Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $m$ and $n$ so that the given function has the range $[-3, 5]$ Find $m$ and $n$ real numbers so that $f(x) = \frac{3x^2 + mx + n}{x^2 + 1}$ takes all and only the values from the interval $[-3, 5]$.
I started by solving the following double inequality:
$$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$$
From the left inequality I got $6x^2 + mx + n + 3 \geq 0$. The coefficient of $x^2$ is positive so the discriminant $\delta$ has to be less than or equal to $0$.
$\delta = m^2 - 24n - 72 \leq 0$
From the right inequality I got $2x^2 -mx + 5 - n \geq 0$. By using the same technique as the one used to solve the left inequality I got:
$\delta = m^2 - 40 + 8n \leq 0$
By combining these two resulted inequalities I got $m \leq 48$, that is $m \in [-4\sqrt3, 4\sqrt3]$.
Now I need to find the values of $m$ and $n$ so that $f$ is surjective, because it must take all the values in the given interval.
I don't know yet how to proceed, so I would appreciate any help from you guys!
Thank you!
| Background: A parabola graph of $f(x)=ax^2 + bx + c$ will have it's maximum/minimum value at its vertex, $x = -\frac b{2a}$.
Review: because $f(x) = a(x + \frac b{2a})^2 - a(\frac b{2a})^2 +c$. As $(x + \frac b{2a})^2 \ge 0$ with equality holding only if $x = - \frac b{2a}$, $f(x)$ achieves it's max min value of $- \frac {b^2}{4a}+c$ and $x =- \frac b{2a}$.
$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$
(and achieves points where equality holds; for the rest of this post, if I write $F(x) \ge a$ I'm going to take it to mean both $F(x) \ge a$ and there exists some (at least one) $y$ so that $F(y) = a$.)
so $-3x^2 - 3 \le 3x^2 + mx + n$
so $0 \le 6x^2 + mx + n+3$
So $6x^2 + mx + (n+3)$ acheives its minimum at $x = -\frac {m}{12}$ and the minimum value is $-\frac{m^2}{24} + n + 3 = 0$
Also
$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$
so $3x^2 + mx + n \le 5x^2 + 5$
so $0 \le 2x^2 - mx - n+5$
So $2x^2 - mx - n+5$ acheives its minimum at $x = \frac {m}{4}$ and the minimum value is $-\frac{m^2}{8} - n + 5 = 0$
So it becomes a matter of solving for $m,n$ where $-\frac{m^2}{8} - n + 5 = 0$ and $-\frac{m^2}{24} + n + 3 = 0$
So $n= 5 - \frac{m^2}{8} = \frac{m^2}{24} - 3$
$\frac{m^2}{24}+\frac{m^2}{8}=8$
$4m^2 = 8*24$
$m = \pm \sqrt{48} = \pm4\sqrt{3}$
$n = 5 - 6 = 2 - 3 = - 1$
Thank's to FreezingFire for pointing out the type that led to a numerically incorrect result.
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Let $x$, $y$, and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum value of $9x+12y+8z.$ Let $x$, $y$, and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum value of $9x+12y+8z.$
We aren't given anything other than what $x^2+y^2+z^2$ is. How can I manipulate that equation into something useful? I'm stuck. Any solutions are highly appreciated.
| By the Cauchy-Schwarz inequality
$$9x+12y+8z \color{red}\leq \sqrt{9^2+12^2+8^2}\sqrt{x^2+y^2+z^2} = \color{red}{17} $$
and equality is achieved at $(x,y,z)=\frac{1}{17}\left(9,12,8\right).$
| {
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Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .
| The left inecuality:
Use Titu's Lemma and $a^2+b^2+c^2\ge ab+bc+ca$
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=\frac {a^2}{ab+ac}+\frac {b^2}{bc+ba}+\frac {c^2}{ca+bc}\ge$$
$$\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}=\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+1\ge\frac12+1=\frac32$$
The left inecuality is Nesbitt's inequality
| {
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About the Fermat quotients with base $2$ On Wikipedia, about Fermat quotients, it says: "Eisenstein discovered that the Fermat quotient with base $2$ could be expressed, $\mod p\ \ $ $p$ odd prime, in terms of the sum of the reciprocals of the numbers lying in the first half of the range ${1, p − 1 }$":
$$ -2 \cdot \frac{2^{ p − 1 }-1}{p} \equiv 1+\frac{1}{2}+\cdot\cdot\cdot\cdot+\frac{1}{(p - 1)/2} \mod p $$
How is this proven?
| We use the following seven binomial-coefficient identities, accompanied by brief proofs. The main proof is farther down.
The first identity, $$2^p = \sum_{k = 0}^p \binom pk, \tag1$$ is found by replacing $x$ and $y$ each with $1$ in the binomial theorem $$(x + y)^p = \sum_{k = 0}^p \binom pk x^ky^{p - k}.$$
The second and third identities are $$1 = \binom p0 \qquad 1= \binom pp. \tag{2, 3}$$ Indeed, given a set with $p$ elements, there is precisely $1$ way to select $0$ elements and precisely $1$ way to select all $p$ elements.
The fourth and fifth identities, $$\binom pk = \binom{p - 1}{k - 1}\frac pk \qquad \binom{p - 1}{k} = \binom pk - \binom{p - 1}{k - 1}, \tag{4, 5}$$ are verified by replacing the binomial coefficients with the their combinatorial definition.
The identities above do not actually require that $p$ be prime, but the next two do. The sixth identity is $$\binom pk \equiv 0 \pmod p \text{ for } 1 \le k \le p - 1. \tag6$$ To prove it, use the combinatorial definition $$\binom pk = \frac{p!}{k!(p - k)!}$$ and note that because $p$ is prime and $1 \le k \le p - 1$, none of the factors in the denominator divides out the factor $p$ in the numerator. Hence, $\binom pk$ is a multiple of $p$ and so congruent to $0$ modulo $p$.
The seventh and final identity, $$\binom{p - 1}{k - 1} \equiv (-1)^{k - 1} \pmod p \text{ for } 1 \le k \le p, \tag7$$ we prove by induction. The claim holds for $k = 1$ because $$\binom{p - 1}{1 - 1} = \binom{p - 1}0 = 1 = (-1)^{1 - 1} \equiv (-1)^{1 - 1} \pmod p.$$ Similarly, the claim holds for $k = p$. Suppose that the claim holds for $k = j$, $1 \le j < p$. Then $$\binom{p - 1}{(j + 1) - 1} = \binom{p - 1}j = \binom pj - \binom{p - 1}{j - 1} \equiv 0 + (-1)(-1)^{j - 1} = (-1)^{(j + 1) - 1} \pmod p$$ where the second equation is the fifth identity, and the congruence uses the sixth identity. Hence, the claim holds for $k = j + 1$ whenever it holds for $k = j$, and so the induction principle guarantees that it holds for every integer $k$, $1 \le k \le p$.
We are now ready for the main proof. We first show that the expression involving the Fermat quotient is congruent modulo $p$ to the $p - 1$-st "alternating" harmonic number. Next, we use straightforward algebra and the congruence identities $a \equiv -(p - a)$ and $p - a \equiv -a$ (mod $p$) to show that that alternating harmonic number, in turn, is congruent modulo $p$ to the $(p - 1)/2$-th harmonic number, the desired result.
\begin{align}
-2 \cdot \frac{2^{p − 1} - 1}p & = - \frac 1p(2^p - 2)\\
& = -\frac 1p \left( \sum_{k = 0}^p \binom pk - \binom p0 - \binom pp \right) \qquad \text{first three identities}\\
& = -\frac 1p \sum_{k = 1}^{p - 1} \binom pk\\
& = -\frac 1p \sum_{k = 1}^{p - 1} \binom{p - 1}{k - 1} \frac pk \qquad \qquad \qquad \quad \; \; \text{fourth identity}\\
& = -\sum_{k = 1}^{p - 1} \binom{p - 1}{k - 1} \frac1k\\
& \equiv -\sum_{k = 1}^{p - 1} \frac{(-1)^{k - 1}}k \pmod p \qquad \qquad \; \; \text{seventh identity} \\
& = -\frac11 + \frac12 - \frac13 + \frac14 - \cdots - \frac1{p - 4} + \frac1{p - 3} - \frac1{p - 2} + \frac1{p - 1} \\
& = \left( -\frac11 - \frac13 - \cdots - \frac1{p - 4} - \frac1{p - 2} \right) + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& \equiv \left( -\frac1{-(p - 1)} - \frac1{-(p - 3)} - \cdots - \frac1{-4} - \frac1{-2} \right) \\
& \quad + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \pmod p \\
& = \left( \frac1{p - 1} + \frac1{p - 3} + \cdots + \frac14 + \frac12 \right) + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& = 2 \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& = 1 + \frac12 + \cdots + \frac1{(p - 3)/2} + \frac1{(p - 1)/2}
\end{align}
| {
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Integral of $\int \frac{\sqrt{9-x^2}}{x}$ I don't understand how the integral
$$\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$$
I keep getting $-3/x +c$ as the answer.
| First thing - it is generally a bad idea to write $\int \dfrac{\sqrt{9-x^2}}{x}$ without the $dx$ part, it takes away the meaning of the integral.
So, solving $\int \dfrac{\sqrt{9-x^2}}{x} \cdot dx$
The key step is to consider the substitution $x = 3\sin \theta$. Thus, $dx = 3\cos \theta \cdot d\theta$
On substitution, the problem becomes $\int \dfrac{\cos \theta}{\sin \theta} \cdot 3\cos \theta \cdot d\theta$ = $\int 3 (\csc \theta - \sin \theta) \cdot d\theta$.
From here, using $\int \csc \theta \cdot d\theta = -\ln (\csc \theta + \cot \theta)$ and $\int \sin \theta \cdot d\theta = - \cos \theta$ and substituting back $x$ would give you the result
$\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$
| {
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How does one prove $\int_0^\infty \frac{\log(x)}{1 + e^{ax}} \, dx = -\frac{\log(2)(2\log(a) + \log(2))}{2a}$ for $a > 0$? Link to WolframAlpha's assertion. Here's my attempt. Using the substitution $t = ax$, we can show the integral is equal to
$$ \frac{1}{a} \int_0^\infty \frac{\log(t)}{1 + e^t}\, dt -\frac{\log(a)}{a} \int_0^\infty \frac{dt}{1 + e^{t} } .$$
The second integral is equal to $\log(2)$ using another substitution $v = e^t$ and partial fractions.
So I'm left with the first integral. I'll switch to complex variables for notation. I make two observations:
(I) The denominator has simple poles when $z = t = (2k-1)\cdot i \pi, \, k \in \mathbb{N}.$
(II) The numerator has a branch point $z = 0$.
| Here is another solution, although it is not as efficient as Dr. MV's solution.
Let
$$I = \int\limits_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x$$ and
\begin{align}
I_{1} &= \int\limits_{0}^{\infty} \frac{x^{b}}{1+e^{ax}} \mathrm{d}x = \int\limits_{0}^{\infty} \frac{x^{b}e^{-ax}}{1+e^{-ax}} \mathrm{d}x \\
&= \sum\limits_{n=0}^{\infty} (-1)^{n} \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x
\end{align}
We designate the last integral on the right as $I_{2}$ and make the substitution $y=(a+na)x$
\begin{align}
I_{2} &= \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x \\
&= \frac{1}{(a+na)^{b+1}} \int\limits_{0}^{\infty} y^{b} e^{-y} \mathrm{d}y \\
&= \frac{\Gamma(b+1)}{(a+na)^{b+1}} \\
&= \frac{\Gamma(b+1)}{a^{b+1}} \frac{1}{(n+1)^{b+1}}
\end{align}
Now $I_{1}$ becomes
\begin{equation}
I_{1} = \frac{\Gamma(b+1)}{a^{b+1}} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{1}{(n+1)^{b+1}} = \frac{\Gamma(b+1)}{a^{b+1}} \eta(b+1)
\end{equation}
and we have
\begin{align}
I &= \lim_{b \to 0} \frac{\partial I_{1}}{\partial b} \\
&= \lim_{b \to 0} \int\limits_{0}^{\infty} \frac{x^{b} \ln(x)}{1+e^{ax}} \mathrm{d}x \\
&= \lim_{b \to 0}\frac{\partial}{\partial b}\frac{\Gamma(b+1)\eta(b+1)}{a^{b+1}} \\
&= \lim_{b \to 0} \frac{a^{b+1} \Big[ \Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) \Big] - \Gamma(b+1)\eta(b+1)a^{b+1}\ln(a)}{\left(a^{b+1}\right)^{2}} \\
&= \lim_{b \to 0} \frac{\Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) - \Gamma(b+1)\eta(b+1)\ln(a)}{a^{b+1}} \\
\tag{a}
&= \frac{1}{a} \left(\Big[\gamma \ln(2) - \frac{1}{2} \ln^{2}(2)\Big] -\gamma \ln(2) - \ln(2)\ln(a) \right) \\
&= -\frac{\ln(2)}{a} \left(\frac{1}{2} \ln(2) + \ln(a) \right)
\end{align}
In step (a) we have
\begin{align}
\lim_{b \to 0} \Gamma^{\prime}(b+1)\eta(b+1) &= \lim_{b \to 0} \Gamma^(b+1)\psi(b+1)\eta(b+1) \\
&= \Gamma(1)\psi(1)\eta(1) \\
&= -\gamma \ln(2)
\end{align}
and
\begin{align}
\lim_{b \to 0} \Gamma(b+1)\eta^{\prime}(b+1) &= \lim_{s \to 1} \eta^{\prime}(s) \\
&= \lim_{s \to 1} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{\ln(n)}{n^{s}} \\
&= \gamma \ln(2) - \frac{1}{2} \ln^{2}(2)
\end{align}
See Dr. MV's solution for a proof of this result.
Notes:
*
*$\Gamma(z)$ is the Gamma function.
*$\eta(s)$ is the Dirichlet eta function.
*$\zeta(s)$ is the Riemann zeta function.
*$\psi(z)$ is the digamma function.
*$\gamma$ is the Euler-Mascheroni constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find k so that polynomial division has remainder 0
Find $k$ so that $x^3-kx^2+3x+7k$ has remainder $0$ when divided by $x+2$.
How do I approach this problem? I know how to do polynomial long division and synthetic division, but I don't know how to apply it in this equation.
| You know the following:
$$
x^3 - kx^2 + 3x + 7k = (x + 2)(ax^2 + bx + c)
$$
Just do the multiplication and find $a$, $b$, and $c$.
$$
(x + 2)(ax^2 + bx + c) = ax^3 + (2a + b)x^2 + (c + 2b)x + 2c
$$
Now we set each coefficient equal:
$$
a = 1 \\
2a + b = -k \rightarrow b = -k - 2\\
c + 2b = 3 \rightarrow c = 3 - 2(-k - 2) \\
2c = 7k \rightarrow 2(3 - 2(-k - 2) = 7k
$$
Solving the last equation gives:
$$
6 + 4k + 8 = 7k \rightarrow 14 = 3k \rightarrow k = \frac{14}{3}
$$
| {
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Closed form of a recurrence relation containing factorials. Let a sequence $\{x_n\}_{n\ge1}$ defined by $x_n=3nx_{n-1}+n!-3^n(n^2-2);\ \forall n\ge2$ with $x_1=10$. Find a closed form of $x_n$.
I tried that $b_n=\frac{x_n-3^n}{n!}$ then the above reduces into $b_n-3b_{n-1}=\frac{3^n}{n!}-9\cdot\frac{3^{n-2}}{(n-2)!}+1$. Now the left is not telescoping. What to do? Is this the right way to approach? I know generating functions but as the problem comes from an olympiad level contest, it must have some manipulative trick.
| Hint. Try with $b_n=\frac{a_n}{n!3^n}$, then the recurrence is
$$b_n=b_{n-1}+\frac{1}{3^n}-\frac{1}{(n-2)!}-\frac{1}{(n-1)!}+\frac{2}{n!}$$
where we used $n^2=n(n-1)+n$.
