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Trouble with $4\times4$ matrix determinant $$
\begin{vmatrix}
1 & -6 & 7 & 5 \\
0 & 0 & 3 & 0 \\
3 & -2 & -8 & 6\\
2 & 0 & 5 & 4\\
\end{vmatrix}
$$
Clearly I want to expand along the second row yielding:
$((-1)^5)3$ times the following matrix
$$
\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 & 6 \\
2 & 0 & 4 \\
\end{vmatrix}
$$
and then breaks down into several smaller matrices:
2 times
$$
\begin{vmatrix}
-6 & 5 \\
-2 & 6 \\
\end{vmatrix}
$$
and 4 times
$$
\begin{vmatrix}
1 & -6 \\
3 & -2 \\
\end{vmatrix}
$$
which should come out to be $-3[(2(-36+10))+(4(-2+18))]$
$-3[(2(-16))+(4(16))]$
$-3(-32+64)=32 \times -3$
but the answer is -36 I don't know what went wrong?
| Another way to approach this 4x4 matrix is to row reduce to a triangular matrix with zeros underneath the diagonal.
\begin{bmatrix}1&-6&7&5\\0&0&3&0\\3&-2&-8&6\\2&0&5&4\end{bmatrix}
Row reducing to the triangular matrix yields:
\begin{bmatrix}1&-6&7&5\\0&16&-29&-9\\0&0&3&0\\0&0&0&\frac{3} {4}\end{bmatrix}
From here, we can multiply the numbers in the diagonal:
-(1 $\cdot$ 16 $\cdot$ 3 $\cdot$ $\frac{3} {4}$) = -36
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Level Curves Problem Show that $x^2+y^2=6$ is a level curve of $f(x,y)=\sqrt{x^2+y^2}-x^2-y^2+2$.
I know that the first equation is a circle but I do not know how to find out if the second one is it too.
Thanks for the help.
(Sorry my English is not good).
| With
$x^2 + y^2 = c, \; \text{ a constant}, \tag 1$
for any $c$, we have
$f(x, y) = \sqrt{x^2 + y^2} - x^2 - y^2 + 2 = \sqrt{x^2 + y^2} - (x^2 + y^2) + 2 = \sqrt c - c + 2; \tag 2$
thus the circle (1) lies in the $\sqrt c - c +2$-level set of $f(x, y)$; note we needn't prove
$f(x, y) = c_0, \; \text{a constant} \tag 3$
describes a circle to establish this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $f$
Find the minimum value of $$f(x)=\frac{\tan \left(x+\frac{\pi}{6}\right)}{\tan x}, \qquad x\in \left(0,\frac{\pi}{3}\right).$$
My approach is as follows. I tried to solve it by segregating it
$$f(x)=\frac{1}{\sqrt{3}\tan x}+\left({\sqrt{3}}+\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}-\tan x},$$ but $f'(x)$ is getting more and more complicated.
| We know that $\tan(x)\tan(y)=1$ then $x+y=\frac{\pi}{2}$ thus we have $f(x)=\frac{1}{\tan(x)\tan(\frac{\pi}{3}-x)}$ thus we need to maximize denominator (lll be referring to it as k)to get minimum value . Now using the fact that $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ we have $\tan(\frac{\pi}{3})=\frac{\tan(\frac{\pi}{3}-x)+\tan(x)}{1-k}$ thus $k=g(x)=1-\frac{1}{\sqrt{3}}(tan(\frac{\pi}{3}-x)+\tan(x))$ thus maximum value is at $g'(x)=\sec^2(\frac{\pi}{3}-x)-\sec^2(x)=0$ in the given domain we have only one solution at $x=\frac{\pi}{6}$ thus the minimum value of $f(x)=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to prove a formula for Gamma function I made some observation for Gamma function
Suppose
$$x=a+i b,a\in \mathbb{R},b\in \mathbb{R}$$
Then
$$
\left| \cos \left(\frac{\pi (a+i b)}{2}\right) \Gamma (a+i b)\right|\to\left| \sqrt{\frac{\pi }{2}} (a+i b)^{a-\frac{1}{2}}\right|
$$
When
$$
a\in [0,1],b\to\infty
$$
How can i prove this?
| Let
$$
x=a +i b,a \in \mathbb{R},b\in \mathbb{R}
$$
Use following formula from
Bateman, Harry (1953) Higher Transcendental Functions, Volumes I, p.47, (6)
$$
\left| \Gamma (x)\right| \to \left| \sqrt{2 \pi } e^{-\frac{1}{2} (\pi b)} x^{a -\frac{1}{2}}\right|,a \in [0,1]
$$
Expand cosine
$$
\left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \sqrt{2 \pi } e^{-\frac{1}{2} (\pi b)} x^{a -\frac{1}{2}} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right)\right|
$$
Due to monotonicity $\left| e^{-\frac{1}{2} (\pi b)} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right) \right|$, and $\left| \sqrt{2 \pi } x^{a-\frac{1}{2}}\right|$, $a \in [0,1]$
Take limit for exponents
$$
\lim_{b\to \infty } \left[e^{-\frac{1}{2} (\pi b)} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right)\right] = \frac{1}{2} e^{-\frac{1}{2} i \pi a }
$$
And then
$$
\lim_{b\to \infty } \left[\left| \frac{1}{2} e^{-\frac{1}{2} i \pi a }\right| \right]=\frac{1}{2}
$$
Hence
$$
\left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \frac{1}{2} \sqrt{2 \pi } x^{a -\frac{1}{2}}\right|
$$
It equals
$$
\left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \sqrt{\frac{\pi }{2}} x^{a -\frac{1}{2}}\right|, a \in [0,1]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If pair of tangents to a circle in the first quadrant is $6x^2-5xy+y^2=0$ and if one point of contact is $(1,2)$, find the radius. The tangents are $2x-y=0$ and $3x-y=0$. Let the radius be $r$ and centre be $(h,k)$
$$r=\frac{|3h-k|}{\sqrt {10}}$$
$$r=\frac{|2h-k|}{\sqrt 5}$$
$$(h-1)^2+(k-2)^2=r^2$$
I invested a considerable amount of effort in solving this equations, but to no result. My method was to square all terms to avoid the modulus, but that complicated things adding the $hk$ term. This also led me to believe that there must be a better way to solve this. Can I know how this problem must be approached?
| The perpendicular to $y=2x$ throught $(1,2)$ is:
$$y=-\frac{1}{2}x+\frac{5}{2}$$
The line bisector of the two lines $y=2x$ and $y=3x$ is:
$$y=(\sqrt2+1)x$$
The other bisector line is:
$$y=(\sqrt2-1)x$$
but in this case the circunference wouldn't be tangent to either $y=2x$ and $y=3x$ lines.
Now, we have to inresect these two lines, or:
$$(\sqrt2+1)x=-\frac{1}{2}x+\frac{3}{2} \leftrightarrow x=5(3-2\sqrt2) \land y=-5+5\sqrt2$$
From this we can compute:
$$r=\sqrt{(14-10\sqrt2)^2+(-7+5\sqrt2)^2}=\sqrt{(5\sqrt{10}-7\sqrt5)^2}=5\sqrt{10}-7\sqrt5$$
Note that we pass from $(14-10\sqrt2)^2+(-7+5\sqrt2)^2$ to $(5\sqrt{10}-7\sqrt5)^2$, simply calculating the first sum and then solve for $a,b$ this system:
$$\left\{\begin{matrix}
a^2+b^2=495
\\ 2ab=-350\sqrt2
\end{matrix}\right.$$
And the solutions are:
$$a=\pm5\sqrt{10} \land y=\mp7\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Using the Divergence Theorem on the surface of a unit sphere
Using the Divergence Theorem, evaluate $\int_S F\cdot dS$ , where $F=(3xy^2 , 3yx^2 , z^3)$, where $S$ is the surface of the unit sphere.
My Attempt
$$ \text{div} F = \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \right) = 3y^{2} + 3x^{2} + 3z^{3} = 3(y^{2} + x^{2} +z^{2}).$$
Let us use spherical coordinates
$$ x = \rho \sin \varphi \cos \theta \quad, y = \rho\sin\varphi \sin\theta \quad, z = \rho\cos\varphi. $$
We get ,
\begin{align*}
\int\limits_{S} \vec{F} \cdot ds &= 3 \int_{0}^{\pi}\int_{0}^{2\pi} \int_{0}^{1} (\rho^{2} \sin^{2}\varphi \sin^{2} \theta + \rho^{2} \sin^{2}\varphi \rho \cos^{2}\theta + \rho^{2} \cos^{2} \varphi ) \rho^{2} \sin\varphi \ d\rho d\theta d\varphi \\
&= 3 \int_{0}^{\pi}\int_{0}^{2\pi} \int_{0}^{1} \rho^{4} (\sin^{3}\varphi (\sin^{2} \theta + \cos^{2}\theta) + \cos^{2}\varphi \sin\varphi) \ d\rho d\theta d\varphi \\
&= \frac35 \int_{0}^{\pi} \int_{0}^{2\pi} \sin^{3}\varphi + \sin\varphi \cos^{2}\varphi \ d\theta d\varphi \\
&= \frac35 \int_{0}^{\pi} \int_{0}^{2\pi} \sin\varphi \ d\theta d\varphi \\
&= \frac{6\pi}{5} \int_{0}^{\pi} \sin\varphi \ d\varphi = \frac{12\pi}{5}.
\end{align*}
Could someone confirm if my reasoning is correct ? I feel like I have missed a step somewhere in the beggining.
| It is correct, in fact, it is even more simple because you could have directly substituted $\rho^{2}=x^{2}+y^{2}+z^{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}} = -6$.
I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.
| I will assume that you meant$$\lim_{x\to1}\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}.$$Multiplying the numerator and the denominator by $\sqrt{x+3}+\sqrt{2x+2}$, this becomes$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$So, define $f(x)$ as $2x-\sqrt{x^2+3}$ and then the limit that you're after is just$$-f'(1)\left(\sqrt{1+3}+\sqrt{2+2}\right)=-4f'(1)$$(which is $-6$ indeed).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that the solution set of $a_n \not= n$, $n \in \mathbb{N}$ is finite. Problem:For fixed positive integers $s, t$, define $a_n$ as the following.
$a_1 = s, a_2 = t$, and $\forall n \ge 1$, $a_{n+2} = \lfloor \sqrt{a_n+(n+2)a_{n+1}+2008} \rfloor$.
Prove that the solution set of $a_n \not= n$, $n \in \mathbb{N}$ is finite.
I have try prove $a_{n}\ge n$,We already know $a_3 \ge 3$ and $a_4 \ge 4$.
We prove by induction that $a_n \ge n$ for all $n\ge 3$.
Assume that $a_n \ge n$ and $a_{n+1} \ge n+1$.
Then
$$(a_{n+2}+1)^2 > a_n + (n+2)a_{n+1} +2008 \ge n+(n+2)(n+1) + 2008 > (n+2)^2$$, so $a_{n+2} \ge n+2$.
Therefore, we have $a_n \ge n$ for $n \ge 3$.then I can't Thanks
| Let $n > 20080$ (large enough).
Let $ b_n = a_n - n$.
We want to show that for some $N$, for any $ n > N$, $ b_n = 0 $.
Claim: $b_{n+2} \leq \frac{ b_{n+1} + \frac{2b_{n+1} + b_n + 2010 } { 20080 }} {2} $
Proof: $ (a_{n+2})^2 \leq n + b_n + (n+2)(n+ 1+b_{n+1}) + 2008 $
$= n^2 + n (4 + b_{n+1} ) + 2b_{n+1} + b_n + 2010 $
$ < n^2 + n (4 + b_{n+1} + \frac{ 2b_{n+1} + b_n + 2010 } { 20080 }) $
$< ( n + 2+\frac{b_{n+1} + \frac{ 2b_{n+1} + b_n + 2010 } { 20080 }} {2} )^2 $
Corollary: There is an $N_1$ such that for any $n \geq N_1$, $b_n \leq 0$.
Given that you've shown $ b_n \geq 0$, we conclude that for $ n > N$, $b_n = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$.
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values.
Thank you very much for helping.
| First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily,
When $x = \dfrac{-1 - \sqrt{8065}}{2}$:
*
*$x^2 + x = 2016$
*$x^2 - 3x + 2 = 2020 + 2\sqrt{8065} = a$ (say)
*$9x^2 - 15x = 18156 + 12\sqrt{8065}$
$$f(2016) + 2f(a) = 18156 + 12\sqrt{8065}$$
When $x = \dfrac{3 + \sqrt{8065}}{2}$:
*
*$x^2 + x = a$
*$x^2 - 3x + 2 = 2016$
*$9x^2 - 15x = 18144 + 6\sqrt{8065}$.
$$f(a) + 2f(2016) = 18144 + 6\sqrt{8065}$$
From the two equations,
$$4f(2016) - f(2016) = 2(18144) - 18156$$
$$\boxed{f(2016) = 6044}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Closed form of the recursive function $F(1):=1,\;F(n):=\sum_{k=1}^{n-1}-F(k)\sin\left(\pi/2^{n-k+1}\right)$ Suppose that $F$ is defined via the recurrence relation
$$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only one thing that I've noticed is that: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$
Edit:
From the comment section:
I was trying to rewrite $\sin\left(\frac{\pi}{2^n}\right)$ by the formula for a double argument and I've ended up with $$\sin\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)}=\frac{1}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)},n\gt 1,$$
but it doesn't seem to help.
$F$ is used in another formula. It should be true for most of the functions
$$g(x)=\sum_{n=1}^\infty \left(\sin\left(x2^{n-1}\right)\sum_{k=1}^n\left(F(k)g\left(\frac{\pi}{2^{n-k+1}}\right)\right)\right),\;x\in\left(0,\frac{\pi}{2}\right)$$
First four values of $F$ are:
\begin{align*}F(1)&=1\\F(2)&=-\sin\left(\frac{\pi}{4}\right)\\F(3)&=-\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{4}\right)\\F(4)&=-\sin\left(\frac{\pi}{16}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)-\sin^3\left(\frac{\pi}{4}\right)\end{align*}
These terms look like if they created some pattern, but the fifth term which is
\begin{align*}F(5)=-\sin\left(\frac{\pi}{32}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{16}\right)-3\sin^2\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{\pi}{4}\right)\end{align*}
meses the pattern up.
| The $F(n+1)$ is the coefficient of $x^n$ in the power series expansion of
$$
\dfrac{1}{{\displaystyle\sum\limits_{n = 0}^\infty {\sin \left( {\dfrac{\pi }{{2^{n + 1} }}} \right)x^n } }}.
$$
Consequently,
$$
F(n + 1) = \sum\limits_{\substack{k_1 + 2k_2 + \cdots + nk_n = n \\
k_1 ,k_2 , \ldots ,k_n \in \mathbb{Z}_{ \ge 0} }} { \frac{{(k_1 + k_2 + \cdots + k_n )!}}{{k_1 !k_2 ! \cdots k_n !}}\prod\limits_{j = 1}^n {( - 1)^{k_j } \sin ^{k_j } \left( {\frac{\pi }{{2^{j + 1} }}} \right)}} ,
$$
for $n\geq 0$. I do not think there is an explicit formula simpler than this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\Psi: V_1\longrightarrow V_2$ such that $\Psi(F)=\Phi \circ F$. Let $V_1=Lin(\mathbb{R_{<=2}[x]},\mathbb{R_{<=2}[x]})$ and $V_2=Lin(\mathbb{R_{<=2}[x]},\mathbb{R})$ two vector spaces and let
$$\Phi \in V_2, \qquad \Phi(p(x))=p'(1).$$
Consider
$$\Psi: V_1\longrightarrow V_2\qquad \Psi(F)=\Phi \circ F.$$
How can I find the dimension of the Kernel of $\Psi$ and a basis?
Then, consider $W=\{F\in V_1 : F(x − 1) = 0\}$. What's the dimension of $W$?
| Consider the base $\mathcal B =\{x^2,x,1\}$ of $\mathbb R_{\leq 2}[x]$ and consider the coordinates isomorphism:
$$
\mathbb R_{\leq 2}[x] \rightarrow \mathbb R^3,\qquad a_2x^2+a_1x+a_0\mapsto (a_2,a_1,a_0)^T
$$
By this isomorphism an element of $\text{Lin}(\mathbb R_{\leq 2}[x],\mathbb R_{\leq 2}[x])$ is just a $3\times 3$ matrix, and an element of $\text{Lin}(\mathbb R_{\leq 2}[x],\mathbb R)$ is a $1\times 3$ matrix.
In particular how does appear the element $\Phi$ read in coordinate? We have $\Phi(a_2x^2+a_1x+a_0) = 2a_2 + a_1$, so it is represented by the matrix:
$$
\mathcal M(\Phi) = \left(\begin{matrix}
2 & 1 & 0
\end{matrix}
\right)
$$
And how does appear the function $\Psi$? We have $\Psi(F)=\Phi\circ F$; since $F$ is represented by a $3\times 3$ matrix we have:
$$
\Psi\left(
F
\right) = \left(\begin{matrix}
2 & 1 & 0
\end{matrix}
\right) \circ \left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right)=
\left(
\begin{matrix}
2a_{1,1} + a_{2,1} & 2a_{1,2} + a_{2,2} & 2a_{1,3} + a_{2,3}
\end{matrix}
\right)
$$
Now it's easy to find $\ker\Psi$ and a base of it:
\begin{gather}
\ker\Psi=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right) \in V_1 \ \text{such that} \ \ \ \begin{matrix} 2a_{1,1} + a_{2,1}=0,\\ 2a_{1,2} + a_{2,2}=0,\\ 2a_{1,3} +a_{2,3}=0
\end{matrix}
\right\}\\
=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
-2a_{1,1} & -2a_{1,2} & -2a_{1,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right)\in V_1\right\}
\end{gather}
So $\ker\Psi$ has dimesion $6$. An explicit base is really easy to see now.
To find a base of $W$ again consider coordinates. The polynomial $x-1$ is represented by the vector $(0,1,-1)^T$, so:
$$
\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right) \left(\begin{matrix}
0\\
1\\
-1
\end{matrix}\right) = \left(\begin{matrix}
0\\
0\\
0
\end{matrix}\right)
$$
And with the same proof as above you will find
$$
W=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,2}\\
a_{2,1} & a_{2,2} & a_{2,2}\\
a_{3,1} & a_{3,2} & a_{3,2}
\end{matrix}
\right)\in V_1
\right\}
$$
So also $W$ has dimension equal to $6$.
I tried to do this exercise without coordinates, but I did not find a simple way. To see the dimensions of the subspaces you can think to how many conditions are the properties of the subspace. For example in the second part, $F(x-1)=0$ gives you $3$ conditions and the dimensions is $9-3=6$. This is not always true because sometimes some conditions are equivalent, but you can use this "fact" in order to have an idea of the result.
| {
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"question_score": "3",
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Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$
multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have
$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$
For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.
What I tried is subbing $1-2x=y$ which gives
$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$
$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$
I think I made it more complicated. Any help would be appriciated.
| $$\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\text{Li}_4(2)+\text{Li}_2(2)\log^22-3\text{Li}_3(2)\log2+6\operatorname{Li}_4\left(\frac12\right)+\frac{21}4\ln2\zeta(3)-\frac{\pi^2}{8}\log^22+\frac{1}{4}\log^42-\frac{29\pi^4}{288}$$
I being known,I deduce
$$\int_0^1\frac{x\ln(1+x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=-\frac{1}{16}\operatorname{Li}_4\left(\frac12\right)+\frac{21}{64}\ln2\zeta(3)-\frac{41}{768} \pi ^2 \log ^2(2)-\frac{1}{96}\log^42+\frac{1609\pi^4}{92160}-\frac{3}{2}\text{Li}_4(2)-\frac{1}{2}\text{Li}_2(2)\log^22+\frac{3}{2}\text{Li}_3(2)\log2$$
Sorry,i couldn't deduct this integral.
$$3\text{Li}_4(2)+\text{Li}_2(2)\log^22-3\text{Li}_3(2)\log2=-3\operatorname{Li}_4\left(\frac12\right)-\frac{21}8\ln2\zeta(3)-\frac{1}{8}\log^42+\frac{\pi^4}{15}$$
| {
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show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y)^2}{2(x^2+y^2)+5z^2}$$
following I want use C-S,But I don't Success
| SOS helps.
For $a^2+b^2+c^2=1$ after $x=\sqrt3a$, $y=\sqrt3b$ and $z=\sqrt3c$ we need to prove that:
$$\frac 1{5-6ab}+\frac 1{5-6bc}+\frac 1{5-6ca}\leq 1$$ or$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{5-6ab}\right)\geq0$$ or
$$\sum_{cyc}\frac{2-6ab}{5-6ab}\geq0$$ or $$\sum_{cyc}\frac{3(a-b)^{2}+2c^{2}-a^{2}-b^{2}}{5-6ab}\geq0$$ or
$$\sum_{cyc}\left(\frac{3(a-b)^{2}}{5-6ab}+\frac{(c-a)(a+c)}{5-6ab}-\frac{(b-c)(b+c)}{5-6ab}\right)\geq0$$ or
$$\sum_{cyc}\frac{3(a-b)^{2}}{5-6ab}+\sum_{cyc}(a-b)\left(\frac{a+b}{5-6bc}-\frac{a+b}{5-6ac}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^{2}\left(\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\right)\geq0.$$
Now, let $a\geq b\geq c.$
Thus, $$S_{c}=\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\geq$$
$$\geq\frac{1}{5-6ac}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}=\frac{5-8bc-2ac}{(5-6ac)(5-6bc)}=$$
$$=\frac{4(b-c)^{2}+(a-c)^{2}+4a^{2}+b^{2}}{(5-6ac)(5-6bc)}\geq0;$$
$$S_{b}=\frac{1}{5-6ac}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}\geq$$
$$\geq\frac{1}{5-6bc}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=\frac{5-8ab-2bc}{(5-6ab)(5-6bc)}=$$
$$=\frac{4(a-b)^{2}+(b-c)^{2}+4c^{2}+a^{2}}{(5-6ab)(5-6bc)}\geq0$$
and $$S_{a}+S_{b}=\frac{1}{5-6bc}-\frac{2(b+c)a}{(5-6ab)(5-6ac)}+\frac{1}{5-6ac}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=$$
$$=\frac{1}{5-6ac}-\frac{2(b+c)a}{(5-6ab)(5-6ac)}+\frac{1}{5-6bc}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=$$
$$=\frac{5-8ab-2ac}{(5-6ab)(5-6ac)}+\frac{5-8ab-2bc}{(5-6ab)(5-6bc)}\geq0.$$
Id est, $$\sum_{cyc}(a-b)^{2}\left(\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\right)=\sum_{cyc}(a-b)^2S_c\geq$$
$$\geq S_b(a-c)^2+S_a(b-c)^2\geq S_b(b-c)^2+S_a(b-c)^2=(b-c)^2(S_b+S_a)\geq0.$$
| {
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Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.
| Also, we can use a Tangent Line method:
$$\sum_{cyc}(a^3-a^2)=\sum_{cyc}(a^3-a^2-\ln{a})\geq0$$ because easy to see that
$$a^3-a^2-\ln{a}\geq0:$$
$$(a^3-a^2-\ln{a})'=3a^2-2a-\frac{1}{a}=$$
$$=\frac{3a^3-3a^2+a^2-a+a-1)}{a}=\frac{(a-1)(3a^2+a+1)}{a},$$
which gives $a_{min}=1$ and we are done!
| {
"language": "en",
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Need help with calculus II series I am working on a problem that my professor isn't really explaining well, so i decided to ask here.
The following is the question
$ f(x) =\sum_{n=1}^\infty \frac{\mathrm{(-1)}^{n+1}\mathrm{(x-5)}^{n}}{(n\mathrm{5}^{n})} $
I am asked to find the interval of convergence of the following
$f(x)$
$f'(x)$
$f''(x)$
$\int f(x) $
So for $ f(x) $ i used the ratio test $ (\frac{a_n+1}{a_n} )$ and got that -> $ 1 - \frac{x}{5} < 1 $ Is this the interval of convergence? I am totally clueless.
My main question was for $ f'(x) $ I have the first derivative as follows
$ f'(x) = \frac{\mathrm{(-1)}^{n+1}\mathrm{(x-5)}^{n-1}}{\mathrm{5}^{n}} $
Using the ratio $ (\frac{a_n+1}{a_n} )$ test on this one got me -> $ - \frac{x-5}{5} $ -> $ \frac{x}{5} - 1 $
But this will never be greater than 1.
I am not sure what I am doing wrong here.
Any help is appreciated
| For the series $\sum_{n = 1}^\infty a_n$, where there is an $N$ such that $a_n \neq 0$ for all $n \geq N$, the ratio test has you calculate
$$ L = \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| $$
If $L < 1$, the series converges absolutely. If $L = 1$ or the limit fails to exist, the test is inconclusive. If $L > 1$ then the series is divergent.
