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Irreducibility and Splitting Fields Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors.
Edited:
This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. Suppose char $F\neq 2,3$. Suppose also that $f$ is neither irreducible nor splits in $F$. Then $f$ is reducible which implies that $a_1 \in F$. i.e. $f(x)=(x-a_1)g(x)$, where $g(x)\in F[x]$ is irreducible with deg $g=2$. Let $K$ be the splitting field of $g$. The $K$ is Galois over $F$.
So if $\sigma \in $ Aut($K/F$), then $\sigma (a_1)=a_1$ since $\sigma $ fixes $F$ and $a_1 \in F$. Since $\sigma$ permutes the roots of $f$, WLOG suppose $\sigma (a_2)=a_3$. Then
$\sigma(\triangle) = \sigma((a_1-a_2)(a_1-a_3)(a_2-a_3))=-\triangle$.
But $\triangle^2=D(f)=81$, so $\triangle = \pm 9 \in F$, so $\triangle \in F$. Therefore,
$9=-9$ $\implies 1=-1 \implies$ char $F =2$ which is a contradiction. So $f$ is either irreducible or splits in $F$.
Next suppose char $F=2$. Well, I'm not exactly sure what I can say about $f$.
I would like to know if my approach is correct and also what to do in the second case. Thanks.
ADDED:
If char $F=2$, then $f=x^3+x+1$. Suppose $b$ is a root of $f$. Then $b^2 $ is also a root, since $f(b^2)=(b^2)^3+b^2+1=(b+1)^2+b^2+1=2b^2+2=0$. Is it enough to conclude that $f$ splits?
| Here is one way to do it using almost no theory, just playing with algebra.
Suppose the polynomial has a root $a$ in $F$. If you divide $x^3 - 3x + 1$ by $x-a$ the quotient is the polynomial
$$
x^2 + ax + a^2 - 3
$$
in $F(a)[x]$. From the quadratic formula (assuming characteristic $\neq 2$ here, for the moment, I guess) you can see that this will have its roots in $F(a)$ if and only if $12-3a^2$ is a square in $F(a)$.
You may know that any element in $F(a)$ can be written in the form $pa^2 + qa + r$ for some $p,q,r$ in $F$. The idea is to square this symbolic expression, rewrite it as a polynomial in $a$ of degree at most $2$, and then see if values $p, q, r$ in $F$ can be found making this equal to $-3a^2 + 0a + 12$.
Using the identity $a^3 = 3a - 1$ (from the fact that $a$ is a root of the given polynomial) and $a^4 = 3a^2 - a$ (from multiplying the previous identity by $a$) one gets
$$
(pa^2 + qa + r)^2 = (3p^2 + 2pr + q^2) a^2 + (-p^2 + 2qr + 6pq) a + (r^2 - 2pq).
$$
So the goal is to find $p,q,r$ satisfying $3p^2 + 2pr + q^2 = -3$, $-p^2 + 2qr + 6pq = 0$, and $r^2 - 2pq = 12$. Since we don't know anything about $F$, we might optimistically look for these $p, q, r$ in $\mathbb{Z}$. This system of $3$ polynomial equations in $3$ integer unknowns can be fed to software, and one finds that e.g. $p = 2$, $q = 1$, and $r = -4$ give a solution (no matter what $F$ is!). So the roots of $x^3 - 3x + 1$ are all in $F(a)$, and we can actually write formulas for them: $a$, $\frac{-a + (2a^2 + a -4)}{2}$, and $\frac{-a - (2a^2 + a - 4)}{2}$, which simplify to $a$, $a^2 - 2$, $-a^2 - a + 2$.
Although we assumed characteristic $\neq 2$ to use the quadratic formula, we can immediately check that the single identity $a^3 - 3a + 1 = 0$ is indeed enough to ensure that
$$
(x-a)(x - (a^2 - 2)) (x - (-a^2 - a + 2)) = x^3 - 3x + 1.
$$
[In detail: expanding, the coefficient of $x^3$ is $1$ on the nose, the coefficient of $x^2$ is $0$ on the nose, and the coefficient of $x$ is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has a remainder of $-3$; similarly the constant term is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has remainder of $1$.]
So if $x^3 -3x + 1$ has one root in $F$, it splits in $F$ for the reason that we can explicitly write the other two roots as polynomials in the one that we already have.
| {
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solving quadratic equations by completing the square
Possible Duplicate:
How to “Re-write completing the square”: $x^2+x+1$
$b^2-4b+4=16$ can someone help me solve this equation. All the steps please.
| if $ax^2+bx^2+c=0$ then
$$
0=a\Big(x^2+\frac{b}{a}x\Big)+c
=a\Big(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\Big)+c
=a\Big(x+\frac{b}{2a}\Big)^2+c-\frac{b^2}{4a}
$$
hence
$$
\Big(x+\frac{b}{2a}\Big)^2=\frac{b^2-4ac}{4a^2}
$$
and
$$
x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
| {
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$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?
| I'm posting this as a community wiki "answer" because I'm not sure it works and the explanation wouldn't show up very well in the comments section I guess. If we factor the equation in $\mathbb{Z}[\sqrt{3}]$ we get $(y^2 - 4\sqrt{3})(y^2 + 4\sqrt{3}) = x^3$. Thus if $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ for some $a + b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$ then we get
$$y^2 - 4 \sqrt{3} = a^3 + 3a^2b\sqrt{3} + 9ab^2 + 3b^3 \sqrt{3} $$
so that we must have $3a^2 b + 3b^3 = -4$ and since $3$ does not divide $4$ we get a contradiction. The problem is that I need to justify somehow that $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ and although I know that $\mathbb{Z}[\sqrt{3}]$ is a unique factorization domain, I would need the two factors $y^2 - 4 \sqrt{3}$ and $y^2 + 4 \sqrt{3}$ to be coprime I think, and I can't seem to justify it or disprove it.
Edit (Alex):
Suppose that we have an integer solution to $x^3 = y^4 - 48$.
Claim: The two factors $y^2\pm 4\sqrt{3}$ are coprime in $\mathbb{Z}[\sqrt{3}]$. Indeed, let $\alpha = v+\sqrt{3}w$ divide both factors. Then, $\alpha$ also divides their difference, which is $8\sqrt{3}$. Without loss of generality assume that $\alpha$ is irreducible. Then, $\alpha$ divides 2 or is (up to a unit) equal to $\sqrt{3}$. Since $\alpha$ also divides $x^3\in \mathbb{Z}$, we deduce that $x^3$ is divisible by either 2 or 3 (or both). This implies that $y$ must also be divisible by either 2 or 3. But then, $x^3$ is also divisible by either 8 or 27 (or both), since it is a cube, and $y^4$ is divisible by 16 or 81. Now, the latter is impossible, since $y^4 - 48 \equiv -3\neq 0\pmod{9}$. The former is also impossible, since it would imply that $x^3=y^4 - 48$ is divisible by 16, so $x$ is divisible by 4, so $x^3$ is divisible by 64 and you would obtain a contradiction modulo 64 as above.
This implies that $y^2+4\sqrt{3}$ is a cube times a unit in $\mathbb{Z}[\sqrt{3}]$, $y^2+4\sqrt{3}=u\cdot(a+\sqrt{3}b)^3$. The units in this ring form an abelian group with torsion $\pm 1$ and with free part of rank one, generated by $2+\sqrt{3}$. The torsion are all cubes, so we can incorporate them into the bracket. Of course, cubes of a non-torsion unit can also be incorporated into the bracket, so it suffices to show that $y^2+4\sqrt{3}\neq (2+\sqrt{3})^{\delta}\cdot(a+\sqrt{3}b)^3$ for $\delta\in{0,1,2}$. This is done through case by case analysis, Adrián has already done the $\delta=0$ case.
| {
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Help to find cubics with square discriminant If the discriminant $b^2-4c$ of the quadratic $x^2 + bx + c$ is a square then it factors. For every discriminant $d^2$ we have can parametrize them all $(b,c) = (d + 2 h,h(d+h))$. edit I realized now that the quadratic case is trivial because it the discriminant is a square iff it factors, so the two variable parametrization is $(x-a)(x-b)$ so it might not represent what is happening with the cubic.
I was hoping to do a similar parametrization for the cubics $x^3 - ax + b$ with square discriminant $4a^3 - 27b^2 = d^2$ but factoring in the Eisenstein integers does not seem to make the problem any easier.
Are there any other promising approaches I could try?
I noticed the problem is simple when $d=0$. In that case we have $a = 3 m^2$, $b = 2 m^3$. Also when $d = b$ we also get a simple parametrization, $a = 7m^2$, $b = 7m^3$ but I don't think these will help to get the general case.
| I would not focus on parametrization as such. Instead, note that $4 a^3 = d^2 + 27 b^2.$ This gives an easily stated restriction on the prime factorization of $a.$ It is necessary and sufficient that $a \geq 0$ and, whenever any prime $p | a$ and $p \equiv 2 \pmod 3,$ then the exponent of $p$ must be even. So $a = 2$ or $a = 5$ or $a=10$ are impossible. Without the cube, there would be a restriction on the prime 3 as well, but it turns out not to matter because you have $a^3.$ Without the $4$ in $4a^3,$ there would be a competition between $d^2 + 27 b^2$ and the other forms in the genus, $4 u^2 \pm 2 u v + 7 v^2.$ But
$$
\left( \begin{array}{cc}
4 & 0 \\
1 & 1
\end{array}
\right)
\; \cdot \;
\left( \begin{array}{cc}
1 & 0 \\
0 & 27
\end{array}
\right)
\; \cdot \;
\left( \begin{array}{cc}
4 & 1 \\
0 & 1
\end{array}
\right) = \; \;\; \;
\left( \begin{array}{cc}
16 & 4 \\
4 & 28
\end{array}
\right)
$$
Now, try $a = 4 = 2^2,$ so the exponent on 2 is even, you get $4a^3 = 256,$ and you get $d = 16, b = 0,$ which seems too easy. i will put some more below, maybe skip squares... But, given a legal $a$ as described, the number of pairs $d,b$ is finite.
$$a = 1,4a^3 = 4, ( d = \pm 2, b = 0), $$
$$a = 3, 4a^3 = 108, ( d = \pm 9, b = \pm 1), ( d = 0, b = \pm 2), $$
$$a = 7, 4a^3 = 1372, ( d = \pm 20, b = \pm 6), ( d = \pm 7, b = \pm 7), $$
$$a = 12, 4a^3 = 6912, ( d = \pm 72, b = \pm 8), ( d = 0, b = \pm 16), $$
$$a = 13, 4a^3 = 8788, ( d = \pm 70, b = \pm 12), ( d = \pm 65, b = \pm 13). $$
Well, given all possible representations $a = s^2 + s t + t^2,$ one may construct all $d,b,$ an annoying task unless $a$ is prime with $a \equiv 1 \pmod 3.$ This is, essentially, what factoring in the Eisenstein integers will give you.
Actually, that last bit was needlessly pessimistic. With a little special treatment of the primes 2,3, I can see how to create all possible representations of $a = j^2 + 3 k^2,$ create all possible representations of $4a^3 = m^2 + 3 n^2,$ then just keep the ones when $3 | n.$ Rather involved but easy enough to program.
| {
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How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$ I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way.
I am interested if there is a way only using elementary methods to prove it.
Using the software Mathematica confirmed this inequality is correct.
| With $abc=1$, we can set :$a=\frac{y}{x}; b=\frac{z}{y}; c=\frac{x}{z}$. We have:
$\frac{x}{z}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\le 3+\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$
$\Leftrightarrow 3xyz+x^3+y^3+z^3\ge xy(x+y)+yz(y+z)+zx(z+x)$.
It's true by Schur's inequality. Equality occurs if only if $a=b=c=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
| To say that $3n^3+6n$ is divisible by $9$ is equivalent to saying that $n^3+2n=n(n^2+2)$ is divisible by $3$. Now split to three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$ (explain why this is true.) Now you have to check just 3 cases.
| {
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Proof of inequality I have problems with proving inequality :
$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$
where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.
Thanks.
| We need to prove that
$$\frac{(a+b+c)(a^2+b^2+c^2)}{10}+\frac{2}{5}abc<\frac{(a+b+c)^3}{20}$$ or
$$(a+b+c)^3>8abc+2(a+b+c)(a^2+b^2+c^2)$$ or
$$\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)>8abc+2\sum_{cyc}(a^3+a^2b+a^2c)$$ or
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0$$
Done!
| {
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Linear Algebra: solving for x I was wondering why in this exercise there isn't a minus $(x^2-2x-3)$. I thought that for this type of exercise you must use $(ad-bc)$.
Problem 50 (p 131) Solve for $x$.
$\begin{vmatrix} x-2& -1\\ -3 & x \end{vmatrix}=0$
$(x-2)\cdot x + (-3)\cdot(-1)=0$
$x^2-2x+3=0$
$x-3=0$ or $x+1=0$
$x=3$ or $x=-1$
| This is a mistake. To calculate the determinant of a $2\times 2$ matrix you use $ad-bc$ which gives you $(x-2)x-(-1)(-3)=x^2-2x-3=(x+1)(x-3)=0$. Hence $x=3$ or $x=-1$.
| {
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Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
| $$
\begin{aligned}
\lim _{x\to \infty }\left(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5}\right)
& = \lim _{t\to 0}\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)
\\& = \lim _{t\to 0}\left(\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)\cdot \frac{\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}+\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)}{\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}+\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}}\right)
\\& = \lim _{t\to 0}\left(\frac{\sqrt[3]{t+1}-\sqrt[3]{-t+1}}{t\left(\sqrt[6]{-t+1}+\sqrt[6]{t+1}\right)}\right)
\\& = \lim _{t\to 0}\left(\frac{\left(1+\frac{1}{3}t+o\left(t\right)\right)-\left(1-\frac{1}{3}t+o\left(t\right)\right)}{t\left(2+o\left(1\right)\right)}\right)
\\& = \color{red}{\frac{1}{3}}
\end{aligned}
$$
Solved with substitution $t=\frac{1}{x}$ and Taylor expansion
| {
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How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms? So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were doing problems, we came about one that I have no idea how to solve. It seems to be a straight forward "here's an equation, solve for $x$." Just one problem, I ended up not being able to solve for $x$, making me feel embarrassed since I was trying to help her and I am supposed to be the one who is "good at math." Anyway, it would be a grreat help if anyone can help me out on this. The equation is....
$$4\sqrt{x-3} - \sqrt{6x-17} = 3$$
The answer comes out to be $x = 7$ because I did it on my calculator. I first tried to just square each side to get rid of the square roots. But once I "FOILed" the left side, there were still square roots and things didn't look good for me. By the way, I am sorry I don't know how to format this equation correctly.
| There is a trick for problems like this that I really love to do.
\begin{align}
4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(multiply)} \\
\text{Let } 4\sqrt{x-3} + \sqrt{6x-17} &= y \\
\hline
16(x-3)-(6x-17) &= 3y \\
16x-48 - 6x + 17 &= 3y \\
y &= \dfrac{10x -31}{3} \\
\hline
4\sqrt{x-3} + \sqrt{6x-17} &= \dfrac{10x -31}{3} &\text{(so now we have this)} \\
4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(subtract them)}\\
\hline
2\sqrt{6x-17} &= \dfrac{10x -40}{3} \\
\sqrt{6x-17} &= \dfrac{5x - 20}{3} \\
3\sqrt{6x-17} &= 5x - 20 \\
54x - 153 &= 25x^2 - 200x + 400 \\
25x^2 - 254x + 553 &= 0 \\
(x-7)(25x - 79) &= 0 \\
x &\in \left\{ 7, \dfrac{79}{25}\right\}
\end{align}
Check for extraneous roots.
When $x = 7$
$$ (4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot 2 - 5 = 3$$
When $x = \dfrac{79}{25}$
$$ (4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot \dfrac 25 - \dfrac 75
= \color{red}{\frac 15}$$
So we reject $x = \dfrac{79}{25}$ as an extraneous root and accept $x=7$,
| {
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Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$
| Firstly, what you have is a limit of a finite sum, and the limit is a series:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}:=\sum_{k=1}^\infty \frac{k+1}{3^k}$$
Now, to the question you asked: there are a few tricks you can use:
1) You know that $\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}-1$ (geometric progression).
2) That implies, by deriving both sides, that $\sum_{k=1}^nkx^{k-1}=\frac{-(n+1)x^{n}(1-x)+(1-x^{n+1})}{(1-x)^2}=\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}$
3) From here you get that $\sum_{k=1}^n(k+1)x^k=\frac{1-x^{n+1}}{1-x}-1+x\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}-x^2+2x}{(1-x)^2}$.
4) Now plug in $x=\frac{1}{3}$. As $n\to\infty$, you have $x^n\to 0$. This implies that $$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}=\frac{-\left(\frac{1}{3}\right)^2+2\cdot\frac{1}{3}}{\left(\frac{2}{3}\right)^2}=\frac{5}{4}$$
| {
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Please help me to find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^2(x-1)}$ around $a=2$ First, sorry if my translations is bad. I need help for this exercise, more precisely , I need to know if my result which I've found is good.
The exercise: Find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^2(x-1)}$ around point $2$. $a=2$ in Taylor series formula: $f(x)=\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$
I tried this: $f(x)=-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x-1}$ then I found $n$-th derivative of each function at point $a=2$:
$f_1(x)=-\frac{1}{x}; f_1^{(n)}(2)=(-1)^{n+1}\frac{n!}{2^{n+1}}$
$f_2(x)=-\frac{1}{x^2}; f_2^{(n)}(2)=(-1)^{n-1}\frac{(n+1)!}{2^{n+2}}$
$f_3(x)=\frac{1}{x-1}; f_3^{(n)}(2)=(-1)^n n!$
Then I wrote this:
$$\begin{align}
f(x)&=\sum\limits_{n=0}^\infty(-1)^{n+1}\frac{(x-2)^n}{2^{n+1}}+\sum\limits_{n=0}^\infty(-1)^{n-1}\frac{(n+1)(x-2)^n}{2^{n+2}}+\sum\limits_{n=0}^\infty(-1)^n(x-2)^n\\
&=\sum\limits_{n=0}^\infty(-1)^{n+1}(x-2)^n\frac{2^{n+2}+n+3}{2^{n+2}}\end{align}$$
I hope someone could help me by telling me if result wich I've found is correct or not.
| The general procedure is correct. And the details look pretty good too. I could now check the details, it would not be difficult.
Instead, I will do the problem in a somewhat different way, which in general is more efficient. It will turn out that a sign error crept into your calculation.
We have
$$f(x)=-\frac{1}{x}-\frac{1}{x^2} + \frac{1}{x-1}$$
and want to express $f(x)$ as a sum of powers of $x-2$. It is useful, though not necessary, to let $y=x-2$. Then $x=y+2$. Substituting for $x$, we obtain
$$f(y+2)=g(y)= -\frac{1}{y+2}-\frac{1}{(y+2)^2} + \frac{1}{y+1}.$$
We want to express $g(y)$ as a sum of powers of $y$. Let's start with the easiest part, $\frac{1}{1+y}$.
It would be a shame to do a whole lot of differentiating when we already know the power series expansion of $1/(1+y)$. Or at least we certainly know the power series expansion of $1/(1-z)$, and then we can put $y=-z$. Thus
$$\frac{1}{1+y}=1-y+y^2-y^3+\cdots=\sum_0^\infty (-1)^ny^n\qquad\qquad\text{(Term $1$)}$$
That was easy. Let's go on to the next easiest term, $1/(y+2)$ (I know there should be a minus sign in front, will take care of it later). We have
$$\frac{1}{2+y}=\frac{1/2}{1+y/2}.$$
It would be a shame not to use the fact that we know the power series expansion of $1/(1-z)$. We get
$$\frac{1}{2+y}=\frac{1}{2}\sum_0^\infty (-1)^n\frac{1}{2^n}y^n=\sum_0^\infty \frac{(-1)^n}{2^{n+1}}y^n,$$
and therefore
$$-\frac{1}{2+y}=\sum_0^\infty \frac{(-1)^{n+1}}{2^{n+1}}y^n.\qquad\qquad\text{(Term $2$)}$$
Finally, we want the expansion of $1/(2+y)^2$ in powers of $y$. We have just obtained the expansion of $1/(2+y)$. Note that $1/(2+y)^2$ is (almost) the derivative of $1/(2+y)$. To be precise, it is the negative of the derivative of $1/(2+y)$. So let us differentiate the series we obtained for $1/(2+y)$ term by term. We find
$$-\frac{1}{(y+2)^2}=\sum_0^\infty \frac{(-1)^{n} n}{2^{n+1}}y^{n-1}=\sum_0^\infty \frac{(-1)^{n+1} (n+1)}{2^{n+2}}y^{n} \qquad\qquad\text{(Term $3$)}$$
Now it is just a matter of adding Terms $1$, $2$, and $3$ together. I get, replacing $y$ by $x-2$, the following:
$$\sum_0^\infty (-1)^n\left(1-\frac{n+3}{2^{n+2}}\right)(x-2)^n.$$
Comment: When we are calculating power series expansions, it is nice to avoid all those differentiations, by recycling standard expansions.
| {
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differential equations A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (half life $1.28 \times 10^9$) and that one of every 9 potassium atom disintegrations yields an argon atom. What is the age of the rock measured from the time it contained only potassium.
please suggest
| After Rahul's comment above, I understand the question now.
