhaps be the idea of two functions $f$ and $g$ being "formally" equivalent if $f$ can be algebraically manipulated to $g$ and vice versa. The other notion would be the more intuitive one where two functions are equal if they have the same domain and map the elements of the domain to the same points in the codomain.
Thanks.
| Note that the notation $(f\circ f)(x)$ gives some intuitive notion that we are going to do $2$ different operations. The first will be to "feed" $x$ to $f$, and then "feed" $f(x)$ to $f$. Indeed we have $(f\circ f)(x) = f(f(x))$. We know that $f$ is not defined for $x = -1$ and therefore the inner $f$ in $f(f(x))$ cannot be "fed" $-1$. However, the outer $f$ is fed values from $f(x)$ and it just so happens that $f(x) = -1 \iff x = -\frac12$.
We can then proceed to write the algebraic manipulations you wrote assuming that $1+x \not= 0, \frac{x}{1+x} \not= -1$ given that otherwise you would be dividing by $0$.
On the other hand, starting from $\frac{1}{2x+1}$ one cannot write
$$\frac{\frac{x}{1+x}}{\frac{1+2x}{1+x}}$$
if we don't explicitly state that $1+x\not=0$, otherwise you would be dividing by $0$. Therefore one can always do the algebraic manipulations, given that one carries the excluded points along.
Therefore, one finds $f(f(x))$ to be defined for $x\not\in \{-\frac12, -1\}$ and for the points where it is defined we have $f(f(x)) = \frac{1}{2x+1}$. Similarly, working backwards like you did, we get
$$\frac{\frac{x}{1+x}}{\frac{1+2x}{1+x}} = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$$
except for the points $x = -\frac12, -1$ because we had to exclude them.
What is more, you are right when you say that two functions are equal if they have the same domain/codomain and if they map the same objects to the same images. Having that in mind, the functions $f(f(x))$ and $\frac1{2x+1}$ are not the same function, unless you restrict the second one to the points where we know $f(f(x))$ is well-defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
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Optimization and maximum geometry What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.
I created two equations:
Square: Area=$x^2$
Triangle: Area= $\sqrt{3 /4}$
However, how can I find the maximum??
I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which is wrong
| Proof with some words
You can create a rectangle inside your triangle. Let $x_B$ and $x_G$ the $x$-coordinates of the blue and green points, respectively. It is clear that $y_B = y_G = 0$. Moreover, it is easy to see that:
$$x_B < 0, x_G >0 ~\text{and}~ x_B = -x_G.$$
The base of the rectangles is $$b = x_G - x_B = 2 x_G.$$
The height of the rectangle is given by the $y$-coordinate of the yellow or red point, let's call them $y_Y$ and $y_R$. Note also that the $x$-coordinates of these two points are:
$$\begin{cases}
x_R = x_B \\
x_Y = x_G
\end{cases}.
$$
Then, using the equation of the diagonal side of your triangle, we have that:
$$\begin{cases}
y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B \\
y_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_G
\end{cases}.
$$
As said, $y_R = y_Y$, indeed:
$$y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B = \frac{\sqrt{3}}{2} - \sqrt{3}x_G = y_Y.$$
Moreover, $h = y_R$, then:
$$h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G.$$
Finally, we found that the rectangle is defined by:
$$\begin{cases}
b = 2x_G\\
h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G
\end{cases},$$
while its area is
$$ A =bh = 2x_G\left(\frac{\sqrt{3}}{2} - \sqrt{3}x_G\right).$$
Your rectangle is a square only when $b=h$. That is, when:
$$2x_G = \frac{\sqrt{3}}{2} - \sqrt{3}x_G \Rightarrow x_G = \frac{\sqrt{3}}{2(2+\sqrt{3})}.$$
Then:
$$b = h = \frac{\sqrt{3}}{2+\sqrt{3}}$$
and the area is $$A = \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^2 = \frac{3}{7+4\sqrt{3}}.$$
It is very important to notice that this square is unique possible by construction. Indeed, we started from a rectangles, and we "proved" that this rectangle is a square only if $b=h= \frac{\sqrt{3}}{2+\sqrt{3}}$. So, the maximum area is
$$A =\frac{3}{7+4\sqrt{3}} = 21 - 12 \sqrt{3} \simeq 0.215390309173472.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Solve this limit $\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$ I am trying to figure out how to make this limit, even with the hopital. I've tried using hopital two times, but the situation 0/0 is still there. I've tried to solve it using wolfram, but I don't the solution. Even with rationalization + hopital nothing comes out. $$\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$$
I wonder if there is some way to solve it, and would really appreciate any suggestion.
| \begin{align}
\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-4x^2}}\frac{2\sqrt{x-2x^2}}{1-4x}, \text{ L'hopital}\\
&= \lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-2x}\sqrt{1+2x}}\frac{2\sqrt{x}\sqrt{1-2x}}{1-4x}\\
&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1+2x}}\frac{2\sqrt{x}}{1-4x}\\
&=\frac{2}{\sqrt{2}}\frac{2\sqrt{\frac{1}{2}}}{1-2}
\\&=-2\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of series in GP $a+ ar+ar^2 + ar^3 +ar^4+ \cdots+ ar^{n-1}=S_n$
$a\left(1+r+r^2 +r^3+\cdots+r^{n-1} \right) = S_n$
Iam trying to get $S_n = \dfrac{a(r^n -1)}{r-1}$
I don't know how to get $1+r+r^2 +r^3+\cdots+r^{n-1} =\dfrac{r^n -1}{r-1}$
Any help will be appreciated $:)$
| Equation (1)
$S_n = a_1 + a_2 + a_3 + a_4 + …....... +a_n$
Putting value of each term,
Equation (2)
$S_n = a + ar + ar^2 + ar^3 + ar^4 + ... + ar^{n-1}$
Multiply equation (2) by r,
Equation (3)
$rS_n = ar + ar^2 + ar^3 + ar^4 + ........... + ar^{n}$
Subtract equation (2) from (3),
$rS_n - S_n = ar^n - a$
$S_n = \frac{a(r^n - 1)}{r - 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)?$ How do we show that?
$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)\tag1$$
$(1)$ it a bit difficult to start with
$$\int_{0}^{\pi}(1+2x)\cdot{\sin(x)[1-\sin^2(x)]\over 1+\cos^2(x)}\mathrm dx\tag2$$
Setting $u=\cos(x)$
$du=-\sin(x)dx$
$$\int_{-1}^{1}(1+2x)\cdot{(u^2)\over 1+u^2}\mathrm du\tag3$$
$$\int_{-1}^{1}(1+2\arccos(u))\cdot{(u^2)\over 1+u^2}\mathrm du\tag4$$
$du=\sec^2(v)dv$
$$\int_{-\pi/4}^{\pi/4}(1+2\arccos(\tan(v)))\tan^2(v)\mathrm dv\tag5$$
$$\int_{-\pi/4}^{\pi/4}\tan^2(v)+2\tan^2(v)\arccos(\tan(v))\mathrm dv=I_1+I_2\tag6$$
$$I_1=\int_{-\pi/4}^{\pi/4}\tan^2(v)\mathrm dv=2-{\pi\over2}\tag7$$
As for $I_2$ I am sure how to do it.
| $J=\displaystyle \int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$
Perform the change of variable $y=\pi-x$,
$\displaystyle J=\int_0^{\pi} (1+2(\pi-x))\dfrac{\sin^3 x}{1+\cos^2 x}dx$
Therefore,
$\begin{align}\displaystyle 2J&=(2+2\pi)\int_0^{\pi}\dfrac{\sin^3 x}{1+\cos^2 x}dx\\
2J&=-(2+2\pi)\int_0^{\pi}\dfrac{\sin^2 x}{1+\cos^2 x}\text{d}(\cos x)\\
2J&=-(2+2\pi)\int_0^{\pi}\dfrac{(1-\cos^2 x)}{1+\cos^2 x}\text{d}(\cos x)\\
\end{align}$
Perform the change of variable $y=\cos x$ in the latter integral,
$\begin{align}\displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1-x^2}{1+x^2}dx\\
\displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1}{1+x^2}dx-(2+2\pi)\int_{-1}^{1}\dfrac{x^2}{1+x^2}dx\\
&=(2+2\pi)\Big[\arctan x\Big]_{-1}^{1}-(2+2\pi)\left(\int_{-1}^1 \dfrac{1+x^2}{1+x^2}dx-\int_{-1}^1 \dfrac{1}{1+x^2}dx\right)\\
&=4(2+2\pi)\dfrac{\pi}{4}-2(2+2\pi)\\
&=2(\pi+1)(\pi-2)
\end{align}$
Therefore,
$\boxed{J=(\pi+1)(\pi-2)}$
| {
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"url": "https://math.stackexchange.com/questions/2108420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac{2x}{|x|(1+x^2)}$
I don't understand how they get this. I checked my answer on online calculator and it is the same.
| You should have developed your result a bit more to obtain the assignment solution.
\begin{align}
-\frac{4x}{(x^2+1)^2\sqrt{1-\frac{(1-x^2)^2}{(x^2+1)^2}}}&=-\frac{4x}{(x^2+1)^2\sqrt{\frac{x^4+1+2x^2-1-x^4+2x^2}{(x^2+1)^2}}} \\
&=-\frac{4x}{(x^2+1)^2\sqrt{\frac{4x^2}{(x^2+1)^2}}}\\
&=-\frac{4x}{(x^2+1)^2\frac{2|x|}{(x^2+1)}}\\
&=-\frac{2x}{|x|(x^2+1) }
\end{align}
Voila!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding a strict Liapunov function I need to show that the equilibrium point $(0,0)$ is asymptotically stable using Liapunov function. That means I shall find some strict Liapunov function.
Its given me the non-linear system:
\begin{cases}
x_1' = -x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2) \\
x_2' = -x_2 -\frac{1}{3}x_2^3
\end{cases}
My attempt: $V(x_1, x_2) = ax_1^2 + bx_2^2$, $a, b > 0$. I would like to determine $a$ and $b$.
Then, $V(0,0) = 0$ and $V(x_1, x_2) > 0$, for all $(x_1, x_2) \neq (0,0)$.