Hence
$$b_N=b_1+\sum_{n=2}^N \left(\frac{1}{3^n}-\frac{1}{(n-2)!}-\frac{1}{(n-1)!}+\frac{2}{n!}\right)\\
=\frac{10}{3}+\sum_{n=2}^N \frac{1}{3^n}-\sum_{n=2}^N \frac{1}{(n-2)!}-\sum_{n=2}^N \frac{1}{(n-1)!}+\sum_{n=2}^N \frac{2}{n!}\\
=\frac{10}{3}+\sum_{n=2}^N \frac{1}{3^n}-\sum_{n=0}^{N-2} \frac{1}{n!}-\sum_{n=1}^{N-1} \frac{1}{n!}+2\sum_{n=2}^N \frac{1}{n!}\\
=\frac{1}{2}-\frac{1}{2\cdot 3^N}+\frac{N+2}{N!}.$$
| {
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Given that $n$ is even, find a closed-form expression for $\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}$ Given that $n$ is even, find a closed-form expression for
$$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k},$$
or, in other words, for the sum
$$2^2\cdot 2\cdot\binom n2 + 2^4\cdot 4\cdot\binom n4 + 2^6\cdot 6\cdot\binom n6 + \cdots + 2^n\cdot n\cdot\binom nn.$$
You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.
I am stuck on this problem! I tried using the identity but it only confuses me more. All solutions are greatly appreciated! Thanks!
| By your identity,we have
$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}=\sum_{k=0}^{n/2}2^{2k}\cdot n\cdot\binom{n-1}{2k-1}=2n\sum_{k=0}^{n/2}2^{2k-1}\cdot\binom{n-1}{2k-1}$
Now
\begin{equation}
-\sum_{k=0}^{n/2}2^{2k-1}\cdot\binom{n-1}{2k-1}=
\sum_{k=0}^{n/2}(-2)^{2k-1}\cdot\binom{n-1}{2k-1}=\frac{1}{2}\big((-2+1)^{n-1}-(2+1)^{n-1}\big)
\end{equation}
you can check the second '=' using the binomial expansion(basically the even terms cancel due to the minus sign before 2). Hence your expression is $n\cdot (3^{n-1}+1)$.
| {
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Second Derivative of Implicit Function My original equation started out as: $x^2 -4xy + 4y + 8 = 0$
I found the first derivative to be $2x -4x\frac{dy}{dx} - 4y + 4\frac{dy}{dx} = 0$, which I am assuming is right.
I then need to find the second derivative in order to find if the turning points on the curve are a minimum or a maximum. I re-arranged my previous equation to get:
$$\frac{dy}{dx} = \frac{4y - 2x}{4-4x}$$
And then I used the quotient rule and subbed in $\frac{dy}{dx}$ where it occured to get:
$$\frac{d^2y}{dx^2} = \frac{8}{(4-4x)^2}[\frac{8y - 8xy - 4x + 4x^2}{4 - 4x} - 1 + 2y]$$
I can leave it in terms of $x$ and $y$ since I can find $y$ from the original equation. Have I done this correctly?
| We are asked to find the second derivative using implicit differentiation for:
$$x^2 -4xy + 4y + 8 = 0 \tag 1$$
Differentiating $(1)$:
$$2 x -4(y + x y') + 4 y' = 0 \tag 2$$
Solving for $y'$ in $(2)$ while noting we can divide out a $2$:
$$y' = \dfrac{x-2y}{2(x-1)} \tag 3$$
Next, we will reuse $(2)$ to find the second derivative, differentiating:
$$2 - 4(y' + y' + x y'') + 4 y'' = 0$$
Simplifying while noting we can divide out a $2$:
$$2(1-x)y'' = -1 + 4 y' = -1 + 4\left(\dfrac{x-2y}{2(x-1)}\right)=-\dfrac{1+x -4y}{(x-1)}$$
Solving for $y''$:
$$y'' = -\dfrac{1+x -4y}{2(x-1)^2}$$
| {
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Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$
Then:
$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$
Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).
On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$ ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.
Why is there a problem and how can I get the correct answer?
Edit: Here is Wolfram's answer:
Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$
| A slick way to compute this considers the closed loop $\gamma(t) = a\sin t + i b\cos t$ given by the ellipse and traversed clockwise. Then
$$\frac 1{2\pi i}\int_\gamma\frac{dw}w = -1$$
independently of $a$ and $b$ (this is computing the number of turns the ellipse makes around $w=0$, which is $-1$). On the other hand, computing the line integral explicitly gives
\begin{align}
\frac{1}{2\pi i}\int_0^{2\pi }\frac{\overline{\gamma(t)}}{|\gamma(t)|^2} \gamma'(t)dt &= \frac{1}{2\pi i}\int_0^{2\pi }\frac{a\sin t - i b\cos t}{a^2\sin^2 t + b^2\cos^2 t} (a\cos t - i b\sin t)dt \\
&= \frac{1}{2\pi i}\int_0^{2\pi }\frac{(a^2-b^2)\cos t\sin t-iab }{a^2\sin^2 t + b^2\cos^2 t} dt \\
\end{align}
The real part of this integral is
$$ -\frac{ab}{2\pi }\int_0^{2\pi }\frac{dt}{a^2\sin^2 t + b^2\cos^2 t} $$
which gives your result (and the imaginary part is of course zero).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
show that $\frac{n(n+2)}{2}-\frac{1}{8}\ln{(n+1)}<\sum_{k=1}^{n}\sqrt{k(k+1)}$ Let $n$ be postive integer,show that
$$\dfrac{n(n+2)}{2}-\dfrac{1}{8}\ln{(n+1)}<\sum_{k=1}^{n}\sqrt{k(k+1)}\tag{(1)}$$
I try use integral
$$\sum_{k=1}^{n}\sqrt{k(k+1)}>\sum_{k=1}^{n}\int_{k-1}^{k}\sqrt{x(1+x)}dx=\int_{0}^{n}\sqrt{x(x+1)}dx$$
but since
$$ \int\sqrt{x(x+1)}dx=\dfrac{2x^3+3x^2+x-\sqrt{x+1}\sqrt{x}\sinh{(\sqrt{x})}}{4\sqrt{x(x+1)}}$$
it seem ugly than inequality $(1)$ LHS,so How prove $(1)?$
| The inequality holds for $n=1$. In order to prove it by induction, it is enough to show that
$$\forall n\geq 1,\qquad n+\frac{1}{2}-\frac{1}{8}\log\left(1+\frac{1}{n}\right) < \sqrt{n(n+1)} \tag{1}$$
where:
$$ n+\frac{1}{2}-\sqrt{n(n+1)} = \frac{1}{8}\int_{0}^{1}\frac{dx}{\sqrt{n^2+n+\frac{x}{4}}}\leq \frac{1}{8}\sqrt{\int_{0}^{1}\frac{dx}{n}\int_{0}^{1}\frac{n\,dx}{n^2+n+\frac{x}{4}}}\tag{2}$$
holds by the Cauchy-Scharz inequality. That gives
$$ n+\frac{1}{2}-\sqrt{n(n+1)} \leq \frac{1}{4}\sqrt{\log\left(1+\frac{1}{4n(n+1)}\right)}\tag{3} $$
that is actually stronger than $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Way to prove $\sin 3^{\circ}$ value I was looking wikipedia ,suddenly I saw this
$$\sin {{3}^{{}^\circ }}=\cos {{87}^{{}^\circ }}=\frac{2\left( 1-\sqrt{3} \right)\sqrt{5+\sqrt{5}}+\left( 1+\sqrt{3} \right)\left( \sqrt{10}-\sqrt{2} \right)}{16}$$
can someone help me that : how to find $\sin3^{\circ}$ or how to prove it .
Is there a Algebraic expressions for this kind of problem ?
Thanks in advanced.
| First
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}5=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
And
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac{1}{2}\left(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}\right)=\frac{1}{4}$$
So, you have the system
$$\left\{\begin{eqnarray}\cos\frac{\pi}{5}\cos\frac{3\pi}{5}&=&-\frac14\\
\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}&=&\frac12\end{eqnarray}\right.$$
Hence $\cos\frac{\pi}5$ and $\cos\frac{3\pi}5$ are the roots of $x^2-\frac12x-\frac14$. For this trinomial, $\Delta=\frac54$ and
$$x=\frac{1\pm\sqrt{5}}{4}$$
Since $\cos\frac{\pi}5>0$ and $\cos\frac{3\pi}5<0$, this means
$$\cos\frac{\pi}5=\frac{1+\sqrt{5}}{4}$$
And
$$\cos\frac{2\pi}{5}=-\cos\frac{3\pi}{5}=\frac{\sqrt{5}-1}{4}$$
Then
$$\sin\frac{\pi}{10}=\sin\left(\frac{\pi}{2}-\frac{2\pi}{5}\right)=\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4}$$
And
$$\cos\frac{\pi}{10}=\sqrt{\frac{1+\cos\frac{\pi}{5}}{2}}=\sqrt{\frac{5+\sqrt5}{8}}$$
Next
$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}+1)$$
$$\sin\frac{\pi}{12}=\sin\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}-1)$$
Finally
$$\sin 3^\circ=\sin\frac{\pi}{60}=\sin\left(\frac{\pi}{10}-\frac\pi{12}\right)\\=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}+1)\right)-\left(\sqrt{\frac{5+\sqrt5}{8}}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}-1)\right)\\
=\frac{2(1-\sqrt3)\sqrt{5+\sqrt{5}}+(1+\sqrt{3})(\sqrt{10}-\sqrt{2})}{16}$$
Computability of $\cos\frac{\pi}{n}$ and $\sin\frac{\pi}{n}$ by radicals is related to constructible polygons. A more difficult result arises for $n=17$, see for instance this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this:
Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
| You could also use that:
$$(k+2)(k+4) = (k+3)^2 - 1$$
and the known sum:
$$ \sum^{n}_{k=1}{k^2} = \frac{n(n+1)(2n+1)}{6} $$
$$ \sum^{n}_{k=1}{(k+2)(k+4)} = \sum^{n}_{k=1}{((k+3)^2 - 1)} = \sum^{n}_{k=1}{(k+3)^2} - \sum^{n}_{k=1}{1} $$
$$ \sum^{n+3}_{k=4}{k^2} - n = \sum^{n+3}_{k=1}{k^2} - \sum^{3}_{k=1}{k^2} - n $$
$$ \frac{(n+3)(n+4)(2n+7)}{6} - 14 - n $$
The rest is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Let $n \in \mathbb{Z}^+$, prove the identity $ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$ Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$
First of all $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$
$$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$
$$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k} \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$
We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$
For the second sum
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Integrating both sides from $0$ to $x$, we see that
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$
Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$
Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
| Suppose we seek to show that
$$\sum_{k=1}^{n-1} {n\choose k}
\frac{k n^{n-k}}{k+1} = \frac{n(n^n-1)}{n+1}.$$
Multiply by $n+1$ to get
$$\sum_{k=1}^{n-1} {n+1\choose k+1}
k n^{n-k} = n(n^n-1).$$
Extend to $k=n$ to obtain
$$\sum_{k=1}^{n} {n+1\choose k+1}
k n^{n-k} = n^{n+1}.$$
In order to prove this we introduce
$$n^{n-k} =
\frac{(n-k)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \exp(nz) \; dz.$$
This yields
$$\frac{1}{(n-k)!} n^{n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \exp(nz) \; dz.$$
or
$${n+1\choose k+1} n^{n-k} =
\frac{(n+1)!}{(k+1)!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \exp(nz) \; dz.$$
Observe that this vanishes when $k\gt n$
so we may extend the sum to infinity, getting
$$\frac{(n+1)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \exp(nz)
\sum_{k\ge 1} \frac{k}{(k+1)!} z^k
\; dz.$$
The sum term is
$$\sum_{k\ge 1} \frac{z^k}{k!}
- \sum_{k\ge 1} \frac{z^k}{(k+1)!}
= \exp(z) - 1 - \frac{1}{z} (\exp(z)-z-1)
\\ = \exp(z) - \frac{1}{z} \exp(z) + \frac{1}{z}.$$
Extracting the coefficients we thus obtain
$$(n+1)!
\left( \frac{(n+1)^n}{n!} - \frac{(n+1)^{n+1}}{(n+1)!}
+ \frac{n^{n+1}}{(n+1)!} \right)
\\ = (n+1)^{n+1} - (n+1)^{n+1} + n^{n+1} = n^{n+1}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find values of $a$ and $b$ that make matrix orthogonal
Given the matrix
$$A=\begin{bmatrix}1/2&a\\b&1/2\\ \end{bmatrix}$$
find the values of $a$ and $b$ that make it orthogonal.
So far I have tried using dot product $$(1/2)a+(1/2)b=0$$ and we can conclude that $a=-b$ and $b=-a$. I also tried the following theorem
$$A^T=A^{-1}$$
so
$$\begin{bmatrix}1/2&b\\a&1/2\\ \end{bmatrix}=
\begin{bmatrix}
(2+\frac{4ab}{1-4ab})&\frac{-4a}{1-4ab}\\
\frac{-4b}{1-4ab}&\frac{2}{1-4ab}\\ \end{bmatrix}$$
Can someone tell if I'am on the right track and point me in the right direction? Thanks!
| Orthogonal matrix means $$AA^T=I$$
Hence,
$$\begin{bmatrix}
\frac{1}{2} & a\\
b& \frac{1}{2}
\end{bmatrix}\begin{bmatrix}
\frac{1}{2} & b\\
a& \frac{1}{2}
\end{bmatrix}=\begin{bmatrix}
\frac{1}{4}+a^2 & \frac{1}{2}(a+b)\\
\frac{1}{2}(a+b)& \frac{1}{4}+b^2
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 1\end{bmatrix}$$
which implies $$a=-b=\pm\frac{\sqrt{3}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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I need help writing the expression $5(\sin 2x- \cos 2x)$ in terms of sine only. I need help writing the expression $5(\sin 2x- \cos 2x)$ in terms of sine only. Using the double angle identity formula, I was able to get it to $5(\sin 2x - 1 + 2\sin^2x)$, but not sure how to move forward. I know the answer is $$5\sqrt{2}\sin\left(2x+ \frac{7\pi}{4}\right)$$ but I'm not sure how to get there. Thanks.
| Write
$5(\sin2x-\cos2x) = a \sin(bx+c)$
Then
$5\sin2x-5\cos2x = a \sin(bx)\cos(c)+a \sin(c)\cos(bx)$
Choose $b=2$, $a\cos(c)=5$ and $a\sin(c)=-5$. So
$a^2\cos^2(c)+a^2\sin^2(c) = 25+25 = 50 \Rightarrow a^2=50 \Rightarrow a = \pm 5\sqrt{2}$
Choose $a = 5\sqrt{2}$
$\Rightarrow\cos(c) = \dfrac{5}{5\sqrt{2}}=\dfrac{\sqrt{2}}{2}$ and
$\sin(c) = \dfrac{-5}{5\sqrt{2}}=-\dfrac{\sqrt{2}}{2}$
Hence $c = \dfrac{7\pi}{4}$
Then
$5(\sin2x-\cos2x) = 5\sqrt{2} \sin\left(2x+\dfrac{7\pi}{4}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $ \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x$ I am trying to evaluate the exact value of the following definite integral:
\begin{align} \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x \end{align}
Since $ \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following:
\begin{align}
\frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3}
\end{align}
\begin{align}
\int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}_0 = 1
\end{align}
\begin{align}
\therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3}
\end{align}
Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?
| Your technique only works for integer multiples. After that, you're implicitly assuming that the integral will scale linearly, and that's not the case. So instead of $\frac{91}{3} \cdot 1$, it should be $$30\cdot 1 + \int_0^\frac{\pi}{6}|\cos x| dx = 30 + \sin\left(\frac{\pi}{6}\right) = \frac{61}{2}$$
A caveat - the period of $|\cos x|$ is in fact $\pi$, not $\frac{\pi}{2}$, so if the integer quotient had been odd (for instance, if the original integral had been from 0 to $\frac{47\pi}{3}$), you would need to start the integral for the remainder from $\frac{\pi}{2}$ (a region where $\cos x$ is negative), e.g.,
$$\int_0^{\frac{47\pi}{3}}|\cos x|dx = 31\cdot1+\int_\frac{\pi}{2}^\frac{2\pi}{3}|\cos x|dx = 31 - \int_\frac{\pi}{2}^\frac{2\pi}{3}\cos x dx = 31 - \frac{\sqrt{3}}{2} + 1 = 32 - \frac{\sqrt{3}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Discrete Math Proof of Inequalities In the following completed proof, in the highlighted steps, it seems as though the proof is defending its claim by showing that $2(\sqrt{k+2}-1) < (\text{the lower highlighted part in the proof})$. Shouldn't it be showing instead that $2(\sqrt{k+2}-1) \ge (\text{the lower highlighted part in the proof})$, in order for the proof to be true? Why or why not?
