Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the partial sum of the series and the limit of it $\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$ Find the partial sum of the series and the limit of it:
$\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$
I did $\lim_{n\to\infty}\frac{1}{2^n}\tan\frac{1}{2^n} = 0$ so we can not say that the series is divergent.
I tried to use telescopic sum but I do not know what to do with that $\tan$. Can you give me any hint how to write it?
UPDATE
So I did what @lab bhattacharjee recommended and I got that: $\frac{1}{2}\tan x = \frac{1}{\tfrac{1}{\tan\frac{x}{2}}-\tan\frac{x}{2}}\iff\tan x = \frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}=\frac{1+\tan\frac{x}{2}+\tan\frac{x}{2} - 1}{(1+\tan\frac{x}{2})(1-\tan\frac{x}{2})} = \frac{1}{1-\tan\frac{x}{2}}-\frac{1}{1+\tan\frac{x}{2}}.$
What should I do next?
| Hint :
$$\cot x-\tan x =2\cot 2x\iff\dfrac12\tan x=? $$
Put $x=1/2^n, n=1,2,3,\cdots m$ and add to find the partial sum
Finally set $m\to\infty $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having trouble in figuring out the best approach to simplify $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.
| Solving your equation $$\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$$ for $c$ we get
$$c=-\frac{ab}{a+b}$$ then we get
$$\frac{1}{a}+\frac{1}{b}-\frac{a+b}{ab}=0$$
| {
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"url": "https://math.stackexchange.com/questions/2969596",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integral $\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$ The following integral appeared on the $8$th Open Mathematical Olympiad of the Belarusian-Russian University. $$I=\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$$ I used power series: $$x-\sin x = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}\rightarrow I=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n+1)!} \int_0^\infty \frac{x^{2n-2}}{x^2+4}dx$$ Taking the inner integral and substituting $\displaystyle{x^2=4t \rightarrow dx=\frac{dt}{\sqrt{t}}}$ gives: $$\int_0^\infty \frac{x^{2n-2}}{x^2+4}dx=4^{n-2}\int_0^\infty \frac{t^{n-1-\frac12}}{t+1}dt$$
$$=4^{n-2} B\left(n-\frac12, 1-n+\frac12\right)=4^{n-2}\Gamma\left(n-\frac12\right)\Gamma\left(1+\frac12-n\right)$$
And using Euler's reflection formula: $$\Gamma\left(n-\frac12\right)\Gamma\left(1+\frac12-n\right)=\pi \csc\left({n\pi-\frac{\pi}{2}}\right)=-\pi\sec(n\pi)=(-1)^{n+1}\pi$$
$$I=\pi\sum_{n=1}^\infty \frac{4^{n-2}}{(2n+1)!}=\frac{\pi}{32} \sum_{n=1}^\infty \frac{2^{2n+1}}{(2n+1)!}=\frac{\pi}{32}(\sinh 2 -1)$$ I did not found the official solution, but the answer given $\displaystyle{\frac{\pi}{32}\left(\frac{e^2-1}{e^2}\right)},\,$ doesn't match. Can you help me find my mistake? And maybe share some different methods to solve this integral?
| Let $J(a)=\displaystyle\int_{0}^{\infty}\frac{ax-\sin ax}{x^3(x^2+1)}dx$. Differentiating by $a$ three times (which is admissible under the integral sign, because the resulting integrals converge uniformly for $a$ in any finite interval), we get $$J'''(a)=\displaystyle\int_{0}^{\infty}\frac{\cos ax}{x^2+1}dx=\frac{\pi}{2}e^{-|a|}.$$ With $J(0)=J'(0)=J''(0)=0$ this gives $J(a)=\dfrac{\pi}{2}\Big(1-|a|+\dfrac{a^2}{2}-e^{-|a|}\Big)\operatorname{sgn}(a)$.
The answer is $\dfrac{J(2)}{16}=\dfrac{\pi}{32}(1-e^{-2})$ as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two different color rooks are placed on a chessboard so they don't attack each other. What is a number of ways we can do that?
Of course we have $64$ choices for the first one and then $49$ choices for the second one. So we have $64\cdot 49$ ways to put them. But some of the configurations are essentially the same if we say rotate chessboard or reflect across some diagonal. So we should divide $64 \cdot 49$ with some number. But which? Is it $4$ since we have $4$ rotations which takes chessboard to it self? And then also by $4$ since we have $4$ reflections across the line? So the final result should be
$$64\cdot 49\over 16$$
But I feel this is not correct, since if we have $7\times 7$ ''chessboard'' then we would get the result: $$49\cdot 36\over 16$$
which is not an integer.
| Hmm, I think I found a proper number.
We can (essentialy) put white rook only on one of a marked cells in pictures (A) or (B):
(A):
\begin{array} {|r|r|r||r|r|r||r|r|}
\hline
\color{white}{*}& * &*& * & \color{white}{*} &\color{white}{*} & \color{white}{*} & \color{white}{*} \\
\hline
& & *& * & & & & \\
\hline
& && * & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
\end{array}
We have $6$ choises for the white rook and of course $49$ for the black rook. So the answer in this case is $6\cdot 49=294$.
(B)
\begin{array} {|r|r|r||r|r|r||r|r|}
\hline
*& && & \color{white}{*} &\color{white}{*} & \color{white}{*} & \color{white}{*} \\
\hline
& * & & & & & & \\
\hline
& &*& & & & & \\
\hline
& & & * & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
& & & & & & & \\
\hline
\end{array}
In this case we have $4$ choises for the white rook and $21$ for the black rook not on diagonal and $7$ if it is on diagonal. So the answer in this case is $4\cdot 28=112$.
Overall we have 406 ways to put two nonatacking rooks on a chessboard.
| {
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Help Factoring A Difficult Expression Given A = $\begin{bmatrix}a & b & c \\c & a & b \\b & c & a \\\end{bmatrix}$, B =$\begin{bmatrix}a & b & c \\a^2 & b^2 & c^2 \\a^4 & b^4 & c^4 \\\end{bmatrix}$
Suppose $C =AB=\begin{bmatrix}
a^2 + a^2b + a^4c & ab + b^3 + b^4c & ac +bc^2 + c^5 \\
\\
ca + a^3 +a^4b & cb + ab^2 + b^5 & c^2 + ac^2 +bc^4 \\
\\
ba + a^2c + a^5 & b^2 + b^2c + ab^4 & bc + c^3 + ac^4
\end{bmatrix}$
I have been asked to show that the following holds:
$det(C)= -abc(b-c)(a-c)(a-b)(a^2-ab+b^2 -bc +c^2 -ac)(a+b+c)^2$
The way i proceeded to go about this is using the fact that $det(AB) = det(A)det(B)$.Now finding the determinant of A and B i have the following.
$det(A)= a^3 -3abc +b^3 +c^3$
$det(B) = abc(c^3b - cb^3 - c^3a + ca^3 + b^3a - ba^3)$
The problem i have is factoring out the necessary factors from the product of the determinants. I have tried loads of methods now and i have got nowhere and am a bit depressed by it. Any ideas would be really appreciated.
Best Regards
| Actually, we have
$$
\det(A)=(a^2 - ab - ac + b^2 - bc + c^2)(a + b + c),
$$
$$
\det(B)=(a + b + c)(a - b)(a - c)(b - c)abc.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Showing that $2^{2^{2^x}}<100^{100^x}$ for large $x$ I wish to show that $2^{2^{2^x}}<100^{100^x}$ for $x$ sufficiently large. I have taken logs (base 10) of both sides to get $2^{(2^x-1)}\log_{10} 2$ and $100^x$. It is not immediately clear how I can prove that the second term here is larger than the first.
Any help appreciated.
| HINT
Since $\log$ function is strictly increasing we have
$$2^{2^{2^x}}<100^{100^x}\iff \log_{10}\left(2^{2^{2^x}}\right)<\log_{10}\left(100^{100^x}\right) \iff 2^{2^x}\log_{10}2<2\cdot 100^x$$
then again
$$2^{2^x}\log_{10}2<2\cdot 100^x \iff \log_{10}\left(2^{2^x}\log_{10}2\right)<\log_{10}\left(2\cdot 100^x\right)$$
$$2^x\log_{10}2 + \log_{10}(\log_{10}2)<\log_{10}2+2x$$
| {
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Prove that $f$ is additive if $f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$
Say $f:\mathbb{R}\to \mathbb{R}$ is non-constant such that $$f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$$
Prove that $f(x+y)=f(x)+f(y)$.
If we put $a=f(0)$ and $y=0$ we get $$f(x)^2+af(x)= f(x)^2+a^2 \implies af(x) = a^2$$
If $a\ne 0$ then $f(x)=a$ is constant function which can not be, so $a=0$. Now if we put $x=y$ we get $$f(x)f(2x)=2f(x)^2$$
From here I have no more idea what to do.
Edit after Lulu's comment: If we put also $y=-x$ we get $$2f(x)^2= f(x)f(2x) = f(x)^2+f(-x)^2\implies \boxed{f(x)^2=f(-x)^2}$$
| This is only part of an answer, as I can't prove that $f$ is even or odd.
First, we want to show that $f$ is either even or odd, given $f(x)^2 = f(-x)^2$.
Say there are two points $x$ and $y$ such that $f(x) = f(-x)$ and $f(y) = -f(-y)$. Then, $f(x)f(x-y) + f(x+y) = f(x)^2+f(y)^2$, and, by plugging in $-x,y$, we get $$f(-x)f(-x-y) + f(y)f(y-x) = f(-x)^2 + f(y)^2 = f(x)^2 + f(y^2)$$ so we have $$f(x)f(-x-y) + f(y)f(y-x) = f(x)f(x-y) + f(y)f(x+y)$$
We now take cases on if $x+y$ and $y-x$ are odd or even.
If they are both even, then we have $f(x+y) = f(x-y)$ or $f(x) = f(y)$ by factorization, which is constant. If they are both odd, Then $f(x) = -f(y)$ or $f(x+y) = f(x-y)$, which still both are contradictions.
Therefore, one has to be even and one has to be odd. I don't see any way to continue from here; so I skip to the part where they are either even or odd.
Now, we show that $f(x)$ is odd.
We know that $f(x)f(x−y)+f(y)f(x+y)=f(x)^2+f(y)^2$, and by plugging in $-y$ in for $y$, we get $f(x)f(x+y) + f(y)f(x-y) = f(x)^2 + f(-y)^2 = f(x)^2 + f(y)^2$, since $f(-y)^2 = f(y)^2$ and therefore $$f(x)(f(x+y)-f(x-y)) = f(y)f(x+y) - f(-y)f(x-y)$$
Now, we know that $f$ is either odd or even (or assume it for now). However, if $f$ is even, then we have $$f(x)(f(x+y)-f(x-y)) = f(y)(f(x+y) - f(x-y))$$ So either $f(x) = f(y)$, contradiction, or $f(x+y) - f(x-y) = 0 \implies f(x+y) = f(x-y)$, and because $x+y$ and $x-y$ can range all real numbers, which is also a contradiction.
Now, if $f(x+y)$ is not $f(x)+f(y)$, then $f(x-y) \neq f(x) - f(y)$, and by replacing $x$ with $y$, we get $f(y-x) \neq f(y) - f(x)$, so $f(x-y)+f(y-x) \neq 0$, which is a contradiction of the fact that $f$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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On integer solutions of $2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$
$\textsf{Background}$
From the double-angle formula $\cos2\alpha=2\cos^2\alpha-1$, we can get $$\cos15^\circ=\sqrt{\frac{1+\cos30^\circ}2}=\frac{\sqrt{2+\sqrt3}}2$$ but we also know that it is equivalent to $\dfrac{\sqrt6+\sqrt2}4$.
This is an example of an equality such that $$2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$$ after some rearranging. We can write $y$ in terms of $x$ without much bother: $$y=2+\frac x2+2\sqrt{1+\sqrt{2x}}-\sqrt{2x(1+\sqrt{2x})}.$$ But when are $x$ and $y$ integers? Here is a plot of the curve:
Some obvious solutions are $(0,0)$, $(0,4)$, $(2,3)$ and $(32,0)$. Of course, $x=2k$ for some integer $k$, leaving us with $$y=2+k+2\sqrt{1+2\sqrt k}-2\sqrt{k(1+2\sqrt k)}$$ Hence this boils down to finding $k$ such that $$(1-\sqrt k)\sqrt{1+2\sqrt k}$$ is an integer. Any advances on this?
| The stated condition leads easily to
$$(\sqrt x + \sqrt{2y} - 2\sqrt2)^2 = 8 - 4\sqrt y.$$
Hence $0 \le y \le 4.$ Trying each value of $y$ in turn, one finds that the only solutions with $x$ also an integer are the four already stated.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.
Here's what I did:
I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:
$\frac{-x^2+2x-3}{x^3-x^2+x-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$
I then found the values for $A$, $B$ and $C$:
$x^2(A+B)+x(-B+C)+A-C=-x^2+2x-3$
$
\left\{
\begin{array}{c}
A+B=-1 \\
-B+C=2 \\
A-C=-3
\end{array}
\right.
$
$A=-1$, $B=0$, $C=2$
So now back again with the integral:
$\int \frac{-1}{x-1}dx+\int \frac{2}{x^2+1}dx=-\int \frac{1}{x-1}dx + 2 \int \frac{1}{x^2+1}dx=-\ln|x-1|+2|x^2+1|=2\ln|\frac{x^2+1}{x-1}|+C$
But why the correct answer is $2 \arctan(x)-\ln|x-1|+C$? Where does the $arctan$ come from?
| The $\arctan$ comes as a result of the antiderivative
$$\int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C$$
You calculated that antiderivative wrong: the antiderivatives of fractions are a little more nuanced than just throwing the denominator in a natural logarithm.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^5-k)$ is true. Then there is $m \in \mathbb{Z}^+$ such that $$k^5 - k = 5m.$$ We must show that this statement is true for $n = k+1$, i.e. show that there is $\ell \in \mathbb{Z}^+$ such that $$(k+1)^5 - (k+1) = 5\ell.$$ Note that $(k+1)^5 - (k+1)$ expands as $k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$.
We can try to find a polynomial $P(x)$ such that $$(k^5-5) + P(x) =(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$$ so as to try to add $P(x)$ to both sides of the assumption. We find that
$$\begin{align}P(x) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - (k^5-5) \\
&=5k^4 + 10k^3 + 10k^2 + 5k \end{align}$$ and we can also observe that since $k\in\mathbb{Z}^+$, we have that $\frac{1}{5}P(x) = k^4 + 2k^3 + 2k^2 + k$ is a positive integer. Thus we add this to both sides of our assumption $$ \begin{align}
k^5 - k &= 5m \\
(k^2 - k) + P(x) &= 5m + P(x) \\
(k+1)^5 - (k+1) &= 5\left(m + \tfrac{1}{5}P(x)\right)
\end{align}.$$ Since $\frac{1}{5}P(x),m \in \mathbb{Z}^+$, it follows that $m + \tfrac{1}{5}P(x) \in \mathbb{Z}^+$. Thus $5 \left| \big[ (k+1)^5 - (k+1) \big] \right.$
By PMI, $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$.
My questions
*
*Is this proof valid?
*What other ways can this be proved by induction? The polynomial expansions took a while to deal with, so I was wondering if there are any alternate methods. (Just FYI, I only have college first-year-level knowledge)
| Here is another way. For every $n \in \mathbb{Z}^{+}$, we have
(here we use $a^2 -b^2 = (a-b)(a+b)$)
\begin{align}
n^5 - n = n(n^4 -1) & = n(n^2-1)(n^2+1) \\
&= n(n-1)(n+1)(n^2 -4 +5) \\
& = n(n-1)(n+1)(n^2 -4) + 5n(n-1)(n+1) \\
& = n(n-1)(n+1)(n-2)(n+2) + 5n(n-1)(n+1) \\
& = (n-2)(n-1)n(n+1)(n+2) + 5n(n-1)(n+1).
\end{align}
In the last line, the first term is a product of 5 consecutive
integers, so it must be divisible by $5$. Moreover, clearly $5$ divides the second one :)
| {
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Integrating a 2D Gaussian over a linear strip How do I show that
$$\int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2 \sigma ^2}} \left(\text{erf}\left(\frac{2 d-\sqrt{2} x}{2 \sigma
}\right)+\text{erf}\left(\frac{2 d+\sqrt{2} x}{2 \sigma }\right)\right)}{2 \sqrt{2
\pi } \sigma }dx=\text{erf}\left( \frac{d}{\sqrt{2}\sigma}\right)?$$
Mathematica couldn't do it, though I've verified this numerically. The following argument shows that these are equal, but I want to show this analytically, i.e. using direct integration methods.
(Let
$$P(x,y) = \frac{1}{2\pi \sigma^2} \exp\left(- \frac{x^2+y^2}{2\sigma^2}\right),$$
which is clearly spherically symmetric. I want to integrate this function over the region
$$R = \{ |x-y| \leq d \mid (x,y) \in \mathbb{R}^2\}$$
which is just a diagonal linear strip of width $d$.
(a) I could evaluate
$$ \int_{-\infty}^{\infty}\int_{x-\sqrt{2}d}^{x+\sqrt{2}d} P(x,y)\,dx\,dy = \int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2 \sigma ^2}} \left(\text{erf}\left(\frac{2 d-\sqrt{2} x}{2 \sigma
}\right)+\text{erf}\left(\frac{2 d+\sqrt{2} x}{2 \sigma }\right)\right)}{2 \sqrt{2
\pi } \sigma }dx \tag{1}$$
(b) or I could note the spherical symmetry of $P(x,y)$, rotate my region $R$ to the $y$-axis, and get
$$\int_{-\infty}^{\infty} \int_{-d}^d P(x,y)\,dx\,dy = \text{erf}\left( \frac{d}{\sqrt{2}\sigma}\right)\tag{2}.)$$
| Direct $x$-integration is not possible. However, expressing the erf functions by (direct) indefinite integrals does the job, after exchanging integration orders:
Notice$$
\text{erf}\left(\frac{2 d\pm\sqrt{2} x}{2 \sigma }\right)= \int
\frac{2 e^{-\frac{( \sqrt{2}d \pm x)^2}{2 \sigma^2} }}{\sqrt{\pi} \sigma} \text{d} d
$$
Hence we can write
$$
\int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2 \sigma ^2}} \left(\text{erf}\left(\frac{2 d-\sqrt{2} x}{2 \sigma
}\right)+\text{erf}\left(\frac{2 d+\sqrt{2} x}{2 \sigma }\right)\right)}{2 \sqrt{2
\pi } \sigma }dx = \\
\int \text{d} d \int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2 \sigma ^2}} \left( \frac{2 e^{-\frac{( \sqrt{2}d + x)^2}{2 \sigma^2} }}{\sqrt{\pi} \sigma} + \frac{2 e^{-\frac{( \sqrt{2}d - x)^2}{2 \sigma^2} }}{\sqrt{\pi} \sigma} \right)}{2 \sqrt{2
\pi } \sigma }dx = \\
2 \int \frac{e^{-d^2/(2 \sigma^2)}}{\sqrt{2 \pi} \sigma} \text{d} d =\text{erf}\left( \frac{d}{\sqrt{2}\sigma}\right)
$$
Done. $\qquad \square$
| {
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Area bounded by $y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and $y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$
The area of the region between the curves
$y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and
$y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$ bounded by the lines $x=0$ and
$x=\frac{\pi}{4}$ is:
a) $\int_{0}^{\sqrt{2}-1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
b) $\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
c) $\int_{0}^{\sqrt{2}+1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
d) $\int_{0}^{\sqrt{2}+1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
My Attempt:
So I started with,
$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx$$
$$\int_{0}^{\frac{\pi}{4}}\frac{2\sin{x}}{\sqrt{\cos{x}}(\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}})}dx$$
Putting $\cos{x}=u$
$$\int_{1}^{\frac{1}{\sqrt{2}}}\frac{-2}{\sqrt{u}(\sqrt{1+\sqrt{1-u^2}}+\sqrt{1-\sqrt{1-u^2}})}du$$
Putting $u=\frac{1}{w}$ and further $w-1=t$,
I end up with,
$$\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)(\sqrt{(1+t)+\sqrt{(1+t)^2-1}})+(\sqrt{(1+t)-\sqrt{(1+t)^2-1}})}dt$$
But I am not able to reduce this to any of the given options? Am I miscalculating something or this is reducible?
Any hints would be helpful.