In order to decide if $V(x_1, x_2)$ is a strict Liapunov function, I wish that $\left<\nabla V(x_1, x_2), (x_1', x_2')\right> < 0$.
\begin{align}
\left<\nabla V(x_1, x_2), (x_1', x_2')\right> &= \left<(2ax_1, 2bx_2) ( -x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2), -x_2 -\frac{1}{3}x_2^3)\right>\\
&= 2ax_1(-x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2)) + 2bx_2(-x_2 -\frac{1}{3}x_2^3) \\
&= -2a(x_1^2 + \frac{1}{3}x_1^4) -2b(x_2^2 + \frac{1}{3}x_2^2) -2ax_1^3\sin(x_2)
\end{align}
I do not know what to do with the term that involves $\sin(x_2)$...
| As i can remember, we have to prove that $\left<\nabla V(x_1, x_2), (x_1', x_2')\right> < 0$ over some neighborhood $B$ of $(0,0)$.
We call $sg(x)$ the sign function defined by:$$sg(x)=1\quad if\ x>0$$
$$sg(x)=-1\quad if\ x<0$$
First, we have \begin{align*}
-1\leq\sin(x_2)\leq 1\\
-2a\leq-2a\sin(x_2)\leq 2a\\
-2a|x_1^3|\leq-2ax_1^3\sin(x_2)\leq2a|x_1^3|\\
-2a(x_1^2 + \frac{1}{3}x_1^4)-2ax_1^3\sin(x_2)\leq2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)
\end{align*}
Now let's study the sign of the expression $(E)= 2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)$
\begin{align*}
2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)&=2ax_1^2(sg(x_1)x_1-1+\frac{1}{3}x_1^2)\\
&=\frac{2}{3}2ax_1^3(x_1^2+3sg(x_1)x_1-3)
\end{align*}
$\Delta=((3sg(x_1))^2-4(-3))=9+12=21 \implies \sqrt{\Delta}\simeq4.58$
Then, $x_1'=\frac{-3sg(x_1^3)-\sqrt{\Delta}}{2}$ and $x_1''=\frac{-3sg(x_1^3)+\sqrt{\Delta}}{2}$
When $x_1\in I=]x_1',x_1''[$, $(E)<0$
Notice that whether the sign of $x_1$ is $(+)$ or $(-)$,the interval I is always a neighborhood of $(0,0)$.
Now that we have $(E)<0$ we add to it the last term of our first equation $-2b(x_2^2+\frac{1}{3}x_2^2)$ which is obviously negative.
Thus we conclude that $$\left<\nabla V(x_1, x_2), (x_1', x_2')\right><0\quad\forall(x_1,x_2)\in B\setminus\{(0,0)\} $$
$$B=\{(x_1,x_2)\in I\times J,\ I=]x_1',x_1''[,\ J=]-\epsilon,+\epsilon[,\ \epsilon>0 \} $$
| {
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"source": "stackexchange",
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a−b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(a−b)^2 = a^2 −b^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(a−b)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.
As for the part where $b=a$, $(a−b)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".
P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.
|
which resulted in $a^2 - 2a(0) + 0^2 = 2a$
$a^2 - 2a(0) + 0^2 = a^2$, not $2a$.
or if I do not substitute the $b^2$ I end up with $a^2 + b^2$.
Why would you not substitute the $b^2$? If you're substituting $b=0$ then you need to do it in all occurrences of $b$. This includes $b^2$. When you do this you'll see that the equation $(a-b)^2 = a^2 - b^2$ reduces to $a^2 = a^2$, which is certainly a true statement for all values of $a$.
Also, you're overcomplicating the $a=b$ case. No need to introduce a new variable $x$. If $a = b$, then you can simply substitute either one in for the other. Let's replace $a$ with $b$. Then we have
$$ (a-b)^2 = (b-b)^2 = 0^2 = 0$$
and on the other hand we have
$$ a^2 - b^2 = b^2 - b^2 = 0 $$
This shows that $(a-b)^2 = a^2 - b^2$ if $a = b$.
I should also point out that the work you've done (or at least the work you've shown us) only proves one direction. You're asked to prove the following:
$$ (a-b)^2 = a^2 - b^2 \text{ if and only if } b = 0 \text{ or } a = b $$
But what you've done so far is:
$$ (a-b)^2 = a^2 - b^2 \text{ if } b = 0 \text{ or } a = b $$
In other words, you need to handle the "only if" part. Do this by assuming $(a-b)^2 = a^2 - b^2$ and showing that the only two possibilities are $b=0$ or $a=b$. At least one of the other current answers offers guidance on this matter.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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} |
Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
| Let $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\epsilon}{4}$. If $n\geq N$, then we get
$$\begin{align}\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3 \right|&
=\left|\frac{n^3+3n}{2n^4-n+1}\right|\\&=\frac{n^3+3n}{2n^4-n+1}\\
&<\frac{n^3+3n}{2n^4-n}\\
&\leq\frac{n^3+3n}{2n^4-n^4}\\
&\leq\frac{n^3+3n^3}{n^4}\\
&=\frac{4}{n}\leq\frac{4}{N}<\epsilon.
\end{align}$$
Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}{4\sqrt{x+1}^ 3}$$
$$f'''(x)=\frac{3}{8\sqrt{x+1}^5}$$
Applying $x^{}_0$ we have:
$$f(1)=\sqrt{2}$$
$$f'(1)=\frac{1}{2\sqrt{2}}$$
$$f''(1)=\frac{-1}{4\sqrt{2}^ 3}$$
$$f'''(1)=\frac{3}{8\sqrt{2}^5}$$
And we can build the Taylor polynomial
$$T(x)=\sqrt2 + \frac{1}{2\sqrt{2}}(x+1)+\frac{-1}{2!·4\sqrt{2}^3}(x+1)^2+\frac{3}{3!·8\sqrt{2}^5}(x+1)^3+R(\xi)$$
Is everything right until here?
What I don't understand is how can I check that $R(\xi) > 10^{-4}$
| $f(x) = \sqrt{1 + x}$
$f'(x) = \frac{1}{2\sqrt{1+x}}$
$f(0) = 1$
$f'(0) = 1/2$
$f(x + h) = f(x) + h f'(x) +.....$
$\sqrt{1.1} = f(.1) = f(0 + .1) \approx1 + 0.1 / 2 = 1.05$
you got confused by the fact they are using x + 1 in the function - i made a start for you, with two terms it is getting closer to the answer
do you see the important part for you? $\sqrt{1.1} = \sqrt(1 + .1) = f(.1)$
since it is easy to workout f(0), f'(0),f''(0) without any square -roots, you then need to centre it around 0 - i.e. f(0 + .1)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$ If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, prove that $a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$.
I guess that the given expression is true if and only if $a=b=c=0$. Is it true? Or,is there any other alternatives?
| We have $$a^{1/4} + b^{1/4} + c^{1/4} =0$$ $$\Rightarrow (a^{1/4} + b^{1/4} + c^{1/4})^2 = a^{1/2}+b^{1/2} + c^{1/2} + 2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}]=0$$ $$\Rightarrow a^{1/2}+b^{1/2}+c^{1/2}=-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}] $$ $$\Rightarrow (a^{1/2}+b^{1/2}+c^{1/2})^2 =(-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}])^2$$ $$\Rightarrow a+b+c+2 [\sqrt{ab}+\sqrt {bc}+\sqrt {ac}] = 4 [\sqrt {ab} + \sqrt {bc} + \sqrt {ac} +2 [(a^2bc)^{1/4}+(ab^2c)^{1/4}+(abc^2)^{1/4}]] $$ $$\Rightarrow a+b+c-2 [\sqrt {ab}+\sqrt {bc}+\sqrt {ac}] =8 (abc)^{1/4}[a^{1/4}+b^{1/4}+c^{1/4}] =0$$ $$\boxed {a+b+c =2[\sqrt {ab}+\sqrt {bc}+\sqrt {ac}]}$$ Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$
| Can we just charge straight at this?
$y$ is odd. $x=2 \ (\Rightarrow y^2=7)$ is not a solution, so $x$ is an odd prime.
$x(5x-7) = (y-1)(y+1)$, so $x \mid (y-1) $ or $x \mid (y+1)$ ($x$ is prime)
so $kx=y\pm1$, $k$ even
$k\ge4$ is too large: $(kx\pm1)^2\ge (4x-1)^2 $ $= 16x^2-8x+1$ $>5x^2-7x+1$. So only $k=2$, that is $x=\frac 12(y\pm1)$, makes the equality feasible.
Considering the two cases:
*
*(1) $x=\frac 12(y+1)$, $y=2x-1$:
$x(5x-7) = 4x(x-1) \implies x = 3, y=5$
*(2) $x=\frac 12(y-1)$, $y=2x+1$:
$x(5x-7) = 4x(x+1) \implies x = 11, y=23$
Note that I didn't constrain $y$ at any point - the two solutions just happened to have $y$ prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
| $a^4-a^3+a^2+2=(a^2+a+1)(a^2-2a+2)=0$.
We can check also $a=1+i$ and $a=1-i$.
| {
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"answer_count": 4,
"answer_id": 0
} |
Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
| You have mistake.
$\binom{6}{4} . x^2 . \left(\frac 2x\right)^4$
= $15 . x^2 . \frac{16}{x^4}$
= $240 . \frac{1}{x^2}$
Here 240 is coefficient of $x^{-2}$ not $x^2$.
Correct term -
$\binom{6}{2} . x^4 . \left(\frac 2x\right)^2$
= $15 . x^4 . \frac{4}{x^2}$
= $60 . x^2$
So answer is 60.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all the solutions of the system $AX=B$ if $B$ is the difference between the first and the fourth column of $A$. Let
$$A \sim
\begin{pmatrix}
1 &2 &4&1\\
0&0&1&2\\
1&3&1&1\\
0&0&0&0\\
\end{pmatrix}$$
such that the equivalence is achieved by elementary row transformations. Find all the solutions of the system $AX=B$ if $B$ is the difference between the first and the fourth column of $A$.
I hope I understood this: $$B=\begin{pmatrix}
0\\
-2\\
0\\
0
\end{pmatrix}$$
But I don't know what should I do here since I don't know what $A$ is exactly...I know $X$ should be a $4$x$1$ matrix and that's about it.
| Hint -
You have A and B.