Why I think this:
Because the P(k+1) we are trying to show, has the left side greater than the right side.
The rest of the work needs to also try to show that the right side(which is being manipulated to try and look like the right side of the P(k+1) inequality) is less than the original P(k+1) inequality's right side, SO THAT the left side of the P(k+1) inequality will always be bigger.
| You've agreed that if $a > 1$ and $b>0$, the $ab > b$.
Now lets pick particular $a$ and $b$:
$$a = \sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}} \qquad \text{and} \qquad
b = 2\sqrt{k+2}.$$
Thus
$$ab = 2 \sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}}\sqrt{k+2} > 2\sqrt{k+2} = b.$$
Therefore $ab - 2 > b - 2$ implies
$$2 \left(\sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}}\sqrt{k+2}\right) - 2 > 2\sqrt{k+2} - 2 = 2\left(\sqrt{k+2}-1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a reccurrence relation We have the following equality: $y''-2zy'-2y=0$, (the derivatives are according to the variable z).
With the initial conditions: $y(0)=1, y'(0)=0$
Let us suppose that the solution can be written as: $\Sigma_{n≥0} a_n z^n$ with a radius of convergence $r>0$
I am asked to find a reccurence relation on the coefficients $a_n$.
Attempt at finding the solution:
We can notice the three following equalities:
*
*y = $\Sigma_{n≥0} a_n z^n$
*y' =$\Sigma_{n≥0} a_n n z^{n-1}$
*y''=$\Sigma_{n≥0} a_n n(n-1)z^{n-2}$
We can thus rewrite:
$y''-2zy'-2y=0$
$\iff \Sigma_{n≥0} a_n n(n-1)z^{n-2} - 2z\Sigma_{n≥0} a_n n z^{n-1}-2\Sigma_{n≥0} a_n z^n = 0$
$\iff \Sigma_{n≥2} (a_n (n-1)nz^{n-2}-2za_n z^n - 2a_nz^n) - 2a_0 -4a_1z=0$
$\iff \Sigma_{n≥2} (a_n (n-1)nz^{n-2}-2a_nz^n(n+1))-2a_0-4a_1 z =0$
Now I'm not sure in what direction I'm going and I don't know how I can find a recurrence relation on $a_n$
| Using
$$ y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n $$
then
\begin{align}
0 &= y'' - 2 \, x \,y' - 2y \\
&= \sum_{n=0} a_{n} \, n(n-1) \, x^{n-2} - 2 \, \sum_{n=0} n \, a_{n} \, x^{n} - 2 \, \sum_{n=0} a_{n} \, x^n \\
&= \sum_{n=2} n(n-1) \, a_{n} \, x^{n-2} - 2\, \sum_{n=0} (n+1) \, a_{n} \, x^{n} \\
&= \sum_{n=0} (n+1)(n+2) \, a_{n+2} \, x^{n} - 2\, \sum_{n=0} (n+1) \, a_{n} \, x^n \\
&= \sum_{n=0} \left[ (n+1)(n+2) \, a_{n+2} - 2 \, (n+1) \, a_{n} \right] \, x^{n}
\end{align}
Equating the coefficients yields
$$a_{n+2} = \frac{2 \,(n+1) \, a_{n} }{(n+1)(n+2)}$$
The first few coefficients are:
\begin{align}
a_{2} &= a_{0} \\
a_{3} &= \frac{2a_{1}}{3} \\
a_{4} &= \frac{a_{0}}{2} \\
a_{5} &= \frac{4a_{1}}{15}.
\end{align}
By using $y(0) = 1$ and $y'(0)=0$ ( which leads to $a_{0} = 1$ and $a_{1} = 0$) then the general form of the coefficients can be determined, which is:
\begin{align}
a_{2n} &= \frac{a_{0}}{n!} = \frac{1}{n!} \\
a_{2n+1} &= 0
\end{align}
and
$$y(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} = e^{x^{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How continue with this proof? I'm trying to proof the following statement:
$7+6 \cdot (7+7^1+7^2+\dots+7^n) = 7^{n+1}$
I try the next:
$n=k$
$7+6 \cdot (7+7^1+7^2+\dots+7^k) = 7^{k+1}$
$7+6 \cdot (7+7^1+7^2+\dots+7^k)-7 = 7^{k+1}-7$
$\frac{6 \cdot (7+7^1+7^2+\dots+7^k)}{6} = \frac{7^{k+1}-7}{6}$
$7+7^1+7^2+\dots+7^k = \frac{7^{k+1}-7}{6}$
$n=k+1$
$7+7^1+7^2+\dots+7^k+7^{k+1} = \frac{7^{k+2}-7}{6}$
Proof:
$\frac{7^{k+1}-7}{6}+7^{k+1} = \frac{7^{k+2}-7}{6}$
$\frac{7^{k+1}-7+6\cdot7^{k+1}}{6}$
Any idea?
| You asked for a proof, so here's proof by induction.
Your statement is equivalent to:
$$6 \cdot \sum_{k=1}^n 7^k = 7^{n+1} - 7.$$
This is true for $n=1$:
$$6 \cdot 7 = 7^2 - 7.$$
Now look at the $n+1$ case:
$$6 \cdot \sum_{k=1}^{n+1} 7^k = \left[6 \cdot \sum_{k=1}^{n} 7^k\right] + 6 \cdot 7^{n+1} \\ = 7^{n+1} - 7 + 6 \cdot 7^{n+1} \\= 7 \cdot 7^{n+1} - 7 \\= 7^{n+2} - 7.$$
I invoked the inductive assumption on the second line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$ $$a_1 = 2$$
$$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$
I tried to use induction but it looks as if the sequence isn't monotonic.
| we have $a_1=2$
by induction $2\geq a_n>1$.
$$a_{n+1}=a_n-\frac{a_n^2-1}{2a_n}$$
$$a_{n+1}-1=(a_n-1)(1-\frac{a_n+1}{2a_n})$$
$$=\frac{1}{2}(a_n-1)(1-\frac{1}{a_n})$$
$\implies$
$$|a_{n+1}-1| \leq \frac{1}{2}|a_n-1|$$
$....\leq (\frac{1}{2})^n|a_1-1|$
this proves $(a_n)$ tends to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Integration of $\arcsin(f(x))$ $$\int\sin^{-1}\left(\dfrac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$$
$\text {attempt}$ Let $x+1=u$, then the internal function becomes $\frac {2u}{\sqrt{4u^2+9}}$, and $dx=du$. Then I attempted by parts but derivative of $sin^{-1}(f(x))$ is troubling me in second part of IBP. Any better approach, or am I missing on a trick? I only know that the answer contains $\ln$ and $\tan^{-1}$. Thanks!
| Let us use integration by parts
$$u=\sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\implies u'=\frac 6 {4 x^2+8 x+13}$$ and $v=x$. So
$$I=\int \sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\,dx=x\sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)-\int \frac{6x}{4 x^2+8 x+13}\,dx$$ Now, using the roots of the quadatic in denominator an partial fractions $$\frac{6x}{4 x^2+8 x+13}=\frac{\frac{3}{2}-i}{2 x+(2+3 i)}+\frac{\frac{3}{2}+i}{2 x+(2-3 i)}$$ Integrating is now easy : you get two terms $a_i \log(b_i)$ in which $a_i,b_i$ are complex numbers. Expanding, you should find a logarithm and an arctangent.
Otherwise, write $$\frac{6x}{4 x^2+8 x+13}=\frac{6x+6-6 }{4 x^2+8 x+13 }=\frac 34\frac{8x+8-8 }{4 x^2+8 x+13 }$$ $$\frac{6x}{4 x^2+8 x+13}=\frac 34 \frac{(4 x^2+8 x+13)'}{4 x^2+8 x+13 }-\frac 6 {4 x^2+8 x+13 }$$ For the last piece, complete the square and things will become quite obvious.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the minimum value of $(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2$ We assume $\alpha,\beta$ are real numbers, I would like to find the minimum value of
$$(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2.$$
After expanding $(\alpha+5)^2+9\cos^2 \beta-2(\alpha+5)|3\cos \beta|+\alpha^2+4\sin^2 \beta-4\alpha|2\sin \beta|$, I would have some help with this.
|
Find the minimum value of $(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2$
Hint. By expanding as you did then factoring the quadratic expression in $\alpha$ one gets
$$ \small
\begin{align}
(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2&=2\left(\alpha+\frac{5-3|\cos \beta|-2|\sin \beta|}2\right)^2 +\frac{\left(5-3|\cos \beta|+2|\sin \beta|\right)^2}2
\\\\&\ge\frac{\left(5-3|\cos \beta|+2|\sin \beta|\right)^2}2
\\\\&\ge 2.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation of a circle tangent which makes a triangle of area $a^2$ with the coordinate axes What is the equation of the tangent to the circle $x^2+y^2-2ax-2ay+a^2=0$ which makes with the coordinate axes a triangle with area $a^2$?
Attempt: The equation of the circle can be re-written as $(x-a)^2+(y-a)^2=a^2$. Let's say the tangent equation is $\frac{x}{c}+\frac{y}{d}=1$. From the condition that it is a tangent I get
$|\frac{\frac{a}{c}+\frac{a}{d}-1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}|=a$. Simplifying I get $a(c+d-\sqrt{c^2+d^2})=cd$ and the area condition gives $\frac{1}{2}cd=a^2$. But I am not able to solve this two equation and get values of $c$ and $d$.
| In fact the problem is that there is a $\pm$ that you haven't considered:
$$\tag{1}|\frac{\frac{a}{c}+\frac{a}{d}-1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}|=a \ \ \ \Leftrightarrow \ \ \ \pm a(c+d-\sqrt{c^2+d^2})=cd.$$
Thus there are 2 cases:
*
*First case, with the plus sign in (1): setting $S=c+d$ and $P=cd$, your system of 2 equations with 2 unknowns $c,d$ can be transformed into
$$\cases{a(S + \sqrt{S^2 - 2P}) = P\\ P = 2a^2}$$
from which we deduce
$$S+\sqrt{S^2-4a^2}=2a \ \Rightarrow \ \sqrt{S^2-4a^2}=2a-S \ \Rightarrow \ S^2-4a^2=(2a-S)^2 \Rightarrow \ S=2a.$$
We are now with the classical issue: determine two values knowing their sum and their product. The solution is obtained by expressing that $c,d$ are roots of the quadratic:
$$X^2-SX+P=0 \ \ \iff \ \ X^2-2aX+2a^2=0 \ \ \iff \ \ (X-a)^2+a^2=0$$
which has no solutions... unless $a=0$ which is not a realistic solution.
In fact, it is graphically very expectable that there cannot be solutions to this problem: look at the figure below: either the area of the triangle is bigger (case OBC) or it is smaller (case OED) that the area of square $OGAH$:
*
*second case, with the minus sign in (1): Up to you, you will find solutions with $c>0$ and $d<0$ or with $c<0$ and $d>0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Trigonometric inequality $\sin(\pi x)>\cos(\pi \sqrt x)$ Solve the inequality
$\sin(\pi x)>\cos(\pi \sqrt x)$
I don't know where to begin. Hints?
| Thought:
Write $\sin (\pi x ) - \cos( \pi \sqrt{x} ) > 0$. Note $\sin ( \pi x) = \cos ( \pi/2 - \pi x) = \cos \left( \frac{ \pi - 2 \pi x }{2} \right) $. Thus,
$$ \cos \left( \frac{ \pi - 2 \pi x }{2} \right) - \cos( \pi \sqrt{x} ) = - 2 \sin \left( \frac{ \frac{\pi - 2 \pi x}{2} + \pi \sqrt{x} }{2} \right)\sin \left( \frac{ \frac{\pi - 2 \pi x}{2} - \pi \sqrt{x} }{2} \right) =$$
$$ = - 2 \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) $$
And this is at least $0$ iff
$$ \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) < 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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3 equations with 4 unknowns and information about the unknowns' types My dad just sent me a math problem on email and asked if I could solve it, but frankly, I have no idea where to start. I've got 3 equations with 4 unknowns and some information about the unknowns.
First, the equations:
$$2(z-1)-x=55$$
$$4xy-8z=12$$
$$a(y+z)=11$$
I am supposed to determine the 2 greatest real number values of $a$, where $x,y,z$ are positive natural numbers.
The problem is driving me nuts, so any help will be greatly appreciated (even just a hint). Thanks in advance.
| The thing that jumps out at me is
$4xy - 8z = 12$ means
$xy -2z = 3$ .... oh, for a second there, I thought that bounded things.
So well...
it gives a relation between $z$ and $xy$ as $z = \frac{xy -3}2$ which means $xy-3$ is even which means $xy$ is odd so $x$ and $y$ are odd.
We also have $2(z-1)-x=55$ so $2(\frac{xy-3}2 - 1) - x = 55$ so $xy-3 -2 - x = 55$
so $xy - x = 60$ so $x(y-1) = 60 = 4*3*5$ but both $x,y$ are odd $y-1$ is even.
So we have:
$(x,y-1) = \{(15,4)(5,12)(3,20)(1,60)\}$
And $(x,y,z=\frac{xy-3}2) = \{(15,5,36),(5,13,31),(3,21,30),(1,61,29)\}$
So $y + z = \{41, 44, 51, 90\}$
So $a(y+z) = 11 \implies a = \frac{11}{y+z} = \{\frac{11}{41},\frac{11}{44},\frac{11}{51},\frac{11}{90}\}$
so the to greatest are $a = 11/41$ and $x,y,z =15,5,36$ and $a= 1/4$ and $x,y,z = 5,13,31$.
Check:
$2(z−1)−x=55; 2(36 - 1) - 15 = 70-15=55;2(31-1)-5=60-5=55$
$4xy−8z=12;4*15*5 -8*36 = 300 - 288 =12; 4*5*13-8*31= 260 - 248 = 12$
$a(y+z)=11;\frac{11}{41}*(5+36) = 11; \frac 14*(13+31) = \frac{44}4 = 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Easier way to solve $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}\ldots=\frac49s$?
Show that $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\frac49s$, where $s=\sum k^{-2}$
I can divide the LHS in the sum of four series
$$1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\\=\sum_{k\ge 1}\frac1{(6k-1)^2}+\sum_{k\ge 1}\frac1{(6k-5)^2}-\sum_{k\ge 1}\frac1{(6k-2)^2}-\sum_{k\ge 1}\frac1{(6k-4)^2}=\frac49\sum_{k\ge 1}\frac1{k^2}$$
but manipulating the above expression is so tedious to show the equality. Do you know a different way to solve this problem? Thank you.
P.S.: I dont know exactly what kind of tags use for this question.
| I don't know if it is an easy way. But I say it anyway.
To make the required sum, we may start from
$S=\sum\frac{1}{k^2}$
Then, we wanna take all the terms of form $\frac{1}{(3k)^2}$ off.
Then, take $2$ times the terms of the form $\frac{1}{(2k)^2}$ off.
As there is an intersection between the terms of the form $\frac{1}{(3k)^2}$ and $\frac{1}{(2k)^2}$, $2$ times the terms of the form $\frac{1}{(6k)^2}$ is added.
The final result is
$\sum\frac{1}{k^2}-\sum\frac{1}{(3k)^2}-2\sum\frac{1}{(2k)^2}+2\sum\frac{1}{(6k)^2}=S-\frac{1}{9}S-\frac{1}{2}S+\frac{1}{18}S=\frac{4}{9}S$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the range of $\frac{1+\sin^4 x}{\sin^4 x}\cdot \frac{1+\cos^4 x}{\cos^4 x}$
Range of function $$f(x) = \frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}$$
Using $\bf{A.\geq G.M}$ Inequality,
$$1+\sin^4 x\geq 2\sin^2 x \qquad\text{and}\qquad 1+\cos^4 x\geq 2\cos^2 x$$
So $$\frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}\geq \frac{4}{\sin^2 x\cdot \cos^2 x} = \frac{16}{(\sin 2x)^2}\geq 16$$
But the answer given as $\left[25,\infty\right)$.