After $t$ years, exactly $\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$ of the potassium will remain, and exactly $\frac{1}{9}$ of the decays will have produced an argon atom. Thus, we have that
$$\text{argon}=\frac{1}{9}\cdot\text{ decayed potassium}$$
and
$$\text{decayed potassium}=\text{original potassium} - \text{remaining potassium}$$
so
$$\text{decayed potassium}=\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium},$$
so
$$\text{argon}=\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium}$$
Now we set this equal to the remaining amount of potassium, which is
$$\text{remaining potassium }=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\cdot\text{ original potassium}$$
This gives the equation
$$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$
in which we want to solve for $t$. This is just simple manipulations now.
We have
$$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$
which becomes
$$\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}=\frac{1}{10}$$
so $$\frac{t}{1.28\times 10^9}\times\ln\left(\frac{1}{2}\right)=\ln\left(\frac{1}{10}\right)$$
so
$$t=(1.28\times10^9)\frac{\ln\left(\frac{1}{10}\right)}{\ln\left(\frac{1}{2}\right)}=4.252\ldots\times 10^9$$
(see here)
| {
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Trouble setting up double integrals I am studying for an entrance exam and have troubles to set up double integrals appropriately. Actually my biggest problem is that I don't get the notation.
$$ A = \{(x,y) | 0 \leq x + y \leq 1, 0 \leq x-y \leq \pi\} $$
$$ \iint_A e^{x+y} \sin(x-y)\mathrm dx\mathrm dy $$
and for another example:
$$ R = \{(x,y) | 1 \leq x^2 + y^2 \leq 4, y \geq 0\} $$
$$ \iint_R \frac{\mathrm dx\mathrm dy}{(x^2+y^2)^2} $$
Normally I would go draw a picture of something like a line or circle.
When I have an equation like $x+y=z$ for example.
I tried interpreting the inequality as an area so for the first one I came up with the between 0 and the line $y=1-x$ which seemed reasonable for I failed to connect this to the other inequality.
I hope somebody can point out how to set this up or give some hints how to interpret this notation.
Any help is greatly appreciated!
| For the first example. The double inequality $0\leq x+y\leq 1$ means that
$$
\left\{
\begin{array}{c}
0\leq x+y \\
x+y\leq 1
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
y\geq -x \\
y\leq 1-x
\end{array}
\right.
$$
and $0\leq x-y\leq \pi $ means that
$$
\left\{
\begin{array}{c}
0\leq x-y \\
x-y\leq \pi
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
y\leq x \\
y\geq x-\pi.
\end{array}
\right.
$$
So the conditions $0\leq x+y\leq 1$ and $0\leq x-y\leq \pi $ are equivalent
to the system of four inequalities
$$
\left\{
\begin{array}{c}
y\geq -x \\
y\leq 1-x \\
y\leq x \\
y\geq x-\pi.
\end{array}\tag{1}
\right.
$$
The region $A$ is a rectangle limited by the four lines $y=-x$, $y=1-x$, $y=x$, $y=x-\pi $ (see figure).
To evaluate
$$
I:=\iint_{A}e^{x+y}\sin (x-y)\;\mathrm{d}x\mathrm{d}\tag{2}y
$$
we may consider the rotated system of coordinates $x',y'$ with respect to the $x,y$ system, the rotation angle being $\theta =-\pi /4$, as shown in the figure. This corresponds to the following transformation of coordinates
$$
\begin{eqnarray*}
x^{\prime } &=&x\cos \left( -\frac{\pi }{4}\right) +y\sin \left( -\frac{\pi
}{4}\right) =\frac{1}{2}\sqrt{2}x-\frac{1}{2}\sqrt{2}y \\
y^{\prime } &=&-x\sin \left( -\frac{\pi }{4}\right) +y\cos \left( -\frac{\pi
}{4}\right) =\frac{1}{2}\sqrt{2}x+\frac{1}{2}\sqrt{2}y,
\end{eqnarray*}
$$
whose inverse is
$$
\begin{eqnarray*}
x &=&x^{\prime }\cos \left( -\frac{\pi }{4}\right) -y^{\prime }\sin \left( -
\frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}%
y^{\prime } \\
y &=&x^{\prime }\sin \left( -\frac{\pi }{4}\right) +y^{\prime }\cos \left( -
\frac{\pi }{4}\right) =-\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}%
y^{\prime }.
\end{eqnarray*}
$$
Since $\frac{\partial (x,y)}{\partial (x^{\prime },y^{\prime })}=1$, the
integral $I$ is transformed into
$$
\begin{eqnarray*}
I &=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\left( \int_{x^{\prime }=0}^{\pi \sqrt{
2}/2}e^{\sqrt{2}y^{\prime }}\sin (\sqrt{2}x^{\prime })\mathrm{d}x^{\prime
}\right) \mathrm{d}y^{\prime } \\
&=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\sqrt{2}e^{y^{\prime }\sqrt{2}}\mathrm{d}
y^{\prime } \\
&=&e-1,\tag{3}
\end{eqnarray*}
$$
because
$$
\begin{eqnarray*}
x-y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\left( -
\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }\right) =\sqrt{2
}x^{\prime } \\
x+y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\frac{1
}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }=\sqrt{2}y^{\prime }.
\end{eqnarray*}
$$
Alternatively we could split $A$ into three regions, a triangle ($0\le x\le 1/2$), a quadrilateral ($1/2\le x\le π/2$) and a triangle ($\pi/2\le x\le (1+\pi)/2$), and evaluate $I$ in the original variables $x,y$:
$$
\begin{eqnarray*}
I &=&\int_{0}^{1/2}\left( \int_{-x}^{x}e^{x+y}\sin (x-y)\mathrm{d}y\right)
\mathrm{d}x \\
&&+\int_{1/2}^{\pi /2}\left( \int_{-x}^{1-x}e^{x+y}\sin (x-y)\mathrm{d}
y\right) \mathrm{d}x \\
&&+\int_{\pi /2}^{(1+\pi )/2}\left( \int_{x-\pi }^{1-x}e^{x+y}\sin (x-y)
\mathrm{d}y\right) \mathrm{d}x.
\end{eqnarray*}
$$
As for the second example $R$ is the semi-annulus centered at $(0,0)$ with outer radius equal to 2, inner radius 1 and $y\ge 0$. The Jacobian of the transformation of Cartesian to polar coordinates is $\frac{\partial \left( x,y\right) }{\partial \left(
r,\theta \right) }=\sqrt{x^{2}+y^{2}}=r$. Hence
$$
\begin{eqnarray*}
\iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\left( x^{2}+y^{2}\right) ^{2}}
&=&\int_{r=1}^{2}\int_{\theta =0}^{\pi }\frac{1}{r^{4}}r\;\mathrm{d}r\mathrm{
d}\theta \\
&=&\int_{0}^{\pi }\left( \int_{1}^{2}\frac{1}{r^{3}}\mathrm{d}r\right) \mathrm{d}
\theta \\
&=&\int_{0}^{\pi }\frac{3}{8}\mathrm{d}\theta \\
&=&\frac{3}{8}\pi.
\end{eqnarray*}
$$
| {
"language": "en",
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} |
partial fraction question
$ \frac{125x^{2}+x+3}{x^{2}(x-5)} =
> \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$
$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$
$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$
$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$
$3 = -5B \Rightarrow B = \frac{-3}{5}$
$-1 = A+B \Rightarrow A = -1 - B \Rightarrow A = \frac{-5}{5} - \frac{-3}{5} \Rightarrow A=\frac{-8}{5}$
$125 = A+C$
Where I did wrong in calculating of variable $A$, because correct answer is $A = \frac{-8}{25}$, but I get $A = \frac{-8}{5}$.
| HINT $\ $ It's simpler to use the Heaviside cover-up method. First, evaluating your $\rm\:E_2 = 2$nd equation at $\rm\:x = 0\:$ yields $\rm\:3 = -5\:b\:.\:$ Next, differentiating $\rm\:E_2\:$ and evaluating at $\rm\:x = 0\:$ yields $\rm\: 1 = b - 5\:a\:.$ Solve those for $\rm\:a,b\:$. Finally evaluating $\rm\:E_2\:$ at $\rm\:x=5\:$ yields $\rm\: 3133 = 25\:c\:.$
| {
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How to know what is the degree of the remainder The example question is
Find the remainder when $8x^4+3x-1$ is divided by $2x^2+1$
The answer did something like
$$8x^4+3x-1=(2x^2+1)(Ax^2+Bx+C)+(Dx+E)$$
Where $(Ax^2+Bx+C)$ is the Quotient and $(Dx+E)$ the remainder. I believe the degree of Quotient is derived from degree of $8x^4+3x-1$ - degree of divisor. But for remainder? Would it not be
| EDIT to add these short answers.
I believe the degree of Quotient is derived from degree of $8x^4+3x-1$
- degree of divisor.
That's right.
But for remainder?
The degree of the remainder is less than the degree of the divisor, by definition of polynomial division.
*
*Let me start with this specific case. To find the remainder ($Dx+E$) we can expand the RHS of the identity shown in the question $$8x^4+3x-1=(2x^2+1)(Ax^2+Bx+C)+(Dx+E)\tag{1}$$ and collect the terms of the same degree. We get $$8x^4+3x-1=2Ax^4+2Bx^3+(A+2C)x^2+(B+D)x+C+E.\tag{2}$$ The polynomial of the LHS is equivalent to to polynomial of the RHS if and only if the coefficients of the terms of the same degree are equal. Therefore we must have the following system of $5$ equations $$2A=8,\ 2B=0, \ A+2C=0, \ B+D=3, \ C+E=-1, $$ whose solution is $$A=4, \ B=0, \ C=-2, \ D=3, \ E=1. \ $$ Hence we obtain $$8x^4+3x-1=(2x^2+1)(4x^2-2)+(3x+1),\tag{3}$$
where $3x+1$ is the remainder. The degree of $8x^4+3x+1$ is $4$ and the degree of $2x^2+1$ is $2$. The degree of the quotient $(4x^2-2)$ is $2=4-2$. The degree of the remainder is $1<2$, which means that it is less than the degree of the divisor $2x^2+1$. Since $2x^2+1\ne 0$, the algebraic identity $(3)$ is equivalent to $$\frac{8x^4+3x-1}{2x^2+1}=4x^2-2+\frac{3x+1}{2x^2+1}.\tag{4}$$
*Now the general case. By definition of polynomial division, given polynomials $A(x),B(x)$, where the degree of $B(x)$ is greater than $0$, it is always possible to find a polynomial $Q(x)$, called quotient, such that the diference $$R(x)=A(x)-B(x)Q(x)\tag{5}$$ is a polynomial whose degree is less than the degree of $B(x)$. This polynomial $R(x)$, called remainder, is unique. The polynomial $A(x)$ is called the dividend and $B(x)$ the divisor. Let $m$ be the degree of $A(x)$, $n$ the degree of $B(x)$ and $q$ the degree of $Q(x)$. If $m<n$, $Q(x)=0$ and $R(x)=A(x)$. If $m\ge n$, then $q=m-n$. (Note: if $n=0$, then $R(x)=0$.) The identity $(5)$ is equivalent to $$A(x)=B(x)Q(x)+R(x)\tag{6}$$ and for $B(x)\ne 0$ to $$\frac{A(x)}{B(x)}=Q(x)+\frac{R(x)}{B(x)}.\tag{7}$$
*Concerning the computation of the quotient and the remainder, in addition to the method detailed above, we can use the polynomial long division or the synthetic division. The long division technique applied to the present case, results in $$\begin{matrix}
4x^2 - 2\\
\qquad\qquad\qquad 2x^2+1\ \overline{ )\ 8x^4 \; +0x^3 \; +0x^2 \; + 3x - 1 }\\
\qquad\qquad\qquad \underline{ 8x^4 \; +0x^3 \;+ \;\;4x^2}\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; -4x^2\; + 3x - 1 \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; \underline{4x^2\; + 0x - 2}\\
\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad\qquad 3x + 1\\
\end{matrix}\tag{8}$$
| {
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given $a^2+b^2=28ab$ what's $\log_{3}\left (\frac{(a+b)^2}{ab}\right)$? given $a^2+b^2=28ab$ what's $\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$?
$\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$
$\log_{3} \left(\dfrac{a^2+b^2+2ab}{ab}\right)$
$\log_{3} \left(\dfrac{a^2+b^2}{ab}+\dfrac{2ab}{ab}\right)$
$\log_{3} \left(\dfrac{28ab}{ab}+\dfrac{2ab}{ab}\right)$
$\log_{3} 30$
Here I tried using properties but couldn't manage to get trough.
---edit----
$\log_{3} 3 + \log_{3} 10 = 1 + \log_{3} 10$
$\log_{10} 3 = \dfrac{25}{12} = \dfrac{\log_{3} 3}{\log_{3} 10} = \dfrac{1}{\log_{3}10}$
$\dfrac{12}{25} = \dfrac{1}{\log_{3}10} \implies \log_{3}10 = \dfrac{25}{12}$
$\log_{3} 30 = 1 + \log_{3} 10 = 1 + \dfrac{25}{12}=\dfrac{12+25}{12}=\dfrac{37}{12}$
| The unwritten core of the problem is that it is possible to determine
$C = (a+b)^2/ab$ (and hence also its logarithm to the base 3) given the first equation on $a$ and $b$. This would not be true if the condition had been modified to
$a^2 + b^2 = 28ab + 5, \quad$ or
$a^3 + b^3 = 28ab, \quad$ or
$\sqrt{a^2+b^2}=28ab$.
Here the condition can be expressed as $C=30$. But it is not true in general that given some condition on $a$ and $b$, every other function of $a$ and $b$ can be calculated.
| {
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Find the limit of a function $$\lim_{x\to 3}\frac{\sqrt{3x} - 3}{\sqrt{2x-4} - \sqrt{2}}.$$
Letting
$$F(x) = \frac{\sqrt{3x} - 3}{\sqrt{2x-4}-\sqrt{2}},$$
we have
$$F(x) = \frac{\sqrt{3}(\sqrt{x} - \sqrt{3})}{\sqrt{2}(\sqrt{x-2}-1)}.$$
Multiplying numerator and denominator by $\sqrt{x-2} + 1$,
$$F(x) = \frac{
(3)^{1/2} ((x(x-2)^{1/2})+(x)^{1/2}-(3(x-2)^{1/2})-(3)^{1/2})}
{\sqrt{2}(x-3)}.$$
Dividing numerator and denominator by $x$ and substituting $3$ for $x$, I get $\frac{0}{\sqrt{2}} = 0$.
Is it correct? My textbook does not have answer, one of the site gives the answer as $\frac{1}{\sqrt{2}}.$.
| Hint: $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$.
Applying the hint twice,
$$
F(x)=\sqrt{\frac32}\frac{\sqrt{x}-\sqrt3}{\sqrt{x-2}-1}=\sqrt{\frac32}\frac{x-3}{\sqrt{x}+\sqrt3}\,\frac{\sqrt{x-2}+1}{(x-2)-1}=\sqrt{\frac32}\frac{\sqrt{x-2}+1}{\sqrt{x}+\sqrt3},
$$
hence
$$
\lim\limits_{x\to3}\,F(x)=\sqrt{\frac32}\frac2{2\sqrt3}=\frac1{\sqrt{2}}.
$$
| {
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Complete the square of a 4th degree expression $k^4 + 2k^3 -k^2 - 2k + 1$ I was reading the solution of a problem and I can't figure out how to go from here:
$k^4 + 2k^3 -k^2 - 2k + 1$
to:
$((k^2 + k) -1)^2$
I guess it's using the complete-the-square method but I've only used that with quadratic expressions.
| $k^4+2k^3−k^2−2k + 1 = k^4 + (k^2 - 2k^2) + 2k^3 - 2k + 1 \quad \Rightarrow$
$k^4+2k^3−k^2−2k + 1 = (k^4 - 2k^2 + 1) + 2k(k^2 - 1) + k^2$
$= (k^2 - 1)^2 + 2(k^2 - 1)k + k^2 = (k^2 - 1 + k)^2 = ((k^2 + k) - 1)^2$
| {
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Beta Function -- finding a lower bound based on parameters I would like to show that
$$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$
for all $c \geq 2$.
I have plotted it out for $c$ up through 200, and it seems to hold.
Does anyone have any tips on if this can be formally shown? I imagine an expert on the beta function might know...
| I was composing the following when the original answer by @MikeSpivey appeared; his edited answer provides considerably more information. The short calculation below arrives at the desired result without requiring integration by parts but does need knowledge of the relationship between the Beta and Gamma functions.
For $c \geq 2$
$$\begin{align*}
\text{Beta}(c+1, 1/c) &= \frac{\Gamma(c+1)\Gamma(1/c)}{\Gamma(c + 1 + 1/c)}\\
&= \frac{c}{c+1/c}\times \frac{c-1}{c-1+1/c}\times \cdots
\times \frac{2}{2+1/c}\times \frac{1}{1+1/c} \times \frac{1}{(1/c)\Gamma(1/c)}\times \Gamma(1/c).
\end{align*}$$
The product of the last three terms on the right is $c^2/(c+1)$ while all the other terms have value less than $1$. Hence,
$$\text{Beta}(c+1, 1/c) < \frac{c^2}{c+1} \Rightarrow \frac{1}{c}\text{Beta}(c+1, 1/c) < \frac{c}{c+1} = 1 - \frac{1}{c+1}$$
| {
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$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
\geq \dots \geq
=2 \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.
What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.
| Mine is a tad less elegant but arguably a bit clearer and assumes only the complete basics.
0. The task
Prove that for $ \forall n \in \mathbb{N}$ it is true that $n! \leq (\frac{n+1}{2})^n $ :
I. Base steps
(I need four base steps because my final inequality works for $k \ge 4$.)
$$n=0: 1 \leq (\frac{1}{2})^0 = 1 $$
$$n=1: 1 \leq (\frac{2}{2})^1 = 1 $$
$$n=2: 2 \leq (\frac{3}{2})^2 = \frac{9}{4} $$
$$n=3: 6 \leq (\frac{4}{2})^3 = 8 $$
II. Inductive assumption
$$k! \leq (\frac{k+1}{2})^k$$
III. Inductive hypothesis
$$k! \leq (\frac{k+1}{2})^k \implies (k+1)! \leq (\frac{k+2}{2})^{k+1}$$
IV. Proof
I am going from the assumption to the hypothesis. We will be using the fact that $(a \leq c) \land (b \geq c) \implies a \leq b$.
Step 1.
\begin{align}
k! \leq (\frac{k+1}{2})^k && \text{multiply both sides by (k+1)} \tag 1\\
(k+1) k! \leq (k+1) (\frac{k+1}{2})^k && \tag 2\\
(k+1)! \leq (k+1) (\frac{k+1}{2})^k && \text{from the definition of the factorial} \tag 3\\
(k+1)! \leq 2(\frac{k+1}{2}) (\frac{k+1}{2})^k && \text{factoring 2 out of (k+1)} \tag 4\\
(k+1)! \leq 2(\frac{k+1}{2})^{k+1} && \tag 5\\
\end{align}
We have shown that (5) is true. Now, we need to show that $2(\frac{k+1}{2})^{k+1} \leq (\frac{k+2}{2})^{k+1}$, so we prove the initial inequality.
Step 2.
\begin{align}
2(\frac{k+1}{2})^{k+1} \leq (\frac{k+2}{2})^{k+1} && \tag 6\\
\end{align}
$$\begin{equation}\begin{aligned}
(\frac{k+2}{2})^{k+1} &= (\frac{k+2}{2}) (\frac{k+2}{2})^k \\
&= (\frac{k+2}{k+1})^k (\frac{k+1}{2})^k (\frac{k+2}{2}) \\
\end{aligned}\end{equation}\tag{7}$$
\begin{align}
(\frac{k+1}{2})^k (\frac{k+2}{2}) \geq (\frac{k+1}{2})^{k+1} && \tag 8\\
\end{align}
Using the trick already mentioned above by Andre Nicolas we managed to show that the last two terms of equation (7) are greater than the main part of the left hand side of the inequality (6). That leaves us with having to prove that $2 \leq (\frac{k+2}{k+1})^k$.
Step 3.
The last step is to notice that $(\frac{k+2}{k+1})$ is in fact equal to one plus a tiny difference and that the increment would get smaller and smaller with greater values of k. So probably it can be represented as $(1 + \frac{1}{k})^k$. And indeed, $ \lim (\frac{k+2}{k+1}) = e$ so for greater $k$s the value of the expression stays close to ~2.76 which is, in particular, more than 2.
This is where we need our four base steps because $2 \leq (\frac{k+2}{k+1})^k$ holds only for $k \in \mathbb{N}-\{0,1,2,3\} $(at least for $k \geq 0)$. The four test cases get us covered here and allow the induction to work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When do addition and inversion of matrices commute? Can you guys help me out with the following problem:
Problem: Find conditions that the matrices $A$ and $B$ have to satisfy in order for the following to be valid: $(A+B)^{-1} = A^{-1} + B^{-1}$.
