Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdot 5y\cdot\frac{25}{y}\cdot z\cdot \frac{1}{36z}\right)^{1/6} \\
\frac{p}{6} & \geqslant \left(\frac{5}{6}\right)^{1/3}\\
p & \geqslant 6\left(\frac{5}{6}\right)^{1/3}
\end{aligned}$$
$$\implies \text{The minimum value of the expression is } 6\left(\frac{5}{6}\right)^{1/3}
$$
Now, consider $f(x)=3x+\dfrac{1}{15x},\ g(y) = 5y+\dfrac{25}{y}$ and $h(z)=z+\dfrac{1}{36z}.$
$$\begin{aligned}f'(x) &= \frac{\mathrm d}{\mathrm dx}\left(3x+\frac{1}{15x}\right) = 3 - \frac{1}{15x^2}\\f''(x)&=\frac{2}{15x^3}\end{aligned}$$
At the critical points, $f'(x) = 0 \implies x = \dfrac{\pm1}{3\sqrt{5}}.$
$f''\left(\dfrac{1}{3\sqrt5}\right) > 0\implies f(x)$ has a local minima at $x = \dfrac{1}{3\sqrt5}.$
Similarly, the local minima of $g(y)$ is at $y = \sqrt5$ and the local minima of $h(z)$ is at $z=1/6.$
Substituting these values into the original expression, the minimum value of the expression comes out to be
$$
\begin{aligned}
p & = f\left(\dfrac{1}{3\sqrt5}\right)+g\left(\sqrt5\right)+h\left(\frac{1}{6}\right) \\
& = \frac{1}{3} + \frac{52}{\sqrt5}.
\end{aligned}
$$
The answer according to the AM-GM inequality is $\approx5.646$ and according to calculus is $\approx23.588$, which are way off. Also, the functions have only two points of inflection, one is the maxima (for values less than $0$) and the other is the minima (for values greater than $0$). As the question clearly states "for $x,y,z \in \mathbb{R}^+$", the "correct" answer should be $\approx23.588$, shouldn't it?
Why are the answers different? Also, is it possible to figure out the individual $x, y$ and $z$ values for which the expression has the minimum value (in case of AM-GM inequality)?
Note: This question is from a Test and the correct answer according to the "test creators" is $$6\left(\dfrac{5}{6}\right)^{1/3}.$$
$\text{Graph made using Desmos.}$
| Hint: Rewrite the expression as
\begin{eqnarray*}
\frac{1}{\sqrt{5} } \left( \sqrt{3} \sqrt[4]{5} x - \frac{1}{\sqrt{3} \sqrt[4]{5} x } \right)^2 + \frac{2}{\sqrt{5}}
\\ +
5\sqrt{5} \left( \frac{y}{ \sqrt[4]{5}} - \frac{ \sqrt[4]{5}}{y } \right)^2 +10\sqrt{5} \\+
\frac{1}{6 } \left( \sqrt{6} z - \frac{1}{\sqrt{6} z } \right)^2 + \frac{1}{3} .
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/3522119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n n^2 i^n}{(2n)!}$ $$\sum_{n=0}^{\infty} \frac{(-1)^n n^2 i^n}{(2n)!}$$
$i^n$ can be written as $(\frac{1+i}{\sqrt{2}})^{2n}$ but the main problem is the series :
$$\sum_{n=0}^{\infty} (-1)^n \frac{n^2 }{(2n)!}$$
I don't understand why this series can be written as a sum of sine and cosine because if we add $\sin{1} + \cos{1} $ (for example) togheter we would get :
$$ \sum_{n=0}^{\infty} (-1)^n \frac{2n+2}{(2n+1)!} $$
The correct answer should be :
$$-\frac{1}{4} \frac{1+i}{\sqrt[4]{2}} \sin{\frac{1+i}{\sqrt[4]{2}}} -\frac{1}{4} i\sqrt{2} \cos\frac{1+i}{\sqrt[4]{2}} $$
| As $$4n^2=2n(2n-1)+2n$$ write $y^2=-i$
$$\dfrac{(-1)^ni^nn^2}{(2n)!}=\dfrac{y^{2n}(2n(2n-1)+2n)}{4(2n)!}=\dfrac{y^2}4\dfrac{y^{2n-2}}{(2n-2)!}+\dfrac y4\dfrac{y^{2n-1}}{(2n-1)!}$$
Now $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
Find $$\dfrac{e^y+e^{-y}}2,\dfrac{e^y-e^{-y}}2$$
Replace the value of $y^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rayleigh quotient on circular region of radius 2 I' m struggling with the following problem.
We have the eigenvalue problem:
$$u'' + \lambda u = 0$$
with associated boundary condition:
$$u' + 3u = 0$$
Now by using the Rayleigh quotient for $0 \leq r \leq 2$and the trial function:
$$\psi_0(r)= \alpha - r $$
where $\alpha$ is chosen so that $\psi_0(r)$ satisfies the boundary condition at $r=2$.
Find the lowest eigenvalue $\lambda_1$
Own work:
I started finding the correct $\alpha$ that suites the BC, hence:
$$ -1 + 3 (\alpha -2) = 0$$
Hence $\alpha = \frac{7}{3}$
An upper bound to $\lambda_1$ is given by the Rayleigh quotient,
$$J[\psi_0] = \frac {D[\psi_0]}{H[\psi_0]}$$ where
$$D[\psi_0] = \int_{\Omega}|\bigtriangledown \psi_0|^2 dx + \int_{\partial \Omega} \psi_{0}(2)^2 dx \quad H[\psi_0] = \int_\Omega \psi_0^2 dx$$
(in polar form)
$$\bigtriangledown = \frac{\partial}{\partial r} + \frac {1}{r} \frac{\partial}{\partial \theta}$$
Thus $|\bigtriangledown \psi_0| = |\psi'_0(r)|=-1 $ and
$$\int_\Omega f(r)dx = \int_0^2 2\pi r f(r)dr$$
so that:$$J[\psi_0] = \frac {\int_0^2 (r + \frac{1}{9}r)dr}{\int_0^2 r(\alpha -r)^2dr}$$
(Already cancelled the $2\pi$ out and left in $\alpha$)
These integrals can be evaluated as:
$$\int_0^2 \frac{10}{9}r dr = \big[\frac {10}{18} r^2 \big]_{0}^{2} = \frac {20}{9}$$
and
$$\int_0^2 r(\alpha -r)^2dr = \big[\frac{1}{2} r^2 \alpha^2 - \frac{2}{3} \alpha r^3+ \frac{1}{4} r^4 \big]_{0}^{2} = 2\alpha^2 - \frac{16 \alpha}{3}+4 = \frac {22}{9}$$
Which would give $$J[\psi_0]= \frac{10}{11}$$
Which is not correct $\big(\lambda_1 \leq \frac {12}{11} \big)$
Perhaps someone can help me, I think I go wrong by defining $D[\psi_0]$
| \begin{equation}
J(\psi_0) \geq \lambda_1
\end{equation}
Where in this case $\psi_0$ is the trialfunction.
In order to understand the Rayleigh-quotient, now assume the same eigenvalue problem as the example above, but now by using the Rayleigh quotient method.
$$ \Delta u + \lambda u = 0$$
with associated boundary condition:
$$\frac {\partial u}{\partial r} + 3u = 0$$
Now by using the Rayleigh quotient for $0 \leq r \leq 2$ and the trial function:
\begin{equation}
\label{eq:trial}
\psi_0(r)= \alpha - r
\end{equation}
where $\alpha$ is chosen so that $\psi_0(r)$ satisfies the boundary condition at $r=2$.
to find the lowest eigenvalue $\lambda_1$
The correct $\alpha$ that suits the boundary conditions,
$$ -1 + 3 (\alpha -2) = 0$$
Hence $\alpha = \frac{7}{3}$
An upper bound to $\lambda_1$ is given by the Rayleigh quotient \cite{courantvol22008methods},
\begin{equation}
J[\psi_0] = \frac {\mathfrak{D} [\psi_0]}{H[\psi_0]}
\end{equation}
where
\begin{equation}
\label{eq:quadratic}
\mathfrak {D} [\psi_0] = \int_{\Omega}|\bigtriangledown \psi_0|^2 dx + \int_{\partial \Omega} \psi_0(2)^2 ds \quad H[\psi_0] = \int_\Omega \psi_0^2 dx
\end{equation}
In the next chapter, a more detailled discussion will be given concerning these quadratic functionals.
(in polar form)
$$\bigtriangledown \psi =\underline e_r \frac{\partial \psi}{\partial r} + \frac {1}{r} \underline e_{\theta} \frac{\partial \psi}{\partial \theta}$$
Where:
$$\underline e_r = \left( \begin{array}{c} \cos \theta \\ \sin \theta \end{array} \right), \quad \underline e_\theta = \left( \begin{array}{c}- \sin \theta \\ \cos \theta \end{array} \right) $$
Thus $|\bigtriangledown \psi_0| = |\psi'_0(r)|=-1 $ and
$$\int_\Omega f(r)dx = \int_0^2 2\pi r f(r)dr$$
so that:$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) +3 (\alpha - 2)^2 \int_{\partial \Omega} ds }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$
(For readability, the $\alpha$ is left in.)
which leads to:
$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) +3 (\alpha - 2)^2 \ 4\pi }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$
$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) + \frac{4\pi}{3} }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$
These integrals can be evaluated as:
$$2 \pi \int_0^2 (r dr)= \frac {36 \pi}{9}$$
and
$$2 \pi\int_0^2 r(\alpha -r)^2dr = 2 \pi \big[\frac{1}{2} r^2 \alpha^2 - \frac{2}{3} \alpha r^3+ \frac{1}{4} r^4 \big]_{0}^{2} = 2\alpha^2 - \frac{16 \alpha}{3}+4 = \frac {44 \pi}{9}$$
Which would give $$J[\psi_0]= \frac{\frac {36}{9} + \frac {12}{9}} {\frac{44}{9}} = \frac{12}{11}$$
Hence an upper bound for $\lambda_1 = \frac {12}{11}$
| {
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"url": "https://math.stackexchange.com/questions/3523847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simple inequality proof problem of my authorship with fixed sum $a+b+c=3$. Prove that for positive numbers $a,b$ and $c$ such that $a+b+c=3$ following inequality holds:
$\sqrt[3]{\frac{1}{3a^2(8b+1)}}+\sqrt[3]{\frac{1}{3b^2(8c+1)}}+\sqrt[3]{\frac{1}{3c^2(8a+1)}}\ge 1$.
There are at least 3 different solutions that I found, so feel free to solve it even if somebody has already posted a solution.
| Using AM-GM and Cauchy-Schwarz:
$$
\begin{aligned}
\frac{1}{3}\cdot LHS &= \sum \frac{1}{\sqrt[3]{9a\cdot 9a \cdot (8b+1)}} \\
&\geq \sum \frac{3}{9a+9a+8b+1} \\
&= 3\sum \frac{1}{18a+8b+1} \\
&\geq 3\cdot \frac{9}{26(a+b+c)+3} \\
&= \frac{1}{3}
\end{aligned}
$$
Equality occurs when $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$ I am trying to prove that $$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$$ for $n \in \mathbb{N}$.
My approach is to prove this by induction and this is what I got so far:
$$\sum_{k=1}^{2(n+1)} \frac{(-1)^{k-1}}{k} = (\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}) + \frac{(-1)^{(2n+1)-1}}{2n+1} + \frac{(-1)^{(2n+2)-1}}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac{(-1)^{(2n+1)-1}}{2n+1} + \frac{(-1)^{(2n+2)-1}}{2n+2}$$
However, I am not aware how to proceed from here to get to: $$\sum_{k=1}^{n+1} \frac{1}{(n+1)+k}?$$
I would greatly appreciate any advice/solutions!
| Got it, working my way backwards using Maximilian Janisch advice:
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k}=\sum_{k=2}^{n+2} \frac{1}{n+k} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} - \frac{1}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2}$$
Thanks to everyone who helped!
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3}
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.
*If $x$ and $y$ are divisible by $3$:
Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?
| $1995$ I'd divisible by $3$ but not by $9$ hence $x , y $
both simultaneously are not multiple of $3$.
$$ $$ let either of $x,y$ is multiple of $3$ and other is not multiple of $3$ then $x^2-y^2$ is not multiple of $3$ where as $ 1995$. Is multiple of $3$ hence not possible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given that $f(x) = \frac{x^2}{x-2}$ show that $8 \le \int_3^4 f(x)\,dx \le 9$. Consider the function:
$$f : \mathbb{R} \setminus \{ 2\} \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{x^2}{x - 2}$$
I have to show that the following statement is true:
$$8 \le \int_3^4 f(x) \, dx \le 9$$
The first thing I did was to find:
$$\int_3^4 f(x) \, dx = \frac{11}{2} + 4 \ln 2$$
I think this is correct. The problem I have now is to show the following:
$$8 \le \dfrac{11}{2} + 4 \ln 2 \le 9$$
What I did in order to accomplish this was to first show that the result of the integral is first $\ge 8$ and then that it is $\le 9$. So first I tried to show:
$$\frac{11}{2} + 4 \ln 2 \ge 8$$
$$\frac{11}{2} - 8 + 4 \ln 2 \ge 0$$
After a few calculations I got to the point where I have to show:
$$8 \ln 2 \ge 5$$
$$\ln 2^8 \ge \ln e^5$$
$$ 256 \ge e^5 \tag 1$$
So I have to show that $256 \ge e ^5$. In a similar manner, working with the other part of the inequality (to trying to show that the result of the integral is $\le 9$) I have to show:
$$ 256 \le e^7 \tag 2$$
So I am stuck at proving $(1)$ and $(2)$. If I approximate $e \approx 2.71$ there is no problem, the inequalities are clearly true, but this feels a bit sloppy. Is there another way to prove $(1)$ and $(2)$ (or even the whole given integral inequality from the start) a little more rigorously, not relying on approximations?
| A very quick way to show this is by finding the area under the curve in the interval from 3 to 4
The plot of $ f(x) $ looks like this- here is the plot.
From the plot it's very clear that area under the curve in $ [3, 4] $ is bounded between 8 and 9
Also, this function can be re-written as:
$$
f(x)=\frac{x^{2}}{x-2} = \frac{x^{2}-4+4}{x-2}= \frac{x^{2}-4}{x-2} + \frac{4}{x-2} \
=x+2+ \frac{4}{x-2}$$
For the values of $ f(x) $ in $ [3,4], f(3)=5+4=9 $ and $ f(4)= 6+2=8$, function integral in the range $[3,4]$ is calculated as
$$\begin{aligned}
f(x)&=\frac{x^{2}}{x-2}\\
&= \frac{x^{2}-4+4}{x-2}\\
&= \frac{x^{2}-4}{x-2} + \frac{4}{x-2}\\
&=x+2+ \frac{4}{x-2}\end{aligned}$$
$$\begin{aligned}
&\int_{3}^{4} x+2+ \frac{4}{x-2}\\
=& \left[\frac{x^{2}}{2}+2x+4\ln(x-2)\right]_3^4\\
=& \left[\frac{4^{2}}{2}+8+4\ln2\right]-\left[\frac{3^{2}}{2}+6+0\right] \\=& 16+ 4\ln2-4.5-6 \\
=& 5.5+4\ln2 (\because \ln2\approx .69)\\
\approx& 8.3
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to go from $C^2y^2+(C^2-M^2)x^2+Gx+H=0$ to an ellipse/ hiperbola? The following arrives on the context of conic sections.
(All letters denote a constant $\in \mathbb{R}$ except for $x$ and $y$).
Michael Spivak´s Calculus, arrives at
$$C^2y^2+(C^2-M^2)x^2+Gx+H=0$$
Then procedes to say that, for $\;C^2\neq M^2$, $\; C^2y^2+(C^2-M^2)x^2+Gx+H=0$ is either a ellipse or an hiperbola and that the values for $G$ and $H$ are trivial.
An ellipse and a hiperbola being objects described by the relation
$$\frac{x^2}{a^2}\pm \frac{y^2}{a^2-c^2}=k\text{,}$$
Trying to ceck this, I divided by $\;C^2(C^2-M^2)$ getting
$$\frac{x^2}{C^2}+\frac{y^2}{(C^2-M^2)}=\frac{-(Gx+h)}{C^2(C^2-M^2)}$$
My problem with this is that $\frac{-(Gx+h)}{C^2(C^2-M^2)}$ is not a constant (thus it can´t equal $k$).
I want to express $\; C^2y^2+(C^2-M^2)x^2+Gx+H=0$ on it´s ellipse/ hiperbola form.
What am I doing wrong? How could I someow get rid of $x$ on the right hand side so that its just a constant and not a function?
| $$C^2y^2+\underbrace{(C^2-M^2)x^2+Gx}_{\text{make this a perfect square}}+H=0$$
$$\begin{align*}
(C^2-M^2)x^2+Gx&=(C^2-M^2)\left(x^2+\frac G{C^2-M^2}x+\frac{G^2}{4(C^2-M^2)^2}-\frac{G^2}{4(C^2-M^2)^2}\right)\\[1ex]
&=(C^2-M^2)\left(\left(x+\frac G{2(C^2-M^2)}\right)^2-\frac{G^2}{4(C^2-M^2)^2}\right)\\[1ex]
&=(C^2-M^2)\left(x+\frac G{2(C^2-M^2)}\right)^2-\frac{G^2}{4(C^2-M^2)}\\[1ex]
\end{align*}$$
Then the conic section has equation
$$C^2y^2+(C^2-M^2)\left(x+\frac G{2(C^2-M^2)}\right)^2=\frac{G^2}{4(C^2-M^2)}-H$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ to show that $\mathrm{det} A = 1$? How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ with $$a_{ij}=\begin{cases} 1& i<j,\\ 0 & i=j,\\ -1& i>j,\end{cases}$$
to show that $\mathrm{det} A = 1$ ?
| Subtract the second from the first row, the third from the second row, ..., the $n$th row from the $n-1$st row and you will get the following matrix (omitted entries are all $0$):
$$\begin{pmatrix}
1 & 1 & 0 & 0 & 0&\dots & 0 & 0&0\\
0 & 1 & 1 & 0 & 0 &\dots & 0 & 0&0 \\
0 & 0 & 1 & 1 & 0 & \dots & 0 & 0&0\\
& & & \ddots & \ddots \\ &&&&\ddots&\ddots
\\ & & & &&\ddots&\ddots
\\&&&&&&\ddots&\ddots
\\
0 & 0 &0&0&0&\dots&0&1&1 \\
-1 & -1 & -1&-1&-1&\dots & -1 &-1 &0
\end{pmatrix}$$
now add the first, third, fifth, ..., $n-1$st row to the last row and you get
$$\begin{pmatrix}
1 & 1 & 0 & 0 & 0&\dots & 0 & 0&0\\
0 & 1 & 1 & 0 & 0 &\dots & 0 & 0&0 \\
0 & 0 & 1 & 1 & 0 & \dots & 0 & 0&0\\
& & & \ddots & \ddots \\ &&&&\ddots&\ddots
\\ & & & &&\ddots&\ddots
\\&&&&&&\ddots&\ddots
\\
0 & 0 &0&0&0&\dots&0&1&1 \\
0 & 0& 0&0&0&\dots & 0 &0 &1
\end{pmatrix}$$
Nice! Now we have an upper diagonal matrix so the determinant is just the product of the diagonal entries which are all $1$ so we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Tangents are drawn to the parabola $y^2=4x$ from the point P (6,5) to the touch the parabola at Q and R.
$C_1$ is the circle which touches the parabola at Q and $C_2$ is the circle which touches the parabola at R. Both circles pass through the focus of the parabola. Find the radius of circle $C_2$
The equation of tangent to the parabola
$$y=mx+\frac am$$
$$5=6m+\frac 1m$$
$$m=\frac 12 , \frac 13$$
Therefore, equation of tangents will be
$$x-2y+4=0$$ and $$x-3y+9$$
The point of intersections with the parabola $y^2=4x$ were found out to be $(4,4)$ and $(9,6)$
Let R be $(9,6)$. Hence circle $C_2$ passes through (9,6) and focus (1,0)
This data isn’t enough to find the radius of the circle. How do I get more information?
| Find the equation of both the circles using $S+kL=0$.
$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles.
Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$
Circle through $(9,6)$ is $(x-9)^2+(y-6)^2+k(3y-x-9)=0$
Since it passes through the focus $(1,0)$, therefore, putting the values of $X$ and $Y$ we get $K=10$.
Hence the equation comes out to be:
$(x-9)^2+(y-6)^2+10(3y-x-9)=0$
or $x^2+y^2-28x+18y+108=0$
Similarly, for the second circle through $(4,4)$: $(x-4)^2+(y-4)^2+k(2y-x-4)=0$
Since it passes through $(1,0)$ we get $K$ as $5$. We have got both the circles and we can find the radius.
Using this method we get the centre of circles as $(13/2,-1)$ and $(14,-9)$.
Therefore using the centres and the point $(1,0)$ we can get the radius using the distance formula.
$r_1$ comes out to be $\sqrt{125/4}$ and $r_2$ comes out to be $\sqrt{250}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induction:
(1) Base: $\tau(1)$.
$$
\begin{align}
a_2 &=\frac{(2+1)a_1-2+1}{2+1-(2-1)a_1} \\
&=\frac{3a_1-1}{3-a_1}.
\end{align}
$$
(2) Assumption: Let
$$ a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1} $$
hold for some $n\in\mathbb N$.
(3) Step $\tau(n+1)$:
$$
\begin{align}
a_n &= \frac{3\cdot\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}-1}{3-\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}} \\
&= \frac{3\cdot2^{n-1}a_1+3a_1-3\cdot2^{n-1}+3-2^{n-1}-1+2^{n-1}a_1-a_1}{3\cdot 2^{n-1}+3-3\cdot 2^{n-1}a_1+3a_1-2^{n-1}a_1-a_1+2^{n-1}-1} \\
&= \frac{(4\cdot 2^{n-1}+2)a_1-4\cdot 2^{n-1}+2}{4\cdot 2^{n-1}+2-(4\cdot 2^{n-1}+2)a_1} \\
&= \frac{(2^n+1)a_1-2^n+1}{2^n+1-(2^n+1)a_1}
\end{align}
$$
Is this correct and is there a more efficient method than induction?
| This is my solution
Let
$$ \alpha_n = V_n + \beta $$
$$a_{n+1} = V_{n+1} + \beta$$
then
$$V_{n+1} = \frac{3V_n + V_n\beta + (\beta^2 - 1)}{3-V_n-\beta} $$
At here we choose $\beta = 1$ as result from $\beta^2 - 1 = 0$
So we got $$V_{n+1} = \frac{4V_n}{2-V_n}$$
Continue we let
$$V_n = \frac{1}{u_n}$$
then
$$\frac{4\frac{1}{u_n}}{2-\frac{1}{u_n}} = \frac{1}{u_{n+1}}$$
$$ <=> \frac{1}{2}u_n - \frac{1}{4} = u_{n+1}$$
From here we need to find $u_1 $based on $v_1$ ,$\beta$ and $a_1$ then find $u_n$,$V_n$. Finally we will find out $\alpha_{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
Now it is $e^{(0\cdot \infty)}$ which we can not conclude about the limit
| L'hopital
$\lim_{x\to 7}{(\frac{(x-8)(x-10)\ln(\frac{x}{7})}{x-7})}=$
$\lim_{x\to 7}\frac{(x-8)(x-10)\frac 7x\frac 17 + (2x-18)\ln \frac x7}1=$
$\frac{(-1)(-3)}7 + (-4)*0=\frac 37$.
So $\lim_{x\to 7}e^{ln(\frac{x}{7})^{(\frac{(x-8)(x-10)}{x-7})}}=\lim_{x\to 7}e^{(\frac{(x-8)(x-10)}{x-7})ln(\frac{x}{7})}=e^{\frac 37}$
That is.... assuming you did everything right. I confess I didn't check your work at all.
| {
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"url": "https://math.stackexchange.com/questions/3536953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
| $$
\begin{aligned}
I &=\int \frac{1}{\left(x^{3}+1\right)^{2}} d x \\
&=-\frac{1}{3} \int \frac{1}{x^{2}} d\left(\frac{1}{x^{3}+1}\right) \\
&=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int \frac{1}{x^{3}\left(x^{3}+1\right)} d x \\
&=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int\left(\frac{1}{x^{3}}-\frac{1}{x^{3}+1}\right) d x \\
&=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}+\frac{1}{3 x^{2}}+\frac{1}{3} \int \frac{d x}{x^{3}+1}
\end{aligned}
$$
By my post,
$$
\therefore I=\frac{1}{18}\left[\frac{6 x}{\left(x^{3}+1\right)}+\ln \left|\frac{(1+x)^{2}}{1-x+x^{2}}\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution I so close to can solve this problem but I don't find the correct response:
$$\int\frac{dx}{2x^2+5}
$$
Always get the answer:
$$ \frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}}
$$
But the correct answer have more one square root multiplying the square root of $5:$
(this is the correct answer):
$$
\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}
$$
I'm using the follow propertie to try solve the problem:
$$
\int{\frac{dx}{u^2+a^2}} = \frac{1}{a}\arctan{\frac{u}{a}}
$$
--My steps for the solution:
Before to apply the properties get square root from $~2x^2~$ and $5$, staying that way:
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}
$$
So when I apply the previous properties get my wrong answer
| Your error stems from applying the chain rule incorrectly. You basically substitute $u=x\sqrt{2}\Rightarrow du=\sqrt{2}\cdot dx$.