In both of your examples, you have lost/ignored the absolute value.
For your first example, as I type this, you don't have a sum, but I will guess you meant
$$ f(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} (x-5)^n}{n 5^n} \text{.} $$
The numerator is only zero when $x = 5$, and that sum is easy to do, so we apply the ratio test to determine what happens when $x \neq 5$. We compute \begin{align*}
L &= \lim_{n \rightarrow \infty} \left| \frac{\frac{(-1)^{(n+1)+1} (x-5)^{(n+1)}}{(n+1) 5^{(n+1)}}}{\frac{(-1)^{n+1} (x-5)^n}{n 5^n}} \right| \\
&= \lim_{n \rightarrow \infty} \left| \frac{(-1)^{(n+1)+1} (x-5)^{(n+1)}}{(n+1) 5^{(n+1)}} \cdot \frac{n 5^n}{(-1)^{n+1} (x-5)^n} \right| \\
&= \lim_{n \rightarrow \infty} \left| \frac{(-1)^{(n+1)}(-1) (x-5)^{n}(x-5)}{(n+1) 5^{n}5} \cdot \frac{n 5^n}{(-1)^{n+1} (x-5)^n} \right| \\
&= \lim_{n \rightarrow \infty} \left| \frac{(-1) (x-5)}{(n+1) 5} \cdot \frac{n}{1} \right| \\
&= |(-1) (x-5)| \lim_{n \rightarrow \infty} \left| \frac{n}{5(n+1)} \right| \\
&= \frac{1}{5} |x-5| \text{.}
\end{align*}
The ratio test assures us the series converges if $L < 1$, so when \begin{align*}
\frac{1}{5} |x-5| &< 1 \\
|x-5| &< 5 \\
-5 < x-5 &< 5 \\
0 < x &< 10 \text{.}
\end{align*}
The ratio test is inconclusive when $L = 1$, that is at $x = 0$ and $x = 10$, so we inspect those individually:
*
*$x = 0$: \begin{align*}
\sum_{n=1}^\infty \frac{(-1)^{n+1} (0-5)^n}{n 5^n}
&= \sum_{n=1}^\infty \frac{(-1)^{n+1} (-1)^n 5^n}{n 5^n} \\
&= \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n} \\
&= \sum_{n=1}^\infty \frac{-1}{n} \\
&= - \sum_{n=1}^\infty \frac{1}{n} \text{,}
\end{align*}
which is minus a diverging $p$-series (in particular, it is minus the harmonic series, which diverges).
*$x = 10$: \begin{align*}
\sum_{n=1}^\infty \frac{(-1)^{n+1} (10-5)^n}{n 5^n}
&= \sum_{n=1}^\infty \frac{(-1)^{n+1} 5^n}{n 5^n} \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \text{,}
\end{align*}
which is an alternating series, whose terms monotonically decrease with limit zero. By the alternating series test, this series converges. (We could also recognize this as the alternating harmonic series, which converges.)
Therefore, the interval of convergence of $f(x)$ is $(0,10]$.
You should be able to adapt the above to your other series. If you have difficulties, ask in comments.
| {
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Find the all positive integer solutions $(a,b)$ to $\frac{a^3+b^3}{ab+4}=2020$.
Find the all positive integer solutions of given equation
$$\frac{a^3+b^3}{ab+4}=2020.$$
I find two possible solutions, namely $(1011,1009)$ and $(1009,1011)$, but the way I solve the equation was messy and I don't know if there are any other solutions.
Source: Turkey $1.$ TST for IMO $2020$
| Write for ease $n=2020$ and let $c=a+b$. As $b=c-a$ we get a following quadrtatic equation on $a$: $$(3c+n)a^2-(3c+nc)a+c^3-4n=0$$
So it discriminat must be a perfect square $d^2$ (as it has solution in $\mathbb{Z}$):
$$d^2 = -3c^4+2nc^3+n^2c^2+48nc+16n^2\;\;\;\;\;(*)$$ from here we get $$\boxed{2n\mid d^2+3c^4}$$
Now what can we say about $c$?
*
*If $5\nmid c$ then $c^4\equiv_5 1$ so $d^2+3\equiv _5 0$ which is not possible. So $5\mid c$.
*Since $8\mid d^2+3c^2$, $d$ and $c$ nust be the same parity. Say both are odd. Since for each odd $x$ we have $x^2\equiv_8 1$ we get $$ 0\equiv _8 d^2+3c^4 \equiv_8 1+3$$ A contradiction. So $c$ and $d$ are even. Since $8\mid 3c^4$ we have $8\mid d^2$ so $4\mid d$.
*If $101\nmid c$ then $$d^2c^{-4} \equiv_{101} -3\implies \Big({-3\over 101}\Big)=1$$
But $$\Big({-3\over 101}\Big) = \Big({-1\over 101}\Big)\Big({3\over 101}\Big) = 1\cdot \Big({101\over 3}\Big)(-1)^{{3-1\over 2}{101-1\over 2}} = -1$$
A contradiction again, so $101\mid c$
So $$\boxed{1010\mid c}$$
Now suppose $c>n$. From $(*)$ we get: \begin{align}3c^4&\leq 2nc^3+n^2c^2+48nc+16n^2\\
&< 2(c-1)c^3+(c-1)c^2+64c^2\\
& = 3c^4-4c^4+65c^2
\end{align}
and now we have $4c^3<65c^2$, a contradiction. So $c\leq 2020$.
So $c\in\{1010,2020\}$ and we check both values manualy...
| {
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Show that this inequality is true Show that $\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{999998}{999999} > \frac{1}{100}$.
I tried to take another multiplication $\frac{3}{5} \cdot \frac{6}{8} \cdot \frac{9}{11} \cdot ... \cdot \frac{999996}{999998}$ so that we would have their multiplications equal to $\frac{2}{999999}$. And if we assume that first one equals to $x$, second one equals $y$ we would have an inequality like $x \gt y$ and $x^2 \gt y \cdot x$ so that we can prove that $ x \gt \frac{1}{1000}$ but I can't make it for $\frac{1}{100}$.
| Let $$A_n=\sqrt[3]{3n+1}\cdot\prod_{k=1}^n\left(1-\frac1{3k}\right).$$
The claim is that
$$\tag1 A_{333333}>1.$$
We compute
$$\begin{align}\left(\frac{A_{n}}{A_{n-1}}\right)^3&=\frac{(3n+1)(3n-1)^3}{(3n-2)27n^3}\\
&=1+\frac{6n-1}{(3n-2)27n^3}\\&>1+\frac{6n-4}{(3n-2)27n^3}\\&=1+\frac2{27n^3}>1\end{align}$$
or $A_n>A_{n-1}$. By induction,
$$A_{333333}>A_1=\sqrt[3]{4}\cdot \frac23 =\sqrt[3]{\frac{32}{27}}>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670193",
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"source": "stackexchange",
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Find the value of $a \in R$ such that $\langle x_n \rangle$ converges to a positive real number when $x_n=\frac{1}{3}\frac{4}{6}...\frac{3n-2}{3n}n^a$ Find the value of $a \in \mathbb{R}$ such that $\langle x_n \rangle$ converges to a positive real number when $x_n=\dfrac{1}{3}\dfrac{4}{6}\cdots\dfrac{3n-2}{3n}n^a$
Here is my approach. First of all, let $a_n=\dfrac{x_n}{n^a}=\dfrac{1}{3}\dfrac{4}{6}\cdots\dfrac{3n-2}{3n}$
Then, let $b_n=\dfrac{2}{4}\dfrac{5}{7}\cdots\dfrac{3n-1}{3n+1}$ and $c_n=\dfrac{3}{5}\dfrac{6}{8}\cdots\dfrac{3n}{3n+2}$
Since $0<a_n<b_n$ and $0<a_n<c_n$
$0<a_n^3<a_nb_nc_n=\dfrac{1}{3}\dfrac{2}{4}\dfrac{3}{5}\cdots\dfrac{3n-2}{3n}\dfrac{3n-1}{3n+1}\dfrac{3n}{3n+2}=\dfrac{2}{(3n+1)(3n+2)}$
Therefore,
$0<x_n^3<\dfrac{2n^{3a}}{(3n+1)(3n+2)}$
So, since the limit of $x_n^3$ should be greater than 0 and less than some positive real number,
the limit of $\dfrac{2n^{3a}}{(3n+1)(3n+2)}$ should be neither 0 nor infinity. Therefore, 3a=2, a=2/3.
Anything wrong? Or better idea?
| Expand $$\log(3n-2) - \log(3n) = \log(1 - \frac23 \frac1n) = - \frac23 \frac1n + \frac49 \epsilon_n \frac1{n^2}$$
where $\epsilon_n$ is a bounded sequence.
Then :
$$\log x_n = -\frac23 \sum_{k = 1}^n \frac1k + a \log n + \frac49 \sum_{k = 1}^n \epsilon_k \frac1{k^2}$$
Le last sum converges : $$\big|\epsilon_k \frac1{k^2} \big| \leq (\max_m |\epsilon_m|) \times \frac1{k^2}$$
The harmonic serie satisfies :
$$\sum_{k = 1}^n \frac1k = \log n - \gamma + \epsilon^1_n$$
where $\epsilon^1_n$ converges to $0$.
(to prove that write $\log n = \sum_{k = 1}^n \log(k+1) - \log k = \sum_{k = 1}^n \log(1+\frac1k)$ and expand)
Thus, $\log x_n$ converges to a finite limite iff :
$$-\frac23 \sum_{k = 1}^n \frac1k + a \log n = (a - \frac23) \log n + \frac23 \gamma + \frac23 \epsilon^1_n$$
converges to a finite limit. Finally $a = 2/3$, as you expected.
| {
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Does $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n } = 1$? Question:
Find the limit
\begin{equation}
A = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }
\end{equation}
The series originated from the asymptotic analysis in this question. I can show that it converges. Numerical evaluation in Mathematica suggests that it is close to $1$.
I am just curious, is it possible to prove one of the following?
*
*$A > 1$
*$A = 1$
*$A < 1$
Perhaps one can think about the integral
\begin{equation}
\int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx
\end{equation}
Update 1:
Let $A_N = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }$, numerical evaluations up to $10,000$ shows $A_N < 1$
We can also plot the difference $1 - A_N$ as a function of $N$. To check the estimation $1- A_N \sim 0.3 / \ln N $ by one of the comments, we plot $( 1- A_N) \ln N$
update 2
Following Crostul's answer, I did an exercise to prove that the integral is also equal to $1$ in the limit $N \rightarrow \infty$
Relation between $A_N$ and the integral:
\begin{equation}
A_N = \sum_{n=1}^{N} \frac{1}{N} \frac{1}{ (1 + \frac{1}{N}) \ln ( N+ 1 ) - \frac{n}{N} \ln( \frac{n}{N} N ) }
\end{equation}
so to find $A_{N\rightarrow \infty}$, we may look at the limit
\begin{equation}
\lim_{N \rightarrow \infty} \int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx
\end{equation}
We can simplify the denominator
\begin{equation}
( 1 + \frac{1}{N} ) \ln ( N+1 ) - x \ln ( xN ) = \ln N ( 1 + \frac{b_N}{N} -x - \frac{x \ln x}{\ln N} )
\end{equation}
where
\begin{equation}
b_N = 1 + ( N + 1 )\frac{\ln (1 + \frac{1}{N})}{\ln N } = 1 + \frac{1}{\ln N} + o(\frac{1}{N} )
\end{equation}
So we study
\begin{equation}
I_N = \frac{1}{\ln N} \int_0^1 \frac{1}{ 1 + \frac{b_N}{N} - x - \frac{x \ln x}{\ln N} } dx
\end{equation}
Now use Crostul's relaxing trick
Define $f(x) = x + \frac{x \ln x}{ \ln N}$, the denominator is $f(1 + \frac{1}{N} ) - f(x)$. On one hand we have
\begin{equation}
f(1 + \frac{1}{N} ) - f(x) \ge f(1 + \frac{1}{N} ) - x \ge 1 + \frac{1}{N} - x
\end{equation}
where the second inequality holds for large enough $N$.
On the other hand, by mean value theorem
\begin{equation}
\frac{f( 1 + \frac{1}{N} ) - f(x) }{1 + \frac{1}{N} - x} = f'( y) \le f'( 1 + \frac{1}{N} ) = 1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }
\end{equation}
Hence
\begin{equation}
\frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }} \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx \le I_N \ln N \le \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx
\end{equation}
Both integrals on the left and right are $\ln N$. Therefore
\begin{equation}
\frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }} \le I_N \le 1
\end{equation}
Taking the limit $N\rightarrow$, we have $\lim_{N\rightarrow \infty } I_N = 1$ .
| You are right: the limit is $1$.
Here it is the full proof:
For all $n \in \{1, \dots N \}$ we have the following inequality:
$$(N+1) \log (N+1)- n \log n \ge (N+1) \log (N+1)- n \log (N+1) =\\ = \log(N+1) (N+1-n)$$
Thus
$$\sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \le \sum_{n=1}^N \frac{1}{\log(N+1) (N+1-n)} = \frac{H_{N}}{\log(N+1)}$$
where $H_N$ denotes the $N$-th harmonic number.
Since
$$\lim_{N \to + \infty} \frac{H_N}{\log(N+1)} =1$$
we have the estimate
$$\limsup_{N \to + \infty} \sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \le 1$$
in other words, if the limit exists, it is smaller or equal than $1$.
Proving the other inequality is harder.
Denote by $f(x)=x \log x$. We can compute its derivative $$f'(x)= \log x +1$$
By the Mean Value Theorem, for all $n \in \{1, \dots N \}$ we have
$$\frac{(N+1)\log(N+1)-n \log n}{N+1-n}= \frac{f(N+1)-f(n)}{(N+1)-n} = f'(c_n) =\log c_n +1 \le \log(N+1)+1$$
where $c_n$ is some real number in the interval $(n, N+1)$
Thus we have the estimate
$$\sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \ge \sum_{n=1}^N \frac{1}{N+1-n} \cdot \frac{1}{\log(N+1)+1} = \frac{H_{N+1}}{\log(N+1)+1}$$
where $H_{N+1}$ denotes the ($N+1$)-th harmonic number.
Since
$$\lim_{N \to + \infty} \frac{H_{N+1}}{\log(N+1)+1} =1$$
we have the estimate
$$\liminf_{N \to + \infty} \sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \ge 1$$
In other words, we have proved the other inequality: and the limit is indeed $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone? Observe that $x_1=1$ and $x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$.
Thus, $x_2>x_1$. In general, we also have $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)>\dfrac{1}{\sqrt{n}}\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\ldots+\dfrac{1}{\sqrt{n}}\right)=1$.
Thus, $x_n\geq 1$ for all $n\in \mathbb{N}$. Also, we have,
$x_{n+1}=\dfrac{1}{\sqrt{n+1}}\left(\sqrt{n}x_n+\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}$.
Is it true that $x_{n+1}>x_n$?
Edit :
Thanks to the solution provided by a co-user The73SuperBug. Proving $x_{n+1}-x_n>0$ is equivalent to proving $x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$. This is explained below :
\begin{equation}
\begin{aligned}
&x_{n+1}-x_n>0\\
\Leftrightarrow & \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}-x_n>0\\
\Leftrightarrow & \left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)x_n-\dfrac{1}{n+1}<0\\
\Leftrightarrow & x_n<\dfrac{1}{(n+1)\left(1-\dfrac{\sqrt{n}}
{\sqrt{n+1}}\right)}\\
\Leftrightarrow & x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}.
\end{aligned}
\end{equation}
| A generalization I couldn't pass by. Let $\color{blue}{S_n=\frac1n\sum_{k=1}^{n-1}f\big(\frac{k}{n}\big)}$ where $f:(0,1)\to\mathbb{R}$ is strictly convex: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a<b,0<t<1.$$ If we put $a=k/(n+1),b=(k+1)/(n+1),t=k/n$ for $0<k<n$ here, we obtain $$f\Big(\frac{k}{n}\Big)<\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big),$$ which, after summing over $k$ (to have "$<$" still, we must assume $n>1$), gives $$\sum_{k=1}^{n-1}f\Big(\frac{k}{n}\Big)<\sum_{k=1}^{\color{red}{n}}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=\color{red}{1}}^{n}\frac{k-1}{n}f\Big(\frac{k}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^{n}f\Big(\frac{k}{n+1}\Big),$$ i.e. $\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$. Returning to the question, if we put $f(x)=1/\sqrt{x}$, we get $x_n=S_n+1/n$ and $n^2 x_n<(n^2-1)x_{n+1}+1$; since $x_{n+1}>1$, the latter implies the needed $x_n<x_{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3$
Show that $\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3$
Performing the operations $C_1 \rightarrow C_1-bC_3$
and $C_2 \rightarrow C_2+aC_3$, I got
$$\Delta =\begin{vmatrix} 1+a^2+b^2&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}$$
So I got the term $1+a^2+b^2$, but I am not able to pull it out. What should I do next?
| $$\begin{aligned}\Delta &=\begin{vmatrix} 1+a^2+b^2&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)\begin{vmatrix} 1&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)^2\begin{vmatrix} 1&0&-2b \\ \ 0& 1 &2a \\\ b &-a &1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)^2[(1-a^2-b^2 + 2a^2) + b(2b)]\\
&= (1+a^2+b^2)^3
\end{aligned}$$
As you can factor a coefficient in a column and then develop the determinant along the first column.
| {
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Showing that for $x > e^{2.5102}, 0 \le \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506x}{\ln x}\rfloor \le 1$ Showing that for $x > e^{2.5102}, 0 \le \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506x}{\ln x}\rfloor \le 1$
Does this argument work:
(1) $\dfrac{1.25506x}{\ln x}$ is increasing for $x > e$ with since:
*
*$\dfrac{d}{dx}(\dfrac{1.25506x}{\ln x}) = \dfrac{1.25506\ln(x)-1.25506}{(\ln x)^2}$
*$1.25506\ln(x) - 1.25506$ is positive
(2) For $x > e^{2.51012}$, since $\dfrac{1.25506x}{\ln x}$ is increasing for $ x > 1$:
$$0 < \dfrac{1.25506(x+1)}{\ln(x+1)} - \dfrac{1.25506x}{\ln x} < \dfrac{1.25506(x+1)}{\ln x} - \dfrac{1.25506x}{\ln x} = \dfrac{1.25506}{\ln x} < 0.5$$
(3) There exists integers $a,b$ such that $0 \le a < 1$ and $0 \le b < 1$ such that
$$\lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506(x)}{\ln(x)}\rfloor = \dfrac{1.25506(x+1)}{\ln(x+1)} - a - \dfrac{1.25506(x)}{\ln(x)} + b$$
(4) Since $-1 < b - a < 1$, it follows that:
$$-1 < \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506(x)}{\ln(x)}\rfloor < 0.5 + 1 = 1.5$$
Does the conclusion follow? Did I make a mistake? Is there a better way to establish the same conclusion?
| (1)(2) are correct though you have a typo in (2). It should be $x\gt e$ instead of $x\gt 1$.
(3) is not correct. If $a,b$ are integers such that $0\le a\lt 1$ and $0\le b\lt 1$, then $a=b=0$ for which
$$\left\lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\right\rfloor - \left\lfloor\dfrac{1.25506(x)}{\ln(x)}\right\rfloor = \dfrac{1.25506(x+1)}{\ln(x+1)} - a - \dfrac{1.25506(x)}{\ln(x)} + b$$
does not necessarily hold since the RHS is not necessarily an integer.
Let $f(x):=\dfrac{1.25506(x)}{\ln(x)}$.
After getting $0\lt f(x+1)-f(x)\lt 0.5$ in (2), we can separate it into two cases :
*
*If there exists an integer $N$ such that $f(x)\lt N\le f(x+1)$, then $\lfloor f(x)\rfloor=N-1$ and $\lfloor f(x+1)\rfloor=N$ so $\lfloor f(x+1)\rfloor-\lfloor f(x)\rfloor=N-(N-1)=1$.
*If there exists an integer $N$ such that $N\le f(x)\lt f(x+1)\lt N+1$, then $\lfloor f(x)\rfloor=\lfloor f(x+1)\rfloor=N$ so $\lfloor f(x+1)\rfloor-\lfloor f(x)\rfloor=N-N=0$.
Therefore, $0\le \lfloor f(x+1)\rfloor -\lfloor f(x)\rfloor\le 1$ follows.
Added :
If you meant $a,b$ are real numbers, not integers, then what you did is correct.
In conclusion, you didn't make any mistakes. You just had a few typos.
| {
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Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$ For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$
$$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$
My solution$:$
Let $x=a^2,\,y=b^2,\,z=c^2$ then $a+b+c=1,$ we need to prove$:$ $$\sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq 1\Leftrightarrow \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq a+b+c$$
By AM-GM$:$ $$\text{LHS} = \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a \sqrt{2a^2(b^2+c^2)}} \geqq \sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)} \geqq a+b+c$$
Last inequality is true by SOS$:$
$$\sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)}-a-b-c=\sum\limits_{cyc} {\frac {{c}^{2} \left( a-b \right) ^{2} \left( a+b \right) \left( 2\, {a}^{2}+ab+2\,{b}^{2}+{c}^{2} \right) }{a \left( 2\,{a}^{2}+{b}^{2}+{c }^{2} \right) b \left( {a}^{2}+2\,{b}^{2}+{c}^{2} \right) }} \geqq 0$$
PS:Is there any another solution for original inequality or the last inequality$?$
Thank you!
| Another way.
Since $$\left(yz-\frac{1}{2}xy-\frac{1}{2}zx, zx-\frac{1}{2}yz-\frac{1}{2}xy,xy-\frac{1}{2}zx-\frac{1}{2}yz\right)$$ and
$$\left(\frac{1}{x\sqrt{2(y+z)}},\frac{1}{y\sqrt{2(z+x)}},\frac{1}{z\sqrt{2(x+y)}}\right)$$ have the same ordering, by AM-GM and Chebyshov we obtain:
$$\sum_{cyc}\frac{x^2+yz}{x\sqrt{2(y+z)}}-1=\sum_{cyc}\left(\frac{x^2+yz}{x\sqrt{2(y+z)}}-\sqrt{x}\right)=$$
$$=\sum_{cyc}\frac{x^2+yz-x\sqrt{2x(y+z)}}{x\sqrt{2(y+z)}}\geq \sum_{cyc}\frac{x^2+yz-\frac{1}{2}x(2x+y+z)}{x\sqrt{2(y+z)}}=$$
$$=\sum_{cyc}\frac{yz-\frac{1}{2}xy-\frac{1}{2}zx}{x\sqrt{2(y+z)}}\geq\frac{1}{3}\sum_{cyc}\left(yz-\frac{1}{2}xy-\frac{1}{2}zx\right)\sum_{cyc}\frac{1}{x\sqrt{2(y+z)}}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684495",
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"source": "stackexchange",
"question_score": "6",
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Am I going correctly? Let $f(x)$ be a polynomial in $x$ and let $a, b$ be two real numbers where $a \neq b$
Show that if $f(x)$ is divided by $(x-a)(x-b)$ then the remainder is
$\frac{(x-a) f(b)-(x-b) f(a)}{b-a}$
MY APPROACH:-
Let $Q(x)$ be quotient so that :
$(x-a)(x-b)Q(x)+{ Remainder }=f(x)$
L.H.S, $(x-a)(x-b) \cdot a(x)+ \cfrac{(x-a)f(b)-(x-b) f(b)}{b-a}$
So if we take $x=a$ :
$ \text { Remainder }=\left.f(x)\right|_{x=a} $
$0+\frac{0-(x-b) f(b)}{(b-a)}=f(a) \\
f(b)=\frac{-(b-a) f(a)}{(x-b)}=\frac{(a-b){f}(a)}{x-b}$
again if we take $x=b,$ then $0+\frac{(x-a) f(b)}{b-a}=f(b)$
Substituting, $f(b)$ value in the equation then
$\frac{(x-a) f(b)}{b-a}=\frac{(a-b) f(a)}{x-b}$
| I think you have the right idea but have not set out your argument very clearly; it looks as if you assume your conclusion about half way through. A better lay out might be:
When dividing $f(x)$ by $(x-a)(x-b)$ you obtain,
$$ f(x) = (x-a)(x-b) q(x) + r(x) $$ where $r(x)$ is a polynomial of degree one or less. Thus $r(x) = \alpha x + \beta$ for values $\alpha$ and $\beta$.
Then substitute the value $x=a$ and $x=b$ in turn to derive,
$$ f(a) = r(a) = \alpha a + \beta, \quad f(b) = r(b) = \alpha b + \beta. $$
You an now solve for $\alpha$ and $\beta$ giving the result you wanted,
$$\alpha =\frac{f(b)-f(a)}{b-a} \text{ and } \beta = \frac{bf(a) - af(b)}{b-a}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688396",
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"source": "stackexchange",
"question_score": "2",
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Another series involving $\log (3)$ I will show that
$$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$
My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums?