Solve AX = B.
$$\begin{pmatrix}
1 &2 &4&1\\
0&0&1&2\\
1&3&1&1\\
0&0&0&0\\
\end{pmatrix} \cdot \begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix} = \begin{pmatrix}
0\\
-2\\
0\\
0
\end{pmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $G=\mathbb R$ and $x*y\equiv x+y+x^3 y^3$, is there an inverse? I am given a set $G = \mathbb{R}$ and $x*y = x+y+x^3y^3$. I need to find out whether there exists an inverse. Please note that this is not a group since * is not associative. The identity element $e$ is $0$. So I did the following:
For the inverse to exists we must have the following:
$x*y = x+y+x^3y^3 = e = y*x = 0$
So, $x+y+x^3y^3$ must equal to $0$. Imagine we fix $x$, then we get a cubic and we know that it has either 1 real solution or 2. For the inverse to exists, which is unique, we need 1 real root. Now, from here I am confused how to show if whether there are always 1 real root or not. I tried considering derivative, that is:
$\frac{dy}{dx} = \frac{-1-3x^2y^3}{1+3x^3y^2}$
For $x>0$: $\frac{dy}{dx}$ is always negative and so the graph is decreasing and, thus will cross x-axis once and there will be 1 real root.
For $x = 0$, the inverse is $0$
As for $x<0$: I cannot say much since $\frac{dy}{dx}$ becomes $\frac{-1-3x^2y^3}{1-3|x^3|y^2}$ and it can be both negative and positive.
So, would that mean that the inverse does not exist for all $x$? And so there is no inverse in $G$?
I hope my reasoning makes sense and I would appreciate any help!
Thanks!
| I think you have already said the answer.
Since $x + y + x^3y^3 = 0$ is a cubic, it has at least one real root.
For any $x$ there exists a $y$ such that $x*y = y*x = 0$
However, we have not proven that for any $x$ there exists a unique $x^{-1}$
Rather than using implicit differentiation, continue with the assumption that $x$ is constant.
$\frac {\partial}{\partial y} (x+y+x^3y^3) = 0\\
1+3y^2 x^3 = 0\\
y = \pm \frac 1 {\sqrt {3|x|^3}}$
Substitute this value of y back in, and suppose $x>0$.
If $x - \frac 4 {3x\sqrt {3x}}, x + \frac 4 {3x\sqrt {3x}}$ have oppose signs. then there are 3 real roots.
$x + \frac 4 {3x\sqrt {3x}} > 0$ for all positive $x$
$x - \frac 4 {3x\sqrt {3x}} < 0\\
3x^2\sqrt {3x} < 4\\
27x^5 < 16\\
x < (\frac {16}{27})^{\frac 15}$
$x\in (-\frac {16}{27}^{\frac 15}, \frac {16}{27}^{\frac 15}) \implies $ there are three $y's$ such $x + y + x^3y^3 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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synthetic division question. Synthetic division is possible when the Divisior is in the form of $x+a$ or $x-a$. but what if the divisor is in the form of $x^2+a$, $x^2-a$, $x^3-a$,... and higher powers. how can we perform synthetic division in such cases.
Thanks
| For the particular case of divisors in the form $x^n-a$ it is possible to replace the long division with $n$ synthetic divisions.
For a polynomial $P(x)$ being divided by $x^3-a$ for example, group the powers of $x$ in $P$ according to the remainder $\bmod 3$ and write it as:
$$
P(x) = P_0(x^3) + x P_1(x^3) + x^2P_2(x^3)
$$
Use synthetic division to calculate the quotients and remainders of the following:
$$
P_k(x) = (x-a)Q_k(x) + r_k \quad\quad \text{for} \;\; k=0,1,2
$$
Then:
$$
P(x)=(x^3-a)Q(x) + R(x)
$$
where $Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3)$ and $R(x)=r_0+r_1x+r_2x^2\,$.
[ EDIT ] Following is a fully worked out example for $P(X)=x^4-6x^3+16x^2-25x+10$ (the polynomial was borrowed from another, unrelated question) being divided by $x^3-2$.
*
*Group the powers:
$$P(X)=x^4-6x^3+16x^2-25x+10= (-6x^3+10) + x\cdot (x^3-25) + x^2 \cdot 16$$
$$
\iff
\begin{cases}
\begin{align}
P_0(x) & = -6x+10 \\
P_1(x) & = x - 25 \\
P_2(x) &= 16
\end{align}
\end{cases}
$$
*
*Divide $P_k$ by $x-2$ and determine $Q_k,r_k$ by synthetic division:
$$
\begin{cases}
\begin{alignat}{3}
P_0(x) & = -6x+10 && = -6(x-2) - 2\\
P_1(x) & = x - 25 && = (x-2) - 23\\
P_2(x) & = 16 && = 16
\end{alignat}
\end{cases}
$$
$$
\iff
\begin{cases}
\begin{align}
Q_0(x) & = -6 \,,\;\; r_0 = -2\\
Q_1(x) & = 1 \,,\;\; r_1 = - 23\\
Q_2(x) & = 0 \,,\;\; r_2 = 16
\end{align}
\end{cases}
$$
*
*Calculate $Q,R$:
$$
Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3) = -6 +x+ x^2 \cdot 0 = x-6\\
R(x)=r_0+r_1x+r_2x^2=16x^2-23 x-2\,$$
*
*Verify that indeed:
$$x^4-6x^3+16x^2-25x+10=(x^3-2)(x-6)+ 16x^2-23 x -2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
| Hint:
$$(x+1)^5-13(x+1)^3+36(x+1)=0$$
$$\left[(x+1)^4-13(x+1)^2+36\right](x+1)=0$$
$$\left((x+1)^2-9\right)\left((x+1)^2-4\right)(x+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
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"answer_count": 5,
"answer_id": 0
} |
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)
I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as $2 + \frac {1}{r+2}$.
$\sqrt {5} \approx2.2360679775$
Equating $\frac {1}{r+2}$ and $0.2360679774$ , I get $r=2.23606797929$.
So, $\frac {2r+5}{r+2}$ is a still a better approximation than $r$.
How to proceed ?
Any hints/ideas/pointers ?
| Observe if $r<\sqrt{5}$, then there exists $\epsilon>0$ such that
\begin{align}
r+\epsilon<\sqrt{5} \ \ \Leftrightarrow& \ \ (r+\epsilon)^2 <5\\
\Leftrightarrow& \ \ r^2-5+2\epsilon r+ \epsilon^2=P(\epsilon)<0
\end{align}
which is true if we sketch $P(\epsilon)$ as a function of $\epsilon$. In particular, we see that the positive root of $P(\epsilon)$ is
\begin{align}
\epsilon = \frac{-2r+\sqrt{20}}{2} = -r+\sqrt{5}= \frac{-r^2+5}{r+\sqrt{5}}>\frac{-r^2+5}{r+2}=\epsilon_0>0
\end{align}
since $\sqrt{5}>2$. Hence
\begin{align}
r<r+\epsilon_0 = r+\frac{-r^2+5}{r+2} = \frac{r^2+2r-r^2+5}{r+2} = \frac{2r+5}{r+2}.
\end{align}
Remark: The idea is to find $\epsilon_0>0$ rational such that $(r+\epsilon_0)^2<5$.
| {
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"url": "https://math.stackexchange.com/questions/2125373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope there'a an exploitable link between them. They fall easily under Muirhead, yet I cannot figure out how to prove them using the Rearrangement Inequality.
Any hints greatly appreciated.
| $(a,b,c)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered.
Thus, by Rearrangement $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a\cdot\frac{1}{a}+b\cdot\frac{1}{b}+c\cdot\frac{1}{c}=3,$$ which gives
$$a^2c+b^2a+c^2b\geq3abc$$
Similarly we'll get $$a^2b+b^2c+c^2a\geq3abc$$ and after summing we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$ I'm having trouble with the following derivation:
Q:
We can use Euler's Theorem ($e^{i\theta} = \cos\theta + i\sin\theta$), where $e$ is the base of the natural logarithms, and $i = \sqrt{-1}$, together with the binomial theorem as above, to derive a number of trigonometric identities. E.g., if we consider $(e^{i\theta})^2$, we can evaluate it two different ways. First, we can multiply exponents, obtaining $e^{i \, 2\theta}$ and then applying Euler's formula to get $\cos(2\theta) + i \sin(2\theta)$, or we can apply Euler's formula to the inside, obtaining $(\cos\theta + i \sin\theta)^2$, which we then evaluate via the binomial theorem:
\begin{align}
\cos(2\theta) + i \sin(2\theta) &= (\cos\theta + i \sin\theta)^2 \\
&= (\cos\theta)^2 + 2 i \cos\theta \sin\theta + i^2 (\sin\theta)^2 \\
&=(\cos\theta)^2 + 2 i \cos\theta \sin\theta − (\sin\theta)^2
\end{align}
Equating real and imaginary parts gives us
\begin{align}
\cos(2\theta) &= (\cos\theta)^2 − (\sin\theta)^2 \\
\sin(2\theta) &= 2 \cos\theta \sin\theta
\end{align}
We can then rewrite the first of these identities, using $1=(\sin\theta)^2+(\cos\theta)^2$ to get $(\cos\theta)^2=1−(\sin\theta)^2$, whence the familiar
$$\cos(2\theta)=1−2(\sin\theta)^2$$
Use this same approach to show $\cos(3\theta)=(4\cos\theta)^3−3\cos\theta$.
A:
This is my work so far:
\begin{align}
e^{i \, 3\theta} &= \cos(3\theta) + i \sin(3\theta) = (\cos\theta)^3 + 3 i (\cos\theta)^2 \sin\theta - 3\cos\theta (\sin\theta)^2 - i(\sin\theta)^3 \\
\cos(3\theta) &= (\cos\theta)^3 - 3(\sin\theta)^2 \cos\theta \\
\sin(3\theta) &= 3\sin\theta(\cos\theta)^2 - (\sin\theta)^3
\end{align}
But now I'm unsure how to get $\cos(3\theta) = (4 \cos\theta)^3 − 3\cos\theta$ from what I've derived.
| I believe your claim is actually incorrect. You should have the identity
\begin{align}
\cos 3\theta = 4\cos^3\theta -3\cos\theta
\end{align}
not
\begin{align}
\cos 3\theta = (4\cos\theta)^3 -3\cos\theta.