Thank you.
| HINT
We could use Tangent half-angle substitution to obtain
$$\sin x=\frac{2t}{1+t^2}\quad \cos x=\frac{1-t^2}{1+t^2}\quad t=\tan \frac x 2 \implies dt=\frac12(1+t^2)dx$$
then
$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx=\int_0^{\sqrt2 -1} \left(\sqrt{\frac{(1+t)^2}{1-t^2}}-\sqrt{\frac{(1-t)^2}{1-t^2}}\right)\frac2{1+t^2}dt=\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total score on one turn.
My working:
$\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k).$
where, $P(X=k) = \frac{1}{6}^{k-1}\frac{5}{6}$
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{2}k\frac{1}{6}^{k-1} = \frac{5}{6}(1+2(\frac{1}{6})) = \frac{10}{9}$
EDITED
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{\infty}k\frac{1}{6}^{k-1} = \frac{5}{6}\frac{1}{(1-\frac{1}{6})^{2}} = \frac{6}{5}$
But on the answer sheet, it is $\frac{21}{5}$
I would appreciate if someone can point out my mistake. Thanks!
| Let the expected value be X. If you throw a six, your expected value is 6+X, otherwise the score is whatever you threw. Therefore
X = (1 + 2 + 3 + 4 + 5 + (6+X)) / 6
(5/6) X = 21/6
X = 21/5
| {
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"url": "https://math.stackexchange.com/questions/2989179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$a,b,c,d,e$ are zeroes of $6x^5+5x^4+4x^3+3x^2+2x+1$ find $(a+1)(b+1)(c+1)(d+1)(e+1)$
If $a,b,c,d,e$ are zeroes of the polynomial
$$6x^5+5x^4+4x^3+3x^2+2x+1$$
find the value of $(a+1)(b+1)(c+1)(d+1)(e+1)$.
According to me in order to solve this problem one should first factorize the given polynomial in the form of:
$$(x-a)(x-b)(x-c)(x-d)(x-e)$$
and equate the two and then compare the coefficients to get the following equations:
$(a+b+c+...)=-5$
$(ab+ac+ad+bd+.....)=4$
$(abc+abd+abe+...)=-3$
$(abcd+abce+abde+.....)=2$
$abcde=-1$
Now, adding all the equations and with some factorization we get:
$$(a+1)(b+1)(c+1)(d+1)(e+1)=-2$$
but my book says the answer is $0.5$.
Where am I going wrong and what is the actual method to solve this, with the correct answer?
| It is just $$-{1\over 6}p(-1)= -(-6+5-4+3-2+1) = {1\over 2}$$
Proof: $$p(-1) = 6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e)$$
$$ = -6(1+a)(1+b)(1+c)(1+d)(1+c)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$
My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$
I cannot prove the final part.
Please help me. Thanks in advance.
| $$
\begin{align}
&1+\color{#C00}{2\cos\left(\frac{2\pi}7\right)}+\color{#090}{2\cos\left(\frac{4\pi}7\right)}+\color{#00F}{2\cos\left(\frac{6\pi}7\right)}\\
&=1+\color{#C00}{\cos\left(\frac{2\pi}7\right)}+\color{#090}{\cos\left(\frac{4\pi}7\right)}+\color{#00F}{\cos\left(\frac{6\pi}7\right)}+\color{#00F}{\cos\left(\frac{8\pi}7\right)}+\color{#090}{\cos\left(\frac{10\pi}7\right)}+\color{#C00}{\cos\left(\frac{12\pi}7\right)}\\
&=\operatorname{Re}\left(1+e^{2\pi i/7}+e^{4\pi i/7}+e^{6\pi i/7}+e^{8\pi i/7}+e^{10\pi i/7}+e^{12\pi i/7}\right)\\
&=\operatorname{Re}\left(\frac{e^{14\pi i/7}-1}{e^{2\pi i/7}-1}\right)\\
&=0
\end{align}
$$
Therefore,
$$
\cos\left(\frac{2\pi}7\right)+\cos\left(\frac{4\pi}7\right)+\cos\left(\frac{6\pi}7\right)=-\frac12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Factoring the polynomial $3(2x+3)^2 + 7(2x+3) - 6$
Factor $3(2x+3)^2 + 7(2x+3) - 6$
What I did:
With the substitution $X=2x+3$:
\begin{align}
3(2x+3)^2 + 7(2x+3) - 6
&= 3X^2-9X+2X-6 \\
&= 3X(X-3)+2(X-3) \\
&= (X-3)(3X+2) \\
&=((2x+3)-3)(3(2x+3)+2) \\
&=(2x+0)(6x+9+2) \\
&=(2x)(6x+11)
\end{align}
| Your method is correct. You made a sign error:
$$...=(X+3)(3X-2)=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the inequality yields
$$1+\frac{x}{2}-\frac{x^2}{8}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<1+\frac{x}{2}$$
which is equivalent to
$$\begin{cases}
\cancel{1+\frac{x}{2}-\frac{x^2}{8}}\cancel{<1+\frac{x}{2}-\frac{x^2}{8}}+\frac{x^3}{16 (\xi+1)^{5/2}}\quad(1)\\
\cancel{1+\frac{x}{2}}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<\cancel{1+\frac{x}{2}}\qquad\,\,\,\,\,\, (2)
\end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$\frac{x^3}{16 (\xi+1)^{5/2}}<\frac{x^2}{8}$$
But $$\displaystyle{\max_{0\leq\xi\leq x}\Bigg\{\frac{1}{16 (\xi+1)^{5/2}}\Bigg\}}=\displaystyle{\min_{0\leq\xi\leq x}{\Big\{16 (\xi+1)^{5/2}}\Big\}}$$
which occurs at $\xi=0$, and thus, for $(2)$, it is left to prove that
$$\frac{x^3}{16}<\frac{x^2}{8}$$
but this happens for $x<0\,\vee\,0<x<2$.
Instead, if I use $\xi=x$ then inequality $(2)$ becomes
$$\frac{x^3}{16 (x+1)^{5/2}}<\frac{x^2}{8}$$
which holds for $-1<x<0\,\vee\,x>0$, and thus coincides with the restriction given by the problem.
Would it be correct to take $\xi=x$ rather than $0$? Is this approach correct at all?
| At $x=0$, all three members equal $1$. Then we can take the derivative and
$$\frac12-\frac x4<\frac1{2\sqrt{x+1}}<\frac12.$$
Again, we have equality at $x=0$ and
$$-\frac14<-\frac1{4(x+1)^{3/2}}<0.$$
This final bracketing is obvious.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
} |
Prove that number $\underbrace{11 \ldots1}_{100} \underbrace{22 \ldots2}_{100}$ is product of two consecutive numbers Prove that number $\underbrace{11 \ldots1}_{100}$$\underbrace{22 \ldots2}_{100}$ is product of two consecutive numbers
$\begin{align}\underbrace{11 \ldots1}_{100} \underbrace{22 \ldots2}_{100}&=10^{199}+10^{198}+\ldots+10^{100}+2(10^{99}+10^{98}+\ldots+10+1)\\&=(10^{100}+2)(10^{99}+10^{98}+\ldots+10+1)=(10^{100}+2)\frac{10^{100}-1}{10-1}\end{align}$.
Is this good path or not?
| Insert $1=\frac{1}{3}*3$
$$
(10^{100}+2)\frac{1}{3} 3\frac{10^{100}-1}{10-1}\\
x= \frac{10^{100}+2}{3}\\
y=3\frac{10^{100}-1}{10-1} = \frac{10^{100}-1}{3}\\\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Residue of order 3 - Find the Laurent Series for the function
\begin{align}
f(z) = \frac{1}{(z^2+4)^3}
\end{align}
about the isolated singular pole $z = 2i$. What is the pole order? What is the residue at the pole?
My attempt:
\begin{align}
f(z) &= \frac{1}{(z^2+4)^3}\\
&= \frac{1}{(z+2i)^3(z-2i)^3}\\
\end{align}
Here we see $z=2i$ is a 3rd order pole.
A Laurent series is defined with respect to a particular non-analytic point $z_0$ and a path of integration C. The path of integration must lie in an annulus surrounding $z_0$ and so
\begin{align}
f(z) &= \sum_{n=-\infty}^\infty a_n(z-z_0)^n
% = \sum_{n=0}^\infty a_n(z-z_0)^{n}
% + \sum_{n=1}^\infty b_n(z-z_0)^{n}
\end{align}
where
\begin{align}
a_n &= \frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{n+1}}\\
% && \text{Regular Part}&\\
% b_n &= \frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{-n-1}} \label{eq:laurentb}
% && \text{Principle Part} &
\end{align}
We find the $a_n$ term using $z_0=2i$,
\begin{align}
a_n &= \frac{1}{2\pi i}\oint_C \frac{\frac{1}{(z+2i)^3(z-2i)^3}}{(z-2i)^{n+1}}\\
a_n &= \frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^3(z-2i)^{n+4}}
\end{align}
and from Cauchy's Integral Formula we can find the residue $(n=-1)$ term,
\begin{align}
a_{(-1)}&=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(z-2i)^{-1+4}}\\
&=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(z-2i)^{3}}\\
&=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(\sqrt{z}+\sqrt{2i})}
\frac{1}{(z-2i)(\sqrt{z}-\sqrt{2i})}????
\end{align}
I can't seem to get anywhere near a correct answer ($Res(f;2i) = -3i/512$). I'm supposed to use the Laurent expansion at $z=2i$ and ultimately find the $z^-1$ coefficient, but I'm so lost.... I've also tried partial fraction expansion, but am running in circles. Engineering student here, so be nice ;)
| Integration is completely unnecessary here. More generally, never integrate if you can differentiate. And most of the times, geometric series is more than enough.
$$
\begin{aligned}
\frac{1}{(z+2i)^3}&=\frac{1}{2}\frac{\mathrm d^2}{\mathrm dz^2}\frac{1}{(z-2i)+4i}\\
&=\frac12\frac{\mathrm d^2}{\mathrm dz^2}\left[-\frac i4+\frac{1}{16}(z-2i)+\frac{i}{64}(z-2i)^2+\cdots\right]\\
&=\frac{i}{64}-\frac{3}{256}(z-2i)-\frac{3i}{512}(z-2i)^2+\cdots
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
f(x)&=\frac{1}{(z-2i)^3}\left[\frac{i}{64}-\frac{3}{256}(z-2i)-\frac{3i}{512}(z-2i)^2+\cdots\right]\\
&=\frac{i}{64}(z-2i)^{-3}-\frac{3}{256}(z-2i)^{-2}\color{red}{-\frac{3i}{512}(z-2i)^{-1}}+\cdots
\end{aligned}
$$
Using the formula
$$
\frac{1}{1-z}=\sum_{n\ge0}z^n
$$
you should be able to write down a general formula for $a_n$, using the steps described above. I leave this to you.
| {
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"url": "https://math.stackexchange.com/questions/3001836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
| We can write your sum as
$$
\eqalign{
& f(n) = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k \cr} \right){1 \over k}} = \cr
& = \sum\limits_{k = 0}^{n - 1} {\left( { - 1} \right)^{\,k} \left( \matrix{
n \cr
k + 1 \cr} \right){1 \over {k + 1}}} = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^{\,k} \left( \matrix{
n \cr
k + 1 \cr} \right){1 \over {k + 1}}} = \cr
& = \sum\limits_{k = 0}^\infty {t_k } \cr}
$$
and we can express it in terms of a Hypergeometric function, since
$$
\eqalign{
& t_0 = \left( \matrix{
n \cr
1 \cr} \right) = n \cr
& {{t_{k + 1} } \over {t_k }} = - {{n^{\,\underline {\,k + 2\,} } } \over {\left( {k + 2} \right)\left( {k + 2} \right)!}}
{{\left( {k + 1} \right)\left( {k + 1} \right)!} \over {n^{\,\underline {\,k + 1\,} } }} = \cr
& = - {{\left( {n - 1 - k} \right)} \over 1}{{\left( {k + 1} \right)} \over {\left( {k + 2} \right)\left( {k + 2} \right)}}
= {{\left( {k - n + 1} \right)\left( {k + 1} \right)} \over {\left( {k + 2} \right)\left( {k + 2} \right)}} \cr}
$$
Then
$$
f(n) = n\;{}_3F_2 \left( {\left. {\matrix{
{ - n + 1,\;1,\;1} \cr
{2,\;2} \cr
} \;} \right|\;1} \right)
$$
Alternatively, we have that
$$
\eqalign{
& f(n + 1) = \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n + 1 \cr
k \cr} \right){1 \over k}} = \cr
& = \left( {\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k \cr} \right){1 \over k}} + \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k - 1 \cr} \right){1 \over k}} } \right) = \cr
& = \left( {\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k \cr} \right){1 \over k}} + {1 \over {n + 1}}\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n + 1 \cr
k \cr} \right)} } \right) = \cr
& = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k \cr} \right){1 \over k}} - {1 \over {n + 1}}\left( {0^{\,n + 1} - 1} \right) = \cr
& = f(n) + {1 \over {n + 1}} \cr}
$$
i.e.:
$$
\left\{ \matrix{
f(0) = 0 \hfill \cr
f(1) = 1 \hfill \cr
f(n + 1) - f(n) = \Delta f(n) = {1 \over {n + 1}} \hfill \cr} \right.
$$
or
$$
\left\{ \matrix{
g(n) = n!f(n) \hfill \cr
g(0) = 0 \hfill \cr
g(1) = 1 \hfill \cr
g(n + 1) = \left( {n + 1} \right)f(n) + n! \hfill \cr} \right.
$$
and this is the recurrence satified by
$$g(n)=\left[ \matrix{ n+1 \cr 2 \cr} \right]$$
where $\left[ \matrix{ n \cr m \cr} \right]$ represents the (unsigned) Stirling number of 1st kind.
Thus
$$ \bbox[lightyellow] {
f(n) = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \binom{n}{k}{1 \over k}}
= {1 \over {n!}}\left[ \matrix{ n + 1 \cr 2 \cr} \right]
}$$
Also refer to OEIS seq. A000254 .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Study the convergence of a series I am studying the convergence of the series $$\sum\limits_{n>=0} \left(\sqrt[n]{n} - 1\right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.
| I assume you meant $n \geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have
$$n-1=
\left(n^{\frac{1}{n}}\right)^{n}-1=\\
\left(n^{\frac{1}{n}}-1\right)\left(n^{\frac{1}{n}(n-1)}+n^{\frac{1}{n}(n-2)}+...+n^{\frac{1}{n}2}+n^{\frac{1}{n}}+1\right)\geq ...$$
using AM-GM
$$... \geq\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}(n-1)}\cdot n^{\frac{1}{n}(n-2)}\cdot ...\cdot n^{\frac{1}{n}2}\cdot n^{\frac{1}{n}}}+1\right)=\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}(n-1+n-2+...+1)}}+1\right)=
\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}\frac{(n-1)n}{2}}}+1\right)=\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt{n}+1\right)$$
Or
$$0< n^{\frac{1}{n}}-1\leq \frac{n-1}{(n-1)\sqrt{n}+1}<\frac{1}{\sqrt{n}} \tag{1}$$
From $(1)$ we see that $\lim\limits_{n\rightarrow\infty} \left(n^{\frac{1}{n}}-1\right)=0$, which means
*
*$0< \left(n^{\frac{1}{n}}-1\right)^n<\frac{1}{n^{\frac{n}{2}}}$, thus $0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n \leq \sum\limits_{n\geq1} \frac{1}{n^{\frac{n}{2}}}$. The latter is converging by ratio test $$\lim\limits_{n\rightarrow\infty}\frac{\frac{1}{(n+1)^{\frac{n+1}{2}}}}{\frac{1}{n^{\frac{n}{2}}}}=
\lim\limits_{n\rightarrow\infty}\frac{n^{\frac{n}{2}}}{(n+1)^{\frac{n+1}{2}}}=
\lim\limits_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^{\frac{n}{2}}\cdot \sqrt{n+1}}=\frac{1}{\sqrt{e}\cdot \lim\limits_{n\rightarrow\infty}\sqrt{n+1}}=0$$
*or simpler version from the definition of the limit, for $\forall n> N(\varepsilon) \Rightarrow 0<\left(n^{\frac{1}{n}}-1\right)<\varepsilon <1$, thus $0<\left(n^{\frac{1}{n}}-1\right)^n<\varepsilon^n$ and $$0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n <
\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n + \sum\limits_{n=N(\varepsilon)+1} \varepsilon^n=
\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n + \frac{e^{N(\varepsilon)+1}}{1-\varepsilon}$$
we just need to find the very first $N(\varepsilon)$ for $0<\varepsilon<1$ and $\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n$ is a finite number as well as $\frac{e^{N(\varepsilon)+1}}{1-\varepsilon}$. As a result $$0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n<\infty$$
i.e. converging.
| {
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What am I doing wrong finding $\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{1/x^2}$? It has an answer here, but I'd like to know where my solution went wrong.
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}} $$
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x +x\cdot 3^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}\cdot \frac{1+x\cdot 3^x}{x\cdot2^x-x\cdot 3^x}\cdot\frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x^2}\cdot \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{2^x-3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{(1+1)^x-(2+1)^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{1+x+o(x)-1-x2-o(x)}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{-1}{1+x\cdot3^x}} $$
$$e^{-1}$$
The answer in the book is $\frac{2}{3}$.
| You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=\sum_{k=0}^{\infty}\binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=\pm 1$ with certain restrictions on $x$. For details see this blog post.
But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$\lim_{x\to 0}\frac{a^x-1}{x}=\log a$$ and thus we get $$\lim_{x\to 0}\frac{2^x-3^x}{x(1+x3^x)}=\lim_{x\to 0}\frac{2^x-1}{x}-\frac{3^x-1}{x}=\log 2-\log 3=\log(2/3)$$ and the desired limit is $e^{\log(2/3)}=2/3$.
Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
| Notice that $|x-y|\leq |x|+|y|$ implies that:
$$\frac{|z|}{|x-y|} \geq \frac{|z|}{|x|+|y|}.$$
Then:
$$S \geq \frac{|a|}{|b|+|c|} + \frac{|b|}{|a|+|c|} + \frac{|c|}{|a|+|b|} = \\
=\frac{A}{B+C} + \frac{B}{A+C} + \frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S \geq \frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S \geq \frac{D + A^3 +B^3 + C^3 + ABC}{D} > \frac{D}{D} = 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$
I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
| @Kemono Chen proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
Divide both sides by $y$ then integrate between $0$ and $1$ we get
$$\color{red}{\frac12\mathcal{I}}=\frac12\int_0^1\frac{\tan^{-1}(y)\ln(1+y^2)}{y}dy=\int_0^1\int_0^y\frac{\ln(1+yx)}{y(1+x^2)}dxdy$$
$$=\int_0^1\frac{1}{1+x^2}\left(\int_x^1\frac{\ln(1+xy)}{y}dy\right)dx=\int_0^1\frac{\operatorname{Li}_2(-x^2)-\operatorname{Li}_2(-x)}{1+x^2}dx$$
$$\overset{IBP}{=}\int_0^1\tan^{-1}(x)\left(\frac{2\ln(1+x^2)}{x}-\frac{\ln(1+x)}{x}\right)dx\\=\color{red}{2\mathcal{I}}-\int_0^1\frac{\tan^{-1}(x)\ln(1+x)}{x}dx$$
which can be written as
$$\int_0^1\tan^{-1}(x)\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Distance from focus to nearest point in ellipse Consider an ellipse with semi-axes $a$ (major) and $b$ (minor). For such an ellipse the distance of focus to the centre is:
$f = \sqrt{a^2-b^2}$
Now, the distance from the focus to the nearest point on the ellipse is along the major semi-axis a, thus this distance is:
$r_1 = a - f = a - \sqrt{a^2-b^2}$
Two simple questions now:
*
*How can we prove this is the shortest distance?
*Can we somewhat prove that the following is always true:
$\frac{a - \sqrt{a^2-b^2}}{b} < 1$
| For 1) I wanted to do a very simple thing. A top part of ellipse has the equation
$y = b\sqrt{1-\frac{x^2}{a^2}}$
The squared distance from focus to any point is then given by
$d^2 = \left(x - \sqrt{a^2-b^2}\right)^2 + b^2\left(1-\frac{x^2}{a^2}\right)$.
If now we compute the derivative:
$\frac{\mathrm{d}}{\mathrm{d}x} \left(d^2\right) = 2 \left[ \left(x - \sqrt{a^2-b^2}\right) - \left(\frac{b}{a}\right)^2 x \right] = 2 \left[ x \left( 1 - \frac{b^2}{a^2} \right) - \sqrt{a^2-b^2} \right] = 0$
from that we should get:
$x = \frac{\sqrt{a^2-b^2}}{1 - \frac{b^2}{a^2}}$
But there must be some mistake here since for minimum $x = a$ and for maximum $x = -a$. Many thanks in advance.
| {
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Solve without a calculator: If $x+\sqrt{x}=13$ then $x+\frac{13}{\sqrt{x}}=?$
$$x+\sqrt{x}=13$$
$$x+\frac{13}{\sqrt{x}}=?$$
I tried to square(and also triple in another attempt) both sides of both of the equations hoping that I would find some expression to plug in. It didn't help.