Denote the series by $S$ and let $S_n$ be its $n$th partial sum.
\begin{align}
S_n &= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 3} + \frac{1}{6k + 5} - \frac{1}{2k + 1} \right )\\
&= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 2} + \frac{1}{6k + 3} + \frac{1}{6k + 4} + \frac{1}{6k + 5} + \frac{1}{6k + 6} \right )\\
& \quad - \sum_{k = 0}^n \left (\frac{1}{2k + 1} + \frac{1}{2k + 2} \right ) - \frac{1}{2} \sum_{k = 0}^n \left (\frac{1}{3k + 1} + \frac{1}{3k + 2} + \frac{1}{3k + 3} \right )\\
& \qquad + \frac{1}{2} \sum_{k = 0}^n \frac{1}{k + 1}\\
&= H_{6n + 3} - H_{2n + 2} - \frac{1}{2} H_{3n + 3} + \frac{1}{2} H_{n + 1}.
\end{align}
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n \frac{1}{k}$. Since $H_n = \log (n) + \gamma + o(1)$ where $\gamma$ is the Euler-Mascheroni constant we see that
$$S_n = \log (6n) - \log (2n) - \frac{1}{2} \log (3n) + \frac{1}{2} \log (n) + o(1) = \frac{1}{2} \log (3) + o(1).$$
Thus
$$S = \lim_{n \to \infty} S_n = \frac{1}{2} \log (3).$$
| Simpler, I do not know (except with integration as @J.G. answered).
Otherwise, in order to have have a good approximation of the partial sums, you can start with
$$s_p=\sum_{n=0}^p \frac 1{a n+b}=\frac{\psi \left(p+1+\frac{b}{a}\right)-\psi
\left(\frac{b}{a}\right)}{a}$$
and consider that, for large values of $p$,
$$s_p=\frac{\log
\left({p}\right)-\psi ^{(0)}\left(\frac{b}{a}\right)}{a}+\frac{a+2 b}{2 a^2 p}-\frac{a^2+6 a b+6 b^2}{12a^3 p^2}+\frac{a^2 b+3 a b^2+2 b^3}{6 a^4
p^3}+O\left(\frac{1}{p^4}\right)$$ which, for your case, would give
$$S_p=\frac 12{\log (3)}-\frac{1}{54 p^2}+\frac{1}{27 p^3}+\cdots$$
| {
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"source": "stackexchange",
"question_score": "2",
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An interesting property of a particular set of triples - multiplying two and adding the other always gives 1
Find all triples of real numbers such that multiplying any two in a
triple and adding the third always gives $1$.
When will this be the case? How can we find all such triples?
So far, I've let the numbers be $a$, $b$ and $c$.
Therefore, $ab+c$ etc must $= 1$ but how can I restrict the possibilities to find all possible triples?
I seem to think this has something to do with 1s and 0s, eg. $0,0,1$ or $0,1,1$
Many thanks guys!!
| If $ab+c=ac+b=bc+a=1$, then $a(b-c)+(c-b)=0$. Factoring out the common factor $b-c$, we get $(a-1)(b-c)=0$. Hence, either $a=1$ or $b=c$. Likewise, either $b=1$ or $a=c$, and either $c=1$ or $a=b$.
Suppose that $a=1$. Then, $b+c=bc+1=1$, so $bc=0$. Hence, one of $b$ and $c$ must be $0$ and the other must be $1$, giving the solutions $(1,0,1)$ and $(1,1,0)$.
Likewise, if $b=1$, then similar steps will give the solutions $(0,1,1)$ and $(1,1,0)$.
Finally, if $c=1$, then similar steps will give the solutions $(0,1,1)$ and $(1,0,1)$.
If none of $a$, $b$, and $c$ are equal to $1$, then they must all be equal. This reduces to solving the equation $a^2+a=1$, or $a^2+a-1=0$. By the quadratic formula, the two resulting solutions are $a=\frac{-1+\sqrt{5}}{2}$ and $a=\frac{-1-\sqrt{5}}{2}$.
Hence, the $5$ solutions are $(0,1,1)$, $(1,0,1)$, $(1,1,0)$, $(\frac{-1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2})$, and $(\frac{-1-\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2})$. Using $\phi$, the golden ratio, the last $2$ solutions can be written as $(\phi-1,\phi-1,\phi-1)$ and $(-\phi,-\phi,-\phi)$ respectively.
| {
"language": "en",
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Show $n I_n I_{n-1} = \frac{\pi}{2}$ where $I_n = \displaystyle\int_0^{\pi/2} \cos^n x dx$. Consider the sequence $(I_n)_{n \in \mathbb{N}}$:
$$ I_0 = \frac{\pi}{2} $$
$$ I_n = \int_0^{\pi/2} \cos ^n x dx$$
For this sequence I have to prove that the following is true:
$$n I_n I_{n-1} = \frac{\pi}{2}$$
for $n \in \mathbb{N}^*$. This is what I've done so far:
$$I_n = \int_0^{\pi/2} \cos^n x dx = \int_0^{\pi/2} \cos x \cos ^{n-1} x dx$$
$$= \int_0^{\pi/2} (\sin x)' \cos ^{n-1} x dx = \sin x \cos ^{n-1} x \Bigg|_0^{\pi/2} + (n-1)\int_0^{\pi/2} \cos ^{n-2}x \sin^2x dx$$
$$ = 0 - 0 + (n-1)\int_0^{\pi/2} \cos ^{n-2} x (1 - \cos ^2 x) dx $$
$$ = (n-1) \int_0^{\pi/2} (\cos ^ {n-2} x - \cos^n x) dx $$
$$= (n-1)(I_{n-2} - I_{n})$$
So that means we have:
$$ I_n = nI_{n-2} - nI_n - I_{n-2} + I_n $$
$$nI_n = (n-1) I_{n-2}$$
And if we multiply with $I_{n-1}$ we get:
$$n I_n I_{n-1} = (n-1) I_{n-2} I_{n-1}$$
But that's as far as I got. I don't see how I could show that the above is equal to $\frac{\pi}{2}$.
| I'll continue from where you got to :
$$nI_{n}I_{n-1} = (n-1)I_{n-2}I_{n-1} $$
Now,
$$nI_{n}I_{n-1} = (n-1)I_{n-2}I_{n-1}$$
$$(n-1)I_{n-1}I_{n-2} = (n-2)I_{n-3}I_{n-2}$$
$$(n-2)I_{n-2}I_{n-3} = (n-3)I_{n-4}I_{n-3}$$
$$.....$$
$$(2)I_2I_1 =I_0I_1$$
Multiply all these equations: You can see most of the terms get cancelled.....you end up with:
$$nI_nI_{n-1} = I_0I_1 $$
$$=\int_0^{\frac{\pi}{2}} \cos^0x\,dx\int_0^{\frac{\pi}{2}} \cos x\,dx$$
$$nI_nI_{n-1} = \left(\frac{\pi}{2}\right) (1)$$
Alternatively,
$$nI_nI_{n-1}=n\left(\int_0^\frac{\pi}{2} \cos^nx\,dx \right)\left(\int_0^\frac{\pi}{2} \cos^{n-1}x\,dx \right) =A$$
Use the substitution :
$$\cos^2x=t$$
$$\cos x = t^{\frac{1}{2}}$$
$$-\sin x \,dx = \frac{1}{2} t^{-\frac{1}{2}}\,dt$$
$$\,dx=-\frac{1}{2} t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}}\,dt $$
$$$$
$$\therefore A = n \left( \int_0^1 \frac{1}{2} (t)^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} (t)^\frac{n}{2} \,dt \right) \left( \int_0^1 \frac{1}{2} (t)^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} (t)^\frac{n-1}{2} \,dt \right)$$
$$A=\frac{n}{4}\left( \int_0^1(t)^{\frac{n+1}{2} -1} (1-t)^{\frac{1}{2}-1}\,dt \right) \left( \int_0^1(t)^{\frac{n}{2} -1} (1-t)^{\frac{1}{2}-1}\,dt \right)$$
$$=\frac{n}{4}\beta\left( \frac{n+1}{2},\frac{1}{2} \right).\beta\left( \frac{n}{2},\frac{1}{2} \right)$$
Here, $\beta(x,y)$ is the Beta Function.
Now, we can use the relation between the beta function and the Gamma function ($\Gamma (n)$) : $$\beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
$$$$
$$A = \frac{n}{4} \frac{\Gamma\left( \frac{n+1}{2} \right) \Gamma\left( \frac{1}{2} \right) }{\Gamma\left( \frac{n}{2} +1 \right)}.\frac{\Gamma\left( \frac{n}{2} \right) \Gamma\left( \frac{1}{2} \right)}{\Gamma\left( \frac{n+1}{2} \right)} $$
The $\Gamma\left( \frac{n+1}{2} \right)$ gets cancelled out from num and denom.
The $\Gamma\left( \frac{1}{2} \right)$ terms have a known value of $\sqrt\pi$
The rest : $\frac{\Gamma\left( \frac{n}{2} \right)}{\Gamma\left( \frac{n}{2} +1 \right)}$ can be reduced by the property of the Gamma function: $=\frac{\left( \frac{n}{2} -1 \right)!}{\left( \frac{n}{2} \right)!} = \frac{2}{n}$
So, we are left with :
$$A=\frac{n}{4} \left(\Gamma\left(\frac{1}{2} \right)\right)^2 \frac{2}{n}$$
$${A}=\frac{(\sqrt\pi)^2}{2} = \frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$ For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$$
My proof$:$
$$4(a+b+c) \prod (a+b)^2 (\text{LHS}-\text{RHS})$$
$$=\prod (a+b) \Big[\sum\limits_{cyc} (ab+bc-2ca)^2+(ab+bc+ca)\sum\limits_{cyc} (a-b)^2 \Big]$$
$$+(a+b+c)(a-b)^2 (b-c)^2 (c-a)^2 \geqq 0$$
| From
$$ \sum \frac{a}{b+c} -\frac{3}{2} - \left(\sum \frac{ab+bc+ca}{(a+b)^2} -\frac{9}{4}\right) = \frac14 \sum \frac{(a-b)^2}{(a+b)^2} \geqslant 0.$$
We can see, the inequality also true for all $a,\,b,\,c$ are real numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$f(x)=(\sin(\tan^{-1}x)+\sin(\cot^{-1}x))^2-1, |x|>1$
Let $f(x)=(\sin(\tan^{-1}x)+\sin(\cot^{-1}x))^2-1, |x|>1$. If $\frac{dy}{dx}=\frac12\frac d{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt3)=\frac{\pi}{6}$, then $y(-\sqrt3)=?$
$$f(x)=(\frac{x}{\sqrt{x^2+1}}+\frac{1}{\sqrt{x^2+1}})^2-1=\frac{2x}{1+x^2}$$
$$\frac{dy}{dx}=\frac12\frac d{dx}(\sin^{-1}(\sin(2\tan^{-1}x)))$$
$$y=\tan^{-1}x+c$$
Using, $y(\sqrt3)=\frac{\pi}{6}$, I get, $c=-\frac\pi6$. Thus, $y(-\sqrt3)=-\frac\pi2$. But the answer is given as $\frac{5\pi}6$.
| $\sin(\cot^{-1}x)=\sin\left(\dfrac\pi2-\tan^{-1}x\right)=\cos(\tan^{-1}x)$
$$\implies f(x)=\left(\sin(\tan^{-1}x)+\sin(\cot^{-1}x)\right)^2-1=\sin2\left(\tan^{-1}x\right)$$
Now $\sin^{-1}\left(\sin(2\tan^{-1}x )\right)=\begin{cases} \pi-2\tan^{-1}x &\mbox{if } 2\tan^{-1}x>\dfrac\pi2\iff x>\tan\dfrac\pi4 \\
-\pi-2\tan^{-1}x& \mbox{if } 2\tan^{-1}x<-\dfrac\pi2 \end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.
| This is a linear ODE then
$$
y=y_h+y_p\\
y'_h + \frac{x}{1+x^2}y_h = 0\\
y'_p + \frac{x}{1+x^2}y_p = x
$$
the homogeneous is separable giving
$$
y_h = \frac{c_0}{\sqrt{1+x^2}}
$$
now using the method of constants variation due to Lagrange we make $y_p = \frac{c_0(x)}{\sqrt{1+x^2}}$ and substituting we obtain
$$
\frac{c_0'(x)}{\sqrt{x^2+1}}-x=0
$$
giving
$$
c_0(x) = \frac{1}{3} \left(x^2+1\right)^{3/2}
$$
and finally
$$
y = \frac{c_0}{\sqrt{1+x^2}}+\frac{1}{3} \left(x^2+1\right)^{3/2}\frac{1}{\sqrt{1+x^2}} = \frac{c_0}{\sqrt{1+x^2}}+\frac{1}{3} \left(x^2+1\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do you find the coefficient of $x$ in $(x + 1)^2$? I want to learn how can I find out the coefficient of the variable $x$ in the expression $(x + 1)^2$. It is a case of a perfect square expansion.
| This is the distributive property of the multiplication: $(a+b)c=ac+bc$.
What is $(x+a)^2$? Well,
$$(x+a)^2=(x+a)(x+a)$$
Let's write one of the factors $(x+a)=c$, and distribute:
$$(x+a)^2=(x+a)c=xc+ac$$
Replacing $c$ back to $x+a$ we get
$$(x+a)^2=(x+a)(x+a)=x(x+a)+a(x+a)$$
And we distribute again
$$(x+a)^2=x(x+a)+a(x+a)=x^2+xa+ax+a^2$$
At the end we get
$$(x+a)^2=x^2+2ax+a^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| $$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
$$3\cdot 2^{3x+1}+2\cdot 3^{3x+1}=7\cdot 2^{2x}\cdot 3^{x}+7\cdot 3^{2x}\cdot 2^{x}$$
Divide both the sides by $2^{3x+1}$, we get
$$3+3 \left(\frac{3}{2}\right)^{3x}=\frac72 \left(\frac{3}{2}\right)^{x}+\frac72 \left(\frac{3}{2}\right)^{2x}$$
Let $\left(\frac{3}{2}\right)^x=t\quad \forall \ \ t>0$
$$3+3t^3=\frac{7}{2}t+\frac{7}{2}t^2$$
$$6t^3-7t^2-7t+6=0$$
$$(t+1)(3t-2)(2t-3)=0$$
$$t=\frac32 \implies \left(\frac32\right)^x=\frac{3}{2}\iff x=1$$
$$t=\frac 23\implies \left(\frac32\right)^x=\frac{2}{3}\iff x=-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find $A$ given that $2I + A +A^2 = B$ where $B$ is given. How to find a matrix $A$ such that the following holds:
$$2I + A +A^2 = B,$$ where the matrix $B$ is given. I tried with char poly of $B$ but not getting any idea.
Note that it is also given that $B$ is invertible.
P.S. $B = \begin{pmatrix}-2&-7&-4\\ \:12&22&12\\ \:-12&-20&-10\end{pmatrix}$.
| Here is an ad hoc method:
It is straightforward (if tedious) to find null spaces of
$B-2I, B-4I, (B-4I)^2$ and determine the Jordan form.
With $V=\begin{bmatrix} 3 & -1 & 1 \\ -6 & 1 & 0 \\ 6 & -1 & -1 \end{bmatrix}$ we see that
$V^{-1}BV = \begin{bmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 2 \end{bmatrix}$, and since $2+0+0^2 = 2$ we can look for $A$ of the form
$A=\begin{bmatrix} \lambda & \alpha & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Then from $A^2+A+2I=B$ we get $\lambda^2+\lambda +2 = 4$ and so $\lambda \in \{-2,1\}$.
Then we need $2 \alpha \lambda + \alpha = 1$ from which we get
$\alpha = {1 \over 2 \lambda +1}$.
Hence $\lambda =-2, \alpha = -{1 \over 3}$ and $\lambda =1, \alpha = {1 \over 3}$ are two solutions (or rather $V A V^{-1}$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
| Why? Just write the function as follows:
$$f(x) = \frac{(x+1)+1}{(x+1)^2} = \frac{x+1}{(x+1)^2}+ \frac{1}{(x+1)^2}
$$
later,
$$f(x)= \frac{1}{(x+1)}+\frac{1}{(x+1)^2}=u+u^2 = (u+0.5)^2-0.25 = \left(\frac{1}{x+1}+0.5\right)^2-0.25$$
Since here you should be able to continue, just see the variation of $x$ and transform to $f(x)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Factorize $(a+b+c)^5-(a+b-c)^5-(a+c-b)^5-(b+c-a)^5$ We can see that $a=0$ makes expression $0$, similarly others make expression $0$.This implies that its factorization has $abc$ in it.I substituted $a+b+c=s$ tried to find remaining factors.From here I did not progress any further
Later I went on finding this on wolframalpha and it's factorization is $80abc(a^2+b^2+c^2)$. How can we get its factorization by hand without actually expanding whole thing?Is there any slick way?
| Let $a+b=x,\,a-b=y$ \begin{align*}
(a + b + c)^5 - (a + b - c)^5 - (a + c - b)^5 - (b + c - a)^5=\\
(x+c)^5+(c-x)^5-(y+c)^5+(y-c)^5=\\
c^5 + 5 c^4 x + 10 c^3 x^2 + 10 c^2 x^3 + 5 c x^4 + x^5+\\
c^5 - 5 c^4 x + 10 c^3 x^2 - 10 c^2 x^3 + 5 c x^4 - x^5+\\
-c^5 - 5 c^4 y - 10 c^3 y^2 - 10 c^2 y^3 - 5 c y^4 - y^5+\\
-c^5 + 5 c^4 y - 10 c^3 y^2 + 10 c^2 y^3 - 5 c y^4 + y^5=\\
20 c^3 (x^2-y^2)+10c(x^4-y^4)=\\
20 c^3 (x-y)(x+y)+10c(x-y)(x+y)(x^2+y^2)=\\
80 c^3 b a + 80 c b a (a^2+b^2)=\\
80 abc(c^2+a^2+b^2)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solving this probability paragraph Let $B_n$ denotes the event that n fair dice are
rolled once with $P(B_n)=1/2^n$ where n is a natural number.
Hence $B_1,B_2,B_3,..B_n$are pairwise mutually exclusive events as n approaches infinity.
The event A occurs with atleast one of the event
$B_1,B_2,B_3,..B_n$ and denotes that the numbers appearing on the dice is S
If even number of dice has been rolled,then show that probability that $S=4$ is very close to $1/16$
next show that probability that greatest number on the dice is 4 if three dice are known to have been rolled is $37/216$
Finally,if $S=3$, then prove that
$P(B_2/S)=24/169$
my approach :
well I tried using the conditional probability formula in part one and baye's theorem in the final one but I am unable to get to the correct answer.kindly help me out,all help is greatly appreciated.
|
probability that greatest number on the dice is 4 if three dice are known to have been rolled is $37/216$
This is about a conditional probability $|B_3$. We have $P=P’-P’’$, where $P’=\left(\frac 46\right)=\frac {64}{216}$ is a probability that the greatest number on a dice is at most $4$ and $P’’=\left(\frac 36\right)=\frac {27}{216}$ is a probability that the greatest number on a dice is at most $3$.
If even number of dice has been rolled,then show that probability that $S=4$ is very close to $1/16$
I tried two interpretations for an event $A$ which is $S=4$, but obtained the following answers.
If $A$ means that at least one thrown dice has $4$ then we have
$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)=
\frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$
$$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \left(1-\left(\frac 56\right)^{2k}\right)\frac 1{2^{2k}}=
1-\frac{\sum_{k=1}^\infty\left(\frac{5}{12}\right)^{2k}}{\sum_{k=1}^\infty \frac 1{2^{2k}}}=$$
$$1-\frac{\frac{\left(\frac{5}{12}\right)^2}{1-\left(\frac{5}{12}\right)^2}}{\frac {\frac 1{2^2}}{1-{\frac 1{2^2}}} }=
1-\frac{\frac{1}{\left(\frac{12}{5}\right)^2-1}} {\frac 1{\frac {2^2}1-1}}=
1-\frac{2^2-1}{{\left(\frac{12}{5}\right)^2-1}}=$$ $$1-\frac{3\cdot 5^2}{{12^2-5^2}}=1-\frac{75}{119}=\frac{44}{119}.$$
If $A$ means that all thrown dices have $4$ then we have
$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)=
\frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$
$$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \frac 1{6^{2k}}\cdot\frac 1{2^{2k}}=
\frac{\frac 1{12^2} \cdot\frac {1}{1-\frac 1{12^2}}}{\frac 1{2^2} \cdot\frac {1}{1-\frac 1{2^2}} }=
\frac {\frac{1}{12^2-1}}{\frac{1}{2^2-1}}=\frac 3{143}.$$
if $S=3$, then prove that $P(B_2/S)=24/169$
$P(B_2|A)=\frac{P(A\cap B_2)}{P(A)}$, but because of the above there is a problem how to interpret $A$.
| {
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"source": "stackexchange",
"question_score": "2",
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Find the condition where the product of two factorials is minimized I have to find the maximum value of a combination ${N \choose H} = \dfrac{N!}{H!(N-H)!}$ where $1 \leq H \leq N \leq 16$. My reasoning was to find the maximum value you have to maximize the numerator $N!$ and minimize the denominator $H!(N-H)!$. The maximum value of numerator occurs when $N = 16$. Then the term becomes, $\dfrac{16!}{H!(16-H)!}$. As, $1 \leq H \leq 16$, I can manually check for which $H$ value the term is maximized. But is there any other way to find out that the denominator is minimized when $H = 8$?
I tried to break now into three conditions, namely when $H > (16-H)$, $H < (16 - H)$ and $ H = 16 - H$, but I couldn't establish the relation between the three conditions. Any help would be appreaciated.
| From the complementary combination $\displaystyle\binom{N}{H} = \binom{N}{N-H}$ we conclude that
$\displaystyle\binom{N}{H}$ is symmetric with respect to $\displaystyle H=\bigg\lceil{\frac{N}{2}}\bigg\rceil$
Now we show that $\displaystyle\binom{N}{H}$ is monotonically increasing for $\displaystyle 0\le H \le \bigg\lceil{\frac{N}{2}}\bigg\rceil$ keeping $N$ fixed. Equivalently we will show $$\text{If } \displaystyle 0\le H_1 < H_2\le \bigg\lceil{\frac{N}{2}}\bigg\rceil \text{ then } \binom{N}{H_1} <\binom{N}{H_2}$$
Proof: Let $H_2 = H_1 + k$ with $k\ge1$. Now
$$\begin{equation*}
\begin{aligned}
\binom{N}{H_2} &= \frac{N\cdot(N-1)\cdots(N-H_2+1)}{1\cdot2\cdot3\cdots H_2}\\
&=\frac{N\cdot(N-1)\cdots (N-H_1+1)\cdot(N-H_1)\cdot(N-H_1-1)\cdots(N-H_1-k+1)}{1\cdot 2\cdot 3\cdots H_1\cdot(H_1+1)\cdots(H_1+k)} \\
&=\binom{N}{H_1}\cdot\frac{(N-H_1)(N-H_1-1)\cdots(N-H_1-k+1)}{(H_1+1)(H_1+2)\cdots(H_1+k)}\\
&=\binom{N}{H_1}\cdot\frac{N-H_1}{H_1+k}\cdot\frac{N-H_1-1}{H_1+k-1}\cdots\frac{N-H_1-k+1}{H_1+1}
\end{aligned}
\end{equation*}$$
We need to show now every fraction of RHS $\ge1$ and at least one fraction $>1$
Notice that, every fraction of RHS is of the form $\displaystyle \frac{N-H_1-r}{H_1+k-r}$
where $0\le r \le k-1$
$$\begin{equation}\displaystyle
\begin{aligned}
\frac{N-H_1-r}{H_1+k-r}&=\frac{N-H_2+k-r}{H_2-r}\\
&\ge \frac{N-H_2+1}{H_2-r} \text{ as } k-r\ge1 \\
&\ge \frac{N-\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil+1}{\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil-r} \\
&=\frac{\bigg\lfloor\displaystyle\frac{N}{2}\bigg\rfloor+1}{\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil-r}
\end{aligned}
\end{equation}$$
If $r > 0$ this fraction $>1$ and if $r=0$ this fraction $\ge1$ and we are done. $\blacksquare$
| {
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Modulus operation to find unknown If the $5$ digit number $538xy$ is divisible by $3,7$ and $11,$ find $x$ and $y$ .
How to solve this problem with the help of modulus operator ?
I was checking the divisibility for 11, 3:
$5-3+8-x+y = a ⋅ 11$ and $5+3+8+x+y = b⋅3$ and I am getting more unknowns ..
| From modulus 11,
$$
53800 + 10x + y \equiv 5 - 3 + 8 - x + y \equiv -1-x+y \equiv 0 \pmod {11}\\
\implies y\equiv 1+x \pmod {11}
$$
but $y$ and $x$ are digits, so $0\le y\le 9$ and $1\le 1+x \le 10$, so it must hold that $y=1+x$.
