Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks ver... | For a start:
Let $z=xy+1$, then $$z(z+x+1)=a^2\implies \boxed{z^2+z(x+1)-a^2=0}$$
Since discriminant is perfect square we have $$d^2 =(x+1)^2+4a^2$$ which means that (Pythagorean triples) $$d=k\cdot (u^2+v^2),\;\; x+1 =k\cdot(u^2-v^2),\;\; a = k\cdot uv$$ for some positive integers $u,v,k$ and $\gcd(u,v)=1$. Since $x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Prove that $f(x)=x$ is the only function satisfying certain properties
Let $\mathbb{N}$ be the set of positive integers. Prove that $f(x)=x$ is the only function from $\mathbb{N}\to\mathbb{N}$ for which $(x^2 + f(y)) | (f(x)^2 + y)$ for all $x,y \in \mathbb{N}.$
It's straightforward to show that $f(x)=x$ satisfies th... | Your approach looks fine. Just for fun, here is another way to prove things.
Since $f:\mathbb{N}\to\mathbb{N}$, $x+f(x)$ is positive for all $x\ge1$, so
$$(x^2+f(y))\mid(f(x)^2+y)\implies x^2+f(y)\le f(x)^2+y\implies f(y)-y\le(f(x)-x)(f(x)+x)$$
Setting $x=y$, we find that $f(y)-y\le(f(y)-y)(f(y)+y)$, which tells us $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A surprising results with a interesting inequality let $x,y,z>0$ such $x+y+z=3$, Find the minimum of the value
$$F=\dfrac{1}{x}+\dfrac{1}{xy}+\dfrac{1}{xyz}$$
I use wolfapha found the surprising results $F_{min}=\dfrac{3+\sqrt{5}}{2}$
see links
maybe use AM-GM inequality,but I can't it,so How to prove $F\ge\dfrac{3+\s... | If we allow for 1-variable calculus, then
*
*Fix $y+z = k$. Show that the minimum of $\frac{1}{y} + \frac{1}{yz} = \frac{z+1}{z(k-z) } $ subject to $ y> 0, z> 0$ has value
$\frac{1}{ (\sqrt{k+1} -1 ) ^2 }$ and occurs at $z = \sqrt{k+1} - 1$.
*So $ \frac{1}{x} + \frac{1}{xy} + \frac{1}{xyz} = \frac{1}{x} ( 1 + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to solve $3\sec(x)-2\cot(x)>0$ (trigonometric inequality)? I am struggling to find the solution, here is what I already tried (it's for a pre-calculus class so no calculus):
$3\sec(x)-2\cot(x)=0 \rightarrow 3(\frac{1}{\cos(x)})=2\left(\frac{\cos(x)}{\sin(x)}\right) \rightarrow 3\sin(x)=2\cos(x)^2 \rightarrow 3\sqrt... | \begin{align*}
3\sec x - 2\cot x & > 0\\
\frac{3}{\cos x} - \frac{2\cos x}{\sin x} & > 0\\
\frac{3\sin x - 2\cos^2x}{\sin x\cos x} & > 0\\
\frac{3\sin x - 2(1 - \sin^2x)}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 3\sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin^2x + 4\sin x - \sin x - 2}{\sin x\cos x} & > 0\\
\frac{2\sin x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Find the greatest value of $P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}$ The problem
Given that $x,y,z > 0$ and $xz-yz-yx=1$
Find the greatest value of
\begin{align} P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}\end{align}
The inequality is assymetric as you can see. I literally got noth... | You found $1+z^2=xy-yz−zx+z^2=(y-z)(x-z)$
Similarly for $1+x^2$ you get $(x-z)(x+y)$
And for $1+y^2$ you get $(y-z)(x+y)$
Therefore, you can replace:
\begin{align} P= \frac{2x^2}{1+x^2} -\frac{2y^2}{1+y^2}+\frac{3z^2}{1+z^2}\end{align}
with
\begin{align} P = \frac{2x^2}{(x-z)(x+y)} -\frac{2y^2}{(y-z)(x+y)}+\frac{3z^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313823",
"timestamp": "2023-03-29T00:00:00",
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Verification for proof by strong induction of $a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a + 1)$ I'm trying to prove by strong induction that
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a + 1),$$
for $n \geq 1$.
By strong induction, I'd like to know if my solution is valid. What I did was:
*
*proof f... | To prove the induction step, you should prove that
$$a^{n+1}-1=(a-1)(a^n+a^{n-1}+\ldots+a+1),$$
given that
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1).$$
Of course this follows easily from the fact that
$$(a^{n+1}-1)-(a^n-1)=a^{n+1}-a^n=(a-1)a^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$ Consider the following polynomial in $x$:
$$p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n.$$
I want to show that $p(x)\geq 0$ for real $x$. It is trivial to show that $p(-1) = 0$, and by graphing $... | Changing your notation slightly, we have
$$
p_{2k}(x) = \frac{2}{2k+1}\left(b_0 + b_1 x + b_1 x^2 + \cdots + b_{2k}x^{2k}+ b_{2k+1} x^{2x+1}\right),
$$where $b_n = k$ if $n$ is even and $b_n=k+1$ if $n$ is odd. We will prove the claim by induction on $k$. The base case $n=0$ gives $p_0(x) =0$ and we also have $p_2(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}$ Simplify $$\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}$$ if $x=\dfrac{2mn}{n^2+1}$ and $m>0,n>1$.
The solution of the authors starts as follows:
$$\dfrac{\left(\sqrt{m+x}+\sqrt{m-x}\right)^2}{(\sqrt{m+x})^2-(\sqrt{m-x})^2}=\dfrac{m+x+2\sqrt{m^2-x^... | \begin{align*}
\bigg(\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}-\sqrt{m-x}}\bigg)
\bigg(\dfrac{\sqrt{m+x}+\sqrt{m-x}}{\sqrt{m+x}+\sqrt{m-x}}\bigg)
\\
\\
=\dfrac{\big(\sqrt{m+x}+\sqrt{m-x}\big)^2}
{(m+x)-(m-x)}\\
\\
=\dfrac{2 \sqrt{m - x} \sqrt{m + x} + 2 m}
{2x}\\
\\
=\dfrac{\sqrt{m^2 - x^2} + m}
{x}
,\space x=\dfrac{2m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove that $ \frac{n+1}{4n^2+3}$ is a Cauchy sequence? By definition, I need to show that for any $\epsilon \gt 0$ there exists $N \in \mathbb{N}$ such that for any $m,n \gt N$:
$ \lvert a_n - a_m \rvert \lt \epsilon$
So I write,
$ |\frac{n+1}{4n^2+3} - \frac{m+1}{4m^2+3} |\leq | \frac{n+1}{4n^2+3}| + |\frac{m+1... | $4n^2+3\geq 4n^2$. Therefore, $\frac{n+1}{4n^2+3} \leq \frac{2n}{4n^2}= \frac{1}{2n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Golden ratio in complex number squares
In the Argand diagram shown below the complex numbers
$– 1 + i, 1 + i, 1 – i, – 1 – i$ represent the vertices of a square ABCD.
The equation of its diagonal BD is $y = x$. The complex number $k + ki$
where $– 1 < k < 0$ represents the point E which is in the fourth quadrant and
l... | (i)
The slope of the line $EH$ is $\dfrac{k+1}{k+2}$. So, the equation of the line $CF$ is given by $y+1=-\dfrac{k+2}{k+1}(x-1)$. Since $F(f+i)$ is on this line, one has
$$1+1=-\dfrac{k+2}{k+1}(f-1)\implies f=\frac{-k}{k+2}$$
(ii)
Using the fact that $|EF|=|EC|$, one has
$$\begin{align}|EF|^2=|EC|^2&\implies (f-k)^2+(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum $\sum\limits_{k=0}^\infty (-2)^k\frac{k+2}{k+1}x^k$ $$\displaystyle\sum_{k=0}^\infty (-2)^k\dfrac{k+2}{k+1}x^k.$$
I showed that this series converges when $|x|<\dfrac{1}{2}$ because $$\displaystyle\lim_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|=2|x|.$$
Now I have to find the sum result. I tried so far t... | So we want to find the value of
$$\sum_{k=0}^\infty \frac{(k+2)(-2x)^k}{k+1}$$
As you already determined, we know
$$\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^kx^{k+1}}{k+1}$$
If you multiply both sides by $x$ you get
$$x\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+2}}{k+1}$$
Now, the interesting step is to take the deriv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Reflect a point across a line using affine transform I'm trying to reflect a point (6,12) across the line y=7 using an affine transformation. Logically, the line y=7 should be halfway the Y coordinate 12 and the Y coordinate of the answer, so (12+Y)/2=7, or 12+Y=14, or Y=2, so the answer is (6,2).