Then, I tried to simply bring the second equation to a common denominator.It didn't help neither.
Then I found the $x$ value ($9.86$) from the first equation by using a calculator and then plugged in to the second equation and got this expression
$$\frac{493}{50}+65\sqrt{\frac{2}{493}}$$
Now, Is this question solvable without a calculator?
| If $x + \frac{13}{\sqrt x} = a$, then:
$$13a = (x+\sqrt x)(x + \frac{13}{\sqrt x})$$
$$= x^2 + 13\sqrt x + x\sqrt x + 13$$
Now use the equality $x + \sqrt x = 13$:
$$13a = x^2 + (x + \sqrt x)\sqrt x + x \sqrt x + (x +\sqrt x) $$
$$(x + \sqrt x)a = x^2 + x\sqrt x + x + x \sqrt x + x + \sqrt x $$
$$(x + \sqrt x)a = (x + \sqrt x)x + \sqrt x(x + \sqrt x) + (x + \sqrt x) $$
$$a = x + \sqrt x + 1 $$
$$a = 13 + 1 = 14$$
You can get to the answer much quicker using Math Lover's answer however.
| {
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Find all real numbers $x,y,z\in [0,1]^3$ such that $(x^2+y^2)\sqrt{1-z^2}\ge z$.... Such that:
$$(x^2+y^2)\sqrt{1-z^2}\ge z$$
and
$$(z^2+y^2)\sqrt{1-x^2}\ge x$$
and
$$(x^2+z^2)\sqrt{1-y^2}\ge y$$
Since $x,y,z$ $\in ]0,1[^3$
then , there are some real numbers $a,b,c$ such that
$\cos a=x, \cos b=y , \cos c=z$
After some manipulations , we find that :
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2 b}\ge \frac{1}{\tan c}$$
.... same for other inequalities
I don't know what i must do now
| The inequality is symetric.
so we can suppose that $x\ge y \ge z$
the second inequality becomes
$$2x^2\sqrt{1-x^2}\ge x$$
$$2x\sqrt{1-x^2}\ge 1$$
$$4x^2-4x^4-1\ge 0$$
$$-(2x^2-1)^2\ge 0$$
$$2x^2-1=0$$
$$x=\frac{1}{\sqrt{2}}$$
By the same way , after remplacing $x$ by its value
we will find that $x=y=z=\frac{1}{\sqrt{2}}$
| {
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
| $\displaystyle \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\cot (2x)\tan (x)=\dfrac{\cos 2x\cdot \sin x}{2\sin x\cos^2 x}=\dfrac{\cos 2x}{2\cos^2x}$
$\displaystyle\lim_{x\to0} \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\displaystyle\lim_{x\to0}\dfrac{\cos 2x}{2\cos^2x}=\dfrac12 $
| {
"language": "en",
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Is it true that there aren't any three different numbers $x,y,z$ such that $x^3+x \equiv y^3+y \equiv z^3+z \pmod p $? Let $p$ be a prime number. Is it true that there aren't any three different numbers $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p $$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$ ?
If not, what are the conditions of $p$ so that the statement is true for prime number $p$ ?
I tried with $p=3,7$ and both of them are correct, so I think that $p \equiv 3 \pmod 4$ may satisfy the statement.
My other attempt: Assume by contradiction, there exist $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p$$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$. Then $$x^2+xy+y^2 \equiv y^2+yz+z^2 \equiv z^2+zx+x^2 \pmod p$$
thus $$x+y+z \equiv 0 \pmod p.$$
Here I am stuck. How can I solve this problem ?
(Sorry for my English)
| If $p=n^2+1$ (such as $5,17,37\dots$), then $x=0,\ y=n,\ z=(p-n)$ will solve your equivalence with all three terms congruent to $0$.
$x^3+x=0;\ y^3+y=n(n^2+1)=np\equiv 0 \mod{p}; z^3+z=(p-n)(p^2-2np+n^2+1)=(p-n)(p^2-2np+p)=p(p-n)(p-2n+1)\equiv 0\mod{p}$
$p$ cannot be of the form $n^2+1$ if $p\equiv 3\mod{4}$.
| {
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How to show $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has $p-2$ incongruent solutions: $2, 3,..., p-1.$ when $p$ is an odd prime? Let $p$ be an odd prime. How would I prove that the congruence $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2, 3,..., p-1?$
I found this question posted, but the answer doesn't quite make sense to me.
Here is my attempt:
Proof
As $p$ is an odd prime, by Fermat's Theorem,
\begin{align*}
x^{p-1}&\equiv1\pmod{p}\\
&\implies x^{p-1}-1\equiv0\pmod{p}
\end{align*}
and by Lagrange's Corollary, with $d=p-1, p-1\mid p-1$, this congruence has exactly $p-1$ solutions.
\begin{align*}
x^{p-1}-1 \pmod{p}&\equiv0\pmod{p}\\
&\equiv (x-1)(x^{p-2}+\cdots+x^2+x+1) \pmod{p}\\
\end{align*}
Assume $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$...
And then I get stuck.
The proof I linked above used:
Now, $x^{p-1}-1\equiv 0\pmod{p}$ has exactly $p-1$ incongruent solutions modulo $p$ by Lagrange's Theorem.
Note that $g(1)=(1-1)f(1)=0\equiv 0\pmod{p}$, so $1$ is a root of $g(x)$ modulo $p$. Hence, the incongruent roots of $g(x)$ modulo $p$ are $1,2,3,\dots,p-1$.
But every root of $g(x)$ other than $1$ is also a root of $f(x)$ (This is the part I'm concerned about. Is it clear that this is the case?), hence $f(x)$ has exactly $p-2$ incongruent roots modulo $p$, which are $2,3,\dots,p-1$. $\blacksquare$
But I don't understand the part about roots. Nor do I quite get the use of creating $g(x)$ and $f(x)$. Could someone explain the proof using a different method?
| To phrase the yellow block differently; the equation $x^{p-1}-1\equiv0\pmod{p}$ has precisely $p-1$ incongruent solutions by Lagrange's theorem. These are precisely the nonzero residues modulo $p$, i.e. the congruence classes $1$, $2$,..., $p-1$. We can factor the left hand side of this equation as follows:
$$x^{p-1}-1=(x-1)(x^{p-2}+x^{p-1}+\cdots+x^2+x+1).$$
In particular this shows that if $x$ is a solution and $x-1\not\equiv0\pmod{p}$ then we must have
$$x^{p-2}+x^{p-1}+\cdots+x^2+x+1\equiv0\pmod{p}.$$
This tells us that $2$, $3$, ..., $p-1$ are solutions to the congruence.
To see that there are no other (incongruent) solutions, note that the only remaining congruence classes are $0$ and $1$. And plugging them in yields
$$0^{p-2}+0^{p-1}+\cdots+0^2+0+1\equiv1\pmod{p},$$
$$1^{p-2}+1^{p-1}+\cdots+1^2+1+1\equiv p-1\pmod{p},$$
so these are not solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Another Limit Conundrum
For what values of $a$ and $b$ is the following limit true?
$$\lim_{x\to0}\left(\frac{\tan2x}{x^3}+\frac{a}{x^2}+\frac{\sin bx}{x}\right)=0$$
This question is really confusing me. I know that $\tan(2x)/x^3$ approaches infinity as $x$ goes to $0$ (L'Hôpital's). I also understand that $\sin(bx)/x$ goes to $b$ as $x$ approaches $0$. However, I am not sure how to get rid of this infinity with the middle term. Any ideas?
Thanks!
| Using the usual Taylor series for
$$y=\frac{\tan(2x)}{x^3}+\frac{a}{x^2}+\frac{\sin (bx)}{x}=\frac{\tan(2x)+ax+x^2\sin (bx)}{x^3}$$ The numerator write
$$\left(2 x+\frac{8 x^3}{3}+\frac{64 x^5}{15}+O\left(x^7\right) \right)+a x+x^2\left(b x-\frac{b^3 x^3}{6}+\frac{b^5 x^5}{120}+O\left(x^7\right) \right)$$ that is to say
$$(a+2) x+\left(b+\frac{8}{3}\right) x^3+\left(\frac{64}{15}-\frac{b^3}{6}\right)
x^5+O\left(x^7\right)$$ making $$y=\frac{a+2}{x^2}+\left(b+\frac{8}{3}\right)+\left(\frac{64}{15}-\frac{b^3}{6}\right)
x^2+O\left(x^4\right)$$ So, in order to have a limit equal to $0$, you need $a=-2$, $b=-\frac{8}{3}$ and then
$$y=\frac{3008 }{405}x^2+O\left(x^4\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt:
we have $x+2y=100-5z$
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
| An alternative way.
Given $x+2y+5z=100$ and it is clear that $0\le z\le20$.
For any possible values of $z$, $x+2y=100-5z$.
Let us take $p=100-5z\ge0$. Solving the equation $x+2y=p$,
$(-p,p)$ is a solution. The general solution of $(x,y)$ is $$x=-p+2q,\ \ y=p-q,\ \ q\in\mathbb{Z}$$
If $p=2k$, then $k=\dfrac p2\le q\le p=2k$.
So, there are $k+1=\dfrac p2+1$ solutions for $(x,y)$
So, we have the following numbers as follows
$$p=100,95,90,85,80,75,......,15,10,5,0$$and$$k+1=51,48,46,43,41,38,......,8,6,3,1$$
The total number of solutions are $$4(10+20+30+40)+5(8+6+3+1)+51=541$$
| {
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An Inconsistency in Numerical Approximation Consider the expression
$$
10^5 - \frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1.
$$
Note that the approximation $10^5+1 \approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - \frac{10^{10}}{1+10^5} \approx 10^5 - \frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - \frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
| The first approximation is fine. The second on is not, because, $10^5$ and $\dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10\,001$ is close to $10\,000$, then $1$ is close to $0$.
| {
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"url": "https://math.stackexchange.com/questions/3017585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
$$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$
$$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$
So the limit of it is $-\infty$
Have I done this well or I missed sth?
| If I'm not mistaken, the actual sum is
$$ \frac{1 - \ln(2)}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$ The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no solution for the equation System, vice versa if $y=0$ and $a<0$.
For the general case I have assumed that $b\neq 0$ and therefore from $II$ we get $y=\frac{b}{2x}$, substituting with $I \Longrightarrow x^2-(\frac{b}{2x})^2=a\iff 4x^4 - b^2 = 4x^2a \iff x^4 -ax^2 - \frac{b^2}{4}=0 (*)$
Completing the square
$(*)\iff x^4 - 2\frac{ax^2}{2}-\frac{b^2}{4} \iff x^4 - 2\frac{ax^2}{2} +\frac{a^2}{4}-\frac{b^2}{4}-\frac{a^2}{4}\iff (x^2 - \frac{a}{2})^2+\frac{-b^2-a^2}{4}=0$
$\Rightarrow x^2-\frac{a}{2}=\sqrt{\frac{b^2+a^2}{4}} \Rightarrow x^2=\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2} \Rightarrow x = \pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}}$
$II\Longrightarrow y= \frac{b}{2(\pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}})}$
Can somebody tell my whether this Solutions are Right or not because when I have tried to verify it I end up with that term:
$\frac{5a^2+a(5\sqrt\frac{b^2+a^2}{4})(4\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2})}{4}$ which should be equal to $a$
Thank you for your time.
| Actually, the statement is true even if $b=0$: just take $(x,y)=\left(\pm\sqrt a,0\right)$ if $a\geqslant0$ and $(x,y)=\left(0,\pm\sqrt{-a}\right)$.
Otherwise, your approach is fine, but not your computations. You should have obtained, when $b\neq0$,$$\pm\left(\sqrt{\frac{a+\sqrt{a^2+b^2}}2},\frac b{|b|}\sqrt{\frac{-a+\sqrt{a^2+b^2}}2}\right),$$where, of course,$$\frac b{\lvert b\rvert}=\begin{cases}1&\text{ if }b>0\\-1&\text{ otherwise.}\end{cases}$$
| {
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Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$ I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$
My method:
Let
$$ I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$
Thus,
\begin{align}
I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\
&= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2}
\end{align}
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$
Now
$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t)$$
And finally,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
| $$ I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$
is a fairly simple hypergeometric series, namely $\frac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};-1\right)$. Since
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n,\qquad \arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n} x^{2n+1} $$
we clearly have $I=\frac{\pi}{2}\,\text{arcsin} \color{red}{\text{h}}(1) = \color{red}{\frac{\pi}{2}\log(1+\sqrt{2})}$.
By enforcing the substitution $x\mapsto\frac{1-x}{1+x}$ (involution) and exploiting the Maclaurin series of $\frac{1}{x}\left(\frac{\pi}{4}-\arctan(1-x)\right)$ I got the mildly interesting acceleration formula
$$ \frac{\pi}{2}\log(1+\sqrt{2})=\small{\sum_{k\geq 0}(-1)^k\left[\frac{2^{6k}}{(4k+1)(8k+1)\binom{8k}{4k}}+\frac{2^{6k+2}}{(4k+2)(8k+3)\binom{8k+2}{4k+1}}+\frac{2^{6k+3}}{(4k+3)(8k+5)\binom{8k+4}{4k+2}}\right]}. $$
In this case we have that a $\phantom{}_2 F_1(\ldots,-1)$ decomposes as a linear combination of three $\phantom{}_6 F_5(\ldots,-1/4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
| Let $1/n=h$
$$\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$
$$=\lim_{h\to0^+}\dfrac{1+\sqrt{1+h^4}-2}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\lim_{h\to0^+}\dfrac{1+h^4-1}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^4}+1}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\dfrac1{(\sqrt1+1)(\sqrt{1+\sqrt1}+\sqrt2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0$ Consider the inequality
$$
\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+y\leq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
| If you let $x = \sin^2 a, y = \sin^2 b$ with $0<b<a<\pi/2$, then your inequality is equivalent to:
$$\sin a\cos b - \sin b\cos a +\sin^2b\sin a - \sin^2a\sin b >\dfrac{\sin^2 a-\sin^2 b}{3}$$ or $$\sin(a-b)>(\sin a-\sin b)\left(\frac 13+\sin a\sin b\right)$$
or $$\dfrac{\cos\left(\frac {a-b}2\right)}{\cos\left(\frac{a+b}{2}\right)}>\frac{1}{3}+\frac 12(\cos(a-b)-\cos(a+b)).$$
Here, the remaining requirement is the same as $\sin a\leq\cos b = \sin(\pi/2-b)$, which is the same as: $$a+b\leq\dfrac{\pi}{2}.$$
Now let $1>\cos\left(\frac{a-b}{2}\right) = s>\cos\left(\frac{a+b}{2}\right)=t\geq\dfrac{1}{\sqrt{2}}.$ Then, the equivalent inequality is:
$$\frac st>\frac 13+\frac 12(2s^2-1 - 2t^2+1) = \frac 13+s^2-t^2$$
But this one is trivial as:
$$\frac 13+s^2-t^2<\frac 13+1-\frac 12 = \frac 56 <1<\frac st.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^2\equiv 5 \pmod{1331p^3}$ Let $p$ be given by $p=2^{89}-1$ and note that it is a Mersenne Prime. The problem is to find the number of incongruent solutions to
$$
x^2\equiv 5 \pmod{1331p^3}
$$
I began the problem by splitting it up into the congruences
$$
x^2\equiv 5 \pmod{1331} $$and$$
x^2\equiv 5 \pmod{p^3}
$$
I found that $x\equiv 4,7\pmod{11}$ are solutions to $x^2\equiv 5\pmod{11}$ and then use Hansel's Lemma all the way up to get that $x\equiv 1258, 73\pmod{1331}$ are solutions to the equation $\pmod{1331}$.
I think all I have to do is solve the second equation and use the Chinese Remainder Theorem at the end but I am stuck because I have no idea where to begin in solving $x^2\equiv 5\pmod{p}$ as p is such a large number.
Any help is appreciated!
| Consider Legendere symbol $\left(\frac{5}{p}\right)$. By quadratic reciprocity
$$\left(\frac{5}{p}\right)\Big(\frac{p}{5}\Big)=(-1)^{\left(\frac{5-1}{2}\right)\left(\frac{p-1}{2}\right)}=1 \implies \left(\frac{5}{p}\right)=\left(\frac{p}{5}\right).$$
But $p=2^{89}-1 \equiv 2(2^2)^{44}-1 \equiv 1 \pmod{5}$. Thus
$$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)=\left(\frac{1}{5}\right)=1.$$
Thus $5$ is indeed a QR modulo $p$. Since $p$ is a prime thus $x^2 \equiv 5 \pmod{5}$ will have two non-congruent solutions. Now you can apply Hensel to see if you will continue to have two solutions as you lift from $p$ to $p^3$.
If you have two solutions for $p^3$ as well, then in all you will have $4$ solutions (combining with two from the previous congruence with $11^3$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Strange sum that always end up with 9 If we have any number, example 4896, and sum all digits
sum = 4+8+9+6 = 27
and than substract this number from the original number, we always get a number that is divisible by 9:
4896-27=4869 -> 4869/9 = 541.
Moreover now everytime i sum the digits from new number (4869), i will get number divisible by 9:
(4+8+6+9=27[divisible])
71 - (7+1) = 63
485 - (4+8+5) = 468/9=52 , 4+6+8 = 18(divisible)
Similarly interesting is with 2 digit numbers where with the same process resulting number is always 9
45 -> 45-9 = 36 (3+6=9)
87 -> 87-15 = 72 (7+2=9)
Its quite beyond my comprehension. Can anyone explain this phenomena?
| I will give a formal proof for numbers with $4$ digits. You can think about how to generalize it (Edit: or refer to @fleablood's answer who gave a formal proof for arbitrarily many digits).
Let $n=abcd$ be a number where $0\leq a,b,c,d\leq 9$ are it's digits. For example if $n=4821$ then $a=4,b=8,c=2,d=1$.
Another way to write $n$ is as $n=a\cdot 1000 + b\cdot 100 +c \cdot 10 +d$.
On the other hand the sum of digits is $a+b+c+d$.
so $$n - \text{The sum of digits} = a\cdot 1000 + b\cdot 100 + c\cdot 10 + d -a-b-c-d$$
The right hand side turns out to be $a\cdot 999 + b\cdot 99 + c\cdot 9$ which is divisible by $9$.
The second thing that interested you is that the sum of digits of the new number is always divisible by $9$. For this you should refer to @Ross Milikan answer.
| {
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How to factor $-x^3+x^2-2$ Is there an easy way to factor $$f(x)=-x^3+x^2-2\;?$$
I have checked step-by-step calculators that all use theorems I am not very familiar with.
It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?
| $$-x^3+x^2-2=-1-x^3+x^2-1$$
$$=-(1+x^3)+(x^2-1)$$
$$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
$$=(x+1)(x-1-(1-x+x^2))$$
$$=(x+1)(-x^2+2x-2)$$
Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
$$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
$$=(x+1)[-(1-x+x^2)+(x-1)]$$
$$=(x+1)(-1+x-x^2+x-1)$$
$$=(x+1)(-x^2+2x-2)$$
| {
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Finding Bounds in proof of Stirling's Formula In a proof of Stirling's Formula, my professor claims that
$\frac 12 $($ln$k + $ln$(k+1)) $\le$ $\int_k^{k+1}$$ln$x $dx$ $\le$ $\frac 12$($ln$k + $ln$(k+1)) + $\frac 1{k^2}$.
I can see that the first inequality is true by concavity of the natural log function (hence the area of the trapezoid beneath natural log is less than the integral of the function itself), but can someone provide insight into the second inequality??