My solution. The first condition is that $A$ and $B$ have to have the same dimensions. Then by expanding the two equalities: $(A+B) \cdot (A^{-1} + B^{-1})=I $ and $(A^{-1} + B^{-1}) \cdot (A+B)=I$, I deduced the second condition: $A$ and $B$ have both to be invertible, and third condition is: $A^{-1} \cdot B=B \cdot A^{-1}$. Is my answer correct?
| For third condition you have that:
$(A+B)(A+B)^{-1}=(A+B)(A^{-1}+B^{-1})$
$(A+B)^{-1}(A+B)=(A^{-1}+B^{-1})(A+B)$
$\Rightarrow (A+B)(A^{-1}+B^{-1})=(A^{-1}+B^{-1})(A+B)\Rightarrow $
$\Rightarrow AA^{-1}+AB^{-1}+BA^{-1}+BB^{-1}=A^{-1}A+A^{-1}B+B^{-1}A+B^{-1}B \Rightarrow$
$\Rightarrow AB^{-1}+BA^{-1} =A^{-1}B+B^{-1}A$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Can you explain this please $T(n) = (n-1)+(n-2)+…1= \frac{(n-1)n}{2}$
Possible Duplicate:
Proof for formula for sum of sequence 1+2+3+…+n?
Can you explain this please
$$T(n) = (n-1)+(n-2)+…1= \frac{(n-1)n}{2}$$
I am really bad at maths but need to understand this for software engineering.
| For example, if we wish to add the numbers
$$
1+2+3+4+5+6
$$
you can pair them as such
$$
(1+6)+(2+5)+(3+4)=7+7+7=21.
$$
But, here $n=7$ and there are $\frac{n-1}{2}=3$ pairs of $n=7$. Thus we have that the sum is $\frac{n-1}{2}\cdot n=3\cdot 7=21$. In general, you can pair the sum
$$
1+2+3+\cdots +(n-2)+(n-1)
$$
as pairs $(n-1+1), (n-2+2), (n-3+3), \ldots$ and so forth. This will give you $\frac{n-1}{2}$ pairs of $n$ if $n-1$ is even. If $n-1$ is odd, leave off $n-1$ and do it for $1+\cdots +(n-2)$ and add on $n-1$ to obtain the same formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of geometric mean of N radii of an ellipse Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?
| The answer has been substantially revised. In particular, the previous revision quoted a wrong value for the integral $I(a,b)$, which resulted in an overall incorrect answer. I hope that this revision is correct. :-)
The limit is $\Big(\frac{a+b}{2} \Big)^2$ and is not $ab$ (as claimed). I break down the solution into multiple steps for ease of understanding.
(1) A Riemann sum and integral. We first convert the product to a sum by taking logs:
$$
\frac{1}{N} \sum_{n=1}^N \ln \left( a^2 \cos^2 (2\pi n/N) + b^2 \sin^2 (2\pi n/N) \right).
$$
This is the Riemann sum of the function $$f: [0,1] \to \mathbb R : x \mapsto \ln \left( a^2 \cos^2 (2\pi x) + b^2 \sin^2 (2\pi x) \right) ,$$
corresponding to the uniform partition of $[0,1]$ into $N$ parts. Since $f$ is integrable (being a continuous function over a compact interval), as $N \to \infty$, this sum tends to
$$
\int_0^1 \ln \left( a^2 \cos^2 (2\pi x) + b^2 \sin^2 (2\pi x) \right) \ dx =
\frac{1}{2 \pi} \int_0^{2\pi} \ln \left( a^2 \cos^2 y + b^2 \sin^2 y \right) \ dy .
$$
Call this integral $I(a,b)$.
(2) Evaluating $I(a,b)$. We use the idea of differentiating under the integral sign.
$$
\begin{align*}
\frac{\partial }{\partial a} I(a,b)
&= \frac{1}{2 \pi} \int_0^{2\pi} \frac{\partial}{\partial a} \ln \left( a^2 \cos^2 y + b^2 \sin^2 y \right) \ dy
\\ &=
\frac{1}{2 \pi} \int_0^{2\pi} \frac{2a \cos^2y}{a^2 \cos^2 y + b^2 \sin^2 y } \ dy .
\\ &=
\frac{a}{\pi} \int_0^{2\pi} g(y) \ dy ,
\end{align*}
$$
where $g(y) = \frac{\cos^2y}{a^2 \cos^2 y + b^2 \sin^2 y }$. Since $g(y) = g(\pi + y)$, we can write this integral as $\frac{2a}{\pi} \int_0^{\pi} g(y) \ dy$. Similarly, since $g(y) = g(\pi - y)$, we can further simplify it to $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy$.
(3) Evaluating $\int_0^{\pi/2} g(y)dy$. Using Wolfram|Alpha, we can find the indefinite integral
$$
\int g(y) \ dy = \int \frac{\cos^2 y}{ a^2 \cos^2 y + b^2 \sin^2 y} \ dx = \frac{ay - b \ \arctan \Bigl( \frac{b \tan y}{a} \Bigr)}{a(a^2 - b^2)} \color{\Grey}{+ \mathrm{const}}.
$$
By plugging in the limits $0$ and $\pi/2$, we get
$$
\int_0^{\pi/2} g(y) \ dy = \frac{ \left. ay - b \; \arctan \Bigl( \frac{b \tan y}{a} \Bigr) \right|_{0}^{\pi/2}}{a(a^2 - b^2)} = \frac{(a-b) \frac{\pi}{2}}{a(a^2 - b^2)} = \frac{\pi}{2a(a+b)}.
$$
Therefore, $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy = \frac{2}{a+b}$.
(4) Back to $I(a,b)$. We have $\frac{\partial}{\partial a} I(a,b) = \frac{2}{a+b}$. Similarly, $\frac{\partial}{\partial b} I(a,b) = \frac{2}{a+b}$. Combining these two observations, we conclude
$$
I(a,b) = 2 \ln (a+b) + \mathrm{const}.
$$
When $a=b=1$, the integral is $0$, which gives the constant to be $- 2\ln 2$. Therefore
$$
I(a,b) = 2\ln \Bigl( \frac{a+b}{2} \Bigr).
$$
(5) Final answer. To get the final answer, we only need to exponentiating this answer. That is, the limit mentioned in the question is equal to $\Bigl(\frac{a+b}{2} \Bigr)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Prove that $\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}$ is bounded above by $\frac{13}{20}$ I want ask a question about a sum. The exercise is as follows:
Prove the following inequality for every $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$
| $$
\begin{align}
\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} & \leq \sum\limits_{k=1}^n \frac1{k(k+3)}\\
& = \sum\limits_{k=1}^n \frac13 \left(\frac1k - \frac1{k+3} \right)\\
& = \frac13 \left( 1 + \frac12 + \frac13 - \frac1{n+1} - \frac1{n+2} - \frac1{n+3} \right)\\ & \leq \frac13 \frac{11}{6}\\
& = \frac{11}{18}
\end{align}
$$
(I noticed it just now. It is the same as Zarrax's and David Mitra's comments)
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
| Let's replace an explicit number $-3$ with a symbol $z$, so that we consider:
$$
\sum_{k=1}^n z^{k-1} \binom{2n}{2k-1} = \sum_{m=0}^{2n} \left( \frac{1-(-1)^m}{2} \right)z^{(m-1)/2} \binom{2n}{m} = \frac{\left(1+\sqrt{z}\right)^{2n} - \left(1-\sqrt{z}\right)^{2n}}{2 \sqrt{z}}
$$
Now substitute $z=-3$. Notice that $1 \pm \sqrt{-3} = 2 \exp\left( \pm i \frac{\pi}{3} \right)$, so it follows that the smallest such an integer is $n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Volume of a cut sphere The sphere $x^2 + y^2 + z^2 = 4$ is cut by the plane $z = 1/2$. How do you calculate the volume of two parts of the sphere using integrals? Thank you!
| The standard setup is
$$
\begin{align}
\int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}\int_{-\sqrt{4-z^2-y^2}}^\sqrt{4-z^2-y^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z
&=\int_\frac{1}{2}^2\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}2\sqrt{4-z^2-y^2}\;\mathrm{d}y\;\mathrm{d}z\\
&=\int_\frac{1}{2}^2\pi(4-z^2)\;\mathrm{d}z\tag{1}
\end{align}
$$
and
$$
\begin{align}
\int_{-2}^\frac{1}{2}\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}\int_{-\sqrt{4-z^2-y^2}}^\sqrt{4-z^2-y^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z
&=\int_{-2}^\frac{1}{2}\int_{-\sqrt{4-z^2}}^\sqrt{4-z^2}2\sqrt{4-z^2-y^2}\;\mathrm{d}y\;\mathrm{d}z\\
&=\int_{-2}^\frac{1}{2}\pi(4-z^2)\;\mathrm{d}z\tag{2}
\end{align}
$$
You might only need the last integral of each, but I started at ground-zero.
| {
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Use induction to prove a product of sums of squares is a sum of squares For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that
$$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$
My initial approach is
Base Case: $(a_1^2+b_1^2) = a_1^2+b_1^2$ which is true. (Although it is trivial)
Prove the statement is true when $n=2$: We have $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$$
(Thanks André Nicolas for pointing it out)
So if $a,b,c,d$ are integers, $ac,bd,ad,bc$ are all integers and integers are closed under addition and subtraction. Hence $(ac-bd),(ad+bc)$ are integers.
Inductive Hypothesis: $\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$ is true
Inductive Step: $$\prod_{i=1}^{n+1}(a_i^2+b_i^2) = \prod_{i=1}^{n}(a_i^2+b_i^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2)$$
Where $c$ and $d$ are integers.
But when we apply $n=2$, we have $(c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (e^2 + f^2)$ where $e$ and $f$ are integers.
Hence, by the principle of induction, the statement we needed to prove is true.
| The following identity is useful:
$$(a^2+b^2)(x^2+y^2)=(ax\pm by)^2 + (ay \mp bx)^2.\qquad\qquad(\ast)$$
This identity is connected with the multiplication of complex numbers. Note that
$$(a+ib)(x+iy)=ax-by+i(ay+bx).$$
The identity $(\ast)$ can be thought of as asserting that the product of the norms of $a+ib$ and $x+iy$, namely $\sqrt{(x^2+y^2)(a^2+b^2)}$, is equal to the norm of their product, namely $\sqrt{(ax-by)^2 +(ay+bx)^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using power series, compute: $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}$
Possible Duplicate:
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
I need to compute the sum of $$\sum_{n=1}^{\infty}\frac{n}{2^n}$$
using power series.
Any hints of how should I do that?
| So you want to compute $$S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots$$
Now consider $$ f(x)= \frac{x}{2} + \frac{x^2}{2^2} + \frac{x^3}{2^3} + \cdots = \displaystyle\frac{\frac{x}{2}}{1 - \frac{x}{2}}$$ From here evaluate the value of $f'(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Odd numbers expressed as : $x^2-y^2$ How to prove following statement :
Conjecture:
An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$
if and only if $n$ is a prime number .
If $x-y=m$ , where $m>1$ then $m \mid n$
Proof :
$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$
$\Rightarrow n=m\cdot (2\cdot y+m) \Rightarrow m \mid n$
Therefore , if $m \neq 1$ it follows that $n$ is a composite number , but how to prove that every odd
composite number ,other than $1$ , has representation : $x^2-y^2$ , where $x-y>1$ ?
| HINT $\ $ (Non) uniqueness follows from the composition law for differences of squares
$\rm\qquad\qquad\ (a^2-b^2)\ (A^2-B^2)\ =\ (a\:A+b\:B)^2-(a\:B+A\:b)^2$
$\rm\qquad\qquad\ \phantom{(a^2-b^2)\ (A^2-B^2)}\ =\ (a\:A-b\:B)^2-(a\:B-A\:b)^2$
E.g. composing $\rm\ 7 = 4^2 - 3^2\ $ with $\ 11 = 6^2 - 5^2\ $ yields for $\rm\: 7\cdot 11\:$ the following $\,2\,$ rep's
$\rm\qquad\qquad\ (4^2-3^2)\ (6^2-5^2)\ =\ (4\cdot 6+3\cdot 5)^2-(4\cdot 5+6\cdot 3)^2\ =\ 39^2 - 38^2$
$\rm\qquad\qquad\ \phantom{(4^2-3^2)\ (6^2-5^2)}\ =\ (4\cdot 6-3\cdot 5)^2-(4\cdot 5-6\cdot 3)^2\ =\ 9^2 - 2^2$
Remark $ $ Just like the e Brahmagupta–Fibonacci identity for composition of sums of squares can be viewed as arising from the multiplicativity of the norms of Gaussian integers, the above can be viewed as arising from norm multiplicativity of split-complex numbers (aka double or perplex numbers), $\,a+b\:\!j\,$ where $\,j^2 = 1.\ \ $
| {
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"timestamp": "2023-03-29T00:00:00",
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$Tx = \frac{x}{2} +\frac{1}{x}$ is a contraction on $M = [1,\infty)$ in $(\mathbb{R},|\cdot|)$? I cannot seem to find a contraction factor such that $$Tx = \frac{x}{2}+\frac{1}{x}$$ is a contraction on the whole set $[1,\infty)$ in the complete normed space $(\mathbb{R}, |\cdot|)$.
My argument for $x,y\in [1,\infty)$:
\begin{align} d(Tx, Ty) &= \left|\frac{x}{2}+\frac{1}{x} - (\frac{y}{2}+\frac{1}{y})\right|
\newline
\newline
&= \left|\frac{x-y}{2}+\frac{1}{x} - \frac{1}{y} \right|
\newline
\newline
&=\left|\frac{x-y}{2}+\frac{y-x}{xy} \right|
\newline
\newline
&\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{xy} \right|
\newline
\newline
&\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{2} \right|,\quad \forall x,y \geq \sqrt{2}
\newline
\newline
&=\frac{1}{2}|x-y| +\frac{1}{2}|x-y|
\newline
\newline
&= |x-y|.
\end{align}
Hence, restricting $T$ to $[\sqrt{2},\infty)$ yields a non-expansive mapping. $T$ also has a fixed point at $x=\sqrt{2}$. However, the book in which the exercise was found states that $T$ is a contraction with a minimal contraction factor $\lambda \in [0,1)$. Can I take better estimates to show that $T$ is a contraction on $[1,\infty)$?
| When you wrote
$$\left|\frac{x-y}{2}+\frac{y-x}{xy} \right|
\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{xy} \right|,$$
you didn't take advantage of the fact that $x-y$ and $y-x$ have opposite signs, so there is cancellation that will give you a better estimate. Instead you could note that
$$\left|\frac{x-y}{2}+\frac{y-x}{xy} \right|=|x-y|\left|\frac{1}{2}-\frac{1}{xy}\right|,$$
and that $$-\frac{1}{2}\leq \frac{1}{2}-\frac{1}{xy}<\frac{1}{2}$$
when $x,y\geq 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Is there a quick way to compute the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$? Is there a way to find the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$?
Just analyzing the roots shows that the polynomial is separable, so I suppose the splitting field would be a Galois extension. However, the roots are not easy to get a handle on, so it's not obvious to me what roots would need to be adjoined to $\mathbb{Q}$ to get the degree.
What is the right way to do this? Thanks.
| Just some culture, in case of curiosity. If you take any prime $q$ that has an integral expression as
$$ q = 2 x^2 + x y + 4 y^2, $$ such as $2, \; 5, \; 7\; 19,$ then
$$ z^3 + z + 1 \equiv 0 \pmod q $$ has no roots in integral $z.$
For the special case of $31,$ we have
$$ z^3 + z + 1 \equiv (z -3)(z - 14)^2 \pmod {31}. $$
For any other prime $$ p = x^2 + x y + 8 y^2, $$ such as $47, \; 67, \; 131, \; 149,$ then
$$ z^3 + z + 1 \equiv 0 \pmod p $$ has three distinct roots in integral $z$ and factors as three distinct linear factors.
For any prime $r$ with $(-31|r) = -1,$ such as $3, \; 11, \; 13, \; 17,$ then
$$ z^3 + z + 1 \equiv 0 \pmod r $$ has a single non-repeated root, so the cubic factors as a linear times a quadratic.
Well, why not. It turns out that every integer $n,$ positive or negative or $0,$ has an expression in integers as
$$ n = x^2 + x y + 8 y^2 + z^3 + z,$$ where we deliberately strip off the 1.
The difficult question is, what integers $n$ have an expression in integers as
$$ n = 2 x^2 + x y + 4 y^2 + z^3 + z \; ?$$
Certainly not all, $n = \pm 1$ do not work. The first few, in absolute value, that do not work are $$ \pm 1, \; \pm 869, \; \pm 25171, \; \pm 21118439, \; \pm 611705641, $$ these being the odd integers $u$ with
$$ 27 u^2 - 31 v^2 = -4. $$ The first few even values of $u$ are
$$ 30 = 3^3 + 3, $$
$$ 729090 = 90^3 + 90, $$
$$ 17718345150 = 2607^3 + 2607, $$
so these are easily expressed as $2 x^2 + x y + 4 y^2 + z^3 + z $ with both $x,y = 0.$
Well, somebody did mention the discriminant of a cubic, we have
$$ \mbox{disc}_z \left( z^3 + z + 1 \right) = -31 $$
and
$$ \mbox{disc}_z \left( z^3 + z + u \right) = -4 - 27 u^2 = -31 v^2.$$
So there.
| {
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Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$ Q. Prove
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$
My try: expand and got:
$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$
Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ... need to know how to apply the formulas more
| Let $u=5^x$. Then $u^2=25^x$ and
$$
\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(u/5+5u)^2}{u^2/25+25u^2}=\frac{u^2(1/5+5)^2}{u^2(1/25+25)}=\frac{(1/5+5)^2}{(1/25+25)}=\frac{338}{313}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
incremental computation of standard deviation How can I compute the standard deviation in an incremental way (using the new value and the last computed mean and/or std deviation) ?
for the non incremental way, I just do something like:
$$S_N=\sqrt{\frac1N\sum_{i=1}^N(x_i-\overline{x})^2}.$$
mean = Mean(list)
for i = 0 to list.size
stdev = stdev + (list[i] - mean)^2
stdev = sqrRoot( stdev / list.size )
| In case someone has to "decrement" and not only "increment" the standard deviation $\sigma$ (for example, when a result $x_i$ in the set is incorrect and needs to be removed or recalculated), you can use this formula:
$
\sigma_{\text{without } x_i} = \sqrt{\frac{n}{n -1} \left[ \sigma_n^2 - \frac{(\bar{x}_n-x_i)^2}{n-1} \right]}
$
Here is the derivation:
\begin{equation} \label{varianceDecrementale}
\begin{split}
\sigma^2_{\text{without } x_i} & = \sum_{j\in \{0, 1, \cdots, i-1, i+1, \cdots, n\}} \frac{x_j^2}{n-1} - \bar{x}_{n \text{ without } x_i}^2 \\
& = \frac{x_0^2 + x_1^2 + \cdots + x_{i - 1}^2 + x_{i + 1}^2 + \cdots + x_n^2}{n - 1} - \bar{x}_{n \text{ without } x_i}^2 \\
& = \frac{n}{n -1} \left[ \frac{x_0^2 + x_1^2 + \cdots + x_{i - 1}^2 + x_{i}^2 + x_{i + 1}^2 + \cdots + x_n^2}{n} - \frac{x_{i}^2}{n}\right] - \bar{x}_{n \text{ without } x_i}^2 \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} \right] - \bar{x}_{n \text{ without } x_i}^2 \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} \right] - \frac{n^2}{(n-1)^2} \left[ \bar{x}_n - \frac{x_i}{n} \right]^2 \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} - \frac{n}{n-1} \left[ \bar{x}_n - \frac{x_i}{n} \right]^2 \right] \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} - \frac{n}{n-1} \left[ \bar{x}_n^2 - 2 \bar{x}_n \frac{x_i}{n} + \frac{x_i^2}{n^2} \right] \right] \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} - \frac{n}{n-1} \bar{x}_n^2 + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right]
\end{split}
\end{equation}
We can remark that
\begin{equation}
\begin{split}
-\frac{n}{n-1} \bar{x}_n^2 & = -\frac{n}{n-1} \bar{x}_n^2 + \bar{x}_n^2 - \bar{x}_n^2 \\
& = -\bar{x}_n^2 - \frac{n}{n-1} \bar{x}_n^2 + \frac{n - 1}{n - 1} \bar{x}_n^2 \\
& = -\bar{x}_n^2 + \frac{-n \bar{x}_n^2 + n \bar{x}_n^2 - \bar{x}_n^2}{n-1} \\
& = - \left (\bar{x}_n^2 + \frac{\bar{x}_n^2}{n-1}\right)
\end{split}
\end{equation}
thus
\begin{equation}
\begin{split}
\sigma^2_{\text{without } x_i} & = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} - \frac{n}{n-1} \bar{x}_n^2 + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right] \\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \frac{x_{i}^2}{n} - \left (\bar{x}_n^2 + \frac{\bar{x}_n^2}{n-1}\right) + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right]\\
& = \frac{n}{n -1} \left[ \sum_i^n \frac{x_i^2}{n} - \bar{x}_n^2 - \frac{x_{i}^2}{n} - \frac{\bar{x}_n^2}{n-1} + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right]\\
& = \frac{n}{n -1} \left[ \sigma^2_n- \frac{x_{i}^2}{n} - \frac{\bar{x}_n^2}{n-1} + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right] \\
& = \frac{n}{n -1} \left[ \sigma^2_n - \frac{x_{i}^2}{n} - \frac{\bar{x}_n^2}{n-1} + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right] \\
& = \frac{n}{n -1} \left[ \sigma^2_n - \frac{x_{i}^2(n-1)}{n(n-1)} - \frac{\bar{x}_n^2}{n-1} + 2 \bar{x}_n \frac{x_i}{n - 1} - \frac{x_i^2}{n (n - 1)} \right] \\
& = \frac{n}{n -1} \left[ \sigma^2_n - \frac{x_{i}^2{n}}{{n}(n-1)} + {\frac{r^2}{n(n-1)}} - \frac{\bar{x}_n^2}{n-1} + 2 \bar{x}_n \frac{x_i}{n - 1} - {\frac{x_i^2}{n (n - 1)}} \right] \\
& = \frac{n}{n -1} \left[ \sigma^2_n - \frac{x_i^2}{n-1} - \frac{\bar{x}_n^2}{n-1} + 2\bar{x}_n\frac{x_i}{n-1} \right]\\
& = \frac{n}{n -1} \left[ \sigma^2_n - \frac{(\bar{x}_n-x_i)^2}{n-1} \right]
\end{split}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "79",
"answer_count": 8,
"answer_id": 1
} |
Simplest proof that $\zeta(s) \to \infty$ as $s \to 1$? For homework I had to prove the divergence of the series $1/(k\log^p k)$ for all real $p$ (it is simple to do so via integration.) However a more elegant means would be to appeal to the behavior of the zeta function $\zeta(s)=\sum\limits_{k=1}^\infty 1/k^s$ on the real line. What is the most elementary proof that $\lim\limits_{s\to 1}\zeta(s)=\infty$?
| It seems that the most elegant proof is Euler's proof via comparison
\begin{align*}
\zeta (1) &= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + \cdots \\
&> 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}} \right) + \cdots \\
&= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \cdots \\
&= \infty.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/103684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplify _Elementary Calculus_ section 1.6 problem 25 Let's try this again. We're still on problem 25 in section 1.6 of Elementary Calculus.
$$\frac{3-\sqrt{c+2}}{c-7}$$
My first thought is (again) to multiply by $3+\sqrt{c+2}$:
$$=\frac{(3-\sqrt{c+2})(3+\sqrt{c+2})}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}$$
This looks "simplified" to me, so I proceed to substitute $c=7+\epsilon, \epsilon \in \mathbb{R}^*, \epsilon \approx 0$:
$$=\frac{9-(9+\epsilon)}{3(7+\epsilon)+(7+\epsilon)\sqrt{7+\epsilon+2}-7\sqrt{9+\epsilon}-21}$$
$$=\frac{-\epsilon}{21+3\epsilon+7\sqrt{9+\epsilon}+\epsilon\sqrt{9+\epsilon}-7\sqrt{9+\epsilon}-21}$$
$$=-\frac{\epsilon}{3\epsilon + \epsilon\sqrt{9+\epsilon}}$$
Knowing the answer is $-\frac{1}{6}$, it seems likely that this somehow reduces to $-\frac{\epsilon}{6\epsilon}$ (apart from some error), but I don't see how to get from $3\epsilon+\epsilon\sqrt{9+\epsilon}$ to $3\epsilon+3\epsilon=6\epsilon$.