\end{align}
Observe
\begin{align}
e^{i3\theta} =&\ (\cos\theta + i\sin\theta)^3 = \cos^3\theta+3(i\sin\theta)\cos^2\theta+3(i\sin\theta)^2\cos\theta +(i\sin\theta)^3\\
=&\ \cos^3\theta -3\sin^2\theta \cos\theta+i(3\sin\theta\cos^2\theta-\sin^2\theta).
\end{align}
Hence taking the real part yields
\begin{align}
\cos 3\theta =&\ \cos^3\theta -3\sin^2\theta \cos \theta\\
=&\ \cos^3\theta - 3(1-\cos^2\theta)\cos \theta\\
=&\ 4\cos^3\theta -3\cos\theta.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.
How do I prove this directly?
| HINT
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$
Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$
Note that the second part has cancelling terms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.
My Attempt
This can be rewritten as
$$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$
Now consider
$$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$
Taking the derivative
$$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$
By symmetry we have
$$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$
Using W|A I got
$$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$
By integeration we have
$$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$
Let $x = 2/a$
$$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
Question
I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
| Ok I was able to find the integral
$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
First note that
$$\int \frac{\log(1+x^2)}{x^2}\,dx = 2 \arctan(x) - \frac{\log(1 + x^2)}{x}+C$$
Using integration by parts
$$I = \frac{\pi^3}{12}+2\int^\infty_0\frac{\arctan(x)\log(1 + x^2)}{(1+x^2)x}\,dx$$
For the integral let
$$F(a) = \int^\infty_0\frac{\arctan(ax)\log(1 + x^2)}{(1+x^2)x}\,dx$$
By differentiation we have
$$F'(a) = \int^\infty_0 \frac{\log(1+x^2)}{(1 + a^2 x^2)(1+x^2)}\,dx $$
Letting $1/a = b$ we get
$$\frac{1}{(1 + a^2 x^2)(1+x^2)} = \frac{1}{a^2} \left\{ \frac{1}{((1/a)^2+x)(1+x^2)}\right\} =\frac{b^2}{1-b^2}\left\{ \frac{1}{b^2+x^2}-\frac{1}{1+x^2} \right\}$$
We conclude that
$$\frac{b^2}{1-b^2}\int^\infty_0 \frac{\log(1+x^2)}{b^2+x^2}-\frac{\log(1+x^2)}{1+x^2} \,dx = \frac{b^2}{1-b^2}\left\{ \frac{\pi}{b}\log (1+b)-\pi\log(2)\right\}$$
Where we used that
$$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+g^2x^2}\,dx = \frac{\pi}{cg}\log \frac{ag+bc}{g}$$
By integration we deduce that
$$\int^1_0 \frac{\pi}{a^2-1}\left\{ a\log \left(1+\frac{1}{a} \right)-\log(2)\right\}\,da = \frac{\pi}{2}\log^2(2)$$
For the last one I used wolfram alpha, however it shouldn't be difficult to prove.
Finally we have
$$\int^\infty_0\frac{\log\left(x^2+1
\right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi
\log^2(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}}{n}\leq x^{-\frac{3}{2}} $$
But $$\int_0^1x^{-\frac{3}{2}}dx=\frac{-2}{\sqrt{x}}\biggr|_0^1$$ which doesn't seem right. Any help will be appreciated.
| A worse bound, a slightly different approach:
$$ \frac{n\sqrt x}{1+n^2x^2} \le \frac{n\sqrt x}{\sqrt{1+n^2x^2}} = n \sqrt{ \frac{x}{1+n^2x^2}} = \frac{1}{\sqrt{x}} \sqrt{\frac{1}{1+\frac{1}{n^2x^2}}} \le \frac{1}{\sqrt{x}} $$
Where we used tha facts that $x \ge \sqrt{x} $ if $x \ge 1$ and $\sqrt{\frac{1}{1+x}} \le 1 $ on $[0,+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
| Let $$S_n= \sum_{n=1}^{\infty} \frac{n}{3^n}\tag1$$
$$\frac13S_n=\sum_{n=1}^{\infty} \frac{n}{3^{n+1}}=\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}-\sum_{n=1}^{\infty}\frac1{3^{n+1}}\tag2$$
$(1)-(2)$,
$$\begin{align}\frac23S_n&=\sum_{n=1}^{\infty} \frac{n}{3^n}-\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\
&=\sum_{n=1}^{\infty} \frac{n}{3^n}-\sum_{n=2}^{\infty} \frac{n}{3^{n}}+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\
&=\frac13+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\
&=\sum_{n=1}^{\infty}\frac1{3^{n}}=\frac12\end{align}$$
Thus,
$$S_n=\frac34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Showing matrices in $SU(2)$ are of form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.
I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex numbers $a,b$ such that $|a|^2+|b|^2 = 1$.
To this end, we put $A:= \begin{pmatrix} r & s \\ t & u\end{pmatrix}$ and impose the two properties.
This yields \begin{align}\operatorname{det}(A) &= ru-st \\ &= 1 \ ,\end{align}
and
\begin{align}
AA^\dagger &= \begin{pmatrix} r & s \\ t & u\end{pmatrix} \begin{pmatrix} r^* & t^* \\ s^* & u^* \end{pmatrix} \\&= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \ .\\
\end{align}
The latter gives rise to
\begin{align}
|r|^2+|s|^2 &= 1 \\
&= |t|^2+|u|^2 \ ,
\end{align}
and
\begin{align}
tr^*+us^* &= 0 \\
&= rt^*+su^* \ .
\end{align}
At this point, I don't know how to proceed. Any hints would be appreciated.
@Omnomnomnom's remark
\begin{align}
A A^\dagger &= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} |r|^2+|t|^2 & sr^* +ut^* \\ rs^*+tu^* & |s|^2 + |u|^2\end{pmatrix} = A^\dagger A \ ,
\end{align}
gives rise to
$$
|t|^2 = |s|^2 \\
|r|^2 = |u|^2
$$
and
$$
AA^\dagger :\begin{pmatrix}
rt^* +su^* = sr^* +ut^* \\
tr^*+us^* = rs^*+tu^*
\end{pmatrix}: A^\dagger A $$
At this point, I'm looking in to find a relation between $t,s$ and $r,u$ respectively.
| We have $tr^\ast=-us^\ast$ so $\left| r\right|^2 \left| t\right|^2 = \left| s\right|^2 \left| u\right|^2$ and $\left| r\right|^2 -\left| r\right|^2\left| u\right|^2 = \left| s\right|^2 \left| u\right|^2$ so $\left| r\right|^2 =\left| u\right|^2$. Hence $r,\,u$ have the same modulus, as do $s,\,t$.
If $tu\ne 0$ define $k:=\dfrac{r^\ast}{u}=-\dfrac{s^\ast}{t}$ so $u=\dfrac{r^\ast}{k},\,1=\dfrac{r^\ast r+s^\ast s}{k}$ and $k=1$. Hence $u=r^\ast$ and similarly $s^\ast=-t$.
If $u=0$ $st=-1$ with $\left| s\right|=\left| t\right|=1$ so $s^\ast=-t$, and $\left| r\right|=\left| u\right|=0$ so $u=r^\ast$.
If $t=0$ then $ru=1$ so $u=r^{-1}=r^\ast$ and $s^\ast=0=-t$ because $\left| s\right|=\left| t\right|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
| $\int \frac{x^2(x-2)}{(x-1)^2} dx$
$\int {x^2(x-1-1)\over(x-1)^2}dx$
$\int {x^2(x-1)-x^2\over(x-1)^2}dx$
$\int ({x^2\over(x-1)}-{x^2\over(x-1)^2})dx$
$\int {x^2\over(x-1)}dx-\int{x^2\over(x-1)^2}dx$
$t=x-1 => dt=dx$
$\int {(t+1)^2\over(t)}dt-\int{(t+1)^2\over(t)^2}dt$
$\int {(t^2+1+2t)\over(t)}dt-\int(1+{1\over t})^2dt$
$\int (t+{1\over t}+2)dt-\int(1+{1\over t})^2dt$
$\int t dt+\int{1\over t}dt+\int2dt-\int(1+{1\over t})^2dt$
${t^2\over2}+\ln t+2t-t+{1\over t}-2\ln t+C$
${(x-1)^2\over2}+(x-1)+{1\over x-1}-\ln |x-1|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Trick to this square root equations Okay, so this is a high school level assignment:
$$
\sqrt{x+14}-\sqrt{x+5}=\sqrt{x-2}-\sqrt{x-7}
$$
Here's a similar one:
$$
\sqrt{x}+\sqrt{x-5}=\sqrt{x+7}+\sqrt{x-8}
$$
When solving these traditionaly, I get a polynomial with exponent to the 4th which I cannot solve (I can guess the solutions via free term and divide the polynomial, accordingly, but I don't think that is the intended method here).
Is there a trick to any of these two tasks that would prevent exponents from getting out of control? How would a highschooler solve them?
| write like this
$$\sqrt{x+14}+\sqrt{x-7}=\sqrt{x-2}+\sqrt{x+5}$$
square both sides:
$$2x+7+2\sqrt{(x+14)(x-7)}=2x+3+2\sqrt{(x-2)(x+5)}\\
2+\sqrt{(x+14)(x-7)}=\sqrt{(x-2)(x+5)}$$
square again
$$4\sqrt{(x+14)(x-7)}+x^2+7x-94=x^2+3x-10\\
\sqrt{(x+14)(x-7)}=21-x \to (x+14)(x-7)=(21-x)^2$$
Can you finish?
Don't forget to test the result in the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
$\lim_{z \to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}$ Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$
Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\sin(\pi)=1$$
$$z^4=\exp(4\pi i/3)=\cos(4\pi/3)+i\sin(4\pi/3)=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$$
So,
\begin{equation*}
\begin{aligned}
\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16} & = \dfrac{1+8}{-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}+4\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)+16} \\
& = \dfrac{9}{\dfrac{27}{2}+i\frac{3\sqrt{3}}{2}} \\
& = \dfrac{6}{9+i\sqrt{3}} \\
& = \dfrac{9}{14}-i\dfrac{\sqrt{3}}{2} \\
\end{aligned}
\end{equation*}
But, when I check my answer on wolframalpha, their answer is $$\dfrac{245}{626}-i\dfrac{21\sqrt{3}}{626}.$$
Can someone tell me what I am doing wrong?
| First, note that:
$$z^3=\cos(\pi)+i\sin(\pi)=\color{red}{-1}$$
And also, you've evaluated $z$ correctly, but the substitution into your limit is wrong.