So
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}=\int\frac{\frac{du}{\sqrt{2}}}{u
^2+(\sqrt{5})^2}=\frac{1}{\sqrt{2}}\int\frac{du}{u
^2+(\sqrt{5})^2}=$$
$$=\frac{\arctan{\frac{u}{\sqrt{5}}}}{\sqrt{5}\mathbf{\sqrt{2}}}+C=\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help. Thanks.
| This is going to get ugly real quick.
$$\int{\frac{5x}{\left(2x^2-3\right)\sqrt{3x^2-2x+1}}}\,\mathrm dx\equiv 5\sqrt3\int\frac x{\left(2x^2 - 3\right)\sqrt{(3x - 1)^2 + 2}}\,\mathrm dx$$
Let $u = 3x - 1\implies\mathrm dx = \dfrac13\mathrm du$ and $x = \dfrac{u + 1}3, x^2 = \dfrac{(u + 1)^2}9$. Therefore,
$$\int\frac x{\left(2x^2 - 3\right)\sqrt{(3x - 1)^2 + 2}}\,\mathrm dx\equiv\int\dfrac{u + 1}{\left(2(u^2 + 2u) - 25\right)\sqrt{u^2 + 2}}\,\mathrm du$$
Next, substitute $u = \sqrt2\tan(v)\implies\mathrm du = \sqrt2\sec^2(v)\,\mathrm dv$. So,
$$\begin{align}\int\dfrac{u + 1}{\left(2(u^2 + 2u) - 25\right)\sqrt{u^2 + 2}}\,\mathrm du&\equiv\int\dfrac{\sqrt2\sec^2(v)\left(\sqrt2\tan(v) + 1\right)}{\left(2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25\right)\sqrt{2\tan^2(v) + 2}}\,\mathrm dv \\ &\stackrel{\sec^2(v) = 1 + \tan^2(v)}=\int\dfrac{\sec(v)\left(\sqrt2\tan(v) + 1\right)}{2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25}\,\mathrm dv\end{align}$$
Perform tangent half-angle substitution to get
$$\int\dfrac{\sec(v)\left(\sqrt2\tan(v) + 1\right)}{2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25}\,\mathrm dv\equiv\int\dfrac{1 + \left(\tan^2\left(\frac v2\right)\right)\left(\frac{\sqrt{2^3}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)}\right) + 1}{\left(1 - \tan^2\left(\frac v2\right)\right)\left(2\left(\frac{\sqrt{2^5}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)} + \frac{8\tan^2\left(\frac v2\right)}{\left(1 - \tan^2\left(\frac v2\right)\right)^2}\right) - 25\right)}\,\mathrm dv$$
Finally, let $w = \tan\left(\dfrac v2\right)\stackrel{\sec^2(v) = 1 + \tan^2(v)}\implies\mathrm dv = \dfrac2{1 + w}\,\mathrm dw$. Therefore,
$$\int\dfrac{1 + \left(\tan^2\left(\frac v2\right)\right)\left(\frac{\sqrt{2^3}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)}\right) + 1}{\left(1 - \tan^2\left(\frac v2\right)\right)\left(2\left(\frac{\sqrt{2^5}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)} + \frac{8\tan^2\left(\frac v2\right)}{\left(1 - \tan^2\left(\frac v2\right)\right)^2}\right) - 25\right)}\,\mathrm dv\equiv\int\dfrac{2\left(w^2 - \sqrt{2^3}w - 1\right)}{25w^4 + \sqrt{2^7}w^3 - 66w^2 - \sqrt{2^7}w + 25}\,\mathrm dw$$
From here on, you need to factorize the denominator and obtain partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to show that this series converges? I found in a book the following exercise:
Show that the series $$\sum_{n=1}^{+\infty} \sin\left(\frac{n^2+n+1}{n+1}\right)$$ converges.
At first sight, I tried to break the fraction and rewrite the series as $$\sum_{n=1}^{+\infty} \sin\left(n+\frac{1}{n+1}\right)$$
This actually behaves (almost) like $$\sum_{n=1}^{+\infty} \sin(n)+C$$ where $$C=\sum_{n=1}^{+\infty} \frac{\cos(n)}{n+1}$$
How can this series converge? Am I missing something here?
| Suppose that
$$\lim_{n\to\infty}\sin\left(\frac{n^2+n+1}{n+1}\right)=\lim_{n\to\infty}\sin\left(n+\frac{1}{n+1}\right)=0$$
We have
$$\sin\left(n+1+\tfrac{1}{n+2}\right)=\sin\left(n+\tfrac{1}{n+1}\right)\cos\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)+\cos\left(n+\tfrac{1}{n+1}\right)\sin\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right) $$
Since
$$\lim_{n\to\infty}\sin\left(n+\tfrac{1}{n+1}\right)=0 \\
\cos\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)\to\cos(1)\\
\sin\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)\to\sin(1)\ne 0$$
it follows that
$$\lim_{n\to\infty}\cos\left(n+\frac{1}{n+1}\right)=0 $$
However,
$$\sin^2\left(n+\frac{1}{n+1}\right)+\cos^2\left(n+\frac{1}{n+1}\right)=1 $$
and we get a contradiction. Therefore, the limit of the summand is not $0$, and the series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a_i^2\right ) \left ( \sum_{i=1}^{n}b_i^2 \right ) }$$
My attempt:
$n=3$
$a_1=\sqrt{ab}$, $a_2=\sqrt{bc}$, $a_3=\sqrt{ac}$
$b_1=\frac{\sqrt{a}}{\sqrt{b}}$, $b_2=\frac{\sqrt{b}}{\sqrt{c}}$, $b_3=\frac{\sqrt{c}}{\sqrt{a}}$
Plugging it in,
$$ab+bc+ac+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a^2 + b^2 + c^2$$
There are $3$ unwanted fractions. Is there any way to remove them?
| $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2>0.$$
We can get it also, by C-S:
$$(1^2+1^2+1^2)(a^2+b^2+c^2)\geq(a+b+c)^2,$$ which is our inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
| Have you tried considering the Lagrangean $$\mathcal L(a, b, c) = a^2 + b^3 + c^4 + \lambda\left(a + b^2 + c^3 - \frac{325}9\right)?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$
What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.
Starting with the result
$$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$
Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives
$$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$
From the following generating function for the harmonic numbers
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$
replacing $x$ with $-x^2$ leads to
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$
Substituting this result into (2) yields
$$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$
or, after integrating the first of the integrals by parts twice
$$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$
I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult.
Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way.
Update
Thanks to Ali Shather, the first of the integrals can be found. Here
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\
&=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\
&=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}.
\end{align}
And using the result I calculated here, namely
$$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$
gives
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\
& \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2.
\end{align}
| A Second (Magical) Solution by Cornel Ioan Valean
To get a different solution, we start from Dan Fulea's integral in this post where Cornel provided with an extremely simple solution, and by simply rearranging it, we have that
$$\frac{\pi^4}{64}$$
$$=\frac{\pi^2}{4}\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle \pi^2/8}-\pi \int_0^1\frac{\arctan(t)\operatorname{arctanh}(t)}{t}\textrm{d}t+\int_0^1\frac{\arctan^2(t)\operatorname{arctanh}(t)}{t}\textrm{d}t,$$
and since the middle integral is provided in the first Cornel's solution and the transformation from the last integral to series is again shown in the first Cornel's solution together with the values of the resulting auxiliary series, we arrive at the desired value of the series,
$$\color{red}{\sum _{n=1}^{\infty} (-1)^{n-1}\frac{ H_n H_{2 n}}{n^2}}$$
$$\color{red}{=2 G^2-2\log(2)\pi G-\frac{1}{8}\log^4(2)-\frac{21}{8}\log(2)\zeta(3)+\frac{1}{4}\log^2(2)\pi ^2+\frac{773}{5760}\pi ^4}$$
$$\color{red}{-4 \pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-3 \operatorname{Li}_4\left(\frac{1}{2}\right)},$$
and the solution is complete.
End of story
A note: The connection with Dan Fulea's integral was observed later, and this latter way is definitely more wonderful to consider.
| {
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"url": "https://math.stackexchange.com/questions/3552194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Calculate PMF of discrete random varialbe
$p(x,y)=\frac{xy^2}{30}$ be the joint PMF of discrete random variables X and Y with the support $S=\{(x,y):x\in\{1,2,3\}, y\in\{1,2\}\}$
Compute $E(X-2Y)$
Question is can I expand to be $E(X)-2E(Y)$? (by linearity)?
| We compute the marginal PMF of $X$ by summing the joint PMF over the possible values of $Y$:
\begin{align}
\mathbb P(X=1) = \mathbb P(X=1,Y=1) + \mathbb P(X=1,Y=2) = \frac{1\cdot1^2}{30} + \frac{1\cdot 2^2}{30} = \frac16\\
\mathbb P(X=2) = \mathbb P(X=2,Y=1) + \mathbb P(X=2,Y=2) = \frac{2\cdot1^2}{30} + \frac{2\cdot 2^2}{30} = \frac13\\
\mathbb P(X=3) = \mathbb P(X=3,Y=1) + \mathbb P(X=3,Y=2) = \frac{3\cdot1^2}{30} + \frac{3\cdot 2^2}{30} = \frac12,
\end{align}
and similarly the marginal PMF of $Y$:
\begin{align}
\mathbb P(Y=1) &= \mathbb P(X=1,Y=1) + \mathbb P(X=2,Y=1) + \mathbb P(X=3,Y=1)\\
&= \frac{1\cdot 1^2}{30} + \frac{2\cdot1^2}{30} + \frac{3\cdot1^2}{30}\\
&= \frac 15\\
\mathbb P(Y=2) &=\mathbb P(X=1,Y=2) + \mathbb P(X=2,Y=2) + \mathbb P(X=3,Y=2)\\
&= \frac{1\cdot 2^2}{30} + \frac{2\cdot2^2}{30} + \frac{3\cdot2^2}{30}\\
&= \frac 45.
\end{align}
The expectation of $X$ is given by
\begin{align}
\mathbb E[X] &= 1\cdot\mathbb P(X=1) + 2\cdot\mathbb P(X=2) + 3\cdot\mathbb P(X=3) \\
&= 1\cdot\frac16+2\cdot\frac13+3\cdot\frac12\\
&= \frac73,
\end{align}
and the expectation of $Y$ by
\begin{align}
\mathbb E[Y] &= 1\cdot\mathbb P(Y=1) + 2\cdot\mathbb P(Y=2)\\
&= 1\cdot\frac15 + 2\cdot\frac45 = \frac95.
\end{align}
It follows from linearity of expectation that
$$
\mathbb E[X-2Y] = \mathbb E[X] - 2\mathbb E[Y] = \frac73 -2\cdot\frac95 = -\frac{19}{15}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554732",
"timestamp": "2023-03-29T00:00:00",
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Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=1^{\infty}$$
Then :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=\lim\limits_{n\to +\infty}e^{g(x)(f(x)-1)}$$
So :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{\frac{n^{4}}{n^{3}}}}{e^{\frac{(n+1)^{4}}{(n+1)^{3}}}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{n}}{e^{n+1}}$$
$$=\frac{1}{e}$$
Is my approach wrong ?
is this called partial limit calculation ?
| Hint:
$$\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}\right)^{(n+1)}}$$
$$=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}\right)^{n}\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 6
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Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$ Must I verify that Left hand side(LHS)= Right hand side(RHS)or can I prove that RHS= LHS?
I don’t know how to prove from LHS=RHS. How to separate the $4r^3+r$ into two terms, i.e. $\displaystyle\Bigl(r+\frac{1}{2}\Bigr)^4-\Bigl(r-\frac{1}{2}\Bigr)^4$
After verifying, I will need to find $$\sum_{r=1}^n (4r^3+r)$$
This is a summation of finite series question.
| Using the first identity,
$$\sum_{r=1}^n(4r^3+r)=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=1}^n\left(r-\frac{1}{2}\right)^4
\\=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=0}^{n-1}\left(r+\frac{1}{2}\right)^4$$
and by telescoping, this is
$$\left(n+\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4.$$
To establish the identity, factor the differences of squares
$$\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4\\
=\left(\left(r+\frac{1}{2}\right)^2+\left(r-\frac{1}{2}\right)^2\right)
\left(\left(r+\frac{1}{2}\right)+\left(r-\frac{1}{2}\right)\right)\left(\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)\right)\\
=\left(2r^2+\frac24\right)\cdot2r\cdot1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Integrate $\int_0^\infty \frac{1}{(1+x^2)^n} dx$ How to find $$\displaystyle \int_0^\infty \frac{1}{(1+x^2)^n}dx\quad ?$$
For $n=1$, we have $(\arctan x) |_0^\infty = \frac{\pi}{2}.$
I tried to integrate by parts to get recurrent formula:
$$
\int_0^\infty \frac{1}{(1+x^2)^n}dx
= \left. \frac{x}{(1 + x^2)^n} \right|_0^\infty ( = 0) + 2n\int_0^\infty \frac{x^2}{(x^2 + 1)^{n+1}}$$
| Hint:
$$I_n:=\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{x}{(x^2+1)^n}+2n\int_0^\infty\frac{x^2}{(x^2+1)^{n+1}}dx
\\=\left.\frac{x}{(x^2+1)^n}\right|_0^\infty+2n\int_0^\infty\frac{x^2+1-1}{(x^2+1)^{n+1}}dx=2n(I_n-I_{n+1})$$
gives the recurrence
$$I_{n+1}=\frac{2n-1}{2n}I_n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \quad \text { by replacing } 3 \theta \text { by }(2 \theta+\theta)$.
So, I used this to simplify the equation to
$2 \cos 3 \theta +1 = 0$
$\implies \cos 3 \theta =\frac{-1}{2}$
Since $\cos^{-1} \frac{-1}{2} = \frac{2\pi}{3} + 2 \pi n$,$\implies \theta =\frac{2 \pi}{9},\frac{8 \pi}{9},\frac{-8 \pi}{9}$ or $\frac{-2 \pi}{9}$. However, the graph seems to be showing another root which is $\frac{4 \pi}{9}$. Why did I miss this root? How should I find more angles that satisfy the equation in a given range. In general, how does one find all angles that satisfy an equation even after adding $2 \pi n$
| $\cos^{-1}(x)$’s range is $[0 , \pi]$ so if we calculate $\cos^{-1}(-\frac{1}{2})$ in our calculator, we are going to get $\frac{2\pi}{3}$
However, the solution to $\cos(x)=-\frac{1}{2}$ are $\pm\frac{2\pi}{3}+2n\pi$ with integer $n$. Since $x=3\theta$, $\theta=\pm\frac{2\pi}{9}+n\frac{6\pi}{9}$.
You missed $\theta=-\frac{2\pi}{9}+\frac{6\pi}{9}$ and $\frac{2\pi}{9}-\frac{6\pi}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use binomial expansion to show the following inequality I wanted to show $\left(1+\frac{x}{n}\right)^n\leq\left(1+\frac{x}{n+1}\right)^{n+1}$ using binomial expansion for $n\geq 2$ and $x\geq 0$.
So my idea was to expand both using binomial expansion and try to compare term-wise. The $k^{th}$ term of $\left(1+\frac{x}{n}\right)^n$ is $n\choose k $$\left(\frac{x}{n}\right)^k$, and similar for the other expansion the $k^{th}$ term is $n+1\choose k $$\left(\frac{x}{n+1}\right)^k$. I tried to compute the difference and tried to divide one by the other, neither yields any sufficient results. How should I approach this question?
| Let $f(x,n):=(1+x/n)^n$, then we need to show $f(x,n+1)\geq f(x,n)$ for $x\geq0,n\geq2$. Observe that the coefficient of $x^k$ in $f(x,n)$ is
$$\frac{1}{n^k}\binom{n}{k}=\frac{1}{k!}\cdot\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dotsm\frac{n-k+1}{n}.$$
Correspondingly the coefficient of $x^k$ in $f(x,n+1)$ is$$\frac{1}{(n+1)^k}\binom{n+1}{k}=\frac{1}{k!}\cdot\frac{n+1}{n+1}\frac{n}{n+1}\frac{n-1}{n+1}\dotsm\frac{n-k+2}{n+1}.$$
To solve the problem, all we need to do is to show that the latter is always greater than or equal to the former. To do so, we pair up each of the $k$ factors (excluding $1/k!$) by noting that
$$\begin{split}\frac{n}{n}&=\frac{n+1}{n+1}\\
\frac{n-1}{n}&\leq\frac{n}{n+1}\\
\frac{n-2}{n}&\leq\frac{n-1}{n+1}\\
&\vdots\\
\frac{n-k+1}{n}&\leq\frac{n-k+2}{n+1}
\end{split}$$
and multiplying all the inequalities together yields the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given that $x_1, x_2, x_3$ are the roots of the polynomial $x^3-2x^2+3x+5=0$ find $(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$. Consider the polynomial:
$$x^3-2x^2+3x+5=0$$
where $x_1, x_2$ and $x_3$ are the roots of the above polynomial. Now, consider the following determinant, which is defined using the above given roots:
$$\Delta = \begin{vmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
x_1^2 & x_2^2 & x_3^2 \\
\end{vmatrix}$$
And what is asked of me is to find $\Delta^2$.
After a bit of manipulation I found the following:
$$\Delta = (x_2-x_1)(x_3-x_1)(x_3-x_2)$$
Interestingly enough, this type of matrix has a special name: Vandermonde matrix and instead of doing that bit of manipulation after which I arrived at the above expression for $\Delta$, I could've used the formula given on that wikipedia page. Anyways...
So, I have to find:
$$\Delta^2 = (x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$$
The problem is that I cannot find any of the roots. I used the rational root theorem and found that there are no rational roots. None of the divisors of the free term, $5$, give $0$ when plugged into the polynomial. I tried all options: $\{\pm 1, \pm 5 \}$ and they all give something $\ne 0$.
So then I used the notation:
$$f(x) = x^3-2x^2+3x+5$$
found the derivative:
$$f'(x) = 3x^2-4x+3$$
and I observed that $f'(x) > 0$ for all $x \in \mathbb{R}$. So the function $f$ is strictly increasing, so we can have at most one solution to $f(x) = 0$. Because of what I showed above, this solution cannot be rational. So I concluded that we have one rational root and two complex (and conjugate, since $f \in \mathbb{R}[X]$) roots. But this is as far as I got. I cannot find them. And I tried finding $\Delta^2$ without finding the roots, but I couldn't solve that either.
| $\Delta$ is not symmetric, but $\Delta^2$ is, so it can be expressed in terms of $a=x_1+x_2+x_3$, $b=x_1x_2+x_2x_3+x_3x_1$ and $c=x_1x_2x_3$. Indeed, we have:
$$\Delta^2 = a^2 b^2 + 18 abc - 4 b^3 - 4 a^3 c - 27 c^2$$
The simplest way I know to prove this identity, is like this: let $x=x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $y=x_1x_2^2+x_2x_3^2+x_3x_1^2$. Then:
$$\Delta^2=(x-y)^2=(x+y)^2-4xy$$
It's pretty simple to notice that $x+y=ab-3c$ and for $xy$, expanding:
$$xy=c(x_1^3+x_2^3+x_3^3)+(x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3)+3c^2$$
and for the sum of cubes we have the well-known factorization:
$$x_1^3+x_2^3+x_3^3 = 3c+a(a^2-3b)$$
and similarly:
$$x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3=3c^2+b(b^2-3ca)$$
Replacing back all of this:
$$
\begin{aligned}
xy &= c[3c+a(a^2-3b)]+[3c^2+b(b^2-3ca)]+3c^2\\
&= b^3 - 6 a b c + 9 c^2 + ca^3
\end{aligned}
$$
and thus:
$$
\begin{aligned}
\Delta^2 &= (ab-3c)^2-4(b^3 - 6 a b c + 9 c^2 + ca^3 )\\
&= a^2b^2+18abc-4b^3-4a^3c-27c^2
\end{aligned}
$$
And we can determine $a,b,c$ from Vieta's ($a=2, b=3, c= -5$). In the end $\Delta^2=-1127$.
| {
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"url": "https://math.stackexchange.com/questions/3562434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A formula of Ramanujan for $\cot\sqrt {w\alpha} \coth\sqrt{w\beta} $ While trying to answer this question I stumbled on a paper by Bruce C. Berndt which contains the following formula by Ramanujan $$\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}=\frac{1}{2w}+\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\beta} \right) \tag{1}$$ which is supposed to hold for all positive numbers $\alpha, \beta$ with $\alpha\beta=\pi^2$.
Berndt mentions that this formula is wrong and missing a term. The corrected version stands as $$\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}=\frac{1}{2w}+\frac{1}{2}\log\frac{\beta}{\alpha}+\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\beta} \right) \tag{2}$$ for $\alpha>0<\beta,\alpha\beta=\pi^2$. Bruce gives some references which contain a proof of the above formula or its equivalents.
Bruce himself derives the above identity by using a change of variables in the following identity established by R. Sitaramchandrarao $$\pi^2xy\cot (\pi x) \coth (\pi y) =1+\frac{\pi^2}{3}(y^2-x^2)-2\pi xy\sum_{n=1}^{\infty} \left(\frac{y^2\coth (\pi n x/y)} {n(n^2+y^2)}-\frac{x^2\coth(\pi n y/x)}{n(n^2-x^2)}\right) \tag{3}$$ Ramanujan gave a similar (but wrong) formula and Sitaramachandrarao fixed it to arrive at $(3)$.
The derivation of $(2)$ from $(3)$ is not that difficult. The RHS of $(3)$ is modified using the identities $$\frac{y^2} {n(n^2+y^2)}=\frac{1}{n}-\frac{n}{n^2+y^2},\frac{x^2}{n(n^2-x^2)}=\frac{n}{n^2-x^2}-\frac{1}{n}$$ and $$\coth z =1+\frac{2}{e^{2z}-1}$$ The derivation also involves a transformation formula for logarithm of Dedekind eta function. However the proof of $(3)$ is omitted in Berndt's paper.
Unfortunately I have not been able to find those references online which contain a proof for $(2)$ or $(3)$. It is also mentioned that the formula could be proved using Mittag-Leffler expansion but I am barely a novice in complex analysis.
It is desirable to find a direct proof of the above result $(2)$ (or $(3)$) which avoids complex analytic methods. I tried to multiply the partial fractions of $\cot a$ and $\coth b$ but I could not manage to get the desired result.
| I thought it would be worthwhile to at least mention how to use the Mittag-Leffler pole expansion theorem to show that $$ \begin{align}\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta} &=\frac{1}{2w}+ \frac{b-a}{6} +\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\beta} \right) \\ &-2 \sum_{m=1}^{\infty} \frac{1}{m} \left(\frac{1}{e^{2m \alpha}-1}- \frac{1}{e^{2m \beta}-1}\right). \end{align} $$
Using the princicpal branch of the square root, and under the assumption that $\alpha$ and $\beta$ are positive parameters such that $\alpha \beta = \pi^{2}$, let $$f(w) = \frac{\pi}{2} \cot (\sqrt{w \alpha}) \coth (\sqrt{w \beta}) - \frac{1}{2w}.$$
The above function is meromorphic with simple poles at $w = \frac{m^{2}\pi^{2}}{\alpha} = m^{2} \beta$ and $w = -\frac{m^{2}\pi^{2}}{\beta} = -m^{2} \alpha$, where $m$ is a positive integer.
(Separately, $\cot(\sqrt{w \alpha})$ and $\coth(\sqrt{w \beta})$ have branch points at the origin, but their product has a simple pole at the origin.)