I know $\frac 1{k^2}$ must be a bound for the area beneath $ln$x but above the trapezoid described by the leftmost expression, but it is not clear how to check that this is true.
| My attempt:
\begin{align}
\int_{\ln(k)}^{\ln(k+1)} &= \int_0^1 \ln(k+x) dx = \int_0^{1/2} \ln(k+x) + \int_{1/2}^{1} \ln(k+1 - x) dx \\
&= \int_0^{1/2} \ln(k) + \frac{x}{k} - \frac{x^2}{2k^2} + o(\frac{1}{k^2}) dx \\
&\quad + \int_{1/2}^1 \ln(k+1) - \frac{x}{k+1} - \frac{x^2}{(k+1)^2} + o(\frac{1}{(k+1)^2}) dx \\
&= \frac{1}{2} (\ln(k) + \ln(k+1)) + \frac{1}{8k} + \frac{1}{8(k+1)} - \frac{1}{2(k+1)} \\
&\quad - \frac{1}{48k^2} + \frac{1}{48(k+1)^2} - \frac{1}{6k^2} + o(\frac{1}{k^2})
\end{align}
Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.
| {
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Find all integers x for which $ |x^3 + 6x^2 + 2x − 6| $ is prime. I think that this means that $|x^3 + 6x^2 + 2^x − 6| \equiv 0 \bmod p$. Ignoring the absolute value first, I get $
x(x^2+2)+6(x^2−1) \equiv 0 \mod p \Rightarrow x(x^2+2) \equiv -6(x^2−1) \bmod p$. But I'm not sure how to proceed.
| Following your thoughts of grouping the terms into:
$$x(x^2+2)+6(x^2−1)$$
consider the following:
*
*The second term, $6(x^2-1)$ is always divisible by $6$ and consequently is divisible by $3$.
*For the first term $x(x^2+2)$, it is also divisible by 3. Why? $x$ is either $0,-1,1 \pmod 3$. Hence, $x^2 \equiv 0,1 \pmod 3$. If $x^2 \equiv 0 \pmod 3$, then $3|x^2 \implies 3|x$. On the other hand, if $x^2 \equiv 1 \pmod 3$, then $x^2 + 2 \equiv 0 \pmod 3$.
Now, what can you conclude?
| {
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Half of Vandermonde's Identity We know Vandermonde's Identity, which states
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose r}$
or at least
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\Theta \left( \frac{1}{2} {m+n \choose r}\right)$
true?
Maybe someone here has even some general insight on other step widths?
Thank you!
| In general, having the ogf (z-Transform)
$$
F(z) = \sum\limits_{0\, \le \;n} {a_{\,n} \,z^{\,n} }
$$
then
$$
{1 \over m}\sum\limits_{0 \le \,k\, \le \,m - 1} {\left( {z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} } \right)^{\,j}
F(z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} )}
= \sum\limits_{0\, \le \;n} {\,a_{\,m\;n - j} \,z^{\,n} }
$$
But unfortunately, the truncated binomial expansion
$$
\sum\limits_{0\, \le \;k} {\left( \matrix{ n \cr r - k \cr} \right)\,z^{\,k} }
$$
does not have in general ($r<n$) a compact closed expression.
We can go either through the Hypergeometric version
$$
\sum\limits_{\left( {0\, \le } \right)\;k\,\left( { \le \,\,r} \right)} {
\binom{m}{k} \binom{n}{r-k}\,z^{\,k} }
= \binom{n}{r} \;{}_2F_{\,1} \left( {\matrix{
{ - m,\; - r} \cr
{n - r + 1} \cr
} \;\left| {\,z} \right.} \right)
$$
or through the double ogf
$$
\eqalign{
& G(x,y,n,m) = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {
\binom{m}{j}\,\binom{n}{k-j} y^{\,j} } } \right)x^{\,k} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {
\binom{m}{j}\left( {x\,y} \right)^{\,j} \sum\limits_{\left( {j\, \le } \right)\,k\,\left( { \le \,n} \right)\,} { \,\binom{n}{k-j}x^{\,k - j} } } = \cr
& = \left( {1 + xy} \right)^{\,m} \left( {1 + x} \right)^{\,n} \cr}
$$
Then for instance we have
$$
\eqalign{
& \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;} {
\left( \matrix{ m \cr 2j \cr} \right)\,\left( \matrix{ n \cr k - 2j \cr} \right)} } \right)x^{\,k} } = \cr
& = {1 \over 2}\left( {G(x,1,n,m) + G(x, - 1,n,m)} \right) = \cr
& = {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {\left( {1 + x} \right)^{\,m} + \left( {1 - x} \right)^{\,m} } \right) = \cr
& = {1 \over 2}\left( {1 + x} \right)^{\,n + m} + {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {1 - x} \right)^{\,m} = \cr
& = {1 \over 2}\left( {1 + x} \right)^{\,n + m} + {1 \over 2}\left( {1 + x} \right)^{\,n - m} \left( {1 - x^{\,2} } \right)^{\,m} = \cr
& = {1 \over 2}\left( {1 + x} \right)^{\,n + m} + {1 \over 2}\left( {1 - x^{\,2} } \right)^{\,{{n + m} \over 2}}
\left( {{{1 + x} \over {1 - x}}} \right)^{\,{{n - m} \over 2}} \cr}
$$
which clearly indicates what is the difference between
$$
{1 \over 2}\binom{n+m}{r}
\quad vs\quad \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;} {
\binom{m}{2j} \, \binom{n}{r-2j} }
$$
Of course the complement will be
$$
\eqalign{
& \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;} {
\binom{m}{2j+1} \,\binom{n}{k - \left( {2j + 1} \right)}} } \right)x^{\,k} } = \cr
& = {1 \over 2}\left( {G(x,1,n,m) - G(x, - 1,n,m)} \right) = \cr
& = {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {\left( {1 + x} \right)^{\,m} - \left( {1 - x} \right)^{\,m} } \right) \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that if $a,b,c \in \mathbb{R^+}\text{ and } abc=8\text{ then } {ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge6$ Question:
Prove that if $a,b,c \in \mathbb{R^+}\text{ and } abc=8\text{ then } {ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge6$
My Approach:
Now we have:
$${ab+4\over a+2}={2\times (ab+4)\over2\times (a+2)}={2ab+8\over2(a+2)}={2ab+abc\over2(a+2)}={ab(2+c)\over2(a+2)}$$
Now similarly we can acheive:
$${bc+4\over b+2}={bc(2+a)\over2(b+2)};{ca+4\over c+2}={ca(2+b)\over2(c+2)}$$
Using AM-GM we get:
$${ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}$$
$$={ab(2+c)\over2(a+2)}+{bc(2+a)\over2(b+2)}+{ca(2+b)\over2(c+2)}$$
$$\ge\sqrt[3]{{ab(2+c)\over2(a+2)}\times{bc(2+a)\over2(b+2)}\times{ca(2+b)\over2(c+2)}}$$
$$=\sqrt[3]{(abc)^2\over2^3}$$
$$=\sqrt[3]{8^2\over8}=2$$
Therefore, we get:
$${ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge2$$
However, the question wants me to prove that its greater than or equal to $6$ and when I try to plug in I always get a value larger than or equal to $6$. So where did I go wrong and how can I fix my mistake. Thank you in advance.
| Because you wrote $$\tfrac{ab(2+c)}{2(a+2)}+\tfrac{bc(2+a)}{b+2)}+\tfrac{ca(2+b)}{2(c+2)}
\ge\sqrt[3]{{ab(2+c)\over2(a+2)}\cdot{bc(2+a)\over2(b+2)}\cdot{ca(2+b)\over2(c+2)}},$$ but it should be
$$\tfrac{ab(2+c)}{2(a+2)}+\tfrac{bc(2+a)}{b+2)}+\tfrac{ca(2+b)}{2(c+2)}
\ge3\sqrt[3]{{ab(2+c)\over2(a+2)}\cdot{bc(2+a)\over2(b+2)}\cdot{ca(2+b)\over2(c+2)}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$ Prove the inequality $$\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$$
for $x>0$.
My attempt: Let $$f(x)=\ln {(1+\frac{1}{x})}-\frac{2}{2x+1}$$
Then $$f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2}$$
$$f''(x)=\frac{1}{x^2}-\frac{1}{(x+1)^2}-\frac{8}{(2x+1)^3}>0$$
Then the function $f$ is convex. There exists a minimal point $x_0$ such that $f(x)\geq f(x_0)$. However, there's no critical point $x_0$ such that $f'(x_0)=0$, and $\lim_{x \rightarrow \infty} \sup {f'(x)}=0$. Then I want to show that $f(x)>0$, how do I continue my proof?
I have been trying another approach using Cauchy's MVT by letting
$$f(x)=\ln {x}$$ $$g(x)=\frac{1}{2x+1}$$ such that $$\frac{f(x+1)-f(x)}{g(x+1)-g(x)}=\frac{f'(c)}{g'(c)}$$ where $c \in (x,x+1)$ but failed.
As what I did is
$$\ln {(1+\frac{1}{x})}=\frac{1}{c} \cdot \frac{(2c+1)^2}{2} \cdot \frac {2}{(2x+1)(2x+3)}$$
I can't simply do the inequality
$$\frac{1}{c} \cdot \frac{(2c+1)^2}{2}>\frac{1}{x} \cdot \frac{(2x+1)^2}{2}$$
as $c>x$ because $\frac{1}{c} < \frac{1}{x}$ but $\frac{(2c+1)^2}{2} > \frac{(2x+1)^2}{2}$.
Edited: Of course, I know that $$\ln {(1+ \frac{1}{x})}>\frac{x}{1+x}$$
for $x>-1$. I just need to prove that $$\frac{x}{x+1}>\frac{2}{2x+1}$$ But I hope to find out another approach using calculus method.
| Here is a more natural way to prove this. Note that:
$$\ln\bigg(1+\frac{1}{u}\bigg) = \int\limits_u^{u+1} \frac{1}{x} dx$$
i.e. the area under the graph of $\frac{1}{x}$ between $u$ and $(u+1)$. This explains why $\frac{2}{2u+1}$ is such a good approximation in the first place -- it comes from approximating the area as a rectangle of width $1$, and height $f\big(\big(u+\frac{1}{2}\big)\big)=\frac{1}{\big(u+\frac{1}{2}\big)}.$
Let $v = \big(u+\frac{1}{2}\big)$ for convenience. Then, an easy way to show that $\frac{2}{2u+1} = \frac{1}{v}$ underestimates the area is to show that:
$$\Bigg( \frac{1}{v-h} - \frac{1}{v} \Bigg) \ge \Bigg( \frac{1}{v} - \frac{1}{v+h} \Bigg)$$
where $h > 0$.
This is clearly true. Thus, the inequality holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $\sqrt{x-a}+\sqrt{x-b}$:
$$\begin{align*}
(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}&=(a-b)(\sqrt{x-a}+\sqrt{x-b})\\
&=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}
\end{align*}$$
Now I have the following:
$$(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
Simplifying the RHS as I was out of ideas at that point:
$$x\sqrt{x-a}-a\sqrt{x-a}+x\sqrt{x-b}-b\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
I noticed that all one of the common factors $\sqrt{x-a},\sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $\sqrt{x-b}$ terms on one side and $\sqrt{x-a}$ terms on the other.
$$\sqrt{x-b}(x-a)=\sqrt{x-a}(2a-b-x)$$
I tried to then square both sides, but that led to quite a mess.
$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$
I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.
The $a>b$ hint is interesting, but I have no clue what implication it may have here.
I think the $(x-a)\sqrt {x-a}$ patterns may mean something, perhaps I could do something with $a\sqrt a=\sqrt{a^3}$, but at this point it is probably a dead end.
I appreciate any help.
| Another way is as follows:
*
*Set $\boxed{x = a + t(a-b)}$ for $t \geq 0$
$$\begin{eqnarray*}
\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}} & = & a-b \\
& \Leftrightarrow & \\
\frac{t(a-b)\sqrt{t(a-b)}+(t+1)(a-b)\sqrt{(t+1)(a-b)}}{\sqrt{t(a-b)}+\sqrt{(t+1)(a-b)}} & = & a-b \\
& \Leftrightarrow & \\
\frac{t\sqrt{t}+(t+1)\sqrt{t+1}}{\sqrt{t}+\sqrt{t+1}} & = & 1 \\
& \Leftrightarrow & \\
(t(\sqrt{t+1} + \sqrt{t})+\sqrt{t+1})(\sqrt{t+1}-\sqrt{t}) & = & 1 \\
& \Leftrightarrow & \\
t+ t+1 - \sqrt{t(t+1)} & = & 1 \\
& \Leftrightarrow & \\
2t & = & \sqrt{t(t+1)} \\
& \Leftrightarrow & \\
t =\frac{1}{3} & \mbox{ or } & t= 0 \\
& \stackrel{x = a + t(a-b)}{\Leftrightarrow} & \\
\boxed{x = a + \frac{1}{3}(a-b)} &\mbox{ or } & \boxed{x= a}
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve similar right triangles, given one's hypotenuse, the other's base, and the sum of their heights. I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.
$\theta$ is wanted—given this lovely figure
given that the two triangles are similar, and given $a$, $b$, and $h$.
I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.
| It is pretty easy to see from the geometry of the figure that
$b \cot \theta + a \cos \theta = h, \tag 1$
whence,
$b \dfrac{\cos \theta}{\sin \theta} + a\cos \theta = h; \tag 2$
now using $\cos^2 \theta + \sin^2 \theta = 1$, i.e. $\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$,
$b \dfrac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} + a \sqrt{1 - \sin^2 \theta} = h; \tag 3$
we choose the positive sign on $\pm \sqrt{1 - \sin^2 \theta}$ since the angle $\theta$ appears to be acute; next, we square:
$b^2 \dfrac{1 - \sin^2 \theta}{\sin^2 \theta} + 2ab \dfrac{1 - \sin^2 \theta}{\sin \theta} + a^2 (1 - \sin^2 \theta) = h^2; \tag 4$
we multiply by $\sin^2 \theta$:
$b^2 (1 - \sin^2 \theta)+ 2ab \sin \theta (1 - \sin^2 \theta) + a^2 \sin^2 \theta (1 - \sin^2 \theta) = h^2 \sin^2 \theta, \tag 5$
which may be written as a quartic equation in $\sin \theta$:
$-a^2 \sin^4 \theta -2ab \sin^3 \theta + (a^2 - b^2 - h^2)\sin^2 \theta + 2ab \sin \theta + b^2 = 0, \tag 6$
or
$a^2 \sin^4 \theta + 2ab \sin^3 \theta + (h^2 - a^2 - b^2)\sin^2 \theta - 2ab \sin \theta - b^2 = 0. \tag 7$
This is about as far as we can push things using elementary algebra and trigonometry. To find $\sin \theta$, we must solve this quartic, which may be done according to this wikipedia page.
| {
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"url": "https://math.stackexchange.com/questions/3049005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| $$\sqrt[4] {28+16\sqrt 3}=\sqrt[4] {(\sqrt {12})^2 +(\sqrt {16})^2 +2\sqrt {16\cdot 12}}=\sqrt {4+2\sqrt 3}=\sqrt {(\sqrt 3)^2 +(1)^2 +2\sqrt {3\cdot 1}}=\sqrt 3 +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
| Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Mistake in solving an equation involving a square root I want to solve $2x = \sqrt{x+3}$, which I have tried as below:
$$\begin{equation}
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\left(x - \frac12 \right)^2 = 1 \\
x = \frac32 , -\frac12
\end{equation}$$
This, however, is incorrect.
What is wrong with my solution?
| You made a mistake when completing the square.
$$x^2-\frac{1}{4}x = \frac{3}{4} \color{red}{\implies\left(x-\frac{1}{2}\right)^2 = 1}$$
This is easy to spot since $(a\pm b)^2 = a^2\pm2ab+b^2$, which means the coefficient of the linear term becomes $-2\left(\frac{1}{2}\right) = -1 \color{red}{\neq -\frac{1}{4}}$. This means something isn’t correct...
Note that the equation is rewritten such that $a = 1$, so you need to add $\left(\frac{b}{2}\right)^2$ to both sides and factor. (In other words, divide the coefficient of the linear term $x$ by $2$ and square the result, which will then be added to both sides.)
$$b = -\frac{1}{4} \implies \left(\frac{b}{2}\right)^2 \implies \frac{1}{64}$$
Which gets
$$x^2-\frac{1}{4}x+\color{blue}{\frac{1}{64}} = \frac{3}{4}+\color{blue}{\frac{1}{64}}$$
Factoring the perfect square trinomial yields
$$\left(x-\frac{1}{8}\right)^2 = \frac{49}{64}$$
And you can probably take it on from here.
Edit: As it has been noted in the other answers (should have clarified this as well), squaring introduces the possibility of extraneous solutions, so always check your solutions by plugging in the values obtained in the original equation. By squaring, you’re solving
$$4x^2 = x+3$$
which is actually
$$2x = \color{blue}{\pm}\sqrt{x+3}$$
so your negative solution will satisfy this new equation but not the original one, since that one is
$$2x = \sqrt{x+3}$$
with no $\pm$.
| {
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"url": "https://math.stackexchange.com/questions/3052746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$
How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?
With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integral
Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=\int_0^1\arctan x\ln(1+x)\frac{dx}x\text{ and }I_2=\int_0^1\arctan x\ln(1+x)\frac{dx}{1+x}$$ into the form of integral Pisco gave.
Integrating by parts to the second integral converts $I_2$ into $\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$.
But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=\frac{1-t}{1+t}$ and got $$\frac{\ln\frac{2}{t+1} \arctan\frac{1-t}{1+t}}{1-t^2}$$ which is not what I want.
| $$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}\ dx$$
Apply IBP for the second integral, we get,
$$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-\frac{3\pi}{8}\ln^22+\frac32\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx$$
The first integral was calculated here :
$$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2-3\Im\operatorname{Li}_3(1+i)$$
And the second integral was calculated here:
$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2G\ln2$$
Substituting these two results, we get $\quad\displaystyle \boxed{I={\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}}$
| {
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"url": "https://math.stackexchange.com/questions/3054741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$
| Write $x=\tan a$ and $y = \tan b$ for some (angles) $a,b$. They exsist since $\tan $ is surjective function. Pluging in starting equation we get:
$$ {\sin a+1\over \cos a}\cdot {\sin b+1\over \cos b}=1$$
and after rearranging we get $$\sin(a)+ \sin(b) = \cos (a+b) - \cos 0$$
which is equivalent to $$2\sin{a+b\over 2}\cos{a-b\over 2} = -2\sin {a+b\over 2}\sin{a+b\over 2}$$
Case 1. $\sin {a+b\over 2}=0$ then $a+b = 2\pi k$ so $$x=\tan a = \tan (2\pi k-b) = -\tan b = -y$$
Case 2. $\sin {a+b\over 2}\ne 0$, then $$\cos {a-b\over 2}+\sin{a+b\over 2}=0$$
Factorising this we get:
$$2\cos ({\pi\over 4}-{b\over 2})\cdot \cos({a\over 2}-{\pi\over 4})=0$$
Now we have to choises again.
1.st: $${\pi\over 4}-{b\over 2} = {\pi\over 2}+\pi k\implies b =-{\pi\over 2}+2\pi k $$
so $y$ does not exist.
2.nd case... we get $x$ does not exsist.
So $x=-y$ and thus second expresins is also $1$.
| {
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Find the residue, state the nature of the singularity, find the constant term in $1/\sin(ze^z)$ at $z=0$
Find the residue, state the nature of the singularity, find the constant term in series $1/\sin(ze^z)$ at $z=0$.
We can rewrite the function $\frac{1}{\sin(ze^z)}$ as $\frac{ze^z}{\sin(ze^z)}\cdot\frac{1}{ze^z}$. What I did next I think might not be true and that's why I'm writing this post.
Since $\lim\limits_{w\rightarrow0}\frac{w}{\sin w}=1$ and $\lim\limits_{z\rightarrow 0}ze^z=0$ and $[{d\over dz}e^z]_{z=0}=1$ basically I said that at zero we can just look at $\frac{1}{ze^z}={1\over z}+\sum\limits_{n=1}^\infty\frac{(-1)^n}{n!}z^{n-1}$ as the Laurent series of our original function.
But I feel like this is far from rigorous...