Therefore, you should be using:
$$\lim_{x\to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}=\frac{\color{red}{-1+8}}{\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)+4\color{red}{\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}+16}$$
Which gives you the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ then the sum of the first $mn$ terms of the sequence is $(mn+1)/2$
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ respectively then prove that the sum of the first $mn$ terms of the sequence is $(mn+1)/2$.
My Attempt ;
$$\textrm t_{m}=\dfrac {\textrm 1}{\textrm n}$$
$$\textrm a + \textrm (m-1)d =\dfrac {1}{n}$$
And,
$$\textrm t_{n}=\dfrac {1}{m}$$
$$\textrm a+\textrm (n-1)d=\dfrac {1}{m}$$
What do I do further?
| Let $u_t = u_0 + a \cdot t$
$$ u_m = \frac{1}{n} $$
$$ u_n = \frac{1}{m} $$
Therefore $$ u_n - u_m = a \cdot \left( n-m \right)= \frac{1}{m} - \frac{1}{n} $$
$$ a=\frac{1}{m \cdot n} $$
So the initial term is$$ u_n = u_0 + \frac{n}{m \cdot n} = u_0 + \frac{1}{m} = \frac {1}{m}$$
$$ u_0 = 0 $$
We have $$ u_t = \frac{t}{m\cdot n} $$
We want to determine $$\sum_{t=0}^{m\cdot n} u_t = \sum_{t=0}^{m\cdot n} \frac{t}{m \cdot n}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \sum_{t=0}^{m\cdot n} t$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \frac{(m \cdot n)\cdot(m \cdot n +1)}{2}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{m \cdot n +1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does there exist a triple of $distinct$ numbers $a,b,c$ such that $(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$?
Does there exist a triple of distinct numbers $a,b,c$ such that $$(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$$ ?
SOURCE : Inequalities (PDF) (Page Number 4 ; Question Number 220.1)
I tried expanding the brackets and I ended up with this messy equation :
$$-5 a^4 b + 5 a^4 c + 10 a^3 b^2 - 10 a^3 c^2 - 10 a^2 b^3 + 10 a^2 c^3 + 5 a b^4 - 5 a c^4 - 5 b^4 c + 10 b^3 c^2 - 10 b^2 c^3 + 5 b c^4 = 0$$
There is no hope of setting $a=b$ or $a=c$ as the question specifically asks for distinct numbers. So, at last I started collecting, grouping, factoring and manipulating the terms around but could find nothing. Wolfram|Alpha gives a solution as :
$$c=\dfrac{1}{2}\big(\pm\sqrt{3}\sqrt{-(a-b)^2} + a+b\big)$$
How can this solution be found?
Another thing I notice about the solution is that it contains a negative term inside the square root, so does that mean that the solution involves complex numbers and that there is no solution for $\big(a,b,c\big)\in \mathbb {R}$ ?
I am very confused about how to continue. Can anyone provide a solution/hint on how to 'properly' solve this problem ?
Thanks in Advance ! :)
| Assume that $a,b,c$ are distinct.
Let $a-b=x \neq 0,\; b-c=y \neq 0$. Note that $a-c=x+y \neq 0$
Note that the equation becomes $$x^5+y^5=(x+y)^5$$
So $$(x+y)^5-x^5-y^5=5xy(x^3+2x^2y+2xy^2+y^3)=0$$
Note that this becomes $$xy(x+y)(x^2+y^2+xy)=0 \iff x^2+xy+y^2=0$$
Using the quadratic formula, we can find $x,y$. Note that there are only non-real solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)
| Another way.
Let $c=\max\{a,b,c\}$.
Hence, $c\geq1$, $a+b\leq2$ and
$$(2a^2+3)(2b^2+3)\geq\left(\frac{(a+b)^2}{2}+3\right)^2$$
because it's just $$(a-b)^2(12-a^2-6ab-b^2)\geq0$$ and
$$12-a^2-6ab-b^2=12-(a+b)^2-4ab\geq12-2(a+b)^2\geq12-8>0.$$
Thus, it remains to prove that $$\left(\frac{(a+b)^2}{2}+3\right)^2(2c^2+3)\geq125$$ or
$$\left(\frac{(3-c)^2}{2}+3\right)^2(2c^2+3)\geq125$$ or
$$(c-1)^2(2c^4-20c^3+93x^2-190c+175)\geq0,$$
which is obvious.
Done!
I have else proofs, but I think the last is the best.
| {
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"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
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} |
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sqrt{x+2}}}{(x-2)\sqrt{1+\sqrt{x+2}+\sqrt3}}$
I am stuck here.
| $$\lim\limits_{x\to 2^{\pm}}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2} = \left[\frac{\sqrt{3-\sqrt{3}}}{0^{\pm}}\right]=\pm\infty$$
As you can see, the left- and right-side limits are different, so the limit $\lim\limits_{x\to 2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2}$ does not exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since partial derivatives are not equal.
Any help would be appreciated.
| To put it into the form you requested:
$$
-(1+x^2) \,dy + (2xy + x^2 + x^4) \,dx = 0 \implies \frac{dy}{dx} - \frac{2xy + x^2 + x^4}{1+x^2} = 0 \\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = \frac{x^2 + x^4}{1+x^2}\\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = x^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$
My Try (Edited from Comments):
By Cauchy Schwarz, we have that $$ (a^2+1)(1+3) \ge \left(a+\sqrt {3}\right)^2 \rightarrow \frac{a}{\sqrt{a^2+1}} \le \frac{2a}{a+\sqrt{3}} $$ I need a new method
| Another way.
Since $ab+ac+bc\leq\frac{1}{3}(a+b+c)^2\leq1$, by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{1+a^2}}\leq\sum_{cyc}\frac{a}{\sqrt{ab+ac+bc+a^2}}=\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq$$
$$\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac{3}{2}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\left|z^3 + {1 \over z^3}\right| \le 2$ then $\left|z + {1 \over z}\right| \le 2$
$\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2$ prove that $\displaystyle \left|z + {1 \over z}\right| \le 2$
$$\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3}\right) = \left(z + {1\over z}\right)\left(z^2 - 1 + {1\over z^2} \right)\left(\overline z + {1\over \overline z}\right)\left(\overline z^2 - 1 + {1\over\overline z^2} \right)$$
$$=\left|z + {1\over z}\right|^2\left|z^2 - 1 + {1\over z^2} \right|^2 \le 2$$
$$\therefore \left|z + {1\over z}\right| \le \sqrt{2}$$ where $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$.
But I am not able to prove $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$, need some help on this.
| It's all about identities.
Note that $\left( z + \frac 1z\right)^3 = z^3 + \frac 1{z^3} + 3\left(z + \frac 1z\right)$.
Apply the triangle inequality:
$$
\left|\left( z + \frac 1z\right)^3 \right| \leq \left|z^3 + \frac 1{z^3}\right| + 3\left|\left(z + \frac 1z\right)\right|
$$
Using what you know:
$$
\left|\left( z + \frac 1z\right) \right|^3 - 3\left|\left(z + \frac 1z\right)\right| \leq \left|z^3 + \frac 1{z^3}\right| \leq 2
$$
Suppose that $x = \left|\left( z + \frac 1z\right) \right|$, then $x^3 - 3x \leq 2$ is true, along with $x \geq 0$.
To solve this, note that $x^3-3x = x(x^2-3)$, which are both increasing functions for $x\geq 0$. So, $x^3-3x$ is also increasing. Hence, we only need to find when $x^3 - 3x=2$, which happens at $x=2$. Hence, we can conclude, by the increasing property, that $0 \leq x \leq 2$ is true, but this is the conclusion of the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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List all the elements of $A = \{ n \in \mathbb{Z} \mid \frac{n^2-n+2}{2n+1} \in \mathbb{Z}\}$ I was given the following set $A = \{ n \in \mathbb{Z} \mid \frac{n^2-n+2}{2n+1} \in \mathbb{Z}\}$
I have to list all the elements of $A$.
I started using the Euclidean division:
$$n^2-n+2=(2n+1)(\frac{1}{2}n-\frac{3}{4}) + \frac{11}{4}$$
In order to eliminate the fractions, I multiplied the whole expression with 4:
$$4(n^2-n+2)=(2n+1)(2n-3) + 11$$
I´m stuck here, maybe I´m on the wrong way. Could anyone give me some hints how can I approach this problem. Thank you in advance.
| The answer based on the hint of @lhf
$$4\frac{n^2-n+2}{2n+1}= 2n-3+ \frac{11}{2n+1}$$
$$4\frac{n^2-n+2}{2n+1} \in \mathbb{Z} \Rightarrow \frac{11}{2n+1} \in \mathbb{Z}$$
$$\Leftrightarrow 2n+1 \in D_{11} \Leftrightarrow 2n+1 \in \{-11, -1, 1 , 11\}$$
$$\Leftrightarrow 2n \in \{-12, -2, 0 , 10\}$$
$$\Leftrightarrow n \in \{-6, -1, 0 , 5\}$$
$$\Rightarrow A = \{-6, -1, 0 , 5\}$$
Another approach:
$$\frac{n^2-n+2}{2n+1} \in \mathbb{Z}$$
$$\Leftrightarrow n^2-n+2 \equiv 0[2n+1]$$
$$\Rightarrow 4(n^2-n+2) \equiv 0[2n+1]$$
$$\Leftrightarrow (2n+1)(2n-3) + 11 \equiv 0[2n+1]$$
$$\Rightarrow 11 \equiv 0[2n+1]$$
$$\Rightarrow (2n+1) \mid 11$$
$$\Leftrightarrow (2n+1) \in D_{11}$$
Another approach:
$$4(n^2-n+2)=(2n+1)(2n-3) + 11$$
According to the successive division:
$$4(n^2-n+2)\wedge (2n+1)=(2n+1) \wedge 11$$
We have $$(\forall n \in \mathbb{Z}) 4 \wedge (2n+1)= 1$$
Since $(\forall n \in \mathbb{Z}) n^2 - n + 2 \in \mathbb{Z^*}$, then
$$4 \wedge (2n+1)= 1 \Rightarrow 4(n^2-n+2)\wedge (2n+1) = (n^2-n+2)\wedge (2n+1)$$
$$\Leftrightarrow (n^2-n+2)\wedge (2n+1) = (2n+1) \wedge 11$$
We have $ \frac{n^2-n+2}{2n+1} \in \mathbb{Z}$, then $(2n+1) \mid (n^2-n+2)$
$$\Leftrightarrow (n^2-n+2)\wedge (2n+1) = \mid2n+1\mid$$
$$\Leftrightarrow (2n+1) \wedge 11 = \mid2n+1\mid$$
$$\Leftrightarrow (2n+1) \in D_{11} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to validate the following inequality? $\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$
| Assume that $z \neq 0$ and $x, y$ are not simultaneously zero. (For example, if $z$ is zero, $b$ can be any arbitrary negative number.)