Please help me. How can I solve it using an inequality or any other way. Thanks.
| Answering to the question of why the OP method gives a lower bound for the codomain, consider (e.g.) the following:
$$
\begin{gathered}
\left( {z^{\,2} + 1} \right) \geqslant 2z\quad \left( {y^{\,2} + 1} \right) \geqslant 2y\quad \Rightarrow \quad \left( {z^{\,2} + 1} \right)\left( {y^{\,2} + 1} \right) \geqslant 4\,z\,y\quad \Rightarrow \hfill \\
\Rightarrow \quad \left( {z^{\,2} + 1} \right)\left( {y^{\,2} + 1} \right) = 4\quad \left| {\;\left| z \right| = 1\; \wedge \;\left| y \right| = 1} \right. \hfill \\
\end{gathered}
$$
Here, since each single inequality bears the $ \geqslant$ sign, implies that the equality is attained for some values of the variable.
When multiplied together , keeping the variables distinct , then surely for all
the $(y,z)$ combining the above values, the equality will be attained.
However, when in it we place $(y(x), z(x))$ we are going to restrict the values of
$y$ and $z$ to stay on the curve parametrized in $x$, which might not allow
$y$ and $z$ to attain "contemporarily" the equality (allow me to say that you might have $( = )\, \times \, ( > )$).
In particular, the substitution
$$
\left\{ \begin{gathered}
y = 1/\sin ^{\,2} x \hfill \\
z = 1/\cos ^{\,2} x \hfill \\
\end{gathered} \right.
$$
will produce the situation which is well evidentiated in the picture.
| {
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"url": "https://math.stackexchange.com/questions/1985787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Asymptotic approximation of Catalan Numbers The nth Catalan number is :
$$C_n = \frac {1} {n+1} \times {2n \choose n}$$
The problem 12-4 of CLRS asks to find :
$$C_n = \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) $$
And Stirling's approximation is:
$$n! = \sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ \Theta \left(\frac {1} {n}\right) \right)} $$
So, the nth catalan number becomes :
$$C_n = \frac {2n!}{(n+1)(n!)^2} $$
That, after applying Stirling's approximation becomes:
$$C_n = \left( \frac {1}{1+n} \right) \left( \frac {4^n}{\sqrt{\pi n}} \right) \frac {1}{\left( 1+\Theta \left(1/n\right) \right)}$$
And then, it becomes hopeless. The Asymptotic bound comes in the denominator, not in the numerator.
What should be done now?
Any help appreciated.
Moon
| We can use Stirling's approximation formula
\begin{align*}
n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)
\end{align*}
to prove:
The following is valid
\begin{align*}
C_n=\frac{1}{n+1}\binom{2n}{n}= \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) \tag{1}
\end{align*}
We obtain using (1)
\begin{align*}
\frac{1}{n+1}\binom{2n}{n}&=\frac{1}{n+1}\cdot\frac{(2n)!}{n!n!}\\
&=\frac{1}{n+1}\cdot\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}\left(1+O\left(\frac{1}{n}\right)\right)\\
&\qquad
\cdot \left(\frac{1}{\sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ O \left(\frac {1} {n}\right) \right)}}\right)^2\\
\\
&=\frac{1}{n+1}\cdot\frac{4^n}{\sqrt{\pi n}}\cdot
\frac{\left(1+O\left(\frac{1}{n}\right)\right)}{\left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)}\tag{2}\\
&=\frac{1}{n\left(1+O\left(\frac{1}{n}\right)\right)}\cdot\frac{4^n}{\sqrt{\pi n}}
\left(1+O\left(\frac{1}{n}\right)\right)^3\tag{3}\\
&=\frac{1}{n}\cdot\frac{4^n}{\sqrt{\pi n}}
\left(1+O\left(\frac{1}{n}\right)\right)^4\\
&=\frac{4^n}{\sqrt{\pi}n^{3/2}}
\left(1+O\left(\frac{1}{n}\right)\right)\tag{4}\\
\end{align*}
and the claim follows.
Comment:
*
*In (2) we do some cancellation
*In (3) we use the geometric series expansion
\begin{align*}
\frac{1}{1+O\left(\frac{1}{n}\right)}=1+O\left(\frac{1}{n}\right)
\end{align*}
*In (4) we use
\begin{align*}
\left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)=1+O\left(\frac{1}{n}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Wrong reasoning yields $1=0$ I was thinking about this sequence
\begin{equation}\tag{1}
n=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\ldots}}}}}
\end{equation}
It is well known that this converges to n. For example:
$$3=\sqrt{9}=\sqrt{6+3}=\sqrt{6+\sqrt{9}}=\sqrt{6+\sqrt{6+3}}=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$
The problem is when $n=0$ and $n=1$, because when we apply formula $(1)$, in both cases we get
$$\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{\ldots}}}}}$$
This yields
$$0=1$$
Can you tell me where the mistake is in the reasoning?
| If we look at $(1)$, say we set
$$x=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\dots}}}}$$
Because this goes on forever, we clearly have that
$$x^2-(n^2-n)=x$$
Solving this using the quadratic formula yields $x=1-n$ or $x=n$. At this point, we simply choose whatever makes since (i.e. if $n>1$, then $x$ is clearly positive so we take $x=n$). In our case of $n=1$, it makes sense to have $x=1-n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving inequality: $|\sqrt{x} - \sqrt{y}| \leq \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$ I am trying to prove the inequality, $|\sqrt{x} - \sqrt{y}| \leq \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$, holds for $x,y > 0$, and then using that to show that the function $f : \mathbb{R}^{+} \to \mathbb{R}, x \mapsto \sqrt{x}$ is continuous. I'm not positive that my proof of the inequality is correct, and also whether my relation to the function is sufficient enough.
My proof:
Let $x \geq y$. Then, $|\sqrt{x} - \sqrt{y}| = \sqrt{x} - \sqrt{y}$, and $|x - y| = x - y$. Hence,
\begin{align*}
2\text{min}\{\sqrt{x}, \sqrt{y}\} &= 2\Big(\frac{1}{2}\Big(\sqrt{x} + \sqrt{y} - |\sqrt{x} - \sqrt{y}|\Big)\Big) \\
&= \sqrt{x} + \sqrt{y} - \sqrt{x} + \sqrt{y} \\
&= 2\sqrt{y}.
\end{align*}
Thus, $\sqrt{x} - \sqrt{y} \leq \frac{x - y}{2\sqrt{y}}$. Consequently,
\begin{equation*}
0 \leq \sqrt{x} - \sqrt{y} \Rightarrow \sqrt{y} \leq \sqrt{x} \Rightarrow y \leq x \Rightarrow \frac{y}{\sqrt{y}} \leq \frac{x}{\sqrt{y}} \Rightarrow \frac{y}{2\sqrt{y}} \leq \frac{x}{2\sqrt{y}} \Rightarrow 0 \leq \frac{x-y}{2\sqrt{y}}.
\end{equation*}
Therefore, $0 \leq \sqrt{x} - \sqrt{y} \leq \frac{x - y}{2\sqrt{y}}$, for $x \geq y$.
Now let $y \geq y - x$. Then, $|\sqrt{x} - \sqrt{y}| = \sqrt{y} - \sqrt{x}$ and $|x - y| = y - x$. Hence,
\begin{align*}
2\text{min}\{\sqrt{x}, \sqrt{y}\} &= 2\Big(\frac{1}{2}\Big(\sqrt{x} + \sqrt{y} - |\sqrt{x} - \sqrt{y}|\Big)\Big) \\
&= \sqrt{x} + \sqrt{y} - (\sqrt{y} - \sqrt{x}) \\
&= \sqrt{x} + \sqrt{y} - \sqrt{y} + \sqrt{x} \\
&= 2\sqrt{x}.
\end{align*}
Thus, $\sqrt{y} - \sqrt{x} \leq \frac{y - x}{2\sqrt{x}}$. Consequently,
\begin{equation*}
0 \leq \sqrt{y} - \sqrt{x} \Rightarrow \sqrt{x} \leq \sqrt{y} \Rightarrow x \leq y \Rightarrow \frac{x}{2\sqrt{x}} \leq \frac{y}{2\sqrt{x}} \Rightarrow 0 \leq \frac{x-y}{2\sqrt{x}}.
\end{equation*}
Therefore, $0 \leq \sqrt{x} - \sqrt{y} \leq \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$, for $ y \geq x$.
Now, let there be a function $f : \mathbb{R}^{+} \to \mathbb{R}, x \mapsto \sqrt{x}$. This function is continuous for a $\sqrt{x} \in \mathbb{R}^{+}$ iff for each $\epsilon > 0$ there exists a $\delta > 0$ s.t. for all $\sqrt{y} \in \mathbb{R}^{+}$ one has $|\sqrt{x} - \sqrt{y}| < \delta$. Let $\delta = \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$. Then $|\sqrt{x} - \sqrt{y}| < \frac{|x-y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$. We proved this is true for cases $x,y > 0, x \geq y$ and $y \geq x$. Then we can conclude that $|f(\sqrt{x}) - f(\sqrt{y})| < \epsilon$, where $\epsilon = \frac{|f(\sqrt{x}) - f(\sqrt{y})|}{2\text{min}\{f(\sqrt{x}), f(\sqrt{y})\}}.$
| Or if $x$ and $y$ are not both $0$,
$$|\sqrt{x} - \sqrt{y}|= \frac{|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| }{|\sqrt{x} + \sqrt{y}| }\\ = \frac{|x - y| }{|\sqrt{x} + \sqrt{y}| }.$$
Since, $x \mapsto \sqrt{x}$ is non-negative, we have $|\sqrt{x} + \sqrt{y}| = \sqrt{x} + \sqrt{y} \geqslant 2 \min(\sqrt{x},\sqrt{y}), $ and
$$|\sqrt{x} - \sqrt{y}| \leqslant \frac{|x - y|}{2 \min(\sqrt{x},\sqrt{y}) }.$$
A more expedient way to prove continuity of $f(x) = \sqrt{x}$ (at $x=0$ as well) is to use
$$|\sqrt{x} - \sqrt{y}|^2 \leqslant |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x - y|.$$
If $|x - y| < \delta = \epsilon^2$, then $|\sqrt{x} - \sqrt{y}| < \epsilon.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\frac{x}{1-x}\geq0$ implies $0\leq x<1$? If $$\frac{x}{1-x}\geq{0}$$
How do we prove that $$0\leq{x}<1$$ using the laws of linear inequalities ?
Confusion:
From $\frac{x}{1-x}\geq{0}$$\implies$$x\geq{0}$ and $x\neq1$
and
From $\frac{x}{x-1}\leq{0}$$\implies$$x\leq{0}$ and $x\neq1$
| If $\frac ab \ge 0$ then either:
1)$a\ge 0$ and $b >0$
Or
2)$a <0$ and $b <0$
If $a=x $ and $b=1-x $ then 2)is impossible, as $x <0 \implies 1-x >1>0$.
So 1) must be true.
So $x\ge 0$ and $1-x>0$ so $1>x\ge 0 $.
OR
To address your confusion:
$\frac {x}{1-x} \ge 1$.
If $1-x <0\implies x >1$ then $\frac {x}{1-x}*(1-x)\le 0*(1-x) $ (remember, the inequality sign flips when multiplying both sides by negative) and $x \le 0$. This contradicts $x>1$.
if, however, $1-x\ge 0\implies x \le 1$ we have $\frac {x}{1-x}*(1-x)\ge 0*(1-z) $ and $x \ge 0$. We can't have $x=1$ so $x <1$ and $x \ge 0$ so $0\le x < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
From $32\cdot (\frac{1-.5^n}{1-.5})$ to $\frac{64(2^n-1)}{2^n}$ I have to prove that $32\cdot (\frac{1-.5^n}{1-.5})$ is equal to $\frac{64(2^n-1)}{2^n}$
I know that they are equal because when I put them in my calculator the graphs are the same. However, I'd like to do this manually.
So far I have done:
$$32\cdot (\frac{1-.5^n}{1-.5}) = 32 \cdot (\frac{.5}{.5} + \frac{.5^n}{.5}) = 32 \cdot (\frac{.5\cdot.5^n}{.5}) = \frac{32\cdot.5\cdot.5^n}{.5}= \frac{8^n}{.5} $$
How do I solve this?
| Don't be afraid of writing $.5 = \frac 12$
$32* (\frac{1-.5^n}{1-.5})=$
$2^5(\frac{1 - \frac 12^n}{\frac 12})=$
$2^6(1 - \frac 12^n)=$
$2^6(1 - \frac 12^n)\frac{2^n}{2^n}=$
$2^6\frac{2^n - 1}{2^n}=$
$64\frac{2^n - 1}{2^n}=$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ We know $ \alpha $ and $ \beta $ are roots of $ax^2+bx+c = 0 $.
also $S_{n-1} = \alpha^{n-1} + \beta^{n-1}$ , $S_{n} = \alpha^{n} + \beta^{n}$ and $S_{n+1} = \alpha^{n+1} + \beta^{n+1}$.
How we can find relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ ?
Note : We know that relation has $a$ , $b$ and $c$.
| $\alpha$ and $\beta$ satisfy the equation:
$ax^2+bx+c=0$
Thus they satisfy the following equation by multiplying with $x^{n-1}$:
$ax^{n+1}+bx^n+cx^{n-1}=0$
Thus:
$a\alpha^{n+1}+b\alpha^n+c\alpha^{n-1}=0$
$a\beta^{n+1}+b\beta^n+c\beta^{n-1}=0$
And by addition:
$aS_{n+1}+bS_n+cS_{n-1}=0$
It generalizes trivially to any polynomial of degree $\geq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find coefficient of ordinary generating function I'm having a hard time understanding how to find the coefficient of a generating function. I'm working through the example from my book (as shown in the photo). I see the answer is C(18, 15) - C(4,1)*C(12,9) + C(4,2)*C(6,3) and I understand why we care about the coefficient of C(18,15) or (x^15) but why do we want to subtract and add the coefficients from these other terms specifically? Why wouldn't the answer just be C(18,15)?
| In the following we use $\binom{n}{k}:=C(n,k)$ to denote the binomial coefficient $n$ choose $k$. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^{27}]&\left(x^3+x^4+\cdots+x^8\right)^{4}\\
&=[x^{27}]x^{12}\left(1+x+\cdots+x^5\right)^{4}\tag{1}\\
&=[x^{15}]\left(1+x+\cdots+x^5\right)^{4}\tag{2}\\
&=[x^{15}]\left(1-x^6\right)^4\left(1+x+x^2+\cdots\right)^4\tag{3}\\
&=[x^{15}]\left(1-\binom{4}{1}x^6+\binom{4}{2}x^{12}-\binom{4}{3}x^{18}+\binom{4}{4}x^{24}\right)\\
&\qquad\qquad\cdot
\left(1+\binom{4}{1}x+\binom{5}{2}x^2+\binom{6}{3}x^3+\cdots\right)\tag{4}\\
&=[x^{15}]\left(1-\binom{4}{1}x^6+\binom{4}{2}x^{12}\right)\cdot
\left(1+\binom{4}{1}x+\binom{5}{2}x^2+\binom{6}{3}x^3+\cdots\right)\tag{5}\\
&=\left([x^{15}]-\binom{4}{1}[x^9]+\binom{4}{2}[x^3]\right)\cdot
\left(1+\binom{4}{1}x+\binom{5}{2}x^2+\binom{6}{3}x^3+\cdots\right)\tag{6}\\
&=\binom{18}{15}-\binom{4}{1}\binom{12}{9}+\binom{4}{2}\binom{6}{3}\tag{7}\\
&=816-4\cdot220+6\cdot20\\
&=56
\end{align*}
and the claim follows.
Comment:
*
*In (1) we factor out $x^3$ and since we have to take the forth power we get $x^{12}$.
*In (2) we use the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
*In (3) we apply Theorem 2.2.1 of the book. This theorem effectively uses the summation formula for a finite geometric series
\begin{align*}
1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x}
\end{align*}
here with $n=5$ and it also uses the binomial series expansion with $\alpha =-4$ to obtain
\begin{align*}
\frac{1}{(1-x)^4}&=\sum_{n=0}^\infty \binom{-4}{n} (-x)^n\\
&=\sum_{n=0}^\infty \binom{n+3}{n}x^n\\
&=1+\binom{4}{1}x+\binom{5}{2}x^2+\binom{6}{3}x^3+\cdots
\end{align*}
Putting all together we get the right hand side of (3)
\begin{align*}
\left(1+x+\cdots+x^5\right)^{4}&=\left(\frac{1-x^6}{1-x}\right)^4\\
&=\left(1-x^6\right)^4\left(1+\binom{4}{1}x+\binom{5}{2}x^2+\binom{6}{3}x^3+\cdots\right)
\end{align*}
*In (4) we expand both, polynomial and series.