From modulus 3,
$$
53800 + 10x + y \equiv 5 + 3 + 8 + x + (x+1) \equiv 2+2x \equiv 0 \pmod {3}\\
\implies x\equiv -1 \pmod {3}
$$
so $x=2,5,8$.
Eventually, from modulus 7,
$$
53800 + 10x + y \equiv 5 + 3x + (x+1) = 6+4x \equiv 0 \pmod {7}\\
\implies x\equiv 2 \pmod {7}
$$
so $x=2,9$.
The only choice is thus $x=2$, $y=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\frac{d\theta }{ d \tan \theta}=\frac{ 1}{ 1+ \tan^2 \theta}$ I suppose that
$$
\frac{d\theta }{ d \tan \theta}=\frac{d \arctan x }{ d x}= \frac{1}{1+x^2}=\frac{ 1}{ 1+ \tan^2 \theta}
$$
So is
$$
\frac{d\theta }{ d \tan \theta}=\frac{ 1}{ 1+ \tan^2 \theta}
$$
correct? And
$$
\frac{d (\theta) }{ d \tan \frac{\theta}{2}}=\frac{ 2}{ 1+ \tan^2 \frac{\theta}{2}} \; ?
$$
| Follows is the way I look at the derivation of
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{1+ \tan^2 \theta} \tag 0$
and related identities; starting with
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{\dfrac{d\tan \theta}{d\theta}} \tag 1$
we use the definition $\tan \theta = \sin \theta / \cos \theta$ and the quotient rule for derivatives to obtain:
$\dfrac{d\tan \theta}{d\theta} = \dfrac{d}{d\theta}\dfrac{\sin \theta}{\cos \theta} = \dfrac{(\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta)}{\cos^2 \theta}$
$= \dfrac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \dfrac{\cos^2 \theta}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta} = 1+ \tan^2 \theta; \tag 2$
now by (1),
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{1+ \tan^2 \theta}; \tag 3$
having obtained this result, we may calculate $\dfrac{d\theta}{d\tan (\theta/2)}$, additionally invoking the chain rule; we set
$u(\theta) = \dfrac{\theta}{2}, \tag4$
whence
$\dfrac{du(\theta)}{d\theta} = \dfrac{1}{2}, \tag 5$
and
$\dfrac{d\tan (\theta/2)}{d\theta} = \dfrac{d\tan u(\theta)}{d\theta} = \dfrac{d\tan u(\theta)}{du} \dfrac{du(\theta)}{d\theta}$
$=\dfrac{1}{2}(1 + \tan^2(u(\theta)) = \dfrac{1 + \tan^2 (\theta/2)}{2}, \tag 6$
and thus
$\dfrac{d\theta}{d\tan (\theta/2)} = \dfrac{2}{1 + \tan^2 (\theta/2)}. \tag 7$
$OE\Delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Transform expression for Taylor series $f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$
For $x_0=1$ calculate Taylor series.
So the above expression has already been proven by induction.
$T_{f,x_0}= \sum^\infty_{k=0}\frac{f^{(k)}(x_0)}{k!}\cdot(x-x_0)^k$
$\begin{align}
T_{f,1} &= \sum^\infty_{k=0}\frac{f^{(k)}(1)}{k!}\cdot(x-1)^k \\
&= \sum^\infty_{k=0} \frac{(-1)^{k+1}}{2^k}\cdot \prod^{k-1}_{j=1}(2j-1)\cdot \frac{1}{\sqrt{1^{2k-1}}} \cdot\frac{1}{k!}\cdot(x-1)^k \\
&= \sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}(2j-1) \cdot \frac{1}{k!} \cdot(-1) \cdot \frac{(-1)^k}{2^k} \cdot (x-1)^k \\
&=\sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}[(2j-1) \cdot \frac{1}{j}] \cdot (-1) \cdot (\frac{(-1)\cdot(x-1)}{2})^k \\
&=\sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}(2-\frac{1}{j}) \cdot (-1) \cdot (\frac{1-x}{2})^k
\end{align}$
Is there a way to transform this expression further
| I have the feeling that there is something in particular when suddenly $k!$ disappears.
Starting from the beginning
$$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\prod^{k-1}_{j=1}(2j-1)\frac{1}{\sqrt{x^{2k-1}}}=\frac{(-1)^{k+1}}{2^k\sqrt{x^{2k-1}}}\prod^{k-1}_{j=1}(2j-1)$$
Now
$$\prod^{k-1}_{j=1}(2j-1)=\frac{2^{k-1} }{\sqrt{\pi }}\Gamma \left(k-\frac{1}{2}\right)$$ makes
$$\frac {f^{(k)}(1)}{k!}=\frac{(-1)^{k+1} }{2 \sqrt{\pi }\,k!}\Gamma \left(k-\frac{1}{2}\right)$$
$$T_{f,1}=\frac{1 }{ 2\sqrt{\pi }}\sum_{k=0}^\infty \frac{(-1)^{k+1} }{k!}\Gamma \left(k-\frac{1}{2}\right)(x-1)^k=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(x-1)^k$$ which is in fact a very, very simple function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
-8 \\
1
\end{bmatrix} = \begin{bmatrix}
3 \\
8
\end{bmatrix}$$
and
$$M\begin{bmatrix}
1 \\
5
\end{bmatrix} = \begin{bmatrix}
-8 \\
7
\end{bmatrix}.$$
Evaluate $$M\begin{bmatrix}
1 \\
1
\end{bmatrix}.$$
What is the question essentially asking? Isn't this just a matrix equation?
| We know that
$$M\begin{pmatrix}
-8 & 1 \\
1 & 5
\end{pmatrix} = \begin{pmatrix}
3 &-8\\
8 & 7
\end{pmatrix}.$$
Multiply that equation from right by
$$
\begin{pmatrix}
-8 & 1 \\
1 & 5
\end{pmatrix}^{-1}=\frac{1}{-41}\begin{pmatrix}
5 & -1 \\
-1 & -8
\end{pmatrix}
$$
to get $M$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
"infinitely oscillating" manifold Find out whether $M:=\{\left(x\cdot\cos\left(\frac{1}{x}\right),\,x\cdot\sin\left(\frac{1}{x}\right)\right)|\,x>0\}$ is a submanifold of $\mathbb{R}^2$ or not.
My guess is that it's not since I've tried to construct a homeomorphism from $\mathbb{R}$ to $M$ $\left(x\mapsto \left(x\cdot\cos\left(\frac{1}{x}\right),\,x\cdot\sin\left(\frac{1}{x}\right)\right)\right)$ and failed in proving bijectivity. I then tried to construct multiple charts from $\mathbb{R}$ and $\mathbb{R}^2$ but it hasn't worked out so far because of the way $M$ looks near the origin.
Thank you very much in advance.
| It is a submanifold.
Consider the map $f : (0,\infty) \to \mathbb R^2, f(x) = (x\cos(\frac{1}{x}),x\sin(\frac{1}{x}))$. We have $M = f((0,\infty))$. The map $f$ is injective since $\lVert f(x) \rVert = \sqrt{x^2\cos^2(\frac{1}{x}) + x^2\sin^2(\frac{1}{x})} = \sqrt{x^2} = x$ which implies $f(x_1) \ne f(x_2)$ for $x_1 \ne x_2$. Thus $\tilde f : (0,\infty) \stackrel{f}{\to} M$ is a bijection. Moreover, $g = \lVert - \rVert_M : M \to (0,\infty)$ is continuous and we have $g \circ \tilde f = id$, therefore $\tilde f$ is homeomorphism.
We have $f'(x) = (\cos(\frac{1}{x}) + \frac{1}{x}\sin(\frac{1}{x}), \sin(\frac{1}{x}) - \frac{1}{x}\cos(\frac{1}{x}))$ and we get
$$\lVert f'(x) \rVert = \sqrt{\cos^2(\frac{1}{x}) + \frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x}) + \frac{1}{x^2}\sin^2(\frac{1}{x}) + \sin^2(\frac{1}{x}) - \frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x}) + \frac{1}{x^2}\cos^2(\frac{1}{x})} \\ = \sqrt{1 + \frac{1}{x^2}} \ne 0 .$$
Thus $f'(x) \ne 0$ so that $f$ is an immersion. Hence $f$ is a smooth embedding and $M$ is a smooth submanifold of $\mathbb R^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating gradient, how to do this?
Given $f(x,y)$, a function that has continuous partial derivatives in
every point.
such that $\nabla f(0,-18)=-2i+3j$
We define a new function $g(x,y)=f(xy+x^2,xy-y^2)$ calculate $\nabla
g(3,-3)$
How I tried to solve this?
I need to find:
$$\nabla g(3,-3) = g_x'(3,-3)i+g_y'(3,-3)j=f(xy+x^2,xy-y^2)_x'(3,-3)i+f(xy+x^2,xy-y^2)_y'(3,-3)j$$
and I got stuck here; I don't have f to calculate the partial directive for it...
| We can make use of the multivariable chain rule (see this answer) which states that given a function $h: \mathbb{R}^n \to \mathbb{R}^m$ and a function $f: \mathbb{R}^m \to \mathbb{R}$ we have
$$
\nabla (f \circ h) = \left(h'\right)^T \cdot \left(\nabla f \circ h\right)
$$
where "$\cdot$" represents matrix multiplication.
In our case, we see that if we define $h(x,y) = (xy+x^2,xy-y^2)$ then it follows from the definition of $g$ that
$$g(x,y) = (f \circ h)(x,y)= f(xy+x^2,xy-y^2) \tag{1}$$
which means that calculating $\nabla g(3,-3)$ is equal to calculating $\nabla (f \circ h)(3,-3) $.
Now, by definition, we know that
$$
h' = \begin{pmatrix}
\frac{\partial}{\partial x}(xy+x^2) & \frac{\partial}{\partial y}(xy+x^2)\\
\frac{\partial}{\partial x}(xy-y^2) & \frac{\partial}{\partial y}(xy-y^2)
\end{pmatrix} = \begin{pmatrix}
y+2x & x\\
y & x-2y
\end{pmatrix}
$$
which means that the transposed matrix $\left(h'\right)^T$ is $\begin{pmatrix}
y+2x & y\\
x & x-2y
\end{pmatrix}
$. If we then evaluate this matrix at $(3,-3)$ we get
$$
\left(h'\right)^T(3,-3) = \begin{pmatrix}
-3+2(3) & -3\\
3 & 3-2(-3)
\end{pmatrix} = \begin{pmatrix}
-3+6 & -3\\
3 & 3+6
\end{pmatrix} = \begin{pmatrix}
3 & -3\\
3 & 9
\end{pmatrix}
$$
On the other hand, we see that
$$
h(3,-3) = (3(-3)+3^2, 3(-3) -(-3)^2) = (-9+9, -9-9) = (0,-18)
$$
which tells us that
$$
\left(\nabla f \circ h\right)(3,-3) = \nabla f\left(h(3,-3)\right) = \nabla f\left(0,-18\right) = \begin{pmatrix} -2 \\3 \end{pmatrix}
$$
using the convention of the gradient as a column vector.
Finally, putting all this together tells us that
$$
\nabla g(3,-3) = \nabla (f \circ h)(3,-3) = \left[\left(h'\right)^T(3,-3)\right] \cdot \left[\left(\nabla f \circ h\right)(3,-3)\right] = \begin{pmatrix}
3 & -3\\
3 & 9
\end{pmatrix} \cdot \begin{pmatrix} -2 \\3 \end{pmatrix} = \begin{pmatrix} -2(3)+3(-3) \\-2(3)+3(9) \end{pmatrix}= \begin{pmatrix} -6-9 \\-6+27 \end{pmatrix}= \begin{pmatrix} -15 \\21 \end{pmatrix} = -15 \boldsymbol{\hat\imath} + 21 \boldsymbol{\hat\jmath}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $
My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) \iff $$
$$f(x) = (1+x)\ln(1+x) + (1-x)\ln(1-x) \quad $$
We know that the basic Taylor series for $\ln(1+x)$ is
$$ \ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} \quad (1)$$
As far as $\ln(1-x)$ is concerned
$$y(x) = \ln(1-x) \iff y'(x) = \frac{-1}{1-x} = - \sum_{n=0}^\infty x^n \text{ (geometric series)} \iff$$
$$y(x) = \int -\sum_{n=0}^\infty x^n = - \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \quad (2)$$
Therefore from $f(x), (1), (2)$ we have:
$$ f(x) = (1+x)\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - (1-x)\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \iff$$
$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$
Why I hesitate
It all makes sense to me up to this point. But the exercise has a follow up sub-question that requires to find:
$$ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $$
I am pretty sure that this sum is somehow connected with the previous power series that we've found, but I can't find a way to calculate it, so I assume that I have made a mistake.
Any ideas?
| $$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$
Supposing the above is right. We want to change the $n+2$'s to $n+1$'s. To do this, write, by letting $m+1 = n+2$,
$$ \sum_{n=0}^\infty \frac{2x^{n+2}}{n+1} = \sum_{m=1}^\infty \frac{2x^{m+1}}{m} = \sum_{n=1}^\infty \frac{2x^{n+1}}{n},$$
where, in the last step, we simply changed the dummy variable $m$ to $n$.
I haven't read it very carefully, but sometimes you can get $2n+1$ in the denominator when you're only summing over odd integers.
SPOILER
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Am I misapplying the chain rule when differentiating $x^{5x+7}$ with respect to $x$? The problem I am attempting to solve is:
\begin{align}
y=x^{5x+7} \\
\text{Find $\frac{dy}{dx}$}
\end{align}
Here is my working so far:
$$\begin{align}
\text{let }u &= 5x+7 \\
\frac{dy}{dx}&=\frac{dy}{du} \cdot \frac{du}{dx} \\
\frac{dy}{du}&=ux^{u-1}=(5x+7)x^{5x+6} \\
\frac{du}{dx}&=5 \\
\therefore \frac{dy}{dx}&=(25x+35)x^{5x+6}
\end{align}$$
(Additionally, I'm having trouble using the align environment in Mathjax. This question wasn't formatted very well. If someone could give me some help in the comments, then I would be very thankful.)
| If $u= 5x+7$ and $y = x^{5x+7} = x^u$ then if you attempt to solve $\frac {dy}{dx}=\frac{dy}{du}\frac {du}{dx}$ then you must solve $\frac {dy}{du} = \frac {dx^u}{du}$ but $x$ is dependent on $u$ so you'd have to solve $\frac {dx^u}{du} = \frac {dx^u}{dx}\frac {dx}{du}$ but to solve $\frac {dx}{du}$ we have $x$ is dependant on $u$ and .... and we'd have an infinite loop.
(By the way, if we treat $x$ as constant we'd have $\frac {dx^u}{du}=\ln(x)x^u$; not $\frac {dx^u}{dx} = ux^{u-1}$ which is what we'd have if we treated $u$ as a constant and differentiating in respect to $x$.)
To do the properly we need to express $y =x^{5x+7}$ entirely in terms of $u$
So if $u = 5x +7$ then $x = \frac {u-7}5$ and so $y =x^{5x+7} = x^u = (\frac {u-1}5)^u$ which ..... just doesn't help.
Better we do: let $y = x^{5x + 7} = e^{\ln x(5x+7)}$ the let $u = \ln x(5x+7)$ so $y=e^u$ and we can do
$\frac {dy}{dx} = \frac {dy}{du}\frac {du}{dx}$.
$\frac {dy}{du}=\frac{de^u}{d^u} = e^u= y$ was easy enough. And we can use the product rule to figure $\frac {du}{dx}$.
Let $w= \ln x$ and $v = 5x+7$ then $\frac {du}{dx} =\frac {dv\cdot w}{dx} = \frac{dv}{dx} w + \frac {dw}{dx}v$.
And $\frac {dv}{dx} = \frac {d(5x+7)}{dx}= 5$ and $\frac {dw}{dx}= \frac {d\ln x}{dx} = \frac 1x$ so
$\frac {du}{dx} = (5\ln x + \frac 1x(5x + 7))=(5\ln x + 5 +\frac 7x)$ and do
$\frac {dy}{dx}=\frac {dy}{du}\frac {du}{dx} = \frac {dy}{du}(\frac {dv}{dx}w + \frac {dw}{dx}v) = e^u(5\ln x + 5 +\frac 7x)= e^{5x+7}(5\ln x + 5 + \frac 7x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
show this $\tan{x}\tan{y}\tan{z}=1$ in an acute.$\Delta ABC$,if $x,y,z$ such$$\cos{A}=\cos{y}\sin{z},\cos{B}=\cos{z}\sin{x},\cos{C}=\cos{x}\sin{y}$$show that
$$\tan{x}\tan{y}\tan{z}=1$$or $$\sin{x}\sin{y}\sin{z}=\cos{x}\cos{y}\cos{z}$$
I want use $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
so we have
$$\sum_{cyc}\cos^2{y}(1-\cos^2{x})+2\cos{x}\cos{y}\cos{z}\sin{x}\sin{y}\sin{z}=1$$
then
$$\sum_{cyc}\cos^2{x}-\sum_{cyc}\cos^2{x}\cos^2{y}+2\cos{x}\cos{y}\cos{z}\sin{x}\sin{y}\sin{z}=1$$
then I can't
| Let
$$
\tan x\tan y\tan z=t
$$
then
$$
\sin x\sin y\sin z=t\cos x\cos y\cos z
$$
so we get
$$
\sum_{cyc}\cos^2x-\sum_{cyc}\cos^2x\cos^2y+2t\cos^2x\cos^2y\cos^2z=1
$$
then
$$
\begin{aligned}
& (2t-1)\cos^2x\cos^2y\cos^2z \\\\[1ex]
=& 1-\sum_{cyc}\cos^2x+\sum_{cyc}\cos^2x\cos^2y-\cos^2x\cos^2y\cos^2z \\\\
=& \left(1-\cos^2x\right)\left(1-\cos^2y\right)\left(1-\cos^2z\right) \\\\[1ex]
=& \sin^2x\sin^2y\sin^2z
\end{aligned}
$$
therefore
$$
2t-1=\frac{\sin^2x\sin^2y\sin^2z}{\cos^2x\cos^2y\cos^2z}=\tan^2x\tan^2y\tan^2z=t^2
$$
so $$t=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$ is greater than $\frac{2}{3}$. I have to prove that the product
$$\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$$
is greater than $\displaystyle\frac{2}{3}$.
I've tried to write $k^{3}-1$ as $(k-1)(k^{2}+k+1)$ or another ways but I couldn't finish it.
| $\begin{array}\\
f(n)
&=\prod_{k=2}^{n}\dfrac{k^{3}-1}{k^{3}+1}\\
&=\dfrac{\prod_{k=2}^{n}(k^{3}-1)}{\prod_{k=2}^{n}(k^{3}+1)}\\
&=\dfrac{\prod_{k=2}^{n}(k-1)(k^2+k+1)}{\prod_{k=2}^{n}(k+1)(k^2-k+1)}\\
&=\dfrac{\prod_{k=1}^{n-1}k}{\prod_{k=3}^{n+1}k}\dfrac{\prod_{k=3}^{n+1}((k-1)^2+(k-1)+1)}{\prod_{k=2}^{n}(k^2-k+1)}
\qquad\text{(this is the only clever step)}\\
&=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-2k+1+k-1+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-k+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{(n+1)^2-(n+1)+1)}{2^2-2+1}\\
&=\dfrac{2}{n(n+1)}\dfrac{n^2+2n+1-n-1+1}{3}\\
&=\dfrac23\dfrac{n^2+n+1}{n^2+n}\\
&=\dfrac23(1+\dfrac1{n^2+n})\\
&\gt \dfrac23
\qquad\text{for all } n\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Geometry Proof to Find Maximum area of $\triangle PIE$
Circle $\omega$ is inscribed in unit square $PLUM,$ and points $I$ and $E$ lie on $\omega$ such that
$U, I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE.$
I'm not sure if there is a solution possible without trigonometry.
Also, for my diagram in my solution, I'm not sure how to center it. Sorry about that.
| Let $O$ be the center of circle $\omega.$ Let $X$ be the foot of the altitude from $P$ to $IE,$ and let $Y$ be the foot from $O$ to $IE.$ Denote line segment $\overline{YO}$ as length $d,$ the radius as $r,$ and $\angle XUP$ as $\theta.$
$\textbf{Claim:}$ The greatest area of $\triangle PIE$ is $\frac{1}{4}.$
$\textbf{Proof:}$ To find the area of $\triangle PIE,$ we can find the lengths of $\overline{IE}$ and $\overline{PX},$ and then use the formula of the area of a triangle to conclude. Let's start by finding lengths $\overline{IO}$ and $\overline{YO},$ and then apply the Pythagorean Theorem to get our $\overline{IY},$ then multiply by two to get the base of $\triangle PIE.$ Clearly, segment $\overline{IO}$ is the radius of the circle, which has length $1/2.$ Then, by taking $\sin \theta,$ we have $$\sin \theta = \frac{\overline{YO}}{\overline{UO}}=\frac{d}{\sqrt{2}/2} \implies d = \frac{\sqrt{2}}{2} \cdot \sin \theta.$$
Similarly with $\overline{PX},$ we take $\sin \theta$ and get $$\sin \theta = \frac{\overline{PX}}{\overline{UP}} = \frac{\overline{PX}}{\sqrt{2}} \implies \overline{PX} = \sqrt{2} \cdot \sin \theta.$$
Thus, after finding these two lengths, we know the largest possible area of $\triangle PIE$ is $$\displaystyle{\max\left(\frac{1}{2} \cdot \overline{PX} \cdot \overline{IE}\right) = \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{r^2-d^2}\right)}.$$
Note that $\overline{IE} = 2\sqrt{r^2-d^2}$ by the Pythagorean Theorem, where clearly in the diagram $\overline{IO}$ is the radius and $\overline{YO}$ is distance $d.$ Simplifying our equation above to lowest terms, we get:
\begin{align*}
\max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{r^2-d^2}\right) &= \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot 2\sqrt{\left(\frac{1}{2} \cdot \frac{1}{2}\right)-\left(\frac{1}{2} \cdot \sin^2 \theta\right)}\right) \\
&= \max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \sin \theta \cdot \sqrt{1 - 2 \sin^2 \theta} \right). \\
\end{align*}
Then, let's subsititute $\alpha = \sin \theta.$ Thus, to maximize the area of $\triangle PIE,$ all we need to do is find the maximum of $\max \left(\frac{1}{2} \cdot \sqrt{2} \cdot \alpha \cdot \sqrt{1-2\alpha^2}\right).$
This implies we need to find the value of $\alpha$ that suffices to maximize the following equation: $$\max \left(\alpha^2 \left(1-2\alpha^2\right)\right) = \max \left(\alpha^2 - 2\alpha^4 \right).$$
Taking the derivative of $\alpha^2 - 2\alpha^4,$ we get $$\frac{\mathrm{d}}{\mathrm{d}\alpha} \left(\alpha^2 - 2\alpha^4\right) = 2\alpha - 8\alpha^3.$$
The equation $2\alpha - 8\alpha^3 = 0$ is maximized when $\alpha = \frac{1}{2},$ which we hence get the largest area of $\triangle PIE$ as $\frac{1}{2} \cdot \frac{1}{2} = \boxed{\frac{1}{4}},$ as desired. $\qquad\blacksquare$
$\textbf{Claim:}$ There is such $\theta$ that achieves the maximum area stated above.
$\textbf{Proof:}$ We have found that the maximum of $\sin \theta = \frac{1}{2},$ thus meaning that the $\theta = 30^{\circ},$ in which that is where the maximum area of $\triangle PIE$ occurs. Hence, proven. $\qquad\blacksquare$
| {
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How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the values of $a$ and $b$ as
$$\frac{4x^2}{144}+\frac{9y^2}{144}=1$$
$$\frac{x^2}{6^2}+\frac{y^2}{4^2}=1$$
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, gives
$a=6$ & $b=4$. From here I have no idea how to solve further?
| Let the top-right corner be at $(x,y)$. Squaring the aspect ratio, we have the system
$$\begin{cases}4x^2+9y^2=144,\\4x^2=9y^2\end{cases}$$
the solution of which is $x^2=18,y^2=8$.
Area $$4\sqrt{18\cdot8}=48.$$
| {
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Question about finding the value of an infinite sum What is the value of:
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$
I realised that
$$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \frac{3}{4}\frac{\pi^2}{6} \qquad \Rightarrow \qquad \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \frac{1}{8} \cdot \pi^2$$
So
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2} + \sum_{k=0}^\infty \frac{1}{(4k+3)^2} = \frac{\pi^2}{8}$$
But I cannot find the value of the second summation.