Consider the equation $$b = -\sqrt{\frac{x + y}{z}}a + \sqrt{\frac{2(x + y + z)}{z}}.$$
Let $a^* = \sqrt{\frac{2(x + y + z)}{x + y}}$ and $b^* = \sqrt{\frac{2(x + y + z)}{z}}$, i.e., $a^*$ is the $a$-intercept and $b^*$ is the $b$-intercept. The least value of $a$, $b$ (hence $a + b$) is then occurring at $(a^*, 0)$ if $a^* < b^*$ and $(0, b^*)$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find value of $\sin 4x$ Given $\tan x = \frac { 1+ \sqrt{1+x}}{1+ \sqrt{1-x}}$. i have to find value of $\sin 4x$. i write $\sin 4x=4 \frac{ (1-\tan^2 x)(2 \tan x)}{(1+\tan^2 x)^2}$ but it seems very complicated to do this? Any other methods?
Thanks
| Write $\tan x=\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}$
As $-1\le y\le1$ for real $\tan x$
WLOG let $y=\cos2u$ where $0\le2u\le\pi$
$$\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}=\dfrac{1+\sqrt2\cos u}{1+\sqrt2\sin u}=\dfrac{\dfrac1{\sqrt2}+\cos u}{\dfrac1{\sqrt2}+\sin u}=\dfrac{2\cos\dfrac{45^\circ +u}2\cos\dfrac{45^\circ-u}2}{2\sin\dfrac{45^\circ +u}2\cos\dfrac{45^\circ-u}2}=\cot\dfrac{45^\circ +u}2$$
So, $\tan x=\tan\left(90^\circ-\dfrac{45^\circ +u}2\right)$
$\implies x=n180^\circ+90^\circ-\dfrac{45^\circ +u}2$ where $n$ is any integer
$\implies\sin4x=-\sin(90^\circ+2u)=-\cos2u$
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $7^{13} \mod 40$ I wanted to compute $7^{13} \mod 40$. I showed that
$$7^{13} \equiv 2^{13} \equiv 2 \mod 5$$
and
$$7^{13} \equiv (-1)^{13} \equiv -1 \mod 8$$.
Therefore, I have that $7^{13} - 2$ is a multiple of $5$, whereas $7^{13} +1$ is a multiple of $8$. I wanted to make both equal, so I solved $-2 + 5k = + 8n$ for natural numbers $n,k$ and found that $n = 9, k = 15$ gave a solution (just tried to make $3 + 8n$ a multiple of $5$. Therefore, I have that
$$7^{13} \equiv -73 \equiv 7 \mod 40.$$
Is this correct? Moreover, is there an easier way? (I also tried to used the Euler totient function, but $\phi(40) = 16$, so $13 \equiv -3 \mod 16$, but I did not know how to proceed with this.)
| $$\phi(40)=16 \to 7^{16}\equiv 1$$mod 40 $$7^{16}\equiv 1 \\7^{13}7^3\equiv 1\\
343.7^{13}\equiv 1 \\
(320+23).7^{13}\equiv 1\\
23.7^{13}\equiv 1\\
23.7^{13}\equiv 1+40 \equiv 1+80\equiv 1+120\equiv 1+160\\
23.7^{13}\equiv 161 \\23.7^{13}\equiv 7(23)$$ divide by $23$ , $(23,40)=1
$
so
$$23.7^{13}\equiv 7(23) \to 7^{13}\equiv 7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can numbers in a sequence be equal after performing some operations I have a sequence of 21 numbers (-10, -9, -8, ..., -1, 0 , 1, 2, ... , 10). I can take any pair $(a, b)$ from the sequence and replace it with pair $((3a - 4b)/5 ), (4a + 3b)/5)$. Is it possible that all numbers in a sequence will be equal after performing some operations?
| Note that after any replacement, all terms of the sequence are still rational numbers.
Suppose in a given step, you replace the pair $\small{\displaystyle{(a, b)}}$ by the pair
$\small{\displaystyle{\left(\frac{3a - 4b}{5}, \frac{4a + 3b}{5}\right)}}$.
Observing that
\begin{align*}
\left(\frac{3a - 4b}{5}\right)^2 + \left(\frac{4a + 3b}{5}\right)^2
&=
\frac{9a^2 - 24ab + 16b^2}{25} + \frac{16a^2 + 24ab + 9b^2}{25}\\[6pt]
&=\frac{25a^2+25b^2}{25}\\[6pt]
&=a^2 + b^2
\end{align*}
it follows that, after any replacement, the sum of the squares of the terms remains the same.
Suppose that after some number of replacements, the terms are all equal to $x$ say.
Since the sum of the squares must be equal to the original sum of the squares, we get
\begin{align*}
&x^2 + x^2 + x^2 + \cdots + x^2 = (-10)^2 + (-9)^2 + (-8)^2 + \cdots + 8^2 + 9^2 + {10}^2\\[6pt]
\implies\; &21x^2=2(1^2+2^2+3^2 + \cdots + 10^2)\\[6pt]
\implies\; &21x^2=2\left(\frac{(10)(10+1)(2(10)+1)}{6}\right)\\[6pt]
\implies\; &21x^2=(10)(11)(7)\\[6pt]
\implies\; &x^2 = {\small{\frac{110}{3}}}\\[6pt]
\end{align*}
contradiction, since $\large{\frac{110}{3}}$ is not the square of a rational number.
It follows that it's not possible, after some number of replacements, to get all equal terms.
| {
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An integral of rational function with third power of cosine hyperbolic function Prove
$$\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx =
\frac{2}{\pi^3}\left(\pi \cosh\left(\frac{\pi}{4} \right)-4\sinh\left(
\frac{\pi}{4}\right) \right)$$
Attempt
Note that
$$\cosh\left( x+\frac{\pi}{4}\right) = \cosh(x)\cosh\left(\frac{\pi}{4} \right)+\sinh(x)\sinh\left( \frac{\pi}{4}\right)$$
Then the integral could be rewritten as
$$I = \cosh\left(\frac{\pi}{4} \right)\int_{-\infty}^{\infty}\frac{\mathrm{sech} ^2(x)}{(5 \pi^2 + 8 \pi x + 16x^2) }dx\\+\sinh\left(\frac{\pi}{4} \right)\int_{-\infty}^{\infty}\frac{\sinh(x)}{(5 \pi^2 + 8 \pi x + 16x^2) \cosh(x)^3}dx$$
You can then integrate by part the second integral
$$\int^{\infty}_{-\infty}\left[\frac{\cosh\left(\frac{\pi}{4} \right)}{(5 \pi^2 + 8 \pi x + 16x^2)}-\frac{ 4\sinh\left( \frac{\pi}{4}\right)(\pi+ 4 x)}{(5 \pi^2 + 8 \pi x + 16 x^2)^2}\right]\mathrm{sech}^2(x)\,dx $$
Integrating again
$$I=-\int^{\infty}_{-\infty}\left[\frac{(8 (4 x + \pi) (32 x + 8 \pi) \sinh(\pi/4))}{(16 x^2 + 8 \pi x + 5 \pi^2)^3} - \frac{(16 \sinh(\pi/4)}{(16 x^2 + 8 \pi x + 5 \pi^2)^2}\\ - \frac{((32 x + 8 \pi) \cosh(\pi/4)}{(16 x^2 + 8 \pi x + 5 \pi^2)^2} \right]\tanh(x)\,dx$$
Note that
$$\tanh(x) = 8 \sum_{k=1}^\infty \frac{x}{(1 - 2 k)^2 \pi^2 + 4 x^2}$$
Consider $R(x)$ a rational function then
$$\int^{\infty}_{-\infty}R(x) \tanh(x) = 8 \sum_{k=1}^\infty \int^{\infty}_{-\infty}R(x)\frac{x}{(1 - 2 k)^2 \pi^2 + 4 x^2} \,dx$$
Any integral of that form could be found (I think) using the residue theorem then the resulting sum can be evaluated using the Digamma function.
Question
*
*Although I think this approach will result in the correct answer I feel that a contour method will be so much easier, any idea ?