Thanks for your help again!
| From this step:
$$=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$$
you have
$$=\frac{-(c-7)}{(c-7)(3+\sqrt{c+2})}$$
Cancel the common factor and the rest should follow straightforwardly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- two methods different conclusion. I tried to use couple of inequalities in this way:
$0\leq\left|\frac{(x^3-y^2)^2}{x^4+y^4}\right| \leq \frac{(x^3)^2}{x^4}=x^2$, so now I can use the squeeze theorem and conclude that this limit is 0, but when I choose a route such as $y=Kx$ I get that the limit is $\frac {K^4}{1+K^4}$, which depends on K, so apparently there's no limit when $(x,y) \to (0,0)$.
Where's my mistake?
Thanks a lot.
| The second result is the correct one. To see this, write $\frac{(x^3-y^2)^2}{x^4+y^4}=\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}+\frac{y^4}{x^4+y^4}$. Since
$$\left|\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}\right|\leq x^2+|x|\frac{x^4+y^4}{x^4+y^4}=x^2+|x|,$$
we just have to study $\lim_{(x,y)\to (0,0)}\frac{y^4}{x^4+y^4}$. It doesn't exist since putting $g(x,y)= \frac{y^4}{x^4+y^4}$ we have $g(x,0)=0$ and $g(0,y)=0$ for $x,y\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to analyze convergence and sum of a telescopic series? I can't find a generic form Hello I would like to know what is the generic form to calculate a telescopic series.
http://www.dummies.com/how-to/content/how-to-analyze-a-telescoping-series.html
here uses a sum that startes at $n=1$, but what if it started at $n=2$ or $n=3$?
the $n$ index has any influence over the sum? I think yes, but no one explains this..
my series is:
$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$
how do I solve this really step by step for new people to those series
| If a telescoping sum starts at $n=m$, then
$$
\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right) =a_{m}-a_{N+1}
$$
and the telescoping series is thus
$$
\begin{eqnarray*}
\sum_{n=m}^{\infty }\left( a_{n}-a_{n+1}\right) &=&\lim_{N\rightarrow
\infty }\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right) \\
&=&a_{m}-\lim_{N\rightarrow \infty }a_{N+1}=a_{m}-\lim_{N\rightarrow \infty
}a_{N+1} \\
&=&a_{m}-\lim_{N\rightarrow \infty }a_{N}.
\end{eqnarray*}
$$
Of course the series converges if and only if there exists $\lim_{N\rightarrow \infty }a_{N}.$
The case $m=1$ is
$$
\begin{eqnarray*}
\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) &=&a_{1}-a_{N+1} \\
\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right) &=&a_{1}-\lim_{N\rightarrow
\infty }a_{N}.
\end{eqnarray*}
$$
The difficult part is to write a series $
\sum_{n=m}^{\infty}u_n$ in the form $
\sum_{n=m}^{\infty}\left( a_{n}-a_{n+1}\right)$, when possible. Concerning Wilf-Zeilberger method (also called "creative telescoping") see answers to the question Mathematical Telescoping.
Example: write $\frac{2n+1}{n^{2}\left( n+1\right) ^{2}}=\frac{1}{n^{2}}-
\frac{1}{\left( n+1\right) ^{2}}$ to evaluate
$$
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{2n+1}{n^{2}\left( n+1\right) ^{2}}
&=&\sum_{n=1}^{\infty }\left( \frac{1}{n^{2}}-\frac{1}{\left( n+1\right) ^{2}
}\right) \\
&=&1-\lim_{N\rightarrow \infty }\frac{1}{N^{2}}=1-0=1.
\end{eqnarray*}
$$
Added in response to the edited question. The standard technique is to
expand $\frac{4}{(n+1)(n+2)}$ into partial fractions. Write
$$
\begin{eqnarray*}
\frac{4}{(n+1)(n+2)} &=&\frac{A}{n+1}+\frac{B}{n+2} \\
&=&\frac{(n+2)A+(n+1)B}{(n+1)(n+2)} \\
&=&\frac{\left( A+B\right) n+2A+B}{(n+1)(n+2)}
\end{eqnarray*}
$$
and find the constants $A$ and $B$. The following system must hold
$$
\left\{
\begin{array}{c}
A+B=0 \\
2A+B=4
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
A=4 \\
B=-4.
\end{array}
\right.
$$
So
$$
\frac{4}{(n+1)(n+2)}=\frac{4}{n+1}-\frac{4}{n+2}
$$
and, since $a_{n}=\frac{4}{n+1}$, you get
$$
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{4}{(n+1)(n+2)} &=&\sum_{n=1}^{\infty }\left( \frac{
4}{n+1}-\frac{4}{n+2}\right) \\
&=&\frac{4}{1+1}-\lim_{n\rightarrow \infty }\frac{4}{n+1} \\
&=&2-0=2.
\end{eqnarray*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Location of a root of a cubic polynomial For $\alpha\in(0,\frac12)$, $\beta\in(0,\infty)$, $N\in\mathbb N\backslash\{0\}$ and $n\in\{0,\ldots,N\}$, how can I prove that exactly one zero of the cubic polynomial
$$
(N+2\beta)x^3-(N+n+3\beta)x^2+(n+\beta+N\alpha-N\alpha^2)x+n\alpha^2-n\alpha
$$
lies in $[\alpha,1-\alpha]$?
| Evaluating at $\alpha$, we have:
$$\begin{align*}
f(\alpha) &= (N+2\beta)\alpha^3 - (N+n+3\beta)\alpha^2 + (n+\beta+N\alpha-N\alpha^2)\alpha + n\alpha^2 - n\alpha\\
&= (N+2\beta - N)\alpha^3 + (-N-n-3\beta+N+n)\alpha^2 + (n+\beta-n)\alpha\\
&= 2\beta\alpha^3 - 3\beta\alpha^2 + \beta\alpha\\
&= \alpha\beta(2\alpha^2 -3\alpha + 1).
\end{align*}$$
Evaluating at $1-\alpha$ gives
$$\begin{align*}
f(1-\alpha) &= (N+2\beta)(1-3\alpha + 3\alpha^2-\alpha^3) - (N+n+3\beta)(1-2\alpha+\alpha^2)\\
&\qquad \mathop{+} (n+\beta+N\alpha-N\alpha^2)(1-\alpha) + n\alpha^2 - n\alpha\\
&= (-N-2\beta +N)\alpha^3 + (3N+6\beta - N-n-3\beta -N-N+n)\alpha^2\\
&\qquad \mathop{+}(-3N-6\beta+2N+2n+6\beta-n-\beta+N-n)\alpha\\
&\qquad \mathop{+} (N+2\beta-N-n-3\beta+n+\beta)\\
&= -2\beta\alpha^3 +3\beta\alpha^2-\beta\alpha\\
&= -\alpha\beta(2\alpha^2 - 3\alpha + 1).
\end{align*}$$
So, unless $2\alpha^2-3\alpha+1$ is $0$, the two values have opposite signs. But the roots of $2x^2-3x+1$ are $1$ and $\frac{1}{2}$, so $\alpha$ cannot be a root.
Thus, there is at least one root for the polynomial in $[\alpha,1-\alpha]$ (in fact, in $(\alpha,1-\alpha)$.
Since $f(x)$ has opposite signs on $\alpha$ and on $1-\alpha$, if $f(x)$ has more than one (distinct) root on $[\alpha,1-\alpha]$, then it must have three distinct roots in the interval (why?). Can all three roots be in that interval?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $ 1 + \sum_{i=0}^{n-2} [ \prod_{k=0}^{i} (n-k)/(r-k) ] $ Is there a way in which the expression
$$ 1 + \sum_{i=0}^{n-2} \prod_{k=0}^{i} \dfrac{n-k}{r-k} $$
may be simplified?
It would simplify some equations I am working with.
| First, using recurrence equation for Euler's $\Gamma$-function:
$$
\prod_{k=0}^m \frac{n-k}{r-k} = \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)}
$$
Then, changing summation variable as $m \to n-2-m$:
$$
1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)} = 1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-(n-2-m))} \cdot \frac{\Gamma(r-(n-m-2))}{\Gamma(r+1)} = \\
1 + \frac{\Gamma(n+1)}{\Gamma(r+1)} \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)}
= 1 + \frac{\Gamma(n+1)}{\Gamma(r+1)} \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)}
$$
Now the last sum is just a hypergeometric series:
$$
\frac{\Gamma(n+1)}{\Gamma(r+1)} \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)} =
\frac{\Gamma(n+1) \Gamma(1+r-n)}{\Gamma(r+1)} \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2) \Gamma(1+r-n)} = \\
\frac{1}{\binom{r}{n}} \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \frac{n}{r-n+1} - \frac{1}{\binom{r}{n}}
$$
The last equality follows because $s_{m+1} - s_m = \binom{m+r-n+1}{m+1}$, where
$$
s_m = \frac{m+1}{r-n+1}\binom{m+r-n+1}{m+1}
$$
And therefore the original sum telescopes
$$ \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \sum_{m=0}^{n-2} (s_{m+1}-s_m) = s_{n-1} - s_0
$$
It now remains to compute $s_{m+1}-s_m$:
$$
s_{m+1} = \frac{m+2}{r-n+1} \binom{m+r-n+2}{m+2} = \frac{m+2}{r-n+1} \cdot \frac{m+r-n+2}{m+2} \binom{m+r-n+1}{m+1} = \\ s_m + \binom{m+r-n+1}{m+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/111803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What is the minimum possible value of $(a + b + c)$? $a$, $b$ and $c$ are real positive numbers satisfying
*
*$\dfrac{1}{3} \le ab + bc + ca \le 1$ and
*$abc \ge \dfrac{1}{27}$
What is the minimum possible value of $(a + b + c)$?
| By the AM-GM inequality, $\sqrt[3]{abc}\leq \frac{a+b+c}{3}$.
Since $abc \geq \frac{1}{27}$, this implies that $\sqrt[3]{abc}\geq \frac{1}{3}$.
So $\frac{1}{3} \leq \frac{a+b+c}{3}$ and hence $1 \leq a+b+c$.
This minimum value is achieved in the symmetric case where $a=b=c=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the value of a? Let $a$ and $b$ be two positive numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?
| According to the given information we have,
$$4\sqrt{ab} = 5(\frac{2ab}{a+b})$$
$$(a+b) =\frac{5}{2}\sqrt{ab})$$
$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} = \frac{5}{2}$$
Let , $t$ $ =$ $\sqrt{\frac{a}{b}}$
$$[t+\frac{1}{t}=\frac{5}{2}].....................Eq(1)$$
A clever person will immediately infer that $t=\frac{1}{2}$
But if its a subjective question we have to justify that also, so
$$({t+\frac{1}{t}})^2=\frac{25}{4}$$
$$t^2 +\frac{1}{t^2} = \frac{17}{4}$$
NOW,
$$(t-\frac{1}{t})^2= t^2+\frac{1}{t^2} -2 =\frac{9}{4}$$
$$[t-\frac{1}{t}=\frac{3}{2} ] .....................Eq(2)$$
neglecting the negative value as we know that$ L.H.S.>0$ , since $,t>0$
From Eq(1) and Eq(2) we have $t=4$, hence $a=4b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
| Note that this works
to find the derivative
of $x^{1/n}$
where $n$
is a positive integer.
We use
$a^n-b^n
=(a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k}
$.
$\begin{array}\\
\frac{(x+h)^{1/n}-x^{1/n}}{h}
&=\frac{(x+h)^{1/n}-x^{1/n}}{h}
\frac{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{(x+h)-x}{h}
\frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{h}{h}
\frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{1}{\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}}\\
\end{array}
$
As $h \to 0$,
$\begin{array}\\
\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}
&\to \sum_{k=0}^{n-1} x^{k/n} x^{(n-1-k)/n}\\
&= \sum_{k=0}^{n-1} x^{(n-1)/n}\\
&= n x^{(n-1)/n}\\
&= n x^{1-1/n}\\
\end{array}
$
Therefore
$(x^{1/n})'
=\frac1{n x^{1-1/n}}
=\frac1{n} x^{1/n-1}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to prove that $\frac{\cos\alpha}{\cos\beta}=a/b$ If $\alpha \not= \beta$, and $$ a\tan \alpha+b\tan\beta=(a+b)\tan\frac{\alpha+\beta}{2}$$ then can we prove that $\frac{\cos\alpha}{\cos\beta}=\frac{a}{b}$?
Seems like I am stuck on this one.
| This is true if we assume $\displaystyle \sin\left(\frac{\alpha -\beta}{2}\right) \neq 0$ and $\displaystyle b \neq 0$.
Let $\displaystyle \frac{a}{b} = c$
Rewrite your equation as
$$c\tan \alpha + \tan \beta = (c+1)\tan(\frac{\alpha+\beta}{2})$$
$$ c\left(\tan \alpha - \tan\left(\frac{\alpha+\beta}{2}\right)\right) = \tan\left(\frac{\alpha+\beta}{2}\right) - \tan \beta$$
Using $\displaystyle \tan x + \tan y = \frac{\sin(x+y)}{\cos x \cos y}$ we get
$$ c \cdot \frac{\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos \alpha \cos \left(\frac{\alpha+\beta}{2}\right)} =\frac{\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos \beta \cos \left(\frac{\alpha+\beta}{2}\right)}$$
and so
$$ \frac{\cos \alpha}{\cos \beta} = c = \frac{a}{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/114723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Limits: dividing by the highest power does not work in this case I believe to solve expressions like:
$$\lim_{x\to \infty} (2x^2 + 1)$$
we need to divide each term in the numerator and denominator by the highest power. In the previous case, the highest power is $x^2$, so we get
$$\lim_{x \to \infty} \left(2 + \frac{1}{x^2} \right) = 2+0=2$$
Now for:
$$\lim_{n \to \infty} \frac{-1}{\sqrt{n} + 2}$$
The answer given is $\frac{-1}{\infty}=0$
On the other hand, what I did was
$$\lim_{x \to \infty} \frac{-1}{1 + \frac{2}{\sqrt{n}}}=\frac{-1}{1}=-1$$
What's wrong?
| What you are saying in your calculations is $2x^2 +1 = (2 + 1/x^2)$, which is wrong. What you want is $2x^2 +1 = x^2(2 + 1/x^2)$, thus
$$\lim_{x \to \infty} 2x^2 +1 = \lim_{x \to \infty} x^2(2+ 1/x^2) = \lim_{x\to \infty} 2x^2 = \infty$$
for your second limit you are making the same mistake: $\sqrt{n} + 2 \neq 1+ 2/\sqrt{n}$, but $\sqrt{n} + 2 = \sqrt{n}(1+2/\sqrt{n})$. Thus,
$$\lim_{x \to \infty} \frac{-1}{\sqrt{n}+2} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}(1+2/\sqrt{n})} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}} = 0$$
This method is usually better suited for indeterminate cases, like $\lim_{x \to \infty} \frac{2x^2 +1}{3x^2 + 4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Which formula do I use to integrate $ \int {\sqrt{x^2 + 81} \over 2} \,dx $ I am having trouble with a question really need help please.
$$
\int {\sqrt{x^2 + 81} \over 2} \,dx
$$
I thought about taking the square root off and turning the question into $\frac 12 \int (x^2 +81)^{1/2}\, dx$ but then wondered if I could use the quotient rule.
| $$ \int \sqrt{x^2+81} dx $$
Integration by parts formula
$$ \int u(x)v^{'}(x) dx = u(x)v(x) - \int v(x)u^{'}(x) dx$$
Therefore assume that $v^{'}(x) = 1$ in this case
Denote the integral
$$ I = \int \sqrt{x^2+81} dx$$
$$
\begin{align*}
I &= x \sqrt{x^2+81} - \int \frac{x^2}{\sqrt{x^2+81}} dx\\
&= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\
&= x \sqrt{x^2+81} - I + 81 \int \frac{1}{\sqrt{x^2+81}} dx
\end{align*}
$$
Therefore
$$ 2I = x \sqrt{x^2+81} +81 \int \frac{1}{\sqrt{x^2+81}} dx$$
the rest you should do it yourself.
I was told that we are not supposed to give complete solution for
homework questions.
IMPROVED EXPLANATION: (BY REQUEST)
$u(x) = \sqrt{x^2+81}$ and
$v(x) = x$, therefore $v^{'}(x) = 1$
$$ u^{'}(x) = \frac{1}{2}\left(x^2+81\right)^{-\frac{1}{2}} \times 2x = \frac{x}{\sqrt{x^2+81}}$$
$$
\begin{align*}
\int \sqrt{x^2+81} dx &= \int u(x) v^{'}(x) \\
&= u(x)v(x) - \int u^{'}(x) v(x) dx\\
&= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\
&= x \sqrt{x^2+81} - \int \sqrt{x^2+81} \hspace{3pt} dx + \int \frac{81}{\sqrt{x^2+81}} dx\\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$ I have following integral and it should be simple, however whatever substitution I use and no matter how many times I integrate it by parts (or combine both) I never get the correct solution (or any alternative solution):
$$\int \frac{dx}{(x^2 + a^2)^2}$$
I'm looking for what is on the Wolfram|Alpha in the alternative solutions section:
$$\frac{\arctan(\frac{x}{a})}{2a^3} + \frac{x}{2a^2(a^2 + x^2)} $$
| $$\int\dfrac{1}{(x^2+a^2)^2}dx$$
Hint : Does there exist a function which when differentiated gives something similar to $1\over (x^2+1)$ ?
Put $\tan^{-1} \frac{x}{a} = t$. This gives $\dfrac{1}{1+(\dfrac{x}{a})^2}.\dfrac{1}{a}.dx = dt = \dfrac{a}{a^2+{x}^2}.dx$
$$\int\dfrac{1}{(x^2+a^2)^2}dx=\int \dfrac{1}{a.( (a\tan t)^2+a^2)}dt=\int \dfrac{1}{a.( a^2\tan^2 t+a^2)}dt=\int \dfrac{1}{a^3(\tan^2 t+1)}dt=\dfrac{1}{a^3}\int \cos^2 t$$
which will lead to the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?
Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$
Side work:
$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = -3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\sqrt{9\csc^2{\theta} - 9}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\tan{\theta} \ d\theta$
$-\int \frac{3 \cdot \sqrt{\csc^2{\theta} - 1}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\cot{\theta}\csc{\theta}\cot{\theta}}{3 \csc^3{\theta}} \ d\theta$
$-\frac{1}{3} \int \cos^2{\theta} \ d\theta$
$-\frac{1}{3} \int \frac{1}{2} \left(1 + \cos{\left(2\theta\right)}\right) \ d\theta$
$-\frac{1}{6} \int 1 + \cos{\left(2\theta\right)} \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{6} \int \cos{\left(2\theta\right)} \ d\theta$
Sidework:
$u = 2\theta$
$du = 2 \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{3} \int \cos{u} \ du$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{u} + C$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{2\theta} + C$
Sidework:
$\sin{\left(2\theta\right)} = 2 \sin{\theta} \cos{\theta}$
$\sin{\theta} = \frac{3}{x}$
$\cos{\theta} = \frac{\sqrt{x^2 - 9}}{x}$
$\theta = \sin^{-1}{\left(\frac{3}{x}\right)}$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2}{3} \cdot \frac{3}{x} \cdot \frac{\sqrt{x^2 - 9}}{x} + C$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2\sqrt{x^2 - 9}}{x^2} + C$
Thank you for your time.
| Let $I = \int \frac{\sqrt{x^2 – 9}}{x^2}$
Easiest is a hyperbolic substitution:
Put $x = 3\cosh{u}, u \le 0$, so $dx = 3\sinh(u)du$
Then \begin{align}
I &=\int\frac{3\sinh(u^2)}{9 \cosh^2(u)}du\\\\
&= \int\tanh^2(u)du\\\\
&= \int\left(1–\mathrm{sech}^2(u)\right)du\\\\
&= u–\tanh(u)+c\\\\
&= u-\frac{3\sinh(u)}{3\cosh(u)}+c\\\\
&= \cosh^{-1}\left(\frac{x}{3}\right)-\frac{\sqrt{x^2–9}}{x}+c\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Is there a way to create a summation function for this? I have an algorithm that generates terms as follows:
$$\begin{align*}
\text{Term 1}: &2\times 3\times\left(\frac{1}{2}+\frac{1}{3}\right) + 4\\
\text{Term 2}: &2\times 3\times 4\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + 3\times 5\\
\text{Term 3}: &2\times 3\times 4\times 5\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right) + 3\times4\times 6\\
\text{Term 4}: &2\times 3\times 4\times 5\times 6\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right) + 3\times4\times 5\times 7
\end{align*}$$
and so on.
The pattern in the first half is obvious; the pattern in the second half is a little trickier. It "splits" the last multiplier into (multiplier-1)*(multiplier+1) so 4 changes into 3*5 and then the 5 on that one changes into 4*6 and so on for the next term.
I am trying to find a clever way to say "sum terms 1-N" given this pattern
| The last term is $$\frac{(n+3)!}{2\times (n+2)},$$
or
$$\frac{(n+1)!(n+3)}{2}$$
where $n$ is the number of the term you are in.
So the $n$th term is:
$$\begin{align*}
t_n &= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \frac{(n+3)!}{2\times(n+2)}\\
&= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \frac{(n+1)!(n+3)}{2}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing point is the orthocenter Given a rectangle WXYZ, let R be a point on its circumscribed circle. Show that, out of the orthogonal projections of R onto WX, XY, YZ, and ZW; one out of these 4 points is the orthocenter of the triangle created by the other three.
| Continuing with the notation and analytic-geometric approach from my comment,
$$
\eqalign{
R_1R_2 \perp R_3R_4 \iff
&
\{y=r\sin\theta\} \perp
\{x=r\cos\theta\}
\\\\&\text{True (horizontal/vertical lines)}
\\\\
R_1R_3 \perp R_2R_4 \iff
&
\overline{(a,r\sin\theta)(r\cos\theta,b)} \perp
\overline{(-a,r\sin\theta)(r\cos\theta,-b)}
\\\iff
&-1=
\frac{b-r\sin\theta}{r\cos\theta-a}\cdot\frac{-b-r\sin\theta}{r\cos\theta+a}
=\frac{r^2\sin^2\theta-b^2}{r^2\cos^2\theta-a^2}
\\&\qquad=\frac{a^2\sin^2\theta-b^2\cos^2\theta}{b^2\cos^2\theta-a^2\sin^2\theta}
\\\\&\text{True}
\\\\
R_1R_4 \perp R_2R_3 \iff
&
\overline{(a,r\sin\theta)(r\cos\theta,-b)} \perp
\overline{(-a,r\sin\theta)(r\cos\theta,b)}
\\\iff&-1=
\frac{-b-r\sin\theta}{r\cos\theta-a}\cdot\frac{b-r\sin\theta}{r\cos\theta+a}
\\\\&\text{...True?}
}
$$
Is this approach accesible for you?
Can you try the last one?
The facts we need are:
$$
\matrix{
\cos^2\theta+\sin^2\theta=1\qquad&\text{true for all }\theta
\\\\
a^2+b^2=r^2\qquad
&\matrix{\text{do you see why I set}\\\text{it up this way above?}}
\\\\
m_{PQ}=\frac{y_P-y_Q}{x_P-x_Q}
&\text{slope formula}
\\\\
-(s-t)=t-s
&\text{basic algebra}
\\\\
(s+t)(s-t)=s^2-t^2
&\text{basic algebra}
\\\\
x=r\cos\theta
&\text{is a circle with radius }r
\\
y=r\sin\theta
&\text{centered at the origin}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$? I have tried something to solve the series
$$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$
My approach is :
$$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n.$$
Differentiating the above equation
$$n(1+x)^{n-1} = \binom{n}{1} + \binom{n}{2}x + \cdots + n\binom{n}{n}x^{n-1}$$
Also,
$$
\left(1+\frac{1}{x}\right)^n =\binom{n}{0} + \binom{n}{1}\frac{1}{x} + \binom{n}{2}\left(\frac{1}{x}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{x}\right)^n$$
Multiplying above two equation I get,
$$\begin{align*}
&{n(1+x)^{n-1}\left(1 + \frac{1}{x}\right)^n}\\
&\quad= \left(
\binom{n}{1}^2 + 2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2 + \cdots + n\binom{n}{n}^2\right)\left(\frac{1}{x}\right) + \text{other terms}
\end{align*}$$
So I can say that coefficient of $\frac{1}{x}$ in expansion of $n(1+x)^{n-1}(1+\frac{1}{x})^n$ will give me the required answer.
Am I doing it correct,please correct me if I'm wrong ?
If I'm right,please tell me how to calculate the coefficient of $\frac{1}{x}$ ?
Based on the answers,I tried to implement the things in a C++ code.
I tried implementing the code using extended euclidean algorithm so that the problem of truncated division can be eliminated but still not abled to figure out why am I getting wrong answer for n>=3. This is my updated code : http://pastebin.com/imS6rdWs I'll be thankful if anyone can help me to figure out what's wrong with this code.
Thanks.
Solution:
Finally abled to solve the problem.Thanks to all those people who spent their precious time for my problem.Thanks a lot.This is my updated code :
http://pastebin.com/WQ9LRy6F
| First we address the overflow issue. Note that $10^9+7$ is relatively prime to all the numbers that come up in a naive calculation of $\binom{2n}{n}$. Indeed $10^9+7$ happens to be prime. So when we are calculating, we can always reduce modulo $10^9+7$ as often as necessary to prevent overflow.
Now to the identity. We have $n$ boys and $n$ girls. We want to choose $n$ people. The number of ways this can be done is $\binom{2n}{n}$. But we can choose $0$ boys and $n$ girls, or $1$ boy and $n-1$ girls, and so on. We can choose $k$ boys and $n-k$ girls in $\binom{n}{k}\binom{n}{n-k}$ ways, or equivalently in $\binom{n}{k}^2$ ways. This gives the standard derivation of the identity
$$\binom{2n}{n}=\sum_{k=0}^n \binom{n}{k}^2.$$
Note now that the $\binom{2n}{n}$ choices are all equally likely. The expected number of boys is, by symmetry, equal to $\frac{n}{2}.$ But the probability that there are $k$ boys is $\frac{\binom{n}{k}^2}{\binom{2n}{n}}$, and therefore the expected number of boys is
$$\sum_{k=0}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}.$$
The term corresponding to $k=0$ is $0$, so can be omitted, and we get
$$\sum_{k=1}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}=\frac{n}{2},$$
which is essentially our identity.
Remark: There is a very nice book on bijective proofs called Proofs that Really Count. A title that so far doesn't seem to have been used is Mean Proofs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
Show by substitution that $\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm dx = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm dx$ How do you show
$$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x$$
without integrating by parts, but only using substitution?
| For any integral of the form
$$I = \int_0^{\pi} x f(\sin x) \mathrm{d}x $$
since $\sin(\pi-x) = \sin(x)$, using substitution $x = \pi-u$, or $u=\pi- x$, one may reduce this integral to the following using
when $x=0, u=\pi$, and when $x=\pi, u=0$, and also $ \mathrm{d}x = \mathrm{-d}u$
$$
\begin{align*}
I = \int_0^{\pi} (\pi - u) f(\sin u) \mathrm{d}u
&= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - \int_{\pi}^{0} u f(\sin u) \mathrm{(-d}u) \\
&= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - \int_0^{\pi} u f(\sin u) \mathrm{d}u \\
&= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - I\\
2I &= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u\\
I & = \frac{\pi}{2}\int_0^{\pi} f(\sin u) \mathrm{d}u = \frac{\pi}{2}\int_0^{\pi} f(\sin x) \mathrm{d}x \tag{A}
\end{align*}
$$
Apply $(A)$ for $\displaystyle{f(\sin x) = \frac{\sin x}{2-\sin^2 x} = \frac{\sin x}{1+\cos^2 x}}$
Although it was not asked, to further notice what is the value of the integral
by substituting further $\cos x = t, -\sin x \mathrm{d}x = \mathrm{d}t \hspace{5pt} cos(\pi) = -1, cos(0) = 1$
$$
\begin{align*}
\frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x &= \frac{\pi}{2} \int_{1}^{-1} \frac{\mathrm{-d}t}{1+t^2}\\
&= \frac{\pi}{2} \int_{-1}^{1} \frac{\mathrm{d}t}{1+t^2}\\
&= \frac{\pi}{2} \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) \\
&= \frac{\pi}{2} \left( \frac{\pi}{4} - \frac{-\pi}{4} \right) = \frac{\pi}{2} \left(\frac{\pi}{2}\right) = \frac{{\pi}^2}{4}\\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Showing the identity: $\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$ My knowledge of trigonometry is still insufficient to resolve this problem. Any help would be greatly appreciated.
$$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$
| Hoping this is not homework:
$$\cot x - 8\cot 8x = \cot x -8\cot 8x= \cot x-8\frac{\cot^24x -1}{2\cot 4x}$$
$$= \cot x-4\frac{\cot^24x -1}{\cot 4x}= \cot x -4\cot 4x +4\tan 4x$$
$$= \cot x -4 \frac{\cot^22x -1}{2\cot 2x}+4\tan 4x=\cot x -2\cot 2x + 2\tan 2 x + 4 \tan 4x$$
$$=\cot x -2 \frac{\cot^2x -1}{2\cot x}+2\tan 2x + 4\tan 4x = \cot x - \cot x + \tan x + 2\tan 2x + 4\tan 4x $$
$$=\tan x + 2\tan 2x + 4\tan 4x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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find the largest of the three angle of the triangles?
The sides of a triangle are given to be $$ x^2 +x +1 ,$$ $$2x+1,$$ $$x^2-1$$Then find the largest of the three angle of the triangles .
Source:ISI entrance exam sample question 2016
I tried by applying hero's formula and then equating to the formula
$$ 0.5 bc \sin\theta $$.
But using these methods , then i have to try 3 times for each angle .
And still i cant get the value of x.
I think i can do it throuh by randomly assigning a value to x and then get the value of the angle opposite to the longest side.
Thanks in advance.
| Note that :
$
\begin{cases}
x^2+x+1 > 2x+1 > x^2-1, & \text{if } x\in(1,1+\sqrt 3) \\
x^2+x+1 > x^2-1 > 2x+1, & \text{if } x\in(1+\sqrt 3,+\infty)
\end{cases}$
Hence , the largest angle of the triangle is angle opposite to the side $x^2+x+1$ . Let us denote that angle as $\alpha$ , then according to Cosine rule we can write following equality :
$(x^2+x+1)^2=(2x+1)^2+(x^2-1)^2-2(2x+1)(x^2-1)\cdot \cos \alpha $
| {
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"url": "https://math.stackexchange.com/questions/126774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$ $$\sum_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$$
I have this answer, please let me know if there is a more beautiful proof.
My answer:
at first, we prove two inequalities:
*
*If $x\in (x,\pi)$ then $\sin x \leq x$
*When $x\in(0,\frac{\pi}{2})$, $\sin x \geq \frac{2x}{\pi}$
1) first, let $y = \sin x -x $
$y^{\prime} = \cos x -1 \leq 0$
so $\sin x - x \leq \sin 0 -0 = 0$
which can be rewritten as
$\sin x \leq x$
2) Let $y=\sin x - \frac{2x}{\pi}$
thus $y^{\prime} = \cos x - \frac{2}{\pi}$ because $x\in (0, \frac{\pi}{2})$
so y at first decreases and then increases on the boundary of $x \in (0,\frac{\pi}{2})$
so $ \sin x - \frac{2}{\pi}\leq \max \{{\sin 0 - \frac{2}{\pi}0, \sin (\frac{\pi}{2} - \frac{2}{\pi}\frac{\pi}{2}) \}}$
so $\sin x \leq \frac{2x}{\pi}$
Then select $M\in N$
$\frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m+1} + \ldots + \frac{\sin ((m+n)x)}{m+n} \leq \frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m} + \ldots + \frac{\sin ((m+n)x)}{m} $
=> $\frac{1}{2M} \times \frac{\sin ((m-\frac{1}{2})x) - \sin ((n+\frac{1}{2})x)}{\sin \frac{x}{2}} < \frac{1}{m \times \sin \frac{x}{2}} \times \sin x + \frac{\sin 2x}{2} + \ldots + \frac{\sin ((m-1)x)}{m-1} < x + \frac{2x}{2} + \ldots + \frac{(m-1)x}{m-1} $
so just need to prove that
$(m-1)x + \frac{1}{m \times \sin \frac{x}{2}} \leq 2\sqrt{\pi}$
select M which satisfies
$ \frac{\sqrt{\pi}}{x} \leq m < \frac{\sqrt{\pi}}{x} + 1 $
so $ (m-1)x < [ \frac{\sqrt{\pi}}{x} \times x = \sqrt{\pi} ] $
thus
$\frac {1}{m \times \sin(\frac{x}{2})}\leq[ \frac{1}{\sqrt{\pi}}\times \frac{2}{\frac{ \sin (0.5x)}{0.5x}} = \frac{1}{\sqrt{\pi}} \times \frac{2}{\frac{\sin 0.5x}{0.5x}} ]$
because $x\in (0, \pi)$ thus $\frac{x}{2} \in (0, \frac{\pi}{2})$
$ (m-1)x + \frac{1}{m \times \sin(0.5x)} \leq 2\sqrt{\pi} $
thanks, for viewing and commenting.
ps. I'm still learning latex and mathematics, so my answer isn't pretty to read, nor is the latex I wrote.
| The best it is possible to state is:
$$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$
Call $f_N(x)=\sum_{n=1}^N \frac{\sin(nx)}{n}$: it is a $2\pi$-periodic function converging to $\frac{\pi-x}{2}$ in $L_2\left([0,2\pi]\right)$.
Since:
$$\frac{d f_N(x)}{dx}= \frac{\cos\left(\frac{N+1}{2}x\right)\sin\left(\frac{N}{2}x\right)}{\sin\left(\frac{x}{2}\right)},$$
we know that $f_N(x)$ has $2n$ stationary points in $[0,2\pi]$, local maxima in $x=\frac{(2k+1)\pi}{N+1}$, the first one occurring in $x_N=\frac{\pi}{N+1}$. Once we prove that the value of $f_N(x)$ in any other local maximum is less than $f_N(x_N)$, and that $f_N(x_N)$ is an increasing sequence (I still must find a convincing proof of this two facts, but they look not too hard to deal with and strongly supported by computer inspection) the best bound we can hope in is:
$$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\lim_{N\to +\infty}\sum_{n=1}^{N}\frac{\sin\left(\frac{\pi x}{N+1}\right)}{n},$$
where the RHS a Riemann sum associated with:
$$\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots<\frac{13}{7},$$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| The complex roots of $x^5-1$ are:
$$
\begin{align}
x_1&=1\\
x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\
x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\
x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\
x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5
\end{align}
$$
using Vieta's formulas you get
$$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$
which yields your first equation.
From now on, let $\varphi=\frac{\pi}5$ (for brevity).
We know that
$\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e.
$$2\cos^2\varphi-\cos\varphi=\frac12.$$
Now from $\cos2\varphi=\cos\varphi-\frac12$ you get
$$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$
(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Factorize polynomial over $GF(3)$ I want to factorize $x^{11}-1$ over $GF(3)$ but I'm stuck at
$(x-1)(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$
I have tried to do it trial and error but failed. Is
$$
x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
$$
already irreducible over $GF(3)$?
| The polynomial
$$
\phi_{11}(x):=\frac{x^{11}-1}{x-1}=\sum_{i=0}^{10}x^i\in GF(3)[x]
$$
is not irreducible. Because $11\mid 3^5-1=242$, there is a primitive eleventh root of unity in the field $GF(3^5)$. This is a quintic extension field of $GF(3)$, so the minimal polynomials of all the primitive eleventh roots of unity are of degree $5$. Thus $\phi_{11}(x)$ must be a product of two irreducible quintic factors.
Finding the factors takes a bit of work. We repeatedly use the fact that, if $\beta$ is a root of a polynomial $f(x)\in GF(3)[x]$, then so
is its image $F(\beta)=\beta^3$ under the Frobenius automorphism.
Let $\alpha$ be one of the roots of $\phi_{11}(x)$.
It has conjugates $F(\alpha)=\alpha^3$, $F(\alpha^3)=\alpha^9$, $F(\alpha^9)=\alpha^{27}=\alpha^5$ (here we use the fact that $\alpha^{11}=1$, so as $27=2\cdot11+5$ we get $\alpha^{27}=\alpha^5$) and $F(\alpha^5)=\alpha^{15}=\alpha^4$ (same trick). We can stop here as $F(\alpha^4)=\alpha^{12}=\alpha$, so we won't get any more conjugates ($F$ generates the relevant Galois group). Therefore the minimal polynomial of $\alpha$ is
$$
m_{\alpha}(x)=(x-\alpha)(x-\alpha^3)(x-\alpha^9)(x-\alpha^5)(x-\alpha^4)=x^5+ax^4+cx^3+bx^2+dx+e\in GF(3)[x].
$$
The other eleventh roots of unity are the reciprocals of these, so
$$
m_{1/\alpha}(x)=(x-\alpha^{10})(x-\alpha^8)(x-\alpha^2)(x-\alpha^6)(x-\alpha^7),
$$
and
$$
\phi_{11}(x)=m_{\alpha}(x)m_{1/\alpha}(x)
$$
is the desired factorization.
Let us next compute the coefficient $e=(-1)^5\alpha\cdot\alpha^3\cdot\alpha^9\cdot\alpha^5\cdot\alpha^4=-\alpha^{22}=-1$. To make further progress, we use the fact that the roots of $m_{1/\alpha}(x)$ are the reciprocals of the roots of $m_{\alpha}(x)$. Thus the polynomial $x^5m_{\alpha}(1/x)=ex^5+dx^4+cx^3+bx^2+ax+1$ has the same roots (with the same multiplicities) as $m_{1/\alpha}(x)$, so they must be scalar multiples of each other.
Taking into account the known constant term $e=-1$ we get
$$
m_{1/\alpha}(x)=x^5-dx^4-cx^3-bx^2-ax-1.
$$
There are four unknowns $a,b,c,d\in GF(3)$. We get a bunch of equations tying these together by expanding $m_{\alpha}(x)m_{1/\alpha}(x)=\phi_{11}(x)$. There may be a systematic way of solving the resulting system, but I took the easy way out, and guessed $a=0$ (there are three possible values of $a$, so guessing won't take much time). This leads to a solution $d=-1$, $b=-1$, $c=1$ and to the factorization
$$
\phi_{11}(x)=(x^5-x^3+x^2-x-1)(x^5+x^4-x^3+x^2-1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\cos \cos 1 - \sin \sin \sin 1$ positive? How can I show that
$$\cos \cos 1 - \sin \sin \sin 1$$
is positive?
This is motivated by this question. If
$$\begin{align}
f(x) &= \cos \cos \cos \cos(\pi/2 + ix) - \sin \sin \sin \sin(\pi/2+ix)\\
&= \cos \cos \cos \sinh x - \sin \sin \sin \cosh x,
\end{align}$$
then it looks like $f(x)$ has a zero in the interval $(0,1)$. This would imply that
$$
\cos \cos \cos \cos(z) - \sin \sin \sin \sin(z)
$$
has infinitely-many zeros in the strip $0 < \Im(z) < 1$.