At $w= m^{2} \beta$, the residue of $f(w)$ is $$\begin{align} \lim_{w \to m^{2} \beta} \frac{\pi}{2}\frac{\coth (\sqrt{w \beta})}{\left( \tan(\sqrt{w \alpha}\right)'} &= \lim_{w \to m^{2} \beta} \, \frac{\pi}{2}\frac{2\coth (\sqrt{w \beta}) \, \sqrt{w \alpha}}{\alpha \sec^{2} (\sqrt{w \alpha)}} \\ &= \frac{\pi \coth(m \beta) m \pi }{\alpha} \\ &= m \beta \coth(m \beta). \end{align}$$
Similarly, at $w = - m^{2} \beta $, the residue of $f(w)$ is $$\begin{align} \lim_{w \to -m^{2} \alpha} \frac{\pi}{2}\frac{\cot (\sqrt{w \alpha})}{\left( \tanh(\sqrt{w \beta}\right)'} &= \lim_{w \to -m^{2} \alpha} \, \frac{\pi}{2}\frac{2\cot (\sqrt{w \alpha}) \, \sqrt{w \beta}}{\beta \operatorname{sech}^{2} (\sqrt{w \beta)}} \\ &= \frac{-\pi i \coth(m \alpha)i m \pi }{\beta } \\ &= m \alpha \coth(m \alpha). \end{align} $$
And the Laurent expansion of $\frac{\pi}{2} \cot (\sqrt{w \alpha}) \coth (\sqrt{w \beta})$ about the origin is $$\frac{\pi}{2 \sqrt{\alpha \beta}} \frac{1}{w} + \frac{\pi(\beta-\alpha)}{6 \sqrt{\alpha \beta}} + \mathcal{O}(w) = \frac{1}{2w} + \frac{\beta- \alpha}{6} + \mathcal{O}(w). $$
The most basic version of the Mittag-Leffler pole expansion theorem states that if $f(w)$ is a meromorphic function with simple poles at $w= a_{1}, a_{2}, \ldots$ (where $0 < |a_{1}| < |a_{2}| < \ldots$ ) with associated residues $b_{1}, b_{2}, \ldots$, then
$$f(w) = \lim_{w \to 0} f(w) + \sum_{m=1}^{\infty} \left(\frac{b_{m}}{w-a_{m}} + \frac{b_{m}}{a_{m}} \right) $$ provided that $f(s)$ is bounded on circles centered at the origin that stay away from the poles.
This formula results from evaluating $$\lim_{M \to \infty} \oint_{C_{M}} \frac{f(s)}{s(s-w)} \mathrm ds $$ where $C_{M}$ is a circle centered at the origin that encloses $M$ poles, and $w$ is inside the circle.
Applying the Mittt-Leffler pole expansion theorem to
$$\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}-\frac{1}{2w}, $$ we get
$$ \begin{align} \frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}-\frac{1}{2w} &= \frac{b-a}{6} + \sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\beta} \right) \\ &+ \sum_{m=1}^{\infty} \frac{1}{m} \left(\coth(m \beta) - \coth(m \alpha) \right), \end{align} $$
where $$ \begin{align} \sum_{m=1}^{\infty} \frac{1}{m} \left(\coth(m \beta) - \coth(m \alpha) \right) &= \sum_{m=1}^{\infty} \frac{1}{m} \left(\coth(m \beta) -1 -\left(\coth(m \alpha)-1 \right) \right) \\ &= - 2 \sum_{m=1}^{\infty} \frac{1}{m} \left(\frac{1}{e^{2 m \alpha}-1}-\frac{1}{e^{2m \beta}-1} \right). \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}{4n}$.
| The telescoping sum trick, with ${1\over k^2-k}={1\over k-1}-{1\over k}$, works, but you need to split off the first term:
$$\sum_{k=n}^{3n-1}{1\over k^2}={1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2}\lt{1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2-k}={1\over n^2}+\left({1\over n}-{1\over3n-1}\right)={5\over4n}-\left({1\over4n}+{1\over3n-1}-{1\over n^2} \right)$$
and
$${1\over4n}+{1\over3n-1}-{1\over n^2}={(3n-1)n+4n^2-4(3n-1)\over4(3n-1)n^2}={7n^2-13n+4\over4(3n-1)n^2}\gt{7n(n-2)+4\over4(3n-1)n^2}\gt0$$
for $n\ge2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i}{2}t+1=0$$
Solving that we get
$$t=\frac{(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2}) ^2-4}} {2} = \frac{3+i \pm \sqrt{-8+6i} } {4} $$
Converting 3+i to polar we get $\sqrt{10} e^{0.3218i}$
Converting $\sqrt{-8+6i}$ to polar we get $\sqrt{10} e^{-0.3218i}$
So $t=\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} $
Which means $e^{iz} =\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} = \frac{\sqrt{10}} {2} (\frac{e^{0.3218i} \pm e^{-0.3218i} }{2})$
And I dont know where to go from there
| Note that the solutions to $t^2-\frac{3+i}{2}t+1=0$ can be simplified as,
$$t=\frac{3+i \pm \sqrt{-8+6i} } {4} =\frac{3+i \pm (1+3i) } {4}$$
or,
$$e^{iz}=1+i=\sqrt2 e^{i(\frac\pi4+2\pi n)} = e^{\frac12\ln2 +i(\frac\pi4+2\pi n)}$$
$$e^{iz}=\frac12(1-i) =\frac1{\sqrt2} e^{-i(\frac\pi4+2\pi n)} = e^{-\frac12\ln2 -i(\frac\pi4+2\pi n)} $$
Thus, the solutions are $z=\pm (\frac\pi4+2\pi n-\frac i2\ln2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integral $\int (\sin x + 2\cos x)^5 dx$ Do I have to expand manually $(\sin x + 2\cos x)^5$?
I don't think I can do integration by parts or substitution with this function.
Edit: Try to make one term, what I know is using $\sin^2 x + \cos^2 x=1$, but I think I still have to expand the polynomial.
| One way is to recall the angle addition identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$ If we suppose there exists a suitable constant $k$ and angle $\theta$ such that $$\sin x + 2 \cos x = k (\sin x \cos \theta + \cos x \sin \theta),$$ then we have $$k \cos \theta = 1, \\ k \sin \theta = 2,$$ hence $$k^2 (\cos^2 \theta + \sin^2 \theta) = 1^2 + 2^2 = 5,$$ hence $k = \sqrt{5}$ and we may take $\theta = \tan^{-1} 2$. Then we have $$\int (\sin x + 2 \cos x)^5 \, dx = \int \left( \sqrt{5} \sin (x + \theta) \right)^5 \, dx = 5^{5/2} \int \sin^5 (x + \theta) \, dx.$$ The rest is handled by the usual methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$ How can we show that:
$$I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$$
The logarithm in the denominator is the main trouble for me here.
I have obtained this integral from the following infinite product:
$$P=e^I=\prod_{k=1}^\infty \frac{(3k-2)^2(6k-1)}{(3k-1)^2(6k-5)}=\sqrt[3]{2}$$
Which can probably be proved by Gamma functions, though I don't have the proof right now.
Is there a way to evaluate the integral without going back to the product in question? Preferably by some semi-elementary means.
A more simple, but related integral is:
$$\int_0^1 \frac{(x-1) dx}{\ln x}=\ln 2$$
I'm sure it's here somewhere, so maybe the same methods of solution would work for the more complicated case.
Update:
I found a way to transform the integral which makes it look even more complicated. If:
$$f(s)=\frac{2 }{s+1}\left(\frac{\, _2F_1\left(1,\frac{s+1}{2};\frac{s+3}{2};-\frac{1}{2} \left(i \sqrt{3}+1\right)\right)}{3+i \sqrt{3}}+\frac{\, _2F_1\left(1,\frac{s+1}{2};\frac{s+3}{2};-\frac{1}{2} \left(1-i \sqrt{3}\right)\right)}{3-i \sqrt{3}}\right)$$
Then:
$$I=\int_0^1 (f(s)-f(s+1)) ds$$
| By using the integral representation
\begin{equation}
\frac{x-1}{\ln x}=\int_0^1x^s\,ds
\end{equation}
we can express
\begin{align}
I&=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}\\
&=\int_0^1 \,ds\int_0^1 \frac{x^s\left( 1-x \right)}{1+x^2+x^4}\,dx
\end{align}
or, with the notation $J(s)=\int_0^1 \frac{x^s}{1+x^2+x^4}\,dx$,
\begin{equation}
I=\int_0^1 \left( J(s)-J(s+1) \right)\,ds
\end{equation}
By changing $x=y^2$ and denoting $j=\exp(2i\pi/3)$,
\begin{align}
J(s)&=\frac{1}{2}\int_0^1\frac{y^{\frac{s-1}{2}}}{1+y+y^2}\,dy\\
&=\frac{1}{2j(1-j)}\int_0^1y^{\frac{s-1}{2}}\left[\frac{j}{1-jy}-\frac{j^2}{1-j^2y}\right]\,dy\\
&=\frac{1}{\sqrt{3}}\Im\int_0^1y^{\frac{s-1}{2}}\frac{j}{1-jy}\,dy
\end{align}
Expansion of $(1-jy)^{-1}$ gives
\begin{align}
J(s)&=\frac{1}{\sqrt{3}}\Im\int_0^1 \sum_{n\ge0}j^{n+1}y^{\frac{s-1}{2}+n}\,dy\\
&=\frac{1}{\sqrt{3}}\Im\sum_{n\ge0}\frac{j^{n+1}}{\frac{s+1}{2}+n}
\end{align}
As $j^3=1$, $\Im j=\sqrt{3}/2$ and $\Im j^2=-\sqrt{3}/2$
\begin{align}
J(s)&=\frac{2}{\sqrt{3}}\Im\sum_{p\ge0}\left[\frac{j}{s+1+6p}+\frac{j^2}{s+3+6p}\right]\\
&=\sum_{p\ge0}\left[\frac{1}{s+1+6p}-\frac{1}{s+3+6p}\right]\\
\end{align}
We can use the representation
\begin{equation}
\psi(a)-\psi(b)=\sum_{k=0}^\infty\left[\frac{1}{k+b}-\frac{1}{k+a}\right]
\end{equation}
where $\Psi(.)$ is the digamma function, to express
\begin{equation}
J(s)=\frac{1}{6}\left[\Psi\left( \frac{s+3}{6} \right)-\Psi\left( \frac{s+1}{6} \right)\right]
\end{equation}
We have thus to evaluate
\begin{align}
I&=\frac{1}{6}\int_0^1\left[\Psi\left( \frac{s+3}{6} \right)-\Psi\left( \frac{s+1}{6} \right)-\Psi\left( \frac{s+4}{6} \right)+\Psi\left( \frac{s+2}{6} \right)\right]\,ds\\
&=\ln\left[\frac{\Gamma(2/3)}{\Gamma(1/2)}\right]-\ln\left[\frac{\Gamma(1/3)}{\Gamma(1/6)}\right]-\ln\left[\frac{\Gamma(5/6)}{\Gamma(2/3)}\right]+\ln\left[\frac{\Gamma(1/2)}{\Gamma(1/3)}\right]\\
&=\ln\left[\frac{\Gamma^2(2/3)\Gamma(1/6)}{\Gamma^2(1/3)\Gamma(5/6)}\right]
\end{align}
But, using the duplication formula for the Gamma function
\begin{equation}
\frac{\Gamma^2(2/3)}{\Gamma^2(1/3)}=\frac{2^{-2/3}}{\pi}\Gamma^2(5/6)
\end{equation}
and the reflection formula
\begin{equation}
\Gamma(5/6)\Gamma(1/6)=\frac{\pi}{\sin\pi/6}
\end{equation}
we deduce
\begin{equation}
I=\frac{\ln2}{3}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572686",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???
|
Applying simple substitutions: $\sqrt[3]{a}=m, \sqrt[3]{b}=n, \sqrt[3]{c}=k$, we have
$$(mnk)^3=1, m+n+k=0 \\ mnk=1, m+n+k=0 \\ m^3+n^3+k^3=m^3+n^3-(m+n)^3 =-3mn(m+n)=\frac{-3}{k} \times (-k)=3.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
= \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
| $\begin{array}\\
f(n)
&=\dfrac{n!}{n^n}\\
&=\dfrac{\prod_{k=1}^n k}{n^n}\\
&=\prod_{k=1}^n (k/n)\\
&=\prod_{k=0}^{n-1} ((n-k)/n)\\
&=\prod_{k=0}^{n-1} (1-k/n)\\
g(n)
&=\ln(f(n))\\
&=\sum_{k=0}^{n-1} \ln(1-k/n)\\
&<\sum_{k=0}^{n-1} -k/n
\qquad \ln(1-x) < -x\\
&=-\dfrac{n(n-1)}{2n}\\
&=-\dfrac{n-1}{2}\\
&=\dfrac12=\dfrac{n}{2}\\
\text{so}\\
f(n)
&\le e^{-(n-1)/2}\\
&=e^{1/2}(e^{-1/2})^n\\
\text{so}\\
\end{array}
$
$\sum_{n=1}^{\infty} f(n)
\le \dfrac{e^{1/2}}{1-e^{-1/2}}
$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Distance from $P(4,3)$ to tengent point on curve $x^2 + y^2 -2x-4=0$
Tangent line to curve $$x^2 + y^2 -2x-4=0$$ at A passes $P(4,3)$.
Find distance A and P.
Tangent of the line is $\frac{2-2x}{\sqrt{2x+4-x^2}}$
A(k,l) means
$l = \sqrt{2k+4-k^2}$
And
$l = \frac{2-2k}{\sqrt{2k+4-k^2}}$
So, I get $k=\frac{3\pm\sqrt{21}}{2}$
$AP = \sqrt{(k-4)^2 + (l-3)^2}$
Its so complicated, am i wrong somewhere?
| The required answer is $\sqrt{4^2+3^2-2•4-4}$ = $\sqrt{13}$
Use the formula $\sqrt{S1}$ Or proceed by using Pythagorean theorem as the other answer.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$$\lim_{x\to \pi/3} \left[\dfrac{\sin^2(x) - \sin^2\left(\dfrac{\pi}{3}\right)}{x^2 -\left(\dfrac{\pi}{3}\right)^2}\right]$$
| We can rewrite the limit in the following way:
$$\lim_{x \to \frac{\pi}{3}} \frac{\sin^2{x} - \sin^2{\frac{\pi}{3}}}{x^2 - \left(\frac{\pi}{3}\right)^2} =
\frac{2\sin{\frac{\pi}{3}}}{2\frac{\pi}{3}} lim_{x \to \frac{\pi}{3}} \frac{\sin{x} - \sin{\frac{\pi}{3}}}{x - \frac{\pi}{3}}=
\frac{\sin{\frac{\pi}{3}}}{\frac{\pi}{3}} lim_{x \to \frac{\pi}{3}} \frac{2\sin{(x-\frac{\pi}{3})}\cos{(x+\frac{\pi}{3})}}{x - \frac{\pi}{3}} =
\frac{\sin{\frac{\pi}{3}}}{\frac{\pi}{3}}2 \cos{\frac{2\pi}{3}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$ The natural action of $GL_2(\mathbb{R})$ on $\mathbb{R}^2$ is given by $A\cdot v=Av$, for $A\in GL_2(\mathbb{R})$ and $v\in\mathbb{R}^2$. Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$.
So I got that the orbit and stabilizer of $\begin{bmatrix}1\\0\end{bmatrix}$ is given by
\begin{align*}
Av=\lbrace\begin{bmatrix}a\\c\end{bmatrix}\rbrace
\qquad \textrm{Stab}(v)=\lbrace\begin{bmatrix}1&b\\0&d\end{bmatrix}\rbrace
\end{align*}
And the orbit and stabilizer of $\begin{bmatrix}1\\1\end{bmatrix}$ is given by
\begin{align*}
Av=\lbrace\begin{bmatrix}a+b\\c+d\end{bmatrix}\rbrace
\qquad \textrm{Stab}(v)=\lbrace\begin{bmatrix}1-b&b\\1-d&d\end{bmatrix}\rbrace
\end{align*}
I'm a little unsure about the stabilizers because the definitions just kind of seem to end at the second line with the equation.
| If $G$ is acting on $X$, and $x \in X$, the stabilizer is
$$
G_x=\{ g \in G | gx=x \}
$$
Hence in our case, we are looking for all the matrices $A \in \mathrm{GL}_2 (\mathbb{R})$ such that $Av=v$, where $v$ is either $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ or $\begin{pmatrix} 1 \\ 1\end{pmatrix}$.
Now, if $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$, we have $\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} 1 \\ 0\end{pmatrix}=\begin{pmatrix} a \\ c\end{pmatrix}$ and this is equal to $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ if and only if $a=1,c=0$, while $b$ can assume any value $\in \mathbb{R}$ and $d \in \mathbb{R} \setminus \{ 0 \}$ (otherwise the matrix is not invertible).
Similarly, $\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} 1 \\ 1\end{pmatrix}=\begin{pmatrix} a+b \\ c+d\end{pmatrix}$ and this is equal to $\begin{pmatrix} 1 \\ 1\end{pmatrix}$ if and only if $a+b=c+d=1$ and thus $a=1-b,c=1-d$. Furthermore $b \neq d$ (in order to for the matrix to be invertible).
I don't know if this answer your question.
| {
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"source": "stackexchange",
"question_score": "2",
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Egyptian fraction representation of $1$ where all denominators of the fractions are odd. Question: Is there an Egyptian fraction representation for $1$ where all the fractions have odd denominators?
I tried to generate one below:
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{23}+\frac{1}{721}+\frac{109}{106711605}.$$
The last term can be further decomposed to: $$\frac{1}{979007}+\frac{158}{1.04471\cdot 10^{14}}.$$
or, it is impossible for any collection of $\frac{1}{n}$ where $n$ is odd to produce $1$?
| Technique: Find an odd abundant number--more precisely, an odd semi-perfect number: ($945$ is the smallest).
Write it as a sum of its factors: $945=315+189+135+105+63+45+35+27+21+7+3$.
Then divide this equation by the original number: $945$, keeping the terms on the right separated:
$1=\frac13+\frac15+\frac17+\frac19+\frac{1}{15}+\frac{1}{21}+\frac{1}{27}+\frac{1}{35}+\frac{1}{45}+\frac{1}{135}+\frac{1}{315}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find game matrix and optimal strategies for each player. A, B simultaneously show a number of fingers to each other (both hands implies 0-10), money(£) is awarded as follows:
$ \bullet $ If the number of extended fingers is equal, then no exchange
$ \bullet $ If the number of extended fingers between each player differs by two, the player with the lowest number of fingers gives £2 to the other player.
$ \bullet $ Otherwise the player with the lowest number of extended fingers receives £3 from the other player.
I began with A's payoff matrix with each row/column representing the choice of $\{0,1, ... , 10\}$ resulting in $11$ choices each:
$$
\begin{bmatrix}
0 & 3 & -2 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3\\
-3 & 0 & 3 & -2 & 3 & 3 & 3 & 3 & 3 & 3 & 3\\
2 & -3 & 0 & 3 & -2 & 3 & 3 & 3 & 3 & 3 & 3\\
-3 & 2 & -3 & 0 & 3 & -2 & 3 & 3 & 3 & 3 & 3\\
-3 & -3 & 2 & -3 & 0 & 3 & -2 & 3 & 3 & 3 & 3\\
-3 & -3 & -3 & 2 & -3 & 0 & 3 & -2 & 3 & 3 & 3\\
-3 & -3 & -3 & -3 & 2 & -3 & 0 & 3 & -2 & 3 & 3\\
-3 & -3 & -3 & -3 & -3 & 2 & -3 & 0 & 3 & -2 & 3\\
-3 & -3 & -3 & -3 & -3 & -3 & 2 & -3 & 0 & 3 & -2\\
-3 & -3 & -3 & -3 & -3 & -3 & -3 & 2 & -3 & 0 & 3\\
-3 & -3 & -3 & -3 & -3 & -3 & -3 & -3 & 2 & -3 & 0\\
\end{bmatrix}
$$
Luckily, I found that each $\{3, 5, 6, 7, 8, 9, 10\}$ were inferior to $\{0\}$ and $\{4\}$ was inferior to $\{1\}$, leaving simplified matrix with strategies $\{0, 1, 2\}$:
$$
\begin{bmatrix}
0 & 3 & -2\\
-3 & 0 & 3\\
2 & -3 & 0\\
\end{bmatrix}
$$
However, I wasn't able to find a nash equilibrium, and hence requesting to find optimal strategies for each player A, B.
| In a mixed-strategy equilibrium, each player must be indifferent among the pure strategies to which she assigns non-zero probability.
In the present case, if all three strategies are assigned non-zero probabilities, this leads to (with the probability for strategy $i$ denoted by $p_i$)
$$
0p_0+3p_1-2p_2=-3p_0+0p_1+3p_2=2p_0-3p_1+0p_2\;.
$$
Together with the normalization constraint $p_0+p_1+p_2=1$, this is a system of three linear equations in three variables. Equating the sum of the first and third expression to the second yields
$$
2p_0-2p_2=-3p_0+3p_2
$$
and thus $p_0=p_2$. Thus the second expression is $0$. (With hindsight, we could have concluded from the symmetry of the game that the expected value of each strategy is $0$.) Then either equation yields $p_1=\frac23p_0$, and normalization then implies $p_0=p_2=\frac38$ and $p_1=\frac14$. The fact that all three probabilities come out as non-zero shows that the assumption that they are all non-zero was self-consistent.
| {
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"url": "https://math.stackexchange.com/questions/3586181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$ Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for the left side but I couldn’t work it out. I would prefer a proof without never ending computations (a.k.a. Brute Force / opening up the brackets) because I know how to do it this way already. A proof using C-B-S, Holder, Titu’s Lemma or their generalizations or other well-known inequalities would be ideal. Thank you!
| There is also the following proof (not mine) by SOS.
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}=\frac{\sum\limits_{cyc}\left((a^2-ab)abc+a(a-b)^2(b-2c)^2\right)}{abc(a+b+c)^2}\geq0.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Ratio between the area of a region and its transformation by a change of variables. The square $T:\{a<x<a+h, b<y<b+h\}, \text{with } a>0,b>0,h>0$ is tranformed into the region $S$ of the plane $uv$ by the following change of variables:
$$
u=\frac{y^2}{x}
$$
$$
v=x^{1/2}y^{1/2}
$$
Find $\frac{area(S)}{area(T)}$.
This is what I have thought:
Using the formula for the change of variables, I have that
$$
area(S)=\iint_Sdudv=\iint_T \left|\frac{\partial(u,v)}{\partial(x,y)}\right|dxdy=
$$
$$
=\int_a^{a+h}\int_b^{b+h}\frac{3}{2}x^{-\frac{3}{2}}y^{\frac{3}{2}}dydx=-\frac{6}{5}(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}})((b+h)^{5/2}-b^{5/2})
$$
Therefore, $\frac{area(S)}{area(T)}=\frac{1}{h^2}(-\frac{6}{5}(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}})((b+h)^{5/2}-b^{5/2}))$.
However, the solution in my book is:
$$\frac{area(S)}{area(T)}=\frac{6}{5}\frac{b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}}{(\sqrt{a+h}+\sqrt{a})(\sqrt{b+h}+\sqrt{b})\sqrt{a(a+h)}}$$
I have tried to do some algebraic transformations to my solution but I do not arrive to the given solution. What am I doing wrong?
Thanks for your help.
| Your solution is the same as the solution in your book.
You have
$$\frac{area(S)}{area(T)}=\frac{1}{h^2}\bigg(-\frac{6}{5}\bigg(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}\bigg)\bigg((b+h)^{5/2}-b^{5/2}\bigg)\bigg)$$
Firstly, we get
$$\begin{align}\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}&=\frac{\sqrt a-\sqrt{a+h}}{\sqrt a\sqrt{a+h}}
\\\\&=\frac{(\sqrt a-\sqrt{a+h})(\sqrt a+\sqrt{a+h})}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}
\\\\&=\frac{-h}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}\end{align}$$
Secondly, letting $x=\sqrt{b+h},y=\sqrt b$, we get
$$\begin{align}&(b+h)^{5/2}-b^{5/2}
\\\\&=x^5-y^5
\\\\&=(x-y)(x^4 + x^3 y + x^2 y^2 + x y^3 + y^4)
\\\\&=(\sqrt{b+h}-\sqrt b)\bigg((b+h)^2+(b+h)\sqrt{b(b+h)}+(b+h)b+b\sqrt{b(b+h)}+b^2\bigg)
\\\\&=(\sqrt{b+h}-\sqrt b)\bigg((b+h)^2+(b+h)\sqrt{b(b+h)}+(b+h)b+b\sqrt{b(b+h)}+b^2\bigg)\times\frac{\sqrt{b+h}+\sqrt b}{\sqrt{b+h}+\sqrt b}
\\\\&=\frac{h}{\sqrt{b+h}+\sqrt b}\bigg(b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}\bigg)\end{align}$$
So, your solution can be written as
$$\frac{area(S)}{area(T)}=\frac{1}{h^2}\bigg(-\frac 65\bigg)\bigg(\frac{-h}{\sqrt{a(a+h)}(\sqrt a+\sqrt{a+h})}\bigg)\times\frac{h}{\sqrt{b+h}+\sqrt b}\bigg(b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}\bigg)$$
$$=\frac{6}{5}\cdot\frac{b^2+(b+h)^2+b(b+h)+(2b+h)\sqrt{b(b+h)}}{(\sqrt{a+h}+\sqrt{a})(\sqrt{b+h}+\sqrt{b})\sqrt{a(a+h)}}$$
which is the solution in your book.
| {
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"source": "stackexchange",
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Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}\int_0^1(1-x)^2dx-2\int_0^1x^2(1-x)dx\right]\\
&=4×\left[\int_0^1(1-x)^2dx-2\int_0^1x^2dx+2\int_0^1x^3dx\right]\\
&=4×\left[-\frac{1}{6}[(1-x)^3]_0^1-\frac{2}{3}[x^3]_0^1+\frac{2}{4}[x^4]_0^1\right]\\
&=4×\left[\frac{1}{6}-\frac{2}{3}+\frac{1}{2}\right]\\
&=0
\end{align}
I can't find my fault please check this..
| HINT
The proposed integral can be expressed as follows:
\begin{align*}
I = \int_{-1}^{0}\int_{-x-1}^{x+1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x + \int_{0}^{1}\int_{x - 1}^{-x + 1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which value of x $\sum_{n=1}^\infty \frac{n^{nx}}{n!}$ converges I want to know for which values of $x$ this series
$$\sum_{n=1}^\infty \frac{n^{nx}}{n!}$$ converges.