Wikipedia has example... | To reflect about the plane $\mathbf{n} (\mathbf{r} - \mathbf{r_0} ) = 0 $, ($\mathbf{n}$ is a unit vector), the affine transformation is
$ T(\mathbf{p}) = \mathbf{r_0} + A (\mathbf{p} - \mathbf{r_0} ) $
where
$T = \mathbf{I} - 2 {\mathbf{n}\mathbf{n}}^T $
In this case, the plane is $y = 7$, so $\mathbf{n} = \begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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:Find the area of the shaded region $ABC$ in the figure below For reference:Calculate the area of the shaded region; if: $AP = 4\sqrt2; PB = 4$ and $HN = 4$.(Answer: $\frac{32}{51}(5\sqrt2+4)$)
My progress
$HN(2r-HN)=BH^2\quad\Rightarrow{}\quad 4(2r-4)=4 \therefore R=\dfrac{5}{2}\\
S_{ABC} = \frac{a.b.c}{4R}=\frac... | Based on confirmation that $PA$ is tangent, by using power of point of $P$, $BC = 4$ and using similar triangles, $AC = AB \sqrt2$.
If $OH = x, OB = ON = 4 - x$, and applying Pythagoras in $\triangle OBH$
$x^2 + 2^2 = (4-x)^2 \implies x = OH = 3/2, OB = 5/2$
As $\angle BAC = \angle BOH = \theta $ (say)
$\cos \theta = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Investigate the convergence or divergence of the sequence $a_1 = \sqrt{c}, a_2 = \sqrt{c + a_1}, a_{n+1} = \sqrt{c + a_n}$ $$a_1 = \sqrt{c}, a_2 = \sqrt{c + a_1}, a_{n+1} = \sqrt{c + a_n}, c > 0$$
I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How c... | Claim: $a_n$ is increasing and bounded above by $M = \dfrac{1+\sqrt{1+4c}}{2}$. Also $\displaystyle \lim_{n\to \infty} a_n = M$
Proof: By induction on $n \ge 1$. $a_2 = \sqrt{c+\sqrt{c}}> \sqrt{c} = a_1$. Assume $a_n > a_{n-1}$, we show $a_{n+1} > a_n$. But $a_{n+1} = \sqrt{c+a_n}> \sqrt{c+a_{n-1}}= a_n$ by the inducti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When does this equality hold: $(A + B)^2 = A^2 + B^2$? Where $A$ and $B$ are matrices. Given
$A =
\begin{bmatrix}
2 & -1 \\
2 & -1
\end{bmatrix}
$
and
$
B =
\begin{bmatrix}
a & 1\\
b & -1
\end{bmatrix}
$
, what are the values of $a$ and $b$ if $(A+B)^2 = A^2 + B^2$?
I think if the equality holds then $AB + BA = 0$.
$... | Yes. There are no such values.
For $AB+BA=0$, we must have $$a=2,b=4$$ and $$4a-b+2=0$$ that is simply not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $ Solve $$ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $$
using shift $x=e^{z}$ and differential operator $Dz=\frac{d}{dz}$
What does $Dz = d / dz$ mean?
I did this but I don't know how to continue. Please help.
$$ (e^{z})^{3}{y}'''+e^{z}{y}'-y = e^{z}\ln(e^{z}) \\\\(e^{3z}){y}'''+e^{z}... | Let $Y(z) = y(x) = y(e^z)$. Then Chain Rule and Product Rule give
\begin{align*}
\frac{dY}{dz} & = \frac{dy}{dx}\frac{dx}{dz} = \frac{dy}{dx}e^z = x\frac{dy}{dx} \\
\frac{d^2Y}{dz^2} & = \frac{d}{dz}\left(\frac{dy}{dx}e^z\right) = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^ze^z = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^{2z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}.$ I want to evaluate the following integral ($\theta_0>0$)
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta
\end{equation*}
So I thought about $d(\cos\theta)=-\si... | We first restrict the angle $\theta$ in the range, $0<\theta_0<\theta<\pi, $
then rewrite the integral by double-angle formula
$$1-\cos \theta=2\sin^2 \frac{\theta}{2} \text{ and } 1+\cos \theta=2\cos^2 \frac{\theta}{2} ,$$
\begin{aligned}
& I=\int_{\theta_{0}}^{\pi} \sqrt{\frac{1-\cos \theta}{\cos \theta_{0}-\cos \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Powers of a Toeplitz matrix I'm searching a closed formula to compute the powers of the following matrix
\begin{equation*}F\triangleq \begin{bmatrix}
1 & T & \frac{T^2}{2}\\
0 & 1 & T\\
0 & 0 & 1
\end{bmatrix}\end{equation*}
where $T$ is a given parameter. For example, if I'm not wrong the first 5+1 powers are
\begin{e... | Your matrix can be written as an exponential of a nilpotent matrix:
$$F=e^{TN},\qquad N=\left(\begin{array}{ccc} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right).$$
Note that $N^2=\left(\begin{array}{ccc} 0 & 0 & 1 \\
0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ and $N^3=0$. Now,
$$F^k=e^{kT N}=\mathbf 1+kTN+\frac{k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotic expansion in 0 of a series I need to find a 2 term asymptotic expansion of the serie of functions $f=\sum u_n$
where for all real $x$, $u_n(x) = 1/(n^2+x^2)$. Obviously, the function being continuous (which is easy to prove by uniform convergence over $\mathbb R$) the limit is $\pi^2/6$.
I have no idea on wh... | According to Find the sum of $\sum \frac{1}{k^2 - a^2}$ when $0<a<1$, the following identity holds
$$\sum_{n=1}^\infty \frac1{n^2-z^2}=\frac1{2z^2}-\frac{\pi\cot\,\pi z}{2z}.$$
After letting $z=ix$, we find
$$\sum_{n=1}^\infty \frac1{n^2+x^2}=\frac{\pi\coth(\pi x)}{2x}-\frac1{2x^2}.$$
Finally, after expanding the Hyper... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do ... | One way to go is to use $a^3-b^3=(a-b)(a^2+ab+b^2)$.
So with $a=1$ and $b=\sqrt[3]{t}$:
$$\frac{1-t}{1-\sqrt[3]{t}} = \frac{1-(\sqrt[3]{t})^3}{1-\sqrt[3]{t}}=1^2+1.\sqrt[3]{t}+(\sqrt[3]{t})^2$$
And with $a=1$ and $b=-\sqrt[3]{t}$:
$$\frac{1+t}{1+\sqrt[3]{t}} = \frac{1-(-\sqrt[3]{t})^3}{1+\sqrt[3]{t}}=1^2-1.\sqrt[3]{t}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show this inequality $ a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$ Given the real $ a $. Prove that
$$a^{14}-a^{13}+a^8-a^5+a^2-a+1>0$$
I tried to factor it as
$$(a-1)\Bigl(a^{13}+a^5(a^2+a+1)+a\Bigr)+1$$
I think it should be written as a sum of squares.
Any idea will be appreciated.
| Notice it a bunch of $+ a^{even}$ and $-a^{odd}$ in descending order of powers so.....
Case 1: $a \le 0$ then $a^{even} \ge 0$ and $-a^{odd} \ge 0$ so $a^{14} - a^{13}+a^8-a^5+a^2 -a + 1=|a|^{14}+|a|^{13}+ |a|^8 +|a|^5 + |a|^2+|a| + 1\ge 1$.
Case 2: $a\ge 1$ then $a^{big} \ge a^{small}$ so $a^{big}-a^{small} >\ge 0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number theoretic problems leading to elliptic curves I am looking for basic number theoretic problems that can be addressed by the use of elliptic curves.