One way to show that such a zero exists would be to show that $f(1) < 0 < f(0)$, the right-side of which is the current question. I don't know how to show the left side either, but now I'm interested in this question for its own sake.
| Recall that
$$
1-\frac{1}{2}x^2 \;<\; \cos x \;<\; 1-\frac{1}{2}x^2 + \frac{1}{24}x^4
$$
for $x>0$. From the upper bound, we see that $\cos 1 < 13/24$. Since $\cos$ is decreasing, it follows that
$$
\cos(\cos 1) \;>\; \cos\left(\frac{13}{24}\right) \;>\; 1-\frac{1}{2}\left(\frac{13}{24}\right)^2 \;>\; 0.85.
$$
Next, recall that
$$
\sin x < x
$$
for all $x>0$. Since $\sin x$ is increasing, this gives us
$$
\sin\sin\sin 1 \;<\; \sin \sin 1 \;<\; \sin 1.
$$
But we also know that
$$
\sin x \;<\; x - \frac{1}{6}x^3 + \frac{1}{120}x^5
$$
for all $x>0$. It follows that
$$
\sin\sin\sin 1 \;<\; \sin 1 \;<\; \frac{101}{120} \;<\; 0.85.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Positive integers expressed in the form : $\frac{a^b+c}{a+c}$ Note : This problem has no specific source .
Is it true that :
Every positive integer $n$ greater than $1$ can be expressed in the form :
$n=\frac{a^b+c}{a+c}$ , where $a,b>1$ , and $c \in \mathbb{Z} \backslash \{0\}$
I am able to prove following :
$\forall a , \exists ~b,c$ such that : $a+c \mid a^b+c$ , where $b>1$
Proof :
$n=\frac{a^b+c}{a+c}=\frac{a^b-a}{a+c}+1$ , therefore it is sufficient to prove :
$a+c \mid a(a^{b-1}-1)$
If $a+c \mid a$ then $b$ can be any positive integer .
If $a+c \not \mid a$ then we have to prove : $a+c \mid a^{b-1}-1$
or , in other words :
$a^{b-1} \equiv 1 \pmod {a+c}$
Now , for every $a$ there is a number $c$ such that : $\gcd(a,a+c)=1$
According to Euler Theorem :
$a^{\varphi(a+c)} \equiv 1 \pmod {a+c}$
Hence , we can write :
$b-1=\varphi(a+c) \Rightarrow b=1+\varphi(a+c)$
Since $\varphi(a+c)\geq 1$ it follows $b>1$
Q.E.D.
But , this is only necessary condition .
| Let $a = n$, $b = 3$, $c = n^2$. This does it.
| {
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Evaluating $\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$ Evaluating $$\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$$
Full Question
Provided Answer
But how do I get from
$$\sum \frac{1}{2^{k-1}+1} - \frac{1}{2^k+1} = \frac{1}{2^{\color{red}0-1}+1}...$$
Why is the summation removed. And why is $k=0$?
In the next line it becomes $n$ again? Then $\lim$ was introduced? Why?
| It is a finite telescoping sum. For any sequence $\rm\{ x_k\}_{k=1}^\infty$ we have
$$\rm \begin{array}{c l} \sum_{k=a}^b (x_k-x_{k+1}) & \rm =(x_a-\color{Maroon}{x_{a+1}})+(\color{Maroon}{x_{a+1}}-x_{a+2})+\cdots+(x_{b-1}-\color{Purple}{x_b})+(\color{Purple}{x_b}-x_{b+1} ) \\ & \rm =x_a-x_{b+1}. \end{array}$$
Notice the repeated cancellation? We could also shift the index back by one, as here:
$$\sum_{k=0}^n\left(\frac{1}{2^{k-1}+1}-\frac{1}{2^k+1}\right)=\frac{1}{2^{0-1}+1}-\frac{1}{2^n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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inequality-positive semidefinite matrices Let $A$ and $B$ be positive semidefinite matrices of the same size. If the largest eigenvalues of $A$ and $B$ are less than equal to $1$. probe that: $$AB+BA\geq- \frac{1}{4}I$$
In the hint, it says use the fact: $0\leq (A+B-\frac{1}{2}I)^{2}$.
In order for $ (A+B-\frac{1}{2}I)^{2}$ to be positive semidefinite, the matrix $ (A+B-\frac{1}{2}I)$ has to be positive semidefinite. The hint given above confused me because I know that $A+B$ is definitely positive semidefinite since it is the sum of two positive semidefinite matrices, but we can't be sure that $A+B-\frac{1}{2}I$ is positive semidefinite. Does anyone know how to prove that: $0\leq (A+B-\frac{1}{2}I)^{2}$?
Now assume the given hint is true, then: $0\leq A\leq I$ and $0\leq B\leq I$. It follows that:$$0\leq (A+B-\frac{1}{2}I)^{2}=A^{2}+B^{2}+\frac{1}{4}I+A(B-I)+B(A-I)\leq A+B+\frac{1}{4}I+A(B-I)+B(A-I)$$
I need to prove that:$A(B-I)+B(A-I)\leq 0$. This will be true if the product of a positive semidefinite matrix and a negative semidefinite matrix is a negative semidefinite matrix. Is this true?
| you have $0 \le (A + B - \frac 12 I)^2$ as the left hand side is a square:
\[ \textstyle
\langle (A + B - \frac 12 I)^2x, x\rangle = \langle (A+B-\frac 12I)x, (A+B-\frac 12I)^*x\rangle =\langle (A+B-\frac 12I)x, (A+B-\frac 12I)x\rangle = \|(A + B -\frac 12I)x\|^2 \ge 0
\]
And for the second part we have $A^2 \le A$ as you wrote. I think you perhaps didn't remember you wanted to have $AB + BA + \frac 14I \ge 0$, for
\begin{align*}
\left(A + B - \frac 12\right)^2 &= A^2 + B^2 + \frac 14 I - A - B + AB + BA\\\
&\le A + B - A - B + A B + BA + \frac 14I\\\
&= AB + BA + \frac 14I.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$
Prove that for all integers $n$, $n \geq 1$,
$$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$
How would I go about proving this?
| $$\begin{eqnarray*} (1 + 3 + 5 + ... + (2n - 1)) &=& (1 + 2 + 3 + ... + 2n) - (2 + 4 + ... + 2n)\\
&=& \sum_{i=1}^{2n}{i} - 2\sum_{i=1}^{n}{i}\\
&=& \frac{(2n)(2n+1)}{2} - 2\left(\frac{n(n+1)}{2}\right)\\
&=& n(2n+1) - n(n+1)\\
&=& n(2n + 1 - n - 1) = n^{2}\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 9,
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Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$ Somewhere in the provided answer:
$$\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}$$
How did they get that? What I have:
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}$$
$$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}{\frac{x}{\sqrt{2}}}$$
So I have an extra $\frac{1}{\sqrt{2}}$ ... I probably had some stupid mistakes?
| You made a mistake in the last step. To see why, let $u = \frac{x}{\sqrt{2}}$, $du = \frac{dx}{\sqrt{2}}$.
\begin{align*}
\frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}} &= \frac{1}{\sqrt{2}} \int \frac{\sqrt{2}du}{\sqrt{1-u^2}} \\
&= \int \frac{du}{\sqrt{1-u^2}} \\
&= \arcsin{u} + c \\
&= \arcsin\left({\frac{x}{\sqrt{2}}}\right) + c
\end{align*}
Basically, you made an implicit variable substitution, but forgot that $dx$ also changes when you change the variable.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction Proof and divisibility by $2^n$ I'm trying to use induction to prove that for every integer $n > 0$ there exists an $n$-digit integer A(n) that is divisible by $2^n$ and that consists entirely of digits “1” and “2”.
Does anyone have a clue where to start ?
I found examples that work on small numbers and the initial step of the induction proof is not an issue. I need to prove it for $n+1$ now.
Thanks for any advise !
| Define the $n$-digit number as follows. Let $f(1)=2$. Suppose that we have defined $f(n)$, and $2^n$ divides $f(n)$. Then define $f(n+1)$ as follows.
If $2^{n+1}$ divides $f(n)$, then $f(n+1)$ is obtained by putting a $2$ in front of the decimal expansion of $f(n)$. Or, to put it another way, $f(n+1)=2\cdot 10^n+f(n)$. If $2^{n+1}$ does not divide $f(n)$, then $f(n+1)$ is obtained by putting a $1$ in front of the decimal expansion of $f(n)$, that is, $f(n+1)=10^n+f(n)$.
We show that $2^{n+1}$ divides $f(n+1)$.
Suppose first that $2^{n+1}$ divides $f(n)$. Then $f(n+1)=2\cdot 10^n+f(n)$. Note that $2\cdot 10^n$ is divisible by $2^{n+1}$, which shows that $f(n+1)$ is divisible by $2^{n+1}$.
Suppose next that $2^{n+1}$ does not divide $f(n)$. Then $f(n)\equiv 2^n\pmod{2^{n+1}}$ (the remainder when we divide $10^n$ by $2^{n+1}$ is $2^n$). But we have $10^n \equiv 2^n \pmod{2^{n+1}}$. It follows that $f(n+1)=10^n+f(n)\equiv 2^n+2^n\equiv 2^{n+1}\equiv 0\pmod{2^{n+1}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of the series $\sum \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)}$ I need to determine whether the following series converges or diverges:
$$ \sum_{n=1}^\infty \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)} $$
My first idea was to show that $a_n$ tends to zero, and so it does.
My next plan was to rewrite the sum as
$$ a_n = \frac{1}{1+n\sqrt{n}} - \frac{1}{(n+1)\left( 1 + n\sqrt{n} \right)} $$
Where the firt one can be compared to a p-series, and the second one is smaller than the first one so it also converges. Is this correct? Is there an easier way to show that the series converge?
| For deciding convergence, Siminore's comparison test answers affirmatively.
If for real $x>0$ we wish to approximate the sum over $\mathbb{Z}^+$ of
$$
\eqalign{
f(x)
&=\frac{x}{(x+1)(x^{3/2}+1)}
=\left(1-\frac1{1+x}\right)\left(1+x^{3/2}\right)^{-1}\\
&=\frac1{1+x^{3/2}}
-\frac1{\left(1+x\right)\left(1+x^{3/2}\right)}
=\frac1{\left(1+x^{-1}\right)\left(1+x^{3/2}\right)}
}
$$
then its antiderivative is
$$
F(x)
= \arctan\left(\sqrt{x}\right)
+ \frac16 \, \log
\frac{
\left(x - \sqrt{x} + 1\right)^4
}{
\left(x + 1\right)^3
\left(\sqrt{x} + 1\right)^2
}
$$
and a rough lower bound on the sum woud be
$$
S = \sum_{x=1}^\infty f(x) >
I = \int_1^\infty f(x)\,dx
= \frac{\pi}4 + \frac56 \, \log2
\approx 1.363021
$$
since $f(x)$ is strictly decreasing for $x > 0.73173541$
since its global maximum is at the root $t\approx0.85541534$ of
the denominator $3t^5+t^3-2$ (for $t=\sqrt x$) of
$$
f'(x)=
-\frac{ 3x^{5/2}+x^{3/2}-2 }{
2\,\left(x+1\right)^2
\left(x^3+2x^{3/2}+1\right)}
\,.
$$
We can think of our sum $S$ as a left-endpoint (improper) Riemann sum.
A correspondingly rough upper bound would be $f(1)+I=\frac14+I\approx1.613021$.
A better approximation would be the average of these two:
$$
S \approx \frac18 + \frac{\pi}4 + \frac56 \, \log2 \approx 1.488021 \,.
$$
The antiderivative $F$ above can be derived by setting
$x=t^2$ (so $dx=2t\,dt$) and
$$
g(t)=f(t^2)=\frac{t^2}{(t^2+1)(t^3+1)}
$$
so that
$$
\eqalign{
t\,g(t)&
=\tfrac13\frac{2t-1}{t^2-t+1}
-\tfrac12\frac{t-1}{t^2+1}
-\tfrac16\frac1{t+1}
\\&
=\tfrac13\frac{2t-1}{t^2-t+1}
-\tfrac14\frac{2t}{t^2+1}
-\tfrac16\frac1{t+1}
+\tfrac14\frac1{t^2+1}
}
$$
and, for $t>0$ (to dispense with absolute values inside the natural logarithm),
$$
\eqalign{
F(x)&
= \int f(x)\,dx
=2\int t\,g(t)\,dt
\\&
= \frac23\ln\left(t^2-t+1\right)
- \frac12\ln\left(t^2 +1\right)
- \frac13\ln\left(t +1\right)
+ \frac12\arctan{t}
}
$$
which yields the above in $x$.
| {
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"url": "https://math.stackexchange.com/questions/139875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maclaurin Series
Integrate using first three terms of appropriate series...
$$\int_0^1 \sin x ~dx.$$
So I use
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}$$
for my three terms and if I integrate just that I get the answer which is $.3103$.
However the solution book is showing a negative is taken outside the integral then just sorta disappears.
Am I missing something here or is this a typo? (What follows is what is in the book):
$$\begin{align*}
\int_0^1\sin x^2\,dx &= \int_0^1\left(x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!}\right)\,dx\\
&= -\int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx\\
&= \left.\left(\frac{1}{3}x^3 - \frac{x^7}{42} + \frac{x^{11}}{1320}\right)\right|_0^1\\
&= \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} = 0.3103.
\end{align*}$$
| Yes, the minus sign is a typo and should not be there.
Note as well that you don't actually have equalities. Since
$$\sin x\approx x - \frac{x^3}{3!} + \frac{x^5}{5!},$$
and we don't have actual equality, we have that
$$\int_0^1\sin(x^2)\,dx \approx \int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx.$$
Likewise, in the final line we actually have
$$\frac{1}{3} - \frac{1}{42} + \frac{1}{1320}\approx 0.3103,$$
not equality.
| {
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"answer_count": 1,
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} |
Is my Riemann Sum correct? For the integral: $$\int_{-1}^{5} \left( x^{2} -4 \right) dx$$
My calculations:
$$\begin{align*}\Delta x &= \frac6n\\\\
x_i &= -1 + \frac{6i}n\\\\
f(x_i) &= 1 + \frac{36i^2}{n^2} -4\\\\
A&=72
\end{align*}$$
I'm unsure if this is correct as it is my first attempt at doing this type of problem.
| Your $\Delta x$ and $x_i$ are fine, but you've gone astray after that: $$f(x_i)=x_i^2=\left(-1+\frac{6i}n\right)^2=1-\frac{12i}n+\frac{36i^2}{n^2}\;.$$ Your Riemann sum is then
$$\begin{align*}
\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\frac6n&=\frac6n\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\\
&=\frac6n\left(\sum_{i=1}^n1-\frac{12}n\sum_{i=1}^ni+\frac{36}{n^2}\sum_{i=1}^ni^2\right)\\
&=\frac6n\left(n-\frac{12}n\cdot\frac{n(n+1)}2+\frac{36}{n^2}\cdot\frac{n(n+1)(2n+1}6\right)\\
&=\frac6n\left(n-6(n+1)+\frac{6(n+1)(2n+1)}n\right)\\
&=\frac6n\cdot\frac{-5n^2-6n+12n^2+18n+6}n\\
&=6\left(7+\frac{12}n+\frac6{n^2}\right)\;,
\end{align*}$$
which converges to $42$ as $n\to\infty$.
| {
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} |
Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\frac{-1}{2}=\\
&\frac{\sqrt{x^2-1}}{x\sqrt{x^2-1}} + \frac{x^2(x^2-1)^\frac{-1}{2}}{\sqrt{x^2-1}}=\\
&\frac{1}{x} + \frac{x^2}{x^2-1}= \frac{x^2 - 1 +x^3}{x^3 - x}
\end{align}$$
From here I am not sure what to do. This is not the right answer and I do not know what to do.
| $\log (x\sqrt{x^{2}-1})=\log x+\log(\sqrt{x^{2}-1})=\log x+ \frac{1}{2}\log({x^{2}-1})$
Let $y=\log x+ \frac{1}{2}\log({x^{2}-1}).$
Now differentiate with respect to $x$, we get,
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{2}\frac{1}{x^{2}-1}2x$
$\therefore \frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^{2}-1}$=$\frac{x^{2}-1+x^{2}}{x(x^{2}-1)}$=$\frac{2x^{2}-1}{x(x^{2}-1)}.$
| {
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What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions? We have the following evaluations:
$$\begin{aligned}
&\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\
&\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\
&\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right)
\\&\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\
&\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = \,?\\ \end{aligned}$$
The paper Borwein and Bradley's Apery-like Formula for $\zeta(4n+3)$ gives the 3rd and 5th in terms of the Dirichlet L-functions, but does anyone know how to evaluate the 5th one in terms of the Hurwitz zeta function $\zeta(s,a)$?
Postscript: (A few hours later)
After Anon gave his answer, I did a little more sleuthing and found the case p = 7 in the Mathworld article on central binomial coefficients (which also had p = 5). The paper I cited was a bit old (1999) and the authors weren’t aware it was already found a year earlier by Plouffe. Hence,
$$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n^5 \, \binom{2n}n} = -\frac{19}{3}\zeta(5)+\frac{2}{3}\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot3^2}\left(\zeta(4,\tfrac{1}{3})-\zeta(4, \tfrac{2}{3}) \right)\\
&\sum_{n=1}^\infty \frac{1}{n^7 \, \binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6, \tfrac{2}{3}) \right)\\
\end{aligned}$$
With this “pattern”, I used an integer relations algorithm to try to find p = 9, 11, 13. No luck so far.
| The paper gives (second-to-last page, right-hand column)
$$\sum_{n\ge1}\frac{1}{n^5\binom{2n}{n}}=\frac{9\sqrt{3}\pi}{8}\color{Purple}{L\left(4,\left(-3\atop\circ\right)\right)}+\frac{\pi^2\zeta(3)}{9}-\frac{19\zeta(5)}{3}. \tag{1}$$
You seek a way to write the $L$-function in purple as a linear combination of Hurwitz $\zeta$ functions.
More generally, let $\chi$ be a Dirichlet character of modulus (period) $m$, and define a 'Kronecker' delta
$$\delta_m(k)=\begin{cases}1 & k\equiv0\bmod m \\ 0 & k\not\equiv 0\bmod m\end{cases}. \tag{2}$$
Notice then that $\delta_m(a-b)$ is $1$ if and only if $a\equiv b\bmod m$. We can therefore decompose $\chi$ as
$$\chi(n)=\sum_{k=0}^{m-1} \chi(k) \delta_m(n-k). \tag{3}$$
Furthermore, the Hurwitz zeta function at $a/m\in[0,1)$ decomposes as
$$\begin{array}{c l} \zeta\left(s,\frac{a}{m}\right) & =\sum_{n=1}^\infty\frac{1}{(n+a/m)^s} \\
& =m^s\sum_{n=1}^\infty\frac{1}{(mn+a)^s} \\
& =m^s\sum_{n\ge1} \frac{\delta_m(n-a)}{n^s}.\end{array} \tag{4}$$
Therefore, we have
$$\begin{array}{c l} L(s,\chi) & =\sum_{n\ge1}\frac{\chi(n)}{n^s} \\
& =\sum_{n\ge1}\frac{1}{n^s}\sum_{k=0}^{m-1}\chi(k)\delta_m(n-k) \\
& =\sum_{k=0}^{m-1}\chi(k)\sum_{n\ge1}\frac{\delta_m(n-k)}{n^s} \\
& =\frac{1}{m^s}\sum_{k=0}^{m-1}\chi(k)\zeta\left(s,\frac{k}{m}\right). \end{array} \tag{5}$$
This formula is listed on Wikipedia's Hurwitz $\zeta$ and Dirichlet $L$-function articles. In particular,
$$L\left(4,\left(\frac{-3}{\circ}\right)\right)=\frac{\zeta\left(4,\frac{1}{3}\right)-\zeta\left(4,\frac{2}{3}\right)}{81} \tag{6}$$
because $\left(\frac{-3}{1}\right)=1$ and $\left(\frac{-3}{2}\right)=-1$ (and $\chi(0)=0$ for all Dirichlet characters). Also see here.
| {
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"question_score": "30",
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How did they get this result? Please, explain these computations:
1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{3}}{2} = \cos x$
How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?
2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{2}}{2} = \cos x$
How did we get $ \frac{\sqrt{2}}{2}$ from $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 $?
| You have:
*
*$$\cos^2 x = -(\frac{1}{2})^2 + 1$$
$$\cos^2 x = -\frac{1}{4} + 1=\frac{3}{4}$$
Which is:
$$\sqrt \cos^2 x = \sqrt \frac{3}{4}$$
$$\cos x = \pm \frac{\sqrt 3}{2}$$
Same works for the second one.
| {
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Are there any integer solutions to $x^2 - (n^2 - 2)y^2 = -1$? I was just wondering if there are any integer solutions to the Diophantine equation:
$x^2 - (n^2 - 2)y^2 = -1 \ \ $ for $n > 2$
I don't think there are any but can't prove why.
| Note that if $d$ is divisible by a prime $p$ of the form $4k+3$, then the equation $x^2-dy^2\equiv -1$ cannot have a solution, for $x^2\equiv -1\pmod{p}$ does not have a solution.
If $n>2$ is odd, then $n^2-2\equiv -1\pmod{4}$, so $n^2-2$ is divisible by a prime of the form $4k+3$.