Any suggestions?
| The sum can be expressed in terms of Catalan's constant. In the following video, the sum is computed in a step by step manner: https://www.youtube.com/watch?v=r2OJtsHNDZA.
| {
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Interesting Partition Questions There is a good question here.
My question is;
"x is a positive integer and
$\lfloor x\rfloor$ denote the largest integer smaller than or equal to $x$. Prove that $\lfloor n / 3\rfloor+1$ is the number of partitions of $n$ into distinct parts where each part is either a power of two or three times a power of two."
There is a Theorem related with this question.
Theorem:
$
p(n \mid \text {parts in } N)=p(n \mid \text { distinct parts in } M) \quad \text { for } n \geq 1
$
where $N$ is any set of integers such that no element of $N$ is a power of two times an element of $N,$ and M is the set containing all elements of $N$ together with all their multiples of powers of two.
Can anyone help? thanks.
| Let’s use a generating function.
If $p(n)$ is the number of partitions of $n$ into numbers of the form $2^k$ or $3\cdot 2^k$, then we have the following generating function:
$$\sum_{n=0}^\infty p(n)x^n = \prod_{k=0}^\infty (1+x^{2^k})(1+x^{3\cdot 2^k})$$
Recall the following identity, which follows from the fact that every nonnegative integer has a unique binary representation:
$$\prod_{k=0}^\infty (1+x^{2^k})=1+x+x^2+...=\frac{1}{1-x}$$
From this, it follows that our generating function is given by
$$\sum_{n=0}^\infty p(n)x^n=\frac{1}{(1-x)(1-x^3)}$$
On the other hand, we have that
$$\begin{align}
\sum_{n=0}^\infty (\lfloor n/3\rfloor +1)x^n
&= 1+x+x^2+2x^3+2x^4+2x^5+3x^6+... \\
&= (1+x+x^2)(1+2x^3+3x^6+4x^9+...) \\
&= \frac{1+x+x^2}{(1-x^3)^2} \\
&= \frac{1}{(1-x)(1-x^3)}
\end{align}$$
Well, whaddaya know?! The two generating functions are equal to each other! Thus, we have the desired result:
$$p(n)=\lfloor n/3\rfloor +1$$
QED! Thanks for the fun problem!
| {
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Find the number of functions $4f^3(x)=13f(x)+6$
$f(x)$ is a function satisfying
$$4f^3(x)=13f(x)+6$$
$\forall\ x\in [-3,3]$ and is discontinuous only and only at all integers in $[-3,3]$. If $N$ denotes the unit digit of the number of all such functions, find $N$.
Answer: $6$.
Solving the cubic for $f(x)$ gives us
$$f(x)=-\frac{3}{2},-\frac{1}{2}\ \text{or}\ 2$$
Then I figured, for $f$ to be discontinuous at integers, we must define it piecewise(i.e casewise) on the integers, with the value of the function changing at integers.
Something like$f(x)=$
$$
\begin{cases}
-0.5 & -3\leq x<-2 \\
-1.5 & -2\leq x<-1 \\
-0.5 & -1\leq x<0 \\
2 & 0\leq x<1 \\
-1.5 & 1\leq x<2 \\
2 & 2\leq x<3 \\
-1.5 & x=3\\
\end{cases}
$$
So I thought, this problem is equivalent to filling $7$ blanks with $3$ distinct objects, and no two adjacent blanks contain the same object.
Hence, the first blank can be filled in $3$ ways, the second in $2$ ways, the third in $2$ ways and so on, until the seventh blank in $2$ ways.
So the number of ways is $$3\times2\times2\times2\times2\times2\times2=192$$
whose unit digit is not $6$.
Could anyone tell me where I went wrong?
Thanks in advance!
| The following are the properties that fully characterize your function $f$:
*
*For every $x$, $f(x)\in \{2,-1/2,-3/2\}$
*$f$ must be constant on every open interval $(n,n+1)$ for $n=-3,-2,-1,0,1,2$
*among $f(n-1,n)$, $f(n)$, $f(n,n+1)$ there must be at least two different values for $n=-2,-1,0,1,2$
*$f(3) \ne f(2,3)$, $f(-3)\ne f(-3,-2)$
As you did, you can start by fixing $f(-3)$ and $f(-3,-2)$ in 6 ways. Then you can fix $f(-2)$ and $f(-2,-1)$ in $6 + 2 =8$ ways. The same 8 ways for $f(-1)$ and $f(-1,0)$, etc.
The last choice is $f(3)$ that can be made in 2 ways.
For a total of $6\times 8^5\times 2$ ways. Since $2^5\equiv 2\pmod{10}$, we have $6\times 8^5\times 2 = 3 \times 2^{17}\equiv 3 \times 2^{5} \equiv 6\pmod {10}$
| {
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Proving that this formula all over the positive integer gives us this sequence Firstly, we have this sequence : $1,1,1,1,1,1,1,1,1,1,2,2,2,...$ which is the sequence of the number of digits in decimal expansion of $n$.
Secondly, we have this formula : $$a_n=\Bigl\lceil\log_{10}(n+1)\Bigr\rceil-\Bigl\lceil\frac{n}{n+1}\Bigr\rceil+1$$ where $n\ge0$
This formula seems to gives us this sequence.
How to prove this ?
| Firstly, for $n \ne 0$, $n$ has $d$ digits if and only if $10^{d - 1} \le n < 10^d$. One way to find $d$ from here is to take logarithms, using $d - 1 \le \log_{10}(n) < d$, so $d = \lfloor \log_{10}(n) \rfloor$.
However, this expression doesn't work so well when $n = 0$, so we have to be a bit creative. Adding $1$, we get $10^{d - 1} + 1 \le n + 1 < 10^d + 1$. But a strict inequality of natural numbers of the form $a < b + 1$ can be equivalently written as $a \le b$, so in fact we can write $10^{d - 1} < n + 1 \le 10^d$. Then $d - 1 < \log_{10}(n + 1) \le d$, so $d = \lceil \log_{10}(n + 1) \rceil$. This expression doesn't freak out quite so much when $n = 0$.
The last bit is just a clever trick so that $a_0 = 1$. If $n = 0$, then $n/(n + 1) = 0$, so $\lceil n/(n + 1) \rceil = 0$. Otherwise, $0 < n/(n + 1) < 1$, so $\lceil n/(n + 1) \rceil = 1$.
| {
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$\frac{x^2}{by+cz}=\frac{y^2}{cz+ax}=\frac{z^2}{ax+by}=2$ If $$\frac{x^2}{by+cz}=\frac{y^2}{cz+ax}=\frac{z^2}{ax+by}=2$$ then find the value of $$\frac{c}{2c+z}+\frac{b}{2b+y}+\frac{a}{2a+x}.$$
I think all the terms need to be manipulated in some way to get the corresponding terms from the expression whose value needs to be found. For example we have to go from $\frac{x^2}{by+cz}$ to $\frac{a}{2a+x}$ in some way or maybe from $\frac{x^2}{by+cz}$ to $\frac{c}{2c+z}$ or something like that. It's very hard to tell from which term to which term I need to go because all the variables are used. So I just tried to make a system of equations.
$$x^2=2(by+cz)$$
$$y^2=2(cz+ax)$$
$$z^2=2(ax+by)$$
$$x^2+y^2+z^2=4(ax+by+cz)$$
$$(x-a)^2+(y-b)^2+(z-c)^2=a^2+b^2+c^2+2(ax+by+cz)$$
But I don't know how to proceed from here.
| Hint:
$$x(x+2a)=x^2+2ax=2(ax+by+cz)$$
$$\dfrac1{2a+x}=?$$
$$\dfrac a{2a+x}=\dfrac{ax}{2(ax+by+cz)}$$
| {
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Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my lights go off.
Any leads? Thanks in advance.
| Although it is not clear what range of values is acceptable for the variables, I believe I have a sense of what you are trying to do here. This is called denesting square roots.
In particular, if $a,b,c$ are positive real numbers such that $a^2-b^2 c$ is non-negative, then it holds that
$$\sqrt{a+ b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2 c}}{2}}+ \sqrt{\frac{a-\sqrt{a^2-b^2 c}}{2}}.$$
It can be easily proven by squaring both sides and using the difference of squares factorization. There is a way of coming up with it naturally by using the same idea behind finding complex square roots, though it might not be an entirely rigorous approach.
This yields
\begin{align*}
\sqrt{a+b+\sqrt{2ab+b^2}}&=\sqrt{\frac{a+b+|a|}{2}}+\sqrt{\frac{a+b-|a|}{2}}\\
&= \begin{cases} \sqrt{a+\frac{b}{2}}+\sqrt{\frac{b}{2}} &\text{ if } a\ge 0\\ \sqrt{\frac{b}{2}}+\sqrt{a+\frac{b}{2}} &\text{ if } a<0\end{cases}.
\end{align*}
which works because $$(a+b)^2-1^2\cdot (2ab+b^2)=a^2$$ is non-negative. Both cases yield the same answer.
| {
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System of equations and recurrence relation I am trying to find the general solution for $N$ of the following system of equations
$$
\begin{cases}
(x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\
{x_n}^2 + {y_n}^2 = 1
\end{cases}
$$
with the initial values $x_0 = 1$ and $y_0 = 0$ and the following
*
*$\theta$ is a constant and $0 \leqslant \theta \leqslant 2$
*$N$ is a constant and we want to find the terms $(x_N, y_N)$
Using substitution with respect to $N$, we have
$$
\begin{align}
x_0 = 1 \quad & ; \quad y_0 = 0 \\
x_1 = -\frac{\theta^2 - 2N^2}{2N^2} \quad & ; \quad y_1 = -\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2} \\
x_2 = \frac{\theta^4 - 4N^2\theta^2 + 2N^4}{2N^4} \quad & ; \quad y_2 = \frac{(\theta^3 - 2N^2\theta) \sqrt{4N^2 - \theta^2}}{2N^4} \\
x_3 = -\frac{\theta^6 - 6N^2\theta^4 + 9N^4\theta^2 - 2N^6}{2N^6} \quad & ; \quad y_3 = -\frac{(\theta^5 - 4N^2\theta^3 + 3N^4\theta) \sqrt{4N^2 - \theta^2}}{2N^6}
\end{align}
$$
By using substitution, it becomes very difficult with $N \geqslant 2$.
| The second equation expresses that the points $(x_n,y_n)$ remain on the unit circle, and the first, that the successive points form chords of constant length, subtending an angle $\alpha=2\arcsin\frac\theta{2N}$.
Hence $$(x_n,y_n)=(\cos n\alpha,\sin n\alpha).$$
By the way, the system has in fact $2^n$ distinct solutions, as from every intermediate point, the chord can be drawn in two directions.
| {
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Evaluating $\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx$ How can i evaluate this integral, maybe differentiation under the integral sign? i started expressing the integral as the following,
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$
But i dont know how to keep going, ill appreciate any solutions or hints.
| Solution using harmonic series
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x^3\right)}{x}\:dx}_{x^3\to x}-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$=-\frac23\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx=\frac23\zeta(2)$$
$$\int _0^1\frac{\ln \left(1+x+x^2\right)}{1+x}\:dx\overset{IBP}{=}\ln(2)\ln(3)-\int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx$$
For the latter integral, set $a=\frac{2\pi}{3}$ in the identity
$$\sum_{n=1}^{\infty}x^{n-1} \cos(na)=\frac{\cos(a)-x}{1-2x\cos(a)+x^2}, \ |x|<1$$
we have
$$-2\sum_{n=1}^{\infty}x^{n-1} \cos(n\frac{2\pi}{3})=\frac{2x+1}{1+x+x^2}$$
$$\Longrightarrow \int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\int_0^1 x^{n-1}\ln(1+x)dx$$
$$=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\left(\frac{H_n-H_{n/2}}{n}\right)$$
$$=-2\Re\sum_{n=1}^\infty \left(e^{i\frac{2\pi}{3}}\right)^n\left(\frac{H_n-H_{n/2}}{n}\right)$$
And finally we use the generating functions
$$\sum_{n=1}^\infty x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\text{Li}_2(x)$$
$$\sum_{n=1}^\infty x^n\frac{H_{n/2}}{n}=i\pi\frac{\ln(1-x^2)-\ln(-x^2)}{x^2}$$
$$+\frac{\ln(x-1)\ln(-x^2)-\ln(x-1)\ln(1-x^2)}{x^2}$$
$$+\frac{\text{Li}_2\left(\frac{1-x}{1+x}\right)-\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2\left(\frac{1}{1-x}\right)}{x^2}$$
I found the second generating function with help of Mathematica after I converted it to integral;
$$\sum_{n=1}^\infty x^n\frac{H_{n/2}}{n}=-\int_0^1\frac{xy^2\ln(1-y^2)}{1-xy}dy$$
| {
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Solve for $x$ in $\sin^{-1}(1-x)-2\sin^{-1}x =\frac{\pi}{2}$ Let $x=\sin y$
$$\sin^{-1}(1-\sin y)-2\sin^{-1}\sin y=\frac{\pi}{2}$$
$$\sin^{-1}(1-\sin y)-2y=\frac{\pi}{2}$$
$$1-\sin y =\sin (\frac{\pi}{2}+2y)$$
$$1-\cos2y=\sin y$$
$$\sin y(2\sin y-1)=0$$
$$x=0,~ \frac 12$$
Clearly, $x=\frac 12$ isn’t correct, because it doesn’t satisfy the original expression
Why was an extraneous root obtained in this solution? I want to know the reason behind it.
|
Find $x$ satisfying $\sin^{-1}(1-x)-2\sin^{-1}x =\frac{\pi}{2}$.
Common mistakes
While going step-by-step, if you go from
*
*$(\sin^{-1}a)=y\stackrel{\text{to}}{\longrightarrow}\sin(\sin^{-1}a)=a=\sin (y\pm 2n\pi)$, and from
*$\sin y=a\stackrel{\text{to}}{\longrightarrow}\sin^{-1}(\sin y)=\begin{cases}2n\pi+y&y\in\text{I, IV quadrant}\\(2n-1)\pi-y&y\in\text{II, III quadrant}\end{cases}=\sin^{-1}a$,
there occurs an uncertainty$\color{red}{^1}$ of $m\pi$ in $y$. So, the new value of $y$ also begins to satisfy the original equation $(\sin^{-1}a)=y$ or $\sin y=a$.
Your mistakes apart from not doing the domain test
Let $x=\sin y$, then
\begin{align*}
\sin^{-1}(1-\sin y)-2\sin^{-1}\sin y&=\frac{\pi}{2}\\
\end{align*}
$\color{red}{\text{CASE I: $y\in$ I, IV quadrant}}$
\begin{align*}
\sin^{-1}(1-\sin y)-2\color{red}{(y+2n\pi)}&=\frac{\pi}{2}\tag{1}\\
\color{red}{\sin^{-1}(1-\sin y)}&\color{red}{=}\color{red}{\frac{\pi}{2}+2(y+2n\pi)\Rightarrow (y+2n\pi)\in\left[-\frac{\pi}{2},0\right]}\Rightarrow y\in\text{IV quadrant}\tag{2}\\
1-\sin y &=\sin \left(\frac{\pi}{2}+2y\color{red}{+4n\pi}\right)\tag{3}\\
1-\cos2y&=\sin y\\
\sin y(2\sin y-1)&=0\Rightarrow \sin y=0,\ \sin y=\frac 12
\end{align*}
but $\sin y=\frac 12$ is not allowed by the range test that bounds $y$ in IV quadrant in which sine is negative.
$\color{red}{\text{CASE II: $y\in$ II, III quadrant}}$
\begin{align*}
\sin^{-1}(1-\sin y)-2\color{red}{[(2n-1)\pi-y]}&=\frac{\pi}{2}\tag{4}\\
\color{red}{\sin^{-1}(1-\sin y)}&\color{red}{=}\color{red}{\frac{\pi}{2}+2[(2n-1)\pi-y]\Rightarrow ((2n-1)\pi-y)\in\left[-\frac{\pi}{2},0\right]\Rightarrow y\in\left[(2n-1)\pi,\frac{4n-1}{2}\pi\right]\Rightarrow y\in\text{III quadrant}}\tag{5}\\
1-\sin y &=\sin \left(\frac{\pi}{2}\color{red}{+2[(2n-1)\pi-y]}\right)\tag{6}\\
1-\cos2y&=\sin y\\
\sin y(2\sin y-1)&=0\Rightarrow \sin y=0,\ \sin y=\frac 12
\end{align*}
but $\sin y=\frac 12$ is not allowed by the range test that bounds $y$ in III quadrant in which sine is negative.
Your mistake apart from the domain test (that gives the answer instantly) is the range test. :p
My solution
Since taking sine both sides produces no uncertainty for both below LHS and RHS can't deviate by even $2\pi$, we get
\begin{align*}
\sin^{-1}(1-x)-\sin^{-1}x&=\frac{\pi}{2}+\sin^{-1}x\\
\sin[\sin^{-1}(1-x)-\sin^{-1}x] &=\sin\left(\frac{\pi}{2}+\sin^{-1}x\right)\\
\sin[\sin^{-1}(1-x)]\cdot\cos[\sin^{-1}x]-\sin(\sin^{-1}x)\cdot\cos[\sin^{-1}(1-x)] &=\cos(\sin^{-1}x)\\
(1-x)\sqrt{1-x^2}-x\sqrt{2x-x^2} &=\sqrt{1-x^2}&\left(\because \sin^{-1}a\in\left[-\frac{\pi}{2},+\frac{\pi}{2}\right]\Rightarrow \cos(\sin^{-1}a)=\color{red}{+}\sqrt{1-a^2}\right)\\
-x\left(\sqrt{1-x^2}+\sqrt{2x-x^2}\right)&=0 &\\
-x\left(\sqrt{1-x^2}+\sqrt{2x-x^2}\right)&=0 &\\
\Rightarrow x&=0&(\because x\in[0,1])
\end{align*}
since both $\sqrt{1-x^2}$ and $\sqrt{2x-x^2}$ can't vanish simultaneously. The former requires $x=\pm 1$ that the latter does not.
$\color{red}{^1}$This occurs due to defining the range of an inverse function as some part of the domain of the original function.
| {
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Can we apply here the Cayley–Hamilton theorem? We have the matrix \begin{equation*}A:=\begin{pmatrix}3 & 1 & 0 & -1& -1 \\ 0 & 2 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 2 & 0 \\ 1 & 0 & 0 & -1 & 1\end{pmatrix}\in M_5(\mathbb{R})\end{equation*}
The characteric polynomial is \begin{equation*}P_A(\lambda)=(2-\lambda)^5\end{equation*}
The eigenvalue $\lambda=2$ has the algebraic multiplicity $5$.
The eigenspace is \begin{equation*}\left \{\begin{pmatrix}e\\ 0\\ c\\ 0\\ e\end{pmatrix}: c, e\in \mathbb{R}\right \}=\left \{e\begin{pmatrix}1\\ 0\\ 0\\ 0\\ 1\end{pmatrix}+c\begin{pmatrix}0\\ 0\\ 1\\ 0\\ 0\end{pmatrix}: c, e\in \mathbb{R}\right \}\end{equation*}
So the geometric multiplicity of the eigenvalue $\lambda=2$ is $2$.
How can we calculate $(A − 2I_5)^3$ ?
Can we apply here the Cayley–Hamilton theorem?
| Here, dimension of eigenspace corresponding to the eigenvalue 2 is 2, There are two Jordan canonical forms possible! So, in one Jordan canonical form, there can be two blocks possible which are of order 3 and order 2 and in another of Jordon canonical form, there can also be possiblity for two blocks of order 4 and 1,
But through the first case max order of block is 3, hence there be possiblity of minimal polynomial of order 3, hence we can use Cayley Hamilton's theorem! For the first case ,Jordon canonical form, $$
\begin{pmatrix}
2 & 1 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 2 \\
\end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
| Substitute $u=\sinh t$ to integrate
\begin{align}
& \int \dfrac{u^3}{(u^2+1)^3}du= \int \frac{\sinh^3t}{\cosh^5t}dt\\
=&\int\tanh^3td(\tanh t)=\frac14\tanh^4t+C= \frac{u^4}{4(u^2+1)^2}+C
\end{align}
| {
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How to integrate $ \int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx$? How to integrate $$ \int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx\,\,?$$
The given answer is $$ \color{brown}I=-\frac{1}{\sqrt{33}}\cdot \tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-3}}{\sqrt{11}\cdot (x-3)}\bigg)+\mathcal{C}.$$
I tried by different substitutions i.e $\frac{x^2 - 2x -1}{x-3} = t$, but I am not getting my desired answer.
$ORIGINAL$ $QUESTION$:
This question was asked in our test and the given answer was option D ,i.e none on the given options were correct.
| $$I=\int \frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}\,dx$$ This can be simplifies using
$$\frac{x-2}{7x^2-36x+48}=\frac 1{7(a-b)}\left(\frac{a-2 } {x-a }+\frac{2-b } {x-b } \right)$$ where
$$a=\frac{2}{7} \left(9-i \sqrt{3}\right) \qquad \text{and} \qquad b=\frac{2}{7} \left(9+i \sqrt{3}\right) $$ which makes that we are facing two integrals
$$I_c=\int \frac {dx} {(x-c)\sqrt{x^2-2x-1}}$$
Complete the square and let $x=1+\sqrt 2 \sec(t)$ which gives
$$I_c=\int \frac{dt}{(1-c) \cos (t)+\sqrt{2}}$$ Now, using the tangent half-angle subtitution
$$I_c=2\int\frac{du}{\left(c+\sqrt{2}-1\right) u^2-c+\sqrt{2}+1}=\frac{2 }{\sqrt{-c^2+2 c+1}}\tan ^{-1}\left(u\frac{\sqrt{c+\sqrt{2}-1}
}{\sqrt{-c+\sqrt{2}+1}}\right)$$ and so on ....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Unclear problem with $n$-th power matrix and limit
Find $$\lim\limits_{n \to \infty} \frac{A_n}{D_n}$$ where
$$\begin{pmatrix}
19 & -48 \\
8 & -21 \\
\end{pmatrix} ^{\! n} =
\begin{pmatrix}
A_n & B_n \\
C_n & D_n \\
\end{pmatrix}$$
$n$ - is the power of a matrix, but what is $A_n, B_n, C_n, D_n$ then? Is it a corresponding element of a matrix in the $n$-th power?
How is this type of problem called? And what is the way to solve that problem?
| Here is an unconventional approach: we have
$$
M = \pmatrix{19 & -48\\ 8 & -21}.
$$
We find that the eigenvalues satisfy
$$
\det(M - xI) = x^2 + 2x - 15 = 0 \implies x = -5,3.
$$
By the Cayley Hamilton theorem, the powers of $M$ satisfy the recurrence
$$
M^n + 2M^{n-1} -15 M^{n-2} = 0
$$
From the theory of constant coefficient homogeneous linear difference equations, it follows that $M^n$ has the form
$$
M^n = (-5)^n P + 3^n Q
$$
for some matrices $P,Q$. We can solve for $P,Q$ using the "initial conditions" of $n=0,1$. We have
$$
P + Q = M^0 = \pmatrix{1&0\\0&1}, \quad (-5)P + 3Q = M^1 = \pmatrix{19 & -48\\8 & -21}.
$$
Subtracting the second equation from $3$ times the first yields
$$
3P - (-5)P + 0Q = 3\pmatrix{1&0\\0&1} - \pmatrix{19 & -48\\8 & -21} \implies\\
8P = \pmatrix{-16 & -48\\8 & 24} \implies P = \pmatrix{-2&-6\\1&3}.
$$
We use the first equation to find
$$
Q = \pmatrix{1&0\\0&1} - P = \pmatrix{3&6\\-1&-2}.
$$
So, we have
$$
M^n = (-5)^n \pmatrix{-2&-6\\1&3} + 3^n\pmatrix{3&6\\-1&-2}.
$$
| {
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Solving $2\sin\left(2x\right)=3\left(1-\cos x\right)$ Background - this was part of a homework packet for students looking to skip HS pre-calc. There is a text book they use as well, but this particular problem was not in it.
$$2\sin\left(2x\right)=3\left(1-\cos\left(x\right)\right)$$
My first step was to eliminate the double angle.
$$4\sin\left(x\right)\cos\left(x\right)=3\left(1-\cos\left(x\right)\right)$$
and distribute on right side
$$4\sin\left(x\right)\cos\left(x\right)=3-3\cos\left(x\right)$$
And this is where I am stuck.
Now, I can see that $0$ and $2\pi$ are solutions, but with a graph of both sides, one more. The worksheet instructions do not say whether or not graphing is allowed, although the particular chapter in the book for this seems to rely heavily on calculator work.