I know few of problems such as congruent number problem, or the problem of extending a Diophantine triple $\{a,b,c\}$ to a quadruple $\{a,b,c,x\} $satisfying $ax+1 =... | Um, all of them? (That's an exaggeration, but only slightly.)
For example consider the $n=3$ case of Fermat's Last Theorem. You're looking for nontrivial integer solutions to $a^3 + b^3 = c^3$. Dividing by $c^3$ and setting $x = a/c$ and $y = b/c$, we get the new equation $x^3 + y^3 = 1$, where now we are looking for r... | {
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"source": "stackexchange",
"question_score": "2",
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How is the summation being expanded? I am trying to understand summations by solving some example problems, but I could not understand how is the second to last line being expanded? I would really appreciate if you could explain me how is it being expanded.
\begin{align}
&\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=1}^{j}1... | I take it that what has to be explained is this (I've introduced parentheses on the left hand side for clarity):
\begin{multline*}
\sum_{i=1}^{n-1}\left(n^2 + n - i^2 - i\right) = \\
(n-1)n^2 + (n-1)n - \left(\frac{n(n+1)(2n+1)}6 - n^2\right) - \left(\frac{n(n+1)}2 - n\right).
\end{multline*}
This equation results from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$? What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$.
Note that $f'(x)=\exp(\frac{1}{2}c^2x^2)c^2x$ and $f''(x)=\exp(\frac{1}{2}c^2x^2)(c^2x)^2+\exp(\frac{1}{2}c^2x^2)c^2$. Then $f'(0)=0$ and $f''(0)=c^... | Just use Taylor expansion as if it were $e^X$ where $X = \frac{1}{2}c^2x^2$ hence:
$$e^X = \sum_{k = 0}^{+\infty} \frac{X^k}{k!}$$
$$e^{\frac{1}{2}x^2x^2} = \sum_{k = 0}^{+\infty} \frac{\left(\frac{1}{2}c^2x^2\right)^k}{k!} =\sum_{k = 0}^{+\infty}\frac{1}{2^k k!} (cx)^{2k}$$
The first terms are given by
$$1+\frac{c^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger . I have been asked that which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger without resorting to calculator.
My Attempt:
I tried taking $f(x)$ to be
$f(x)=\sin 4x-\tan 3x$
$f'(x)=4\cos 4x-3\sec^23x=\cos 4x(4-3\sec^2... | Working with radians, you want to compare
$$\sin \left(\frac{\pi }{6}-\frac{\pi}{90}\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8}-\frac{\pi }{120}\right)$$ Let $\epsilon=\frac \pi {120}$ and we shall compare
$$\sin \left(\frac{\pi }{6}-\frac 4 3\epsilon\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Proving that a given mapping is a inner product How to prove that a mapping $\langle\cdot,\cdot\rangle:\mathbb{R_{3}[x]}\times\mathbb{R_{3}[x]}\rightarrow\mathbb{R}$ is a inner product, where $\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx?$
First, for $a,b\in\mathbb{R}$ and $f,g,h\in\mathbb{R_{3}[x]}$, we have
$\la... | The answers given already are the most elegant, but for completeness let me give another way (continuing the approach in the post). Let $f = ax^3 +b x^2+cx +d.$ Then
\begin{align*}
\langle f, f \rangle &= \frac{2}{7} a^2 + \frac{4}{5} ac + \frac{2}{3} c^2 + \frac{2}{5} b^2 + \frac{4}{3} bd + 2d^2 \\
&= \frac{2}{3} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a decomposition of a $2\times 2$ diagonal matrix into the sum of two diagonal matrix which satisfy some conditions on the coefficients Let $M=\begin{pmatrix} \alpha&0\\
0&\beta
\end{pmatrix} $ be an arbitrary $2\times 2$ real constant diagonal matrix with $\alpha\neq 0$ and $\beta\neq 0$. I want to find a decompos... | If $\alpha$ and $\beta$ are both nonnegative then for any $t > 1$,
$$
\begin{pmatrix}
\alpha & 0 \\
0 & \beta
\end{pmatrix}
=
\begin{pmatrix}
t\alpha & 0 \\
0 & t\beta
\end{pmatrix}
+
\begin{pmatrix}
(1-t)\alpha & 0 \\
0 & (1-t)\beta
\end{pmatrix}.
$$
A similar argument works if both are nonpositive.
It's not hard ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating a limit with exponents There is a limit I want to calculate, and I've calculated it as follows:
$$\lim_{x \to \infty} \frac{7^{-x+1}-2\cdot 5^{-x}}{3^{-x}-7^{-x}} = \lim_{x \to \infty} \frac{\frac{7}{7^x}-\frac{2}{5^x}}{\frac{1}{3^x}-\frac{1}{7^x}}$$
If I now multiply the denominator and numerator by $3^x$,... | Your answer is correct. In order to be sure it's nice to write :
$$\lim_{x \to +\infty} \dfrac{7 \dfrac{3^x}{7^x} - 2 \dfrac{3^x}{5^x}}{1 - \dfrac{3^x}{7^x}} = \dfrac{7 \times 0 - 2 \times 0}{1 - 0} = \dfrac{0}{1} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to u... | There is only a single real solution:
For $x\geq -1$ :
$$\left[\sqrt{3} \sqrt{1+x} - \sqrt{2} \sqrt{2+x}\right]^2 + \left[(1+x) - \sqrt{2} \sqrt{1+x^2}\right]^2 \geq 0$$
$$\left[7 + 5 x - 2 \sqrt{6(1+x)(2+x)} \right] + \left[ 3 + 2 x + 3 x^2 - 2 (1+x) \sqrt{2(1+x^2)} \right] \geq 0$$
$$ 10 + 7x + 3 x^2 \geq 2 \sqrt{6(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Deriving the formula for $\cos^{-1}x+\cos^{-1}y$ I am trying to prove the following:
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\ge0\\2\pi-cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\le0\end{cases}$$
Let $\cos^{-1}x=A, \cos^{-1}y=B.... | In first case you found that
$$0\leq A+B\leq \pi$$ which becomes
$$0\leq\cos^{-1}x+\cos^{-1}y\leq \pi$$
Note the first inequality above is true for all $x,y\in[-1,1]$
Now the second inequality becomes $$\cos^{-1}x+\cos^{-1}y\leq \pi$$
$$\cos^{-1}x\leq\pi-\cos^{-1}y$$
$$\cos^{-1}x\leq\cos^{-1}(-y)$$
Now $f(x)=\cos^{-1}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}... | Along the line of @robjohn's solution, consider the change of variables $u=\frac{1}{1+x^2}$. Then
$$du=-2 u^2\Big(\frac{1}{u}-1\Big)^{1/2} dx=-\frac12 u^{3/2}(1-u)^{-1/2}dx$$
and so
$$\int^\infty_0\frac{dx}{(1+x^2)^n}=\frac12\int^1_0u^{n-\tfrac12-1}(1-u)^{\tfrac12-1}\,du=\frac12B\big(n-\tfrac12,\frac12\big)=\frac12\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Rolling dice game - who first will roll number 3 Let's consider very easy game with players A and B - they roll a dice starting with player A. If any of players roll a three, then he wins. I want to calculate probability that player B wins.
Intuition
Intuition is that $P(\textrm{player B wins}) < P(\textrm{player A win... | Denoting the ultimate probability of $A$ winning by $a$,
either $A$ wins immediately, or both fail and we are back to square $1$
So we have $a = \frac16 + \frac{5}6\frac56\cdot{a}$
which yields $a = \frac6{11},\;\; b = \frac5{11}$
You should be able to now spot the error in your first formulation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? How to calculate integral $\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? I got this integral by using Abel-Plana formula on series $\sum_{n=0}^\infty \frac{1}{(n+1)^2}$. This integral can be splitted into two integrals with bounds from 0 to 1 and from 1 to infi... | Recall Binet's second $\ln \Gamma$ formula:
$$\int_{0}^{\infty} \frac{\arctan (t z)}{e^{2\pi t}-1} \, dt = \frac{1}{2} \ln \Gamma \left(\frac{1}{z}\right) + \frac{1}{2}\left(\frac{1}{z}-\frac{1}{2}\right) \ln (z) + \frac{1}{2z} - \frac{1}{4} \ln (2 \pi)$$
Consider the following integral:
$$\int \frac{t}{(e^{2\pi t}-1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If a, b, c are positive numbers then prove that $\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq a+b+c$.