If $n$ is divisible by $4$, then again $n^2-2$ is divisible by a prime of the form $4k+3$. But this leaves the possibility $n\equiv 2\pmod{4}$, where $n^2-2$ need not have a prime divisor of the form $4k+3$.
Remark: Will Jagy has settled the problem in general, by observing that the continued fraction of $\sqrt{n^2-2}$ has period $4$. (If $\sqrt{d}$ has continued fraction with even period, then the equation $x^2-dy^2=-1$ has no integer solutions.)
There is an approach that does not use properties of continued fractions, but instead uses basic properties of Pell equations. Note that $x=n^2-1$, $y=n$ is a solution of the Pell equation $x^2-(n^2-2)y^2=1$. If there were solutions of $x^2-(n^2-2)y^2=-1$, there would be a fundamental solution $(a_0,b_0)$, and $(n^2-1,n)$ would be an "even power" of $(a_0,b_0)$, in the sense that
$n^2-1+n\sqrt{n^2-2}=(a_0+b_0\sqrt{n^2-2})^{2k}$ for some positive integer $k$. This is not possible, for if $(a_0+b_0\sqrt{n^2-2})^{2k}=a+b\sqrt{n^2-2}$, then $a \ge n^2-1$, and we cannot have equality.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge? $$\int_0^\infty x/(1+x^2\sin^2x) \mathrm dx$$
I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot.
| Since $-1 \leq \sin{x} \leq 1$, you have $\sin^{2}{x} \leq 1$, and from this you can conclude that
\begin{align*}
1+x^{2}\sin^{2}{x} &\leq 1+2x^{2} \\\ \Longrightarrow \frac{1}{1+x^{2}\sin^{2}{x}} &\geq \frac{1}{1+2x^{2}} \\\ \Longrightarrow \frac{x}{1+x^{2}\sin^{2}{x}}&\geq \frac{x}{1+2x^{2}} \\\
\Longrightarrow \int_{0}^{\infty} \frac{x}{1+x^{2}\sin^{2}{x}} \ dx &\geq \int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx
\end{align*}
Now
\begin{align*}
\int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx &= \frac{1}{4} \cdot \int_{0}^{\infty} \frac{4x}{1+2x^{2}} \ dx \\\ &= \frac{1}{4} \bigl(\log{x}\bigr)_{1}^{\infty} \to \infty
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Diophantine equation $x^y-y^x=11$ How can one find all integer solutions to $x^y-y^x=k$, for a given k?
Example case $x^y-y^x=11$
| The numbers $x^y - y^x$ blow up when $x$ and $y$ get large. Besides the trivial solutions of the form $(k+1, 1)$ the only values of $k < 1000$ giving solutions are:
$$\begin{align} 3^2 - 2^3 &= 1 \\ 2^5 - 5^2 &= 7 \\ 3^4 - 4^3 &= 17 \\ 2^6 - 6^2 &= 28 \\ 2^7 - 7^2 &= 79 \\ 3^5 - 5^3 &= 118 \\ 2^8 - 8^2 &= 192 \\ 4^5 - 5^4 &= 399 \\ 2^9 - 9^2 &= 431 \\ 3^6 - 6^3 &= 513 \\ 2^{10} - 10^2 &= 924 \end{align}$$
For a small theoretical analysis, suppose $x < y$ (if $x > y$, the result is negative). Then $y^x = x^x (1 + \frac{y - x}{x})^x < x^x e^{y-x}$, so $$x^y - y^x \geq x^x (x^{y-x} - e^{y-x}).$$ So for $x \geq 3$ this blows up enormously. So for small $k$, only small values of $x$ and $y$ are possible.
| {
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Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5, 7 $
Please help.
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ and $f'(x) = x^3+1$. Hence, $f^′(x)= 0$ for $x=2$. Therefore in an algebraically closed field of characteristic $3$, $f(x)$ has multiple roots.
| Here's an elementary answer:
a multiple root of $x^4 + x + 6$ in $\mathbf F_p$ is the exact same thing as a root common to $x^4 + x + 6$ and its derivative $4x^3 + 1$. But, on $\mathbf Z$, we have the following formula:
$$4(x^4 + x + 6) = (4x^3 + 1) x + (3x + 24).$$
So, for $p \neq 2$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $3\alpha + 24 = 0$.
So, for $p \neq 2, 3$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $\alpha + 8 = 0$.
So the $p\neq 2, 3$ answering the questions are exactly the prime factors ($\neq 2, 3$) of $f(-8) = 4094 = 2\cdot 23\cdot 89$ (and, for those $p$, the multiple root is $-8$).
You have to check the two remaining cases by hand. The factorisations into prime factors of $f$ on $\mathbf F_2$ and $\mathbf F_3$ are $x(x+1)(x^2+x+1)$ and $x(x+1)^3$, respectively.
So the prime numbers answering the question are 3, 23 and 89 (and the multiple roots are $-1$ (triple), $-8$ and $-8$, respectively).
| {
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"question_score": "4",
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Calculate $\ln(2)$ using Riemann sum.
Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?
Show that
$$\ln(2) = \lim_{n\rightarrow\infty}\left( \frac{1}{n + 1} + \frac{1}{n + 2} + ... +
\frac{1}{2n}\right)$$
by considering the lower Riemann sum of $f$ where $f(x) = \frac{1}{x}$ over $[1, 2]$
I was confused looking at the equality to begin with, since taking $n \rightarrow \infty$ for all of those terms would become $0$ right?
Anyway, I attempted it regardless.
$$\sum_{k=1}^n \frac{1}{n}(f(1 + \frac{k}{n}))$$
$$= \sum_{k=1}^n \frac{1}{n}(\frac{1}{1+ \frac{k}{n}})$$
$$=\sum_{k=1}^n \frac{1}{n + k} = $$ the sum from the question?
I wasn't sure what to do from here. I tried something else though:
$$=\frac{1}{n + \frac{n(n+1)}{2}}$$
$$=\frac{2}{n^2 + 3n}$$ which seemed equally useless if I'm taking $n \rightarrow \infty$ as it all becomes $0$.
| METHOD I
We may recall the celebre limit that yields Euler-Mascheroni constant, namely:
$$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$
Then everything boils down to:
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$
METHOD II
Use one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:
$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$
Taking $k=n,n+1,...,2n$ values to the inequality and then summing all relations, we get all we need in order to apply Squeeze theorem.
METHOD III
We may use Botez-Catalan identity and immediately get that:
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{2n+1}\frac{1}{2n}= $$
$$\lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{n+1}\frac{1}{n}=\ln{2}.$$
The last series' limit is obtained by using Taylor expansion of $\ln(x+1)$ and take $x=1$
The proofs are complete.
| {
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Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
| WLOG assume $a \leq b$, therefore, $a^2 \leq b^2$ and $\frac{1}{b - 1} \leq \frac{1}{a - 1}$. Therefore, by Rearrangement Inequality,
$$\frac{a^2}{b-1} + \frac{b^2}{a - 1} \geq \frac{a^2}{a-1} + \frac{b^2}{b - 1}$$
From here, one can simply follow Valentin's steps,
$$\frac{a^2}{a - 1} + \frac{b^2}{b - 1} = 4 + (a - 1) + \frac{1}{a - 1} + (b - 1) + \frac{1}{b - 1} \geq 4 + 2 + 2 = 8$$
===================
As Chris points out, $(a - 2)^2 \geq 0$ implies $a^2 - 4a + 4 \geq 0$ or $a^2 \geq 4(a - 1)$ and hence $\frac{a^2}{a - 1} \geq 4$. Thereby, cutting short few more steps.
| {
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Limit of $ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}},m\rightarrow\infty$ I would like to show that $$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{2}$$
Using integrals:
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}} \leq m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}}+\frac{m}{(m+1)\sqrt{2m+1}}$$
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} = \frac{\pi}{2}-\arctan \left(1+\frac{1}{m} \right)=\frac{\pi}{4}+o(1)$$
The result I get is:
$$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{4}$$
Where did I go wrong?
| Use this and think about
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}} \leq m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}}+\frac{m}{(m+1)\sqrt{2m+1}}$$
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} = \frac{\pi}{2}-\arctan \left(1+\frac{1}{m} \right)=\frac{\pi}{4}+o(1)$$
| {
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Finding the limit I need to find the limit of this problem. I pretty much know you have to multiply by the conjugate but I get lost after I do that.
$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$$
| $$\begin{eqnarray}
-\lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{1-x} &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\\
&=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\frac{1/\sqrt{x}+1}{1/\sqrt{x}+1}\\
&=& \lim\limits_{x\to 1} \frac{1/x - 1}{(x-1)(1/\sqrt{x}+1)}\\
&=& \lim\limits_{x\to 1} \frac{1}{1/\sqrt{x}+1}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}\\
&=& \frac{1}{2}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}
\end{eqnarray}$$
This limit can be evaluated by noting that substituting $1/x$ for $x$ gives one over the limit, but should give the same value since $x\to 1$ is the same as $1/x\to 1$, thus the limit $L$ satisfies $L=1/L$, so $L=1$. This gives us a final answer of $-1/2$.
| {
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Calculate: $\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$ Calculate the following sum for integers $n\ge2$:
$$\sum_{k=0}^{n-2} 2^{k} \tan \left(\frac{\pi}{2^{n-k}}\right)$$
I'm trying to obtain a closed form if that is possible.
| We have this nice identity $$\tan(\theta) = \cot(\theta)-2 \cot(2 \theta)$$
Making use of this, and denoting $\displaystyle \sum_{k=0}^{m} 2^k \tan(2^k \theta)$ as $S$, we get that \begin{align}S & = \tan(\theta) + 2 \tan(2 \theta) + 4 \tan(4 \theta) + \cdots + 2^m \tan(2^m \theta)\\
& = \cot(\theta) -2 \cot(2 \theta) +2 \cot(2\theta) - \cdots + 2^m \cot(2^m \theta) - 2^{m+1} \cot(2^{m+1} \theta)\\
& = \cot(\theta) - 2^{m+1} \cot(2^{m+1} \theta)
\end{align}
In your case, $\theta = \dfrac{\pi}{2^n}$ and $m= n-2$. Hence, we get the sum to be $$S = \cot \left(\dfrac{\pi}{2^n} \right) - 2^{n-1} \cot \left( 2^{n-1} \cdot \dfrac{\pi}{2^n}\right) = \cot \left(\dfrac{\pi}{2^n} \right) - 2^{n-1} \cot \left( \dfrac{\pi}2\right) = \cot \left(\dfrac{\pi}{2^n} \right)$$
Proof for $\tan(\theta) = \cot(\theta)-2 \cot(2 \theta)$
$$\cot(\theta) - \tan(\theta) = \dfrac{\cos(\theta)}{\sin(\theta)} - \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{\cos^2(\theta) - \sin^2(\theta)}{\sin(\theta) \cos(\theta)}= 2\dfrac{\cos(2\theta)}{\sin(2 \theta)} = 2 \cot(2 \theta)$$
| {
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Integral as a limit of a sum How do I get
$$
\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$
as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
| $$\int _a^b {f(t) dt}=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n f(\frac{k(b-a)}{n}+a)$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{1}\sum \limits_{k=1}^n \frac{1}{k(b-a)+na}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{(b-a)}\sum \limits_{k=1}^n \frac{1}{k+\frac{na}{b-a}}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{(b/a)-1}}$$
$b/a=x$
$$f(x-1)=\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x-1}}$$
$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$
$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$
$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3} (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$
$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation
After solving limits. We get:
$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$
$$f(x-1)=\ln(x)$$
$b/a=x$
$$\int _a^b \frac{1}{t} dt=f(x-1)=\ln(x)=\ln(b/a)$$
| {
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Homogenous polynomials In section 3.1 (3rd paragraph on page 4) in this paper, I cannot understand why $Q$ and $R$ are homogeneous:
(Given $A$, $B$ are homogenous. Capital letters denote homogenous polynmials.)
| The trick is to realize that all terms of $q$ (or $r$) are homogeneous of the same degree BUT you have to introduce a grading with negative degrees in $K(y,z)$.
For example, if you want to divide $yz^2+x^3$ by $xz+y^2$ you first use the euclidean algorithm to obtain $$ yz^2+x^3= (\frac {y}{x}z -\frac {y^3}{x^2}) \cdot (xz+y^2) + (x^3+ \frac {y^5}{x^2} ) $$
Notice that every term of $r=x^3+ \frac {y^5}{x^2}$ above has degree $3$, but that $\frac {y^5}{x^2}$ has degree $3=5-2$ because $\frac {1}{x^2}$ has negative degree $-2$ . Similarly every term of $q=\frac {y}{x}z -\frac {y^3}{x^2}$ has degree $1$.
Getting rid of denominators will just multiply all terms of $q$ and $r$ by a homogeneous polynomial $H$ of degree $d$ and will thus conserve homogeneity of $q$ and $r$ .
In our example you multiply throughout by $x^2$ which has degree $2$ and obtain the final equality
$$ x^2yz^2+x^5= (xyz-y^3) \cdot (xz+y^2) + (x^5+ y^5 ) $$
where $Q=xyz -y^3$ is homogeneous of degree $1+2=3$ and $R=x^5+ y^5 $ is homogeneous of degree $3+2=5$, as required.
| {
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Factoring by grouping: $x^4 - y^4 -4x^2 + 4$ Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms.
Thank you.
| $$x^4-y^4-4x^2+4=(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$
| {
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Simplify $\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}$ Simplify $\displaystyle{\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}}$.
I get this when I was doing another Q,
but I don't know how to further simplify it.
Can anyone help me, please?
| Hint: Let's note $o:=\frac {1+\sqrt{5}}2$, $a:=\sqrt{10-2\sqrt{5}}$ and $b:=\sqrt{5+2\sqrt{5}}$
then $ab=\sqrt{30+10\sqrt{5}}=5+\sqrt{5}$
Compute $(o\cdot a+b)^2$ to conclude.
| {
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Computing large powers $\!\bmod n$ when $n$ has multiple prime factors. How will the congruence modulo works for large exponents? What theorem/s may be used? For example to show that $7^{82}$ is congruent to $9 \pmod {40}$.
| Maybe just start calculating. Note that $7^2=49\equiv 9\pmod{40}$. So $7^4\equiv 81\equiv 1\pmod{40}$. We got lucky.
It follows that $7^{80}=(7^{16})^5\equiv 1\pmod{40}$. Thus
$7^{82}=7^{80}7^2\equiv 49\equiv 9\pmod{40}$.
In general, repeated squaring, and reduction with respect to our modulus (in this case $40$) gets us to high powers fairly quickly.
A more elaborate way is to use Euler's Theorem. If $a$ is relatively prime to $m$, then $a^{\varphi(m)}\equiv 1\pmod{m}.$
Here $\varphi$ is the Euler $\phi$-function. We have $\varphi(40)=16$, so taking $a=7$ we have $a^{16}\equiv 1 \pmod{40}$. It follows that $a^{80}=(a^{16})^5\equiv 1\pmod{40}$. So $a^{82}\equiv a^2=49\pmod{40}$. But $49\equiv 9\pmod{40}$.
Still another way is to factor $40$ as $2^3\cdot 5$. We then work separately modulo $8$ and modulo $5$.
First modulo $8$: We have $7\equiv -1\pmod{8}$, so $7^{82}\equiv (-1)^{82}\equiv 1\pmod{8}$.
Next modulo $5$: We have $7\equiv 2\pmod{5}$, so $7^2\equiv 4\equiv -1\pmod{5}$, and therefore $7^4\equiv 1\pmod{5}$. (We could also get this directly by using Fermat's Theorem.) It follows that $7^{80}\equiv 1\pmod{5}$, and therefore $7^{82}\equiv 4\pmod 5$.
Now we are looking for the numbers that are congruent to $1$ modulo $8$ and to $4$ modulo $5$. For bigger numbers, we can use the Chinese Remainder Theorem. But finding a number congruent to $1$ modulo $8$ and to $4$ modulo $5$ can also be done without much machinery.
In general, the technique to use depends very much on the modulus $m$. If it is huge, and we do not know its prime factorization, then the Binary Method of exponentiation is quite efficient.
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What's $T\left(n\right)$? If $T\left( n \right) = 8T\left( n-1 \right) - 15T\left( n-2 \right); T\left(1\right) = 1; T\left( 2 \right) = 4$,
What's $T\left(n\right)$ ?
I use this method:
Let $c(T(n) - aT(n-1)) = T(n-1) - aT(n-2)$
from $T(n) = 8T(n-1) - 15T(n-2)$, we can get $\begin{cases}c = \frac{1}{3}\\a = 5\end{cases}$ and $\begin{cases}c = \frac{1}{5}\\a = 3\end{cases}$,
then, we get $\frac{T(n)-5T(n-1)}{T(n-1)-5T(n-2)} = 3$,
so, we reach the answer: $T(n) = \frac{3^{n-1}+5^{n-1}}{2}$.
Sorry for my poor English.
| This is second order recurrence equation. Since $T(n)=8T(n-1)-15T(n-2)$, we first solve the quadratic equation: $x^2=8x-15$. Its roots are $3$ and $5$. Therefore, $T(n)$ must be in the form:
$$T(n)=a\cdot 3^n+b\cdot 5^n, n\geq 1$$for some constants $a$ and $b$.
To find $a$ and $b$, since $1=T(1)=3a+5b$ and $4=T(2)=9a+25b$, we get $a=1/6$ and $b=1/10$, i.e.
$$T(n)=\frac{1}{6}\cdot 3^n+\frac{1}{10}\cdot 5^n, n\geq 1$$
| {
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Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The book asks me to show that the sequence $\{C_n\}$, with base case $C_0 = 0$, also satisfies the recurrence relation $nC_n=(n+1)C_{n-1}+2n$ for $n=1,2,\cdots$.
I tried to solve it by induction. For this, I wrote the second recurrence relation for $n+1$:
$$(n+1)C_{n+1} = (n+2)C_{n} + 2n + 2$$
Then, assuming that the first recurrence relation holds for $n$, I tried to substitute $C_n = n + 1 + \frac{2}{n}\sum\limits_{k=0}^{n-1}C_k$ in the above equation, to see if I obtain $C_{n+1} = n + 2 + \frac{2}{n+1}\sum\limits_{k=0}^{n}C_k$.
$$\begin{align*}
(n+1)C_{n+1} &= (n+2)C_{n} + 2n + 2\\
(n+1)C_{n+1} &= (n+2)\left( n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k \right ) + 2n + 2\\
(n+1)C_{n+1} &= n(n+2) + n + 2 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k + 2n + 2\\
(n+1)C_{n+1} &= n^2 + 5n + 4 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
(n+1)C_{n+1} &= (n+1)(n+4) + \dfrac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
C_{n+1} &= n+4 + \dfrac{2(n+2)}{n(n+1)}\sum_{k=0}^{n-1}C_k
\end{align*}$$
From this point, I'm not sure how to proceed.
| As other answers have shown, the approach that you took is not the easiest one, but it can be made to work. Note first that the new recurrence is equivalent (after division by $n$) to
$$C_n=\left(1+\frac1n\right)C_{n-1}+2\;;\tag{1}$$
this will be a little easier to work with. I’ll assume $(1)$ as an induction hypothesis and use the original recurrence to prove the next case of $(1)$:
$$\begin{align*}
C_{n+1}&\overset{(1)}=n+2+\frac2{n+1}\sum_{k=0}^nC_k\\
&\overset{(2)}=n+2+\frac2{n+1}\left(C_n+\sum_{k=0}^{n-1}C_k\right)\\
&\overset{(3)}=n+2+\frac2{n+1}C_n+\frac{n}{n+1}\cdot\frac2n\sum_{k=0}^{n-1}C_k\\
&\overset{(4)}=n+2+\frac2{n+1}C_n+\frac{n}{n+1}(C_n-n-1)\\
&\overset{(5)}=n+2+\frac{n+2}{n+1}C_n-n\\
&\overset{(6)}=\left(1+\frac1{n+1}\right)C_n+2\;,
\end{align*}$$
which is what we wanted. $(1)$ is the original recurrence; $(2)$ is a standard trick when proving things about sums by induction; $(3)$ is just multiplying out and adjusting the coefficient of the summation into a form that matches the original recurrence; $(4)$ uses the original recurrence; and $(5)$ and $(6)$ are just algebra.
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Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| A solution for straightforward mathematicians. :-)
It seems the following.
After multiplication of both sides of the inequality by a common denominator and simplification, we reduce the initial inequality to
$$1+ab^2+bc^2+ca^2\ge 4a^3b^3c^3.$$
Since $$ab^2+bc^2+ca^2\ge 3abc$$ it suffices to check that
$$1+3abc-4(abc)^3\ge 0,$$
that is
$$(1-abc)(2abc+1)^2\ge 0.$$
The last inequality holds because $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$
| {
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Computing a double series. How to compute following double series,
$$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \cfrac{(-1)^{(i+j)}}{i^2+2nij+j^2} $$
where $n\in \mathbb{N}$.
| This double series is not absolutely convergent. You might try a kind of Abel summation:
$\lim_{z \to -1+} f_n(z)$ where
$$f_n(z) = \sum_{i=1}^\infty \sum_{j=1}^\infty \dfrac{z^{i+j}}{i^2 + 2nij + j^2}$$
But I would find a closed form surprising.