This series is defined $ \forall x \in R$.
$a_n=\frac{n^{nx}}{n!}= \frac{(n^x)^{n}}{n!}= \frac{(n^x)}{n} \frac{(n^x)}{n-1}
\frac{(n^x)}{n-2}...\frac{(n^x)}{3} \frac{(n^x)}{2} \frac{(n^x)}{1}= n^{x-1} \frac{n^{x-1}}{1-\frac{1}{n}}...\frac{(n^x)}{3} \frac{(n^x)}{2} \frac{(n^x)}{1}
\sim 0 \Leftrightarrow x<1$.
Applying the ratio test:
$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{(n+1)x}}{(n+1)!} \frac{n!}{(n)^{nx}}=\frac{(n+1)^{(n+1)x}}{n+1} \frac{1}{(n)^{nx}}=\frac{(n+1)^{nx+x-1}}{n^{nx}} \sim \frac{(n)^{nx+x-1}}{n^{nx}}= (n)^{x-1}<1 \Leftrightarrow x-1<0 \Leftrightarrow x<1 $
Is it right?
| For these I always use
$n! \sim (n/e)^n$.
So
$\sum_{n=1}^\infty \frac{n^{nx}}{n!}
\sim \sum_{n=1}^\infty \frac{n^{nx}}{(n/e)^n}
= \sum_{n=1}^\infty (en^x/n)^n
= \sum_{n=1}^\infty (en^{x-1})^n
$
so we want
$en^{x-1} < 1
$
or,
taking logs,
$1+(x-1)\ln(n) < 0$
or
$(x-1)\ln(n) < -1
$.
This is false if
$x \ge 1$,
and is true
if $x < 1$.
To check,
write $x = 1-c$.
the terms are
$\begin{array}\\
\frac{n^{nx}}{n!}
&\sim \frac{n^{n(1-c)}}{(n/e)^n}
&= \frac{n^{n(1-c)}e^n}{n^n}
&= n^{-cn}e^n
&= (n^{-c}e)^n
\end{array}
$
which goes to zero
only if $c > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6t+1=0$$
Or,$$3t^2+6t-1=0$$
On solving I got the roots
$$\dfrac{-3+2\sqrt6}3$$ and $$\dfrac{-3-2\sqrt6}3$$
Then I plugged in the values and found only 2 roots are real use discriminant
| This is a program for finding out. There are at least two real roots since there is one between $-1$ and $0,$ by IVT. Then by continuity and the fact that the polynomial is positive for large values of $|x|,$ there must be another real root, clearly a negative one, since for $x>0,$ we have that the polynomial always rakes positive values.
To find out the status of the other roots, find out how the function behaves by considering its first two derivatives. The first gives $12x^3+18x^2+2x+6,$ which definitely has a real root, clearly negative. We may find out the other roots of this by again taking derivatives, which gives $36x^2+36x+2,$ which has two real roots, both negative. These roots are $$\frac{-3\pm\sqrt 7}{6}.$$ The derivative of this gives $72x+36,$ which is negative for any $x<-1/2.$ Since $\frac{-3-\sqrt 7}{6}$ is less than $-1/2,$ then it follows that the cubic has a maximum value there. At the other root $\frac{-3+\sqrt 7}{6}>-1/2,$ the cubic has a minimum value. Since the cubic is negative for large negative $x$ and positive for large positive $x,$ then the sign of either of the extreme value fixes whether the other two roots are real or not. The maximum occurs near $-1/2,$ in the interval $(-1,0).$ Here the cubic has a clearly positive value. Hence the maximum value of the cubic is positive. Therefore all its roots are real.
This means that the stationary points of the quartic are three in number. If none of them is a saddle point, then likely all the roots are real. Now since the quadratic is positive in the interval $(-\infty, r_1),$ with $r_1$ being its more negative root, then for the root of the cubic in this interval, we deduce that the quartic has a minimum here. Thus the other two roots of the cubic are in $(r_1,0).$ In $(r_1,r_2),$ $r_2$ being the less negative of the roots of the quadratic, the quadratic is negative. Thus the roots of the cubic here indicate that there is a maximum value. The last stationary point of the quartic must then necessarily be a minimum. Finally, if we show that the two minimum values of the quartic are negative (or that one minimum is negative and the maximum positive), then all its roots must be real. To do this, note that the minimum in $(-1,0)$ must be negative since the quartic is negative when $x=-1/2.$ As for the maximum value, it also occurs in $(-1,0).$ This has to be positive since there is a root here. And now the proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $n \text{ mod } r < \frac{n}{2}$ I am trying to prove a section of the Euclidean Algorithm for greatest common factors which states:
$\gcd(m, n) = \gcd(n, r) = \gcd(r, n \text{ mod } r)$, where $r =m \text{ mod } n$
Prove that: $n \text{ mod } r < \frac{n}{2}$ (assuming all variables are integers)
My first approach was to use division into cases to solve this proof but later I thought about using the quotient remainder theorem. If someone could guide me through this proof it would be greatly appreciated. Thank you in advance!
| First, we must assume that $n \not \mid m$, since $n \pmod{0}$ is undefined. Also, assume that $n,m \geq 0$. Note that $r = m \pmod{n} \in \{ 1, 2, \dots , n-1\}$. To show that $ n \pmod{r} < \frac{n}{2}$, we proceed by cases. Suppose that $r < \frac{n}{2}$, then $n \pmod{r} < \frac{n}{2}$ Suppose that $r > \frac{n}{2}$. By the division algorithm, we have that $n = r + n-r : n - r < r$, Since $r > \frac{n}{2} , n-r = n \pmod{r} < \frac{n}{2}$. If $n$ is even, then $r = \frac{n}{2} \implies n \equiv 0 \pmod{r}$. If $n$ is odd, then $r = \frac{n}{2}$ cannot occur. Let me know if you have questions.
EDIT:
The proof case 3 cases
$$
\begin{cases}
r < \frac{n}{2} & \text{ Done because any residue of r is less than } \frac{n}{2}
\\ r = \frac{n}{2} & \begin{cases} n \text{ is even } & \text{ Done because } n \pmod{r} = 0 \\ n \text{ is odd } & \text{ Done because } \frac{n}{2} \text{ is not an integer so r cannot take that value} \end{cases}
\\ r > \frac{n}{2} & \text{ Done because } n = r + n-r : n - r = n\pmod{r} < \frac{n}{2}, \text{ since } r > \frac{n}{2} \end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
equation with quadratic power I was thinking how to solve:
*
*If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$
*Solve $x^{x-x^2+13} = x^2-12$
I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.
| Following the method of ONG SEE HAI HCI , we have $A = 2x+1$
then:
$$x^{(x-1)^2}=x^{x^2+2-(2x+1)}=2x+1$$
then $x^{x^2+2-A} = A $ with only solution $x=\sqrt{A}$.
This is because $x^{x^2+2-A}$ is increasing.
Then $x=\sqrt{A} \rightarrow x^2 = 2x+1 \rightarrow x = 2+ \frac{1}{x} \rightarrow x-\frac{1}{x} = 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Using Laplace transform to solve IVP involving complex roots I am attempting to solve the following IVP :
$$ y'' + y' + \frac54 y =t -(t-\pi /2 )u_{\pi /2}(t) \quad, y(0) =0 ,\quad y'(0) = 0. \tag{1}$$
My reasoning is as follows,
\begin{align*}
\mathcal{L}[y'' + y' + \frac54 y] &= \mathcal{L}[t -(t-\pi /2 )u_{\pi /2}(t)] \\
\mathcal{L}[y''] + \mathcal{L}[y'] + \frac54 \mathcal{L}[y] &= \mathcal{L}[t] - \mathcal{L}[(t-\pi /2 )u_{\pi /2}(t)]\\
&\hspace{-5cm} \text{Letting $Y(s) = \mathcal{L}[y(t)]$,}\\
Y(s)(s^{2} + s + 5/4 ) &= \frac{1}{s^2} -\frac{e^{\frac{-\pi s}{2}}}{s^2}\\
\implies Y(s) = \frac{1- e^{\frac{-\pi s}{2}}}{s^2 (s^{2} + s + 5/4 )}
\end{align*}
Since $s^{2} +s +5/4 $ has complex roots I can't do a simple partial fraction decomposition and hence must go with
\begin{gather*}
\frac{A}{s^{2}} + \frac{Bs + C}{s^2 +s + 5/4} = \frac{1- e^{\frac{-\pi s}{2}}}{s^2 (s^2 + s +5/4)} \tag{2}\\
\implies A(s^2 +s + 5/4) + (Bs+C)(s^2) = 1- e^{\frac{-\pi s}{2}}
\end{gather*}
Now I will be solving for A,B,C
\begin{align*}
&s=0 \implies \frac{5A}{4} = 0 \implies A = 0. \\
&s =1 \implies \frac{5A}{4} -B + C = 1- e^{\pi / 2} \\
&\hphantom{s =1 \implies \frac{5A}{4}} \therefore C-B = 1- e^{\pi / 2}\\
&s =1 \implies \frac{13A}{4} + B + C = 1- e^{-\pi /2}\\
&\hphantom{s =1 \implies} \frac{13A}{4}+ 2B + C-B = 1-e^{-\pi/2} \\
&\hphantom{s =1 \implies} \therefore B = \frac{e^{\pi/2} - e^{-\pi/2}}{2} = \sinh (\pi/2)\\
&\hphantom{s =1 \implies} \therefore C = 1 - e^{-\pi/2} - \sinh(\pi/2)
\end{align*}
From the Laplace transformation table it is known that
$$ \mathcal{L}[e^{-bt}\sin \omega t] = \frac{\omega}{(s+b)^{2} +\omega^{2}} \ \ \text{and} \ \ \mathcal{L}[e^{-bt}\cos(\omega t)] = \frac{s+b}{(s+b)^2 +\omega^2}, \tag{3}$$
hence I must complete the square in the denominator of Equation 2, yielding
$$ \frac{A}{s^2} + \frac{Bs+C}{s^2 + s + 5/4} = \frac{A}{s^2} + \frac{Bs+ C}{\left(s+\frac12 \right)^2 + 1}.$$
Replacing $B,C$ with the previously found values,
\begin{gather}
\frac{Bs+ C}{\left(s+\frac12 \right)^2 + 1} = \frac{\sinh(\pi/2) (s) + 1 - 3^{-\pi /2} - \sinh (\pi /2)}{\left(s+\frac12 \right)^2 + 1}
\end{gather}
Now i try to express this as a linear combination of the equations in Equation 3, using $\alpha$ and $\beta$:
\begin{gather}
\frac{\sinh(\pi/2) (s) + 1 - 3^{-\pi /2} - \sinh (\pi /2)}{\left(s+\frac12 \right)^2 + 1} = \frac{\alpha (1)}{\left(s+\frac12 \right)^2 + 1} + \frac{\beta(s+ \frac12 )}{\left(s+\frac12 \right)^2 + 1} \\
\implies \beta = \sinh (\pi/2) \\
\implies \alpha + \frac{\beta}{2} = 1 + e^{-\pi/2} - \sinh(\pi/2) \\
\therefore \alpha = 1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2)
\end{gather}
Finally,
$$ Y(s) = 1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2) \left(\frac{1}{(s+\frac12 )^2 +1 }\right) +\sinh (\pi/2) \left(\frac{s+\frac12}{(s+\frac12 )^2 +1}\right) $$
Since $y(t) = \mathcal{L}^{-1} [Y(s)]$,
$$ y(t) = (1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2)) e^{-t/2}\sin(t) + (\sinh(\pi/2)) e^{-t/2}\cos(t).$$
All of this seem quite long and I'm not entirely sure if my reasoning is correct, could someone give me a hint or confirm ?
| $$f(s)= \dfrac 1 {s^2(s^2+s+5/4)}$$
Should be decomposed as
$$f(s)= \dfrac A s + \dfrac B{s^2} +\dfrac {Cs+D} {s^2+s+5/4}$$
And both $A,B$ aren't zero
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Law of Cosines and Heron's Formula in inequalities I had the following question:
Suppose $a$, $b$, and $c$ are non-zero real numbers, and $x$, $y$, and $z$ satisfy the equations
$$
bx + ay = c, cx + az = b, cy + bz = a.
$$
Prove that $-1 < x, y, z < 1$ if and only if $a^4 + b^4 + c^4 < 2 (a^2b^2 + b^2c^2 + c^2a^2)$.
After some thoughts, I factorized the inequality to $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0$, which I believe is related to Heron's Formula of area.
I also rewrote the three equations into $a^2=b^2+c^2-2bcx$, $b^2=a^2+c^2-2acy$ and $c^2=a^2+b^2-2abz$, which I believe is related to the Law of Cosine.
I am guessing that $a$, $b$ and $c$ are sides of triangles and $x$, $y$, $z$ are the cosine value to the angle opposite the sides. However, I wasn't able to do this without assuming the side lengths are positive.
| I hope the following can help.
Since $$0<a^2b^2c^2=ab\cdot ac\cdot bc,$$ we can assume that $$bc>0.$$
Thus, since $-1<x<1$ and by your work $x=\frac{b^2+c^2-a^2}{2bc},$ we obtain:
$$-1<\frac{b^2+c^2-a^2}{2bc}<1,$$ which gives
$$(a+b+c)(b+c-a)>0$$ and $$(a+b-c)(a+c-b)>0$$ and from here
$$(a+b+c)\prod_{cyc}(a+b-c)>0$$ or
$$\sum_{cyc}(2a^2b^2-a^4)>0.$$
Also, let $\sum\limits_{cyc}(2a^2b^2-a^4)>0,$ $abc\neq0$, $x=\frac{b^2+c^2-a^2}{2bc},$ $y=\frac{a^2+c^2-b^2}{2ac}$ and $z=\frac{a^2+b^2-c^2}{2ab}.$
We need to prove that $$\{x,y,z\}\subset(-1,1).$$
Indeed, changing $a$ on $-a$ does not change the condition $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$ and what we need to prove.
Thus, we can assume that $a>0$, $b>0$ and $c>0$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find possible coordinates of a triangle that I know only the sides length I want to find the possible coordinates of each vertex of a triangle of which I know only the sides lengths like (3,4,5).
To find the first edge, I let E1(0;0) and the second E2(3;0). but I have a problem to find the 3rd vertex.How can I find the 3rd ?
I know that we can find the angles with trigonometry but it's only for a right triangle, so I'm a little bit lost.
| Let in triangle $ABC$, we know the side lengths $AB=c$, $AC=b$ and $BC=a$. We look for coordinates of vertices $A$, $B$ and $C$.
So, without loss of generality, we can take $B=(0,0)$, and then $C=(0,a)$ because $BC=a$.
We located to vertices $B$ and $C$ and we need only locate vertex $A$. To do so, suppose the circles $$x^2+y^2=c^2\quad \text{and}\quad (x-a)^2+y^2=b^2$$
then the intersection of this two circles gives the location of $A$.
Note that we have $$x^2+y^2=c^2\quad \text{and}\quad x^2+y^2-2ax+a^2=b^2$$
therefore $$c^2-2ax+a^2=b^2$$
or $$x=\frac{c^2+a^2-b^2}{2a}$$
and thus since $x^2+y^2=c^2$, $$y=\pm\sqrt{c^2-\frac{(c^2+a^2-b^2)^2}{4a^2}}$$
and the coordinates of $A$ is
$$\left(\frac{c^2+a^2-b^2}{2a},\pm\sqrt{c^2-\frac{(c^2+a^2-b^2)^2}{4a^2}} \right)$$
well, you can choose one of the signs $\pm$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Could you check my results for entropies? Could you please check my results to the following exericse:
Let $X,Y$ be independent RVs, each with values in $\{0, 1\}$ and probabilities $(1/2,1/2)$. Compute $I(X; Y )$, $I(X + Y ; X)$ and $I(X + Y ; X − Y )$.
Remark: $I$ stand for the mutual information of $X$ and $Y$. It holds
$$I(X;Y) = H(X)+H(Y)-H(X;Y)$$
, where $H$ denotes the entropy.
With this in mind I computed the resprective entropies as:
$$H(X) = H(Y) = 1$$
$$H(X+Y) = H(X-Y) =1.5$$
$$H(X,Y) = 2$$
$$H(X+Y, X) = H(X+Y, Y) = 2.5$$
$$H(X+Y, X-Y) = 3$$
In the end I get $I(X; Y ) = I(X + Y ; X) = I(X + Y ; X − Y ) = 0$.
Is this O.K.?
| Some sanity checks that we can do without getting into the full calculation:
*
*$H(X+Y,X)$ and $H(X+Y,X-Y)$ should both be at most $2$, because $H(X,Y)$ is $2$, and the pairs $(X+Y,X)$ and $(X+Y,X-Y)$ shouldn't have more entropy than $(X,Y)$: after all, we can compute them from $(X,Y)$!
*In fact, knowing $X+Y$ and $X$, we can compute $Y$, and knowing $X+Y$ and $X-Y$, we can compute $X$ and $Y$. So we should have $H(X+Y,X) = H(X+Y,X-Y) = H(X,Y) = 2$.
*$I(X+Y;X)$ should be positive because some values of $X$ definitely convey information about $X+Y$.
*$I(X+Y;X-Y)$ should be positive because some values of $X-Y$ definitely convey information about $X+Y$. (If $X-Y=0$, we know that $X+Y$ is either $0$ or $2$; if $X-Y=\pm1$, we know that $X+Y=1$.)
For computing $H(X+Y)$ we do have to look at the formula. ($H(X-Y)$ will be the same). $X+Y$ takes on values $0, 1, 2$ with probabilities $\frac14, \frac12, \frac14$, so the surprisal of $X+Y$ takes on values $2, 1, 2$ with probabilities $\frac14, \frac12, \frac14$, and we get $H(X+Y) = 2 \cdot \frac14 + 1 \cdot \frac12 + 2 \cdot \frac14 = 1.5$. So your computation checks out.
Now, we get $$I(X+Y;X) = H(X+Y) + H(X) - H(X+Y,X) = 1.5 + 1 - 2 = 0.5$$ and $$I(X+Y;X-Y) = H(X+Y) + H(X-Y) - H(X+Y,X-Y) = 1.5 + 1.5 - 2 = 1.$$
| {
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"url": "https://math.stackexchange.com/questions/3603523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove :$\tan^{-1}\left(e^{i\theta}\right)=\frac{n\pi}{2}+\frac{\pi}{4}-\frac{i}{2}\ln\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)$
Prove:
$$\tan^{-1}\left(e^{i\theta}\right)=\frac{n\pi}{2}+\frac{\pi}{4}-\frac{i}{2}\ln\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)$$
My Attempt :
Is $\tan^{-1}$ defined for imaginary quantity?
$\tan^{-1}\left(e^{i\theta}\right)=\tan^{-1}\left(\cos\theta + i\sin \theta\right)$
Is there a series expansion for $\tan^{-1}x$?
| Take the derivative to get
$$\frac{d}{d\theta}\left(\tan^{-1}\left(e^{i\theta}\right)\right) = \frac{ie^{i\theta}}{e^{i2\theta}+1} = \frac{e^{i\theta}}{2}\left(\frac{1}{e^{i\theta}-i}-\frac{1}{e^{i\theta}+i}\right)$$
Now integrate the right hand side
$$\int \frac{e^{i\theta}}{2}\left(\frac{1}{e^{i\theta}-i}-\frac{1}{e^{i\theta}+i}\right)\:d\theta = \frac{1}{2i}\ln\left(\frac{e^{i\theta}-i}{e^{i\theta}+i}\right)+C = -\frac{i}{2}\ln\left(\frac{e^{i\theta}-e^{i\frac{\pi}{2}}}{e^{i\theta}+e^{i\frac{\pi}{2}}}\right)+C$$
$$= -\frac{i}{2}\ln\left(\frac{e^{i\left(\frac{\theta}{2}-\frac{\pi}{4}\right)}-e^{-i\left(\frac{\theta}{2}-\frac{\pi}{4}\right)}}{e^{i\left(\frac{\theta}{2}-\frac{\pi}{4}\right)}+e^{-i\left(\frac{\theta}{2}-\frac{\pi}{4}\right)}}\right)+C = -\frac{i}{2}\ln\left(i\tan\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\right)+C$$
and we can absorb the $-\frac{i}{2}\ln\left(i\right)$ into the $+C$. From here, since $e^{i\theta}$ is $2\pi$-periodic, plug in $\theta = 2\pi n$ $(n\in\mathbb{Z})$ on both sides to solve for $C$:
$$\tan^{-1}\left(e^{i2\pi n}\right) = -\frac{i}{2}\ln\left(\tan\left(\frac{2\pi n}{2}-\frac{\pi}{4}\right)\right)+C \implies \frac{\pi}{4} = -\frac{i}{2}\ln(-1)+C$$
then use $\ln(-1) = i(\pi + 2\pi n)$ to get that
$$C = \frac{\pi n}{2} - \frac{\pi}{4}$$
giving us our final answer:
$$\tan^{-1}\left(e^{i\theta}\right) = \frac{\pi n}{2} - \frac{\pi}{4} -\frac{i}{2}\ln\left(\tan\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1-c<1.$$
Thus by AM-GM inequality we have $$\frac{(1-a)+(1-b)+(1-c)}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{(1-a)(1-b)(1-c)}\ge 27.$$
Also by AM-GM inequality we have $$\frac{a+b+c}{3}\ge (abc)^\frac{1}{3}\\ \implies \frac{2}{3}\ge (abc)^\frac{1}{3}\\\implies abc\le \frac{8}{27}\\\implies \frac{1}{abc}\ge \frac{27}{8}.$$
This implies that, we have $$\frac{1}{abc(1-a)(1-b)(1-c)}\ge \frac{27^2}{8}\\ \implies \frac{(abc)^2}{abc(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}\\\implies \frac{abc}{(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}.$$
How to proceed after this?
| Your inequality is the same as
$$\sqrt{(1 - a)(1 - b)}\sqrt{(1 - b)(1 - c)}\sqrt{(1 - a)(1 - c)}
\leq {abc \over 8}$$
By the AM-GM inequality you have
$$\sqrt{(1 - a)(1 - b)} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$\sqrt{(1 - b)(1 - c)} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$\sqrt{(1 - a)(1 - c)} \leq {1 - a + 1 - c \over 2} = {b \over 2}$$
Multiplying these together gives the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the minimum value of $T=a^2+b^2+c^2+d^2$ Given reals $a,b,c,d$ such that $$\left\{\begin{matrix}(a+b)(c+d)=2
& & \\(a+c)(b+d)=3
& & \\ (a+d)(b+c)=4
& &
\end{matrix}\right..$$
Find the minimum value of $ T=a^2+b^2+c^2+d^2.$
I noticed that $(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=2ab+2ac+2ad+2bc+2bd+2cd\leq 3(a^2+b^2+c^2)=3T$
by applying the inequality $x^2+y^2\geq2xy$ $\forall x,y\in \mathbb R$. So $T\geq 3$ . However, equality doesn't occur with this method.
Could you help me solve the problem?
The answer is 7.
| By your work and by AM-GM we obtain:
$$a^2+b^2+c^2+d^2=(a+b+c+d)^2-2-3-4\geq4(a+d)(b+c)-9=7.$$
The equality occurs for example, for $a+d=b+c=2$,
which gives that the equality indeed occurs:
$$d=2-a,$$ $$c=2-b,$$
$$(a+b)(4-a-b)=2$$ and
$$(a-b+2)(b-a+2)=3.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find a complex root of $x^{2021}=x^{2020}+1$ while this complex root also satisfies a quadratic equation with integer coefficients? How to find a complex root of $x^{2021}=x^{2020}+1$ while this complex root also satisfies a quadratic equation with integer coefficients? I have no previous experience in solving complex equations so just have no clue on this kind of question... Is there any useful pattern on this?
| Hint: $x^5 - x^4 - 1 = (x^2 - x + 1) ( x^3 - x - 1)$.