If a, b, c are positive numbers then prove that $\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq a+b+c$.
I tried using AM-GM inequality, $$\frac{a+b}{2}\geq \sqrt{ab}$$ $$a^2+b^2+2a... | Hint: Use $\dfrac{x^2+y^2}{x+y} \ge \dfrac{x+y}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without trig sub So I am evaluating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without using trig sub integrals. So far I have
$$u=\sqrt{3-x^2}, x^2=3-u^2,du=-\frac{x}{\sqrt{3-x^2}}dx, dx = -\frac{\sqrt{3-x^2}}{x}$$
So rewriting I get
$$\int \frac{1}{x\sqrt{3-x^2}}dx=-\int\frac{1}{3-u... | Your answer looks fine to me. To double check, you can differentiate your answer, or confer with wolfram, which gives
$$\int \frac{1}{x\sqrt {3-x^2}}=-\frac{1}{\sqrt 3}\tanh^{-1}\sqrt{1-x^2/3}+C,$$
and using the identity
$$\tanh^{-1}t=\frac{1}{2}\ln \left(\frac{1+t}{1-t}\right),t\in(-1,1)$$
we have
$$\begin{align}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
I found the similar question in Quora. There was slightly a mistake.
$$(\tan\theta+\sin\theta)^2-n^2=4\sqrt{mn}$$
$$(\tan\thet... | Let's call $x=\tan\theta-\sin\theta$ then
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
becomes
\begin{align}
x^2+4\sqrt{m}\sqrt{x}-n^2-4\sqrt{m}\sqrt{n}&=0\\
(x^2-n^2)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(x-n)(x+n)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(\sqrt{x}-\sqrt{n})(\sqrt{x}+\sqrt{n})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simpl... | Let $z=r*e^{i\theta}$, So here we have,
$$\bar{z}*z^2*|z|=8\sqrt{2}(-sin\frac{\pi}{5}-icos\frac{\pi}{5})^8$$
$$(e^{-i\theta})(e^{i\theta})^2(r)=8\sqrt{2}(-i(cos\frac{\pi}{5}-isin\frac{\pi}{5}))^8$$
$$(e^{-i\theta})(e^{-\theta^2})(r)=8\sqrt{2}*1*(cos(-\frac{\pi}{5})+isin(-\frac{\pi}{5}))^8$$
$$(cos\theta-isin\theta)(e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Integrals and Hypergeometric function I'm dealing with these two integrals:
$$I_1=\int_{-\pi}^{\pi} \frac{\cos(x)\cos(nx)}{(1+e \cos(x))^3} \mathrm{d}x, \quad I_2=\int_{-\pi}^{\pi} \frac{\sin(x)\sin(nx)}{(1+e \cos(x))^3}\mathrm{d}x$$
is there a way to reduce them to a Hypergeometric form or to solve them ... | From this answer we know that
$$
\int_0^\pi\frac{\cos(mx)}{p-q\cos (x)}\ \mathrm{d}x=\frac{\pi}{\sqrt{p^2-q^2}}\left(\frac{p-\sqrt{p^2-q^2}}{q}\right)^m\quad\hbox{for}\quad |q|<p \tag{1}
$$
We'll come back to this identity to solve both integrals.
For $I_1$ first notice that
$$
\frac{\partial}{\partial p}\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matching pairs of gloves A closet contains $13$ different pairs of gloves. The 26 gloves are randomly arranged into $13$ pairs.
(a) Find the probability that all left-hand gloves are paired with right-hand gloves (not necessarily matching).
(b) Find the probability that all of the gloves are arranged so that each left-... | Your solutions are correct.
There is another way to count the number of elements in the sample space. Suppose we have $n$ pairs of gloves in the closet. Grab any of the $2n$ gloves in the closet. It does not matter which glove we choose. There are $2n - 1$ gloves left in the closet. Pair one of them with the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.
| We have $e^x > 1+x+x^2/2+x^3/6$, so we show that the right side is bigger than $x^2+x$ for positive $x$. Let
$$f(x) = 1+x+x^2/2+x^3/6-(x^2+x) = x^3/6-x^2/2+1.$$
Find the minimum value of $f(x)$ using usual calculus methods:
$$f'(x) = x^2/2-x = 0$$
gives one positive critical point $x=2$. $f''(x) = x-1$ and $f''(2) >0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$
The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve ... | Let $x = \sqrt[3]{40 + 11\sqrt{13}} + \sqrt[3]{40 - 11\sqrt{13}}$. Then we have that
\begin{align*}
x^{3} = 80 + 9x & \Longleftrightarrow x^{3} - 9x - 80 = 0\\\\
& \Longleftrightarrow (x^{3} - 25x) + (16x - 80) = 0\\\\
& \Longleftrightarrow x(x^{2} - 25) + 16(x - 5) = 0\\\\
& \Longleftrightarrow x(x + 5)(x - 5) + 16(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ ... | Note that your sequence can be written as follows
$$t_n = \frac{1}{n}\sum_{j=1}^n\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/2}}dt\; + \frac{1}{\sqrt{n+1}}.$$
Clearly the last term of the summation vanishes as $n\to+\infty$ while the first term is the Cesaro sum of the sequence
$$a_j:=\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How can I find the equation of a parabola with focus and directrix? Focus: (3,5)
Directrix: $y=-x-1$
I'm comfortable with questions where the directrix is simply $y=3$ or something that is constant (i.e. a horizontal line).
I know that I need to find the distance between a general point (x,y) and (3,5) as well as the d... | First, you can find the vertex of the parabola. It lies half way between the focus and the directrix.
The axis of the parabola passes through the focus, and is perpendicular to the directrix. The directrix is
$x + y + 1 = 0$
So the unit vector along the axis is
$u_1 = [1, 1]^T / \sqrt{2} $
The distance between the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor Series of $(\frac{z}{1 - z})^2$ around $z = i$ I am having some trouble trying to find the Taylor series of $f(z) = \left(\frac{z}{1 - z}\right)^2$ around $z = i$. I have started a little bit, but unsure how to complete. Below is my attempt:
Since $f(z) = \left(\frac{z}{1 - z}\right)^2$, we are able to rewrite t... | Observe that for
\begin{equation*}
f(z) = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1
\end{equation*}
you may reshift the indices of the first summation so that
\begin{equation*}
f(z) = \sum_{n = 1}^{\infty} \dfrac{(z - i)^{n}}{(1 - i)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What type of product? $\mathbf{a} \circledast \mathbf{b}:= \mathbf{a}\mathbf{b}^T - \mathbf{b}\mathbf{a}^T$ What do you know about the names and properties of the products of the following definitions?
\begin{align}
&\mathbf{a,b}\in\mathbb{R}^N, \mathbf{C}\in\mathbb{R}^{N\times N}\\
&\mathbf{C} = \mathbf{a} \circledast... | This looks like the representation of the wedge product $a\wedge b$ as a skew-symmetric matrix. (Note that $\mathbf C$ is skew-symmetric.) If $a=\sum a_ie_i$ and $b=\sum b_je_j$, then $a\wedge b = \sum\limits_{i<j} (a_ib_j-a_jb_i)e_i\wedge e_j$.