From this it results that the residue at zero is $1$, which is true for the original function;
the constant term is $-1$, also true for $\frac{1}{\sin(ze^z)}$ and $z_0=0$ is a pole of degree $1$ also true.
| You can find the first few terms of the series, step-by-step
$$ ze^z = z\left(1 + z + \frac{z^2}{2} + \dots \right) = z + z^2 + \frac{z^3}{2} + \dots $$
\begin{align} \sin(ze^z) &= \sin\left(z + z^2 + \frac{z^3}{2} + \dots \right) \\
&= \left(z + z^2 + \frac{z^3}{2} \dots \right) - \frac{1}{3!}\left(z + z^2 + \frac{z^3}{2} +\dots \right)^3 + \dots \\
&= z + z^2 + \frac{z^3}{2} - \frac{z^3}{6} + \dots \\
&= z + z^2 + \frac{z^3}{3} + \dots \end{align}
\begin{align}
\frac{1}{\sin(ze^z)} &= \frac{1}{z+z^2+\frac{z^3}{3}+\dots} \\
&= \frac{1}{z} \frac{1}{1 + z + \frac{z^2}{3} + \dots} \\
&= \frac{1}{z} \left[1 - \left(z + \frac{z^2}{3} + \dots\right) + \left(z + \frac{z^2}{3} + \dots\right)^2 - \dots \right] \\
&= \frac{1}{z}\left[1 - z - \frac{z^2}{3} + z^2 + \dots\right] \\
&= \frac{1}{z}\left[1 - z + \frac{2z^2}{3} + \dots \right]
\end{align}
so the residue is $1$ and the constant term is $-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ B_{n}(x)=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}$ converges uniformly. Let $n\geq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]\subseteq \Bbb{R}.$
MY WORK
Let $[a,b]\subseteq \Bbb{R}$ be arbitrary, $x\in [a,b]$ and $n\in \Bbb{N}$ be fixed. Then,
\begin{align} \left| \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}\right|\leq \left( \dfrac{\left| x\right|} {2}\right)^{n+2k}\leq \left(\dfrac{c}{2}\right)^{n+2k}.\end{align}
where $c=\max\{|a|,|b|\}.$ Now,
\begin{align} \sum^{\infty}_{k=0} \left(\dfrac{c}{2}\right)^{n+2k} =\left(\dfrac{c}{2}\right)^{n}\sum^{\infty}_{k=0} \left(\dfrac{c}{2}\right)^{2k} =\left(\dfrac{c}{2}\right)^{n} \left(\dfrac{4}{4-c^2}\right)<\infty,\;\;\text{where}\;\;c<2.\end{align}
Hence,
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
converges uniformly on $[a,b]\subseteq \Bbb{R}$ with $c<2.$
QUESTION:
The question asks for a proof for any closed interval of $ \Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $c\geq 2$? Or I'm I missing something in the proof?
| Happy New Year! Credits to Jakobian
\begin{align} \left| \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}\right|\leq \dfrac{1} {k!}\left( \dfrac{\left| x\right|} {2}\right)^{n+2k}\leq \dfrac{1} {k!}\left(\dfrac{c}{2}\right)^{n+2k}.\end{align}
Since,
\begin{align} \lim\limits_{k\to\infty}\left|\dfrac{k!} {(k+1)!}\left(\dfrac{c}{2}\right)^{n+2k+2} \left(\dfrac{2}{c}\right)^{n+2k} \right|&=\lim\limits_{k\to\infty}\left|\dfrac{1} {(k+1)}\left(\dfrac{c}{2}\right)^{2} \right|\\&=\dfrac{c^2}{4}\lim\limits_{k\to\infty}\dfrac{1} {(k+1)} \\&=0<1\end{align}
Thus, $\sum^{\infty}_{k=0} \dfrac{1} {k!}\left(\dfrac{c}{2}\right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
converges uniformly on any $[a,b]\subseteq \Bbb{R}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$. Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$. Finally, let $s(X)=\sigma(X)-X$ denote the sum of the aliquot divisors of $X$.
This is a question about Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College, titled Characterization of the Positive Integers with Abundancy Index of the Form $(2x-1)/x$. (A copy of the poster presentation is available via the following hyperlink.)
In the abstract of the paper, it is stated in the fourth sentence that
Rational numbers of the form $(2x-1)/x$ are important since both even and odd perfect numbers have a divisor with abundancy index of this form.
Let $M = 2^{p-1}(2^p - 1)$ be an even perfect number, and let $N = q^k n^2$ be an odd perfect number.
Clearly,
$$I(2^{p-1}) = \frac{2^p - 1}{2^{p-1}} = \frac{2x_1 - 1}{x_1}$$
where $x_1 = 2^{p-1}$. (In other words, $2^{p-1}$ is an even almost perfect number, since it is a power of two.)
However,
$$I(p^k) = \frac{p^{k+1} - 1}{p^{k+1} - p^k}$$
and
$$I(n^2) = \frac{2}{I(p^k)} = \frac{2(p^{k+1} - p^k)}{p^{k+1} - 1}$$
so clearly $p^k$ is not almost perfect (since $p$ must be odd).
Additionally, since
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)} \geq 3$$
(see the paper [Dris, 2012]), then clearly $n^2$ is likewise not almost perfect. (Similarly, it can be proved that $n$ and $q^k n$ are not almost perfect.)
So I think the trivial divisor $1$ of an odd perfect number has the required abundancy index
$$I(1) = 1 = \frac{2\cdot{1} - 1}{1} = \frac{2x_2 - 1}{x_2}$$
where $x_2 = 1$.
Here is my question:
Is there any other divisor $m > 1$ of an odd perfect number $N = q^k n^2$ such that
$$I(m) = \frac{2x - 1}{x}$$
for some positive integer $x$?
| The following answer is based on a theorem in a paper (to appear) communicated to Dris by Holdener (co-authored with Rachfal).
Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$. Suppose that $k>1$.
Consider the proper factor
$$q^{\frac{k-1}{2}} n^2.$$
This has abundancy index
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = I\bigg(q^{\frac{k-1}{2}}\bigg)I(n^2) = I\bigg(q^{\frac{k-1}{2}}\bigg)\cdot\frac{2}{I(q^k)} = \frac{q^{\frac{k+1}{2}} - 1}{q^{\frac{k-1}{2}}(q - 1)}\cdot\frac{2q^k (q - 1)}{q^{k+1} - 1}.$$
Since $k \equiv 1 \pmod 4$, $k+1$ is even, so that
$$q^{k+1} - 1 = \bigg(q^{\frac{k+1}{2}} - 1\bigg)\cdot\bigg(q^{\frac{k+1}{2}} + 1\bigg).$$
Canceling the common factor
$$q^{\frac{k-1}{2}} (q - 1)\bigg(q^{\frac{k+1}{2}} - 1\bigg)$$
in the numerator and denominator of
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg),$$
we get
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = \frac{2q^{\frac{k+1}{2}}}{q^{\frac{k+1}{2}} + 1} = 2 - \frac{1}{\bigg(\frac{q^{\frac{k+1}{2}}+1}{2}\bigg)},$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
how to show $ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})$ converge/diverge $$ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}\left(e^{\frac{1}{n^2}} - 1\right)\left(\sqrt{n^4 - 8}\right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $\sin(\frac{1}{n}) < \frac{1}{n}$. I have tried to show with Cauchy that it diverges.
| Hint. One may observe that,
$$
\lim_{ n\to \infty}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0\ge10$, $n\ge n_0$,
$$
\frac12 \le(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8}) \tag1
$$ then, as $n \ge 10$,
$$
\frac{1}{n}-\frac{1}{6n^3}\le\sin{\frac{1}{n}}\tag2
$$ yielding, for $N\ge n_0$,
$$
\frac{1}{2}\sum_{n_0}^{N}\frac{1}{n\ln n}-\frac{1}{12}\sum_{n_0}^{N}\frac{1}{n^3\ln n}\le\sum_{n_0}^{N} \frac{\sin{\frac{1}{n}}}{\ln n}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
| {
"language": "en",
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"source": "stackexchange",
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} |
Prove $\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$ I have to prove that the following limit is equal to $\sqrt{\pi/2}$:
$$\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$$
In order to calculate this limit, we know that:
$$I_n=\int_0^{\frac{\pi}{2}}\sin^nx\ dx\quad I_{2n}=\frac{1\cdot3\cdot..\cdot(2n-3)(2n-1)}{2\cdot4\cdot..\cdot(2n-2)(2n)}\frac{\pi}{2}\quad I_{2n+1}=\frac{2\cdot4\cdot..\cdot(2n-2)(2n)}{1\cdot3\cdot..\cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$\lim_{n\to\infty}\frac{1}{I_{2n}\sqrt{2n+1}}\frac{\pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!
| Using the double factorial notation we need to find $$\lim_{n\to\infty} \frac {(2n)!!}{(2n-1)!!\sqrt {2n+1}}$$
Now using the relation between double factorial and the factorial, the limit changes to $$\lim_{n\to\infty} \frac {2^{2n}(n!)^2}{(2n)!\sqrt {2n+1}}$$
Using Stirling's approximation for factorials we get $$\lim_{n\to\infty} \frac {2^{2n}\cdot (2\pi n)\cdot \left(\frac ne \right)^{2n}}{\sqrt {2\pi}\cdot\sqrt {2n} \cdot\left(\frac {2n}{e}\right)^{2n} \cdot \sqrt {2n+1}}$$
Hence limit changes to $$\lim_{n\to\infty} \frac {n\sqrt {2\pi}}{\sqrt {2n} \cdot \sqrt {2n+1}}$$
Which easily evaluates to $\sqrt {\frac {\pi}{2}}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| Observe that
$$
\sqrt{8}\times (1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\times \sqrt{3x}
$$ and
$$
\sqrt{8}\times \sqrt{3}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the dimension of the subspace of $\mathbb{R}^4$ spanned by $x_1= (1, 2, -1, 0)$, $x_2=(2, 5, -3, 2)$, $x_3=(2, 4, -2, 0)$, $x_4=(3, 8, -5, 4)$
Find the dimension of the subspace of $\mathbb{R}^4$ spanned by the vectors
$$x_1= \begin{bmatrix}
1 & 2 & -1& 0
\end{bmatrix}^T, \qquad
x_2= \begin{bmatrix}
2 & 5 & -3& 2
\end{bmatrix}^T$$
$$x_3= \begin{bmatrix}
2 & 4 & -2& 0
\end{bmatrix}^T, \quad \text{and} \quad x_4= \begin{bmatrix}
3 & 8 & -5& 4
\end{bmatrix}^T.$$
The answer
The subspace $$Span(x_1,x_2,x_3,x_4)$$ is the same as the column space of the matrix
$$
X = \begin{pmatrix} 1 & 2 & 2 & 3 \\
2 & 5 & 4 & 8 \\
-1 & -3 & -2 & -5 \\
0 & 2 & 0 & 4
\end{pmatrix}
$$
The row echelon form of X is
$$
\begin{pmatrix}
1 & 2 & 2 & 3 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
The first two columns x1,x2 of X will form a basis for the column space of X. Thus, dim $$Span(x_1 , x_2 ,x_3 ,x_4)=2$$
My question
I know how to reduce the row echelon form of X, but I have no idea how to know which coloums form a basis a column space and why not use row space?
| The given vectors are the columns of
$$
A =
\left[\begin{array}{rrrr}
1 & 2 & 2 & 3 \\
2 & 5 & 4 & 8 \\
-1 & -3 & -2 & -5 \\
0 & 2 & 0 & 4
\end{array}\right]
$$
The reduced row echelon form of $A$ is
$$
\operatorname{rref}(A)=\left[\begin{array}{rrrr}
1 & 0 & 2 & -1 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
The first two columns of $\operatorname{rref}(A)$ are the pivot columns. This means that the first two columns of $A$ form a basis of $\operatorname{Col}(A)$.
Additionally, the nonzero rows of $\operatorname{rref}(A^\top)$ form a basis of $\operatorname{Col}(A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Probability of coins in a bag Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)\cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2\cdot(4/25)$ which is $8/25$.
| Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = \frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = \frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = \frac{32}{100} = \frac{8}{25}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Evaluating $\sum_{n=1}^\infty \frac{1}{(n^5)!} \approx 1$ and proving that is irrational. Define $\delta = \sum_{n=1}^\infty \frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $\delta$ is irrational, begin defining $S_n$ as:
\begin{align}
S_n = (n^5)!\delta \: - (n^5)!\sum_{k=1}^n\frac{1}{(k^5)!}
\end{align}
Where $(n^5)!\sum_{k=1}^n\frac{1}{(k^5)!}$ is an integer. Write $\delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so
\begin{align}
S_n &= (n^5)!\delta \: - (n^5)!\sum_{k=0}^n\frac{1}{(k^5)!}\\
&=(n^5)!\left(\delta - \sum_{k=0}^n\frac{1}{(k^5)!} \right)\\
&=(n^5)! \left( \frac{1}{(1^5)!}+\frac{1}{(2^5)!}+...+\frac{1}{(n^5)!}+
\frac{1}{(n+1)^5!}+... - \sum_{k=1}^n\frac{1}{(k^5)!} \right)
\end{align}
expanding the sum on the right it's possible to cancel a few terms:
\begin{align}
S_n &= (n^5)! \left( \frac{1}{(1^5)!}+\frac{1}{(2^5)!}+...+\frac{1}{(n^5)!}+
\frac{1}{(n+1)^5!}+... - \frac{1}{(1^5)!} - \frac{1}{(2^5)!}-...-\frac{1}{(n^5)!}
\right)\\
&=(n^5)! \left( \frac{1}{(n+1)^5!}+\frac{1}{(n+2)^5!}+... \right)\\
&= \frac{(n^5)!}{(n+1)^5!}+\frac{(n^5)!}{(n+2)^5!}+... \\
\end{align}
From this post we have that
\begin{align}
S_n = \frac{(n^5)!}{(n+1)^5!}+\frac{(n^5)!}{(n+2)^5!}+\cdots < \frac{1}{n^5+1}
\end{align}
So we have that $0<S_n<\frac{1}{n^5+1}$. Assume that $\delta = p/q$ then:
\begin{align}
0< (n^5)!p \: - q(n^5)!\sum_{k=1}^n\frac{1}{(k^5)!} < \frac{q}{n^5+1}
\end{align}
So, for a large $n$, we found a integer between $0$ and $1$, meaning $\delta$ is irrational
Wolfram says that $\delta=1$, but then the proof above is wrong.
How to find $\delta$ analytically? Numerically, also by wolf:
$\delta \approx \sum_{n=1}^{10} 1/(n^5)! \approx 1.000000000000000000000000000000000003800390754854743592594...$
|
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
| {
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"question_score": "3",
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Possible to prove that a particular trigonometric expression is always positive? This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $\Delta > 0$ always?
$$\begin{align}
\Delta &\equiv \sin(t) \sin\left(r+ (2 \pi -2 r - t)\frac{\epsilon}4 \right) \sin\left( \frac{2 - \epsilon}2 (\pi-r-t)\right) \\
&\quad {} - \frac{2 - \epsilon}2 \sin(r) \sin\left(t-\frac{\epsilon \; t}{4}\right) \sin(r+t)
\end{align}$$
where $$0<r<\frac{\pi}{4} \qquad 0<t<\frac{\pi}{4} \qquad 0<\epsilon <1$$
Some relevant posts include this one that renders the final form of $\Delta$, which hasn't gotten satisfactory answers.
| Given
$$0<r<\frac{\pi}{4} \qquad 0<t<\frac{\pi}{4} \qquad 0<\varepsilon <1\tag1$$
Easily to see that
$$t<\dfrac\pi2,\quad \dfrac{\varepsilon t}4 <\dfrac\pi{16}.$$
At the same time, sine increases in $\left(0,\dfrac\pi2\right).$
Therefore,
$$\begin{align}
&\sin t >\sin\left(t-\dfrac{\varepsilon t}4\right),\\[4pt]
&\Delta > \sin t\left(\sin\left(r+(2 \pi -2 r - t)\frac{\varepsilon}4 \right) \sin\left(\frac{2 - \varepsilon}2 (\pi-r-t)\right) - \frac{2 - \varepsilon}2\sin r\sin(r+t)\right).
\end{align}$$
On the other hand,
$$\sin\left(\frac{2-\varepsilon}2(\pi-r-t)\right)
= \sin\left((\pi-r-t)-(2\pi-2r-2t)\frac\varepsilon4\right)\\
= \sin\left(r+t+(2\pi-2r-2t)\frac\varepsilon4\right)
= \sin\left(r+t+\dfrac{3\varepsilon t}4+(2\pi-2r-t)\frac\varepsilon4\right).$$
So it is sufficiently to prove inequality $\delta(\varepsilon) >0,$ where
$$\delta(\varepsilon) = \sin(r+\varepsilon\varphi)\sin\left(r+t+\dfrac{3\varepsilon t}4+\varepsilon\varphi\right) - \frac{2 - \varepsilon}2 \sin(r)\sin(r+t),\tag2$$
$$\varphi = \dfrac{2\pi-2r-t}4 \in\left(\dfrac{5\pi}{16},\dfrac\pi2\right),\quad \dfrac{3\varepsilon t}4 <\dfrac{3\pi}{16},\tag3$$
under the conditions $(1).$
Really,
$$\delta(\varepsilon) = \sin(r+\varepsilon\varphi)\sin\left(r+t+\dfrac{3\varepsilon t}4 +\varepsilon\varphi\right) - \frac{2 - \varepsilon}2 \sin(r)\sin(r+t)\\
= \frac12\left(\cos\left(t+\dfrac{3\varepsilon t}4\right) - \cos\left(2r+t+\dfrac{3\varepsilon t}4+2\varepsilon\varphi\right)- \cos(t) + \cos(2r+t) +\varepsilon\sin(r)\sin(r+t)\right)\\
= \frac12\left(\cos\left(t+\dfrac{3\varepsilon t}4\right) - \cos(t) - \cos\left(2r+t+\dfrac{3\varepsilon t}4+2\varepsilon\varphi\right) + \cos(2r+t) +\varepsilon\sin(r)\sin(r+t)\right)\\
= -\sin\dfrac{3\varepsilon t}8 \sin\left(t+\dfrac{3\varepsilon t}8\right) + \sin\left(\varepsilon\varphi+\dfrac{3\varepsilon t}8\right) \sin\left(2r+t+\varepsilon\varphi+\dfrac{3\varepsilon t}8\right) + \dfrac\varepsilon2\sin(r)\sin(r+t).$$
Taking in account that
$$\varepsilon < 1 < 1+\dfrac{\pi-2r-2t}{\pi-r+t} = 2-\dfrac{r+3t}{\pi-r+t},\tag4$$
one can get
$$t+\dfrac{3\varepsilon t}8 < t+\dfrac{3\varepsilon t}8+2r +\varepsilon\varphi, \tag5$$
$$ t+\dfrac{3\varepsilon t}8+2r+\varepsilon\varphi = \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+\dfrac{3\varepsilon t}4+2r+\varepsilon\dfrac{2\pi-2r-t}4\\
= \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r + \varepsilon\dfrac{\pi-r+t}2\\
< \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r
+ \left(2-\dfrac{r+3t}{\pi-r+t}\right)\dfrac{\pi-r+t}2\\
= \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r + \pi-r+t -r- 3t
= \pi-t-\dfrac{3\varepsilon t}8,$$
$$ t+\dfrac{3\varepsilon t}8+2r+\varepsilon\varphi
< \pi-t-\dfrac{3\varepsilon t}8.\tag6$$
From $(5)-(6)$ should
$$\sin\left(t+\dfrac{\varepsilon t}8+2r+\varepsilon\varphi\right) > \sin\left(t+\dfrac{\varepsilon t}8\right),$$
so
$$\delta(\varepsilon) = \sin\left(\dfrac{\varepsilon t}8+\varepsilon\varphi\right) \sin\left(t+\dfrac{\varepsilon t}8+2r+\varepsilon\varphi\right) - \sin\dfrac{\varepsilon t}8 \sin\left(t+\dfrac{\varepsilon t}8\right) + \dfrac\varepsilon2\sin(r)\sin(r+t) > 0.$$
$\mathbf{Proved.}$
| {
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"source": "stackexchange",
"question_score": "5",
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Number of integral values of $c$ in solution set
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-\bigg(\frac{2c}{c-5}\bigg)x+\frac{c-4}{c-5}=0\;\;, c\neq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-\frac{4c}{c-5}+\frac{c-4}{c-5}<0$$
$$\frac{4(c-5)-4c+c-4}{c-5}<0$$
$$\frac{c-24}{c-5}<0\Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone help me whats wrong in my reasoning
| You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probabillity - rolling a dice 20 times, probability of a result gets only once
A dice is rolled $20$ times, with the possible results $\left\{1,2,3,4,5,6\right\}$.
Let $X$ be the number of results, out of the possible 6, which were chosen only once during the 20 rolls.
Calculate $P\left\{X\right\}$
I find it hard to identify the kind of variable it is. It isn't bio nominal nor hyper geometric.
I understand I have to choose 4 rolls out of the 20, and the combination between them is $4!$, giving me -
$$ \frac{\binom{20}{4} \times \binom{6}{4} \times 4!}{6^{20}} $$
For the chosen "results", the chosen "rolls' and the inner combination between them. But how about the other "rolls"? Something is missing here.
| Observe that since there are $20$ rolls, $P(X=6)=0$. So we need only check the probabilities that $X=1,2,3,4,5$. These can be done on a case by case basis.
For $X=5$, the probability is
$$\frac{\binom{6}{5}\cdot\binom{20}{5}\cdot 5!}{6^{20}}$$
since we must choose the the $5$ results which will occur only once, and choose which rolls they occur on in $\binom{20}{5}$ ways, accounting for their orderings.