*In (5) we observe that powers $x^{18}$ and $x^{24}$ of the polynomial do not contribute anything to $[x^{15}]$ and can so be safely skipped.
*In (6) we use the linearity of the coefficient of operator and apply again the rule as we did in (2).
*In (7) we select the coefficients of $[x^{15}], [x^9]$ and $[x^3]$ of the series and obtain the representation in the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of integer solutions to $Ax + By + Cz = D$ How many integer solutions exits for the equations, $Ax + By + Cz = D$, where
$B = (A + 1), C = (A + 2)$, and $x, y, z$ are non-negative i.e $x, y, z, >= 0$
I require a general solution which can be implemented using code as well. We would be given the values of $A, D$.
| So, assuming that $A,\, D$ are non negative integers
$$
\left\{ \matrix{
0 \le x,y,z,A,D\; \in \mathbb Z \hfill \cr
Ax + \left( {A + 1} \right)y + \left( {A + 2} \right)z = D \hfill \cr} \right.
$$
when $A=0$ we are left with $y+2z=D$ and it is easy to see that the number of solutions is
$$
N_{A = 0} = 1 + \left\lfloor {{D \over 2}} \right\rfloor = \left\lceil {{{D + 1} \over 2}} \right\rceil
$$
Taking then $1 \le A$, we can write
$$
\left\{ \matrix{
0 \le x,y,k,A,D \hfill \cr
Ax + \left( {A + 1} \right)y = D - \left( {A + 2} \right)k \hfill \cr} \right.
$$
and since $\gcd \left( {A,A + 1} \right) = 1$ we can apply to the Popoviciu's Theorem which reads
$$ \bbox[lightyellow] {
\eqalign{
& p_{\left\{ {a,b} \right\}} (n) = \left| {\,\left\{ \matrix{
0 \le x,y,a,b,n \in Z \hfill \cr
\gcd (a,b) = 1 \hfill \cr
ax + by = n \hfill \cr} \right.\;} \right| = \cr
& = {n \over {ab}} - \left\{ {{{b^{\,\left( { - 1} \right)} n} \over a}} \right\} - \left\{ {{{a^{\,\left( { - 1} \right)} n} \over b}} \right\} + 1 \cr}
} \tag{1}$$
where:
$\left\{ x \right\} $ denotes the fractional part : $\left\{ x \right\} = x - \left\lfloor x \right\rfloor $
and the exponent in brackets denotes the modular inverse
$$
b^{\,\left( { - 1} \right)} b \equiv 1\;\left( {\bmod a} \right)\quad a^{\,\left( { - 1} \right)} a \equiv 1\;\left( {\bmod b} \right)
$$
In this respect we have
$$
\eqalign{
& A^{\,\left( { - 1} \right)} A \equiv 1\;\left( {\bmod A + 1} \right)\quad \Rightarrow \quad A^{\,\left( { - 1} \right)} = A \cr
& \left( {A + 1} \right)^{\,\left( { - 1} \right)} \left( {A + 1} \right) \equiv 1\;\left( {\bmod A} \right)\quad \Rightarrow
\quad \left( {A + 1} \right)^{\,\left( { - 1} \right)} = 1 \cr}
$$
Therefore the Popoviciu's theorem gives
$$
N_{\,1 \le A} = \sum\limits_{k = 0}^{\left\lfloor {D/\left( {A + 2} \right)} \right\rfloor } {\,\left( {{{D - \left( {A + 2} \right)k} \over {A\left( {A + 1} \right)}}
- \left\{ {{{D - \left( {A + 2} \right)k} \over A}} \right\} - \left\{ {{{A\left( {D - \left( {A + 2} \right)k} \right)} \over {A + 1}}} \right\} + 1} \right)}
$$
The terms above can be simplified to some extent. Let's put
$$d=\left\lfloor {D/\left( {A + 2} \right)} \right\rfloor $$
Then
$$
\eqalign{
& \sum\limits_{k = 0}^d {{{D - \left( {A + 2} \right)k} \over {A\left( {A + 1} \right)}} + 1}
= {{D + A\left( {A + 1} \right)} \over {A\left( {A + 1} \right)}}\left( {d + 1} \right) - {{\left( {A + 2} \right)} \over {A\left( {A + 1} \right)}}{{\left( {d + 1} \right)d} \over 2} = \cr
& = {{\left( {d + 1} \right)} \over {A\left( {A + 1} \right)}}\left( {D - d + A\left( {A - d/2 + 1} \right)} \right) \cr}
$$
and
$$
\eqalign{
& \left\{ {{{D - \left( {A + 2} \right)k} \over A}} \right\} + \left\{ {{{A\left( {D - \left( {A + 2} \right)k} \right)} \over {A + 1}}} \right\} = \cr
& = \left\{ {{{D - 2k} \over A}} \right\} + \left\{ {{{A\left( {D - k} \right)} \over {A + 1}}} \right\} \cr}
$$
Finally we obtain
$$ \bbox[lightyellow] {
\eqalign{
& N_{\,1 \le A} = {{\left( {d + 1} \right)} \over {A\left( {A + 1} \right)}}\left( {D - d + A\left( {A - d/2 + 1} \right)} \right) + \cr
& - \sum\limits_{k = 0}^d {\left\{ {{{D - 2k} \over A}} \right\} + \left\{ {{{A\left( {D - k} \right)} \over {A + 1}}} \right\}} \cr}
} \tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find a binomial term/general formula for recurrence relation We know that Pascal's triangle obeys the recurrence relation $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $
And we can simply $\binom{n}{k}$ by $\frac{n!}{k!\,(n - k)!}$
I have a recurrence relation where
$$ f(n, k) = f(n - 1, k) + f(n - 2, k - 1) $$
How can I get a generel formula for that?
Thanks in advance!
Edited:
Base Case:
$f(n, 1) = n$ and $f(n, k) = 0$ when $n < k$
|
We can solve the recurrence relation using generating functions. We define
\begin{align*}
F(x,y)=\sum_{n=0}^\infty\sum_{k=0}^\infty f(n,k)x^ny^k
\end{align*}
find a closed expression and extract the coefficient $[x^ny^k]F(x,y)=f(n,k)$.
From the stated boundary conditions of the recurrence relation
\begin{align*}
f(n,k)=f(n-1,k)+f(n-2,k-1)\qquad\qquad n\geq 2, k\geq 1\tag{1}
\end{align*}
which are
\begin{align*}
f(n,1)=1\qquad n\geq 1\\
f(n,k)=0\qquad n<k
\end{align*}
it also follows by (1)
\begin{align*}
f(n,0)=1\qquad n\geq 0
\end{align*}
We obtain
\begin{align*}
\sum_{n=2}^\infty&\sum_{k=1}^\infty f(n,k)x^ny^k\\
&=\sum_{n=2}^\infty\sum_{k=1}^\infty f(n-1,k)x^ny^k+\sum_{n=2}^\infty\sum_{k=1}^\infty f(n-2,k-1)x^ny^k\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty f(n,k)x^{n+1}y^k+\sum_{n=0}^\infty\sum_{k=0}^\infty f(n,k)x^{n+2}y^{k+1}\\
&=x\left(F(x,y)-\sum_{n=1}^\infty f(n,0)x^{n}-\sum_{k=0}^\infty f(0,k)y^k\right)+x^2yF(x,y)\\
&=x\left(F(x,y)-\left(\frac{1}{1-x}-1\right)-1\right)+x^2yF(x,y)\\
&=(x+x^2y)F(x,y)-\frac{x}{1-x}
\end{align*}
The LHS is
\begin{align*}
F(x,y)&-\sum_{n=2}^\infty f(n,0)x^n-\sum_{k=0}^\infty f(0,k)y^k-\sum_{k=0}^\infty f(1,k)xy^k\\
&=F(x,y)-\left(\frac{1}{1-x}-1-x\right)-1-(x+xy)\\
&=F(x,y)-\frac{1}{1-x}-xy
\end{align*}
LHS=RHS gives
\begin{align*}
F(x,y)-\frac{1}{1-x}-xy&=F(x,y)(x+x^2y)-\frac{x}{1-x}\\
F(x,y)(1-x-x^2y)&=1+xy\\
F(x,y)&=\frac{1+xy}{1-x-x^2y}
\end{align*}
In order to extract the coefficients of $F(x,y)$ we expand the generating function in powers of $x$ and $y$.
\begin{align*}
F(x,y)&=\frac{1+xy}{1-x}\cdot\frac{1}{1-\frac{x^2}{1-x}y}\\
&=\frac{1+xy}{1-x}\sum_{j=0}^\infty \left(\frac{x^2}{1-x}\right)^jy^j\\
&=\left(1+xy\right)\sum_{j=0}^\infty \frac{x^{2j}}{(1-x)^{j+1}}y^j\\
&=\left(1+xy\right)\sum_{j=0}^\infty x^{2j}\sum_{l=0}^\infty \binom{-(j+1)}{l}(-x)^ly^j\\
&=\left(1+xy\right)\sum_{j=0}^\infty x^{2j}\sum_{l=0}^\infty \binom{j+l}{l}x^ly^j\\
\end{align*}
We extract the coefficient $[x^ny^k]$ and we also use Iverson brackets
\begin{align*}
[[P(x)]]=\begin{cases}
1&\qquad P(x) \ \text{ true}\\
0&\qquad P(x) \ \text{ false}
\end{cases}
\end{align*}
This way we can treat multiple cases in one expression.
We obtain for $0\leq k\leq n$
\begin{align*}
[x^ny^k]F(x,y)&=[x^n][y^k]\left(1+xy\right)\sum_{j=0}^\infty x^{2j}\sum_{l=0}^\infty \binom{j+l}{l}x^ly^j\\
&=[x^n]\left([y^k]+x[y^{k-1}][[k\geq 1]]\right)\sum_{j=0}^\infty x^{2j}\sum_{l=0}^\infty \binom{j+l}{l}x^ly^j\\
&=[x^n]\left(x^{2k}\sum_{l=0}^\infty \binom{k+l}{l}x^l+ x^{2k-1}\sum_{l=0}^\infty \binom{k+l-1}{l}x^l[[k\geq 1]]\right)\\
&=\left([x^{n-2k}][[n\geq 2k]]\sum_{l=0}^\infty \binom{k+l}{l}x^l\right.\\
&\qquad\quad\left.+ [x^{n-2k+1}][[n\geq 2k-1]]\sum_{l=0}^\infty \binom{k+l-1}{l}x^l[[k\geq 1]]\right)\\
&=\binom{n-k}{n-2k}[[n\geq 2k]]+\binom{n-k}{n-2k+1}[[n\geq 2k-1]][[k\geq 1]]\\
&=\binom{n-k}{k}+\binom{n-k}{k-1}[[k\geq 1]]\\
&=\binom{n-k+1}{k}
\end{align*}
In the last two lines we use the convention $\binom{p}{q}=0$ for $0\leq p<q$.
We finally conclude
\begin{align*}
f(n,k)=\begin{cases}
\binom{n-k+1}{k}&\qquad\qquad 0\leq k<n\\
0&\qquad\qquad \text{otherwise}
\end{cases}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001859",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The integral $\ J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$ Here
Is there a general formula for $I(m,n)$?
I asked for a general formula for $\ I(m,n):=\int_0^{\infty} \frac{x^m}{x^n+1}dx$.
Is the formula $$I(m,n)=\frac{\pi}{n\sin((m+1)\frac{\pi}{n})}$$ true for every pair $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2\ $ and not just for non-negative integers with $\ 0\le m\le n-2\ $ ?
I wondered whether there is a similar closed form for
$$J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$$
I figured out that $$I(m,n)=J(m,n)+J(n-2-m,n)$$ holds for $0\le m\le n-2$.
For $\ m=0\ $, the first few values are
$$J(0,0)=\frac{1}{2}$$ $$J(0,1)=\ln(2)$$ $$J(0,2)=\frac{\pi}{4}$$ $$J(0,3)=\frac{2\sqrt{3}\pi+\ln(64)}{18}$$ $$J(0,4)=\frac{\pi+2\coth^{-1}(\sqrt{2})}{4\sqrt{2}}$$ $$J(0,5)=\frac{1}{5}\sqrt{\frac{1}{10}(5+\sqrt{5})}\pi+\frac{1}{20}(\ln(16)+2\sqrt{5}\coth^{-1}(\frac{3}{\sqrt{5}}))$$ $$J(0,6)=\frac{\pi+\sqrt{3}\ln(2+\sqrt{3})}{6}$$
Is there a general formula for $\ J(m,n)\ $ and does it hold for all pairs $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2$ ?
| There does exist a closed form for
$$
J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx \tag1
$$ in terms of the digamma function $\psi(\cdot)$.
Proposition. Let $m=1,2,\cdots$ and $n=1,2,\cdots$. One has
$$
J(m,n)=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right) \tag2
$$
then using
$$
\psi\left(r+1\right)-\psi\left(r \right)=\frac1r,\quad r \in \mathbb{Q}^*,\tag3
$$ and $$
\psi\left(\frac{m}{2n}\right) = -\gamma -\ln(4n) -\frac{\pi}{2}\cot\left(\frac{m\pi}{2n}\right) +2\sum_{k=1}^{n-1} \cos\left(\frac{\pi km}{n} \right) \ln\sin\left(\frac{k\pi}{2n}\right) \quad (m<2n)\tag4
$$ one gets a closed form in terms of a finite number of elementary functions.
Hint. By the change of variable, $x=u^{1/n}$, $dx=\dfrac1n u^{1/n-1}du$, one may write
$$
\begin{align}
J(m,n)&=\int_0^1 \frac{x^m}{x^n+1}dx
\\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}}{1+u}du
\\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}(1-u)}{1-u^2}du
\\&=\frac1{2n} \int_0^1 \frac{v^{\frac{m+n+1}{2n}-1}}{1-v}dv-\frac1{2n} \int_0^1 \frac{v^{\frac{m+1}{2n}-1}}{1-v}dv
\\&=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right)
\end{align}
$$ then one may conclude with Gauss's digamma theorem.
Edit. For any real numbers $a, b$ such that $a>0$ and $b>0$ we have
$$
\int_0^1 \frac{x^a}{x^b+1}\:dx=\frac1{2b} \psi\left(\frac{a+b+1}{2b}\right)-\frac1{2b}\psi\left(\frac{a+1}{2b} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Smith normal form of a specific matrix. What is the smith normal form decomposition $U^{-1}DV^{-1}$ of
$$
\pmatrix{q&1&0&0&0&0&1\\1&q&1&0&0&0&0\\0&1&q&1&0&0&0\\0&0&1&q&1&0&0\\0&0&0&1&q&1&0\\0&0&0&0&1&q&1\\1&0&0&0&0&1&q}
$$
where $q\in\Bbb Z$?
I am very much interested in $U,V$ as well ( the entire decomposition).
| According to http://www-math.mit.edu/~rstan/transparencies/snf.pdf,
the Smith Normal form for a nonsingular matrix $A$ is
$$
\textrm{Diag}(e_1,e_2,e_3,e_4,e_5,e_6,e_7),
$$
where $e_1\cdots e_i = \gcd \big( i\times i\textrm{ minors of }A \big)$.