(You might read more about exterior algebra.)
| {
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"url": "https://math.stackexchange.com/questions/4408155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_{-\infty}^\infty \frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$ I recently attempted to evaluate the following integral
$$\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$$
I started by inserting a parameter, $t$
$$F(t)=\int_{-\infty}^\infty\frac{\ln{(tx^4+x^2+t)}}{x^4+1}dx$$
Where F(0) is the following
$$F(0)... | Utilize the integral
$\int_{-\infty}^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{2\pi}b \ln(a+b)$, along with the shorthands $p=e^{i\frac\pi6}$ and $q= e^{-i\frac\pi4} $
\begin{align}
&\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx\\
=& \int_{-\infty}^\infty\frac{\ln{(x^2+p^2) (x^2+\bar{p}^2)}}{(x^2+q^2)( x^2+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work because I lose equality c... | We have $|z^2 -1|\cdot |z-1|^2 = \lvert p(z) \rvert$ with $p(z) = (z^2-1)(z-1)^2$. Since $p(z)$ is holomorphic, $ \lvert p(z) \rvert$ attains its maximum on the closed disk $D^2 = \{ z \in \mathbb C \mid \lvert z \rvert \le 1\}$ on the boundary $\partial D^2 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. This follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x$ I was wondering if it is possible to compute this integral in closed form:
$$
\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x
$$
I tried making a substitution with $s = x^2$ and also tried e... | Under substitution $\frac{x}{\sqrt{2}} \to x$ we get that your integral is equal to $I =2 \sqrt{2} \int_0^{\infty} \frac{x^4}{(x^2 + \xi)^2}e^{-x^2}\mathrm{d}x$ where $\xi = \frac{a}{2}$. Additionally, since $ \frac{x^4}{(x^2 + \xi)^2} = 1 - \frac{2\xi}{x^2 + \xi} + \frac{\xi^2}{\left(x^2 + \xi \right)^2}$, the problem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the probability density function of the random variable Let $X$ be a random variable with
$$f(x) = \frac{1}{2 \theta} , -\theta < x < \theta$$
Let $Y=\frac{1}{X^2}$.Then what is the probability density function of $Y$?
Case $1:$ let $y < 0$. Then $P(\{Y \le y\}) = 0$
Case $2$: let $0 \le y < \theta$.
Then $P(Y \l... | $\begin{aligned}
F_Y(y) &= P(Y \le y) \\
&= P \left(\frac{1}{X^2} \le y\right) \\
&= 1 - P \left(X^2 \le \frac 1y \right) \\
&= 1- P \left(- \frac 1 {\sqrt{y}} \le X \le \frac 1 {\sqrt{y}}\right) \\
&= 1 - \frac{1}{\theta \cdot \sqrt y} \\
\end{aligned}$
Now differentiating it gives you the density function of $~Y$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the ri... | Given that $a > b > 0$, we take
$$
S = a^3 - b^3, \ \ T = a^2 b - b^2 a = a b (a - b)
$$
Note that we can write
$$
S = a^3 - b^3 = (a - b) (a^2 + a b + b^2)
$$
Thus,
$$
S - T = (a - b) (a^2 + a b + b^2) - a b (a - b) = (a - b)(a^2 + b^2) > 0
$$
since $a > b > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the probability density function of the random variable $\frac{1}{X}$? How to find the probability density function of the random variable $\frac{1}{X}$?
Let $X$ be a random variable with pdf
$f_X(x)= \begin{cases} 0 ; x \le 0 \\ \frac{1}{2} ; 0 < x \le 1 \\ \frac{1}{2x^2} ; 1 < x < \infty \end{cases}$
I wa... | $X$ is a positive r.v. and so is $Y$. Hence, $P(Y\leq y)=0$ for $y \leq 0$.
Also, $P(Y \leq y)=\frac y 2$ for $0<y \leq 1$ and $P(Y \leq y)=1-\frac 1 {2y}$ for $y >1$.
[For $0<y\leq 1$ we have $P(Y \leq y)=P(\frac 1 X \leq y)=P(X\geq \frac 1 y)=\int_{1/y}^{\infty} \frac 1 {2x^{2}}dx=\frac y 2$. I will leave the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can we prove the inequality without opening the parentheses? $(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$$
I tried to use the following inequalities:
$$x^2+y^2+z^2≥xy+yz+xz$$
and The Cauchy–S... | Rewrite the left hand side as follows:
$$
(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)=
\\
=\left(\frac{2}{3}(x+y+z)(xy+yz+zx)\right)\cdot(x^2+y^2+z^2)+
\\
+\left(\frac{1}{3}(x+y+z)(x^2+y^2+z^2)\right)\cdot(xy+yz+zx).
$$
Can you continue now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to expand the numerator (Harmonic numbers) in ter... | $$\color{royalblue}{\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n})}{n2^n \binom{2n}{n}} = \frac{\pi ^2}{36}-\frac{\log ^2(2)}{3}}$$
Proof sketch: Suffices to find $S=\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n2^n \binom{2n}{n}}$. Note that
$$\int_0^1 \frac{x^n(1-x)^n}{x} \log x dx = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proof of $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}}$ Original Question:
Use the proof of Thm 7.2.4 given above to show that if $n \geq 2$, then $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}} < \frac{e}{4^{\frac{1}{n}}}$.
Thm 7.2.4: $\displaystyle\lim_{n\right... | You can use induction. Check the case $n=2$ by hand. Then
\begin{align*}
\frac{{(n + 1)^{n + 1} }}{{(n + 1)!}} & = \frac{{n^n }}{{n!}}\frac{{(n + 1)^n }}{{n^n }} > \left( {\frac{n}{{n + 1}}} \right)^n \frac{{e^n }}{{n + 1}}\frac{{(n + 1)^n }}{{n^n }} = \frac{{e^n }}{{n + 1}} \\ &= \left( {\frac{{n + 1}}{{n + 2}}} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximize $z$ over $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$ Suppose that $x$, $y$, and $z$ are real numbers such that $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$, but this doesn't say anything. I also ... | Hint: Since you tried using CS inequality, here is a way:
$$(1+1)\cdot(x^2+y^2)\geqslant (x+y)^2 \implies 2\cdot(6-z^2)\geqslant (3-z)^2 \\ \implies 1+\sqrt2 \geqslant z \geqslant 1-\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The number of solutions $(x,y)$ of the congruence $n \equiv X^2-XY+Y^2$ (mod $p^\alpha$). Is there general formula of the solutions of the congruence?
\begin{equation}
n\equiv X^2-XY+Y^2 \pmod r,
\end{equation}
where $n\in\Bbb Z$ and $r\in\Bbb N$.
If we define an arithmetic function (two variables or one variable with ... | So far, just an answer when $r=3^a.$
When $r$ is odd, this can be multiplied by $4$ to get $$(2X-Y)^2+3Y^2\equiv 4n\pmod r$$
Then solutions to $Z^2+3Y^2\equiv 4n$ are in $1-1$ correspondence with your congruence.
If $r=3^a,$ then:
*
*If $n\equiv 1\pmod 3,$ there is two solutions for every $Y, $ for $N(n,3^a)=2r$ solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
prove that the sum of the elements in two subsets is the same
Eight consecutive positive integers are partitioned into two subsets such that the sum of the squares in each subset is the same. Prove that the sum of the elements in each subset is also the same, assuming that the smallest element is at least 56.
The met... | Here is a counterexample, if the two partition sets don't have to have the same number of elements, using $\{2,3,4,5,6,7,8,9\}$:
$2^2+3^2+4^2+7^2+8^2=142=5^2+6^2+9^2$.
But $2+3+4+7+8=24$ and $5+6+9=20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in... | Note that $$\frac{1+(-1)^k}{2} = \begin{cases}1 &\text{if $k$ is even}\\ 0 &\text{if $k$ is odd}\end{cases}$$
implies that $$\sum_k a_{2k} = \sum_k \frac{1+(-1)^k}{2} a_k.$$ Taking $a_k = \frac{\binom{2k}{k}}{(k/2)^2 16^{k/2}}$ yields
\begin{align}
\sum_{k=1}^\infty \frac{\binom{4k}{2k}}{k^2 16^k}
&= \sum_{k=1}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$x^4+y^4+z^4=\frac{m}{n}$, find $m+n$. $x^4+y^4+z^4$=$m\over n$
x, y, z are all real numbers,
satisfying $xy+yz+zx=1$ and $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$
m, n are positive integers and their greatest common divisor is 1. Calculate m+n.