For $X=4$, the probability is
$$\frac{\binom{6}{4}\cdot\binom{20}{4}\cdot 4!\cdot (2^{16}-30)}{6^{20}}$$
since we must choose the $4$ results which will occur only once, choose which rolls they occur on, order these $4$ results (in $4!$ ways), and then fill in the remaining $16$ rolls with the other two results. We must be a bit careful here, since we need each of the other results to occur at least twice, or not at all. Denote the remaining results by $x$ and $y$. Since there are $16$ rolls to fill in with $x'$s and $y$'s, there are $2^{16}$ possible outcomes. $15$ of them consist of $15 x'$s and one $y$, and another $15$ consist of $15 y$'s and one $x$. Discarding these $30$ undesirable outcomes leaves $2^{16}-30$.
The argument is similar for the other cases, but the last bit corresponding to the results that don't appear exactly once gets a bit more complicated. For $X=3$, we have have $17$ rolls that must be filled with, say, $x,y,z$ such that neither $x,y,$ nor $z$ appears exactly once. There are $17\cdot 2^{16}$ ways for $x,y,$ or $z$ to appear once (place it in one of $17$ positions then fill the other $16$ rolls with the other two results). And there are $\binom{17}{2}$ ways for two of them to appear only once. Since these are double-counted above, there are $3(17\cdot 2^{16}-\binom{17}{2})$ undesirable cases to discard.
I'll leave the cases $X=1,2$ up to you to compute. For now I'll just denote by $C_{4},C_{5}$ the number of ways to arrange the remaining results without any of them appearing exactly once. Therefore
$$P(X=3)=\frac{\binom{6}{3}\cdot\binom{20}{3}\cdot 3!\cdot [3(17\cdot 2^{16}-\binom{17}{2})]}{6^{20}}$$
$$P(X=2)=\frac{\binom{6}{2}\cdot\binom{20}{2}\cdot 2!\cdot (4^{18}-C_{4})}{6^{20}}$$
$$P(X=1)=\frac{\binom{6}{1}\cdot\binom{20}{1}\cdot (5^{16}-C_{5})}{6^{20}}$$
For completeness, observe that (clearly) $P(X<1)=P(X>6)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070013",
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"source": "stackexchange",
"question_score": "1",
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} |
Solve the sum $\sum _{i=1}^4\:\sum _{j=1}^i\:\left(i\cdot \:j-1\right)$
Is it correct if this sum is solved this way?
| $$\begin{align}
\sum_{i=1}^n\sum_{j=1}^i (ij-1)
&=\sum_{i=1}^ni\left(\sum_{j=1}^ij-1\right)\\
&=\sum_{i=1}^ni\left(\frac {i(i+1)}2-1\right)\\
&=\frac 12\sum_{i=1}^ni(i^2+i-2)\\
&=\frac 12 \sum_{i=1}^n i(i-1)(i+2)\\
&=\frac 12 \sum_{i=1}^n(i+1)i(i-1)+i(i-1)\\
&=\frac 12 \sum_{i=1}^n 6\binom {i+1}3+2\binom i2\\
&=3\binom {n+2}4+\binom {n+1}3\\
&=\binom{n+1}3\left(3\cdot \frac {n+2}4+1\right)\\
&=\frac 14 \binom {n+1}3(3n+10)\\
&=\frac 1{24}(n-1)n(n+1)(3n+10)\end{align}$$
Alternatively,
$$\begin{align}
\sum_{i=1}^n\sum_{j=1}^i (ij-1)
&=\frac 12\left(\sum_{i=1}^ni^3+\sum_{i=1}^ni^2\right)-\sum_{i=1}^n i\\
&=\frac 12 \left[\binom {n+1}2 ^2+\frac 13\binom {n+1}2 (2n+1)-2\binom {n+1}2\right]\\
&=\frac 12\binom {n+1}2 \left[\binom {n+1}2+\frac 13 (2n+1)-2\right]\\
&=\frac 12 \binom {n+1}2\cdot \frac16 \big[3n(n+1)+2(2n+1)-12\big]\\
&=\frac 1{12} \binom {n+1}2\cdot (3n^2+7n-10)\\
&=\frac 1{12}\binom {n+1}2 (n-1)(3n+10)\\
&=\frac 14\binom{n+2}3(3n+10)\\
&=\frac 1{24}(n-1)n(n+1)(3n+10)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Eigenvalue of (some) $ 4 \times 4 $ symmetric matrices $$A=\pmatrix{
0 & 3 & 2 & 0 \\
3 & 0 & 0 & 2 \\
2 & 0 & 0 & 3 \\
0 & 2 & 3 & 0 \\
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
| Observe
\begin{align}
M=
\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}
\end{align}
where $B = 2I_2$ and $A=\begin{pmatrix}0 & 3\\ 3 & 0 \end{pmatrix}$ and $A$ & $B$ commute. Then we see that
\begin{align}
\det\left(M-\lambda I_4 \right) = \det ((A-\lambda I_2)^2-B^2)
\end{align}
where we used the determinant formula for block matrices.
Note that
\begin{align}
(A-\lambda I_2)^2=
\begin{pmatrix}
\lambda^2+9 & -6\lambda\\
-6\lambda & \lambda^2+9
\end{pmatrix}
\end{align}
which means
\begin{align}
\det (M-\lambda I_4) = (\lambda^2+5)^2-36\lambda^2 = (\lambda^2-1^2).(\lambda^2-5^2)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
| This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n \geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + \sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
\begin{align*}
r_n &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + \sqrt{a_n + (n + 4)}}}} \\
&= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + ((n - 1) + 4)}}} \\
&= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 2} + ((n - 2) + 4)}}} \\
&\vdots \\
&= \sqrt{a_0 + \sqrt{a_1 + 5}} \\
&= \sqrt{a_0 + 4} \\
&= \sqrt{(0 + 3)^2} \\
&= 3.
\end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $\lim_{n \to \infty} r_n$, then this should also be an acceptable answer.
Now canonically, the expression $\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}$ (note that the ellipsis appears in a sum) would refer to the sequence $$r'_n = \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_n + 0}}}.$$ But since $r'_n \le r_n$, convergence follows immediately from the monotone convergence theorem.
Edit: For any integer $r \geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = \sqrt{p_0 + \sqrt{p_1 + \cdots + \sqrt{p_n + q_n}}}$$ for all $n \geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{41 + 8}}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:
Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$
For example, the case for $n=3$ can be proven like this:
Let $a,b,c \in \Bbb{R}^+$. Then we have:
$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $
$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $
By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:
$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $
However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.
Here is my attempt:
Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$
Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.
Inductive hypothesis: I assume $P(n)$ is true.
Inductive step:
$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$
$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$
And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.
Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?
| Just apply
$$AM\geq GM\geq HM$$
$$\Longrightarrow \frac{\sum_{i=1}^n a_i}{n}\geq \frac{n}{\sum_{i=1}^n\frac{1}{a_i}}$$
Now the result is immediate.
Here equality holds iff all $a_i's$ are equal.
Hope it helps:)
| {
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"url": "https://math.stackexchange.com/questions/3081320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cauchy Schwarz - Finding minimum of a function For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$. This is a weekend brain teaser in the 2nd week of Calc 3.
| Here is an alternate solution using the Cauchy-Schwarz Inequality:-
We have to minimize $f(x,y,z)=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$ subject to $2x+y+z=5$
From the Cauchy-Schwarz Inequality , we have $$(2\cdot(x-1)+1 \cdot (y+2) +1\cdot z)^2 \leq (2^2+1^2+1^2)((x-1)^2+(y+2)^2+z^2)$$ $$ \therefore (2x+y+z)^2 \leq
6\cdot ( (f(x,y,z)-1)$$
$$ \implies \frac{(2x+y+z)^2}{6} + 1 \leq f(x,y,z) \implies f(x,y,z) \geq \frac{25}{6}+1=\boxed{\frac{31}{6}} $$
Let’s take it a step further , and generalise the result .
We have to find the minimum of $f(x,y,z)=a_1x^2+b_1y^2+c_1z^2+a_2x+b_2y+c_2z+d$ , subject to $a_0x+b_0y+c_0z=l$
First , we complete the squares , and obtain $$f(x,y,z)=(\sqrt{a_1}x+\frac{a_2}{2\sqrt{a_1}})^2+(\sqrt{b_1}y+\frac{b_2}{2\sqrt{b_2}})^2+(\sqrt{c_1}z+\frac{c_2}{2\sqrt{c_1}})^2 +(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1}))$$
Then , by the Cauchy Schwarz Inequality , we have :- $$((\sqrt{a_1}x+\frac{a_2}{2\sqrt{a_1}})\cdot \frac{a_0}{\sqrt{a_1}}+(\sqrt{b_1}y+\frac{b_2}{2\sqrt{b_1}})\cdot \frac{b_0}{\sqrt{b_1}}+(\sqrt{c_1}z+\frac{c_2}{2\sqrt{c_1}})\cdot \frac{c_0}{\sqrt{c_1}})^2 \leq (f(x,y,z)-(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1})))(\frac{a_0^2}{a_1}+\frac{b_0^2}{b_1}+\frac{c_0^2}{c_1}) $$
And from this , we obtain :- $$ f(x,y,z) \geq \frac{(l+\frac{a_2a_0}{2a_1}+\frac{b_0b_2}{2b_1}+\frac{c_0c_2}{2c_1})^2}{(\frac{a_0^2}{a_1}+\frac{b_0^2}{b_1}+\frac{c_0^2}{c_1})}+(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1})$$ Perhaps not the prettiest result , but certainly works when none of the denominators equate to $0$ . And in your case , on substituting the values , the formula happens to equal $\frac{31}{6}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Any thoughts on this integral? $\int \cos^2(x)\cdot\sin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
| Write $c$ for $\cos x$ and $s$ for $\sin x$. Then the integrand is
\begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \\
&= \frac{1}{4} (2cs)^2 (1 - c^2) \\
&= \frac{1}{2} \frac{1}{4} (2cs)^2 (2 - 2c^2) \\
&= \frac{1}{2} \left( \frac{1}{4} (2cs)^2 (1 - 2c^2) \right)
+ \frac{1}{2} \left( \frac{1}{4} (2cs)^2 \right) \\
\end{align}
Now $2cs = \sin 2x$, and $1 - 2c^2 = -\cos 2x$, so from here things should be relatively simple.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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I can't see what is wrong with my answer to this P&C problem:
In how many different ways can 8 different books be distributed among 3 students if each receiver gets at least 2 books?
Here's what I did:
First take 6 books from the lot; that can be done in $\binom{8}{6}$ ways.
Then pick two from the 6 books, which can be done in $\binom{6}{2}$ ways. Pick 2 more from the remaining 4, which can be done in $\binom{4}{2}$ ways. The remaining 2 books are taken by the third person.
The remaining two books can go to one person $\binom{3}{1}$ or two different people $\binom{3}{1}\cdot \binom{2}{1}$.
So final answer is
$$\binom{8}{6}\binom{6}{2}\binom{4}{2}\binom{2}{2}\left(\binom{3}{1}+\binom{3}{2}\binom{2}{1}\right).$$
Where am I going wrong?
| As already explained you are overcounting the correct number: according to your procedure, the same distribution can be obtained in different orders.
Split the enumeration into different cases according to the possible distributions of the books among the three students: say $n_1$ books to the first student, $n_2$ books to the second student and $n_3$ books to the third one.
Since $n_1,n_2,n_3\geq 2$ and $n_1+n_2+n_3=8$, the triple $(n_1,n_2,n_3)$ can be:
1) $(4,2,2)$, $(2,4,2)$ or $(2,2,4)$: we have $3\cdot \binom{8}{4}\cdot \binom{4}{2}$ ways.
2) $(3,3,2)$, $(3,2,3)$ or $(2,3,3)$: we have $3\cdot \binom{8}{3}\cdot \binom{5}{3}$ ways.
Hence the total number is
$$3\cdot \binom{8}{4}\cdot \binom{4}{2}
+3\cdot \binom{8}{3}\cdot \binom{5}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inner product induced by the polarisation identitiy I am required to show that given a Banach space $(X,\|\cdot\|)$ on $\mathbb{R}$ that if the parallelogram identity holds then $X$ is a Hilbert space with the inner product given by the polarisation identity.
Proof:
Let $x+z=\frac{x+y}{2}+z+\frac{x-y}{2}$ and $y+z=\frac{x+y}{2}+z-\frac{x-y}{2}$, then by the parallelogram identity,
\begin{align}
\|x+z\|^{2}+\|y+z\|^{2}=2\bigg(\bigg|\bigg|\frac{x+y}{2}+z\bigg|\bigg|^{2}+\bigg|\bigg|\frac{x-y}{2}\bigg|\bigg|^{2}\bigg),
\end{align}
for all $x,y,z\in X$.
Now I want to show that $(x|z)+(y|z)=(x+y|z)$. Using the polarisation identity,
\begin{align}
(x|z)+(y|z) &= \frac{1}{4}((\|x+z\|^{2}+\|y+z\|^{2})-(\|x-z\|^{2}+\|y-z\|^{2}))\\
&=\frac{1}{4}\bigg(2\bigg|\bigg|\frac{x+y}{2}+z\bigg|\bigg|^{2}-2\bigg|\bigg|\frac{x+y}{2}-z\bigg|\bigg|^{2}\bigg)\\
&=2\bigg(\frac{x+y}{2}\bigg|z\bigg).
\end{align}
I am stuck here, I cannot work out how to remove the half out of the first term in the inner product I am left with.
With the answer provided I would like to complete the proof and have it checked, if that is OK.
Let $y=0$ then,
\begin{align}
\frac{1}{2}(x|z)=\bigg(\frac{x}{2}\bigg|z\bigg),
\end{align}
for all $x,z\in X$. Hence,
\begin{align}
(x|z)+(y|z)=(x+y|z).
\end{align}
Now we show this is true for -1 and any natural number $n$.
\begin{align}
(-x|y) &=\frac{1}{4}(\|-x+y\|^{2}-\|-x-y\|^{2})\\
&= \frac{1}{4}(\|x-y\|^{2}-\|x+y\|^{2}\\
&= -\frac{1}{4}(\|x+y\|^{2}-\|x-y\|^{2} = -(x|y).
\end{align}
Let $n\in\mathbb{N}$, then for $n=$, $1(x|y)=(1x|y)$. Assume $(nx|y)=n(x|y)$ holds for $n$. Now,
\begin{align}
((n+1)x|y) &= (nx+1x|y)\\
&= (nx|y)+(1x|y)\\
&= n(x|y)+1(x|y) = (n+1)(x|y).
\end{align}
Hence $(nx|y)=n(x|y)$ by induction. This extends $(nx|y)=n(x|y)$ to $n\in\mathbb{Z}$.
Finally I wish to show that $(\lambda x|y)=\lambda(x|y)$ for $\lambda\in\mathbb{R}$. From above, for any $n\in\mathbb{Z}$, $(nx|y)=n(x|y)$. Furthermore, for any $n\in\mathbb{Z}$,
\begin{align}
n\bigg(\frac{1}{n}x\bigg|y\bigg) &= \bigg(\frac{1}{n}x\bigg|y\bigg) + \bigg(\frac{1}{n}x\bigg|y\bigg) + \bigg(\frac{1}{n}x\bigg|y\bigg) + \cdots\\
&= (x|y),
\end{align}
which implies,
\begin{align}
\bigg(\frac{1}{n}x\bigg|y\bigg)=\frac{1}{n}(x|y).
\end{align}
Then for any $m,n\in\mathbb{Z}$ with $\frac{m}{n}\in\mathbb{Q}$ we have,
\begin{align}
\bigg(\frac{m}{n}x\bigg|y\bigg) = \frac{m}{n}(x|y).
\end{align}
Hence, by continuity of the norm,
\begin{align}
(\lambda x|y) =\lambda(x|y),
\end{align}
for $\lambda\in\mathbb{R}$.
| If you take your equality with $y=0$, you have shown that
$$
\left(\tfrac12\,x|z\right)=\tfrac12\,(x|z).
$$
And that's exactly what you need, now you can remove the $1/2$ you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. What could be a proof for it?
| If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$\left\lfloor \frac{a-b}{2} \right\rfloor = \frac{a-b}{2} \quad\text{and}\quad \left\lceil \frac{a+b}{2} \right\rceil = \frac{a+b}{2}.$$
Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$\left\lfloor \frac{a-b}{2} \right\rfloor = \frac{a-b}{2} - \frac12 \quad\text{and}\quad \left\lceil \frac{a+b}{2} \right\rceil = \frac{a+b}{2} + \frac12.$$
In either case, it's easy to check that your equation holds.
BTW, as noted by Gareth McCaughan, your equation in fact holds for all real numbers $b$, as long as $a$ is an integer. One fairly simple way to show this is to note that $\frac{a+b}{2} = a - \frac{a-b}{2}.$ Thus, we can rewrite your equation as $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil a - \frac{a-b}{2} \right\rceil = a.$$
Since $a$ is an integer (by assumption), and since $\lceil k + x \rceil = k + \lceil x \rceil$ for any integer $k$, we can extract $a$ from the ceiling term to get $$\left\lfloor \frac{a-b}{2} \right\rfloor + a + \left\lceil - \frac{a-b}{2} \right\rceil = a,$$ and finally, by applying the identity $\lceil -x \rceil = - \lfloor x \rfloor$, rewrite this as $$\left\lfloor \frac{a-b}{2} \right\rfloor + a - \left\lfloor \frac{a-b}{2} \right\rfloor = a.$$ Cancelling the floor terms then just leaves the identity $a = a$.
On the other hand, as also noted by Gareth, your equation cannot hold for any non-integer $a$, since its left-hand side is always an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| $$1=\cos^6x+\sin^4x\leq\cos^2x+\sin^2x=1,$$ which gives the following system.
$$\cos^6x=\cos^2x$$ and $$\sin^4x=\sin^2x.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$? Question:
If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a_n$ is convergent and $\left(\frac{4}{3}\right)^n$ is divergent.
But I somehow cannot find out how will their multiplication behave..
But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.