My question
Can this be solved by manipulation, if so, how? If not, what is the hint to stop and go to the graph?
| Rewrite $2\sin\left(2x\right)=3\left(1-\cos x\right)$ as
$$ 4\sin\frac x2\cos\frac x2\cos x= 3\sin^2\frac x2$$
Then, let $t= \tan\frac x2, \> \cos x = \frac{1-t^2}{1+t^2}$ and factorize
$$\sin\frac x2 \frac{4-3t-4t^2-3t^3}{1+t^2}=0
$$
The factor $\sin\frac x2=0 $ yields $\frac x2 =\pi n$ and $4-3t-4t^2-3t^3=0$ has one real root at
$$t= \frac19\left(-4+\sqrt[3]{584+9\sqrt{4227}}-\sqrt[3]{-584+9\sqrt{4227}}\right)
$$
Thus, the solutions are
$$x=2\pi n,\>\>\> 2\pi n + 2\arctan t$$
| {
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Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
| We use the elementary limit results $\displaystyle \lim_{x\to 0}\frac{\sin x}{x} =1=\lim_{x\to 0} \frac{\tan x}{x}$
Now coming to the main problem we may write it as $$\begin{aligned}\lim_{x\to 0} \frac{\sqrt{1+x^2\left(\frac{\sin x}{x}\right)}-\sqrt{\cos x}}{x^2\left(\frac{\tan x}{x}\right)}&=\lim_{x\to 0} \frac{\sqrt{1+x^2}-\sqrt{\cos x}}{x^2}\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots -\sqrt{1-\underbrace {\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots}}_{q}\right)\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x^2}{2}+O(x^4)-\left(1-\frac{q}{2}+\frac{q^2}{4}+O(q^6)\right)\right)\\&=\lim_{x\to 0} \frac{1}{x^2}\left(1+\frac{x^2}{2}-1+\frac{q}{2}-O(q^4)\right) \\&=\lim_{x\to 0} \frac{1}{x^2}\left(\frac{x^2}{2}+\frac{1}{2}\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6)\right)\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with a differential equation system Given $x' = -x$ and $y' = -4x^3+y$, we want to linearize and show phase portrait at origin.
So I make system $\vec{Y}' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\vec{Y}$ by just scrapping the $-4x^3$ term. But now we have repeated $0$ eigenvalue, so I try to find an eigenvector.
$\left[ \begin{pmatrix} -1 & 0 \\ 0 &1 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} \right]\begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies v_1 = v_2 = 0$. So $\vec{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$?
Unless I am mistaken. What kind of eigenvector is this? I can't think of how to draw a phase portrait, thanks!
| The assertion that the eigenvalues of the matrix $\vec Y'$ are both zero is erroneous. However, we have:
The eigenvectors of the matrix
$\vec Y' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \tag 1$
are $(1, 0)^T$, with eigenvalue $-1$, and $(0, 1)$, with eigenvalue $1$, as is easily checked, e.g.
$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = -1\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag 2$
with a similar calculation for eigenvcector $(0, 1)$. Thus the point $(0, 0)$ is a saddle, as corroborated by the phase portrait which is easily drawn.
| {
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
| After square completion and substituting $u=\frac{x-2}{3}$, there is a simple standard trick to evaluate the integral without trigonometric substitutions:
$$\int \dfrac{dx}{(x^2-4x+13)^2} \stackrel{u=\frac{x-2}{3}}{=}\frac 1{27} \underbrace{\int \frac{1}{(u^2+1)^2}du}_{I(u)}$$
Just rewrite the numerator
$$I(u) = \int\frac{1+u^2-u^2}{(u^2+1)^2}du = \arctan u - \frac 12\underbrace{\int u \frac{2u}{(u^2+1)^2}}_{J(u)}$$
So, only one quick partial integration gives
$$J(u) = -\frac u{u^2+1}+\arctan u$$
Hence,
$$I(u) = \arctan u - \frac 12\left(-\frac u{u^2+1}+\arctan u\right) =\frac 12 \left(\arctan u + \frac u{u^2+1}\right)$$
Finally, substitute back $u=\frac{x-2}{3}$ and you are done:
$$\int \dfrac{dx}{(x^2-4x+13)^2} = \frac 1{27}I(u)= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{\frac{x-2}{3}}{\left(\frac{x-2}{3}\right)^2+1}\right) (+C)$$
$$= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{3(x-2)}{\left(x-2\right)^2+9}\right)(+C)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+acd+bcd)x + 1
\end{align}$$
and $f(x)$ has no negative roots... but how else do I proceed?
There is a trivial solution $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 27$. We just need to prove it's unique.
| Assume $d = \max{a,b,c,d}$. Looking at the inequality:
$$(a+b+c)^2\geq 3(ab+bc+ca)$$
beginning edit by Will:
from Michael,
$$ 82 = 3 (bc+ca+ab) + 3d(a+b+c), $$
from displayed inequality
$$ 82 \leq (a+b+c)^2 + 3d(a+b+c) $$
$$ 82 \leq (28-d)^2 + 3 d (28-d) $$
$$ 82 \leq 784 - 56d + d^2 + 84d - 3 d^2 $$
$$ 0 \leq 702 + 28 d - 2 d^2 $$
$$ 0 \geq 2 d^2 - 28 d - 702 $$
$$ 0 \geq d^2 + 14 d - 351 $$
$$ 0 \geq (d+13)(d-27). $$
As $d >0$ we get
$$ 0 \geq d-27 $$
$$ 27 \geq d $$
end of edit by Will
will give you $d\leq 27.$ Consequently, $abc\geq \dfrac{1}{27}.$
SECOND EDIT by WILL
$$ f = ( ab + bc + ca)^2 - 3abc(a+b+c) $$
$$ 4(b^2 - bc + c^2) f = \left( 2 (b^2 - bc + c^2) a - bc(b+c) \right)^2 + 3b^2 c^2 (b-c)^2 $$
Conclusion: permute the letters, $ f \geq 0$ and $f \neq 0$ unless $a=b=c.$ Real $a,b,c$ otherwise unrestricted
END SECOND EDIT by WILL
From $a+b+c \geq 1$and $abc\geq \dfrac{1}{27},$ we find that $ab+bc+ca\geq \dfrac{1}{3}.$ Then,
$$\dfrac{1}{3}\leq ab+bc+ca = \dfrac{82}{3} - d(28-d)\iff d^2-28d+27 \geq 0$$
This means $(d-27)(d-1)\geq 0$ so $d = 27.$ The rest should follow immediately.
| {
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"timestamp": "2023-03-29T00:00:00",
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Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by setting $x=\pi/4$ in the Fourier series for $f(x) = \pi/2 - x/2$ and then subtracting the extraneous terms (which are a multiple of the Gregory-Leibniz series for $\pi/4$).
I read that this series appears in a letter from Newton to Leibniz. However, I do not have access the letter which appears in this volume.
| Although the question appears to be about how Newton historically did it, I'll convert a popular comment to an answer showing how techniques from his era, similar to those that handle the Gregory series, evaluate the series above:
$$\begin{align}\sum_{n\ge0}\left(\frac{1}{8n+1}+\frac{1}{8n+3}-\frac{1}{8n+5}-\frac{1}{8n+7}\right)&=\sum_{n\ge0}\int_{0}^{1}x^{8n}\left(1+x^{2}\right)\left(1-x^{4}\right)dx\\&=\int_{0}^{1}\frac{1+x^{2}}{1+x^{4}}dx\\&=\int_{0}^{1}\frac{1+x^{2}}{\left(1-x\sqrt{2}+x^{2}\right)\left(1+x\sqrt{2}+x^{2}\right)}dx\\&=\frac{1}{2}\sum_{\pm}\int_{0}^{1}\frac{dx}{1\pm x\sqrt{2}+x^{2}}\\&=\frac{1}{\sqrt{2}}\sum_{\pm}\left[\arctan\left(x\sqrt{2}\pm1\right)\right]_{0}^{1}\\&=\frac{\arctan\left(\sqrt{2}+1\right)+\arctan\left(\sqrt{2}-1\right)}{\sqrt{2}}\\&=\frac{\pi}{2\sqrt{2}}.\end{align}$$
| {
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Evaluating $\iint dx\,dy$ over the region bounded by $y^2=x$ and $x^2+y^2=2x$ in the first quadrant
Identify the region bounded by the curves $y^2=x$ and $x^2+y^2=2x$, that lies in the first quadrant and evaluate $\iint dx\,dy$ over this region.
In my book the solution is like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}} \, dx \, dy\\
&=\int_{x=0}^1 \big[y\big]_{\sqrt x}^{\sqrt{2x-x^2}}\,dx\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&{\begin{aligned}\\
=\int_0^1\sqrt{1-x^2}\,dx-\int_0^1\sqrt{x}&\,dx\text{(applying} \int_0^af(x)\,dx=&\int_0^af(a-x)\,dx \text{ in the first part)}\\
\end{aligned}\\}\\
&=\left[\frac{\sqrt{1-x^2}}{2}+\sin^{-1}x\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac32}\right]_0^1\\
&=\frac{\pi}{2}-\frac12-\frac23(1-0)\\
&=\frac{\pi}{2}-\frac76\\
\end{align}\\
$$
And I did it like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}}dx\,dy\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&=\int_0^1\sqrt{1-(x-1)^2}\,dx-\int_0^1\sqrt{x}\,dx\\
&{\begin{aligned}\\
=&\left[\frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac12\sin^{-1}(x-1)\right]_0^1&-\left[\frac23x^{\frac32}\right]_0^1\\
\end{aligned}\\}\\
&=-\frac{\pi}{4}-\frac23\\
\end{align}\\
$$
Which one is correct?
| Clearly, your area cannot be negative, so your result is immediately incorrect.
The system $$x = y^2 \\ x^2 + y^2 = 2x$$ is readily solved by substitution. We have $$\begin{align}
0 &= x^2 + y^2 - 2x \\
&= x^2 + x - 2x \\
&= x^2 - x \\
&= x(x-1).
\end{align}$$ Hence $x \in \{0, 1\}$ and the full solution set is $$(x,y) \in \{(0,0), (1, -1), (1, 1)\}.$$ In the first quadrant, the area of interest may be expressed as
$$\begin{align}
\int_{x = 0}^1 \int_{y = \sqrt{x}}^\sqrt{2x-x^2} \, dy \, dx
&= \int_{x=0}^1 \sqrt{2x - x^2} - \sqrt{x} \, dx \\
&= \int_{x=0}^1 \sqrt{1 - (1-x)^2} - \int_{x=0}^1 \sqrt{x} \, dx \\
&= \int_{u=0}^1 \sqrt{1-u^2} \, du - \left[\frac{2}{3}x^{3/2}\right]_{x=0}^1 \\
&= \int_{\theta = 0}^{\pi/2} \sqrt{1 - \sin^2 \theta} \cos \theta \, d\theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \cos^2 \theta \, d \theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta - \frac{2}{3} \\
&= \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\theta = 0}^{\pi/2} - \frac{2}{3} \\
&= \left(\frac{\pi}{4} + 0 - 0 + 0\right) - \frac{2}{3} \\
&= \frac{\pi}{4} - \frac{2}{3}.
\end{align}$$ This step-by-step calculation should resolve all doubt. This is because for a fixed $x \in [0,1]$, we note $$y = \sqrt{x} \le \sqrt{2x-x^2}.$$ Alternatively, we may change the order of integration, but this requires us to solve the equation for the circle in terms of $x$. We can do this by completing the square: $x^2 - 2x + y^2 = 0$ implies $$1-y^2 = x^2 - 2x + 1 = (x-1)^2,$$ hence $$x = 1 \pm \sqrt{1-y^2},$$ and we choose the negative root because we require $x < 1$. Therefore, the area can be expressed as $$\int_{y=0}^1 \int_{x=1 - \sqrt{1-y^2}}^{y^2} \, dx \, dy.$$ Both integrals evaluate to $$\frac{\pi}{4} - \frac{2}{3}.$$
As has already been noted, the figure is misleading because the point $(1,1)$ lies directly above the center of the circle at $(1,0)$.
We can also check our solution by noting that the desired area is equal to the area under a parabola $y = x^2$ on $x \in [0,1]$, minus the area of a unit square from which a quarter of a unit circle has been cut out; i.e., this is simply $$\int_{x=0}^1 x^2 \, dx - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3}.$$
| {
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Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).
The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."
I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.
My attempt:
Assume to the contrary that solution of this equation is greater then or equal to 2.
Let $x=\frac 2p$.
Then we have
$ (\frac 2p)^5-2(\frac 2p)^3-3=0 $
$ \frac {2^5}{p^5} - \frac {2^4}{p^3} - 3 = 0$
We now consider two case: when $p=1$ and $p<1$.
when $p=1$:
$ 2^5 - 2^4 - 3 = 32 - 16 - 3 = 13$.
Since $x = 2$ is not a solution this is a contradiction.
When $p<1$:
If $p<1$, then we know $1/p>1.$ This implies $\frac {1}{p^{n+1}} > \frac {1}{p^n}.$
Since $\frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality
$ \frac 1{p^5}(2^5)- \frac 1{p^3} 2^4-3 > \frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds
Since for all $x > 2$ is not a solution this is a contradiction.
Thus, solution $x$ is less then 2.
| Let $x$ be a root and $x>2$.
Thus, $$x^5-2x^3-3=x^5-2x^4+2x^4-4x^3+2x^3-3>0,$$ which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Derive $\mathbf b$ from $\mathbf a = \mathbf b× \mathbf c$ I have an equation:
$$\mathbf a = \mathbf b × \mathbf c,$$
where $\mathbf a$ $\mathbf b$ and $\mathbf c$ are 3-vectors.
How could I derive $b$ from the equation and express it in terms of $\mathbf a$ and $\mathbf c$?
| Note that
*
*if $\mathbf{c}\cdot\mathbf{a}\neq 0$ then there are no solutions.
*if $\mathbf{c}=\mathbf{0}$, then there are no solutions unless $\mathbf{a}=\mathbf0$, in which case every $\mathbf{b}$ works.
*if $\mathbf{c}\neq\mathbf{0}$, then $\mathbf{p}\times\mathbf{c}=\mathbf{0}$ if and only if $\mathbf{p}$ is a scalar multiple of $\mathbf{c}$.
So if $\mathbf{a},\mathbf{c}\neq\mathbf{0}$ and $\mathbf{c}\cdot\mathbf{a}=0$, the solutions are $\mathbf{b}=\mathbf{b_0}+\lambda\mathbf{c}$, some $\mathbf{b}_0$ such that $\mathbf{b}_0\times\mathbf{c}=\mathbf{a}$.
How could we find such a $\mathbf{b}_0$? One way is to select it so that $b^2$ is minimized. That is, $\mathbf{b}_0$ is perpendicular to $\mathbf{c}$. Since $\mathbf{b}_0$ is perpendicular to $\mathbf{a}$ (from $\mathbf{a}=\mathbf{b}\times\mathbf{c}$), we get $\mathbf{b}_0=\mu\mathbf{c}\times\mathbf{a}$. To solve for $\mu$, we take scalar triple product
$$
\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\mathbf{a}\cdot\mathbf{a}=a^2
$$
so
$$
\mu=\frac{a^2}{\lvert\mathbf{c}\times\mathbf{a}\rvert^2}
$$
and so the solutions are $\mathbf{b}=\frac{a^2}{\lvert\mathbf{c}\times\mathbf{a}\rvert^2}(\mathbf{c}\times\mathbf{a})+\lambda\mathbf{c}$ for some scalar $\lambda$.
| {
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Proofs by Induction: are my two proofs correct? I have been trying to understand how proof by mathematical induction works, and I am struggling a bit. But, I think I am understanding it and I just want to verify that what I am doing is correct (and if not, why?)
I have attached a screenshot (as a link) of my problem (black ink) and my work (red ink). My main issue is understanding what the final conclusion should be. What I did was check to see if the left and right side of the problem were equal after assuming $k + 1$ is true, and adding the appropriate terms to both sides, and simplifying.
So, in my final steps of the induction phase, my question is, did I reach the right result?
Prove: $1 + 3 + 6 + \cdots + \dfrac{n(n + 1)}{2} = \dfrac{n(n + 1)(n + 2)}{6}$.
Base: $P(1) = 1$.
Induction:
\begin{align*}
\underbrace{1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2}}_{\dfrac{k(k + 1)(k + 2)}{6}} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{(k + 1)(k + 2)(k + 3)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}
\end{align*}
Prove: $5 + 10 + 15 + \cdots + 5n = \dfrac{5n(n + 1)}{2}$
Base: $P(1) = 5$
Induction:
\begin{align*}
5 + 10 + 15 + \cdots + 5k + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)\\
\frac{5k(k + 1)}{2} + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)
\end{align*}
My problem and my work
| In a proof by mathematical induction, we wish to establish that some property $P(n)$ holds for each positive integer $n$ (or for each integer greater than some fixed integer $n_0$). We must first establish that the base case holds. Once we establish that it holds, we may assume the property holds for some positive integer $k$. We then need to prove that if $P(k)$ holds, then $P(k + 1)$ holds. Then, if our base case is $P(1)$, we obtain the chain of implications
$$P(1) \implies P(2) \implies P(3) \implies \cdots$$
and $P(1)$, which establishes that the property holds for every positive integer.
You should not assume $P(k + 1)$ is true. We must prove that $P(1)$ holds and that if $P(k)$ holds, then $P(k + 1)$ holds for each positive integer $k$.
Let's look at the first proposition.
Proof. Let $P(n)$ be the statement that
$$1 + 3 + 6 + \cdots + \frac{n(n + 1)}{2} = \frac{n(n + 1)(n + 2)}{6}$$
Let $n = 1$. Then
$$\frac{n(n + 1)}{2} = \frac{1(1 + 1)}{2} =\frac{1 \cdot 2}{2} = 1 = \frac{1 \cdot 2 \cdot 3}{6} = \frac{1(1 + 1)(1 + 2)}{6}$$
Hence, $P(1)$ holds.
Since $P(1)$ holds, we may assume $P(k)$ holds for some positive integer $k$. Hence,
$$1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2} = \frac{k(k + 1)(k + 2)}{6}$$
This is our induction hypothesis.
Let $n = k + 1$. Then
\begin{align*}
1 + 3 + 6 + & \cdots + \frac{k(k + 1)}{2} + \frac{(k + 1)(k + 2)}{2}\\
& = \frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} && \text{by the induction hypothesis}\\
& = \frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6}\\
& = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
& = \frac{(k + 1)[(k + 1) + 1][(k + 1) + 2]}{6}
\end{align*}
Thus, $P(k) \implies P(k + 1)$ for each positive integer $k$.
Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each positive integer $k$, $P(n)$ holds for each positive integer $n$.$\blacksquare$
I will leave the second proof to you.
| {
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The quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
QUESTION: Suppose that $m$ and $n$ are integers, such that both the quadratic equations $$x^2+mx-n=0$$ and $$x^2-mx+n=0$$ have integer roots. Prove that $n$ is divisible by $6$.
MY APPROACH:
$\because$ the roots $\in\Bbb{Z}$ therefore, discriminant of the quadratic equations must be a perfect square..
$$\therefore m^2+4n=p^2$$ and $$m^2-4n=q^2$$ for some, $p,q≥0$ and $p,q\in\Bbb{Z}$. Now subtracting these equations we get, $$8n=p^2-q^2$$ $$\implies p^2-q^2\equiv0\pmod{8}$$ Therefore, $p$ and $q$ cannot be of the form $(2×n)$ where $n$ is odd. But this does not seem to help much. So I going back one step, we can write, $$n=\frac{p^2-q^2}{8}$$ But here I am stuck.. I do not know how may I use $8$ with the property of squares to prove that $n$ must be divisible by $6$..
Any help will be much appreciated... Thank you so much :)
| Now as you got $m^2+4n=p^2$ and $m^2-4n=q^2$, we solve further by taking cases.
Case 1: $m$ is even
Therefore let $m=2k$ for some positive integer $k$ and by judging the equation we can see that $p$ and $q$ are even too. Let $p=2a$ and $q=2b$ for some positive integers $a$ and $b$. Substituting the values of $m,p$ and $q$, we get $$k^2+n=a^2 \\ k^2-n=b^2$$
This implies $$a^2+b^2=2k^2 \\ a^2-b^2=2n$$
Now let's assume that $n$ is odd, but that would mean $a^2-b^2$ is not divisible by $4$, so either $a^2 \equiv 1\pmod{4}$ and $b^2\equiv 0 \pmod{4}$ or vice versa (Remember that a square is always $\equiv 0~\text{or}~1\pmod{4}$).
Therefore $a^2+b^2\equiv 1 \pmod{4}$ in both cases. But we have on the other side $a^2+b^2=2k^2$ which is always either $\equiv 0 \pmod{4}$ or $\equiv 2 \pmod{4}$.
Thus, we get a contradiction. This implies $n$ is even.
Now we assume $n$ is not divisible by $3$ i.e either $2n\equiv 1 \pmod{3}$ or $2n\equiv 2\pmod{3}$. Now a square is always either $\equiv 0 ~\text{or}~ 1\pmod{3}$.
Therefore $a^2-b^2=2n\equiv 2\pmod{3}$ is never possible and thus the remaining possibility is $a^2-b^2=2n\equiv 1 \pmod{3}$. This implies $a^2\equiv 1 \pmod{3}$ and $b^2\equiv 0\pmod{3}$. Therefore, $a^2+b^2 \equiv 1\pmod{3}$, but we had from the other equation $a^2+b^2=2k^2$ which is always $\equiv 0~\text{or}~2\pmod{3}$. Thus, we get a contradiction. Hence, $n$ is divisible by $3$ as well. Thus, $n$ is divisible by $6$.
Similar analysis goes for the other case where $m$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the asymptote and the upper bound for $1\big/ \ln\left(\frac{x+1}{x-1}\right)$ as $x\to\infty$. I need to show that $\left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ asymptotically approaches to a line from below that I should determine, and hence, show that $\frac{x}{2} > \left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ for $x>1$. I would be very happy if someone helps. Thank you.
| Fix $x > 1$ and write $\varphi(t) = \frac{1}{x+t}$. Since $\varphi_x$ is strictly convex, it follows from the Jensen's inequality that
\begin{align*}
\frac{1}{2} \log\left(\frac{x+1}{x-1}\right)
= \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t
> \varphi_x \left( \frac{1}{2} \int_{-1}^{1} t \, \mathrm{d}t \right)
= \frac{1}{x}
\end{align*}
On the other hand, again by the strict convexity of $\varphi_x$,
\begin{align*}
\frac{1}{2} \log\left(\frac{x+1}{x-1}\right)
= \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t
< \frac{\varphi_x(-1) + \varphi_x(1)}{2}
= \frac{x}{x^2-1}.
\end{align*}
Altogether, it follows that
$$ \frac{x}{2} - \frac{1}{2x} < \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} < \frac{x}{2} $$
for all $x > 1$.
Addendum. It is easy to check that the above function admits the Laurent expansion of the form
$$ \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} = \frac{x}{2} - \sum_{n=0}^{\infty} \frac{a_{2n+1}}{x^{2n+1}} $$
as $x\to\infty$. Now here comes a hard part: $\texttt{Mathematica}$ seems to suggests $a_{2n+1} > 0$ for all $n \geq 0$. For instance,
$$
a_1 = \frac{1}{6}, \quad
a_3 = \frac{2}{45}, \quad
a_5 = \frac{22}{945}, \quad
a_7 = \frac{214}{14175}, \quad
a_9 = \frac{5098}{467775}, \quad \cdots. $$
It will be interesting to be able to prove it, although I have no good idea to begin with.
| {
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"timestamp": "2023-03-29T00:00:00",
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For What positive values of x ; the below mentioned series is convergent and divergent? $$ \sum \frac{1}{x^{n}+x^{-n}}$$
My attempt
$$
\begin{aligned}
&\begin{aligned}
\therefore & u_{n+1}=\frac{x^{n+1}}{x^{2 n+2}+1} \\
\therefore & \frac{u_{n+1}}{u_{n}}=\frac{x^{n+1}}{x^{2 n+2}+1} \cdot \frac{x^{2 n}+1}{x^{n}} \\
& \frac{u_{n+1}}{u_{n}}=x \frac{1+\left(\frac{1}{x}\right)^{2 n}}{x^{2}+\left(\frac{1}{x}\right)^{2 n}}
\end{aligned}\\
& i f x>1 \quad \therefore \quad \frac{1}{x}<1 \quad \therefore \text { when } n \rightarrow \infty,\left(\frac{1}{x}\right)^{2 n} \rightarrow 0
\end{aligned}\\
$$
What to do next?
| In your attempt you have shown that $\lim \sup \frac {u_{n+1}} {u_n} \leq x \frac {1+0} {x^{2}+0}=\frac 1 x <1$. Hence Ratio Test tells you that the series is convergent for $x>1$. But there is a bettter approach as shown below:
For $x>1$ it is dominated by $\sum \frac 1 {x^{n}}$ which is convergent.
For $0<x<1$ it is dominated by $\sum x^{n}$ which is convergent.