$ $
Hint: $x^3 +1 = (x^2 - x + 1 ) ( x + 1)$.
$ $
Hence, $ x^{6n+k} \equiv x^{k} \pmod{x^2-x+1}$
$ $
Hence $ x^{2021} - x^{2020} - 1 \equiv x^5 - x^4 - 1 \equiv 0 \pmod{x^2-x+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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System of 2 equations Solve the system of equations:
$$
\left\{\begin{matrix}2\sqrt{x+y} = y^2+y-x & \quad(1) \\ x(y^2+y)=(y^4-y^2)^2-2 & \quad(2) \end{matrix}\right.
$$
My attempt:
From $(1)$ I get:
$$(1) \implies (\sqrt{x+y}+y+2)(\sqrt{x+y}-y) = 0$$
So either $\sqrt{x+y}+y+2 = 0 \implies x = y^2+3y+4$ or $\sqrt{x+y}-y = 0 \implies x = y^2-y$. If I replace either of them into $(2)$, It would be very messy since this will become a $8$th-degree polynomial.
How can I solve this problem?
| If we substitute $x=y^2-y$ we can factor $(2)$ as
$$(y^2-2)(y^2+1)(y^4-y^2+1)=0$$
and we get $y=\pm\sqrt2,\pm i,\pm w,\pm w^*$ where $w=e^{i\pi/6}$. Only $y=\sqrt2,i,w,w^*$ lead to equality in $(1)$ under the principal square root. If $x=y^2+3y+4$ we get an irreducible octic in $y$ with two real roots and three complex conjugate root pairs, but all of them do not lead to equality in $(1)$ if the principal square root is taken.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}{4}$
Now I know that values of a and b but how do I get x and y?
Note:
The textbook solution says that if we consider a=1 and b=-2 then x and y are the roots of
$$t^2+t-2=0$$
But shouldn't it be
$$t^2-t-2=0$$
from Vieta's formulas?
I would like a clarification for this or any other solutions would be fine too.
| I like the following way.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, $$8u^3-6uv^2=7$$ and $$4u^2-2v^2+2u+v^2=4,$$
which gives $$v^2=4u^2+2u-4.$$
But since $$v^2\leq u^2,$$ we obtain:
$$4u^2+2u-4\leq u^2$$ or
$$3u^2+2u-4\leq0$$ or
$$\frac{-1-\sqrt{13}}{3}\leq u\leq\frac{-1+\sqrt{13}}{3}.$$
Thus, $$8u^3-6u(4u^2+2u-4)=7$$ or
$$16u^3+12u^2-24u+7=0$$ or $$(2u-1)^2(4u+7)=0,$$ which gives $$u=\frac{1}{2},$$
$$x+y=1,$$ $$xy=-2$$ and we got the answer:
$$\{(-1,2),(2,-1)\}.$$
$$u^2\geq v^2$$ it's
$$\left(\frac{x+y}{2}\right)^2\geq xy$$ or
$$(x-y)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving functional equation $f(x)=3f(x+1)-3f(x+2)$ It is given that a function f(x) satisfy:
$$f(x)=3f(x+1)-3f(x+2)\quad \text{ and } \quad f(3)=3^{1000}$$ then find value of $f(2019)$.
I further wanted to ask that is there some general method to solve such equation. The method that I know to solve such questions is to substitute $x$ with $x+1$ in equation and there by making new equation which is
$$ f(x+1)=3f(x+2)-3f(x+3)$$ then again substitute $x$ with $x+2$ in original equation and make new equation $$f(x+2)=3f(x+3)-3f(x+4)$$ Do this for a couple of times and then on combining the equations, in most of the question we get some relation like f(x) = f(x+a) but that does not work here. Please share your ideas on how to solve such questions.
| $$ \begin{align} f(x)&=f(x-1)-\frac{f(x-2)}{3}, f(3)=3^{1000}\\
f(x)&=f(x-1)-\frac{1}{3}f(x-2)\\
f(x+1)&=f(x)-\frac{1}{3}f(x-1)=(f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}f(x-1)\\
f(x+1) &= \frac{2}{3}f(x-1)-\frac{1}{3}f(x-2)\\
f(x+2)&=f(x+1)-\frac{1}{3}f(x)=(\frac{2}{3}f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}(f(x-1)-\frac{1}{3}f(x-2))\\
f(x+2)&=\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2)\\
f(x+3)&=f(x+2)-\frac{1}{3}f(x+1)=(\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2))-\frac{1}{3}(\frac{2}{3}f(x-1)-\frac{1}{3}f(x-2))\\
f(x+3) &= \frac{1}{9}f(x-1)-\frac{1}{9}f(x-2)
f(x+4)=(\frac{1}{9}f(x-1)-\frac{1}{9}f(x-2))-\frac{1}{3}(\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2))\\
f(x+4)&=(-\frac{1}{9}+\frac{2}{27})f(x-2)=-\frac{1}{27}f(x-2)\end{align}$$
$$\therefore f(x+4)=-\frac{1}{27}f(x-2) \implies f(x+6)=-\frac{1}{27}f(x)$$
$$\implies f(x+6k)=(-\frac{1}{27})^k f(x) \forall k \in N$$
$$f(2019)=f(3+6\cdot 336)=(-\frac{1}{27})^{336} f(3)=\frac{1}{3^{1008}}3^{1000} = \frac{1}{3^8}$$
$$\therefore f(2019) = \frac{1}{3^8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Having problems understanding the correct answer on a classic urn probability question. There are two urns, one urn $U_1$ containing $3$ black balls $B$ and $6$ white balls $B^c$, while the other urn $U_2$, contains $100$ white balls. An urn is selected uniformly at random and then a ball is drawn uniformly at random from the chosen urn. Suppose that the first drawn ball was white and returned to its original urn. What is the probability that another ball drawn from that same urn, will be black?
The answer: $\frac{2}{15}$ makes no sense to me. Even when I see the answer described (so I may need this explained like I am $5$).
I first assumed $P(B^c, U_1) = \frac{1}{3}$ and $P(B|U{_1}) = \frac{1}{3}$. Then the answer would be $\frac{1}{3}x\frac{1}{3}=\frac{1}{9}$.
Second, I reasoned that $P(B^c | U_{1,2}) = \frac{5}{6}$, and drawing from $U_1$ or $P(U_1) = \frac{1}{2}$. So $P(B^c, U{_1}) = \frac{5}{12}$, and $P(B|U{_1}) = \frac{1}{3}$. Then the answer would be $\frac{5}{12}x\frac{1}{3} = \frac{5}{36}$
Answer: you saw that the probability $\frac{5}{6}$ of event $B_1^c$ (first drawn ball was white), is composed of $P(B_1^c, U_1) = \frac{1}{2}$ while $P(B_1^c, U_1^c) = \frac{1}{3}$.
So, by the definition of the conditional probability $P(U_1|B_1^c) = \left(\frac{1}{2}\right)/\left(\frac{5}{6}\right)= \frac{3}{5}$.
This is precisely the probability of using the urn $U_1$ in the experiment described in part (c), so repeating the analysis of part (a) we conclude that the event $B$ of getting black ball when using same urn twice, has the following probability conditioned on $B_1^c$,
$$\begin{align*}
P(B|B_1^c) &=P(B|U_1,B_1^c)P(U_1|B_1^c)+P(B|U_1^c,B_1^c)P(U_1^c|B_1^c) \\
&= 0\times\frac{3}{5}+\frac{1}{3}\times\frac{2}{5} \\
&= \frac{2}{15}.
\end{align*}$$
| If you are looking for an intuitive explanation, you can proceed this way. Suppose that the experiment is run $90$ times. $45$ times we choose urn $2$, and the first ball will be white. $45$ times we choose urn $1$. $30$ times the first ball is white, and $15$ times the first ball is black, but that didn't happen! So we have $75$ possibilities to consider, $30$ where we chose urn $1$ and $45$ where we chose urn $2$. The second ball can only be black if we chose urn $1$. $10$ times it is black and $20$ times it is white. Therefore, the probability that the second ball is balck, given that the first is white is $$\frac{10}{75}=\frac2{15}$$
You can see that this conforms to the definition of conditional probability. $\frac{10}{90}$ is the probability that the first ball is white and the second black, and $\frac{75}{90}$ is the probability that the first ball is white.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613380",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| I use the following trick.
If $f(x)$ is a polynomial, then $f(x)f(-x)$ is a polynomial with only even degree terms, so there exists a polynomial $g(x)$ such that
$$g(x^2)=f(x)f(-x).$$ Then the zeros of $g(x)$ are exactly the squares of the zeros of $f(x)$.
With $f(x)=x^4+px^3+qx^2+rx+1$ we arrive at
$$
g(x)=1 + (2 q - r^2) x + (2 + q^2 - 2 p r) x^2 + (-p^2 + 2 q) x^3 + x^4.
$$
Repeating the dose we look for a polynomial $h(x)$ such that $h(x^2)=g(x)g(-x)$.
Expanding gives us
$$
\begin{aligned}
h(x)&=1 + (4 - 2 q^2 - 4 p r + 4 q r^2 - r^4) x\\
& + (6 + 4 p^2 q - 4 q^2 +
q^4 - 8 p r - 4 p q^2 r + 2 p^2 r^2 + 4 q r^2) x^2\\
& + (4 - p^4 +
4 p^2 q - 2 q^2 - 4 p r) x^3 + x^4.
\end{aligned}
$$
As explained in the comments, because
$$
h(x)=(x-a^4)(x-b^4)(x-c^4)(x-d^4)
$$ the answer is
$$
h(-1)=p^4+2 p^2 r^2-4 p q^2 r+q^4+r^4.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating triple integral over volume common to 2 surfaces In one of my assignments, I have a question which asks you for the triple integral of $z^2$ over the volume common to a given sphere and a cylinder.
I need to transform the equation to cylindrical polar form. Also I am not able to get the limits. In the polar form limit of $\varphi$ would be $2\pi$ but I am not understanding the limit of $\rho$ and $z$.
| The tricky part here is figuring out how the volume is described in cylindrical coordinates (with $\rho$, $\varphi$ and $z$).
The volume is bounded below and above by the sphere. Using the equation of the sphere, $x^2+y^2+z^2=a^2$, we get:
\begin{align*}
&x^2+y^2+z^2=a^2\\
\iff\quad&z^2=a^2-x^2-y^2\\
\iff\quad&z=\pm\sqrt{a^2-x^2-y^2}\\
\iff\quad&z=\pm\sqrt{a^2-\rho^2},
\end{align*}
since $x^2+y^2=\rho^2$. The lower part of the sphere is $z=-\sqrt{a^2-\rho^2}$ and the upper is $z=\sqrt{a^2-\rho^2}$, thus the limits of $z$ will be $-\sqrt{a^2-\rho^2}\leq z \leq\sqrt{a^2-\rho^2}$.
The limits in $\rho$ and $\varphi$ are determined by the cylinder $x^2+y^2=ax$. Using the equation and assuming $a\geq0$, we get
\begin{align*}
&x^2+y^2=ax\\
\iff\quad&\rho^2=a\rho\cos(\varphi)\\
\iff\quad&\rho=a\cos(\varphi)
\end{align*}
The limits of $\rho$ must then be $0\leq\rho\leq a\cos(\varphi)$ (the lower limit $\rho=0$ is by definition of cylindrical coordinates). We also notice that in order for the upper limit to be non-negative (which it must be, otherwise the inequality has no solutions), we require
$$a\cos(\varphi)\geq0\iff\cos(\varphi)\geq0\iff-\frac{\pi}{2}\leq\varphi\leq\frac{\pi}{2}.$$
Notice now that if $a<0$, in order to get a positive radius we want $a\cos(\varphi)\geq0$, which in this case is equivalent to
$$\cos(\varphi)\leq0\iff\frac{\pi}{2}\leq\varphi\leq\frac{3\pi}{2}.$$
Now we have our limits. The differential is transformed by $dxdydz=\rho\ d\rho d\varphi dz$, finally giving us
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{a\cos(\varphi)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\varphi\quad\text{if}\ a\geq0$$
and
$$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\int_0^{a\cos(\varphi)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\varphi\quad\text{if}\ a<0.$$
Notice however that letting $\theta=\varphi-\pi$ in the integral for $a<0$ yields
$$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_0^{a\cos(\theta+\pi)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\theta=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_0^{-a\cos(\theta)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\theta.$$
Since the integrand is odd in terms of $\rho$, and in general for odd $f(x)$ we have
$$\int_0^{-a}f(x)\ dx=-\int_{-a}^0f(x)\ dx=\int_0^a f(x)\ dx,$$
we then obtain
$$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\int_0^{a\cos(\varphi)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\varphi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{a\cos(\theta)}\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}\rho z^2\ dzd\rho d\theta.$$
Therefore the two cases $a\geq0$ and $a<0$ are really the same. We can also simplify further by noticing that the function is even in terms of $z$ and $\varphi$, giving us the final transformed integral as
$$4\int_0^{\frac{\pi}{2}}\int_0^{a\cos(\varphi)}\int_0^{\sqrt{a^2-\rho^2}} \rho z^2\ dzd\rho d\varphi.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be side lengths of a triangle, $a+b+c=1$. Prove that $P=a^3+b^3+c^3+3abc<\frac{1}{4}$.
Let $a,b,c$ be side lengths of a triangle such that $a+b+c=1$. Prove that $$a^3+b^3+c^3+3abc<\frac{1}{4}\,.$$
I solved this question. However, I'd like to know if there's a neater solution that doesn't involve the substitutions that I used.
My solution:
$$P=a^3+b^3+c^3-3abc+6abc$$ $$=(\sum a) (\sum a^2 -\sum ab)+6abc$$ $$=\sum a^2 -\sum ab+6abc$$ $$= (\sum a)^2-3\sum ab+6abc$$ $$=1-3\sum ab+6abc.$$
We have that $abc=4pRr, \sum ab=p^2+r^2+4Rr$ and $p=\frac{1}{2}$ where $p,R,r$ are the semiperimeter, circumradius, and incenter of the triangle, respectively. Plugging these into $P$:
$$P=1-3p^2-3r^2-12Rr+24pRr$$
$$=\frac{1}{4}-3r^2<\frac{1}{4}.$$
| A solution without substitutions.
We need to prove that $$\sum_{cyc}(4a^3+4abc)<(a+b+c)^3$$ or
$$\sum_{cyc}(3a^3-3a^2b-3a^2c+2abc)<0$$ or $$-3\prod_{cyc}(a+b-c)<0,$$ which is obvios.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does this functional equation follow from periodicity? I'm reading a proof where they say that given
$$\phi(x) = \Gamma(x)\Gamma(1-x)\sin \pi x$$
and
$$g(x) = [\log \phi(x)]''$$
then, since $g$ is periodic with period 1, it satisfies the functional equation
$$\frac{1}{4} \left(g\left(\frac{x}{2}\right) + g\left(\frac{x+1}{2}\right)\right) = g(x).$$
I haven't been able to prove this even after expanding out $g(x)$. Is there a quick proof that $g$ satisfies this functional equation?
| I figured it out myself. The only way seems to be by brute force.
$$g(x)=\frac{1}{\phi\left(\frac{x}{2}\right)^2
\phi\left(\frac{x+1}{2}\right)^2}\left[\phi\left(\frac{x}{2}\right) \phi\left(\frac{x+1}{2}\right) \left(\frac{1}{4}
\phi\left(\frac{x+1}{2}\right) \phi''\left(\frac{x}{2}\right)+\frac{1}{4}
\phi\left(\frac{x}{2}\right) \phi''\left(\frac{x+1}{2}\right)+\frac{1}{2}
\phi'\left(\frac{x}{2}\right) \phi'\left(\frac{x+1}{2}\right)\right)-\left(\frac{1}{2}
\phi\left(\frac{x+1}{2}\right) \phi'\left(\frac{x}{2}\right)+\frac{1}{2}
\phi\left(\frac{x}{2}\right)
\phi'\left(\frac{x+1}{2}\right)\right)^2\right]$$
And this simplifies to
$$g(x)=\frac{1}{4}
\left(\frac{f''\left(\frac{x}{2}\right)}{f\left(\frac{x}{2}\right)}-\frac{f'\left(
\frac{x}{2}\right)^2}{f\left(\frac{x}{2}\right)^2} + \frac{f\left(\frac{x+1}{2}\right)
f''\left(\frac{x+1}{2}\right)-f'\left(\frac{x+1}{2}\right)^2}{f\left(\frac{x+1}{2}
\right)^2}\right) = \frac{1}{4}\left[g\left(
\frac{x}{2}\right)+g\left(
\frac{x+1}{2}\right)\right].
$$
Would be interested in knowing if there were a faster and simpler way though...
| {
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3x3 system of equations I have the following system of equations:
\begin{equation}
x+y+kz = 1
\end{equation}
\begin{equation}
x+ky+z = 1
\end{equation}
\begin{equation}
kx+y+z =1
\end{equation}
where $x,y,z \in \mathbf{R}$.
For what values of $k \in \mathbf{R} $
i) the system has a single solution
ii) the system has multiple solutions
iii) the system has no solution
Can anyone help with that please?
I wrote the above system in a matrix form:
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
1 & k & 1 & 1\\
k & 1 & 1 & 1
\end{array}
\end{equation*}
I am not sure how to proceed.. If I subtract the second row from the first (1)-(2) I have the following:
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
0 & 1-k & k-1 & 0\\
k & 1 & 1 & 1
\end{array}
\end{equation*}
Then if I subtract the third row from the first row (1)-(3):
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
0 & 1-k & k-1 & 0\\
1-k & 0 & k-1 & 0
\end{array}
\end{equation*}
| $$\Delta=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-k^3+3k-2=-(k+2)(k-1)^2,$$
$$\Delta_x=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ 1 & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_y=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & 1 & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_z=\left|\left(\begin{array}{cc}1 & 1 & 1\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$ which gives the answer:
i) $k\neq-2$ and $k\neq1$;
ii) $k=1$;
iii) $k=-2$.
| {
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"answer_count": 2,
"answer_id": 1
} |
Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?
$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?
| You have shown that you can see why the limit of the expression should be $3$. Regarding why it shouldn't be $0$, consider how $\displaystyle \frac{x^3-1}{x-1}=\left(x^2+x+1\right)\frac{x-1}{x-1}$ would be evaluated at different $x$-values. At $x=3$, we have $\require{cancel}\displaystyle \left(3^2+3+1\right)\frac{\cancel{2}}{\cancel{2}}$ and likewise, at $x=0.5$, we would have $\require{cancel}\displaystyle \left(0.5^2+0.5+1\right)\frac{\cancel{0.5}}{\cancel{0.5}}$. The pattern should be clear; for any $x\ne1$, the terms in the fraction are nonzero and can cancel with no problem. But this includes all $x$ in the vicinity of $1$, which is exactly what we pass through as we approach $1$. Since a limit is not interested in what happens at $x=1$ but in what happens near $x=1$, and we never need to hit $1$ exactly, we never need to evaluate $\cfrac00$ and it does not affect the answer.
| {
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Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
| If you are able to use theorems from complex analysis (although it is likely you can not) then this integral is easily solved using the residue theorem. Since such an answer might be useful to others, I'll put it here for posterity.
First, note that
$$\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Define $\gamma$ to be the counter-clockwise path traveled on the upper-plain semicircle of radius $R$. That is, $\gamma=\{Re^{i\theta}:0\leq \theta\leq \pi\}$. Turned into a line integral, this is
$$\int_\gamma \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\int_{0}^{\pi} \frac{(Re^{i\theta})^2}{((Re^{i\theta})^2+a^2)((Re^{i\theta})^2+b^2)} Rie^{i\theta}d\theta$$
However, since the numerator has $R^3$ and the denominator has $R^4$, we know that
$$\lim_{R\to\infty}\left(\int_{0}^{\pi} \frac{(Re^{i\theta})^2}{((Re^{i\theta})^2+a^2)((Re^{i\theta})^2+b^2)} Rie^{i\theta}d\theta\right)=0$$
This implies
$$\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
$$=\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_\gamma \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Putting these three path integrals together, we get a simple closed curve (call it $\beta$) which starts at $(-R,0)$, goes to $(R,0)$, and then follows $\gamma$ back to $(-R,0)$. This implies the integral is equal to
$$=\lim_{R\to\infty}\left(\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Since this is a simple, closed, positively-oriented curve, the residue theorem applies. Now, for $R>\text{max}\{a,b\}$ the function
$$\frac{z^2}{(z^2+a^2)(z^2+b^2)}$$
has two singular points inside $\beta$ at $ia$ and $ib$. To calculate the residues at these points, we simply need
$$\text{Res}(ia)=\lim_{z\to ia} (z-ia)\frac{z^2}{(z^2+a^2)(z^2+b^2)}=\lim_{z\to ia} (z-ia)\frac{z^2}{(z-ia)(z+ia)(z^2+b^2)}$$
$$=\lim_{z\to ia}\frac{z^2}{(z+ia)(z^2+b^2)}=\frac{-a^2}{2ia(b^2-a^2)}=\frac{-a}{2i(b^2-a^2)}$$
For $ib$, we get a residue of
$$\text{Res}(ib)=\frac{-b}{2i(a^2-b^2)}$$
Then the residue theorem states
$$\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=2\pi i\left( \text{Res}(ia)+\text{Res}(ib)\right)$$
$$=2\pi i\left(\frac{-a}{2i(b^2-a^2)}+\frac{-b}{2i(a^2-b^2)}\right)=\frac{\pi}{a+b}$$
We conclude
$$\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\lim_{R\to\infty}\left(\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
$$=\lim_{R\to\infty}2\pi i\left( \text{Res}(ia)+\text{Res}(ib)\right)=\lim_{R\to\infty}\frac{\pi}{a+b}=\frac{\pi}{a+b}$$
| {
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"source": "stackexchange",
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If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?
| By C-S $$\sum_{cyc}\frac{a}{b+c-a}=\sum_{cyc}\frac{a^2}{a(b+c-a)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}a(b+c-a)}\geq3,$$
where the last inequality it's just $$\sum_{cyc}(a-b)^2\geq0.$$
| {
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Umbilical points of the ellipsoid. The following is Exercise (11) of Chapter 3 of Curves and Surfaces, 2nd edition, by Montiel and Ros:
Determine the umbilical points of the ellipsoid of equation
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
$$
where $0 < a < b < c$.
My question is:
Why the umbilical points must have $y = 0$? Why can't it happen that $x = 0$ or $z = 0$?
The book provides a solution, that I worked out as follows (without quite understanding the conclusion):
Denote the ellipsoid by $S$. We begin by noting that $S$ is the inverse image of a regular value of the function
$$
f(x, y, z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}.
$$
Indeed, if $f(x, y, z) = 1$ then
$$
\nabla f(x, y, z) = 2 \left (\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right) \neq 0.
$$
It is clear that $S = f^{-1}(\{1\})$.
The gradient gives us a Gauss map defined on $S$:
$$
N(x, y, z) = \frac{\nabla f(x, y, z)}{|\nabla f (x, y, z)|}.
$$
Then, there exists a nonvanishing differentiable function $h$ defined on $S$ such that
$$
h(x, y, z) N(x, y, z) = \left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right).
$$
Using the product rule and the fact that the right-hand side is linear, for all $p = (x, y, z) \in S$ and $v = (v_1, v_2, v_3) \in T_pS$ we have that
$$
(dh)_p(v)N(p) + h(p)(dN)_p(v) = \left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right).
$$
The point $p$ is umbilical if and only if $(dN)_p$ is a multiple of the identity. We claim that this happens if and only if the left-hand side vanishes after scalar multiplication by $N(p) \wedge v$. It is clear that the first term vanishes, since $(N(p) \wedge v) \perp N(p)$. Suppose $(dN)_p$ is a multiple of the identity. Then $h(p)(dN)_p(v) \in \langle v \rangle$ and the second term vanishes as well. On the other hand, assume that
$$
\left(h(p) (dN)_p(v)\right) \cdot (N(p) \wedge v) = 0.
$$
Then $(dN)_p(v)$ lies in the plane spanned by $N(p)$ and $v$. But since $(dN)_p(v) \in T_pS$, if follows that it is a multiple of $v$, and therefore the claim follows.