Assume that $A$ is nonsingular. In fact, it turns out that $q=-2$ is the only value of $q$ such that $A$ is singular. This is because
$$
\det A = (q+2)(q^3-q^2-2q+1)^2.$$
If $q\neq -2$, then we have $e_1=e_2=e_3=e_4=e_5=1$, and
$$
\begin{align}
e_6&=\gcd(q^6-5q^4+6q^2-1,q(q^4-4q^2+3)-1) \\ &= (q^3-q^2-2q+1)\gcd(q^3+q^2-2q-1,q^2+q-1) \\ &=(q^3-q^2-2q+1) \gcd(q^2+q-1,q+1) \\&=(q^3-q^2-2q+1) \gcd(q+1, -1)\\
&=q^3-q^2-2q+1.\end{align}
$$
Finally,
$$
e_6e_7=\det A = (q+2)(q^3-q^2-2q+1)^2.$$
This gives
$$
e_7=(q+2)(q^3-q^2-2q+1).$$
If $q=-2$, then the Smith Normal Form for $A$ contains $0$ in the diagonal. Elementary row and column operations take $A$ to a matrix with all zero on the last row and the last column. Thus, we have $e_1=e_2=e_3=e_4=e_5=1$, and
$$
e_6=(q^3-q^2-2q+1)(q^3+q^2-2q-1)=7, \ \ e_7=0.
$$
| {
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"url": "https://math.stackexchange.com/questions/2004858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Solve the irrational equation I can not solve the equation:$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0.$$ Can someone help me. Thanky veru much.
| $$\sqrt[4]{x^3-2x^2-5x+6}=\sqrt[4]{(x+2)(x-1)(x-3)}\\\sqrt{x^2+5x+6}=\sqrt{(x+2)(x+3)}$$
Both quantities must be zero then we get the system $$\begin{cases}(x+2)(x-1)(x-3)=0\\(x+2)(x+3)=0\end{cases}$$ It is apparent that $(x+2)=0\iff x=-2$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$satisfying $\alpha+\beta+\gamma = \pi$
The $\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$ Where $\alpha,\beta,\gamma\in \mathbb{R}$ satisfying $\alpha+\beta+\gamma = \pi$
$\bf{Options ::}$ $(a)\;\; + ve \;\;\;\;\;\;\; (b)\;\; -ve \;\;\;\;\;\; (c)\;\; 0\;\;\;\;\;\; (d)\;\; -3$
$\bf{My\; Try::}$ Putting $\displaystyle \alpha = -\frac{\pi}{2}\;\;\;\;\;\; ,\beta = -\frac{\pi}{2}\;\;\;\;\;\; ,\gamma = 2\pi$
Then we get $\sin\alpha+\sin \beta+\sin \gamma = -ve $
$\bf{Added::}$ Trying using analytical way: Given $\alpha+\beta+\gamma = \pi$
$$\sin \alpha+\sin \beta+\sin \gamma = 2\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)+2\sin \frac{\gamma}{2}\cdot \cos \frac{\gamma}{2}$$
So $$ = 2\cos \frac{\gamma}{2}\left[\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}$$
Now Using $\displaystyle -1 \leq \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\leq 1$
So we get $$\sin \alpha+\sin \beta+\sin \gamma = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\geq -4$$
But this is not possible bcz $\displaystyle \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\neq -1$ Simultaneously
I did not understand how can i solve it
Help me , Thanks
| With an hopefully obvious notation, $$\sin\alpha+\sin\beta+\sin(\alpha+\beta)=a+b\pm a\sqrt{1-b^2}\pm b\sqrt{1-a^2}.$$
Cancelling the gradient, and rearranging,
$$\begin{cases}\sqrt{1-a^2}\pm\sqrt{1-a^2}\sqrt{1-b^2}\mp ab=0,
\\\sqrt{1-b^2}\mp ab\pm\sqrt{1-a^2}\sqrt{1-b^2}=0.\end{cases}$$
This proves a symmetry, $\sqrt{1-a^2}=\sqrt{1-b^2}$ or $a=\pm b$.
Now,
$$\sqrt{1-a^2}\pm(1-a^2)\mp a^2=0$$ or
$$c\pm c^2\mp(1-c^2)=0$$
yields the positive solutions
$$c=1,a=0
\\c=\frac12,a=\pm\frac{\sqrt3}2.$$
and the minimum is
$$-\frac{3\sqrt3}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof equality of series How to prove equality :
$$\sum_{n=0}^\infty x^n \frac{1+x^{2n+2}}{(1-x^{2n+2})^2} =
\sum_{n=0}^\infty (-1)^n \frac{x^n}{(1-x^{n+1})^2}, \: \forall x \in (-1, 1)$$
Any ideas how to start.
| Start with the identity due to Euler
$$
\prod_{m=0}^\infty\frac{1}{1-x^{2m+1}}=\prod_{m=1}^\infty(1+x^m).
$$
Taking logarithm of both sides, then expanding the logarithms into power series and reversing the order of summation we have
$$
\sum_{n=1}^\infty\frac{x^n}{n(1-x^{2n})}=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n(1-x^{n})}.
$$
Now differentiate termwise to obtain
$$\sum_{n=1}^\infty x^{n-1} \frac{1+x^{2n}}{(1-x^{2n})^2} =
\sum_{n=1}^\infty (-1)^{n-1} \frac{x^{n-1}}{(1-x^{n})^2},
$$
which is equivalent to
$$
\sum_{n=0}^\infty x^n \frac{1+x^{2n+2}}{(1-x^{2n+2})^2} =
\sum_{n=0}^\infty (-1)^n \frac{x^n}{(1-x^{n+1})^2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$ (i)$\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$
(ii)$\binom{n}{1}+2^2\cdot\binom{n}{2}+3^2\cdot\binom{n}{3}+...+n^2\cdot\binom{n}{n}=n(n+1)\cdot2^{n-2}$
I assume that the binomial series can maybe prove the formulas above $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$
| $$f_n(x)=(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$$
$$f_n'(x)=n(1+x)^{n-1}=\binom{n}{1}+2x\cdot\binom{n}{2}+3x^2\cdot\binom{n}{3}+...+nx^{n-1}\cdot\binom{n}{n}$$
Evaluate everything for $x=1$ and you have the answer to (i).
| {
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Solve using Cauchy Schwarz Inequality Use Cauchy–Schwarz inequality to prove that for three positive reals $a, b, c$ such that $a + b + c \leqslant 3$ then $\frac{1}{\sqrt{a}} +\frac{1}{\sqrt{b}} +\frac{1}{\sqrt{c}} \geqslant 3$. (Use C-S for $\sqrt{a} , \sqrt{b} , \sqrt{c}$ and $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c} }$).
How do I solve this?
| From Cauchy Schwarz inequality we have that $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}$ and $\sqrt{a},\sqrt{b},\sqrt{c}$
Then $$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq (\frac{1}{\sqrt{a}} \cdot \sqrt{a}+\frac{1}{\sqrt{b}} \cdot \sqrt{b}+\frac{1}{\sqrt{c}} \cdot \sqrt{c})^2=(1+1+1)^2=9$$
So
$$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$
In another hand for any $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$
because after expending you get that
$$x^2+y^2+z^2\geq xy+yz+zx$$
So back to our question we have that for $x=\sqrt{a},y=\sqrt{b},z=\sqrt{c}$
$$3(a+b+c)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ since a+b+c=3 and getting ride of square then
$$3\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})$$
From first inequality we have that
$$3 (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})\geq (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$
or
$$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq 3$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\tan 3a = \tan 3b = \tan 3c$ if $\sin a + \sin b + \sin c = 0$ and $\cos a + \cos b + \cos c = 0$ Let $a, b$ and $c$ be three angles such that:
$$\sin a + \sin b + \sin c = 0$$
$$\cos a + \cos b + \cos c = 0$$
Prove that $\tan 3a = \tan 3b = \tan 3c$.
I haven't done anything meaningful yet on this problem because I have no idea how I should start.
Thank you in advance!
| We have
$sin(a)=-sin(b)-sin(c)$
and
$cos(a)=-cos(b)-cos(c)$
thus
$1=2+2sin(b)sin(c)+2cos(b)cos(c)$
$\implies$
$cos(b-c)=-\frac 12=cos(a-b)=cos(a-c)$
$\implies$
$b-c=\pm\frac{2\pi}{3}+2k_1\pi$
and
$a-b=\pm\frac{2\pi}{3}+2k_2\pi$
$\implies$
$\tan(3a)=\tan(3b)=\tan(3c)$.
| {
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"url": "https://math.stackexchange.com/questions/2015438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Other ways to calculate this indefinite integral ($\int \frac{2\,dx}{(\cos(x) - \sin(x))^2}$)? I came across the following indefinite integral
$$ \int \frac{2\,dx}{(\cos(x) - \sin(x))^2} $$
and was able to solve it by doing the following:
First I wrote
$$\begin{align*}
\int \frac{2\,dx}{(\cos(x) - \sin(x))^2} &= \int \frac{2\,dx}{1 - 2\cos(x)\sin(x)} \\
&= \int \frac{2\,dx}{1 - \sin(2x)} \\
\end{align*}$$
Then setting $2x = z$,
$$\begin{align*}
&= \int \frac{dz}{1-\sin(z)}\\
&= \int \frac{1+\sin(z)}{\cos^2(z)}\,dz \\
&= \int \sec^2(z) + \tan(z)\sec(z) \,dz \\
&= \tan(z) + \sec(z) \\
&= \tan(2x) + \sec(2x).
\end{align*}$$
The solutions to the problem were given as $\tan(x + \pi/2)$ or $\frac{\cos(x) + \sin(x)}{\cos(x) - \sin(x)}$. I checked that these solutions are in face equivalent to my solution of $\tan(2x) + \sec(2x)$.
My question is, are there other ways to calculate this integral that more "directly" produce these solutions? Actually, any elegant calculation methods in general would be interesting.
| We are going to evaluate the integral by auxiliary angle. $$
\begin{aligned}
\int \frac{2 d x}{(\cos x-\sin x)^{2}} &=\int \frac{2 d x}{\left[\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)\right]^{2}} \\
&=\int \sec ^{2}\left(x+\frac{\pi}{4}\right) d x \\
&=\tan \left(x+\frac{\pi}{4}\right)+C \\ (\textrm{ OR }&=\frac{2 \sin x}{\cos x-\sin x}+C’)
\end{aligned}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Does factoring by sum and difference of cube or sum and difference of squares yield the same answer Q) Factor: $s^6-1$
If I start by doing sum of difference of squares I get
$(s^3)^2 - (1)^2$
$(s^3-1)(s^3+1)$
$(s+1)(s^2-s+1)(s-1)(s^2+s+1)$
If I start by doing sum or difference of cubes I get
$S^6 - 1$
$(s^2)^3 - (1)^3$
$(s^2 - 1) (s^4+s^2 + 1)$
ANS: $(s-1)(s+1) (s^4+s^2 + 1)$
Both the answers are different, as a matter of fact I know that when I started by doing the sum or difference of square method, the answer I got was correct , but what am I doing wrong in the 2nd method then?
| $s^4+s^2+1=(s^2-s+1)(s^2+s+1)$ so it is the same answer.
The way to complete the factoring is:
$s^4+s^2+1=s^4+2s^2+1-s^2=(s^2+1)^2-s^2=(s^2-s+1)(s^2+s+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $27 + abc = (a +b +c)(ab + ac+ bc)$ so $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$ Let $a$, $b$ and $c$ be positive numbers such that $27 + abc = (a +b +c)(ab + ac+ bc)$. Prove that:
$$\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$$
I have that $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a} \geq \frac{a+b+c}{3}$ then if I proof that $a+b+c \geq \frac{9}{2}$ I finish. Also $27 + abc = (a +b +c)(ab + ac+ bc)$ can be factorized to $27=(a+b)(ab + ac + bc + c^2)$.
| Michael Rozenberg's answer is nice; just would like to point out that the inequality $\sum\limits_{cyclic} \frac{a^{2}}{a+2b}\geq \frac{(a+b+c)}{3}$ follows from the Cauchy Schwarz inequality directly.
| {
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Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?
| Since $18=9\cdot 2$, we can rewrite the equation as$$(9\cdot2)^{n+1}=2^{n+1}\cdot 27\tag1$$
Simplifying $(1)$, we can eliminate $2^{n+1}$ to obtain$$3^{2n+2}=3^3\tag2$$
Which can be solved by $2n+2=3$. Solving, we get $n=\frac 12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a_{1} = 1$ and $a_{n} = n(a_{n-1}+1)\;\forall n\geq 2.$ Then $ \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$
Let $a_{1} = 1$ and $a_{n} = n(a_{n-1}+1)\;\forall n\geq 2.$ Then $\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$
$\bf{My\; Try::}$ Given $a_{n} = n(a_{n-1}+1)$
Replace $n\rightarrow n+1\;,$ We get $$a_{n+1}=(n+1)(a_{n}+1)\Rightarrow \frac{a_{n+1}}{n+1} = a_{n}+1$$
Now how can i solve it, Help required, Thanks
| Note that $$a_{r+1}=(r+1)(a_r+1) \implies \frac{a_{r+1}}{r+1}=a_r+1$$
Now $$\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{a_r+1}{a_{r}}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{1}{a_{r}}\cdot \frac{a_{r+1}}{r+1}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{1}{r+1}\cdot \frac{a_{r+1}}{a_{r}}\right)$$
$$=\displaystyle \left(\frac{1}{1+1}\cdot \frac{a_{1+1}}{a_{1}}\right)\left(\frac{1}{2+1}\cdot \frac{a_{2+1}}{a_{2}}\right)\left(\frac{1}{3+1}\cdot \frac{a_{3+1}}{a_{3}}\right)\ldots \left(\frac{1}{r+1}\cdot \frac{a_{r+1}}{a_{r}}\right)\ldots $$
$$=\displaystyle \left(\frac{1}{2}\cdot \frac{\not a_{2}}{a_{1}}\right)\left(\frac{1}{3}\cdot \frac{\not a_{3}}{\not a_{2}}\right)\left(\frac{1}{4}\cdot \frac{\not a_{4}}{\not a_{3}}\right)\ldots \left(\frac{1}{r+1}\cdot \frac{\not a_{r+1}}{\not a_{r}}\right)\ldots $$
Hope, this will serve as a sufficient hint since you wanted a hint only.
| {
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Solve a given trigonometric equation Solve the following equation:
$$\sin x + \sin \left( x + \frac{7\pi}{24} \right) = \frac{\sqrt{2 - \sqrt 2}}{2} + \frac{\sqrt 6 + \sqrt 2}{4}$$
So far, I found out that $\frac{\sqrt{2 - \sqrt 2}}{2} = \sin \frac{\pi}{8}$ and $\frac{\sqrt 6 + \sqrt 2}{4} = \sin \frac{7\pi}{12}$.
Thank you!
| Hint
Expand the second $\sin$, to write your
equation as
$$A\cos(x)+B\sin(x)=1$$
or
$$\cos(x+\alpha)=\frac{1}{\sqrt{A^2+B^2}}$$
second approach
the equation is
$$
\sin(x)+\sin(x+\frac{7\pi}{24})=$$
$$\sin( \frac{\pi}{8})+\sin(\frac{7\pi}{12} )=$$
$$\sin(\frac{\pi}{8})+\sin\left(\pi-(\frac{\pi}{8}+\frac{7\pi}{24})\right)=$$
$$\sin(\frac{\pi}{8})+\sin(\frac{\pi}{8}+\frac{7\pi}{24})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$
I have no idea on how to face this limit. Its value at $0$ seems to be $0$, but its limit equals $\frac{23}{24}$. I tried l'Hôpital but the new expressions were as confusing. I think maybe using squeeze theorem would be useful, but I don't know how to apply it in this case.