For $x=1$ it is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of $\sin^8(x)$ Suppose we have the following function:
$$\sin^8(x)$$
We have to find its anti-derivative
To find the indefinite integral of $\sin^4(x)$, I converted everything to $\cos(2x)$ and $\cos(4x)$ and then integrated. However this method wont be suitable to find the indefinite integral $\sin^8(x)$ since we have to expand a lot. Is there any other way I can evaluate it easily, and more efficiently?
| The development through the double angle formulae is not so long, let me show.
\begin{align}
\sin^8x
&=(\sin^2x)^4=\\
&=\left(\frac{1-\cos2x}{2}\right)^4=\\
&=\frac{1}{16}(1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x)=\\
&=\frac{1}{16}[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+(\cos^22x)^2]=\\
&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+\left(\frac{1+\cos4x}{2}\right)^2\right]=\\
&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+{}\right.\\
&\qquad\qquad\qquad\left.+\frac{1}{4}(1+2\cos4x+\cos^24x)\right]=\\
&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+\right.\\
&\qquad\qquad\qquad\left.+\frac{1}{4}\left(1+2\cos4x+\frac{1+\cos8x}{2}\right)\right]=\\
\end{align}
so we have
\begin{align}
\int\sin^8xdx
&=\frac{1}{16}\left[x-2\sin2x+3\left(x+\frac{1}{4}\sin4x\right)-2\left(\sin2x-\frac{1}{3}\sin^32x\right)+{}\right.\\
&\qquad\left.\frac{1}{4}\left(x+\frac{1}{2}\sin4x+\frac{1}{2}\left(x+\frac{1}{8}\sin8x\right)\right)\right]+C
\end{align}
| {
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Show that the kernel ker(B) is a vector subspace of the domain. I want to know how to show that $\ker(B)$ is a vector subspace of the domain.
| I'll sketch how to obtain the reduced row echelon form:
\begin{align}
&\phantom{{}\rightsquigarrow{}} \begin{pmatrix}
3 & -3 & 1 & 5 & 1 & 5 & 5\\
1 & -1 & 1 & 1 & 1 & 3 & -1\\
2 & -2 & 1 & 3 & 0 & 5 & 2\\
2 & -2 & 0 & 4 & 0 & 2 & 1
\end{pmatrix}\rightsquigarrow
\begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
3 & -3 & 1 & 5 & 1 & 5 & 5\\
2 & -2 & 1 & 3 & 0 & 5 & 2\\
2 & -2 & 0 & 4 & 0 & 2 & 1
\end{pmatrix} \\
&\rightsquigarrow
\begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
0 & 1 & -2 & 2 & -2 & -4 & 8\\
0 & 0 &- 1 & 1 & -2 & -1 & 4\\
0 & 0 & -2 & 2 & -2 & -4 & 3
\end{pmatrix} \rightsquigarrow
\begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
0 & 1 & 0 & 0 & 0 & -2 & 0\\
0 & 0 &- 1 & 1 & -2 & -1 & 4\\
0 & 0 & 0 & 0 & 2 & -2 & -5
\end{pmatrix}
\end{align}
You should ultimately obtain, if I'm not mistaken:
$$ \begin{pmatrix}
1 & -1 & 0 & 2 & 0 & 1 & 0\\
0 & 0 & 1 & -1 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 1 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}$$
which shows that
\begin{cases}
x_1=x_2-2x_4-x_6, \\ x_3=x_4-3x_6,\\x_5=x_6,&x_7=0.
\end{cases}
This defines an isomorphism from $K^3$ ($K$ denotes the base field) onto $\ker B$:
\begin{align}
K^3&\xrightarrow{\quad f\quad} \ker B \\
(x,y,z)\:&|\mkern-7mu\xrightarrow{\quad \enspace\quad}(x-2y-z, x, y-3z, y, z,z,0)
\end{align}
and this isomorphism maps any basis of $K^3$, for instance the canonical basis, onto a basis of $\ker B$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$, find $y^2+2y$
If $$y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$$ what is $y^2+2y$?
Attempt:
We know that for negative and fractional indices,
$$(1+x)^n = 1 + nx + n(n-1)/2!\cdot x^2 + n(n-1)(n-2)/3!\cdot x^3 + \cdots$$
Rewriting the series in question, we get:
$$\frac{2}{5} \left(1 + \frac{1\cdot3}{2!}\cdot \frac{2}{5}+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^2+\cdots\right)$$
I know this looks like the binomial expansion above, but I have no idea how to proceed further.
| Substitute $x=\frac{2}{5}$:
$$y=\sum_{i=1}^\infty\frac{x^i}{i!}(\prod_{k=1}^{i}(2k-1))$$
$$y=\sum_{i=1}^\infty\frac{x^i}{i!}((2i-1)!!)$$
For $|x|<\frac{1}{2}$, this is the Taylor series for $\frac{1}{\sqrt{1-2x}}-1$
$$y=\frac{1}{\sqrt{1-2x}}-1$$
$$y=\frac{1}{\sqrt{1-0.8}}-1$$
$$y=\frac{1}{\sqrt{0.2}}-1$$
$$y=\sqrt{5}-1$$
$$y^2+2y=(\sqrt{5}-1)^2+2(\sqrt{5}-1)$$
$$y^2+2y=5+1-2\sqrt{5}+2\sqrt{5}-2$$
$$y^2+2y=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$ Prove that $$\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$$
This is not a duplicate of this post, the bounds are different and the integral evaluates to a slightly different value. I tried looking at the solution from the linked post but I'm not familiar with harmonic numbers or complex analysis and the real solution is long. I tried IBP but got no where. Any advice for this monster integral (real analysis only please)?
| Changing the bounds makes the integral way simpler, because after letting $x\to \frac{1}{x}$ we can get rid of that $\arctan x$.
$$I=\int_0^{\infty} \frac{\arctan x}{x} \ln\left(\frac{1+x^2}{{(1-x)}^2}\right)dx\overset{x\to \frac{1}{x}}=\int_0^\infty \frac{\arctan \left(\frac{1}{x}\right)}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx$$
$$\Rightarrow 2I=\frac{\pi}{2} \int_0^\infty \frac{1}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{x = \tan \frac{t}{2}}=-\frac{\pi}{2}\int_0^\pi\frac{\ln(1-\sin t)}{\sin t}dt$$
Also from here we know that:
$$I(a)=\int_{0}^{\pi} \frac{\ln(1+\sin a\sin x)}{\sin x}dx=a(\pi -a)$$
$$\Rightarrow I=-\frac12 \frac{\pi}{2}I\left(\frac{3\pi}{2}\right)=-\frac12 \frac{\pi}{2}\left(-\frac{3\pi^2}{4}\right)=\frac{3\pi^3}{16}$$
Another way to deal with the last integral (credits to this answer), is to consider:
$$\mathcal J(a)=\int_0^\frac{\pi}{2}\arctan\left(\frac{\sin x -\tan\frac{a}{2}}{\cos x}\right)dx$$
And differentiate w.r.t. a, obtaining:
$$\mathcal J'(a)=-\frac12\int_0^\frac{\pi}{2}\frac{\cos x}{1-\sin a\sin x}dx=\frac12 \frac{\ln(1-\sin a)}{\sin a}$$
$$\mathcal J(\pi)-\mathcal J(0)=-\frac{\pi^2}{4}-\frac{\pi^2}{8}=\frac12\int_0^\pi\frac{\ln(1-\sin a)}{\sin a}da$$
$$\Rightarrow \int_0^\pi \frac{\ln(1-\sin a)}{\sin a}da=-\frac{3\pi^2}{4}$$
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Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$ $$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$
According to WA. I attempted using
$$\prod_{k=0}^{\infty}\Bigl(1-\frac{x^2}{\pi^2k^2}\Bigr)=\frac{\sin x}{x}$$
But I couldn’t reindex the product appropriately to use it the way I wanted to (factoring out a 4 and then continuing from there). I’d like to have at least a direction to go in or an idea on how to do the product.
| It isn't pretty, but we can prove this by working backwards using Euler's product definition of the gamma function (seen here):
$$
\Gamma(x) = \lim_{n\to\infty} n!(n+1)^x \prod_{k=0}^n (x+k)^{-1}.
$$
When we substitute that in for the right hand side, we get
$$
\begin{align}
\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)} &= \frac{a^{2}-4}{a^{2}}\frac{\left(n+1\right)^{\frac{a+4}{2}}n!^{2}\prod_{k=0}^{n}\left(\frac{a+4}{4}+k\right)^{-2}}{\left(n+1\right)^{\frac{a+4}{2}}n!^{2}\left(\prod_{k=0}^{n}\left(\frac{a+2}{4}+k\right)^{-1}\right)\left(\prod_{k=0}^{n}\left(\frac{a+6}{4}+k\right)^{-1}\right)}\\
&= \frac{a^{2}-4}{a^{2}}\frac{\prod_{k=0}^{n}\left(\frac{a+4}{4}+k\right)^{-2}}{\prod_{k=0}^{n}\left(\frac{a+2}{4}+k\right)^{-1}\left(\frac{a+6}{4}+k\right)^{-1}}\\
&= \frac{a^{2}-4}{a^{2}}\prod_{k=0}^{\infty}\frac{\left(\frac{a+2}{4}+k\right)\left(\frac{a+6}{4}+k\right)}{\left(\frac{a+4}{4}+k\right)^{2}}\\
&= \frac{a^{2}-4}{a^{2}}\prod_{k=1}^{\infty}\frac{\left(\frac{a}{4}+k+\frac{1}{2}\right)\left(\frac{a}{4}+k-\frac{1}{2}\right)}{\left(\frac{a}{4}+k\right)^{2}}\\
&=\frac{a^{2}-4}{a^{2}}\prod_{k=1}^{\infty}\frac{\left(\frac{a}{4}+k\right)^{2}-\frac{1}{4}}{\left(\frac{a}{4}+k\right)^{2}}\\
&= \left(1-\frac{4}{a^{2}}\right)\prod_{k=1}^{\infty}\left(1-\frac{4}{\left(a+4k\right)^{2}}\right)\\
&= \prod_{k=0}^{\infty}\left(1-\frac{4}{\left(4k+a\right)^{2}}\right).
\end{align}
$$
(I omitted the limit of $n$ in the equations because it already takes up so much space.)
| {
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"url": "https://math.stackexchange.com/questions/3776147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve Bernouilli differential equation I think that my question is dumb... but I am stuck in a step.
I need to solve this ode: $$\rho' = \rho(1+\rho^2)$$
I made the variable change $u:=\rho^{-2}$. Then, separating variables and replace $u$ by $\rho^{-2}$, i get $\rho=\frac{1}{\sqrt{e^{-2t}-1}}$.
But, when i check the solution, the result is: $\rho=\frac{1}{\sqrt{e^{-2t}-1+\color{red}{\frac{1}{\rho_{0}^2}e^{-2t}}}}$.
I think i forget in some integration step the term "$+C$"... I don't know
Thanks in advance
| EDIT
Let $y = \rho$ for writing convenience. To begin with, notice that
\begin{align*}
y' = y(1+y^{2}) & \Longleftrightarrow \frac{y'}{y(1+y^{2})} = 1
\end{align*}
Then we have that
\begin{align*}
\frac{1}{y(1+y^{2})} = \frac{(1 + y^{2}) - y^{2}}{y(1+y^{2})} = \frac{1}{y} - \frac{y}{1+y^{2}}
\end{align*}
Finally, we obtain the solution by integrating both sides:
\begin{align*}
\int\frac{\mathrm{d}y}{y(1+y^{2})} = \int1\mathrm{d}x & \Longleftrightarrow \ln|y| - \frac{\ln(1+y^{2})}{2} = x + c\\\\
& \Longleftrightarrow \ln(y^{2}) - \ln(1+y^{2}) = 2x + k\\\\
& \Longleftrightarrow \ln\left(\frac{y^{2}}{1+y^{2}}\right) = 2x + k\\\\
& \Longleftrightarrow \frac{y^{2}}{1+y^{2}} = \exp(2x+k)\\\\
& \Longleftrightarrow y^{2}(1 - \exp(2x+k)) = \exp(2x+k)\\\\
& \Longleftrightarrow y^{2} = \frac{\exp(2x+k)}{1-\exp(2x+k)}\\\\
& \Longleftrightarrow y = \pm\left(\frac{\exp(2x+k)}{1-\exp(2x+k)}\right)^{1/2}
\end{align*}
In order to determine $k$, apply the initial condition $y(0) = y_{0}$.
Hopefully this helps.
| {
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"url": "https://math.stackexchange.com/questions/3776220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ for a natural number $ n \geq 2$? Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ holds for a natural number $ n \geq 2$?
My attempt at solving:
Using $(a+b)^2=a^2+2ab+b^2$; if $(a+b)^2=a^2+b^2$, $2ab=0$ therefore $a$ and/or $b$ must be $0$.
If $a$ and/or $b$ is $0$ then $a^2$ and/or $b^2$ will be $0$.
Therefore $(a+b)^2$ can never equal $a^2+b^2$.
| Suppose $a\neq0$. Then
$$(a+b)^n=a^n\Big(1+\tfrac{b}{a}\Big)^n$$
Then, to is enough to consider the question for which values $x$ is $(1+x)^n=1+x^n$. For then $(a+b)^n=a^n+b^n$ with $b=ax$.
This leads to finding all the roots of
$$
p_n(x):=\sum^{n-1}_{k=1}\binom{n}{k}x^k=0
$$
When $n=2$, $p_2(x)=2x=0$ and so $x=0$
when $n=3$, $p_3(x)= 3x+3x^2=0$ and so $x=0$ and $x=-1$,
when $n=4$, $p_4(x)=4x+6x^2 + 4 x^3=2x(2+3x+2x^2)=0$ and so, $x=0$, $x=\frac{-3+i\sqrt{7}}{4}$ and $x=\frac{-3-i\sqrt{7}}{4}$.
and so on.
Notice that pairs $(a, b)$ that satisfy $(a+b)^n=a^n+b^n$ may be complex pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$ $$\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$$
My approach:
Since it is easy to evaluate $\int{\sec^2x}$ , integration by parts seems like a viable option.
Let $$I_n=\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}}$$
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\int{\frac{\sec x \tan x}{(\sec x+\tan x)^\frac{9}{2}}dx}$$
Evaluating the new integral again using by parts yields
$$\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}+\frac{9}{2}\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}\,dx}$$
$$=\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2} I_n$$
Plugging it back, we obtain
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{81}{4}I_n $$
$$\frac{-77}{4}I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$$
This obviously doesn't match with bprp's answer. Help!
Edit:
How do I convert my answer to the answer obtained by him, if mine is correct
| You missed a simplification after second step:
$$I=\frac{\tan x}{(\sec x+\tan x)}\frac{9}{2}+\frac{9}{2}\int \frac{\sec x \tan x}{(\sec x+\tan x)^{\frac92}}dx$$
Now, take the original expression for $I$
$$ I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{\frac92}}$$
Add this $ \frac{9}{2} $ times this to previous expression
$$ \frac{11}{2} I = \frac{\tan x}{(\sec x+\tan x)}\frac{9}{2}+\frac{9}{2}\int \frac{\sec x \tan x + \sec^2 x}{(\sec x+\tan x)^{\frac92}}dx$$
Consider the second integral:
$$ J = \int \frac{\sec x \tan x + \sec^2 x}{(\sec x+\tan x)^{\frac92}}dx$$
Substitute $ \sec x + \tan x = t$
$$ J = \int t^{-\frac92} dt$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.
My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-2|=2\\
\end{align}\\
$$
Now, $w=\frac{2z+3}{z-4}$
$$\begin{align}\\
&\frac w2=\frac{2z+3}{2z-8}\\
&\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\
&\implies\frac{w}{w-2}=\frac{2z+3}{11}\\
&\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\
&\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\
&\implies z=\frac{4w+3}{w-2}\\
&\implies z-2=\frac{2w+7}{w-2}\\
\end{align}\\
$$
$$\therefore\left|\frac{2w+7}{w-2}\right|=2\\
\implies 2w+7=2w-4
$$
Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways?
| Ak19 answered what I would have answered before I got to it, so here's another possible way of how to do it. These transformations map generalized circles to generalized circles; a generalized circle is either a circle or a line. This particular transformation maps the point $z=0$ on the given circle to $w=-\frac34$, the point $z=2+2i$ on the given circle to $-\frac34-\frac{11}4i$, and the point $z=2-2i$ on the given circle to $-\frac34+\frac{11}4i$, so we can see that it maps the circle to the line $u=-\frac34$.
| {
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"url": "https://math.stackexchange.com/questions/3781558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$n$ is prime iff $\binom{n^2}{n} \equiv n \pmod{n^4}$? Can you prove or disprove the following claim:
Let $n$ be a natural number greater than two , then $$n \text{ is prime iff } \binom{n^2}{n} \equiv n \pmod{n^4}$$
You can run this test here. I have verified this claim for all $n$ up to $100000$ .
| Note that $\displaystyle\binom{n^2}{n} = \frac{1}{(n - 1)!} \frac{n^2 (n^2 - 1) ... (n^2 - (n - 1))}{n} = \frac{1}{(n - 1)!} n (n^2 - 1) ... (n^2 - (n - 1))$
Consider a prime $p > 2$. Then $1, 2, ..., p - 1$ are all invertible modulo $p^4$; thus, so is $(p - 1)!$.
Now consider $\displaystyle\binom{p^2}{p} = \frac{1}{(p - 1)!} p (p^2 - 1) ... (p^2 - (p - 1))$.
Define the polynomial $P(x) = x (x^2 - 1) (x^2 - 2) ... (x^2 - (p - 1))$. We wish to reduce $P(x)$ modulo $x^4$. We note that this will only have an $x$ and $x^3$ term since $P$ is odd. The $x$ term will clearly be $(p - 1)! x$; the $x^3$ term will be $-(p - 1)! x^3 \left(\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1}\right)$. Then mod $p^4$, we have $\displaystyle \binom{p^2}{p} = p - p^3 \left(\frac{1}1 + \frac{1}2 + \cdots+ \frac{1}{p - 1}\right)$ (taking division modulo $p^4$ as well).
Note that when reducing $\mod p$, we have $\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1} = 1 + 2 + ... + (p - 1)$, since every number from $1$ to $p - 1$ is a unit. And this sum is equal to $\frac{p (p - 1)}{2} \equiv 0 \pmod p$, since $p > 2$. Thus, we see that $\frac{1}1 + \frac{1}2 + \cdots + \frac{1}{p - 1}$ will be divisible by $p$ when the division is done $\mod p^4$ as well.
Thus, we have that for all $p>2$ prime, $\displaystyle \binom{p^2}{p} \equiv p \pmod {p^4}$.
I don't have anything going the other direction yet.
| {
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"url": "https://math.stackexchange.com/questions/3781690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Inverse Laplace Transform via Circuit Analysis [HELP] Inverse Laplace Transform $\frac{1}{s^2 + \sqrt{2}s + 1}$
so what I did it changed the denominator to complete the square format which is $\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$, then I can solve for $s$, it will make it as
$$
\left(\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i\right)
\left(\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i\right)
$$
So now, to the sheet of paper is to do Partial Fraction Decomposition of this which is absurd to me because of complex roots it has:
$$
\frac{1}{s^2 + s\sqrt{2} + 1}
= \frac{1}{\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}}
$$
Partial Fraction of Complex root will be
$$
\frac{K}{\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i}
+ \frac{K^*}{\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i}
$$
to follow the formula sheet.
which I got my K = -$i\frac{\sqrt{2}}{2}$ and $K^*$ = $i\frac{\sqrt{2}}{2}$
the problem I get is magnitude and $\theta$ is undefined it makes no sense at all.
| Once we complete the square, we can use the sine formula and Frequency Shift Theorem to evaluate the inverse transform:
If we accept that
$$\mathcal{L}(\sin(at)) = \frac{a}{s^2+a^2}$$
and
$$\mathcal{L}(e^{ct}f(t)) = F(s-c)$$
where $F(s) = \mathcal{L}(f(t))$, we can take our original fraction:
$\begin{align}
\mathcal{L}^{-1}(\frac{1}{s^2+\sqrt{2}s+1}) & = \mathcal{L}^{-1}(\frac{1}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\
& = \mathcal{L}^{-1}(\sqrt{2}\frac{\frac{1}{\sqrt{2}}}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\
& = \sqrt{2}*\exp{\frac{-t}{\sqrt{2}}}*\sin(\frac{t}{\sqrt{2}})
\end{align}$
In that last step, we combined the two formulae above, as our fraction was in the form of $\mathcal{L}(\sin(at))$, but shifted by $c = \frac{-1}{\sqrt{2}}$, creating the '$\exp{\frac{-t}{\sqrt{2}}}$' term in the final answer.
Were you to continue the partial fraction decomposition method directly, you would end up with two exponentials terms that you could manipulate into the same answer above using the identity:
$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\triangle ABC$ with a point $D$ inside has $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$, and $\angle DCB=18^\circ$.
Let $ABC$ be a triangle with a point $D$ inside. Suppose that $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$ and $\angle DCB=18^\circ$. Show that $$\frac{BD}{AB}=\sqrt2.$$
I am requesting a geometric proof (with as little trigonometry as possible). A completely geometric proof would be most appreciated. I have a trigonometric proof below.
Trigonometric Proof
Wlog, let $AB=1$. Note that $\angle ABC=\angle ACB=30^\circ$, so $AC=1$. Then by law of sines on $\triangle ACD$, $$AD=\frac{\sin 12^\circ}{\sin 18^\circ}.$$ By law of cosines on $\triangle ABD$,
$$BD^2=1^2+\frac{\sin^212^\circ}{\sin^2{18^\circ}}-2\frac{\sin 12^\circ}{\sin 18^\circ}\cos 114^\circ.$$
As $\cos 114^\circ=-\sin24^\circ$, we get
$$BD^2=2+\frac{-\sin^218^\circ+\sin^212^\circ+2\sin12^\circ\sin18^\circ\sin 24^\circ}{\sin^218^\circ}.$$
Then from the identities $\sin^2\alpha-\sin^2\beta=\sin(\alpha-\beta)\sin(\alpha+\beta)$ and $\sin(2\alpha)=2\sin\alpha\cos\alpha$, we have
$$BD^2=2+\frac{-\sin 6^\circ\sin 30^\circ+4\sin 6^\circ\cos 6^\circ \sin 18^\circ\sin24^\circ}{\sin^218^\circ}.$$
Because $\sin 30^\circ=\frac12$, we conclude that $BD=\sqrt{2}$ if we can prove
$$8\cos 6^\circ \sin 18^\circ \sin 24^\circ=1.$$
This is true because by the identity $2\sin\alpha\cos\beta=\sin({\alpha+\beta})+\sin(\alpha-\beta)$, we have
$$2\sin 24^\circ \cos 6^\circ =\sin 30^\circ+\sin 18^\circ.$$
Since $\sin 30^\circ=\frac12$, we obtain
$$8\cos 6^\circ \sin 18^\circ \sin 24^\circ =2\sin 18^\circ +4\sin^218^\circ=1,$$
noting that $\sin 18^\circ=\frac{\sqrt5-1}{4}$.
Attempt at Geometric Proof
I discovered something that might be useful. Construct the points $E$ and $G$ outside $\triangle ABC$ so that $\triangle EBA$ and $\triangle GAC$ are similar to $\triangle ABC$ (see the figure below). Clearly, $EAG$ is a straight line parallel to $BC$. Let $F$ and $H$ be the points corresponding to $D$ in $\triangle EBA$ and $\triangle GAC$, respectively (that is, $\angle FAB=\angle DCB=\angle HCA$ and $\angle FAE=\angle DCA=\angle HCG$). Then $\triangle FBD$ and $\triangle HDC$ are isosceles triangles similar to $\triangle ABC$, and $\square AFDH$ is a parallelogram. I haven't been able to do anything further than this without trigonometry.
Here is a bit more attempt. If $M$ is the reflection of $A$ wrt $BC$, then through the use of trigonometric version of Ceva's thm, I can prove that $\angle AMD=42^\circ$ and $\angle CMD=18^\circ$. Not sure how to prove this with just geometry. But this result may be useful. (Although we can use law of sines on $\triangle MCD$ to get $MD$ and then use law of cosines on $\triangle BMD$ to get $BD$ in terms of $AB$ too. But this is still a heavily trigonometric solution, even if the algebra is less complicated than the one I wrote above.)
I have a few more observations. They may be useless. Let $D'$ be the point obtained by reflecting $D$ across the perpendicular bisector of $BC$. Draw a regular pentagon $ADKK'D'$. Geogebra tells me that $\angle ABK=54^\circ$ and $\angle AKB=48^\circ$. This can be proven using trigonometry, although a geometric proof should exist. But it is easy to show that $KD\perp CD$ and $K'D'\perp BD'$.