It might be interesting to try this asymptotically as $n \to \infty$, using
$$ \dfrac{1}{i^2 + 2nij + j^2} = {\frac {1}{2ijn}}-{\frac {{i}^{2}+{j}^{2}}{4{i}^{2}{j}^{2}{n}^
{2}}}+{\frac { ( {i}^{2}+{j}^{2} ) ^{2}}{8 {i}^{3}{j}^{3
}{n}^{3}}} - \dfrac{(i^2 + j^2)^3}{16 i^4 j^4 n^4}+\ldots$$
With Maple's help, I get
$$ f_n(z) \sim \dfrac{(\ln(1-z))^2}{2n} - \dfrac{z \ \text{polylog}(2,z)}{2 (1-z) n^2}
+ \dfrac{z \ \text{polylog}(3,z)}{4 (z-1)^2 n^3} + \dfrac{(\ln(1-z))^2}{4 n^3} +
+ \dfrac{z(z+1) \ \text{polylog}(4,z)}{8 (z-1)^3 n^4 } + \dfrac{3 z \ \text{polylog}(2,z)}{8 (z-1) n^4} + \ldots$$
which as $z \to -1+$ becomes
$$ \dfrac{(\ln 2)^2}{2n} - \dfrac{\pi^2}{48 n^2} + \dfrac{3 \zeta(3) + 16 (\ln 2)^2}{64 n^3} - \dfrac{\pi^2}{64 n^4} + \ldots$$
There must be a pattern here...
EDIT: The coefficient of $n^{-m}$ is
$$ \eqalign{\sum_{i=1}^\infty \sum_{j=1}^\infty &\dfrac{(-1)^{m+1}}{2^m i^m j^m} (i^2 + j^2)^{m-1} z^{i+j}
= \dfrac{(-1)^{m+1}}{2^m} \sum_{k=0}^{m-1} {{m-1} \choose k} \sum_{i=1}^\infty i^{2k-m} z^i \sum_{j=1}^\infty j^{m-2-2k} z^j\cr
&= \dfrac{(-1)^{m+1}}{2^m} \sum_{k=0}^{m-1} {{m-1} \choose k} \operatorname{polylog}(-2k+m,z) \operatorname{polylog}(-m+2+2k,z)\cr}$$
Note that for nonnegative integers $p$,
$ \operatorname{polylog}(-p,z) = z P_p(1-z)/(1-z)^{p+1}$ where $P_p$ is a polynomial of degree $p-1$. If $m$ is even, each term involves a polylog of positive index times a rational function. If $m$ is odd, there is also (for $k=(m-1)/2$) one term
$2^{-m} {{m-1} \choose {(m-1)/2}} \ln(1-z)^2$ since $\operatorname{polylog}(1,z) = -\log(1-z)$.
When we take $z \to -1$,
$\operatorname{polylog}(p,-1) = (2^{1-p}-1) \zeta(p)$ for integers $p \ge 2$, while
$\operatorname{polylog}(-p,-1) = -(2^{p+1}-1) B_{p+1}/(p+1)$, where $B_{p+1}$ is the $(p+1)$'th Bernoulli number (which is $0$ for even $p\ge 2$).
Thus for even $m$, the only terms that survive are those for $k=m/2$ and $m/2-1$,
and the coefficient of $n^{-m}$ is $\displaystyle - 2^{-m} {{m-1} \choose {m/2}} \dfrac{\pi^2}{12}$.
For odd $m$, we get
$$ 2^{-m} {{m-1} \choose {(m-1)/2}} (\ln 2)^2 + 2^{2-2m} \sum_{k=0}^{(m-3)/2} {{m-1}\choose k} (2^{m-2k-1}-1)^2 2^{2k} \dfrac{B_{m-2k-1}}{m-2k-1} \zeta(m-2k)$$
| {
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"url": "https://math.stackexchange.com/questions/169303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$(\sin\theta+\cos\theta)^2=1+\sin2\theta$ 49) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*}
(\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.
Right Side:
\begin{align*}
1+\sin2\theta=1+2\sin\theta\cos\theta
\end{align*}
Thank you!
| Open parentheses and use:
$$(1)\,\,\sin^2x+\cos^2x=1$$
$$(2)\,\,\sin 2x=2\sin x\cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\int 1/(1+x^{2n})\,dx$ for an arbitrary positive integer $n$? How to find
$$\int\dfrac{dx}{1+x^{2n}}$$
where $n \in \mathbb N$?
Remark
When $n=1$, the antiderivative is $\tan^{-1}x+C$. But already with $n=2$ this is something much more complicated. Is there a general method?
| If the integral is taken from $0$ to $\infty$, there is more than one way to evaluate this. One is
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}}
&=\int_0^1\frac{\mathrm{d}t}{1+t^{2n}}+\int_0^1\frac{t^{2n-2}\,\mathrm{d}t}{1+t^{2n}}\\
&=\int_0^1(1-t^{2n}+t^{4n}-t^{6n}+\dots)\,\mathrm{d}t\\
&+\int_0^1(t^{2n-2}-t^{4n-2}+t^{6n-2}+\dots)\,\mathrm{d}t\\
&=1-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{6n+1}+\dots\\
&+\frac{1}{2n-1}-\frac{1}{4n-1}+\frac{1}{6n-1}-\dots\\
&=\frac{1}{2n}\left(\frac{1}{0+\frac{1}{2n}}-\frac{1}{1+\frac{1}{2n}}+\frac{1}{2+\frac{1}{2n}}-\frac{1}{3+\frac{1}{2n}}+\dots\right)\\
&+\frac{1}{2n}\left(-\frac{1}{-1+\frac{1}{2n}}+\frac{1}{-2+\frac{1}{2n}}-\frac{1}{-3+\frac{1}{2n}}-\dots\right)\\
&=\frac{1}{2n}\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac{1}{2n}}\\
&=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{1}
\end{align}
$$
The last step uses the result from "An Infinite Alternating Harmonic Series" on this page.
Another method is to use contour integration to evaluate
$$
\frac12\int_{-\infty}^\infty\frac{\mathrm{d}t}{1+t^{2n}}
=\frac12\oint_\gamma\frac{\mathrm{d}z}{1+z^{2n}}\tag{2}
$$
where $\gamma$ is the path from $-\infty$ to $\infty$ along the real axis (which picks up the integral in question), then circling back counter-clockwise around the upper half-plane (which vanishes). The countour integral in $(2)$ is $2\pi i$ times the sum of the residues of $\frac{1}{1+z^{2n}}$ in the upper half-plane.
The poles of the integrand in $(2)$ are given by
$$
\zeta_k=e^{\frac{\pi i}{2n}(2k+1)}\tag{3}
$$
where $k=0\dots n-1$ represent the roots in the upper half-plane. All the poles are simple, so the residues are
$$
\begin{align}
\mathrm{Res}_{z=\zeta_k}\left(\frac{1}{1+z^{2n}}\right)
&=\lim_{z\to\zeta_k}\frac{z-\zeta_k}{1+z^{2n}}\\
&=-\frac{1}{2n}\zeta_{k}\\
&=-\frac{1}{2n}e^{\frac{\pi i}{2n}(2k+1)}\tag{4}
\end{align}
$$
Thus, we get
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}}
&=-\frac{2\pi i}{4n}\sum_{k=0}^{n-1}e^{\frac{\pi i}{2n}(2k+1)}\\
&=-\frac{\pi i}{2n}e^{\frac{\pi i}{2n}}\frac{1-(-1)}{1-e^{\frac{\pi i}{n}}}\\
&=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
$$\cos^2x-\sin^2x=\sin$$
Edit
$$1-\sin^2x-\sin^2x=\sin x$$
$$2\sin^2 x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$x=-1,\dfrac{1}{2}$$
$$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$
$$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$
| The calculation is almost completely correct. You reached the two possibilities $\sin x=\frac{1}{2}$ and $\sin x=-1$.
We are interested in solutions in the interval $-\pi \lt x\le \pi$.
Certainly $x=\frac{\pi}{6}$ is a solution, since $\sin(\pi/6)=\frac{1}{2}$. But there is another $x$ in our interval whose sine is $\frac{1}{2}$, namely $x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$. A look at the graph of $y=\sin x$ shows this. You can do a partial verification by calculator, by asking it to compute $\sin(5\pi/6)$, the sine of $150^\circ$.
There is only one place $x$ in our interval where $\sin x=-1$, so that part is fully correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.
| Hint:
$$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$
and
$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving a number with $3^n$ equal digits is divisible by $3^n$ Prove a number with $3^n$ equal digits is divisible by $3^n$.
My thoughts about the problem are: a number with $3^n$ equal digits $d$ is equal to $d\frac {10^{3^n} - 1} {9}$. We use Lifting The Exponent lemma, or plain induction.
| This follows from the fact that $x^3-1=(x-1)(x^2+x+1)$ and that $10\equiv1\pmod{3}$. That is,
$$
10^{3^n}-1=\left(10^{3^{n-1}}-1\right)\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\tag{1}
$$
and
$$
\begin{align}
\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)
&\equiv1+1+1\\
&\equiv0\pmod{3}\tag{2}
\end{align}
$$
so that
$$
\left.3\,\middle|\,\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\right.\tag{3}
$$
We start out with
$$
\left.3^2\,\middle|\,\left(10^{3^0}-1\right)\right.\tag{4}
$$
Next, combining $(1)$ and $(3)$ yields that
$$
\left.3^{n+1}\,\middle|\,\left(10^{3^{n-1}}-1\right)\right.\Rightarrow\left.3^{n+2}\,\middle|\,\left(10^{3^n}-1\right)\right.\tag{5}
$$
Therefore, induction on $n$, using $(4)$ and $(5)$, says
$$
\left.3^n\,\middle|\,\frac{10^{3^n}-1}{9}\right.\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Help on differential equation $y''-2\sin y'+3y=\cos x$ $y''-2\sin y'+3y=\cos x$
I'm trying to solve it by power series, but I just can't find the way to get $\sin y'$. Is there any special way to find it?
| Maple finds the two fundamental solutions:
(with $y(0)=0$ and $y'(0)=b$)
$$\eqalign{ y \left( x \right) &=bx+ \left( \sin \left( b \right) +{\frac {1}{2}}
\right) {x}^{2}+ \left( \frac23\,\cos \left( b \right) \sin \left( b
\right) +\frac13\,\cos \left( b \right) -\frac12\,b \right) {x}^{3}\cr+& \left( \frac2
3\, \left( \cos \left( b \right) \right) ^{2}\sin \left( b \right) -
\frac23\,\sin \left( b \right) -{\frac {1}{2}}-\frac14\,\cos \left( b \right)
b+\frac12\, \left( \cos \left( b \right) \right) ^{2} \right) {x}^{4}\cr+&
\left( \frac45\, \left( \cos \left( b \right) \right) ^{3}\sin \left( b
\right) -{\frac {11}{10}}\,\cos \left( b \right) \sin \left( b
\right) +{\frac {3}{20}}\,\sin \left( b \right) b-{\frac {13}{15}}\,
\cos \left( b \right)\right. \cr& \left.+\frac38\,b-\frac25\, \left( \cos \left( b \right)
\right) ^{2}b+\frac45\, \left( \cos \left( b \right) \right) ^{3}
\right) {x}^{5}\cr +&O \left( {x}^{6} \right) \cr}
$$
and (with $y(0)=a$ and $y'(0)=0$)
$$y \left( x \right) =a+ \left( -\frac32\,a+{\frac {1}{2}} \right) {x}^{2}+
\left( -a+{\frac {1}{3}} \right) {x}^{3}-\frac18\,a{x}^{4}+ \left( \frac14\,a
-{\frac {1}{15}}+{\frac {9}{20}}\,{a}^{3}-{\frac {9}{20}}\,{a}^{2}
\right) {x}^{5}+O \left( {x}^{6} \right)
$$
Arbitrarily many terms of the series can be found (but they get complicated).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
| $f(x)=1-x^2$ inserting that gives:
$$
\begin{eqnarray}
x^2 f(x) + f(1-x) &=&x^2(1-x^2) + (1 - (1-x)^2)\\
&=&x^2-x^4 +(1-(1-2x+x^2))\\
&=&x^2-x^4 +(1-1+2x-x^2)\\
&=&x^2-x^4+2x-x^2 \\
&=& 2x -x^4
\end{eqnarray}
$$
as required. But I admit: I had some help...
But you can also write it as
$$
x^4+f(x) x^2 -2x-f(1-x)=0
$$
and try to solve the depressed quartic equation...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Prove $\cos 3x =4\cos^3x-3\cos x$ How would I solve the following double angle identity.
$$\cos 3x =4\cos^3x-3\cos x $$
I know $\,\cos 3x = \cos(2x+x)$
So know I have $\,\cos 2x +\cos x \,$ , Which is $\,(2\cos^2x-1)\cos x$
But I am not sure what to do next.
| I will attempt to answer my own question now.
$$\begin{eqnarray}
\cos(2x)(\cos x)-\sin(2x)(\sin x)& =& (2\cos^2 x-1)(\cos x)-2\sin x\cdot\cos x\cdot\sin x\\
&=&2\cos^3x-\cos x-2\sin^2 x\cos x\\
&=&2\cos^3x-\cos x-2(1-\cos^2 x)(\cos x)\\
&=&2\cos^3x-\cos x-2\cos x+2\cos^3 x\\
&=&4\cos^3 x-3\cos x
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
A three variable binomial coefficient identity I found the following problem while working through Richard Stanley's Bijective Proof Problems (Page 5, Problem 16). It asks for a combinatorial proof of the following:
$$ \sum_{i+j+k=n} \binom{i+j}{i}\binom{j+k}{j}\binom{k+i}{k} = \sum_{r=0}^{n} \binom{2r}{r}$$
where $n \ge 0$, and $i,j,k \in \mathbb{N}$, though any proof would work for me.
I also found a similar identity in Concrete Mathematics, which was equivalent to this one, but I could not see how the identity follows from the hint provided in the exercises.
My initial observation was to note that the ordinary generating function of the right hand side is $\displaystyle \frac {1}{1-x} \frac{1}{\sqrt{1-4x}}$, but couldn't think of any way to establish the same generating function for the left hand side.
| Restating your question, you are seeking to find the generating function of the left-hand-side:
$$
g(x) = \sum_{n=0}^\infty x^n \sum_{i+j+k=n}\binom{i+j}{i} \binom{j+k}{j} \binom{k+i}{k} = \sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty x^{i+j+k} \frac{(i+j)! (i+k)! (j+k)!}{i!^2 j!^2 k!^2}
$$
First, carry out summation over $i$:
$$
g(x) = \sum_{j=0}^\infty \sum_{k=0}^\infty x^{j+k} \frac{(j+k)!}{j!\cdot k!} {}_2F_1\left(1+j, 1+k; 1; x\right)
$$
Now use Euler's transformation ${}_2F_1\left(1+j, 1+k; 1; x\right) = (1-x)^{-j-k-1} \, {}_2F_1\left(-j, -k; 1, x\right)$, which gives
$$
g(x) = \frac{1}{1-x} \sum_{j=0}^\infty \sum_{k=0}^\infty \left(\frac{x}{1-x}\right)^{j+k} \frac{(j+k)!}{j!\cdot k!} {}_2F_1\left(-j, -k; 1; x\right) = \\ \frac{1}{1-x}
\sum_{j=0}^\infty \sum_{k=0}^\infty \left(\frac{x}{1-x}\right)^{j+k} \frac{(j+k)!}{j!\cdot k!} \sum_{r=0}^{\min(j,k)} \binom{j}{r}\binom{k}{r} x^r = \\
\frac{1}{1-x} \sum_{r=0}^\infty x^r \sum_{j=r}^\infty \sum_{k=r}^\infty \binom{k+j}{k} \binom{j}{r}\binom{k}{r} \left(\frac{x}{1-x}\right)^{j+k}
$$
Using
$$
\sum_{j=r}^\infty \sum_{k=r}^\infty \binom{k+j}{k} \binom{j}{r}\binom{k}{r} z^{j+k} =
\sum_{j=r}^\infty \binom{j}{r} z^{j+r} \sum_{k=0}^\infty \frac{(k+j+r)!}{j! r! k!} z^k =\\
\sum_{j=r}^\infty \binom{j}{r} z^{j+r} \binom{j+r}{j} \sum_{k=0}^\infty \frac{(j+r+1)_k}{k!} z^k = \sum_{j=r}^\infty \binom{j}{r} z^{j+r} \binom{j+r}{j} \left(1-z\right)^{-j-r-1} = \\ \frac{z^{2r}}{(1-z)^{2r+1}} \frac{1}{r!^2} \sum_{j=0}^\infty \frac{(j+2r)!}{j!} \left(\frac{z}{1-z}\right)^j = \frac{z^{2r}}{(1-z)^{2r+1}} \binom{2r}{r} \left(1-\frac{z}{1-z}\right)^{-1-2r} = \binom{2r}{r} z^{2r} \left(1-2z\right)^{-2r-1}
$$
we continue:
$$
g(x) = \frac{1}{1-x} \sum_{r=0}^\infty x^r \binom{2r}{r} \left(\frac{x}{1-x}\right)^{2r} \left(1 - 2 \frac{x}{1-x} \right)^{-1-2r} = \\ \frac{1}{1-x} \sum_{r=0}^\infty \binom{2r}{r} \frac{1-x}{1-3x} \left(\frac{x^3}{(1-3x)^2}\right)^r = \\
\frac{1}{1-3x} \sum_{r=0}^\infty \binom{2r}{r}\left(\frac{x^3}{(1-3x)^2}\right)^r = \frac{1}{1-3x} \left(1 - 4 \frac{x^3}{(1-3x)^2}\right)^{-1/2} = \frac{1}{1-3x} \left( \frac{(1-4x)(1-x)^2}{(1-3x)^2}\right)^{-1/2} = \frac{1}{1-x} \frac{1}{\sqrt{1-4x}}
$$
which is exactly the generating function of the right-hand-side:
$$
\sum_{n=0}^\infty x^n \sum_{r=0}^n \binom{2r}{r} \stackrel{n=r+k}{=} \sum_{k=0}^\infty x^r \sum_{r=0}^\infty \binom{2r}{r} x^r = \frac{1}{1-x} \cdot \frac{1}{\sqrt{1-4x}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
} |
Proving an estimate for this integral How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$
| In the interval from $1$ to $2$, $\frac{(1+x)^{1/3}}{x^2}\le \frac{3^{1/3}}{x^2}$.
Thus
$$\int_1^2 \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_1^2\frac{3^{1/3}}{x^2}\,dx= \frac{1}{2}(3^{1/3}).$$
In the interval from $2$ to $\infty$, $(1+x)^{1/3}\le \left(\frac{x}{2}+x\right)^{1/3}$. So
$$\int_2^\infty \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_2^\infty(3/2)^{1/3}x^{-5/3}\,dx= \frac{3}{4}(3^{1/3}).$$
So our full integral is $\le \frac{5}{4}(3^{1/3})$. Since $(5/4)^3 \lt 2$, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Proving $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$ I would like to know how to prove the following assertion :
For every $n>0$: $$\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$$
| Or you could obfuscate the problem. Let $t=\sqrt{n+1}$. Then we have $t>1$ and $\sqrt{n}=\sqrt{t^2-1}$, so
$$
\begin{align*}
&\sqrt{t^2-1}+\frac{1}{t} \ge t\\
\text{iff}\qquad &\sqrt{t^2-1}\ge t-\frac{1}{t}\\
\text{iff}\qquad &t^2-1 \ge t^2-2+\frac{1}{t^2}\\
\text{iff}\qquad &1 \ge \frac{1}{t^2}\\
\text{iff}\qquad &t^2\ \ge 1
\end{align*}
$$
which is true, since $t>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ Find derivative of $$y= \frac{ax+b}{cx+d}$$
I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$
Use it to evaluate:
$$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$$
I figured that here $y=\frac{x+1}{x+3}$ and $$\frac{dy}{dx}=\frac{1}{x+3}-\frac{(x+1)}{(x+3)^2}$$
and using the technique I learned from my last question I did this:
$$\frac{dy}{dx}=\frac{(x+3)}{(x+3)^2}-\frac{(x+1)}{(x+3)^2}=\frac{2}{(x+3)^2}$$
which I could then substitute back, having changed the limits by substituting $1$ into $y$ and then $0$ into $y$:
$$y|_{x=1}=\frac{x+1}{x+3}=\frac{1}{2}$$
$$y|_{x=0}=\frac{1}{3}=\frac{1}{3}$$
$$2\int_0^1{\frac{dy}{dx}}\ln(y)dx=2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
This gives me:
$$2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
$$=2\left[y(\ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2} = 2\left[\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)\right]-\frac{1}{3}\left[\ln\left(\frac{1}{3}\right)-\frac{1}{3}\right]\\$$
$$\ln\left(\frac{1}{2}\right)-1-\frac{2}{3}\ln\left(\frac{1}{3}\right)+\frac{2}{9}$$
The problem is I am supposed to end up with something else. Can anyone spot any issues with this?
EDIT: This is the answer I am supposed to be getting:
$$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$$
| You should have been careful when multiplying by $2$, I think that you should have multiplied by $\frac{1}{2}$ in your formal calculation.
Also, you may need to rewrite your answer a bit to get it right, such as computing $\frac{2}{9} - 1 = -\frac{7}{9}$, which is implicitly required from you by any textbook, I presume.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.