Any help would be appreciated, thanks in advance.
| Add and subtract $\cos x$ in numerator and then proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\notag\\
&= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos x + \cos x - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\notag\\
&= \lim_{x \to 0}\dfrac{1 - \dfrac{x^{2}}{2} - \cos x}{x^{4}} + \dfrac{\cos x - \cos\left(\dfrac{x}{1 - x^{2}}\right)}{x^{4}}\tag{1}\\
&= -\frac{1}{24} + 2\lim_{x \to 0}\dfrac{\sin\left(\dfrac{2x - x^{3}}{2(1 - x^{2})}\right)\sin\left(\dfrac{x^{3}}{2(1 - x^{2})}\right)}{x^{4}}\tag{2}\\
&= -\frac{1}{24} + 2\lim_{x \to 0}\dfrac{\sin\left(\dfrac{2x - x^{3}}{2(1 - x^{2})}\right)}{\dfrac{2x - x^{3}}{2(1 - x^{2})}}\cdot\dfrac{2 - x^{2}}{2(1 - x^{2})}\cdot\dfrac{\sin\left(\dfrac{x^{3}}{2(1 - x^{2})}\right)}{\dfrac{x^{3}}{2(1 - x^{2})}}\cdot\dfrac{1}{2(1 - x^{2})}\notag\\
&= -\frac{1}{24} + 2\cdot 1\cdot 1\cdot 1\cdot\frac{1}{2}\notag\\
&= \frac{23}{24}\notag
\end{align}
We have used Taylor series expansion for $\cos x$ while moving from step $(1)$ to step $(2)$.
| {
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Peter planted a tree. Its growth is given by $h(t) = \frac{a}{1+be^{kt}}$ for $t$ years. Find $a$,$b$ and $k$
Peter planted a tree on his backyard. Over the course of some years he
measured the height of the tree. first directly, and later on using
his knowledge of trigonometry. When he thought he had enough data, he
defined as the model for the growth of the tree in meters, after $t$
years of being planted, the following function:
$$h(t) = \frac{a}{1+be^{kt}}$$
Determine $a$,$b$ and $k$, knowing that:
*
*the tree was 50 cm tall when it was planted
*the maximum height the tree can have is 5 meters (approximate value)
*after 2 years, the tree has twice the height it had when it was planted
(Note: " 5 meters (approximate value)" is an exact quote)
I did:
$50 cm = 0.5m$,
$$0.5 = \frac{a}{1+be^{-k\cdot0}} \Leftrightarrow 0.5 = \frac{a}{1+b}$$
so I conlcude that
$a = \frac{b+1}{2}$ and $b = 2a-1$
Then I used the information on the third item,
$h(2) = 1$
and I tried to isolate $k$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-k\cdot2}} \Leftrightarrow 1 = \frac{b+1}{2(1+be^{-k2})} \Leftrightarrow 1 = \frac{b+1}{2+2be^{-k2}} \Leftrightarrow b+1 = 2+2be^{-2k} \Leftrightarrow \frac{b-1}{2} = be^{2k} \Leftrightarrow \frac{\frac{b-1}{2}}{b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = (\frac{1}{e^2})^k \Leftrightarrow k = \log_{\frac{1}{e^2}}{\frac{b-1}{2b}} \Leftrightarrow k = -\frac{1}{2} \ln(\frac{b-1}{2b}) \Leftrightarrow k = \ln(\sqrt{\frac{2b}{b-1}})$$
Then I tried the third sentence with $k = \ln(\sqrt{\frac{2b}{b-1}})$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-2\ln(\sqrt{\frac{2b}{b-1}})}}$$
This took me an $\infty$ of time to solve only to find out it was wrong, so I will skip the steps and tell you that both sides are always equal to 1 except for $x \in [-1;1]$ which show "ERROR"
How do I solve this?
| We have that
$$
h(t) = \frac{a}{1+be^{kt}}
$$
First information gives us.
$$
h(0) = \frac{a}{1+be^{k\cdot0}} = \frac{a}{1+b} = 0.5 \Leftrightarrow a = \frac{1+b}{2}
$$
Second information gives us the height of the tree after long time.
$$
\lim_{t \to \infty} h(t) \Rightarrow a = 5 \Rightarrow b = 9
$$
Third information gives $h(2) = 2h(0)$
$$
h(2) = 2h(0) = 1 \Leftrightarrow \frac{5}{1+9e^{2k}} = 1 \Leftrightarrow 5 = 1 + 9e^{2k} \\
\Leftrightarrow\\
4 = 9e^{2k} \\
\Leftrightarrow \\
2k = \ln\left(\frac{4}{9}\right) \Leftrightarrow
k = \ln\left(\frac{2}{3}\right)
$$
You wrote that $k = \ln\left(\sqrt{\frac{2b}{b-1}}\right)$, with $b=9$ you get the same $k$ as me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Error finding Fourier series I want to find the Fourier series for
$f(x)=\left\{\begin{array}{rcl} 0 & \mbox{ if } & -\pi \leq x \leq 0 \\ \sin x & \mbox{ if } & 0 < x < \pi \end{array}\right.$
The Fourier series is given by $F=a_0+\sum (a_n\cos nx + b_n \sin nx)$
My problem is that both $a_n$ and $b_n$ vanish when I try to calculate them.
$$a_0=\frac{1}{2\pi}\int_{0}^{\pi}\sin x = -\frac{1}{2\pi}(\cos \pi - cos 0)= \frac{1}{\pi}$$
$$a_n=\frac{1}{\pi}\int_{0}^{\pi}\sin x\cos nx dx = \frac{1}{\pi}\left(-\cos x \cos nx - \frac{1}{n}\int \sin nx \cos x dx\right)$$
$$=\frac{1}{\pi}\left(-\cos x\cos nx- \frac{1}{n} \left( - \frac{1}{n}\cos x\cos nx -\frac{1}{n}\int \sin x \cos nx dx \right)\right)$$
$$=\frac{1}{\pi}\left(-\cos x\cos nx+\frac{1}{n^2}\cos x \cos nx + \frac{1}{n^2}\int_{0}^{\pi}\sin x\cos nx dx\right)$$
$$\Rightarrow \frac{1}{\pi}\int_{0}^{\pi}\sin x\cos nx dx = \frac{1}{\pi}\frac{(1/n^2-1)\cos x \cos nx}{1-1/n^2}|^\pi_0$$
$$=-\frac{1}{\pi}\cos x \cos nx |^\pi_0 =0$$
$$b_n=\frac{1}{\pi}\int_0^\pi \sin x \sin nx=\frac{1}{\pi}\frac{(1/n^2)\cos x\sin nx - (1/n)\sin x \cos nx}{1-1/n^2}|_0^\pi=0$$
| $a_0=\frac{1}{\pi}\quad$ : OK.
$a_1=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\cos(x) dx = 0$
$a_n=-\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\cos(nx) dx = -\frac{1}{\pi}\frac{1+\cos(n\pi)}{n^2-1}= -\frac{1}{\pi}\:\frac{1+(-1)^n}{n^2-1}\qquad \qquad n\geq 2$
$b_1=\frac{1}{\pi}\int_{0}^{\pi}\sin^2(x) dx =\frac{1}{2}$
$b_n=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\sin(nx) dx =0\qquad \qquad n\neq 1$
$$F(x)=\frac{1}{\pi}+\frac{1}{2}\sin(x)-\frac{1}{\pi}\sum_{n=2}^\infty \frac{1+(-1)^n}{n^2-1}\cos(nx)$$
$$F(x)=\frac{1}{\pi}+\frac{1}{2}\sin(x)-\frac{2}{\pi}\sum_{k=1}^\infty \frac{1}{4\:k^2-1}\cos(2kx)$$
Graphs of incomplete Fourier series :
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Maximum protrusion over a folded piece of paper The pages of a book are $x$ cm tall and $y$ cm wide. If a page is folded appropriately, a corner of the page can stick out above the top of the book. What is the maximum amount that a page can protrude above the top of the book without tearing the page or separating it from the binding?
| I'll change the notation since I prefer using $x$ and $y$ as coordinates in the plane.
Let's say the edges of the page are on the lines $x=0$, $x=a$, $y=0$, $y=b$ before folding, and you fold along a line $y = c x + d$ with positive slope that crosses the top and bottom edges of the page, so $c > 0$, $0 \le (b-d)/c \le a$
and $0 \le -d/c \le a$.
$$(x,y) \to \left(\frac{1-c^2}{1+c^2} x + \frac{2c}{1+c^2} y - \frac{2cd}{1+c^2}, \frac{2c}{1+c^2} x + \frac{c^2-1}{1+c^2} y + \dfrac{2d}{1+c^2}\right)$$
We want to maximize
$$ f(c,d) = \frac{2c}{1+c^2} a + \frac{c^2-1}{1+c^2} b + \dfrac{2d}{1+c^2}$$
subject to $c > 0$, $0 \ge d \ge b-ac$.
Since $\partial f/\partial d = 2/(1+c^2) > 0$, the maximum will occur on the upper boundary of the region in $(c,d)$ space, i.e. at $d=0$. Taking the derivative with respect to $c$, we find that the solution is
$$ c = \dfrac{b + \sqrt{a^2+b^2}}{a},\ d=0$$
and the maximum protrusion amount turns out to be
$$ \dfrac{a^2}{b+\sqrt{a^2+b^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Show that the algebraic curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ can be given by a polynomial of degree $6$? I imagine that this should be done in the following way: There is a polynomial $P$ such that:
$$P(x^{\frac{2}{3}}+y^{\frac{2}{3}})=P$$
My first guess (obvious observation?) is that it can't be a polynomial in one variable, otherwise one of the variables will have a rational power. So the number of variables must be $2$. I've tried this:
$$(x^n+y^m)(x^{\frac{2}{3}}+y^{\frac{2}{3}})=(x^n+y^m)$$
$$(x^n+y^m)x^{\frac{2}{3}}+(x^n+y^m)y^{\frac{2}{3}}=x^n+y^m$$
$$x^nx^{\frac{2}{3}}+y^mx^{\frac{2}{3}}+x^ny^{\frac{2}{3}}+y^my^{\frac{2}{3}}-x^n-y^m=0$$
$$x^nx^{\frac{2}{3}}+\overbrace{y^m(x^{\frac{2}{3}}-1)+x^n(y^{\frac{2}{3}}-1)}^{\text{Annoying cross-terms }}+y^my^{\frac{2}{3}}=0$$
So this seems to show that we need to have at least one of the sums $n+2/3,m+2/3$ to be equal to $6$. But what should be the value of the other? And how do we eliminate the annoying cross-terms?
I understand also that $P$ could be $x^{a_1}+y^{b_1}+x^{a_2}+y^{b_2}+\dots +x^{a_t}+y^{b_t}$ for a certain $t$ but I don't know how to look into the entire collection of P's to find for it and also I don't know how to cut the general annoying cross-terms that this would give me. If possible, I'd like to have only some hints.
| Hint:
$$
\begin{align}
1=\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)^3 & = \left(\sqrt[3]{x^2}\right)^3 + 3\;\sqrt[3]{x^2}\;\sqrt[3]{y^2}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)+\left(\sqrt[3]{y^2}\right)^3 \\
& =x^2 + y^2 + 3\;\sqrt[3]{x^2\,y^2}
\end{align}
$$
Regroup, and raise to the $3^{rd}$ power one more time.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Let $\psi = \frac{1+\sqrt{5}}{2} $
$$x + \frac{1}{x} = \psi$$
$$x^2 + 1 = \psi x $$
$$x^2 - \psi x + 1 = 0$$
Use quadratic formula here to solve for $x$. Then plug that into the given expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 4
} |
Differentiable function to find a+b Let function $f(x)= \begin{cases}ax^3+2, &x<\dfrac 12 \\ bx^2+1, & x\ge \dfrac 12\end{cases}$. If the function is differentiable at $x=\dfrac 12$, then $a+b=$???
| $f(n) =
\begin{cases}
ax^3+2 & \text{x<1/2} \\[2ex]
bx^2+1, & \text{x≥1/2}
\end{cases}$
$f(n)' =
\begin{cases}
3ax^2 & \text{left derivative 1/2} \\[2ex]
2bx, & \text{right derivative in 1/2}
\end{cases}$
If f is differentiable at a point x, then f must also be continuous at x.
Since v(x) is a continuous function, then the limit as x approaches t is equal to the value of v(x) at x = t
$\begin{cases}
3a(1/2)^2=b(1/2)^2+1 & \text{left derivative at 1/2} \\[2ex]
2b(1/2)=b(1/2)^2+1, & \text{right derivative at 1/2}
\end{cases}$
$\begin{cases}
a(3/4)=b(1/4)+1 & \text{}\\[2ex]
(1-1/4)b=1, & \text{}
\end{cases}$
$\begin{cases}
a=16/9& \text{}\\[2ex]
b=4/3, & \text{}
\end{cases}$
$a+b=28/9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find and . Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find $$ and $$.
Can anyone see a slick way to do this? The solution is $x=10$ and $y=2$, but I am struggling with the method.
| $$\overbrace{\frac 38 \binom x{y+1}=\binom xy}^\text{given}=\frac {y+1}{x-y}\binom x{y+1}\\
\frac {y+1}{x-y}=\frac 38\\
3x=11y+8\qquad\cdots (1)\\\\
\binom x{\color{orange}{y}}=\binom x{\color{blue}{y+6}}\\
\binom x{\color{blue}{x-y}}=\binom x{\color{orange}{x-y-6}}\\
x-y=y+6\\
x=2y+6\qquad\cdots (2)\\
\text {From }(1),(2), \hspace{6cm}\\
\color{red}{x=10, y=2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
| For $n>1$ we have $$1/n^5+1/(n+1)^5+1/(n+2)^5+...<$$ $$<\frac {1}{n^3}\left(\frac {1}{(n-1)n}+\frac {1}{n(n+1)}+\frac {1}{(n+1)(n+2)}+...\right)=$$ $$=\frac {1}{n^3}\left((\frac {1}{n-1}-\frac {1}{n})+(\frac {1}{n}-\frac {1}{n+1})+(\frac {1}{n+1}-\frac {1}{n+2})+...\right)=$$ $$=\frac {1}{n^3(n-1)}$$ because the sum of the first $k$ terms of the last infinite series above is $\frac {1}{n-1}-\frac {1}{n-1+k}.$
This is not as sharp an upper bound as can be found by deeper methods. We have $1.036<\sum_{n=1}^{n=5}n^{-5}<1.037$ and the sum of the remaining terms is less than $1/(6^3\cdot 5)=1/1080.$ So the sum rounded to 2 decimal places is $1.04$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$ Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$
i know that $ \sum _{k=1} 1/k^2 = \pi^2/6$ how to do prove this
| I will show that
$\sum _{n=1}^{\infty} \frac{1}{(n-\frac{1}{2})^2}
=\dfrac{\pi^2}{2}
$.
$$S
=\sum _{n=1}^{\infty} \frac{1}{(n-\frac{1}{2})^2}
=\sum _{n=1}^{\infty} \frac{4}{(2n-1)^2}
=4\sum _{n=1}^{\infty} \frac{1}{(2n-1)^2}
.$$
So this is 4 times
the sum of the odd terms in
$$\sum _{n=1}^{\infty} \frac{1}{n^2}
.$$
To get the sum of the even terms,
$$\sum _{n=1}^{\infty} \frac{1}{(2n)^2}
=\sum _{n=1}^{\infty} \frac{1}{4n^2}
=\frac14\sum _{n=1}^{\infty} \frac{1}{n^2}
=\frac14\frac{\pi^2}{6}
=\frac{\pi^2}{24}
.$$
Combine these two,
using
$\sum _{n=1}^{\infty} \frac{1}{n^2}
=\frac{\pi^2}{6}
$
like this:
$\frac{S}{4}+\frac{\pi^2}{24}
=\frac{\pi^2}{6}
$
or
$\frac{S}{4}
=\frac{\pi^2}{8}
$
or
$S
=\frac{\pi^2}{2}
$.
So your comment appears correct!
As a check,
$\frac{\pi^2}{6}
\approx 1.5
$
and
$S > \frac1{.5^2}
=4
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Show that a ratio test is inconclusive for a given series, then determine whether the series converges/diverges?
4. Consider the series $\sum a_n$ where $$a_n=\begin{cases}n/2^n&n\text{ odd}\\1/2^n&n\text{ even}\end{cases}$$
*
*a. Show the Ratio Test is inconclusive
*b. Use the Root Test to determine whether the series is convergent or divergent.
Not sure how to do this..
| The series is a combination of geometric series and Gabriel's Staircase:
$$\sum_{k=0}^\infty a_k = \sum_{k=0}^\infty a_{2k}+\sum_{k=0}^\infty a_{2k+1} = \sum_{k=0}^\infty (\frac{1}{2})^{2k} + \sum_{k=0}^\infty (2k+1)(\frac{1}{2})^{2k+1}$$
We split again, rearrange and combine again:
$$= \sum_{k=0}^\infty (\frac{1}{2})^{2k} + \sum_{k=0}^\infty k (\frac{1}{4})^{k} + \sum_{k=0}^\infty (\frac{1}{2})^{2k+1} = \sum_{k=0}^\infty (\frac{1}{2})^{k} + \sum_{k=0}^\infty k (\frac{1}{4})^{k}$$
Now both series converge and we get as the combined value:
$$= \frac{1}{1-\frac{1}{2}} + \frac{1}{4} \frac{1}{(1-\frac{1}{4})^2} = 2 + \frac{1}{4} \frac{16}{9} = \frac{22}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why can a quadratic equation have only 2 roots? It is commonly known that the quadratic equation $ax^2+bx+c=0$ has two solutions given by: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ But how do I prove that another root couldn't exist?