In all of my attempts, I always ended up with one of the following two trigonometric identities:
$$\cos 6^\circ \sin 18^\circ \sin 24^\circ=1/8,$$
$$\cos 36^\circ-\sin18^\circ =1/2.$$
(Of course these identities are equivalent.) I think a geometric proof will need an appearance of a regular pentagon and probably an equilateral triangle, and maybe a square.
| Let $\omega$, $O$ be the circumcircle and circumcenter of $\triangle ABC$, respectively. Let $P,Q,R,S$ be four points on the shorter arc $AC$ of $\omega$ dividing this arc into five equal parts.
First, we shall prove that $\triangle RSD$ is equilateral. Let $D'$ be a point inside $\omega$ such that $\triangle RSD'$ is equilateral. Also, let $E$ be inside $\omega$ such that $\triangle PQE$ is equilateral. Invoking symmetries we see that $\triangle D'SC \equiv \triangle D'RQ \equiv \triangle EQR \equiv \triangle EPA$. Note that $\angle EQR = \angle QRD'=\angle QRS-60^\circ = 168^\circ - 60^\circ = 108^\circ$. Hence $\angle D'QR = 90^\circ - \frac 12\angle QRD' = 36^\circ$ and $\angle EQD'=108^\circ - 36^\circ = 72^\circ$. But also $\angle D'EQ = 180^\circ - \angle EQR = 180^\circ - 108^\circ = 72^\circ$. Hence $ED'Q$ is isosceles with $QD'=ED'$. Again, using symmetries we see that $AED'C$ is an isosceles trapezoid with $AE=ED'=D'C$. We have $\angle ACD'=\angle SCD' - \angle SCA = 36^\circ - 24^\circ = 12^\circ$. Since $AED'C$ is an isosceles trapezoid, it is cyclic and since $AE=ED'=D'C$, it follows that $\angle D'AC = \frac 12 \angle EAC = \frac 12 \angle ACD'=6^\circ$. Hence $D'$ coincides with $D$.
Now comes my favourite part. Some angle chasing shows that $\angle QCE = 18^\circ = \angle DCB$ and $\angle DQC = 24^\circ = \angle BQE$. Hence $D$ and $E$ are isogonal conjugates in $\triangle BQC$. It follows that $\angle CBD = \angle EBQ$.
Choose $T$ on $\omega$ so that $BT$ is a diameter. Clearly, $\triangle BQE$ is symmetric to $\triangle TRD$ with respect to perpendicular bisector of $QR$. In particular, $\angle RTD = \angle EBQ$.
Let $RT$ intersect $BC$ at $X$. Since $\angle CBD = \angle EBQ = \angle RTD$, quadrilateral $BDXT$ is cyclic. Hence $\angle BDT = \angle BXT$. Then some angle chasing shows that $\angle DOB = 102^\circ = \angle BXT = \angle BDT$. This precisely means that the circumcircle of $DOT$ is tangent to $BD$ at $D$. Tangent-secant theorem yields $BD^2=BO\cdot BT = BO \cdot 2BO = 2BO^2$. Hence $$\frac{BD}{AB} = \frac{BD}{BO} = \sqrt 2,$$
as desired.
| {
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"url": "https://math.stackexchange.com/questions/3782069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?
| By the root test, the series converses for all $x$ such that
$$
\limsup_n\sqrt[n]{\frac{1}{n3^n|x-5|^n}}=\frac{1}{3|x-5|}\lim_n\frac{1}{\sqrt[n]{n}}=\frac{1}{3|x-5|}<1
$$
Thus, the series converges for all $x$ such that $|x-5|>\frac{1}{3}$, i.e., all $x$ in $(-\infty,\tfrac{14}{3})\cup(\tfrac{15}{3},\infty)$.
At the point $x=\frac{15}{3}$ the series becomes $\sum_n\frac{(-1)^{n-1}}{n}$ which converges (is an alternating series with decreasing driving term $\frac{1}{n}$). At $x=\frac{14}{3}$ the series becomes $-\sum_n\frac{1}{n}$ which diverges. Therefore, the series converges for all $x\in (-\infty,\tfrac{14}{3})\cup[\tfrac{15}{3},\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $g(x) = \frac{x^3+9}{x^2}$ one-to-one? Let $g(x) =\frac{x^3+9}{x^2}$ restricted to $D(g) = (-\infty,0)$. Is it 1-1?
My approach
$g(x) \text{ 1-1 }: g(x_1)=g(x_2) \iff x_1 = x_2$
Let $x_1, x_2 \in D(g)$ then,
$$\frac{x_1^3+9}{x_1^2} = \frac{x_2^3+9}{x_2^2} \iff x_2^2 \cdot x_1^3 + 9x_2^2 - x_1^2 \cdot x_2^3 + 9x_1^2 =0 \iff\\
\iff x_2^2(x_1^3+9)-x_1^2(x_2^3+9) = 0 \quad (1)$$
In order for $(1)$ to hold,
*
*$x_2 = x_1 = 0$, which can't be because $0 \notin D(g)$ or
*$x_1 = x_2 = \sqrt[3]{-9} \in D(g)$
Therefore $g(x_1)=g(x_2) \iff x_1 = x_2$ holds only for $\sqrt[3]{-9}$ (and not for every $x \in D(g)$)
$$\boxed{ \text{Hence, }g(x)\text{ is NOT 1-1 }}$$
Is this correct?
| I would use a calculus approach:
$$g(x) = \frac{x^3+9}{x^2}=x+\frac{9}{x^2}$$
Differentiating gives us $$1-\frac{9}{x^3}>0$$ which means it is a monotonic function over the negative domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Quasilinear PDE $u_t + (u^2)_x = 0$ cauchy problem The problem I am trying to solve is:
\begin{equation}\label{eq:3.1}
\begin{cases}
\partial_t u + \partial_x(u^2)=0 & x\in \mathbb{R}, t \in (0,\infty]\\
u(x,0)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{cases}
\end{equation}
What I have done is:
We will try to reduce the problem to ODEs on a curve $x(t)$ on the $(t,x)$ plane. The equation can be compared with the canonical form,
\begin{equation}
a\frac{\partial u}{\partial x} +b\frac{\partial }{\partial t} = c,
\end{equation}
where $a = 2u$, $b= 1$ and $c=0$. From the Lagrange-Charpit equations, we have,
\begin{align}\label{eq:3.2}
&\frac{dx}{a}=\frac{dt}{b}=\frac{du}{c} & \text{ substituting we have,}\nonumber\\
\implies &\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}&
\end{align}
using second and the third ratio from the equation we have,
\begin{align}\label{eq:3.3}
&\frac{du}{dt}=0 & \text{integrating we have,} \nonumber\\
\implies&u=B,&
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions,
\begin{equation}\label{eq:3.4}
u(x,0)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{equation}
where the characteristic curve $x(t)$, passes through $(c,0)$. By substitution we have,
\begin{equation}
B=
\begin{cases}
0 & x\leq 0\\
c & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
Therefore solution can be written as
\begin{equation}\label{eq:3.5}
u=
\begin{cases}
0 & x\leq 0\\
c & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
using first and the second ratios from the equation we have,
\begin{align}\label{eq:3.6}
&\frac{dx}{dt}=2u & \text{substituting we have,} \nonumber\\
\implies&\frac{dx}{dt}=
\begin{cases}
0 & x\leq 0\\
2c & 0<x\leq 1\\
2 & x>1.
\end{cases}
&\text{integrating we have,}\nonumber\\
\implies&x=
\begin{cases}
B & x\leq 0\\
2ct+B & 0<x\leq 1\\
2t+B & x>1.
\end{cases}
&\nonumber\\
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions , and that the characteristic curve $x(t)$ passes through $(c,0)$ we have,
\begin{equation}
x=
\begin{cases}
c & x\leq 0\\
2ct+c & 0<x\leq 1\\
2t+c & x>1.
\end{cases}
\end{equation}
Therefore $u$ becomes,
\begin{equation}
u(x,t)=
\begin{cases}
0 & x\leq 0\\
\frac{x}{2t+1} & 0<x\leq 1\\
1 & x>1.
\end{cases}
\end{equation}
I think I am missing something. The solution should have $t$ dependence in the intervals. Thanks.
| This PDE is very similar to Burgers equation, and the solution $u(x,t)$ deduced from the method of characteristics reads $u = f(x-2u t)$ in implicit form, where $f = u(\cdot, t=0)$. Following the steps in the linked post (see also the comments section), we find
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 & & x\leq 0\\
&\tfrac{x}{1+2t} & & 0< x\leq 1+2t\\
&1 & & x> 1+2t
\end{aligned}\right.
$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculation involving determinant of a matrix Suppose I have the following Toeplitz symmetric matrix
\begin{align}
M=\begin{bmatrix}
1 & c & c & x \\
c & 1 & c & c \\
c & c & 1 & c \\
x & c & c & 1
\end{bmatrix}
\end{align}
I want to write an algorithm that takes $c$ as input and calculates the range of $x$ for which matrix $M$ is positive semidefinite.
Currently, I do Gaussian elimination by hand and reduce the problem to checking the determinant of a $2 \times 2$ matrix. But how do I automate the process so I can write a function that takes $c$ and $n$ as inputs, where $n$ is the dimension of $M$, and returns the range of $x$. Thanks!
| For your specific example, done with pen and paper,
$$M_4=\left(
\begin{array}{cccc}
1 & c & c & x \\
c & 1 & c & c \\
c & c & 1 & c \\
x & c & c & 1
\end{array}
\right)$$
$$\Delta_4=\left(4 c^3-5 c^2+1\right)+\left(4 c^2-4 c^3\right) x+\left(c^2-1\right)
x^2$$ With a computer
$$M_5=\left(
\begin{array}{ccccc}
1 & c & c & c & x \\
c & 1 & c & c & c \\
c & c & 1 & c & c \\
c & c & c & 1 & c \\
x & c & c & c & 1
\end{array}
\right)$$
$$\Delta_5=\left(-6 c^4+14 c^3-9 c^2+1\right)+6 \left(c^4-2 c^3+c^2\right) x+\left(-2 c^3+3
c^2-1\right) x^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Solving quintic equations of the form $x^5-x+A=0$ I was on Wolfram Alpha exploring quintic equations that were unsolvable using radicals. Specifically, I was looking at quintics of the form $x^5-x+A=0$ for nonzero integers $A$. I noticed that the roots were always expressible as sums of generalized hypergeometric functions:
$$B_1(_4F_3(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{3125|A|^4}{256}))+B_2(_4F_3(\frac{7}{10},\frac{9}{10},\frac{11}{10},\frac{13}{10};\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3125|A|^4}{256}))+B_3(_4F_3(\frac{9}{20},\frac{13}{20},\frac{17}{20},\frac{21}{20};\frac{3}{4},\frac{5}{4},\frac{3}{2};\frac{3125|A|^4}{256}))+B_4(_4F_3(\frac{-1}{20},\frac{3}{20},\frac{7}{20},\frac{11}{20};\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{3125|A|^4}{256}))$$
where the five roots have $(B_1,B_2,B_3,B_4)\in\{(A,0,0,0),(-\frac{A}{4},-\frac{5A|A|}{32},\frac{5|A|^3}{32},-1),(-\frac{A}{4},\frac{5A|A|}{32},-i\frac{5|A|^3}{32},i),(-\frac{A}{4},\frac{5A|A|}{32},i\frac{5|A|^3}{32},-i),(-\frac{A}{4},-\frac{5A|A|}{32},-\frac{5|A|^3}{32},1)\}$
After observing this, I was left with a lot of questions. First, given $A$, is there a formula I can use to generate the values for $D$, $E$, $H$, and $K$? Second, why do these patterns persist? Third, if I take a different set of quintics that can't be solved using radicals and that differ only in their constant term, does a similar pattern to the roots exist? Fourth, can anyone prove that these patterns that I found will always hold?
Edit: I found the patterns for $D$, $E$, $H$ and $K$. The question has been updated accordingly.
| The answer to your third question is yes! The method uses Bring radicals, whose explicit form in terms of generalized hypergeometric functions can be found using the Lagrange inversion theorem. (In fact since any quintic can be reduced to this form, in principle this method can be used to solve any quintic.) I can answer your second and fourth questions partially by developing this method. But I'm afraid I will only be able to obtain the first solution with coefficients $(A, 0, 0, 0)$.
The idea at its core is quite simple. Basically we rewrite the equation as $x^5 - x = - A$, treat the left hand side as a function $f(x) = x^5 - x$, then try to answer the question "what is $f^{-1}(-A)$." This is then done by expressing $f^{-1}$ as a power series. The Lagrange inversion theorem gives this inverse as
$$
x = \sum_{k=0}^\infty \binom{5 k}{k} \frac{A^{4k+1}}{4k+1}\ .
$$
Unfortunately, this series doesn't converge for all values of $A$. In fact the radius of convergence is $4/(5\times 5^{1/4})\approx 0.535$, so evaluating the series directly would only give us the solution for one integer $A = 0$. This is where the generalized hypergeometric function comes in. We can analytically continue this series to define a function of $A$. The function whose power series (at zero) is
$$
\sum_{n=0}^{\infty}\prod_{k=0}^{n} \frac{(k+a_1)\cdots(k+a_p)}{(k+b_1)\dots(k+b_q)(k+1)} z
$$
is denoted as $_p F_q(a_1,\dots, a_p;b_1,\dots,b_q;z)$. To convert our function $f^{-1}(A)$ into the standard form, we need to compute the ratio between consecutive terms, which is
$$
\begin{align}
& \quad \frac{(5k +5)!A^{4k+5}}{(k+1)!(4k+4)!(4k+5)}\cdot\frac{k!(4k)!(4k+1)}{(5k)!A^{4k+1}}\\
& = \frac{(5k+5)(5k+4)(5k+3)(5k+2)(5k+1)(4k+1)A^4}{(k+1)(4k+4)(4k+3)(4k+2)(4k+1)(4k+5)} \\
& = \frac{5(5k+4)(5k+3)(5k+2)(5k+1)}{4(4k+5)(4k+3)(4k+2)(k+1)}A^4 \\
& = \frac{(k+1/5)(k+2/5)(k+3/5)(k+4/5)}{(k+1/2)(k+3/4)(k+5/4)(k+1)}\left(5\left(\frac{5A}{4}\right)^4\right)\ .
\end{align}
$$
Now since the numerator has four factors and the denominator has three factors besides $(k+1)$, this is $_4F_3$ (times an extra factor of $A$, since the starting term in our series is $A$, not $1$). The parameters are the numbers added to $k$ in each factor, and the argument is $(5^5/4^4)A^4 = (3125/256)A^4$. This gives you the first solution $A \;_3F_4(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{3125}{256}A^4)$.
In order to obtain the other roots, in principle we can find use this root to factor the polynomial and try to solve the resulting quartic. However that involves too much computation and doesn't seem like the neatest way of obtaining the results you got here.
| {
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"url": "https://math.stackexchange.com/questions/3786161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Fourier series of $f(x) = |x|^3$ and evaluating series I found the Fourier serie for the function $$f: [-\pi, \pi], \quad f(x) = |x|^3$$
Coefficents:
$$
a_0 = \frac{\pi^3}{2}
$$
$$
a_n = \frac{6 \pi}{n^2} \cos(n\pi) - \frac{12}{n^4\pi} \cos(n\pi) + \frac{6}{n^4}
$$
$$
b_n = 0
$$
So, the Fourier serie is given by
$$
f(x) = \frac{a_0}{2} + \sum_{n = 1}^{+\infty} a_n \cos(nx)
$$
$$
f(x) = \frac{\pi^3}{4} + \sum_{n = 1}^{+\infty}\left( \frac{6 \pi}{n^2} (-1)^n - \frac{12}{n^4\pi} (-1)^n + \frac{6}{n^4} \right) \cos(nx)
$$
Now, I should evaluate the following serie:
$$
\sum_{n = 1}^{+\infty} \frac{1}{n^4}(\pi^2 n^2 (-1)^n - 2 (-1)^n + 2)
$$
How can I find it with the help of this Fourier series?
| I think the fourier coefficients are
$$n\neq1,\;\;a_n=\frac{6(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{12}{\pi n^4}$$
So when we susntitute $\;x=0\;$ , we get (Dirichlet's convergence theorem)
$$0=\frac{\pi^3}4+6\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{2}{\pi n^4}\right)\implies-\frac{\pi^3}{24}=\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{2}{\pi n^4}\right)\implies$$
$$-\frac{\pi^4}{24}=\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{ n^4}(-1)^n-\frac{2}{ n^4}\right)$$
and there you have your sum...(of course, I assume you know the sum of $\;\sum\limits_{n=1}^\infty\frac1{n^4}\;$ ...)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| Let $A_p:=\left(\frac{1}{N}\sum_{i=1}^Na_i^p\right)^{1/p}$ be the $p$th mean of $(a_i)$.
By the extended AM-GM inequality, $GM\le A_n\le A_{n+1}$. Hence $$GM\times A_n^n\le A_{n+1}\times A_{n+1}^n=A_{n+1}^{n+1}$$ or $$\sqrt[3]{abc}\times\frac{a^n+b^n+c^n}{3}\le\frac{a^{n+1}+b^{n+1}+c^{n+1}}{3}$$
| {
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"source": "stackexchange",
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Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
| Calculus is not required to evaluate the sum. Let $$f(z) = \sum_{k=1}^\infty (k+1)^2 z^k.$$ Then $$z f(z) = \sum_{k=1}^\infty (k+1)^2 z^{k+1} = -z + \sum_{k=1}^\infty k^2 z^k$$ hence $$f(z) - zf(z) = \sum_{k=1}^\infty (k+1)^2 z^k + z - \sum_{k=1}^\infty k^2 z^k = z + \sum_{k=1}^\infty (2k+1) z^k.$$ Now let $$g(z) = \sum_{k=1}^\infty (2k+1) z^k.$$ Then using the same technique, $$g(z) - z g(z) = \sum_{k=1}^\infty (2k+1)z^k - \sum_{k=2}^\infty (2k-1)z^k = z + 2\sum_{k=1}^\infty z^k = z + \frac{2z}{1-z}.$$ Therefore, $$g(z) = \frac{z(3-z)}{(1-z)^2},$$ and $$f(z) = \frac{z(4-3z+z^2)}{(1-z)^3}.$$ All that remains is to select $z = 1/3$ to obtain the value of the desired sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the acceleration vector of elliptical curve. Satisfying Kepler's first law but not second. I am trying to solve problem 16 in section 1.6 of David Bressoud book Second Year Calculus. He gives a hint at the end of the book which says. If $r$ and $\theta$ are related by $\frac{r^2\cos^2(\theta)}{a^2}+\frac{r^2\sin^2(\theta)}{b^2}=1$ and if $r^2\frac{d\theta}{dt}=k$ then $\vec{a}=\frac{-rk}{a^2b^2}\vec{u_r}$. Where $\vec{u_r}$ and $\vec{u_\theta}$ are the local coordinates. He gives a formula for $\vec{a}$ earlier in the text as $\vec{a}=(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})^2)\vec{u_r}+\frac{1}{r}\frac{d}{dt}(r^2\frac{d\theta}{dt})\vec{u_\theta}$. Specifically I need help calculating $\frac{d^2r}{dt}$ without the $\theta$ term. Also $k$ is a constant. My current calculation has lead to $\frac{d^2r}{dt^2}=\frac{k}{r}(\frac{1}{b^2}-\frac{1}{a^2})(\frac{k\cos(2\theta)}{r}+\frac{\sin(2\theta)}{2})$.
| we have $\frac{r^2 \cos^2(\theta)}{a^2}+\frac{r^2 \sin^2 (\theta)}{b^2} = 1$
Differentiating this equation with respect to $t$, gives $$r^2\sin(2\theta)\frac{d\theta}{dt}\left(\frac{1}{b^2}-\frac{1}{a^2}\right)+2\frac{1}{r}\frac{dr}{dt}=0$$
Now replacing $\frac{d\theta}{dt} = \frac{k}{r^2}$, gives
$$\sin(2\theta)\left(\frac{1}{b^2}-\frac{1}{a^2}\right)+2\frac{1}{r}\frac{dr}{dt}=0$$
We need to find the derivative of $\sin 2\theta$ to progress. This equals $$2\frac{k}{r^2}\cos 2\theta$$
Now, $\cos 2\theta = 2\cos^2\theta -1$ and from the equation, using $\cos^2(\theta) + \sin^2 \theta = 1$, we have $$ \cos^2(\theta) = \left(\frac{1}{r^2} - \frac{1}{b^2}\right)\left(\frac{1}{a^2}-\frac{1}{b^2}\right)^{-1}$$ So $$\cos 2\theta = 2 \left(\frac{1}{r^2} - \frac{1}{b^2}\right)\left(\frac{1}{a^2}-\frac{1}{b^2}\right)^{-1} - 1$$
Now from before we have $$\frac{d}{dt}\left(\frac{1}{r}\frac{dr}{dt}\right) = \frac{k}{r^2}\cos 2\theta\left(\frac{1}{a^2}-\frac{1}{b^2}\right)$$
$$\frac{1}{r}\frac{d^2r}{dt^2}-\frac{1}{r^2}\left(\frac{dr}{dt}\right)^2 = \frac{k}{r^2}\cos 2\theta\left(\frac{1}{a^2}-\frac{1}{b^2}\right)$$
We already have an expression for $\left(\frac{dr}{dt}\right)^2$ in terms of $\sin^2 2 \theta = 1-\cos^2 2\theta$. So all that's left to do is to substitute in the expression for $\cos^2 2\theta$.
| {
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Why $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$ is equal to 1? $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
By maths calculator it results 1.
I calculate and results $\sqrt{-\frac{1}{2}}$.
$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
$\sqrt{\frac{-{(3)}^{{\displaystyle\frac12}\times2}}{2^2}+\frac{1^2}{2^2}}=\sqrt{\frac{-3}4+\frac14}=\sqrt{\frac{-3+1}4}=\sqrt{\frac{-2}4}=\sqrt{-\frac12}$
Enlighten me what went wrong?
| Hint:$(-\sqrt{3})^2=(-1)^2(3)^{{\frac{1}{2}}2}=3$
| {
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Are there any identities for the determinant of almost upper triangular matrices of the following form? I've encountered a problem in which I need to compute the determinant of an almost upper triangular matrix of the following form:
$$ A = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\
1 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\
1 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\
1 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\
1 & 0 & 0 & 0 & a_{5,5} & \\
\vdots & & & & & \ddots \\ \\
1 & 0 & 0 & 0 & \dots & 0 & a_{N,N}
\end{pmatrix} $$
All matrix entries below the diagonal are zero, except those in the first column, which are equal to one.
The matrix is infinite, so $N \to \infty$. I wonder whether there are identities that describe the form of the determinant of this matrix. References to relevant articles are appreciated.
| Note that we can write this matrix in the form $A = B + uv^T$, where
$$
B = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\
0 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\
0 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\
0 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\
0 & 0 & 0 & 0 & a_{5,5} & \\
\vdots & & & & & \ddots \\ \\
0 & 0 & 0 & 0 & \dots & 0 & a_{N,N}
\end{pmatrix}, \quad
u = (0,1,\dots,1)^T, \quad v = (1,0,\dots,0)^T.
$$
With the matrix determinant lemma, we find that
$$
\det(A) = \det(B + uv^T) = (1 + v^TB^{-1}u) \det(B)
\\ = (1 + v^TB^{-1}u) \cdot a_{22} a_{33} \cdots a_{NN}.
$$
From there, it suffices to find $v^TB^{-1}u$, i.e. the first entry of $B^{-1}u$.
I don't think that there is a nice explicit form for $v^TB^{-1}u$, but the answer can be computed very efficiently because the matrix is upper triangular.
| {
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Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$
And, $r = 8$
And, $8\cos \theta = 8$ and $\theta = 0$.
So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$
I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?
| $8^{\frac{1}{3}}$=$2(1)^{\frac{1}{3}}=2,2\omega,2{\omega}^2$
here $\omega$ is cube root of unity
| {
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"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$, where did I go wrong in my induction step? Can someone help me out with the induction step?
Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$
Base case n=1: $$4^1 = 2^{1*(1+1)} = 2^2$$
Induction step (to show: $2^{(n+1)*(n+2)} = 2^{n^2+3n+2}$ ) :
$$ \prod_{k=1}^{n+1} 4^k = 4^{n+1} * 2^{n*(n+1)} = \frac{1}{2} * 4^{n+1} * 4^{n*(n+1)} = \frac{1}{2} 4^{n^2+n+n+1}= \frac{1}{2} 4^{n^2+2n+1} = 2^{n^2+2n+1}$$
but now we have
$$ 2^{n^2+3n+2} \neq 2^{n^2+2n+1} $$
Where did I go wrong?
| The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$ and so
$\prod_{k=1}^n4^k=4^{\sum_{k=1}^nk}=4^{\frac{n(n+1)}{2}}=2^{n(n+1)}$
| {
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"url": "https://math.stackexchange.com/questions/3791672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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