From the previous paragraph, we conclude that
$$
\left( \frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right) \cdot (N(p) \wedge v) = 0,
$$
that is, $\left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right)$ lies on the plane spanned by $N(p)$ and $v$. Since these three vectors are linearly dependent, we have that
$$
\begin{vmatrix}
\displaystyle \frac{v_1}{a^2} & \displaystyle \frac{v_2}{b^2} & \displaystyle \frac{v_3}{c^2} \\
v_1 & v_2 & v_3 \\
\displaystyle \frac{x}{a^2} & \displaystyle \frac{y}{b^2} & \displaystyle \frac{z}{c^2}
\end{vmatrix}
= 0, \quad \text{ if } \quad N(p)\cdot v = 0.
$$
But the determinant is a quadratic form in the variables $v_i$ which vanishes on a plane:
$$
q(v) = \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right)v_1 v_2 + \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)v_2 v_3 + \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)v_1 v_3
$$
Hence, the matrix of this quadratic form is
$$
A = \frac12 \begin{bmatrix}
0 & \displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) &
\displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)\\
\displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) & 0 &
\displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)\\
\displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right) & \displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right) & 0
\end{bmatrix}.
$$
So far, so good. Now, the book argues that since this quadratic form vanishes on a plane, the determinant of $A$ must be zero.
Why is this so?
Moving forward, $\det A = 0$ if $x =0$ or $y = 0$ or $z = 0$. The book claims that it can only be $y = 0$. Why?
I have found this similar question, but I did not quite understand what is going on:
Umbilical points of Ellipsoid alternate method
Thanks in advance and kind regards
| I think I got this.
Suppose $x = 0$. From $N(p) \cdot v = 0$ we have that
$$
v_3 = - \frac{y}{b^2} \frac{c^2}{z}v_2.
$$
Then
$$
q(v) = \left( \frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) - \frac{y}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \right) v_1 v_2 = 0 \quad \forall v_1, v_2 \in \Bbb{R}.
$$
Then
$$
\frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) = \frac{y^2}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \implies \frac{z^2}{c^2} = \frac{y^2}{b^2} \left( \frac{a^2 - c^2}{b^2 - a^2} \right) < 0,
$$
which is absurd.
If $y = 0$, by the same argument we arrive at
$$
\frac{x^2}{a^2} = \frac{z^2}{c^2} \left( \frac{b^2 - a^2}{c^2-b^2} \right),
$$
which is feasible.
If in turn $z = 0$, then
$$
\frac{x^2}{a^2} = \frac{t^2}{b^2} \left( \frac{a^2 - c^2}{c^2-b^2} \right) < 0,
$$
which is also absurd. Hence, the only possible option is $y = 0$.
Any comments will be appreciated.
| {
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Pair of tangents from point $(2\sqrt2,1)$ to hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$
From a point $(2\sqrt2,1)$ a pair of tangents are drawn to $$\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$$ which intersect the coordinate axes in concyclic points. If one of the tangents is inclined at an angle of $\arctan\frac{1}{\sqrt{2}}$ with the transverse axis of the hyperbola, then find the equation of the hyperbola and also the circle formed using the concyclic points.
My Attempt
A tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with slope $m$ is given by
$y=mx±\sqrt{a^2m^2-b^2}$
Plugging $(2\sqrt2,1)$ in this equation, I get
$m^2(8-a^2)+m(-4\sqrt2)+(1+b^2)=0$
This equation gives two values of $m$
$m_1=\frac{1}{\sqrt2}$ and $m_2$
$m_1+m_2=\Large\frac{4\sqrt2}{8-a^2}$
And
$m_1m_2=\Large\frac{1+b^2}{8-a^2}$
How do I proceed further? I know we have to use the fact that the points at which the tangents intersect the axes are concyclic. How do I apply this and get the required result or is there another easy way to do this?
| Given the point $(2\sqrt2,1)$ and the slopes $\frac1{\sqrt2}$, $m$, the equations of the two tangent lines are
$$ y-1 =\frac1{\sqrt2}( x-2\sqrt2), \>\>\>\>\>y-1 = m(x-2\sqrt2)$$
which intersect the axes at $A(\sqrt2,0)$, $B(0,-1)$ and $C(2\sqrt2-\frac1m, 0)$, $D(0, 1-2\sqrt2m)$, respectively. Given that $A$, $B$, $C$ abd $D$ are concyclic, we have $\angle ACB = \angle ADB=\theta$, i.e.
$$\tan\theta=\frac {BO}{CO}=\frac {AO}{DO}
\implies \frac1{2\sqrt2-\frac1m}=\frac{\sqrt2}{2\sqrt2m-1}$$
which leads to $m=\sqrt2$. The tangent line equations to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by
$$(y-m x)^2=a^2m^2-b^2$$
Substitute the point $(2\sqrt2,1)$ and the slopes $m=\frac1{\sqrt2},\>\sqrt2$
into the equations to get
$$2a^2-b^2=9,\>\>\>\>\>\frac12a^2 -b^2 = 1$$
Solve to obtain $a^2=\frac{16}{3}$, $b^2=\frac53$ and the equation of the hyperbola
$$\frac{3x^2}{16}-\frac{3y^2}{5}=1$$
From the known axis intersections, the cyclic circle is obtain as,
$$(x-\frac5{2\sqrt2})^2+(y+2)^2=\frac{33}8$$
| {
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Restrictions on a Function If $f(x) =$
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
then for how many values of $x$ is $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!
| Just do it in cases:
Case 1: $f(x) < -4$ then $f(f(x)) = 5$ means
$f(x) + 3 = 5$ so $f(x) =2$ and that's a contradiction.
So it must be that:
Case 2: $f(x) \ge -4$ so $f(f(x)) = 5$ means
$f(x)^2 -4 = 5$
$f(x)^2 = 9$
$f(x) = \pm 3$.
Case 3: $x < -4$ so $f(x) = \pm 3$ means $x+3\pm 3$ meas $x=-6, 0$ but only $-6< -4$ so $x = -6$.
Case 4: $x \ge -4$ so $f(x) = \pm 3$ means $x^2 -4 =\pm 3$ so $x^2 = 1, 7$.
So $x = \pm 1, \pm \sqrt 7$. All of those are acceptable as the are all $< -4$
So there are five solutions.
a) $x=-6$ and $f(f(-6)) = f(-6+3) = f(-3)= (-3)^2-4 = 9-4=5$.
b)c) $x =\pm 1$ and $f(f(\pm 1))= f((\pm 1)^2-4)=f(-3)= 5$.
d)e) $x =\pm \sqrt7$ and $f(f(\pm \sqrt 7)) = f((\pm \sqrt 7)^2-4)=f(7-4)=f(3)=3^2-4 = 9-4 = 5$.
| {
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A Question Lebesgue Dominated Convergence Theorem I have a question about the application of Lebesgue Dominated Convergence Theorem.
$\lim _{n \rightarrow \infty} \int_{0}^{1}\left(1+n x^{2}\right)\left(1+x^{2}\right)^{-n} d x=?$
Firstly i have a reference about this question http://www.ma.man.ac.uk/~mdc/old/341/solutions3.pdf
By using this reference my solution is
Starting from $(1+n x^2)(1+x^2)>1+(n+1) x^2$ to see that $1+n x^2>\frac{1+(n+1) x^2}{1+x^2}$ and so;
$$
\frac{1+n x^2}{(1+x^2)^{n}}>\frac{1+(n+1) x^2}{(1+x^2)^{n+1}}
$$
For $x=0$ then all terms in the sequence of (fn(x)) equal 1 so the limit is $1 .$ If $0<x<1$ and $x=1$ we start with the observation that the binomial expansion gives $(1+x^2)^{n} \geq 1+n x^2+\frac{n(n-1)}{2} x^{4}$ and so
$$
\frac{1+n x^2}{(1+x^2)^{n}} \leq \frac{1+n x^2}{1+n x^2+\frac{n(n-1)}{2} x^{4}} \rightarrow 0
$$
as $n \rightarrow \infty$. Thus the limit is 1 if $x=0$ and 0 elsewhere, that is, 0 a.e. $(\mu)$
We could choose the dominating function to be the n = 3 term, $h(x)=(1+3 x^2) /(1+x^2)^{3}$
so $h$ is integrable. But also since the sequence of functions is decreasing each function, at least for $n \geq 3,$ is integrable. Hence, using the Dominated convergence Theorem we can justify the interchange in
$$
\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{1+n x^2}{(1+x^2)^{n}} d x=\int_{0}^{1} \lim _{n \rightarrow \infty} \frac{1+n x^2}{(1+x^2)^{n}} d x=\int_{0}^{1} 0 dx=0
$$
I want to ask about my solution . is it right or not? Or is there any missing solution? Thanks for your support.
| You can make this easier: since $(1+x^2)^n \ge 1 +nx^2,$ we have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2} =1.$$
So the constant function $1,$ which is certainly integrable on $[0,1],$ is a dominating function for your sequence.
As you said, we also have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2+[n(n-1)]/2\cdot x^4} \to 0$$
on the interval $(0,1].$ By the DCT we're done.
| {
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How do I find $ \int \frac{6}{5\sin^{2}x+4}dx $? One of the homework questions I am working on asks to show that $ \frac{\partial }{\partial x} \left(\tan^{-1}\frac{3\tan x}{2}\right) = \frac{6}{5\sin^{2}x+4} $ which I am able to do. However, I wonder how I should approach the reverse version of this question if it simply asks me to solve the integral
$ \int \frac{6}{5\sin^{2}x+4}dx $
My first instinct was to change $ \sin^{2}x $ into something that involves cos2x but I did not get anywhere since I had cos2x in the denominator and I did not know what to do next.
Could someone please show me how to solve this integral and more importantly how should I approach integrals in similar form like this?
Thank you so much for the help!
| \begin{align*}
I&=\int \frac{6}{5\sin^2x+4}\,\mathrm dx\\
&=\int \frac{6}{9\sin^2x+4\cos^2x}\,\mathrm dx\\
&=\int \frac{6}{9\tan^2x+4}\frac{1}{\cos^2x}\,\mathrm dx\\
&=(let\ t = \tan(x) )\quad\int \frac{6}{9t^2+4}\,\mathrm dt\\
&=(let\ p = \frac {3}{2}t)\quad\int \frac{1}{p^2+1}\,\mathrm dp\\
&=\arctan(p)+C\\
&=(p = \frac{3}{2}\tan(x))\quad \arctan(\frac{3}{2}\tan(x))+C\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Vieta's formulas for quadratic equation problem I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it.
Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divide roots by $a$.
For example:
$$-6x^2+7x+5=0$$
I solve
$$
y^2 + 7y -30 = 0\\
y_1=-10\\
y_2=3
$$
And then divide the roots:
$$
x_1=\frac{-10}{-6}=\frac{5}{3}\\
x_2=\frac{3}{-6}=-\frac{1}{2}
$$
Which gives me a correct answer. But I want to know why it's so. For now what I figured out is only that:
$$
\text{for } a\neq 0 \text{ :}\\
ax^2 + bx + c = 0 \Leftrightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0\\
x_1 + x_2 = -\frac{b}{a} \Leftrightarrow a(x_1 + x_2) = -b\\
x_1 \cdot x_2 = \frac{c}{a} \Leftrightarrow a(x_1 \cdot x_2) = c\\
$$
$$
y^2 + by + ca = 0\\
y_1 + y_1 = -b\\
y_1 \cdot y_2 = ca \Leftrightarrow c = \frac{y_1 \cdot y_2}{a}\\
$$
$$
a(x_1 + x_2) = y_1 + y_1 \Leftrightarrow x_1 + x_2 = \frac{y_1}{a} + \frac{y_2}{a}\\
a(x_1 \cdot x_2) = \frac{y_1 \cdot y_2}{a} \Leftrightarrow x_1 \cdot x_2 = \frac{y_1}{a} \cdot \frac{y_2}{a}\\
$$
Any ideas?
| By the quadratic formula the roots of $aX^2+bX+c$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
The roots of $Y^2+bY+ca$ are $y=\frac{-b\pm \sqrt{b^2-4ca}}{2}$...
We see that $x=\frac1a y$ ...
| {
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"url": "https://math.stackexchange.com/questions/3642361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Kernel and image of a product of two rectangular matrices I cannot find any concise and effective answer to the following problem :
Problem : given two matrices $A$ and $B$ of size $3\times 2$ and $2\times 3$ such that $AB = \left(\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 2 \end{matrix}\right)$ find the kernel and the image of $AB$ and $BA$.
What I have found so far : Ker$(AB)=$ Vect$\left(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right)$ and Im$(AB) = $ Vect$\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right)$. Furthermore $AB \in S_3(\mathbb{R})$ so AB is diagonalizable, and I have $\chi_{AB}=X(X+1)(X-3)$ so :
$\mathrm{Sp}(AB) = \{-1,0,3 \}$ and $E_{-1}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right) \quad E_{0}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right) \quad E_{3}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right) $
Then I had the idea of saying if $X$ is an eigenvector of $AB$ then there exists $\lambda \in \{-1,0,3\}$ such as $ABX = \lambda X$ so $BABX = \lambda BX$ and thus $BX$ is either null or an eigenvector of $BA$ for the eigenvalue $\lambda$. Furthermore $BA$ is a square matrix of size 2 so $BA$ cannot have $3$ distinct eigenvalues, it has at most $2$ eigenvalues that can be found within $\{-1,0,3\}$.
I have stopped here. I don't expect any complete answer, just any idea will be welcomed.
| Notice that rank($AB$) $=2$
This implies $A$ and $B$ are full rank, i.e. $B$ is surjective and $A$ is injective.
Since $A$ is injective, Ker $(AB) = $ Ker $B$
Since $B$ is surjective, Im $(AB) = $ Im $A$
Since $\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right)$ are linearly independent, we see that
Ker $B$ $\cap$ Im $A =$ Ker $AB$ $\cap$ Im $AB = 0$
Thus,
Ker $BA = $ Ker $A = 0$ so that $BA$ is invertible since it is square
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding residue of complex function, the result is different when using Laurent series and residue theorem. Find the residue of
$$f(z)=\dfrac{z^3+2z+1}{(z-1)(z+3)}$$
on simple pole $z=1$.
If I using residue theorem, I have
\begin{align}
\underset{z=1} {\operatorname{Res}} f(z) =
\lim\limits_{z\to 1} (z-1)\dfrac{z^3+2z+1}{(z-1)(z+3)}=\dfrac{1+2+1}{4}=1.
\end{align}
If I using Laurent series method, I have
\begin{align}
f(z)&=\dfrac{z^3+2z+1}{(z-1)(z+3)}\\
&=(z-2)+\dfrac{9z-5}{(z-1)(z+3)}\\
&= -1+(z-1)+\dfrac{9(z+3)-32}{(z-1)(z+3)}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z+3}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z-1+4}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\cdot\dfrac{1}{1+\dfrac{4}{z-1}}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-1)^n \left(\dfrac{4}{z-1}\right)^n\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-4)^n \left(z-1\right)^{-n}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}+\sum\limits_{n=0}^{\infty} (-32)(-4)^n \left(z-1\right)^{-n-2}.
\end{align}
Now I have coefficient of $(z-1)^{-1}$ is $9$, so we can conclude
\begin{align}
\underset{z=1} {\operatorname{Res}} f(z) =9.
\end{align}
My question
When I using residue theorem and Laurent series method, why the result is distinct? What my mistake in my work?
| The "Laurent series method" had an incorrect conclusion going from the fourth to the fifth lines. There was a still an order $1$ term on the far right, it just still had to be pulled out.
We can calculate the Laurent series another way. From partial fraction decomposition we have that
$$\frac{4}{(z-1)(z+3)} = \frac{1}{z-1}-\frac{1}{z+3}$$
which means we can rewrite the function as
$$f(z) = \frac{z^3+2z+1}{4(z-1)} - \frac{z^3+2z+1}{4(z+3)}$$
This time the term on the far right has no singularity at $z=1$ so it will not contribute to the residue. We will only focus on the term on the left.
Next, shift the variables in the numerator to the desired center:
$$\frac{z^3+2z+1}{4(z-1)} = \frac{(z-1)^3+3(z-1)^2+5(z-1)+4}{4(z-1)}$$
$$ = \frac{1}{4}(z-1)^2+\frac{3}{4}(z-1) + \frac{5}{4} + \frac{1}{z-1}$$
which has a residue of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to find directional derivative I am trying to find the directional derivative of the following problem
$F(x,y,z) = 4x^2+ 3y−3xz+ 2z^2$
at the point $(2,1,2)$ in the direction $i−k$;
I worked out the derivatives of $F(x,y,z)$ as
$f_x = 8x -3z$
$f_y = 3$
$f_z = -3x+4z$
But I don't know to do next; can someone explain how to find the directional derivative here please?
Thank you
| Let the gradient $\nabla F$ of $F$ be
$$\nabla F = (f_x, f_y, f_z).$$
Then the gradient evaluated at $(2,1,2)$,
$$\nabla F(2, 1, 2) = (8 \cdot 2 - 3 \cdot 2, 3, -3 \cdot 2 + 4 \cdot 2) = (10, 3, 2)$$
can be dotted with the normalized direction $\frac{1}{\sqrt{2}}(1, 0, -1) = \frac{1}{\sqrt{2}}(i - k)$ to arrive at the directional derivative you are searching for
$$ \nabla F(2, 1, 2) \bullet \frac{1}{\sqrt{2}}(1, 0,-1) = (10, 3, 2)\bullet \frac{1}{\sqrt{2}} (1, 0, -1) = \frac{1}{\sqrt{2}}(10 - 2) = \frac{8}{\sqrt{2}}.
$$
Using the method described by J.W. Tanner, we notice that the same conclusion is found by computing it simply as a weighting of the directional derivatives in directions $x, y$ and $z$,
$$
\begin{align}
\frac{1}{\sqrt{2}}\cdot f_x(2, 1, 2) + 0 \cdot f_y(2, 1, 2) - \frac{1}{\sqrt{2}}\cdot f_z(2,1,2) &= \\ \frac{1}{\sqrt{2}}\cdot 10 + \frac{0}{\sqrt{2}}\cdot 3 - \frac{1}{\sqrt{2}} \cdot 2 &= \\\frac{10 - 2}{\sqrt{2}} &= \\\frac{8}{\sqrt{2}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$
After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??
| As we can see that the term $9(abc)^{\frac{1}{3}}$ is difficult to deal with, we consider the following inequality $$-\sqrt{(a^2-ab+b^2)} -\sqrt{(b^2+c^2-bc)} -\sqrt{(a^2+c^2-ac)} +4(a+b+c)\ge 9(abc)^{1/3}.$$
Then we consider $$\frac{a+b}{2}-\sqrt{(a^2-ab+b^2)}=\frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}},$$the numerator is non-positive, and the denominator $$\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}\ge \frac{a+b}{2}+\frac{a+b}{2}.$$So we have $$\frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}}\ge \frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{a+b}=\frac{3ab}{a+b}-\frac{3}{4}(a+b).$$ Hence $$-\sqrt{(a^2-ab+b^2)} -\sqrt{(b^2+c^2-bc)} -\sqrt{(a^2+c^2-ac)} +4(a+b+c)\ge \sum_{cyc} \frac{3ab}{a+b}+\frac{3}{2}\sum a.$$Then you just need to use AM-GM inequality to prove it is greater than $9(abc)^{1/3}$.
One of my schoolmate of lower grade also finds a solution(before me).With the permission of him, I post his solution as follows:$$2(a+b)-\sqrt{a^2+b^2-ab}=\frac{a+b}{2}+\frac{1}{2}\sqrt{a^2+b^2-ab}+\frac{3(a+b)}{2}-\frac{3}{2}\sqrt{a^2+b^2-ab}$$$$=\frac{a+b+\sqrt{a^2+b^2-ab}}{2}+\frac{3}{2}\frac{3ab}{a+b+\sqrt{a^2+b^2-ab}}\ge 3\sqrt{ab}$$
So sum up and use AM-GM, we can finish the proof.
| {
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"url": "https://math.stackexchange.com/questions/3652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\log_{a} c + \log_{b} c = \log_{a+b} c$ if and only if $1 + \log_{b} a = \log_{a+b} a$ If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$
Prove it!
I have looked at the solution but it is not clear for me.
We will prove that if $$\log_{a} c + \log_{b} c = \log_{a+b} c$$
than it is
$$1 + \log_{b} a = \log_{a+b} a$$
$$\log_{a} c + \log_{b} c = \frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$
So this is only thing that is not clear for me how is this equal.
$$\frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$
| Note that
$$log_b (a)= \frac{ln (a)}{ln(b)}$$
where
$ln(a)$ is the natural log of $a$.
Any arbitrary base can be assumed instead of $e$.
So, if
$$log_a (c)+ log_b (c)= log_{a+b} (c) $$
$$\implies\frac{ln(c)}{ln(a)}+\frac{ln(c)}{ln(b)}=\frac{ln(c)}{ln(a+b)}$$
$$\implies\frac{1}{ln(a)}+\frac{1}{ln(b)}=\frac{1}{ln(a+b)}\ ,\ \ assuming \ \ c\neq1$$
$$\implies1+\frac{ln(a)}{ln(b)}=\frac{ln(a)}{ln(a+b)}$$
$$\implies1+log_b(a)=\frac{ln(a)}{ln(a+b)}$$
$$\implies1+log_b(a)=log_{a+b}(a)$$
The steps are perfectly reversible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
| Just synthetic divide and subtract and add till you reduce to something manageable.
$6x^3 -x^2 -3x + 2010 = $
$6x^3+ 2x^2 - 3x^2 - 3x + 2010=$
$2x(3x^3 + x^2) - 3x^2 - 3x+2010=$
$2x - 3x^2 - 3x + 2010=$
$-3x^2 - x + 2010=$
$-(3x^2 + x) +x - x +2010=$
$-1 + 0 + 2010=$
$2009$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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At what values of the parameter a the antiderivative of the function has at most one local minimum. Function below. At what values of the parameter a the antiderivative of the function has at most one local minimum?
Function:
(x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3))
or you can look here - 1
But what to do next?
I understant that I need to consider f(x) as first derivative and then look when second derivative f'(x)>0 - which means that it's local minimum.
Thank you
| let $F(x)$ be the anti derivative of the given function. Then
$$F'(x) = (x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3)) $$
You can see clearly that $F'(x)=0$ at x = 0.
Now when is F''(x)>0 ?
So,F"(x)>0 only when the function is "increasing" at $x$.
https://www.desmos.com/calculator/mtjvcgkkov
If you go on to this graph and play with $a$.
You will see how the graph behaves at $x=0$, first, the value of the function is $0$ for all values of 'a' at $x=0$. Secondly the function becomes increasing or in other words $F''(x)>0$ for $a<0$.
$$$$
The more conventional way is harder as
$$F''(x) = e^{\frac{\sin (x)}{x^2+3}} \left(\frac{\left(x^2-3 x+2\right) x (x-a)
\left(\left(x^2+3\right) \cos (x)-2 x \sin
(x)\right)^2}{\left(x^2+3\right)^4}+\frac{2 \left(a \left(-3 x^2+6
x-2\right)+x \left(4 x^2-9 x+4\right)\right) \left(\left(x^2+3\right) \cos
(x)-2 x \sin (x)\right)}{\left(x^2+3\right)^2}+\frac{\left(x^2-3 x+2\right)
x (a-x) \left(\left(x^4+15\right) \sin (x)+4 x \left(x^2+3\right) \cos
(x)\right)}{\left(x^2+3\right)^3}-6 (a+3) x+6 a+12 x^2+4\right)$$
and
$$F''(0) = -2a$$
Which is only positive for $a<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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ODE: $ \biggl( e^x + \frac{A\sin x}{y} \biggl) dx + \biggl(\frac{B\cos x}{y^2} + (B-1)e^x \biggl)dy =0$ Question says equation is exact, find A and B.
$$ \biggl( e^x + \frac{A\sin x}{y} \biggl) dx + \biggl(\frac{B\cos x}{y^2} + (B-1)e^x \biggl)dy =0$$
$$\frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $$
$$-\frac{A\sin x}{y^2} = \frac{-B\sin x}{y^2}+(B-1)e^x $$
$$-A = -B+(B-1)e^x $$
if I give $B=1$ equation becomes $A = B$, where I do wrong calculation ?
| At your last step, it should be
$$\frac{(B-A)\sin(x)}{y^2} =(B-1)e^x$$
which holds when
$$\begin{cases}
B-A=0\\ B-1=0
\end{cases}$$
and therefore $B=1$ and $A=B=1$. Then
$$dF=\biggl( e^x + \frac{\sin(x)}{y} \biggl) dx + \biggl(\frac{\cos(x)}{y^2} \biggl)dy$$
where $F(x,y)=e^x-\frac{\cos(x)}{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt forward:
$tan^2(x) + 1 = {1\over cos^2(x)} = y \text{ and } \\1-cos^2(x) = ycos^2(x) - cos^2(x) = cos^2(x)(y-1)$ combined we have:
$$5y{(5^{y-1} - 1) \over y-1}$$
as $y \rightarrow 1$ $5y{(5^{y-1} - 1) \over y-1}$ is not defined. Since both, nominator and denominator are $0$ I tried L'Hopital but ended at:
$5 5^{y-1} + 5y(y-1)5^{y-2}-5$ with $\lim_{y \rightarrow 1} = 5 + 0 - 5 = 0$ and $(y-1)' = 1$
Here I have to stop with no solution. I have also tried to use the quotient rule to differentiate the expression which did not get me anywhere.
| As $x\to 0$,
$$1 - \cos^2(x) = \sin^2(x) \sim x^2$$
Use this to simplify thigs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$? Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, but it doesn't work.