My thinkings so far are as f... | Taking resultants of the polynomials given by
$$
f=5\left(x+\frac{1}{x}\right)-12\left(y+\frac{1}{y}\right),g=5\left(x+\frac{1}{x}\right)-13\left(z+\frac{1}{z}\right), h=xy+yz+zx-1,
$$
set to zero, i.e., with
\begin{align*}
0 & = 5x^2y - 12xy^2 - 12x + 5y,\\
0 & = 5x^2z - 13xz^2 - 13x + 5z,\\
0 & = xy + xz + yz - 1
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the formula for higher integration of Lambert´s W function? First I will present the notation for higher integration, which I copied from Danya Rose
\begin{align}
J_x^n(f(x)) = \int ...\int f(x) dx^n
\end{align}
Here are the higher integrations of the principial branch of the Lambert W function.
\begin{align}
... | This is not an answer.
First remark
$$I_n=\int \big[W(x)\big]^n \,dx=(-1)^n \Big[\Gamma (n+1,-W(x))-\Gamma (n+2,-W(x))\Big]$$
Second remark
I shall use $W=W(x)$ to have more room.
$$J_n=\int J_{n-1} \,dx \qquad \text{with} \qquad J_1=\frac x{W}\big[W^2-W+1 \big]$$
Define
$$K_n=\frac{W^n \,x^{-n}\, J_n-\,n^{-(n+1)}}{W}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4430481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the recursively defined sequence $x_{1} =0$ $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=(1/2) + x_{2n}$ I've posted this question before, but there was a typo which messed up my search for a solution
$x_{1} =0$, $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=1/2 + x_{2n}$
1)Which is the limit of the sequences $x_{2n}$ and $x_{2n+1... | First, you can substitute $x_{2n}$ in $x_{2n+1} = \frac{1}{2} + x_{2n}$ with $x_{2n} = \frac{1}{2} x_{2n-1} $ to get $x_{2n+1} = \frac{1}{2} + \frac{1}{2}x_{2n-1}$.
Define another sequence $y_n = x_{2n+1}$, so $y_0 = x_1 = 0$ and $y_n = x_{2n+1} = \frac{1}{2} + \frac{1}{2} x_{2(n-1) + 1} = \frac{1}{2} + \frac{1}{2}y_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4430976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n \to \infty}a_n$ where $a_1=1$ and $a_{n+1}=a_n+\frac{1}{2^na_n}$. There are some attempts as follows.
Obviously, $\{a_n\}$ is increasing. Thus $a_n\ge a_1=1$. Therefore
$$a_{n+1}-a_n=\frac{1}{2^na_n}\le \frac{1}{2^n},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\le a_1+\sum_{k=1}^{n-1}\frac{... | When written under this form : $\quad 2^na_{n+1}a_n=2^n{a_n}^2+1$
We see that we can get rid of the $2^n$ term by setting $\ b_n=\sqrt{2^n}\,a_n\implies \frac 1{\sqrt{2}}b_{n+1}b_n={b_n}^2+1$
And we reduced the study to $$b_{n+1}=\sqrt{2}\left(b_n+\frac 1{b_n}\right)$$
Which according to this question Study convergence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Show that $E(X_n)=0$ but a.s. The sum of $\frac{1}{n}(X_1+X_2+…X_n)\rightarrow -1$ as $n\rightarrow\infty$ If $X_1,X_2,…$ is a sequence of independent random variables such that,
\begin{cases}
n^2-1 & \text{with probability } \frac{1}{n^2} \\
-1 & \text{with probability } 1-\frac{1}{n^2}
\end{cases}
How... | Use the Borel-Cantelli lemma to conclude that $X_n=-1$ for all but finitely many $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{... | A more general solution:
I’ll state the following propositions that can be trivially proven using induction, sums of geometric series, second derivatives and the use of $\tan^2 x=\sec^2 x-1$.
For $n\in\mathbb{N}$
$$\sec^{2n}(x)=\frac{(-1)^n \, 2^{2n}}{(2n-1)!} \sum_{k=1}^{\infty} (-1)^k e^{2ixk} \prod_{r=0}^{2n-2} (k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt... | there is another way to solve it, I copy it from internet, but I don't know how to find this method.
first to deformation the algebraic expression: $$(2xy-1)^2=(5y+2)(y-2)\iff (2xy-1)^2 = 9y^2-4y^2-8y-4\iff (2xy-1)^2+(2y+2)^2=9y^2$$
then we have $$\left(2x-\frac{1}{y}\right)^2+\left(2+\frac{1}{y}\right)^2=9$$
and $$x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Maclaurin series of $1- \cos^{2/3} x$ has all coefficients positive Experimenting with WA I noticed that the function $1- \cos^{\frac{2}{3}}x$ has the Maclaurin expansion with all coefficients positive ( works for any exponent in $[0, \frac{2}{3}]$). A trivial conclusion from this is $|\cos x|\le 1$, but it implies mor... | Start with the positive series
$$\csc x = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{15120} x^5 + \cdots $$
Take the square and obtain the positive series
$$\csc^2 x =\frac{1}{x^2} + \frac{1}{3} + \frac{1}{15} x^2 + \frac{2}{189} x^4 + \cdots $$
Now take the derivative and get the series
$$- \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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Problem regarding the distance between a circle and a parabola The question is this, copy pasted here:
Consider the circle $C$ whose equation is
$$(x - 2) ^ 2 + (y - 8) ^ 2 = 1$$
and the parabola $P$ with the equation
$$y^ 2 =4x.$$
I have no idea how to model the question, I tried showing that the $x-$coordinates of th... | Let $(x,y) $ be on the parbola, then
$ y^2 = 4 x$
The normal vector is $[ 4, -2 y] $ , while the vector from $(2, 8)$ to $(x,y)$ is $[ x - 2, y - 8 ]$, and we want these two vectors to be aligned; so we can use the determinant and set it equal to zero, namely,
$4 (y-8) + 2 y (x - 2) = 0 $
We have to solve the above equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be... | Notice that we have $2x+3=0$ and you should check that $x=-\frac32$ is a minimum.
$$f'(x) = \frac{12(2x+3)}{(x^2+3x+6)^2}$$ shows that $x=-\frac32$ is the global minimum as the gradient is negative for $x<-\frac32$ and then positive for $x>-\frac32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$ sequence
Sequence $\{a_n\},n\in\mathbb N_+$ with all terms positive integers satisfy $a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$. Find $\{a_n\}$.
I suppose all terms of $\{a_n\}$ are $1$. This problem makes me think of a lot of conclusions, including
*
*$n\in\mathbb{Z}_+$ can be written as... | We claim that all $a_{n}$ are $1$, we will use proof by contradiction to make our claim clear.
Suppose $\{ a_{n}\}$ is some sequence which is not the trivial sequence composed of $1$'s and satisfies your requirements.
$\textbf{Let}$ $v$ $\textbf{be the smallest positive integer so that}$ $a_{v} \neq 1$. Note that $a_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of solution of equation $16 \sin^3x=14+(\sin x+7)^{\frac{1}{3}}$ in $[0,4\pi]$. Number of solutions of equation $16 \sin^3x=14+(\sin x+7)^{\dfrac{1}{3}}$ in $[0,4\pi]$.
My thinking:
*
*Directly satisfy $\sin x=1$.
*I thought of doing $f(x)=f^{-1}(x)\;$ but couldn't proceed further because I can't think of a ... | Let $y=\sin x$. Then $$16(y^3-1)=\sqrt[3]{y+7}-2$$
$$16(y^3-1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=y-1$$
Then $y=1$ or $$16(y^2+y+1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$
$$4(4y^2+4y+4)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$
$$4\left((2y+1)^2+3\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Asymptotic behaviour of $\frac{n^n}{n^n - (n - 1)^n + 1}$ for large $n$ I am interested in the behaviour when $n$ is large of the following function:
$$f(n) := \frac{n^n}{n^n - (n - 1)^n + 1}.$$
The limit of this function as $n$ approaches infinity is
$$\lim_{n \to \infty} f(n) = \frac{e}{e - 1},$$
where $e$ is Napier'... | Here are the details of Greg Martin's answer.