Please help
Thank you.
| EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
\begin{align}
R_n = \frac{(\frac{4}{3})^{n+1}(3-a_{n+1})}{(\frac{4}{3})^{n}(3-a_{n})} &=
\frac{4}{3} \times \frac{3-\frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
\frac{4}{3} \times \frac{\frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \\
&= \frac{4}{3} \times \frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
\frac{4}{3} \times \frac{a_n}{a_n+1}
\end{align}
$$
We know that $a_n < 3$. Since $\frac{x}{x+1}$ is increasing, $\frac{a_n}{a_n+1} < \frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How can $2\cos(x-\frac{\pi}2) = -2\sin(x-\frac{\pi}2)$ How can $2\cos(x-\dfrac{\pi}2) = -2\sin(x-\dfrac{\pi}2)$
I know that $\cos(-x) = \cos(x)$ and that $\cos(\dfrac{\pi}2-x) = \sin(x)$
From these two formulas I can get
$1.$ $2\cos(x-\dfrac{\pi}2) = 2\cos(\dfrac{\pi}2-x)$
$2.$ $2\cos(\dfrac{\pi}2-x) = 2 \sin(x)$
$3.$ $ 2\sin(x) = -2\sin(-x)$
$4.$ How can I get $-2\sin(-x) = -2\sin(x-\dfrac{\pi}2)$
It was part of calculation of limit:
$lim_{x->(\dfrac{\pi}{2})} \dfrac{1-e^{2cosx}}{2cosx} \dfrac{2cos(x-\dfrac{\pi}{2})}{sin(4(x-\dfrac{\pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above
| $$2\cos\left(x-\dfrac\pi2\right)=-2\sin\left(x-\dfrac\pi2\right)$$
$$\implies\tan\left(x-\dfrac\pi2\right)=-1=\tan\left(-\dfrac\pi4\right)$$
$$\implies x-\dfrac\pi2=n\pi-\dfrac\pi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$X \sim Exp(\frac{1}{3}) $ If $Y= \max(X,2)$ Find $E(Y)$ $X \sim Exp(\frac{1}{3}) $
If $Y= \max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
\begin{cases}
\ 2, & \text{if $0<x\le2$ } \\[2ex]
X, & \text{if $2<x<\infty$ }
\end{cases}$
$E(Y)=E(Y|0<x<2)P(0<x\le2)+E(Y|2<x<\infty)P(2<x<\infty)$
$E(Y)=2(1-e^{\frac{2}{3}})+e^{\frac{-2}{3}}\int \dfrac{x}{3}e^{-\frac{x}{3}} dx=2(1-e^{\frac{2}{3}})+\dfrac{e^{\frac{-2}{3}}}{3}\bigg(\bigg(-3xe^{-\frac{x}{3}}\bigg)_{2}^{\infty}-\bigg(9e^{-\frac{x}{3}}\bigg)_{2}^{\infty}\bigg)=2-2e^{\frac{-2}{3}}+\dfrac{e^{\frac{-2}{3}}}{3}\bigg(\bigg(6e^{-\frac{2}{3}}\bigg)+\bigg(9e^{-\frac{2}{3}}\bigg)\bigg)=2-2e^{-\frac{2}{3}}+5e^{-\frac{4}{3}}$
Second way
$E(Y)=\int \max(X,2)f(x)dx=\int_0^{2} 2f(x)dx+\int _{2}^{\infty}xf(x)dx=2(1-e^{\frac{-2}{3}})+\bigg(\bigg(-3xe^{-\frac{x}{3}}\bigg)_{2}^{\infty}-\bigg(9e^{-\frac{x}{3}}\bigg)_{2}^{\infty}\bigg)=2-2^{-\frac{2}{3}}+5e^{-\frac{2}{3}}=2+3e^{-\frac{2}{3}}$
Now I am not sure which of them is correct and why can anyone tell me ?
| This is how I would proceed (similar to your second method):
$E(Y)=\int_0^22\times\frac{1}{3}e^{\frac{-x}{3}}dx + \int_2^\infty x\times\frac{1}{3}e^{\frac{-x}{3}}dx$
$=2(1-e^{\frac{-2}{3}}) +3\int_\frac{2}{3}^\infty t\times e^{-t}dt$
$=2(1-e^{\frac{-2}{3}}) +3\Gamma(2,\frac{2}{3})$
$=2+3e^{\frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Milk" the integral $\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$ I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
\begin{align}
\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx&= \int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}\mathrm dx\\
&=\frac12\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}\mathrm dx\\
&=\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\mathrm dx\\
&=\sqrt{\frac{\pi(a+1)}{2}}\frac{\Gamma(r-\frac12)}{(2a+2)^r\Gamma(r)}
,\end{align}
Which works for $r>\frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\mathrm dx=\frac{\sqrt{\pi}\,\Gamma(r-\frac12)}{2^{2r}\Gamma(r)}.$$
Taking $\frac{d}{dr}$ on both sides,
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\log\left(\frac{x}{x^2+1}\right)\mathrm dx=\frac{\sqrt{\pi}}{2}\frac{d}{dr}\frac{\Gamma(r-\frac12)}{2^{2r}\Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that
$$\int_0^\infty\left(\frac{x}{x^2+1}\right)^{2r}\log^n\left[\frac{x}{x^2+1}\right]\frac{\mathrm dx}{(x^2+1)^2}=\frac{\sqrt\pi}{2^{n+4}}\left(\frac{d}{dr}\right)^n\frac{\Gamma(r+\frac32)}{4^r\Gamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
| First I will like to give some steps and maybe some more insight of this integral.$$I=\int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx+\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$$
With $x\rightarrow \frac{1}{x}$ in the second one we get:
$$\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx=\int_0^1 \left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2\left(\frac{1}{x^s}+1\right)}\mathrm dx$$
Now if we add with first part of the integral that was splited using: $\displaystyle{\frac{1}{x^s+1}+\frac{1}{\frac{1}{x^s}+1}=1}\,$ this being the reason why the $s$ doesn't affect our integral.
$$I=\int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx$$
Again via $x\rightarrow \frac{1}{x}$ we get:
$$I=\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx$$
$$\Rightarrow I=\frac12\int_0^\infty \left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2}dx=\frac14 \int_{-\infty}^\infty \left(\frac{1}{x^2+\frac{1}{x^2}+2a}\right)^r\left(1+\frac{1}{x^2}\right)dx$$
And now by writting $\displaystyle{x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2}$ and do a $x-\frac{1}{x}=t$ we get: $$I=\frac12 \int_{-\infty}^\infty \frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=x\sqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-\frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=\frac12 \int_{-\infty}^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx= \frac12 \int_{-\infty}^\infty \left(\frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}\right)^r \frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $x\rightarrow x-\frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty f(x-\cot x)dx$ to get :$$I=\frac12 \int_{-\infty}^\infty \frac{1}{(x^2+2(a+1))^r}dx=\frac12 \int_{-\infty}^\infty \frac{1}{((x-\cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=\frac12$ and $r=2$ to get: $$\int_0^\infty \frac{1}{((x-\cot x)^2 +1)^2}dx=\sqrt 2 \pi$$
One can do the same thing when there is $x^4$ in the denominator, but that's quite evilish.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=\frac12\int_0^\infty x^{r-1} \frac{1}{(x^2+2ax+1)^r}\left(\sqrt x+\frac{1}{\sqrt x}\right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem.
| {
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"url": "https://math.stackexchange.com/questions/3091108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Summation of combination of binomial coefficient Is there any way to find:
$$\sum_{i=0}^n {\binom{n}{i}i^k}$$
I know that we can find it for small k by using binomial theorem by differentiating both sides and then multiplying both sides by x and repeating till the form presents itself and then putting x =1, but is there a way for any general power( k here) of i?
| I think that the general form of $$S_k=\sum_{i=0}^n {\binom{n}{i}i^k}$$ is "just" an hypergeometric function
$$S_k=n \, _kF_{k-1}(2,\cdots,2,1-n;1,\cdots,1;-1)$$ in which the arguments appear $(k-1)$ times.
Looking for cases, the result seems to be of the form
$$S_k=2^{n-k}\,n \, P_{k-1}(n)$$ where $P$ is a polynomial.
For the very first values of $k$, the results are
$$\left(
\begin{array}{cc}
k & P_{k-1}(n) \\
1 & 1 \\
2 & n+1 \\
3 & n^2+3 n \\
4 & n^3+6 n^2+3 n-2 \\
5 & n^4+10 n^3+15 n^2-10 n \\
6 & n^5+15 n^4+45 n^3-15 n^2-30 n+16 \\
7 & n^6+21 n^5+105 n^4+35 n^3-210 n^2+112 n \\
8 & n^7+28 n^6+210 n^5+280 n^4-735 n^3+28 n^2+588 n-272 \\
9 & n^8+36 n^7+378 n^6+1008 n^5-1575 n^4-2436 n^3+5292 n^2-2448 n n+7936
\end{array}
\right)$$ where
*
*the second coefficient is $\binom{k+1}{2}=\frac{1}{2} k (k+1)$
*the third coefficient is $\binom{\binom{k}{2}}{2}=\frac{1}{8} (k-2) (k-1) k (k+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find limit of function of two variables (if exist) Find $$
\lim_{(x,0)\to (\pi,0)}\frac{3sin^3(x+y)}{y^2+(x-\pi)^2}.
$$
My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)\to(\pi,0)$ this function tends to $0$ (when $n\to \infty$) (for example I checked $(x_n,y_n)=(\frac{1}{n}+\pi,\frac{1}{n}).$ I would be grateful for your hints and ideas.
| I would expect the limit to exist and be zero.
*
*Note that $\sin (t\pm\pi)=\pm\sin t$. So $\sin (x+y)=-\sin (x-\pi+y)$.
*We also have, for small $t$, $|\sin t|\leq t$.
*And, $(a+b)^3\leq 4a^3+4b^3$.
Thus, for $(x,y)$ close to $(\pi,0)$,
\begin{align}
\left|\frac{3\sin^3(x+y)}{y^2+(x-\pi)^2}\right|
&=\left|\frac{3\sin^3(x-\pi+y)}{y^2+(x-\pi)^2}\right|
\leq\left|\frac{3(x-\pi+y)^3}{y^2+(x-\pi)^2}\right|
\leq\left|\frac{12(x-\pi)^3+12y^3}{y^2+(x-\pi)^2}\right|\\ \ \\
&\leq\left|\frac{12(x-\pi)^3 }{y^2+(x-\pi)^2}\right|+\left|\frac{ 12y^3}{y^2+(x-\pi)^2}\right|\\ \ \\
&\leq12|x-\pi|+12|y|,
\end{align}
using that
$$
\left|\frac{ (x-\pi)^2 }{y^2+(x-\pi)^2}\right|\leq1,\ \ \ \left|\frac{ y^2 }{y^2+(x-\pi)^2}\right|\leq1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.
| Multiply by conjugate:
$$\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}=\\
\lim_{x \rightarrow 0} \frac{1-\cos^2 x\cos^2 2x\cdots \cos^2 nx}{x^2\cdot(1+\cos x\cos 2x\cdots \cos nx)}=\\
\frac12 \lim_{x \rightarrow 0} \frac{1-(1-\sin^2 x)(1- \sin^2 2x)\cdots (1-\sin^2 nx)}{x^2}=\\
\frac12 \lim_{x \rightarrow 0} \frac{(\sin^2 x+\sin^2 2x+\cdots +\sin^2 nx)-(\sin^2 x \sin^2 2x +\cdots)+\cdots}{x^2}=\\
\frac12 \left(1+2^2+\cdots +n^2\right)=\frac{n(n+1)(2n+1)}{12}.$$
Note: The expansion of the brackets is performed by the elementary symmetric polynomial rule (see also here) and the limits of the degrees $4$ and higher when divided by $x^2$ will vanish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hyperparameter gradients for Matérn covariance Similarly to this question I am seeking the derivative of a covariance function with respect to its parameters. However, that question is specifically about the squared exponential kernel.
How can I calculate the hyperparameter gradients for the Matérn 5/2 kernel?
$$k_{\mathrm{Matern}}(r) = \sigma^2 \frac{2^{1-\nu}}{\Gamma(\nu)}\bigg(\frac{\sqrt{2\nu}r}{\ell}\bigg)^\nu K_{\nu}\bigg(\frac{\sqrt{2\nu}r}{\ell}\bigg)$$
|
How can I calculate the hyperparameter gradients for the Matérn 5/2 kernel?
If the general Matern kernel is written
$$ \mathcal{K}_M(r) = \sigma^2\frac{2^{1-\nu}}{\Gamma(\nu)}
\left( \frac{r\sqrt{2\nu}}{\ell} \right)^\nu K_\nu\left( \frac{r\sqrt{2\nu}}{\ell} \right) $$
then the $5/2$ kernel is given by
$$ \mathcal{K}_{M,5/2}(r) = \sigma^2\left[1 + \frac{r\sqrt{5}}{\ell} + \frac{5r^2}{3\ell^2} \right]
\exp\left(-\frac{r\sqrt{5}}{\ell}\right) $$
Then the gradient with respect to $\ell$ and $\sigma$ is
\begin{align}
\frac{\partial \mathcal{K}_{M,5/2} }{\partial \ell}
&=
\sigma^2\exp\left(-\frac{r\sqrt{5}}{\ell}\right)\frac{\partial }{\partial \ell}
\left[1 + \frac{r\sqrt{5}}{\ell} + \frac{5r^2}{3\ell^2} \right]
+
\sigma^2\left[1 + \frac{r\sqrt{5}}{\ell} + \frac{5r^2}{3\ell^2} \right]
\frac{\partial }{\partial \ell}
\exp\left(-\frac{r\sqrt{5}}{\ell}\right) \\
&=
\sigma^2\exp\left(-\frac{r\sqrt{5}}{\ell}\right)
\left[ \frac{-r\sqrt{5}}{\ell^2} - \frac{10r^2}{3\ell^3} \right]
+
\sigma^2\left[1 + \frac{r\sqrt{5}}{\ell} + \frac{5r^2}{3\ell^2} \right]
\left(\frac{r\sqrt{5}}{\ell^2}\right) \exp\left(-\frac{r\sqrt{5}}{\ell}\right) \\
&=
\sigma^2\exp\left(-\frac{r\sqrt{5}}{\ell}\right)
\left[
\frac{-r\sqrt{5}}{\ell^2} - \frac{10r^2}{3\ell^3}
+
\frac{r\sqrt{5}}{\ell^2} + \frac{5{r^2}}{\ell^3} + \frac{5\sqrt{5}r^3}{3\ell^4}
\right] \\
&=
\sigma^2\exp\left(-\frac{r\sqrt{5}}{\ell}\right)
\left[
\frac{5{r^2}}{3\ell^3} + \frac{5\sqrt{5}r^3}{3\ell^4}
\right] \\
&=
\frac{5{r^2}\sigma^2}{3\ell^3}
\exp\left(-\frac{r\sqrt{5}}{\ell}\right)
\left[
1 + \frac{r\sqrt{5}}{\ell}
\right] \\
\frac{\partial \mathcal{K}_{M,5/2} }{\partial \sigma}
&=
\frac{2}{\sigma}\mathcal{K}_{M,5/2}(r)
\end{align}
Source: Rasmussen and Williams, Gaussian Processes for Machine Learning.
| {
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"url": "https://math.stackexchange.com/questions/3094825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| Until the last step your attempt is correct. However, the last step (splitting to two limits) is not allowed, because it becomes the indetermined form $\infty-\infty$. Here is an alternative way with splitting:
$$\lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}=\\\lim_{x\to 0}\frac{(2\sqrt{1-x}-2)+(2-\sqrt[3]{8-x})}{x}=\\
\lim_{x\to 0}\frac{2(\sqrt{1-x}-1)}{x}+\lim_{x\to 0}\frac{2-\sqrt[3]{8-x}}{x}=\\
\lim_{x\to 0}\frac{2(1-x-1)}{x(\sqrt{1-x}+1)}+\lim_{x\to 0}\frac{8-(8-x)}{x(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2})}=\\
-1+\frac1{12}=-\frac{11}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$.
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$
*
*make exponent postive
$2x\frac{1}{\sqrt{4-x}} - 3\sqrt{4-x} = 0$
*
*square everything
$4x^2\frac{1}{4-x} - 9(4-x) =0$
*
*combine into one fraction
$\frac{1-9(4-x)[4x^2(4-x)]}{4x^2(4-x)} = 0$
*
*remove like terms
$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$
| If we put $x=4t$, the equation becomes
$$\frac{4t}{\sqrt{1-t}}=6\sqrt{1-t}$$
and
$$2t=3(1-t)$$
which gives
$$t=\frac 35 \; ,\; x=\frac{12}{5}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Writing a number in polar form (help converting $\theta$ to $\pi$)
Write $w = \sqrt{3} - i$ in polar form.
How is $\theta = \frac{-1}{\sqrt3} \textrm{ converted into } \frac{-\pi}{6}$? I understand that $w$ lies in the fourth quadrant of the unit circle, but that's about all I understand. I tried to convert degrees to radians, but I don't think that was the smart decision (I failed).
| The first thing to realize is that if you ever have $x^2 + y^2 = 1$ then there is a $\theta$ so that $x = \cos \theta; y = \sin \theta$ and $\tan \theta = \frac yx$.
And second thing to realise is that if you have any $a,b$ where they don't both equal $0$, you can rewrite $a = (\sqrt{a^2+b^2})a'$ and $b = (\sqrt{a^2 + b^2})b'$ where $a' = \frac a{\sqrt{a^2 + b^2}}$ and $b' = \frac b{\sqrt{a^2 + b^2}}$ and you will then have $a'^2 + b'^2 = 1$.
Which means there is $\theta$ so that $\tan \theta = \frac {b'}{a'} = \frac ba$, $a' = \cos \theta$ and $a = (\sqrt{a^2 + b^2})\cos \theta$ and $b' = \sin \theta$ and $b = (\sqrt{a^2 + b^2})\sin \theta$.
That means for any $a + bi \ne 0$ so that $a+bi$ is a complex number and $a$ is its "real" part and $bi$ is its "imaginary part", we can write:
$a+bi = (\sqrt{a^2 + b^2})(\frac a{\sqrt{a^2 + b^2}} + \frac b{\sqrt{a^2+b^2}}i)$
and that means there exist a $\theta$ so that
$a + bi = (\sqrt{a^2 + b^2})(\cos \theta + i\sin \theta)$.
If we imagine $a+bi$ as the point $(a,b)$ on a complex plane, this should not be surpising. $\sqrt{a^2 + b^2}$ is simply the distance from $(a,b)$ to $(0,0)$ and $\theta$ is simply the angle between $(a,b)$ and $(0,0)$ and the real axis.
And we can always find this $\theta$ by solving $\tan \theta = \frac ba$ and determining the proper quadrant by the positive and negative values of $a,b$.
So if $w = \sqrt 3 - i$ then if $r = \sqrt {\sqrt 3^2 + 1} = 2$ we have
$w = 2(\frac {\sqrt 3}2 - \frac 12 i)$.
And we can solve $\cos \theta = \frac{\sqrt{3}}2$ and $\sin \theta = -\frac 12$ and $\tan \theta = \frac 1 {\sqrt{3}}$. That theta is $-\frac {\pi}6$.
So
$w = \sqrt 3 - i$
$=2(\frac {\sqrt 3}2 - \frac 12 i)$
$= 2(\cos (-\frac \pi 6) + \sin (-\frac \pi 6) i)$.
And .... that's that.
====
Oh! Important post-script!
If you have $w = a + bi$ and you want to find the right $\theta$ and you think, "Okay, I've got to figure out $\sqrt {a^2 + b^2}$ and I have to divide $a$ and $b$ by it..."
You don't. Note: $\tan \theta = \frac {\frac a{\sqrt{a^2+b^2}}}{\frac b{\sqrt{a^2 + b^2}}} = \frac ab$. So $\theta = \arctan \frac ab$ or $\theta = \arctan \frac ab +\pi$.
So given $w = \sqrt 3 - i$ we need to simply solve $\tan \theta = \frac {-1}{\sqrt 3}$ with $\theta$ in the fourth quadrant.
As to WHY that works is because $w = 2(\frac {\sqrt 3}2 - \frac 12 i)$ but we don't have to do the (somewhat) tedious calculation $a' = \frac {\sqrt 3}{\sqrt{(\sqrt 3)^2 + (-1)^2}}$ and $ b' = \frac {-1} {\sqrt{(\sqrt 3)^2 + (-1)^2}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Formula for partial sum of binoms: $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom nk \binom mk$ I am dealing with some expressions containing combinatoric numbers. Does anybody know a formula for this?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$
| We can find a closed form expression in case $m\leq \left\lfloor\frac{n}{2}\right\rfloor$. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}
We obtain for $m\leq \left\lfloor\frac{n}{2}\right\rfloor$:
\begin{align*}
\color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{k}\binom{m}{k}}
&=\sum_{k=0}^m\binom{m}{k}\binom{n}{k}\\
&=\sum_{k=0}^m\binom{m}{k}[z^k](1+z)^n\\
&=[z^0](1+z)^n\sum_{k=0}^m\binom{m}{k}z^{-k}\\
&=[z^0](1+z)^n\left(1+\frac{1}{z}\right)^m\\
&=[z^m](1+z)^{m+n}\\
&\,\,\color{blue}{=\binom{m+n}{m}}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the equation represents a sphere, find center and radius $2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1$
So here's my guess:
$2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1$
Get all variables on one side
$2x^2 - 8x + 16 + 2y^2 + 2z^2 + 24z + 144 = 1 + 16 + 144$ Complete square
$(2x - 8)^2 + 2y^2 + (2z^2 + 24)^2 = 161$ Stopped here
Is it possible to complete the square with the $y$ value too? Would it just be 1?
| Note that $(2x)^2 \ne 2x^2$. The answer is:
$$2(x-2)^2 + 2(y-0)^2 + 2(z+6)^2 = 8 + 0 + 72 + 1$$
Thus, center $(2,0,-6)$, and radius $\sqrt{81/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear recursion with constant coefficients that fullfil $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$ Let $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$
I know that the generating function of $a_n= 3^n+7^n = A(x) = \frac{1}{1 - 3x} + \frac{1}{1 - 7x}.$
The exponential generating function
$$A(x) = \sum_{n=0}^{\infty}\frac{a_n}{n!}x^n$$
of the sequence $(a_n)_{n \in \mathbb{N}}$ is
$A(x)= \sum_{n=0}^{\infty}\frac{3^n}{n!}x^n+ \sum_{n=0}^{\infty}\frac{7^n}{n!}x^n= \sum_{n=0}^{\infty}\frac{(3x)^n}{n!}+ \sum_{n=0}^{\infty}\frac{(7x)^n}{n!}=e^{3x}+e^{7x}.$
I want to find out a linear recursion with constant coefficients, that fullfil $a_n$, but I don't know how its done
| $$a_n-3^n = 7^n$$
so $$ a_{n+1}-3^{n+1}= 7^{n+1}$$
so $$a_{n+1}-3^{n+1}=7\cdot 7^n = 7(a_n-3^n )$$
so $$a_{n+1}-7a_n = -4\cdot 3^n$$
so $$a_{n+2}-7a_{n+1} = -4\cdot 3^{n+1}$$ $$=-12\cdot 3^n$$ $$= 3(-4\cdot 3^n)$$
$$ =3 (a_{n+1}-7a_n )$$
So $$a_{n+2}-10a_{n+1} +21a_n =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104885",
"timestamp": "2023-03-29T00:00:00",
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Precalculus algebra exercise Hi I need to solve this problem and I don’t know how so I’d appreciate a hint.