Since the fractions available have odd numerator and denominator, the same will be true of any product, so even integers aren't reachable. Odd integers can be reached, however.
Let $f(q)$ be the fraction $\frac{4q-1}{2q+1}$. Then $f(1)=1$, $f(4)f(7)=3$ and $3 f(4)=5$. So $1,3,5$ are obtainable.
Suppose now that $m\ge 2$ and we've obtained all odd numbers up to and including $4m-3$. Since $2m+1\le4m-3$ we can use $2m+1$ to obtain
\begin{gather*}
4m-1 = (2m+1) f(m),\\
4m+1=(2m+1) f(3m+1).
\end{gather*}
Hence $4m-1$ and $4m+1$ are obtainable as well, so having obtained $5$ we can then reach all odd positive integers.
Any help appreciated!
| For integer $a ≥ 0$ define $f(a) = (4a + 1) / (2a + 1)$.
Let $g_k(a) = f(a) \cdot f(2^1a) \cdot f(2^2a) \cdot … \cdot f(2^{k-1}a)$. $g_k(a)$ is a telescopic product, and $g_k(a) = (2^{k+1}a + 1) / (2a+1)$.
We have $g_k((b-1) / 2) = (2^k b - 2^k + 1) / b$.
We have $g_k((cy - 1) / 2) = (2^k cy - (2^k - 1)) / cy = 2^k - (2^k - 1) / cy$.
Pick an odd x > 1. Find the largest k such that x+1 is divisible by $2^k$, so $k ≥ 1$. Let $y = (x + 1) / 2^k$, then y is odd, $y < x$, and $2^k y = x + 1$.
Let $c = 2^k - 1$, then $y \cdot g_k((cy - 1) / 2) = 2^k y - 1 = (x + 1) - 1 = x$, so x is formed by multiplying y by k values of the form $f(a)$.
Together with 1 = f(0) this shows by induction that every odd number is a product of numbers of the form f(a). I think the number of values f(a) used according to this proof is k whenever $2^{k-1}<x<2^k$. Not sure if more or fewer values could be used.
| {
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Prove $\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$ with complex numbers (roots of unity) I want to prove that
Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$
I am able to do this by using $w^n+w^{-n}=2\cos \frac{2nπ}{9}$ and hence expanding the LHS but it is laborious and, so, I was wondering is there is a more efficient or elegant method of proving this this identity using the ninth root of unity.
Thanks very much
| Start out as by noticing that
$$
\omega = cis(\frac{2\pi}{9}),\quad \omega^{8}=cis(\frac{-2\pi}{9})\quad \\
\omega^2 = cis(\frac{4\pi}{9}),\quad\omega^7=cis(\frac{-4\pi}{9})
\\ etc.
$$
Now see,
$$ \omega\ + \omega^8 = 2\ cos(\frac{2\pi}{9}) \\
\omega^2\ + \omega^7= 2\ cos(\frac{4\pi}{9})\\
\omega^3 + \omega^6 = 2\ cos(\frac{6\pi}{9})\\
\omega^4\ + \omega^5= 2\ cos(\frac{8\pi}{9})\\
$$
If we multiply these we get:
$$
[2\ cos(\frac{2\pi}{9})][2\ cos(\frac{4\pi}{9})][2\ cos(\frac{6\pi}{9})][2\ cos(\frac{8\pi}{9})] = (\omega\ + \omega^8)(\omega^2\ + \omega^7)(\omega^3 + \omega^6)(\omega^4\ + \omega^5)
$$
If we carefully pull out the $\omega$s, we can get the R.H.S to equal $ \ \omega^{10}(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7)$
We also know that $ \ \omega^{10} = \omega$, so we can simplify it to be $\ \omega(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7) $
Using the fact: $\ \omega\ + \omega^2\ + \omega^3\ + \omega^4\ + \omega^5\ + \omega^6\ + \omega^7\ + \omega^8\ = -1 $, the expression simplifies to -1.
Now simplifying the L.H.S. we get: $16[cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{6\pi}{9})\cos(\frac{8\pi}{9})]$. Knowing that $cos(\frac{2\pi}{3}) = -\frac{1}{2}$, we can finally get that:
$$
-8\ [cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = -1 \\
[cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = \frac{1}{8} \\
[cos(\frac{\pi}{9})\cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})] = \frac{1}{8}
$$
| {
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Why does this function only have 3 rational solution? Why does the function $f(x,y) = x^3 - x - y^2$ only have the three rational roots
$(0,0)$, $(1,0)$, $(-1,0)$?
| We have $f(x,y)=x^3-x-y^2$. We define a pair of roots $(x,y)$ to satisfy $f(x,y)=0$. Then:
$$x^3-x=y^2$$
Thus, we are looking for rational solutions for the above equation. Let $x=\frac{m}{n}$ where $\gcd(m,n)=1$ and $n>0$. Assume $y^2 > 0$.
$$x^3-x=x(x^2-1)=x(x-1)(x+1)=\frac{m(m-n)(m+n)}{n}=y^2$$
We have the numbers $m-n$, $m+n$, $m$ and $n$ to be pairwise relatively prime since $\gcd(m,n)=1$. This shows that:
$$\frac{m(m-n)(m+n)}{n}=y^2 \implies (m-n,m+n,m,n) \text{ are of the form } k^2\text{ or }-k^2$$
where $k$ is an integer. All of these values cannot be positive by Fermat's Right Triangle Theorem. We know that $n$ is positive. If $m+n$ is negative, then $m$ has to be negative and since $m>m-n$, $m-n$ will also be negative. This will be a contradiction since we will get:
$$0>\frac{m(m-n)(m+n)}{n}=y^2$$
as there are three negatives in the numerator. Let $m+n$ to be positive. As we know that $m$ and $m-n$ cannot be both positive and $y^2 \geqslant 0$, we need both $m$ and $m-n$ to be negative. Then, $n-m$ and $-m$ are positive. We have:
$$y^2=\frac{m(m-n)(m+n)}{n}=\frac{(-m)(n-m)(n+m)}{n}=\frac{(n)(n-m)(n+m)}{-m} \cdot \frac{m^2}{n^2}$$
$$\bigg(\frac{yn}{m}\bigg)^2=\frac{(n)(n-m)(n+m)}{-m}$$
which is again a contradiction by Fermat's Right Triangle Theorem.
We must thus have $y^2=0$. This means $x^3-x=0$ whose only roots (which are all rational) are $-1$, $0$ and $1$, proving the necessary.
| {
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Finding $\int _{-\infty }^{\infty }\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx$ with real methods How can i prove with real methods that $\int_{-\infty}^\infty\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx=\frac\pi2(e^{-b}+\cos(b))$? I was able to prove this using complex analysis but i dont know how to attack with without it.
As said in the comments we can try using Feynman's trick for this one:
$$I=\int _{-\infty }^{\infty }\frac{x^3\sin \left(x\right)}{x^4-b^4}\:dx$$
$$I\left(a\right)=\int _{-\infty }^{\infty }\frac{x^3\sin \left(ax\right)}{x^4-b^4}\:dx=\underbrace{\int _{-\infty }^{\infty }\frac{\sin \left(ax\right)}{x}\:dx}_{\pi }+b^4\int _{-\infty }^{\infty }\frac{\sin \left(ax\right)}{x\left(x^4-b^4\right)}\:dx$$
$$I'\left(a\right)=b^4\int _{-\infty }^{\infty }\frac{\cos \left(ax\right)}{x^4-b^4}\:dx$$
$$I''''\left(a\right)=b^4\underbrace{\int _{-\infty }^{\infty }\frac{x^3\sin \left(ax\right)}{x^4-b^4}\:dx}_{I\left(a\right)}$$
$$I''''\left(a\right)-b^4I\left(a\right)=0$$
And i think it boils down to this, i dont know if theres a more simpler or easier approach.
As xpaul showed, the general solution for that differential equation is $$I\left(a\right)=C_1e^{ab}+C_2e^{-ab}+C_3\cos \left(ab\right)+C_4\sin \left(ab\right)$$
But instead of letting $I\left(0\right)=0$ its better to use $I\left(0\right)=\pi $
thus having:
$$I\left(0\right)=C_1+C_2+C_3=\pi $$
$$I'\left(0\right)=C_1b-C_2b+C_4b=-\frac{\pi b}{2}$$
$$I''\left(0\right)=C_1b^2+C_2b^2-C_3b^2=0$$
$$I'''\left(0\right)=C_1b^3-C_2b^3-C_4b^3=-\frac{\pi b^3}{2}$$
with those mistakes fixed this eventually gives:
$$C_1=0,C_2=\frac{\pi }{2},C_3=\frac{\pi }{2},C_4=0$$
So,
$$I\left(a\right)=\frac{\pi }{2}\left(e^{-ab}+\cos \left(ab\right)\right)$$
$$I=\frac{\pi }{2}\left(e^{-b}+\cos \left(b\right)\right)$$
| The characteristic equation of
$$ I''''(t)-b^4I(t)=0 \tag1$$
is
$$ r^4-b^4=0. \tag2$$
(2) has roots $\pm b,\pm bi$ and hence (1) has the general solution
$$ I(t)=C_1e^{bt}+C_2e^{-bt}+C_3\sin(bt)+C_4\cos(bt). $$
Noting
$$I'\left(a\right)=b^4\int _{-\infty }^{\infty }\frac{\cos \left(ax\right)}{x^4-b^4}\:dx$$
one has $I(0)=I''(0)=0$ and
\begin{eqnarray}
I'\left(0\right)&=&b^4\int _{-\infty }^{\infty }\frac{1}{x^4-b^4}\:dx=\frac{b^2}{2}\bigg[\int _{-\infty }^{\infty }\frac1{x^2-b^2}dx-\int _{-\infty }^{\infty }\frac1{x^2+b^2}dx\bigg]=-\frac{b\pi}{2}\\
I'''(0)&=&-b^4\int _{-\infty }^{\infty }\frac{x^2}{x^4-b^4}\:dx=\frac{b^3\pi}{2}.
\end{eqnarray}
So
\begin{eqnarray}
C_1+C_2+C_4=0,\\
C_1b^2+C_2b^2-C_4b^2=0,\\
C_1b-C_2b+C_3b=-\frac{b\pi}{2},\\
C_1b^3-C_2b^3-C_3b^3=-\frac{b^3\pi}{2}
\end{eqnarray}
which implies that
$$ C_1=C_3=0, C_2=C_4=\frac{\pi}{2}. $$
Thus
$$ I=I(1)=\frac{\pi}{2}(e^{-b}+\cos b).$$
| {
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Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way of doing it.
The integral is $$\int\frac{4x^4}{{x^8+1}}\;dx$$
Thanks in advance.
| Not an answer. Maybe this result inspires a by-hand method. The integral is "elementary".
\begin{align*}
&\int \frac{4x^4}{x^8+1} \,\mathrm{d}x \\
&= 4 \left(-\frac{1}{8} \cos \left(\frac{\pi }{8}\right) \log \left(x^2-2 x \sin
\left(\frac{\pi }{8}\right)+1\right)
\right. \\ &\quad \left. {}
+\frac{1}{8} \cos \left(\frac{\pi
}{8}\right) \log \left(x^2+2 x \sin \left(\frac{\pi
}{8}\right)+1\right)
\right. \\ &\quad \left. {}
+\frac{1}{8} \sin \left(\frac{\pi }{8}\right) \log
\left(x^2-2 x \cos \left(\frac{\pi }{8}\right)+1\right)
\right. \\ &\quad \left. {}
-\frac{1}{8} \sin
\left(\frac{\pi }{8}\right) \log \left(x^2+2 x \cos \left(\frac{\pi
}{8}\right)+1\right)
\right. \\ &\quad \left. {}
+\frac{1}{4} \cos \left(\frac{\pi }{8}\right) \tan
^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(x-\cos \left(\frac{\pi
}{8}\right)\right)\right)
\right. \\ &\quad \left. {}
+\frac{1}{4} \cos \left(\frac{\pi }{8}\right) \tan
^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(x+\cos \left(\frac{\pi
}{8}\right)\right)\right)
\right. \\ &\quad \left. {}
-\frac{1}{4} \sin \left(\frac{\pi }{8}\right) \tan
^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(x-\sin \left(\frac{\pi
}{8}\right)\right)\right)
\right. \\ &\quad \left. {}
-\frac{1}{4} \sin \left(\frac{\pi }{8}\right) \tan
^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(x+\sin \left(\frac{\pi
}{8}\right)\right)\right)\right) \text{.}
\end{align*}
Produced by a CAS. There are some "simplifications", but they don't shorten the expression much.
| {
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if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.
| Following the steps given in Wikipedia, we can set the problem as
$$\text{Minimize}[a^2+b^2,\{a,b\}]$$
subject to the three inequality constraints
$$P=16-3a^2 <0$$
$$D=-3 a^4+32 a^2-16 a b \leq 0$$
$$\Delta=-27 a^4-4 a^3 b^3+36 a^3 b-2 a^2 b^2+256 a^2+36 a b^3-512 a b-27 b^4+256 b^2 \leq 0$$
This is a quite simple task, leading to the solutions $a=\pm4$ and $b=\mp4$ and then $a^2+b^2 \geq 32$.
The symmetry, already underlined in other answers is obvious looking at the constraints.
All the above was done using calculus. It could be done without any if, as in many optimization problems of the kind, we assume that the solution will correspond to the saturation of inequality constraints. If this is the case, then
$$D=0 \implies b=2 a\left(1-\frac{3 }{16}a^2\right)$$
$$\Delta=0 \implies a^2(a^2-16)^2(459 a^6-7776 a^4+43776 a^2-65536)=0$$
$a^2=0$ must be rejected since $P<0$ would not be satisfied.
Concerning the cubic in $a^2$, it shows only one real root which is
$$a^2=\frac{32}{51} \left(9-\cosh \left(\frac{1}{3} \cosh ^{-1}(577)\right)\right) \approx 2.32647$$to be rejected since $P<0$ would not be satisfied $(2.32647 < \frac {16}3)$.
All of that makes that the only solution is $a^2=16$ to which corresponds $b^2=16$ too.
One thing I did not have time to do is to look at the shadow prices of this problem.
| {
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How To Prove $\int_0^1 (\ln x)^2 \frac{1+x^2}{1+x^4} \, dx=\frac {3\sqrt{2}\pi^3}{64}$ Question:-Prove that $$\int_0^1( \ln x)^2 \frac{1+x^2}{1+x^4} \, dx=\frac {3\sqrt{2}\pi^3}{64}$$
My attempt:
$$\int_0^1 \left( \ln\frac{1}{x}\right)^2 \frac{1+x^{-2}}{x^{-2}+x^2} \, dx$$
Put $\left(x-\frac{1}{x}\right)=t$, we get
$$\int_{-\infty}^0\left( \ln\frac{(t+\sqrt{t^2+4})}{2}\right)^2 \frac{1}{t^2+2} \, dt$$
After that I got stuck.
Can anybody help me!
| Let $f(x)=\frac{1+x^2}{1+x^4}\ln^2 x$. By the substitution $x\mapsto\frac1x$, it can proved that $I=\int^1_0f(x)dx=\int^\infty_1f(x)dx$. Hence $2I=\int^\infty_0 f(x)dx$.
Employing Feynman's technique,
$$2I=\frac{\partial^2}{\partial a^2}\underbrace{\int^\infty_0\frac{1+x^2}{1+x^4}x^adx}_{J}\bigg\vert_{a=0}$$
We rewrite it as
$$J=\frac12\int^\infty_0\frac{1+z}{1+z^2}z^b dz\qquad b=\frac{a-1}{2}$$ by enforcing $z=x^2$, so as to facilitate the evaluation of $J$ by residue calculus.
We take the branch cut of $z^b$ along the positive real axis and consider a keyhole contour ($C$) integral.
Then, $$\oint_C \frac{1+z}{1+z^2}z^b dz=2\pi i(\operatorname{Res}_{z=i}+\operatorname{Res}_{z=-i})$$
Note that as $b<0$, the integrand is $o(z^{-1})$ and thus the large circular contour integral vanishes as its radius tends to infinity.
The integral above positive real axis is $2J$ while that below is $-e^{-2\pi ib}\cdot2J$.
Hence, $$2(1-e^{2\pi ib})J=2\pi i\left[\frac{1+i}{2i}i^b+\frac{1-i}{-2i}(-i)^b\right]$$
$$2(1-e^{2\pi ib})J=\pi \left[(1+i)e^{\pi ib/2}-(1-i)e^{3\pi ib/2}\right]$$
$$2(e^{-\pi ib}-e^{\pi ib})J=\pi \left[(1+i)e^{-\pi ib/2}-(1-i)e^{\pi ib/2}\right]$$
$$-4i\sin(\pi b)J=\pi \left[e^{-\pi ib/2}-e^{\pi ib/2}+i(e^{-\pi ib/2}+e^{\pi ib/2})\right]$$
$$-4i\sin(\pi b)J=\pi \left(-2i\sin(\pi b/2)+2i(\cos(\pi b/2)\right)$$
$$J=\frac{\pi}{4}(-\csc(\pi b/2)+\sec(\pi b/2))$$
Note that in the last step we used double-angle formula $\sin\theta=2\sin\frac{\theta}2\cos\frac{\theta}2$.
$$I=\frac12\frac{\partial^2 J}{\partial a^2}\bigg\vert_{a=0}=\frac12\frac{\partial^2 J}{(2\partial b)^2}\bigg\vert_{b=-1/2}=\frac12\cdot\frac14\cdot\frac{\pi}{4}\cdot\frac{\pi^2}{4}(-\left(-3\sqrt2)+3\sqrt2\right)=\frac{3\sqrt 2}{64}\pi^3$$
| {
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$\frac{1+\frac{1}{2^{11}}+\frac{1}{3^{11}}+\frac{1}{4^{11}}+...}{1-\frac{1}{2^{11}}+\frac{1}{3^{11}}-\frac{1}{4^{11}}+...}$ I'm not sure how to solve this summation and would like assistance. I believe the answer is ||$\frac{1024}{1023}$|| according to Wolfram. I don't know what the proper approach is (creating two summations and somehow combining or common denominators..?).
| I will assume that the +1 in the denominator is a -1 and go on,
we can write $\frac{1}{1}+\frac{1}{2^{11}}+\frac{1}{3^{11}}+.......=S$
now we can take the even terms to one side and express it as
$\frac{1}{1}$+$\frac{1}{3^{11}}$+$\frac{1}{5^{11}}$=S +$\frac{1}{2^{11}}(S)$
$\frac{1}{1}+\frac{1}{3^{11}}+\frac{1}{5^{11}}=S(\frac{2^{11}-1}{2^{11}})$
now the equation in the question becomes
$\frac{S}{S-2(S(\frac{2^{11}-1}{2^{11}})}$ which gives the answer $\frac {1024}{1023}$
| {
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Diophantine equation with power Find all the integer solutions of:
$2^{n+1} + 41 = m^2$
I am stuck, and I am not sure if I am going on the right path..
adding 8 to both sides:
$2^{n+1} + 41 + 8 = m^2 + 8$
$2^{n+1} + 49 = m^2 + 8$
$2^{n+1} - 8 = m^2 -49$
$2(2^n - 4) = (m+7)(m-7)$
I am not sure on how to keep going, how do I manage to solve this question?
Thank you!
| Some insight can be gained by looking at the equation $\bmod 40$. Note that for $n\ge 1$, the residues of $2^{n+1} \bmod 40$ are $\{4,8,16,32,24\}$ with $4$ occurring only once and the other values cycling. Since $4$ does not yield a solution ($4+41\ne m^2$), we can ignore its single occurrence. Of course, $41\equiv 1 \bmod 40$, so we are looking for solutions to $m^2\equiv \{9,17,33,25\} \bmod 40$. No integer squared ends in the digits $3,7$, so we can further eliminate $17,33$ from consideration, leaving us to examine $m^2\equiv \{9,25\} \bmod 40$. From this, we conclude that $m$ ends in one of the digits $\{3,7,5\}$.
Case 1: $m^2\equiv 25 \bmod 40$. Then $2^{n+1}$ must end in the digits $84$. This occurs for $n+1=20k+14$. $n+1=14,34$ do not yield solutions, and for $n+1=54,74,\dots$ I am not able to calculate, so there might be solutions, but I can't say definitely yes or no.
Case 2: $m^2\equiv 9 \bmod 40$. This implies that $2^{n+1}$ ends in the digit $8$, which occurs when $n+1=4k+3$. The readily found solutions for $n+1=3,7$ are $m=7,13$. Up to $n+1=47$, no further solutions are found.
Added by edit: OP correctly points out in a comment that $n$ must be even. If $n$ is odd, then $n+1$ is even, $2^{n+1}$ is a square, and by rearranging terms, $41$ can be expressed as the difference between two squares. Since $41$ is prime, it can only be expressed as the difference between two squares in one way: $41^2-40^2=(21+20)(21-20)$. In this complexion, $m=21$, but $20$ is not a power of $2$, so $n$ cannot be odd, hence $n+1$ cannot be even, which is required in Case 1 where $n+1=20k+14$. Thus there are no Case 1 solutions, as OP states.
| {
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An equivalent of Jacobi's two square theorem Jacobi's two square theorem:
The number of representations of $n$ as a sum of two squares is $4$ times the difference
between the number of divisors of $n$ congruent to $1$ modulo $4$ and the number of divisors
of $n$ congruent to $3$ modulo $4$.
This is equivalent to
$$\left(\sum_{n\in\mathbb{Z}}q^{n^2}\right)^2=1+4\sum_{n\ge 0}\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right).$$
But why are these two theorems equivalent? What does the sum
$$\left(\sum_{n\in\mathbb{Z}}q^{n^2}\right)^2$$
have to do with ways of writing numbers as sums of two squares?
| You remark about convergence in a comment. Convergence is not relevant. These series are recording combinatorial information in their coefficients. They are formal power series, recording data in their coefficients.
Let us start with a smaller version of the last sum.
\begin{align*}
\sum_{n=-4}^4 q^{n^2}
&= q^{(-4)^2} + q^{(-3)^2} + q^{(-2)^2} + q^{(-1)^2} + q^{0^2} + q^{1^2} + q^{2^2} + q^{3^2} + q^{4^2} \\
&= q^{16} + q^{9} + q^{4} + q^{1} + q^{0} + q^{1} + q^{4} + q^{9} + q^{16} \\
&= q^{0} + 2q^{1} + 2q^{4} + 2q^{9} + 2q^{16} \text{.}
\end{align*}
What do we see? The squares of the range of integers we are keeping appear as exponents. The number of ways of writing one of these squares as a square of an integer appears as the coefficient. So we should be able to see that the sum over all integers, which just extends this sum with more terms of the form $c q^s$ where $s$ is a square and $c=2$ is the number of integers squaring to $s$, records the combinatorial information that there are $c$ ways to express $s$ as the square of an integer.
What happens when we square this series? We construct many terms of the form $c_1 q^{s_1} \cdot c_2 q^{s_2} = c_1 c_2 q^{s_1 + s_2}$, each representing that from the $c_1$ ways to express the square $s_1$ and the $c_2$ ways to express the square $s_2$, there are $c_1 c_2$ ways to express the sum of squares $s_1 + s_2$. Let's perform the squaring of our fragment of the last series.
$$ \left( \sum_{n=-4}^4 q^{n^2} \right)^2 = q^0 + 4q + 4q^2 + 4q^4 + 8q^5 + 4 q^8 + 8 q^{10} + 8 q^{13} + 4 q^{16} + 8 q^{17} + 4 q^{18} + 8 q^{20} + 8 q^{25} + 4 q^{32} $$
This says, among many other things, there are $8$ ways to write $5$ as the sum of two squares of integers from $[-4,4]$. Looking at the source of that term,
$$ 8 q^5 = 2q^1 \cdot 2q^4 + 2q^4 \cdot 2q^1 \text{,} $$
we can read off that the eight ways of writing $5$ in this manner are \begin{align*}
2^2 &+ 1^2, & 2^2 &+ (-1)^2, & (-2)^2 &+ 1^2, & (-2)^2 &+ (-1)^2 \\
1^2 &+ 2^2, & 1^2 &+ (-2)^2, & (-1)^2 &+ 2^2, & (-1)^2 &+ (-2)^2 \text{.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$.