First,
$$f(n) = \frac{1}{1 - (1 - \frac{1}{n})^n + n^{-n}}.$$
Since
$$\log(1 - \frac{1}{n})^n = n\log(1 - \frac{1}{n}) = n(-\frac{1}{n} - \frac{1}{2} \frac{1}{n^2} + O(\frac{1}{n^3})) = - 1 - \frac{1}{2}\frac{1}{n} + O(\frac{1}{n^2}),$$
exponentiating gives
$$(1 - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You... | $ 5x + 3y = 100 $
Has the solution $ x = 20 - 3 t $ and $ y = 5 t $ where $ t= 1, 2, \dots $
Maximum $t $ is $\text{int}(\frac{20}{3}) = 6 $
$xy = 5 t (20 - 3 t) $
Its peak is at $\frac{1}{2} ( 0 + \frac{20}{3} ) = \frac{10}{3} $
Since this not an integer, the maximum is attained at $t = 3$ (because it closer to $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Possible growth rates of a matrix entry with respect to exponentiation Let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, so $A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$. Thus, $(A^n)_{1,1} = 1 = \Theta(1)$, and $(A^n)_{1,2} = n = \Theta(n)$.
Given a constant $c$, if $B = \begin{pmatrix} c/2 & c/2 \\ c/2 & c/... | Due to the Cayley-Hamilton theorem, matrix elements with respect to exponentiation adhere to the linear recurrence defined by the characteristic polynomial of a matrix. And a linear recurrence $A_0, A_1, \dots$ with the characteristic polynomial
$$
P(x) = \prod\limits_{i=1}^k (x-x_i)^{d_i},
$$
has a generic solution
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I calculate the area of a circle centered at (2,2) with radius 2 using double integrals
This is what the graph loosk like.
Obviously I know $\pi r^2$ but if I specifically wanted to find the area using double integrals in polar coordinates, how would I go about it?
My guess is that $\theta$ goes from $0$ to $\... | After getting $r^2=4(r\cos\theta+r\sin\theta)-4$ you can solve for $r$ to get $r_{1,2}(\theta) = 2\left(\sin\theta+\cos\theta\pm\sqrt{2\sin\theta\cos\theta}\right)$
The two solutions are the two intersections of the ray from the origin with the circle:
When using polar coordinates, the area can be obtained as $\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find number of points of discontinuity of $f(x)=\frac x5-\left\lfloor \frac x5\right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$ for $x\in[0,100]$ Find number of points of discontinuity of $$f(x)=\frac{x}{5}-\left\lfloor \frac{x}{5} \right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$$ for $x\in[0,100]$ .
My Attemp... | $\left\lfloor \frac{x}{5}\right\rfloor$ is continuous in the multiples of $5$.
$\left\lfloor \frac{x}{2}\right\rfloor$ is continuous in the multiples of $2$. So $f$ is continuous in the integers that are simultaneously multiples of $5$ and $2$, since $\gcd(2,5)=1$, those are the multiples of $\operatorname{lcm}(2,5)=10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$.
Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$.
Firstly if $k$ is even, then as $k=2t$ for $t \in \Bbb Z$ and so $x^k=(x^t)^2$ and $y^k=(y^t)^2$. These are both perfect squares and... | EDIT: It appears I read too fast and OP wasn't asking for a solution, but rather an explanation of an already given solution. Nevertheless, I think my solution is pretty neat and clean, so I'll leave it up.
As you've shown, $k$ must be odd. Let $p\mid k$ be prime, then
$$\frac{x^k+y^k}{x^p+y^p}=\frac{x^k-(-y)^k}{x^p-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Usage of Zsigmondy's theorem
Find the solutions for the equation $$3^x-5^y=z^2$$ where $x,y,z$ are positive integers.
We find one solution $(x,y,z)=(2,1,2)$ by checking small cases. Initially looking at this modulo $4$ one gets that $$(-1)^x-1^y \equiv 0,1 \pmod 4$$ as squares leave residues $0$ and $1$ modulo $4$. T... | There's 2 forms of Zsigmondy's theorem, one for forms $a^n-b^n$ and one for forms $a^n+b^n$. Your RHS is in the second form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find if the integral $ \int_{0}^{\infty}\frac{\tan^{-1}(x)}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$ converges without solving it I'm trying to solve this integral, but without any success.
Is there a way to tell if the integral is convergent or divergent without actually solving it? Am I missing something?
$$\int_{0}^{\infty... | We want to show $$\int_{0}^{\infty} \dfrac{\arctan x}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$$ converges. We'll consider an initial and final segment separately and bound them both.
Let's look at the factors. For large $x$ we have
$$\arctan(x) \le \pi/2 \qquad \dfrac{1}{(2x+1) } \simeq \frac{1}{2x} \qquad \dfrac{1}{ (x^4-6x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x$? By my post, I had found that
$$
\begin{aligned}
\frac{\pi^{3}}{8}&= \int_{0}^{\infty} \frac{\ln ^{2} x}{x^{2}+1} d x \\&\stackrel{x\mapsto\tan x}{=} \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x \\
&= \int_{0}^{\frac{\pi}{2}}[\ln (\sin x)-\ln (\co... | If you are familiar with beta function and feymman's trick then
$$ B(x,y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt $$
$$ \frac{\partial }{\partial x} B(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$
$$ \frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} B(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x... | You can use
$$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1 $$
to handle the limit. In fact
\begin{eqnarray}
&&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}\\
&=&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{x^4}\bigg[\frac{x}{\ln \left(x+1\right)}\bigg]^4\\
&=&\lim _{x\to 0}\frac{-\sin x+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$ On $[a,b]$ function $f$ is differentiable for arbitrary order, and $f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$.
I first try Taylor expansion at $x_0=\frac{a+b}{2}$, and drop higher order term ($(x-x_0)^3$ terms). But this ... | Let $L(x) = f'\left(\frac{a+b}{2}\right)\left(x - \frac{a+b}{2}\right) + f\left(\frac{a+b}{2}\right)$. Then, $L$ describes the line tangent to $f$ at $x=\frac{a+b}{2}$. Note that:
$$\int_{a}^{b}L(x)\ dx = (b-a)f\left(\frac{a+b}{2}\right)$$
Now, consider $g(x) = f(x)-L(x)$. Note that $g''(x) > 0$ because $L''(x)=0$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding the roots of $x^5+x^2-9x+3$ I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$
I can take it from here employing the quadratic and cubic formula, but how can I get the same answ... | Here's one way to think about it:
$$\begin{align}x^5+x^2-9x + 3 &= x^5-9x+x^2+3\\ &= x(x^4-9)+x^2+3 \\ &= x(x^2-3)(x^2+3)+x^2+3 \\ &= (x^2+3)(x(x^2-3)+1) \\ &= (x^2+3)(x^3-3x+1). \end{align}$$
In words, the trick is to recognize that $x^5-9x$ is "almost" a difference of squares (i.e., up to the factor of $x$ multiplyi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.... | Firstly, recognise the "symmetry" in the integral
$$I = \int_{0}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln \left(\frac{\ln(1/2+x)}{\ln(1/2-x)}\right)\, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln\left(-\ln\left(\frac{1}{2}+x\right)\right)\,dx$$
Now enforce the substitution $-\ln(1/2+x)=u$:
$$\implies I = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Assistance with solving the integral Can you give me an idea how to handle this integral?
$\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\,dx$
| Using the tangent half-angle substitution
$$I=\int_{0}^{2\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx=2\int_{0}^{\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx$$
$$I=8 \sqrt{2} \int_0^\infty \sqrt{\frac{t^2+2}{\left(t^2+1\right)^5}}\,dt=4 \sqrt{2}\int_1^\infty \frac 1 {x^2}\sqrt{\frac{x+1}{(x-1) x}}$$ The antiderivative is nasty but it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | Write $P\equiv \sqrt{x^2-1}$ and $Q\equiv x^3-3x+1$ then the first equation can be written as
$$
Q(x+P)+1=0 \tag{1}
$$
It's clear that $|x|\geq 1$ otherwise there is no solution for $P$. By inspection, $x=1$ is one of the solutions.
Assuming $|x|>1$, multiply $(1)$ by $(P-x)$:
$$
Q(P^2-x^2)+P-x=0
$$
Since $P^2-x^2=-1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
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How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ ... | By the Rational Root Theorem, if such a polynomial has a rational root, it must be $x = \pm 1$. It turns out that 8 of the 16 possible cubics (the ones with even numbers of + and - coefficients) have such a root.