*Maybe there is an easier method considering the nice closed form ?
| $$I=\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\int_{-\infty}^{\infty}\frac{1}{(4x+\pi+2i\pi)(4x+\pi-2i\pi)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\frac{-i}{16\pi}\int_{-\infty}^{\infty}\left(\frac{1}{x+\frac{\pi}{4}-\frac{i\pi}{2}
}-\frac{1}{x+\frac{\pi}{4}+\frac{i\pi}{2}}\right)\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
Make the substitution $x+\pi/4 \to x$
$$I=\frac{-i}{16\pi}\int_{-\infty}^{\infty}\left(\frac{1}{x-\frac{i\pi}{2}
}-\frac{1}{x+\frac{i\pi}{2}}\right)\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
We now split into two integrals and investigate each
$$I_1=\int_{-\infty}^{\infty}\frac{1}{x+\frac{i\pi}{2}}\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
Let $x+\frac{i\pi}{2} \to x$
$$\color{red}{I_1=\int_{\frac{i\pi}{2}-\infty}^{\frac{i\pi}{2}+\infty}\frac{\sinh(x)}{x\cosh^3(π/4 - x)} dx}$$
$$I_2=\int_{-\infty}^{\infty}\frac{1}{x-\frac{i\pi}{2}}\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
Let $x-\frac{i\pi}{2} \to x$
$$\color{red}{I_2=\int_{-\frac{i\pi}{2}-\infty}^{-\frac{i\pi}{2}+\infty}\frac{\sinh(x)}{x\cosh^3(π/4 - x)} dx}$$
We now need to calculate $I_2 - I_1$ and multiply the result by $\frac{-i}{16}$, but I am stumped. There are clear symmetries in $I_1$ and $I_2$, as only the domain of integration changes; the function inside the integral stays the same. If anyone has any suggestions, I will gladly take them.
| {
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"url": "https://math.stackexchange.com/questions/2150148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\frac{\ln(u)}{u-1}$, then I infer that $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, then I have:
$$\sqrt{3-2} \rightarrow 1
\\ 3^2-9 \rightarrow 0
$$
and so
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = 1$$
But this is wrong. So I tried another method:
2:
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = \frac{\ln(\sqrt{x-2})}{(x-3)(x+3)} = \lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{(x-3)} \cdot \lim_{x \rightarrow 3} \frac{1}{x+3} = \frac{1}{6}$$
Which is also wrong.
My questions are: What did I do wrong in each method and how do I solve this?
EDIT: If $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, and $\ln(x) \rightarrow 0^+$, does this mean that $\frac{0^+}{0^+} \rightarrow 1$?
| Just in case you want to see another approach.
Consider $$A=\frac{\ln(\sqrt{x-2})}{x^2-9}=\frac 12\frac{\ln({x-2})}{x^2-9}$$ Now, as Simply Beautiful Art answered, let $x=u+3$ which makes $$A=\frac{\log (1+u)}{2 u (u+6)}$$ Now, use Taylor series around $u=0$ $$\log(1+u)=u-\frac{u^2}{2}+O\left(u^3\right)$$ which makes $$A=\frac{u-\frac{u^2}{2}+O\left(u^3\right)}{2 u (u+6)}=\frac{1-\frac{u}{2}+O\left(u^2\right)}{2 (u+6)}$$ Now, long division $$A=\frac{1}{12}-\frac{u}{18}+O\left(u^2\right)$$ which shows the limit and how it is approached.
| {
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What is the remainder when $15^{40}$ divided by $1309$? I know that $$ 15 \equiv 1\pmod{7}, N \equiv 1\pmod{7},$$ but cannot proceed further.
| Note that $1309=7 \times 11 \times 17$. So $$15^{40} \equiv 1^{40} \equiv 1 \pmod {7}$$
Using $15 \equiv 1 \pmod {7}$. Then, $$15^{40} \equiv \left(15^{10}\right)^4\equiv 1 \pmod {11}$$
And $$15^{40} \equiv (15+17 \times 2)^{40} \equiv 7^{80} \equiv \left(7^{16}\right)^5 \equiv 1 \pmod {17}$$
From Fermat's Little Theorem. So $$15^{40} \equiv 1 \pmod {1309}$$
Using CRT. We are done.
| {
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What is the sum of the solutions to $6x^3+7x^2-x-2=0$ What is the easy way to solve the problem?
The sum of the solutions to $6x^3+7x^2-x-2=0$ is:
$$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$
| Constant is -2.
Using remainder theorem
So you have 1,-1,2,-2.
By putting-1 in x you get to know that -1 satisfies the solution.
So -1is a solution and x+1 is a factor of above equation.
After dividing 6x^2+7x^2-x-2 by x+1 we get,
6x^2 +x-2 as quotient.
Solving this quadratic equation you get 1/2 and -2/3 as solutions.
So -1 + 1/2 + -2/3= -7/6 none of the above.
| {
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Minimize $P=5\left(x^2+y^2\right)+2z^2$ For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$
I have no idea. Can you make a few suggestions?
| Let $x=y=2$ and $z=4$.
Hence, $P=72$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$5(x^2+y^2)+2z^2\geq72\left(\sqrt[3]{\frac{(x+y)(x+z)(y+z)}{144}}\right)^2.$$
It's enough to prove last inequality for non-negative variables.
Let $x+y=tz$.
Since $x^2+y^2\geq\frac{1}{2}(x+y)^2$ and $(x+z)(y+z)\leq\frac{1}{2}(x+y+2z)^2$, it remains to prove that
$$\frac{5}{2}t^2+2\geq72\left(\sqrt[3]{\frac{\frac{t(t+2)^2}{4}}{144}}\right)^2$$ or
$$(5t^2+4)^3\geq9t^2(t+2)^4,$$
which is C-S and AM-GM:
$$(5t^2+4)^3=\left(\frac{(5+4)(5t^2+4)}{9}\right)^3\geq\left(\frac{(5t+4)^2}{9}\right)^3=$$
$$=\left(\frac{(3t+2(t+2))^2}{9}\right)^3\geq\left(\frac{\left(3\sqrt[3]{3t\cdot(t+2)^2}\right)^2}{9}\right)^3=9t^2(t+2)^4.$$
Done!
| {
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Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| We have $7^4 = 49^2 < 50^2 = 4 \times 5^4 < 5^5$. Hence
$$7^{55} < 7^{56} = (7^4)^{14} < (5^5)^{14} = 5^{70} < 5^{72}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$
If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?
Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Let us consider this as a quadratic form respect to $a$.
Applying the quadratic formula $a=\frac{-\cos x\pm\sqrt{5\cos^2 x+4\cos x}}2 $
and substituting $\cos x$ with $1$ and $-1$
we get 3 values of where the graph should touch the x axis $-2,0,1$
How should I proceed now?
| The smallest positive integer is $1$. That doesn't satisfy the condition, because
$$ \sin^2 x + \cos x + 1 > 1 + \cos x $$
has equality when $\sin^2x = 0$, which happens, among other places, at $x=0$.
The next positive integer is $2$. Does that work?
$$ \sin^2 x + 2\cos x + 4 > 1 + \cos x $$
holds if and only if
$$ \sin^2 x + \cos x + 3 > 0 $$
which is easily true -- since $\sin^2 x$ is never less than $0$ and $\cos x$ is never less than $-1$, the left-hand side is always $\ge 2$.
So the answer is $$ \Huge 2 $$
| {
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Find the limit : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$ Find the limit: Without the use of the L'Hôspital's Rule
$$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$
My try:
$u=x-1$
Now:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-\sqrt[n]{n}-\sqrt[n]{n(u+1)-n}}$$
| We can simplify the term of interest and rationalize terms to obtain
$$\begin{align}
\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}&=\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}\,)}\\\\
&=\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}\,\,\sqrt[n]{x-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}\,)}\\\\
&=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{\sqrt[n]{x-1}}{\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}}\right)\\\\
&=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{\sqrt[n]{x-1}}{\frac{x-1}{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}-\sqrt[n]{x-1}}\right)\\\\
&=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{1}{\frac{\sqrt[n]{(x-1)^{n-1}}}{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}-1}\right)\\\\
&\to \left(\frac{\sqrt[n]{n}}{\sqrt[n]{n}}\right)\left(\frac{1}{\frac{0}{\sqrt[n]{n}}-1}\right)=-1
\end{align}$$
| {
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For $a,b,c \in R$ and $a,b,c>0$. Minimize $A=a^3+b^3+c^3$ For $a,b,c \in R$ and $a,b,c>0$ satisfy $a^2+b^2+c^2=27$, minimize $$A=a^3+b^3+c^3$$
| By Power-Mean Inequality,
$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$
$a^3+b^3+c^3 \ge 81$
| {
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Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square. There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.
EDIT: I found solution by symmetric polynomials, which is posted below.
|
EDIT: I found solution by symmetric polynomials (in variables a, b, c)
The following more or less transcribes OP's solution in direct calculations, without explicitly using Newton's relations. From the assumption that $a+b+c=0\,$:
$$
0 = (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)
$$
$$
\implies 2(ab+bc+ca)=-(a^2+b^2+c^2) \tag{1}
$$
$$
\require{cancel}
(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + \cancel{2abc(a+b+c) } \tag{2}
$$
$$
\begin{align}
a^4+b^4+c^4 & = (a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2) \\[5px]
& \overset{(1),(2)}{=} 4(ab+bc+ca)^2 - 2(ab+bc+ca)^2 \\
& = 2 (ab+bc+ca)^2
\end{align}
$$
The latter gives $32(a^4+b^4+c^4)=\big(8 (ab+bc+ca)\big)^2\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.
| By applying Caushy Schwartz inequality we get
( 1 + a + b + c ) ( 1 + 1/a + 1/b + 1/c ) >= 16
Equality holds when a^2 = b^2 = c^2 = 1
Therefore
a = b = c = 1
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\sum\tan{\frac{A}{2}}-\frac{\sqrt{3}-1}{8}\prod\csc{\frac{A}{2}}\le 1$ In $\Delta ABC$,show that
$$\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}}-\frac{\sqrt{3}-1}{8}\csc{\frac{A}{2}}\csc{\frac{B}{2}}\csc{\frac{C}{2}}\le 1$$
I tried also $$\left(\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}}\right)^2\ge 3\left(\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{A}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{B}{2}}\right)=3,$$ but we get there something, which impossible to kill during a competition.
| Easy to show that $\sum\limits_{cyc}\tan\frac{\alpha}{2}=\frac{4R+r}{p}$ and $\prod\limits_{cyc}\sin\frac{\alpha}{2}=\frac{r}{4R}$.
Also, if $M$ is a center gravity of the triangle and $I$ is a center of the inscribed circle in the triangle,
so $MI^2=\frac{p^2+5r^2-16Rr}{9}$, which gives $p\geq\sqrt{16Rr-5r^2}$.
Let $R=2xr$. Hence, we need to prove that
$$\frac{4R+r}{p}-\frac{(\sqrt3-1)R}{2r}\leq1$$ or
$$p\geq\frac{2r(4R+r)}{2r+(\sqrt3-1))R},$$
for which it's enough to prove that
$$\sqrt{16Rr-5r^2}\geq\frac{2r(4R+r)}{2r+(\sqrt3-1))R}$$ or
$$(32x-5)((\sqrt3-1)x+1)^2\geq(8x+1)^2$$ or
$$(x-1)(32(2-\sqrt3)x^2-5(2-\sqrt3)x+3)\geq0,$$
which is obviously true for $x\geq1$.