If
$a^2x^2 + b^2y^2 + c^2z^2 = 0$
$a^2x^3 + b^2y^3 + c^2z^3 = 0$
$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2$
Then $a^4x^3 + b^4y^3 + c^4z^3 = 0$
I think that $a^4x^3 + b^4y^3 + c^4z^3 = 0$ is a factor in an expression which can be found by manipulating the three given equations. I can see that
$\frac 1x - a^2 - \frac 1y + b^2 = 0$
So I tried to add, subtract, multiply given equations.
| Suppose that
$$
\begin{cases}
(ax)^2 + (by)^2 + (cz)^2 = 0\\
a^2x^3 + b^2y^3 + c^2z^3 = 0\\
\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2
\end{cases}
$$
However, for any $a, x \in \mathbb R$, $(ax)^2 \geq 0$. So we have
$$
0 = (ax)^2 + (by)^2 + (cz)^2 \geq 0 + 0 + 0 = 0
$$
Therefore in each inequality, there's equality, so
$$
\begin{cases}
ax = 0\\
by = 0\\
cz = 0
\end{cases}
$$
This implies
$$
a^4x^3+b^4y^3+c^4z^3 = (ax)a^3x^2 + (by)b^3y^2 + (cz)c^3z^2 = 0\cdot a^3x^2 + 0 \cdot b^3y^2 + 0 \cdot c^3z^2 = 0
$$
How funny that we didn't need to use equations (2) and (3).
| {
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"source": "stackexchange",
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How to evaluate $\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$
$$\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$$
How can I deal with n-th root and the second part of the formula which I know is equal to zero?
I do not know really where to start, I tried to used $a^x = e^{x\ln(a)}$ but that doesnt seem to get me nowhere, I still have to deal with the second part which I know is equal to zero, however I feel like I should do something also with this part?
How can I solve it, and what is the general reasoning when you see limit like this?
| You may also proceed using
*
*$(\star)$: $a^n-b^n = (a-b)\sum_{k=0}^{n-1}a^kb^{n-k}$
\begin{eqnarray*} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]
& = & \color{blue}{n^4} \sqrt[n]{1+\frac{4^n\cdot n^n}{n^{4n}}}\left[\left( 4+\frac{4}{n^2} + \frac{1}{n^4} \right)^9 - \left(4+\frac{4}{n^2}\right)^9 \right]\\
& \stackrel{(\star)}{=} & \color{blue}{n^4}\sqrt[n]{1+\frac{4^n}{n^{3n}}} \frac{1}{\color{blue}{n^4} }\sum_{k=0}^8\left(4+\frac{4}{n^2} + \frac{1}{n^4} \right)^k\left(4+\frac{4}{n^2}\right)^{8-k}\\
& = & \sqrt[n]{1+\frac{4^n}{n^{3n}}}\sum_{k=0}^8\left(4+\frac{4}{n^2} + \frac{1}{n^4} \right)^k\left(4+\frac{4}{n^2}\right)^{8-k} \\
& \stackrel{n \to \infty}{\longrightarrow} & 1 \cdot \sum_{k=0}^84^8 \\
& = & \boxed{9\cdot4^8} \\
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find:
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$.
| The key is to find the greatest common divisor of the two differences $\frac{1}{6}$ and $\frac{2}{15}$. How? Exactly like we would for integers.
If we divide $\frac{1}{6}$ by $\frac{2}{15}$, we get $1$ with a remainder of $\frac{1}{6}-\cdot\frac{2}{15} = \frac{5}{30}-\frac{4}{30}=\frac{1}{30}$.
Next, we have the pair $\frac{2}{15}$ and $\frac{1}{30}$. Dividing the former by the latter, we get $4$ with a remainder of zero.
That remainder of zero marks an end to the algorithm. The greatest common divisor is the last number we had before we reached zero - in this case, $\frac1{30}$.
So that's $d$. To find $m$ and $n$, we divide: $m=\frac 16/ \frac{1}{30} = 5$ and $n=\frac{2}{15}/ \frac{1}{30} = 4$.
The Euclidean algorithm works to find the greatest common divisor of any two rational numbers, not just any two integers as usually presented.
And yes, that means that if we write down three different rational numbers, there's always some arithmetic progression that has all three of them as terms. The common difference can be the greatest common divisor, or any further divisor of that number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry problem (proving a relation between sides of a triangle) Let $ABC$ be a triangle with unequal sides. The medians of $ABC$,
when extended, intersect its circumcircle in points $L, M, N$. If $L$ lies
on the median through $A$ and $LM = LN$, prove that: $2BC^{2}=CA^2+AB^2$.
So I first wrote midpoint theorem $AB^2+AC^2=2AX^2 +2XB^2$ (refer to the diagram) and so I tried to put $AX$ in terms of $BX$ or $BC$ but I can't find a way. I'm also not sure where $LM$ and $LN$ come in, so it would be helpful if someone could tell me the line of thought. Thanks!
| Below is a brute force solution. Though there should be more elegant solutions, I have decided to present this one due to the beauty of some resulting algebraic expressions.
First, it should be clarified that (as will be seen) the real meaning of the condition "a triangle with unequal sides" is "a triangle with $AC\ne AB$".
Everywhere below we use notation:
$$
BC=a,\quad CA=b,\quad AB=c,\quad AX=m_a,\quad BY=m_b,\quad CZ=m_c.
$$
First we compute the length of the segment $LM$. For this we need the lengths $OL$, $OM$ and cosine of the angle $\widehat{LOM}$:
$$
OL=OX+XL=\frac{m_a}3+\frac{a^2}{4m_a}=\frac{4m_a^2+3a^2}{12m_a}\\
=\frac{(2b^2+2c^2-a^2)+3a^2}{12m_a}=\frac{a^2+b^2+c^2}{6m_a},
$$
and similarly
$$
OM=\frac{a^2+b^2+c^2}{6m_b}.
$$
Further:
$$
\cos \widehat{LOM}=\frac{OA^2+OB^2-BC^2}{2OA\cdot OB}=\frac{m_a^2+m_b^2-\left(\frac32c\right)^2}{2m_am_b}
$$
Combining together one obtains:
$$
LM=\frac{a^2+b^2+c^2}{6}\left[\frac1{m_a^2}+\frac1{m_b^2}
-\frac2{m_am_b}\frac{m_a^2+m_b^2-\left(\frac32c\right)^2}{2m_am_b}\right]^{1/2}\\
=\frac{(a^2+b^2+c^2)c}{4m_am_b}.
$$
Similarly:
$$
LN=\frac{(a^2+b^2+c^2)b}{4m_am_c}.
$$
Combining two last equalities one obtains:
$$
LM=LN\implies m_bb=m_cc\\
\implies (2a^2+2c^2-b^2)b^2=(2a^2+2b^2-c^2)c^2\implies (2a^2-b^2-c^2)(b^2-c^2)=0.
$$
As $b\ne c$ one is left with
$$
b^2+c^2=2a^2.
$$
The equality $ m_bb=m_cc$ has a simple geometric interpretation. It means that the medians $BY$ and $CZ$ intersect the respective sides $AC$ and $AB$ at equal angles. If $\angle BYC=\angle CZB$ then $b=c$. If $\angle BYC=\angle CZA$ then $b^2+c^2=2a^2$.
The required triangle $ABC$ can be constructed as follows:
Draw a circle $\cal C$ with radius $\frac{\sqrt3}2 BC$ centered at the midpoint $X$ of the segment $BC$. Then arbitrary point of the circle (except for the intersection points with $BC$) can be taken as the third vertex $A$ of the triangle.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find integral of this fraction with a radical. Is my process right? I think this involves a u-sub and then partial fractions? Is this process right? Basically I learned any integral can be solved with using usub and integration by parts... and using partial fractions is just some algebraic manipulation that then relies on some definitions of integration. I just wanted to make sure my process was right before continuing?
The integral in question:
$$ \int \frac{dx}{2 \sqrt{x+3} + x}$$
So u sub first:
$ u = \sqrt{x + 3}$ so $ u^2 = x + 3$ so $ u^2 - 3 = x$ so $\frac{dx}{du} = 2u$ so $dx = 2udu$
So the integral is equivalent to:
$$ \int \frac{2udu}{2u - u^2 - 3}$$
$$= \int \frac{2udu}{u^2 + 2u - 3}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)}$$
So from here I can use partial fractions because we have distinct linear factors in the denominator? Is that right?
So is this partial fraction formula right?
$$ \frac{2udu}{(u + 3)(u - 1)} = \frac{A}{ u + 3} + \frac{B}{u - 1}$$
so
$$ 2u = A(u - 1) + B(u + 3)$$
$$ 2u = Au - A + Bu + 3B$$
$$ 2u = (A+B)u - A + 3B$$
so
$$ 2 = A + B$$ and
$$ 0 = -A + 3B$$
$$2 = 4B$$
So $B = \frac{1}{2}$ and $A = \frac{3}{2}$
So then we got:
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \int \frac{3}{2} \cdot \frac{du}{u + 3} + \frac{1}{2} \cdot \frac{du}{u-1}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \int \frac{3}{2} \cdot \frac{du}{u + 3} + \int \frac{1}{2} \cdot \frac{du}{u-1}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \frac{3}{2} \int \cdot \frac{du}{u + 3} + \frac{1}{2} \int \cdot \frac{du}{u-1}$$
So I guess formally do the right sides still need another usub? thats the only way to solve although I guess we can do that in our head eventually?
$$so \int \frac{du}{u+3}$$
if $z = u + 3$ and $dz = du$
so
$$\int \frac{dz}{z} = \ln{z} = \ln(u+3) $$
So we have:
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \frac{3}{2} \ln {(u + 3)} + \frac{1}{2} \ln {(u - 1)}$$
So finally and remembering that $u = \sqrt{x+3}$
$$ \int \frac{dx}{2 \sqrt{x+3} + x} = \frac{3}{2} \ln {(\sqrt{x+3} + 3)} + \frac{1}{2} \ln {(\sqrt{x+3} - 1)} + C$$
Is that right? We get here through a u-sub, then partial fractions, then another u-sub?
| Notice how $\int \frac{1}{x} dx = \ln{(abs(x))} + C$ and not $\ln{(x)} + C$, for instance, the term $\frac{1}{2}\ln{(u-1)}$ is actually $\frac{1}{2} \ln{(abs(u-1))}$. The final expression becomes:
$\int \frac{dx}{2 \sqrt{x+3} + x} = \frac{3}{2} \ln {(\sqrt{x+3} + 3)} + \frac{1}{2} \ln {(abs(\sqrt{x+3} - 1))} + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \left(x^2(1+x^3)^{\frac{2}{3}}\right)^{-1} dx$ I'm stuck on evaluating this indefinite integral.
$$\int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}$$
I tried doing a u-substitution on the $1+x^3$ term inside the two-thirds power but didn't get anywhere. Any help?
| There is a "dirty" trick having in mind that
*
*$\frac{d}{dx}\left( \sqrt[3]{1+x^3}\right)= \frac{x^2}{\left(1+x^3\right)^{\frac{2}{3}}}$
Then partial integration gives:
\begin{eqnarray*} \int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}\; dx
& = & \int\frac{1+x^3 - x^3}{x^2\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & \int\underbrace{\frac{1}{x^2}}_{u'}\underbrace{\sqrt[3]{1+x^3}}_{v}\; dx - \int\frac{x}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & -\frac{\sqrt[3]{1+x^3}}{x} + \int\frac{1}{x}\cdot \frac{x^2}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx - \int\frac{x}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & -\frac{\sqrt[3]{1+x^3}}{x} (+C)
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$,
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
| Hint: Substitute $x=\tfrac12 u -\tfrac12$
Addendum: Your original problem was $\int_0^1\frac{x\; dx}{(2x+1)^3}$. For this substitution, compute:
Integrand: We have $x=\tfrac12 u -\tfrac12=\tfrac12(u-1)$, so
$$\frac{x}{(2x+1)^3}=\frac{\tfrac12(u-1)}{(2\cdot\frac12(u-1)+1)^3}
=\tfrac12\frac{u-1}{u^3}=\boxed{\tfrac12\left(u^{-2}-u^{-3}\right)}$$
Differential: We have $x=\tfrac12 u -\tfrac12$, so
$$dx = d\left(\tfrac12 u -\tfrac12\right)=\boxed{\tfrac12du}$$
Limits of integration: The limits for $x$ are $0$ and $1$, so we find the corresponding values of $u$:
$$x=0\implies 0=\tfrac12 u -\tfrac12\implies u=\boxed{1}$$
$$x=1\implies 1 = \tfrac12 u -\tfrac12 \implies u=\boxed{3}$$
So, replacing the integrand, differential, and limits with the boxed items above we have
$$\int_0^1\frac{x\; dx}{(2x+1)^3} = \int_1^3\tfrac12\left(u^{-2}-u^{-3}\right) \tfrac12du$$
$$=\tfrac14\int_1^3\left(u^{-2}-u^{-3}\right) du$$
$$=\left.\frac14\left(-u^{-1}+\tfrac12u^{-2} \right)\right]_1^3$$
$$=\tfrac14[(-\tfrac13+\tfrac1{18})-(-1+\tfrac12)]$$
$$=\tfrac14[-\tfrac{5}{18}+\tfrac12]$$
$$=\tfrac14[\tfrac{4}{18}]$$
$$=\boxed{\tfrac{1}{18}}$$
This is the precise procedure you should follow for any substitution. Don't take shortcuts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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calculation of $\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$
Calculation of $$\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$$
My Try: Let $$I = \int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x+1)(x+2)}dx$$
Put $x=\sin^2 \theta$ and $dx = 2\sin \theta \cos \theta$ and changing limits
So $$I = \int^{\frac{\pi}{2}}_{0}\frac{2}{(1+\sin^2 \theta)(2+\sin^2 \theta)}d\theta$$
Could someone help me to solve it , Thanks
| We have $$\frac{I}{2}=\int_0^{\pi/2}\left(\frac{1}{1+\sin^2\theta}-\frac{1}{2+\sin^2\theta}\right)d\theta,$$and with $t=\tan\theta$ we have $\sin^2\theta=\frac{t^2}{1+t^2}$ so $$\int_0^{\pi/2}\frac{d\theta}{k+\sin^2\theta}=\frac{1}{k+1}\int_0^\infty\frac{dt}{\frac{k}{k+1}+t^2}=\frac{\pi}{2\sqrt{k(k+1)}},$$so$$I=\pi\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to find the minimum value of $\dfrac{a_1^2+a_2^2+\dots+a_n^2}{(a_1+a_2+\dots+a_n)^2}$? Motivation: To find the maximum value of $c$ in 2.4-1 Lemma (Linear combinations) of the book Kreyszig - Introductory Functional Analysis with Applications for $X=\mathbb{R^n}$ which reduces to solving the following question:
What is the minimum value of $\dfrac{a_1^2+a_2^2+\dots+a_n^2}{(a_1+a_2+\dots+a_n)^2}$ for $a_i \ge 0$?
For $n=2$, it is easy: let $x=a_1/a_2$ and the minimum occurs when $x=1$ or $a_1=a_2$.
For $n>2$, I tried the techniques that I had learnt from multi-variable Calculus but it gets quite complicated with no light ahead!
| Let $\vec{a} = \begin{bmatrix}a_1 & a_2 & \cdots & a_n\end{bmatrix}^T$ and $\vec{1} = \begin{bmatrix}1 & 1 & \cdots & 1\end{bmatrix}^T$. Using the Cauchy-Schwarz Inequality, we have
\begin{align*}
\|\vec{a}\|^2 \|\vec{1}\|^2 &\ge (\vec{a} \cdot \vec{1})^2
\\
(a_1^2+a_2^2+\cdots+a_n^2)(1^2+1^2+\cdots+1^2) &\ge (a_1\cdot 1+a_2\cdot 1+\cdots+a_n\cdot 1)^2
\\
n(a_1^2+a_2^2+\cdots+a_n^2) &\ge (a_1+a_2+\cdots+a_n)^2
\\
\dfrac{a_1^2+a_2^2+\cdots+a_n^2}{(a_1+a_2+\cdots+a_n)^2} &\ge \dfrac{1}{n}
\end{align*}
Equality holds iff $\vec{a}$ and $\vec{1}$ are parallel, i.e. $a_1 = a_2 = \cdots = a_n$. (Also, we need the $a_i$'s to be non-zero.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Why is $\sum_{i=1}^6 2^i = 2^7-2$?
Why is $$\sum_{i=1}^6 2^i = 2^7-2$$
| $$\sum_{i=1}^6 2^i =2+2^2+2^3+2^4+2^5+2^6$$ $$= 2(1+2+2^2+..+2^5) = 2{2^6-1\over 2-1} =2^7-2$$
We have generaly $$1+x+x^2+...+x^n = {x^{n+1}-1\over x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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find the maximum of the value $\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\frac{a_{i}}{i}\right)^2$ let $a_{i}\ge 0$ and such $$a_{1}+a_{2}+\cdots+a_{n}=1$$
find the maximum of the value
$$\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\dfrac{a_{i}}{i}\right)^2$$
I try to use
From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$,
$$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$
But I can't it.Thanks
| Alternative solution
Problem: Let $a_i\ge 0, \forall i$ with $a_1 + a_2 + \cdots + a_n = 1$. Prove that the maximum of
$$\left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2$$
is $\frac{4(n+1)^3}{27n^2}$.
We may use the approach in my answer to this equation:
An upper bound of product of two inner products
Consider the optimization problem
$$\max_{a_i\ge 0, \forall i; \ \sum_{i=1}^n a_i = 1} \left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2.$$
Let $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$ be a global maximizer.
We claim that if $1 < k < n$, then $a_k^\ast = 0$.
Indeed, if $a_k^\ast > 0$, let
$$a_1' = a_1^\ast + (1 - y) a_k^\ast, \quad a_k' = 0, \quad a_n' = a_n^\ast + y a_k^\ast$$
where $\frac{k-1}{n-1} < y < \frac{n}{k}\cdot \frac{k-1}{n-1}$. We have
$$a_1' + a_k' + a_n' = a_1^\ast + a_k^\ast + a_n^\ast,$$
and
\begin{align}
&a_1' + ka_k' + na_n' - (a_1^\ast + ka_k^\ast + na_n^\ast)\\
=\ & (n-1)\left(y - \frac{k-1}{n-1}\right) a_k^\ast\\
>\ & 0,
\end{align}
and
\begin{align}
&a_1' + \frac{a_k'}{k} + \frac{a_n'}{n} - \left(a_1^\ast + \frac{a_k^\ast}{k}
+ \frac{a_n^\ast}{n}\right)\\
=\ & \frac{n-1}{n}\left(\frac{n}{k}\cdot \frac{k-1}{n-1} - y\right)a_k^\ast\\
>\ & 0.
\end{align}
However, this contradicts the optimality of $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$.
Now, since $a_2^\ast = a_3^\ast = \cdots = a_{n-1}^\ast = 0$, we have
\begin{align}
&\left(\sum_{i=1}^n i a_i^\ast\right)\left(\sum_{i=1}^n \frac{a_i^\ast}{i}\right)^2\\
=\ & (a_1^\ast + n a_n^\ast)\left(a_1^\ast + \frac{a_n^\ast}{n}\right)^2\\
=\ & (1 - a_n^\ast + na_n^\ast)\left(1 - a_n^\ast + \frac{a_n^\ast}{n}\right)^2\\
=\ & \frac{1}{2n^2}[2 + 2(n-1)a_n^\ast][n - (n-1)a_n^\ast]^2\\
\le\ & \frac{1}{2n^2} \left(\frac{2 + 2(n-1)a_n^\ast + n - (n-1)a_n^\ast
+ n - (n-1)a_n^\ast}{3}\right)^3\\
=\ & \frac{4(n+1)^3}{27n^2}
\end{align}
with equality if $a_n^\ast = \frac{n-2}{3(n-1)}$, where we have used AM-GM inequality.
Thus, the desired maximum is $\frac{4(n+1)^3}{27n^2}$
at $a_1 = \frac{2n-1}{3(n-1)}, a_2 = a_3 = \cdots = a_{n-1} = 0, a_n = \frac{n-2}{3(n-1)}$.
| {
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"url": "https://math.stackexchange.com/questions/3128289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.