For a quartic, we get $f(x) = \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1 = \pm 1 \pm 1 \pm 1 \pm 1 \pm 1$, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Factors of central binomial coefficient The central binomial coefficient $\binom{2n}{n}$ is divisible by $n+1$, as seen from the
identity
$$\binom{2n}{n} = (n+1)\binom{2n}{n} -(n+1)\binom{2n}{n+1}.$$
In fact the Catalan numbers
$$C_n=\frac{1}{n+1}\binom{2n}{n}$$
have a very large (over 200 according to R. Stanley's boo... | Here is a different proof of the nice result found by Dark Malthorp.
Let
$$K_n=\frac{3}{(2n-1)(n+1)}\binom{2n}{n}$$
We will use the easily verified identities:
$$ K_n=\frac{6}{n(n+1)}\binom{2(n-1)}{n-1}\hspace{10ex} (1) $$
and
$$K_n=\frac{3}{2(2n-1)(2n+1)}\binom{2(n+1)}{n+1} \hspace{10ex} (2)$$
We will also make use of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Painting the board by selecting $2\times 2$ type squares from the $m\times n$ type board Problem: Initially we have an $m\times n$ board with all unit squares white. As a result, we want to paint any two squares with a common edge, one black and one white. (That is, we want to get it like a chessboard pattern after the... | Hint As you noticed, you need some sort of invariant to prove you cannot succeed. You used the difference $b-w$ for the squares in two columns. Instead, find a modulo invariant for the number of black squares alone in a single column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
h... | (The following is about the other roots in ConMan's answer, but it was too long for a comment.)
Courtesy WA, the original equation can be brought to polynomial form with groebnerBasis[ {u - v - w, x - 2 + (u^2 - 2)^2, v^2 x - x^2 + 1, w^2 x - x + 1}, {x}, {u, v, w} ] which returns and factors:
$$
x^6 - 10 x^5 + 39 x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{... | You can use this transformation:
$$\lim_{x \to 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}
=24\lim _{x \rightarrow 0} \frac{\sin^2{\frac{x}{2}}-\frac{x^{2}}{4}}{x^{4}}
=\frac{24}{16}\lim _{y \rightarrow 0} \frac{\sin^2{y}-y^2}{y^{4}}
=\frac{24}{16}\lim _{y \rightarrow 0} \left(\frac{\sin{y}-y}{y^{3}}.\frac{\sin{y}+y}{y}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to ... | Premise
$0 < |x| < \delta \implies -\delta < x < \delta.$
To Do
Establish a relationship between $0 < \epsilon$ and $\delta$ such that the premise will imply that
$\displaystyle \left|\frac{x}{x+1}\right| < \epsilon \iff -\epsilon < \frac{x}{x+1} < \epsilon.$
In addition to establishing a relationship between $\delta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \si... | Using the tangent half-angle substitution
$$\int \frac{\sin (x)}{1-\sin (x) \cos (x)} \,d x=4\int \frac{t}{t^4+2 t^3+2 t^2-2 t+1}\,dt$$
The denominator has four roots
$$r_1=\frac{1}{2} \left(-1+\sqrt{3}-i \sqrt{2 \left(2-\sqrt{3}\right)}\right)\qquad r_2=\frac{1}{2} \left(-1+\sqrt{3}+i \sqrt{2 \left(2-\sqrt{3}\right)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2... | The trick with this question is to try finding $x^2+\frac{1}{x^2}$, via squaring $x+\frac{1}{x}$:
$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$
$2=x^2+\frac{1}{x^2}+2$
$0=x^2+\frac{1}{x^2}$
The fact that this equals 0 is key to solving the problem, since subtracting $\frac{1}{x^2}$ from both sides and then putting both sides... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) ... | $\sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}$ $($ see here$) $
$\begin{align}S &=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} \\\end{align}$
Put $r=\frac{1}{2}, n=2022$.
Hence $S=2 (1-\frac{1}{2^{2023}})=\frac{2^{2023}-1}{2^{2022}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that If $n,k\in\mathbb{N}$ and $\binom{n}{k}$ is a prime number, then $k=1$ or $k=n-1$. My attempt was the following:
Suppose $p=\binom{n}{k}$ is prime. By definition,
\begin{align}
p&=\binom{n}{k}\nonumber\\
p&=\frac{n}{k}\binom{n-1}{k-1}\nonumber\\
\end{align}
Since $p,\binom{n-1... | Notice that if $k = n$, then $p = \binom{n}{k} = 1$ is not prime. So let's assume $1 < k < n-1$. Then $p > n$ that implies $\gcd(p, n) = 1$ and by the equation $kp = n\binom{n-1}{k-1}$ we have $n \mid kp$. But this implies $n \mid k$ that is impossible because $1 < k < n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $e^{At}$ with a $2\times2$ parametric matrix. I would like to calculate the exponential $e^{At}$ where $A$ is a $2\times2$ matrix:
$$
A=\left[\begin{array}{cc}
0 & 1 \\
a & b
\end{array}\right]
$$
and $a,b\in\mathbb R$. The calculus is based on the powers of this matrix, i.e.
$$e^{At}=I+tA+\frac{1}{2}t^2... | For general $A,$ the trick would be to diagonalize $A.$
This particular matrix is easier if you know the general formula for linear recurrences. $A$ has the property that $A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\ax+by\end{pmatrix},$ so we can use it to define sequences $p_n,q_n$ as $$A^n\begin{pmatrix}1\\0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If a and b are the distinct roots of the equation $x^2+3^{1/4}x + 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$.
If $a$ and $b$ are the distinct roots of the equation $x^2+3^{1/4}x +
3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$
My attempt:
LHS = $(a^{108}+b^{108})-(... | $x^2+3^{\frac{1}{4}}x + 3^{\frac{1}{2}} =0$
$\implies \left(x^2+3^{\frac{1}{2}}\right)^2=\left(-3^{\frac{1}{4}}x\right)^2$
$\implies x^4+2x^2{3}^{\frac{1}{2}}+3=3^{\frac{1}{2}}x^2$
$\implies\left(x^4+3\right)^2=3x^4$
$\implies x^8+6x^4+9=3x^4$
$\implies x^8=-3x^4-9$
$ \begin{align}\implies x^{12}&=-3x^8-9x^4\\&=9x^4+27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| Render the equation as
$x^2+7x+(4-2^y)=0$
and treat this as a quadratic equation for $x$. If this is to have natural number roots the discriminant must be a perfect square, which you can work out:
$\Delta=33+(4×2^y)=m^2$
where wlog $m$ may be rendered positive.
If $y$ is odd, then $2^y\equiv2\bmod3$, and substituting t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Is there any method to compute $\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)$ other than complex numbers? In this couple of days, I need to know the high derivatives of $cos^kx$ whose power $k$ make the differentiation much harder. Then I attempt to
use the identity $$
\cos x=\frac{1}{2}\left(e^{x i}+e^{-x i}\right),... | You could use Faà di Bruno's formula, with $g(x)=\cos(x)$ and $f(x)=x^k$.
Just for an example, with $\frac{d^3}{dx^3}\cos^4(x)$ you need to sum over $(m_1,m_2,m_3)$ such that $m_1+2m_2+3m_3=3$, with each $m_i\geq0$. Such tuples are: $(3,0,0)$, $(1,1,0)$, and $(0,0,1)$. So
$$
\begin{align}
\frac{d^3}{dx^3}\cos^4(x)
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$ for any $\triangle ABC$ I found this problem on Mathematics, Textbook for Class XI by NCERT, ed. January 2021 that uses the sine and cosine formulas along with standard trigonometric identities for proving an identity.
By triangle ABC, the question assumes... | First you need to know The sum to product identities:
$$\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2}) $$
Then the problem is solved:
$$
\begin{equation*}
\begin{aligned}
(b+c)\cos(\frac{B+C}{2})-a\cos(\frac{B-C}{2}) &= R[(\sin B+\sin C)\sin(\frac{A}{2})-\sin A\cos(\frac{B-C}{2})] \\
&= R\sin(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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