Done!
| {
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Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
| Hint:
In general a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested if $a^2-b$ is a perfect square and in this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
In Your case $a^2-b=1$ so....
| {
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"timestamp": "2023-03-29T00:00:00",
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How many combinations of pennies, dimes, nickels, and quarters create 0.32$? I need help solving this. I cannot find the complete number of combinations. I have already found $5$, but I can't find any more.
| Leaving out the pennies in each combo ...
There are 2 combos with a quarter:
25+5 (that is: 1 quarter + 1 nickel ...so 2 pennies...)
25 (So just a quarter ...so 7 pennies ... for combos below, you'll have to figure out how many pennies to add ...)
There is 1 combo with 3 dimes:
10+10+10
There are 3 combos with 2 dimes:
10+10+5+5
10+10+5
10+10
There are 5 combos with 1 dime:
10+5+5+5+5
10+5+5+5
10+5+5
10+5
10
There are 7 combos without dimes or quarters:
5+5+5+5+5+5
5+5+5+5+5
5+5+5+5
5+5+5
5+5
5
(32 pennies)
Total: 18 combos
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
| I write it step by step:
With $\sin A+\sin^2 A=1$ we have
$\sin A=1-\sin^2 A=\cos^2A$
so
\begin{eqnarray}
&& a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0\\
&& a(\cos^2A)^6+b(\cos^2A)^4+c(\cos^2A)^3-1=0\\
&& a\sin^6A+b\sin^4A+c\sin^3A-1=0\\
&& a(1-\cos^2A)^3+b(1-\cos^2A)^2+c\sin A(1-\cos^2A)-1=0\\
&& a(1-\sin A)^3+b(1-\sin A)^2+c\sin A(1-\sin A)-1=0\\
&& a(1-3\sin A+3\sin^2A-\sin^3A+b(1-2\sin A+\sin^2A)+c(\sin A-\sin^2 A)-1=0\\
&& a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\
&& a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\
&& a-3a\sin A+3a(1-\cos^2A)-a\sin A(1-\cos^2A)+b-2b\sin A+b(1-\cos^2A)+c\sin A-c(1-\cos^2A)-1=0\\
&& a-3a\sin A+3a(1-\sin A)-a\sin A(1-\sin A)+b-2b\sin A+b(1-\sin A)+c\sin A-c(1-\sin A)-1=0\\
&& a-3a\sin A+3a-3a\sin A-a\sin A+a\sin^2A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\
&& a-3a\sin A+3a-3a\sin A-a\sin A+a-a\sin A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\
&& (a+3a+a+b+b-c-1)+(-3a-3a-a-a-2b-b+c+c)\sin A=0\\
&& (5a+2b-c-1)+(-8a-3b+2c)\sin A=0
\end{eqnarray}
| {
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What is the limit of this sequence created using Pascal's triangle? Let use the Pascal Triangle to create a sequence. We sum from the first term on the left to the term in the middle
and then we take the inverse root times two. We get then that the terms are
\begin{eqnarray}
&&a_0 := 2 \\
&&a_1 = 2 \frac{1}{1^{1/1}} = 2\\
&&a_2 = 2 \frac{1}{(2+1)^{1/2}} = 1.154700\dots\\
&&a_3 = 2 \frac{1}{(3+1)^{1/3}} = 1.25992\dots\\
&&a_4 = 2 \frac{1}{(6+4+1)^{1/4}} = 1.098200\dots\\
&&a_5 = 2 \frac{1}{(10+5+1)^{1/5}} = 1.14869\dots\\
&&a_6 = 2 \frac{1}{(20+15+6+1)^{1/6}} = 1.0727\dots\\
&&a_7 = 2 \frac{1}{(35+21+7+1)^{1/7}} = 1.10408\dots\\
\end{eqnarray}
The first line of the triangle creates the $a_0$ the second creates the $a_1$ and so on. Then I have my conjecture:
This sequence converges to 1
But I couldn't prove it or disprove it. What I did was tried to relate to the Binomial expansion: Let $0\leq x \leq 1$ so we have that
$$(1+x)^2 = 1 + 2x^2 + x^4 \geq x^2 + 2x^2 = 3x^2$$
So we have that
$$\frac{x}{x+1} \leq \frac{1}{\sqrt{3}}$$
And I tried to translate this to some limit using this $x$. There is a way to solve this problem like this? This series converges?
| The sum of the elements in "half a row" of Pascal's triangle is either $2^{n-1}$ (if we are talking about the $n$-th row with $n$ being odd) or $2^{n-1}+\frac{1}{2}\binom{n}{n/2}$ if $n$ is even, due to $\binom{n}{k}=\binom{n}{n-k}$.
The central binomial coefficient $\binom{2n}{n}$ behaves in the following way
$$ \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}} $$
hence you are essentially asking about
$$ \lim_{n\to +\infty} 2\cdot\frac{1}{(2^{n-1}+E(n))^{1/n}} $$
that is clearly $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many integers between $10000$ and $99999$ ( inclusive) are divisible by $3$ or $5$ or $7$? How many integers between $10000$ and $99999$ (inclusive) are divisible by $3$ or $5$ or $7$ ?
My Try :
Total Integers between $10000$ and $99999$ are $89999$.
$\left\lfloor\frac{89999}{3}\right\rfloor+\left\lfloor\frac{89999}{5}\right\rfloor+\left\lfloor\frac{89999}{7}\right\rfloor$ - $\left\lfloor\frac{89999}{3\times5}\right\rfloor-\left\lfloor\frac{89999}{3\times7}\right\rfloor-\left\lfloor\frac{89999}{5\times7}\right\rfloor$ + $\left\lfloor\frac{89999}{3\times5\times7}\right\rfloor$ =
$48857$
I don't have an answer for this. Am I right here ?
| The number of integers between $3$ and $4$ inclusive divisible by $3$ is not $\lfloor\frac{(4-3+1)}{3}\rfloor=\lfloor\frac{2}{3}\rfloor=0$ but is instead $\lfloor \frac{4}{3}\rfloor - \lfloor\frac{2}{3}\rfloor = 1$.
In the same way, you should not be using $\lfloor \frac{90000}{7}\rfloor$ but instead you should be using $\lfloor\frac{99999}{7}\rfloor - \lfloor\frac{9999}{7}\rfloor$
That is to say, the number of numbers $n$ in the range $a\leq n\leq b$ which is divisible by $7$ are those numbers in the range $1\leq n\leq b$ divisible by seven which are not numbers in the range $1\leq n\leq a-1$ which are divisible by $7$.
| {
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Sum of some natural numbers equal to $n$ In how many ways we can have some natural numbers that their sum is equal to $n$ and none of them is greater than $k$, for given $n$ and $k$?
NOTE: We don't know the number of the elements.
Can anyone help me with this problem? I can't find anything on the internet.
For example, for $n = 5$ and $k = 2$ the answer is $8$:
2 + 2 + 1, 2 + 1 + 2, 1 + 2 + 2, 1 + 1 + 1 + 2, 1 + 1 + 2 + 1
1 + 2 + 1 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1
| How about using generating functions? The number of ways to add $m$ positive integers each of which is less than or equal to $k$ so that their sum is $n$ is the coefficient of $x^n$ in $(x+x^2+\ldots+x^k)^m$. So the coefficient of $x^5$ in $\sum_{m=1}^{5} (x+x^2)^m$ is the answer for your example, which is 8.
| {
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"url": "https://math.stackexchange.com/questions/2168595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I add the terms in the binomial expansion of $(100+2)^6$? So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)$+$\binom {6} {1}$$100^5(2)$+....
I stumbled upon this step. How did they add the humongous numbers? I am really confused.
Kindly help me clear my query.
| $10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$
$= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An integration $ \int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x.$
How can we evaluate
$$\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x?$$
Usually when having as the integrand a logarithmic function, the first thing would be to try to integrate by parts, however in this case there's a product of logarithms and trying to integrate by parts would just produce two harder integrals and it might not be a good approach.
There is also An integral by O. Furdui $\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$ which is quite similar to this one, however the substitution $\sqrt{1+x}-\sqrt{1-x}=2\sin t$ doesn't seem to be useful due to the $\sqrt{1+x}+\sqrt{1-x}$ term.
Is there a better approach to this integral that reduces it to something simpler?
| One may prove that
$$
\int_0^1 x\ln \left( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left( \sqrt {1+x} -\sqrt{1-x}\right)\:\mathrm{d}x=\frac{\ln^2 2}4-\frac{3\ln2}4+\frac1{16}. \tag1
$$
Hint. By using
$$
\begin{align}
4ab&=(a+b)^2-(a-b)^2
\end{align}
$$ giving
$$
\begin{align}
4\ln b \ln c&=\ln^2 (bc)-\ln^2 \left(\frac bc\right)
\end{align}
$$ and by observing that
$$
\begin{align}
&\left ( \sqrt{1+x}+\sqrt{1-x}\right)\left ( \sqrt {1+x} -\sqrt{1-x} \right)=2x
\\
&\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt {1+x} +\sqrt{1-x}}=\frac{x}{1+\sqrt{1-x^2}},\quad 0<x<1,
\end{align}
$$ one gets
$$
\begin{align}
4&\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\:\mathrm{d}x\\=&\int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x
-\int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:\mathrm{d}x. \tag2
\end{align}
$$ Then integrating by parts twice one gets
$$
\int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x=\frac{\ln^2 2}2-\frac{\ln2}2+\frac14, \tag3
$$ by the change of variable $u=\dfrac{x}{1+\sqrt{1-x^2}}$, one has
$$
\begin{align}
\int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:dx&=\int_0^1\frac{4u(1-u^2)}{(1+u^2)^3}\cdot \ln^2 u\:du
\\&=\frac12\int_0^1\frac{(1-v)}{(1+v)^3}\cdot \ln^2 v\:du
\\&= \ln 2,\tag4\end{align}
$$ obtained by parts. By inserting $(4)$ and $(3)$ into $(2)$ one gets $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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