Prove that for all triangles with angles $\alpha, \beta, \gamma$, $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 3}{\sin\alpha + \sin\beta + \sin\gamma}$$
Let $\tan\dfrac{\alpha}{2} = a, \tan\dfrac{\beta}{2} = b, \tan\dfrac{\gamma}{2} = c$, we have that $$\dfrac{\sin\beta}{\cos\beta + 1} = \dfrac{1}{b}, \cos\beta = \dfrac{1 - b^2}{1 + b^2}, \sin\beta = \dfrac{2b}{1 + b^2}$$ and $bc + ca + ab = 1$.
It needs to be proven that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1 - a^2}{1 + a^2} + \dfrac{1 - b^2}{1 + b^2} + \dfrac{1 - c^2}{1 + c^2} + 3}{\dfrac{2a}{1 + a^2} + \dfrac{2b}{1 + b^2} + \dfrac{2c}{1 + c^2}}$$
$$\impliedby \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1}{1 + a^2} + \dfrac{1}{1 + b^2} + \dfrac{1}{1 + c^2}}{\dfrac{a}{1 + a^2} + \dfrac{b}{1 + b^2} + \dfrac{c}{1 + c^2}}$$
$$\impliedby \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\left(\frac{a}{1 + a^2} + \frac{b}{1 + b^2} + \frac{c}{1 + c^2}\right) = \frac{1}{1 + a^2} + \frac{1}{1 + b^2} + \frac{1}{1 + c^2}$$
$$\impliedby \left(\frac{a}{b} + \frac{a}{c}\right)\frac{1}{1 + a^2} + \left(\frac{b}{c} + \frac{b}{a}\right)\frac{1}{1 + b^2} + \left(\frac{c}{a} + \frac{c}{b}\right)\frac{1}{1 + c^2} = 0$$
$$\impliedby \frac{a(b + c)}{bc(c + a)(a + b)} + \frac{b(c + a)}{ca(a + b)(b + c)} + \frac{c(a + b)}{ab( b + c)(c + a)} = 0$$
$$\impliedby \frac{1 - bc}{(1 - ca)(1 - ab)} + \frac{1 - ca}{(1 - ab)(1 - bc)} + \frac{1 - ab}{(1 - bc)(1 - ca)} = 0$$
$$\impliedby (1 - bc)^2 + (1 - ca)^2 + (1 - ab)^2 = 0$$
$$\impliedby bc = ca = ab = 1 \impliedby bc + ca + ab = 3,$$ which is definitely incorrect.
I've surmised that the correct equality is $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 1}{\sin\alpha + \sin\beta + \sin\gamma},$$ but then I wouldn't know what to do first.
| Using your transformation I a get sligtly different equation:
$$a+b+c=\frac{\sum_{cyc}\frac{1}{1+a^2}}{\sum_{cyc}\frac{a}{1+a^2}}$$
$$\sum_{cyc}\frac{a(a+b+c)}{1+a^2}=\sum_{cyc}\frac{1}{1+a^2}$$
$$\sum_{cyc}\frac{a^2+ab+ac-1}{(a+b)(a+c)}=0$$
$$\sum_{cyc}(a^2-bc)(b+c)=0$$
$$\sum_{cyc}a^2b+a^2c-b^2c-bc^2=0$$
Which is true!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Extreme values of $|z|$ when $|z^2+1|=|z-1|$ Problem statement: Find the extreme values of $|z|$ when $$|z^2+1|=|z-1|,\ z\in \mathbb{C}-\{0\}$$
My try:
$$|z^2+1|=|z-1|\implies|z^2+1|^2=|z-1|^2\implies(z^2+1)(\overline{z}^2+1)=(z-1)(\overline{z}-1)$$
$$\implies|z|^4+(z+\overline{z})^2=3|z|^2-(z+\overline{z})\cdots\text{(I have skipped some algebra here..)}$$
Since $(z+\overline{z})$ is a real number, let $a=(z+\overline{z})$.Now,
$$|z|^4-3|z|^2+(a^2+a)=0$$
Using the quadratic formula, we have
$$|z|^2=\frac{3\pm\sqrt{9-4(a^2+a)}}{2}$$
Since the discriminant is strictly non-negative and the minimum value of $(a^2+a)$ is $\frac{-1}{4}$, we have,
$$\frac{-1}{4} \le(a^2+a) \le \frac{9}{4}\implies -1 \le4(a^2+a) \le9$$
Now, considering the two values of $|z|^2$ and accordingly maximising or minimising $(a^2+a)$ using the above inequality, we get
$$\max{|z|}=\sqrt{\frac{3+\sqrt{10}}{2}}\ \ \ \text{and}\ \ \ \min{|z|=0}$$
First of all, I would like to know whether my solution is correct. Secondly, I have trouble minimising $|z|$ because the problem statement says $|z|\ne0$, hence my answer contradicts the problem statement.
I would also like to know if there is any other way of solving this problem of finding extreme values of $|z|$ given an equation in $z$.
Thanks for any answers!!
| (this solution mostly is similar to what you have, which is fine, but you can avoid the quadratic equation and bounding)
Observe that
$$|z^2+1|=|z-1| \iff (z^2 + 1)(\overline{z}^2 + 1) = (z - 1)(\overline{z}-1) \iff |z|^4 - |z|^2 + (z^2 + \overline{z}^2 + z + \overline{z}) = 0$$
Note that for all pairs of reals $(a, b)$, there exists a complex number $c$ such that $c+\overline{c} = a, c\overline{c} = b$ if $a^2 \leq 4b$. This is because $a = 2\text{Re}(c)$ and $b = \text{Re}(c)^2 + \text{Im}(c)^2 \geq \text{Re}(c)^2 = \frac{a^2}{4}$. If $a$, $b$ satisfy this condition, then we can easily contruct $c$ using this, so we're done.
If we let $s = z + \overline{z} = 2\text{Re}(z), p=z\overline{z} = |z|^2$, then $p^2 - 3p + s^2 + s = 0$ with $p$ and $s$ real, so $s^2 \leq 4p$.
This quadratic equation can be rewritten as $(p-\frac{3}{2})^2 + (s+\frac{1}{2})^2 = \frac{5}{2}$. Hence, $p$ is maximised when $s = -\frac{1}{2}$, so $p = \frac{3+\sqrt{10}}{2}$ (which we can easily observe to satisfy $s^2 \leq 4p$). This yields the maxima for $|z|$ as $\sqrt{\frac{3+\sqrt{10}}{2}}$ at $z = -\frac{1}{4} \pm \frac{\sqrt{23 + 8\sqrt{10}}}{4}i$.
Furthermore, $p$ is minimised when $p \to 0^+$, so $z \to 0^+$. Thus, there's no explicit minimum, but there is an infimum of $0$.
Hope that helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How $\int_1^\infty \sin \left( \frac{2}{x^{5/3}} \right)dx $ absolute converge? I have the integral $$\displaystyle{\int_1^\infty {\sin \left( \frac{2}{x^{\frac{5}{3}}} \right)}dx }$$
I know it is converge. but when I check absolute converge I did the following: according to the trigonometric identity :$\cos(2\theta)=1-2\sin^2(\theta)$ I got
$$\left|\sin\left(\frac{2}{x^{\frac{5}{3}}}\right)\right| \geq \sin^2\left(\frac{2}{x^{\frac{5}{3}}}\right)=\frac{1}{2} -\frac{1}{2} \cos\left(\frac{4}{x^{\frac{5}{3}}}\right)$$
$\displaystyle{\int_1^\infty \frac{1}{2}dx }$ is diverge so why I cant conclude that $\displaystyle{\int_1^\infty {\sin \left( \frac{2}{x^{\frac{5}{3}}} \right)}dx }$ also diverge?
| Put $y=\frac 2 {x^{5/3}}$. We get $\int_1^{\infty}| \sin (\frac 2 {x^{5/3}})|dx= 2^{-2/5}\frac 3 5 \int_0^{2} |\sin y| y^{-8/5}dy$. Now use the fact that $\frac {\sin y} y \to 1$ as $ y \to 0$ and $\int_0^{2} y^{-3/5}dy <\infty$ to see that the given inetgral is absolutely convergent.
| {
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Find $g(x)$ if $f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$ $$f:[-2,2] \rightarrow \Bbb R$$
$$\text {and }f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$$
And, let $g(x)$ be equal to $|f(x)|+f(|x|)$
We need to find the value of $g(x)$ (define it).
I begin by finding the value of $|f(x)|$ first :
$$|f(x)|=\begin{cases} |-1|, & -2 \leq x \leq 0 \\ |x-1|, & 0 < x \leq 2 \end{cases}=\begin{cases} 1, & -2 \leq x \leq 0 \\ |x-1|, & 0 < x \leq 2\end{cases}$$
Now, to determine what $|x-1|$ would be, we need to determine whether $(x-1)$ is positive or negative or zero.
If $x-1 \geq 0$, $|x-1| = x-1$ and if $x-1 < 0$, then $|x-1| = -(x-1) = 1-x$
If $x-1 \geq 0$, then $x \geq 1$ and if $x - 1 < 0$, then $x < 1$. We can now split the condition $0 < x \leq 2$ from the earlier definition of $|f(x)|$ as $0<x<1$ and $1 \leq x \leq 2$, where $0<x<1 \implies |x-1| = 1-x$ and $1 \leq x \leq 2 \implies |x-1| = x-1$
So, $$ |f(x)| = \begin{cases} 1, & -2 \leq x \leq 0 \\ 1-x, & 0<x<1 \\ x-1, & 1 \leq x \leq 2 \end{cases}$$
Now, we need to define $f(|x|)$. Here's what I do :
$$f(|x|) = \begin{cases} -1, & -2 \leq x \leq 0 \\ |x|-1, & 0<|x| \leq 2 \end{cases}$$
Now, is the next step that I do here correct or even necessary?
Now, $|x|$ can never be negative but can be zero when $x=0$. So, the condition $-2 \leq |x| \leq 0$ can be replaced by $x = 0$ from which we obtain the following definition for $f(|x|)$:
$$f(|x|) = \begin{cases} -1, & x=0 \\ |x|-1, & 0 < x \leq 2 \end{cases} = \begin{cases} |x|-1, & 0 \leq |x| \leq 2 \end{cases}$$
I did the last one because we observe that $0$ is mapped to $-1$ and $|0|-1 = -1$, so it will still be mapped to $-1$ if we put it in the second condition. So, basically, for any value of $x$ that is a part of domain of $f$, $f(|x|) = |x|-1$.
Now, we add $|f(x)|$ and $f(|x|)$ to obtain $g(x)$.
$$g(x) = |f(x)| + f(|x|) = \begin{cases} |x|-1+1, & -2 \leq x \leq 0 \\ |x|-1+1-x, & 0 < x < 1 \\ |x|-1+x-1, & 1 \leq x \leq 2\end{cases}$$.
In the first condition, the value of $x$ is either negative or $0$, so if it is negative, $|x| = -x$ and else, it is 0 which is also equal to $-0$, which means that in the first condition, $|x|$ can be substituted by $-x$. In the second and, the value of $x$ is always positive, so $|x| = x$. So, we can substitute $|x|$ by $x$ in the second and third conditions and arrive at the definition of $g(x)$ that is :
$$g(x) = \begin{cases} -x-1+1, & -2 \leq x \leq 0 \\ x-1+1-x, & 0 < x < 1 \\ x-1+x-1, & 1 \leq x \leq 2 \end{cases} = \begin{cases} -x, & -2 \leq x \leq 0 \\ 0, & 0 < x < 1 \\ 2x-2, & 1 \leq x \leq 2 \end{cases}$$
Now, I want to know if this process is correct and if there is some alternative, better approach to this problem.
Sorry for the long post, I thought that showing my line of reasoning would make the question better.
Thanks
| Your final answer appears to be correct (though I have not checked all the steps of your reasoning), however you may try to apply symmetry by dividing into three cases: $x=0$, $-2\leq x<0$, and $0<x\leq 2$. Then it would be easy to see how to get $g(x)$, since $|x|$ for $x$ in the second case implies $|x|$ belongs to the third case. The rest is left to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$ I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
My attempt
$$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
or,
$$ S=1 + \left( \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \right)+\left(\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}\right)+\left(\frac{1}{11}-\frac{1}{6}+\frac{1}{13}\right)+\left( \frac{1}{15}-\frac{1}{8}+\frac{1}{17}\right)+ \cdots$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{2n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{(4n-1)(4n)}-\frac{1}{(4n)(4n+1)} \right)$$
Since rearrrangement is allowed now, we have
$$S= 1+\sum_{n=1}^\infty \frac{1}{(4n-1)(4n)}-\sum_{n=1}^\infty\frac{1}{(4n)(4n+1)}$$
or,
$$ S = 1+ I_1 - I_2$$
But even after many trials, I was unable to find the value of either $I_1$ or $I_2$, any help regarding this will be much appreciated.
| Since $S=1+\sum_{n\ge1}\int_0^1x^{4n-2}(1-2x+x^2)dx$, by monotone convergence$$S=1+\int_0^1\frac{x^2(1-x)^2dx}{1-x^4}=1+\int_0^1\frac{x^2(1-x)dx}{(1+x)(1+x^2)}.$$You can do the rest with partial fractions:$$S=1+\left[-x+\ln(x+1)+\frac12\ln(x^2+1)\right]_0^1=\frac32\ln2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by Taylor expanding around $0$; first cosine, then ln. But I get nowhere. I would appreciate any help.
| A possible way to show convergence is to rewrite
$$\cos \frac 1n = \cos \frac{2}{2n} = 1- 2\sin^2 \frac 1{2n} $$
and now use root test and the standard limits $\lim_{t\to 0}(1-t)^{\frac 1t} = \frac 1e$ and $\lim_{t\to 0}\frac{\sin t}{t}=1$:
\begin{eqnarray*}\sqrt[n]{\left(1- 2\sin^2 \frac 1{2n}\right)^{n^3}}
& = & \left(\left(1- 2\sin^2 \frac 1{2n}\right)^{\frac{1}{2\sin^2 \frac 1{2n}}}\right)^{n^2\cdot 2\sin^2 \frac 1{2n}}\\
& \stackrel{n\to \infty}{\longrightarrow} & \left(\frac 1e\right)^{\frac 12}=\frac 1{\sqrt e} < 1
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695778",
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"source": "stackexchange",
"question_score": "2",
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Find $\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$ Need solve the next integral
$$\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$$
Tried something with Laurent’s series, but i can’t conclude anything.
Thanks
| METHODOLGY $1$: CONTOUR INTEGRATION
Enforce the substitution $x\mapsto x^2$ to find that
$$\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx=4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\tag1$$
Let $f(z)$ be given by
$$f(z)=\oint_C \frac{\log^2(z)}{(z^2+1)^2}\,dz\tag2$$
where we choose to cut the plane along the positive real axis and where $C$ is the classical keyhole contour.
We then have from $(2)$
$$\begin{align}
\int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\\\\&+2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\tag3\end{align}$$
The left-hand side of $(3)$ becomes
$$\begin{align}
\int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=-i4\pi\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\\\\& +4\pi^2\int_0^\infty \frac{1}{(x^2+1)^2}\,dx\tag4
\end{align}$$
Note that the imaginary part of the right-hand side of $(4)$ is $-\pi$ times the integral of interest on the right-hand side of $(1)$. Thus, we find that
$$\begin{align}
\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx&=-2\text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\right)\\\\
&-2 \text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\right)\\\\
&=-2\left(-\frac\pi4+\frac{3\pi}{4}\right)\\\\
&=-\pi
\end{align}$$
METHODOLGY $2$: REAL ANALYSIS ONLY
We begin by enforcing the substitution $x\mapsto \tan(x)$ in the integral on the right-hand side of $(1)$ to reveal
$$\begin{align}
4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx&=4\int_0^{\pi/2}\cos^2(x) \log(\tan(x))\,dx\\\\
&=4\int_0^{\pi/2}\cos^2(x) \log(\sin(x))\,dx\\\\&-4\int_0^{\pi/2}\cos^2(x) \log(\cos(x))\,dx\tag5 \\\\
&=4\int_0^{\pi/2}(2\cos^2(x)-1)\log(\sin(x))\,dx\tag6\\\\
&=4\int_0^{\pi/2}\cos(2x)\log(\sin(x))\,dx\tag7\\\\
&=4\left(-\int_0^{\pi/2}\cos^2(x)\,dx\right)\tag8\\\\
&=-\pi
\end{align}$$
as expected.
In going from $(5)$ to $(6)$ we made use of the transformation $x\mapsto \pi/2 -x$ in the second integral on the right-hand side of $(5)$.
In going from $(7)$ to $(8)$, we used integration by parts with $u=\log(\sin(x))$ and $v=\sin(x)\cos(x)$
| {
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there do not exist intgers $a,b,c,d,$ with $k>1$ such that $(a+bw+cw^2+dw^3)^k=1+w$
let $w=e^{\frac{2\pi\cdot i}{5}}$ be a primitive fifth root of unity,Prove that there do not exist intgers $a,b,c,d,$ with $k>1$ such that
$$(a+bw+cw^2+dw^3)^k=1+w$$
I try:let $x=a+bw+cw+dw^3(a,b,c,d\in Z)$ and note that $w+w^{-1}=\dfrac{\sqrt{5}-1}{2}$ and $w^2+w^{-2}=w^3+w^{-3}=-\dfrac{\sqrt{5}+1}{2}$,I deduce
$$|x|^2=(a^2+b^2+c^2+d^2)+\dfrac{\sqrt{5}-1}{2}(ab+bc+cd)-\dfrac{\sqrt{5}+1}{2}(ac+bd+ad)$$
and $$|1+w|^2=\dfrac{3+\sqrt{5}}{2}$$
so we need prove that:there not exist $a,b,c,d,k$ with $k>1$ such that
$$\left((a^2+b^2+c^2+d^2)+\dfrac{\sqrt{5}-1}{2}(ab+bc+cd)-\dfrac{\sqrt{5}+1}{2}(ac+bd+ad)\right)^k=\dfrac{3+\sqrt{5}}{2}$$ then I can't it,Thank you for you help me!
| Too long for a comment
Remark: Proceed along the OP's approach. Not sure if it works.
From the last equation of the OP, we have $(\frac{C}{2} + \frac{D}{2}\sqrt{5})^k = \frac{3}{2} + \frac{1}{2}\sqrt{5}$
where
\begin{align}
C &= 2 a^2-a b-a c-a d+2 b^2-b c-b d+2 c^2-c d+2 d^2, \\
D &= a b-a c-a d+b c-b d+c d.
\end{align}
(Remark: We have $C\ge 0$ and $C^2-5D^2 \ge 0$. See below.)
Then, $(\frac{C}{2} - \frac{D}{2}\sqrt{5})^k = \frac{3}{2} - \frac{1}{2}\sqrt{5}$.
Thus, $(\frac{C^2-5D^2}{4})^k = 1$. Thus, $C^2-5D^2 = 4$. (Remark: This is the same as @dust05's one in @dust05's first comment.)
We have the identity $4(C^2 - 5D^2) = 10p_1^2 + 10p_2^2 + p_3^2 + 5p_4^2$ where
\begin{align}
p_1 &= a b-b^2+c^2-c d, \\
p_2 &= -a^2+a c-b d+d^2, \\
p_3 &= a^2-3 a b-3 a c+2 a d+b^2+2 b c-3 b d+c^2-3 c d+d^2, \\
p_4 &= a^2-a b+a c-2 a d-b^2+2 b c+b d-c^2-c d+d^2.
\end{align}
Since $p_1, p_2, p_3, p_4$ are all integers, from $10p_1^2 + 10p_2^2 + p_3^2 + 5p_4^2 = 16$,
there are only three possible cases:
1) $p_1 = 0, p_2 = 0, p_3^2 = 16, p_4 = 0$:
Solutions: $a=d, b=c, c^2-3cd+d^2=1$, or $a=d, b=c, c^2-3cd+d^2=-1$
2) $p_1^2 = 1, p_2 = 0, p_3^2 = 1, p_4^2 = 1$
Solutions: $a=c, d=0, a^2+ab-b^2=1$, or $a=c, d=0, a^2+ab-b^2= -1$,
or $a=0, b= d, c^2-cd-d^2 = 1$, or $a=0, b= d, c^2-cd-d^2 = -1$
3) $p_1 = 0, p_2^2 = 1, p_3^2 = 1, p_4^2 = 1$
Solutions: $b=0, c=d, a^2 - ac - c^2 = 1$, or $b=0, c=d, a^2 - ac - c^2 = -1$,
or $a=b, c = 0, b^2+bd-d^2 = 1$, or $a=b, c = 0, b^2+bd-d^2 = -1$
We need to prove that for all solution above, $(\frac{C}{2} + \frac{D}{2}\sqrt{5})^k = \frac{3}{2} + \frac{1}{2}\sqrt{5}$ never hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$.
I don't know how to deal with this non-linear differential equation. I tried to consider $\ddot{x}(t+2\pi)+3x(t+2\pi)+x^3(t+2\pi)=0$ but with no success...
I need to prove this in order to solve a problem of dependence on initial conditions.
Can you please help me?
| Letting $y=\dot{x}$ and $g(x)=3x+x^3$ we obtained the following first order planar system
$$
\begin{pmatrix} \dot{y} \\
\dot{x}
\end{pmatrix} = \begin{pmatrix} -3x - x^3\\ y
\end{pmatrix} = \begin{pmatrix} - g(x) \\ y
\end{pmatrix} \tag{1}\label{one}
$$
This is a Hamiltonian system
$$
\begin{pmatrix} \dot{y}\\ \dot{x} \end{pmatrix} =
\begin{pmatrix} -\partial_xH(x,y)\\ \partial_yH(x,y) \end{pmatrix}
$$
where $H(x,y)=\frac32 x^2 +\frac{x^4}{4}+\frac{1}{2}y^2=G(x) +\frac{1}{2}y^2$, where $G(x)=\int^x_0g$. Thus any solution $\mathbf{x}(t)=(x(t),y(t))$ to $\eqref{one}$ satisfies
$$
H(\mathbf{x}(t))=H(\mathbf{x}_0)
$$
Here is a plot of a few level curves of the first integral $H$.
Since $G(x)= \frac32 x^2 + \frac14 x^4$ has a minimum at $x=0$, all solutions near the critical point $(0,0)$ of $\eqref{one}$ are periodic (this comes from Poincaré-Bendixon's theorem) and since $G(x)$ is even, a periodic solution crosses the $x$-axis at points $(\pm b,0)$ and the period of the solution is given by
\begin{aligned}
T_b&=2\int^b_{-b}\frac{1}{\sqrt{2(G(b)-G(x))}}\,dx=\frac{4}{\sqrt{2}}\int^b_0\frac{1}{\sqrt{G(b)-G(x)}}\,dx\\
&=4\sqrt{2}\int^b_0\frac{1}{\sqrt{(b^2+3)^2-(x^2 + 3)^2}}\,dx
\end{aligned}
Since $\sqrt{(b^2+3)^2-(x^2+3)^2}=\sqrt{(b-x)(b+x)(6+b^2+x^2)}$, we have that
$$\frac{4\sqrt{2}}{\sqrt{6+2b^2}}\leq T(b)\leq \frac{8\sqrt{2}}{\sqrt{6+b^2}}<\frac{8}{\sqrt{3}}<2\pi$$
Here is a plot of the period the period of solutions to $\eqref{one}$ as a function of the intercept $b$ as well as per and lower bounds for its decay.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then the original inequality is equal to
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \sqrt{3(ab+bc+ca)}+2(a+b+c)
\end{align*}
By using Cauchy's inequality,we can get
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \frac{1}{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2
\end{align*}
So if we can proof (Since the original equality is true ,then the following equality must be true)
\begin{align*}
(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\geq 2\sqrt{3(ab+bc+ca)}+4(a+b+c)
\end{align*}
or
\begin{align*}
\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)\tag{*}
\end{align*}
then the problem is done.But I can't prove (*).
| Let $ab+ac+bc=1$.
Thus, we need to prove that:
$$\sum_{cyc}\sqrt{a^2+1}\geq\sqrt3+a+b+c.$$
Now, let $a=\tan x$, $b=\tan y$ and $c=\tan z$, where $\{x,y,z\}\subset\left(0^{\circ},90^{\circ}\right).$
Thus, $x+y+z=90^{\circ}$ and we need to prove that $$\sum_{cyc}f(x)\geq\sqrt3,$$ where $$f(x)=\frac{1}{\cos{x}}-\tan{x}.$$
But $$f''(x)=\frac{(1-\sin{x})^2}{\cos^3x}>0$$ and now our inequality follows from Jensen.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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