I think derivation of quadratic formula is not enough....
| If you wrote the theorem out, it would look like this
THEOREM: Let $a,b,$ and $c$ be real numbers with $a \ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
That means
if $ax^2+bx+c=0$, then $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
and
if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 9
} |
Deriving formula for surface area of an ellipsoid I am doing some research on ellipsoids. I am not sure where the formula for the surface area of a prolate ellipsoid comes from. Can anyone please help me with how to derive the formula. I have the formula below
| The method is very standard and appears in most calculus texts.
Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be the ellipse such that $a>b$.
*
*For prolate spheroid, it rotates about the $x$-axis.
\begin{align*}
y &= \frac{b}{a} \sqrt{a^2-x^2} \\
\frac{dy}{dx} &= -\frac{bx}{a\sqrt{a^2-x^2}} \\
ds &= \sqrt{1+\left( \frac{dy}{dx} \right)^2} \, dx \\
&= \sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}} \, dx \\
&= a\frac{\sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}}
{\sqrt{a^2-x^2}} \, dx \\
S &= \int_{-a}^{a} 2\pi y \, ds \\
&= 4b\pi \int_{0}^{a}
\sqrt{1-\left( 1-\frac{b^2}{a^2} \right)\frac{x^2}{a^2}} \, dx \\
&= 4b\pi
\left[
\frac{x}{2} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}+
\frac{a^2}{2\sqrt{a^2-b^2}} \sin^{-1} \frac{x\sqrt{a^2-b^2}}{a^2}
\right]_{0}^{a} \\
&= 2\pi b
\left(
b+\frac{a^2}{\sqrt{a^2-b^2}} \sin^{-1} \frac{\sqrt{a^2-b^2}}{a}
\right) \\
\end{align*}
*
*For oblate spheroid, it rotates about the $y$-axis.
\begin{align*}
x &= \frac{a}{b} \sqrt{b^2-y^2} \\
\frac{dx}{dy} &= -\frac{ay}{b\sqrt{b^2-y^2}} \\
ds &= \sqrt{1+\left( \frac{dx}{dy} \right)^2} \, dy \\
&= \sqrt{1+\frac{a^2y^2}{b^2(b^2-y^2)}} \, dy \\
&= b\frac{\sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}}
{\sqrt{b^2-y^2}} \, dy \\
S &= \int_{-b}^{b} 2\pi x \, ds \\
&= 4a\pi \int_{0}^{b}
\sqrt{1+\left( \frac{a^2}{b^2}-1 \right)\frac{y^2}{b^2}} \, dy \\
&= 4a\pi
\left[
\frac{y}{2} \sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}+
\frac{b^2}{2\sqrt{a^2-b^2}} \sinh^{-1} \frac{y\sqrt{a^2-b^2}}{b^2}
\right]_{0}^{b} \\
&= 2\pi a
\left(
a+\frac{b^2}{\sqrt{a^2-b^2}} \sinh^{-1} \frac{\sqrt{a^2-b^2}}{b}
\right) \\
\end{align*}
The two cases are interchangeable by flipping the roles of $a$ and $b$ together with $\sinh iz=i\sin z$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.
How do I solve?
Thanks for any help!!
| $\lim_\limits{x\to y} \frac{\sin^2 x - \sin^2 y}{x^2-y^2} =\lim_\limits{x\to y} \left(\frac{\sin x - \sin y}{x-y}\right)\left(\frac{\sin x + \sin y}{x+y}\right)\\
(\frac {d}{dy} \sin y)(\frac {\sin y}{y})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Prove the given cubic inequality $a,b,c> 0$ prove $$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
I tried AM-GM but can't reached to solution please help me
| Using the Angel form of the Cauchy-Schwarz inequality: $\displaystyle \sum_{cyclic}\dfrac{a^3}{b+c} = \displaystyle \sum_{cyclic}\dfrac{(a^2)^2}{ab+ac}\ge \dfrac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\ge \dfrac{a^2+b^2+c^2}{2}$ since $a^2+b^2+c^2 \ge ab+bc+ca$ which is clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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rotation of hyperbola How can we rotate the rectangular hyperbola
xy=c ( c is any constant)
Into a form of standard hyperbola that is
(x/a)$^2$ - (y/b)$^2$ = 1
By rotating the hyperbola .
| $xy = c$
let $x = u-v\\ y = u+v$
$xy = u^2 - v^2 = c$
Now the transformation that I have just done has a little bit of spacial compression to it.
If you do a traformation along the lines of $x = au-bv, y = bu+av$ then there will be compression on the order of $\sqrt {a^2 + b^2}$ So, it is not a bad idea to choose $a,b$ such that $a^2 + b^2 = 1$
or $x = \cos \phi u - \sin\phi v\\ y = \sin\phi u + \cos\phi v$
and by trig identity that $cos^2 \phi + sin^2 \phi = 1$
$x = \frac {\sqrt 2}{2} u-\frac {\sqrt 2}{2}v\\ y = \frac {\sqrt 2}{2}u+\frac {\sqrt 2}{2}v\\
xy = \frac {u^2}2 + \frac {v^2}2 = c\\
\frac {u^2}{2c} + \frac {v^2}{2c} = 1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
About the strange identity $\frac{\pi}{4}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\cdot\frac{\zeta(4n+2)}{2^{2n+1}}$ I just wonder if this series
$${\pi\over 4}=\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\cdot{\zeta(4n+2)\over 2^{2n+1}}$$
Is equivalent to
$${\pi \over 4}=\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}?$$
Can anyone have a method to verify these sums?
| The generating function for the values of the $\zeta$ function at even, positive integers is well-known:
$$ \frac{1-\pi x \cot(\pi x)}{2}=\sum_{n\geq 1}x^{2n}\zeta(2n) \tag{1}$$
and that is easy to prove by considering the logarithmic derivative of the Weierstrass product for the sine function. By applying a discrete Fourier transform to $(1)$ we get:
$$ \sum_{n\geq 1}x^{4n+2}\zeta(4n+2) = -\frac{\pi x}{12}\left(2\pi x+3\cot(\pi x)-3\coth(\pi x)\right)\tag{2} $$
and by termwise integration:
$$\sum_{n\geq 1}\frac{x^{4n+2}}{4n+2}\zeta(4n+2)=\frac{1}{12}\left(-\pi^2 x^2+3\log\frac{\pi x}{\sin(\pi x)}-3\log\frac{\pi x}{\sinh(\pi x)}\right)\tag{4} $$
as well as:
$$\sum_{n\geq 1}\frac{x^{2n+1}}{4n+2}\zeta(4n+2)=\frac{1}{12}\left(-\pi^2 x+3\log\frac{\pi\sqrt{x}}{\sin(\pi\sqrt{x})}-3\log\frac{\pi\sqrt{x}}{\sinh(\pi\sqrt{x})}\right)\tag{5} $$
To compute the first series given by the OP, it is enough to evaluate $(5)$ at $x=\frac{i}{2}$.
That ensures
$$ \sum_{n\geq 0}\frac{(-1)^n}{2n+1} = \frac{\pi}{4} = \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\cdot\frac{\zeta(4n+2)}{2^{2n+1}}\tag{6}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Pair of Equations This is an problem I came up on some website or another...
" Two perpendicular lines are represented by the equations $2x - 3y = 6$ and $6x + ky = 4$. What is the value of $k$? TOSS-UP"
Any help? $3$ variables, $2$ equations?? No idea how to solve.
I tried elimination by getting to $ky + 9y = -14$, and then $k(y+9) = - 14$, and then $-14/(y+9) = k$, but I don't know how to do this.
| Equation of line y = mx + c
Where m is slope of line.
Line 1-
$2x-3y=6$
$3y=2x-6$
$y=\frac{2}{3}x-2$
So slope $m_{1}$ = $\frac{2}{3}$
Line 2-
$6x+ky=4$
$ky=-6x+4$
$y=\frac{-6}{k}x+\frac{4}{k}$
So slope $m_{2}$ = $\frac{-6}{k}$
Two lines are perpendicular so $m_{1} \times m_{2}$ = -1
$\frac{2}{3} \times \frac{-6}{k}$ = -1
k = 4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Help to prove $F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$ $0,1,1,2,3,5,8,...$ for $n=0,1,2,3,4...$ it is the n-th Fibonacci numbers.
$$F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$$
How can we show that?
I try:
Expand and simplify to
$F_{n+2}[(F_nF_{n+1})^2+3F_{n-1}^2F_nF_{n+1}+F_{n-1}^4]=F_n^5+F_{n+1}^5$
Any hints would be helpful? Thank you!
I saw two to these identities from mathworld
$F_{n-1}F_{n+1}=F_n^2+(-1)^n$ and $F_n^4-F_{n-2}F_{n-1}F_{n+1}F_{n+2}=1$
May be these can help to simplify the above but I can't see it yet.
A hint from @Rohan
$(a+2b)[b^2(a+b)^2+3a^2b(a+b)+a^4]$
$=(a+2b)[4a^2b^2+2ab^3+3a^3b+a^4+b^4]$
$=(4a^3b^2+2a^2b^3+3a^4b+a^5+ab^4)+(8a^2b^3+4ab^4+6a^3b^2+2ba^4+2b^5)$
$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+2b^5$
| One of the intersting method for obtaining relations between elements of Fibonacci numbers is based on matrix method. I saw this post about a relation between Fibonacci numbers and now the peresent question. The strategy of answers of two questions are the same and is based on this fact that if $F_n=a$ and $F_n+1=b$ then $F_n+2=a+b$ and finish. Now I want to suggest the matrix method for this kind of Question.
Suppose that we want find a closed-form expression for $F_n^m$ where $F_n$ is the $n$th term of Fibonacci numbers and $m$ is non-negative integer number. The number $m$ is odd or even. Let $m=2k+1$ where $k$ is a natural number. We know the $Q$ matrix is in the following form
$$
Q=
\left( \begin {array}{cc}
0&1\\
1&1
\end {array} \right)
$$
And It is well-kown that the $n$th power of $Q$ matrix is as follows
$$
Q^n=
\left( \begin {array}{cc}
F_{{n-1}}&F_{{n}}\\
F_{{n}}&F_{{n+1}}
\end {array} \right)
$$
It is easy to see that
$$
det(Q^n)={(-1)}^n \Longrightarrow F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2}={(-1)}^n
$$
Now, we want to find a closed form expression for $F_n^m=F_n^{2k+1}=F_n^{2k}\,F_n$. At first, we compute the following equation
$$
{(F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2})}^k={(-1)}^{nk}
$$
We can rewrite the above eqation as shown
$$
F_n^{2k}=G(n)+{(-1)}^{nk} \Longrightarrow F_n^{2k+1}=F_n\,G(n)+F_n\,{(-1)}^{nk}
$$
Where $G(n)$ is a function based on the Fibonacci elements. For example, we want to find a closed form experesion for $F_n^5$. At first, we obtain the following equation
$$
{(F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2})}^2={(-1)}^{2n}
$$
$$
{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+{F_{{n}}}^{4}-2\,F_{{n-1}}{F_{{n}}}^{2}
F_{{n+1}}=1
$$
$$
{F_{{n}}}^{4}=2\,F_{{n-1}}{F_{{n}}}^{2}
F_{{n+1}}-{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+1
$$
$$
{F_{{n}}}^{5}=2\,F_{{n-1}}{F_{{n}}}^{3}
F_{{n+1}}-{F_n}{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+F_n
$$
Now, by using the expression of ${F_{{n}}}^{5}$, we conclude that
$$
{F_{{n+1}}}^{5}=2\,F_{{n}}{F_{{n+1}}}^{3}
F_{{n+2}}-{F_{n+1}}{F_{{n}}}^{2}{F_{{n+2}}}^{2}+F_{n+1}
$$
From the summation of the expressions ${F_{{n}}}^{5}$ and ${F_{{n}}}^{5}$, we obtain our results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?
| $$x+y+z=1 \quad \text{and} \quad xyz=1$$
\begin{align}
xy(1-x-y) &= 1 \\
xy - x^2y - xy^2 = 1 \\
xy^2 + x^2y - xy + 1 &= 0 \\
xy^2 + (x^2 - x)y + 1 &= 0 \\
y^2 + (x-1)y + \dfrac 1x &= 0 \qquad\text{{$x\ne 0$ since $xyz=1$.}}\\
y &= \dfrac 12(1-x) \pm \dfrac 12\sqrt{(x-1)^2-\dfrac 4x}
\end{align}
So we need to find non negative real numbers, $n$, for which we can solve
$$(x-1)^2-\dfrac 4x = n^2$$
Clearly then, there exists an $n$ for all negative values of x.
When x is positive, Wolfram alpha tells us that the single real-valued solution to $(x-1)^2-\dfrac 4x = 0$ is $x=\xi$, where
$\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$
A value of $n$ will exists for all $x \ge \xi$.
Working it out, the solution set is
$$\{x,y,z\} =
\left\{x,\;
\dfrac 12(1-x) + \dfrac 12\sqrt{(x-1)^2-\dfrac 4x},\;
\dfrac 12(1-x) - \dfrac 12\sqrt{(x-1)^2-\dfrac 4x}
\;\right\}$$
for all $x \in (-\infty, 0) \cup [\xi, \infty)$
where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 15,
"answer_id": 3
} |
Quadratic equation has integral roots
Find all positive numbers $p$ for which the equation $x^2+px+3p = 0$ has integral roots.
We have by the quadratic formula $$x = \dfrac{-p \pm \sqrt{p^2-12p}}{2}.$$ Thus, $p^2-12p = p(p-12)$ must be a perfect square. How do we continue?
| Let $a,b$ be the integer roots. Then by Vieta's formulas:
$$
\begin{align}
a + b = -p \tag{1} \\
ab = 3p \tag{2}
\end{align}
$$
By $(2)$ the roots must have the same sign, and by $(1)$ they must be both negative. If $p=0$ then the trivial solution is $a=b=0\,$, otherwise dividing $(1) / (2)\,$:
$$
\frac{1}{a}+\frac{1}{b} = -\frac{1}{3} \quad \iff \quad \frac{1}{(-a)}+\frac{1}{(-b)} = \frac{1}{3}
$$
The above gives another trivial solution $a=b=-6 \iff p=12\,$, otherwise it reduces to the problem of expressing the unit fraction $\cfrac{1}{3}$ as the sum of two distinct unit fractions $\cfrac{1}{(-a)}+\cfrac{1}{(-b)}$. The latter is (related to egyptian fractions, and is) known to have a unique solution since $3$ is a prime (see this for example), which is $a=-4,b=-12$ in this case, giving $p=16$.
Summing up the subcases listed above, $p \in \{0, 12, 16\}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Prove that ${\pi\over2}=\left({2\over1}\right)^{1\over2}\left({2^2\over 1\cdot3}\right)^{1\over 4} \cdots$ A general infinite product take that form let say, $F(p)=p_1p_2p_3p_4...$
I notices a pattern that accelerate $F(p)$
Accelerate version
$$F(p)=q_1^{1\over 2}q_2^{1\over 4}q_3^{1\over 8}q_4^{1\over 16}...$$
Where
$q_1=p_1$, $q_2=p_1p_2$, $q_3=p_1p_2^2p_3$, $q_4=p_1p_2^3p_3^3p_4$
$q_5=p_1p_2^4p_3^6p_4^4p_5$ And so on .... The power are binomial coefficients
Let apply this idea to Wallis's product
$${\pi\over2}={2\over 1}{2\over3}{4\over3}{4\over5}{6\over5}\cdots$$
$q_1={2\over1}$, $q_2={2\over1}{2\over3}={2^2\over 1\cdot3}$
$q_3={2\over1}{\left(2\over3\right)^2}{4\over3}={2^3\cdot4\over 1\cdot3^3}$
$q_4={2\over1}{\left(2\over3\right)^3}{\left(4\over3\right)^3}{4\over5}={2^4\cdot4^4\over1\cdot3^6\cdot5}$ and son on...
$${\pi\over2}=\left({2\over1}\right)^{1\over2}\left({2^2\over 1\cdot3}\right)^{1\over 4}
\left({2^3\cdot4\over 1\cdot3^3}\right)^{1\over8}\left({2^4\cdot4^4\over1\cdot3^6\cdot5}\right)^{1\over16}\cdots$$
Question:
How can I prove that
$${\pi\over2}=\left({2\over1}\right)^{1\over2}\left({2^2\over 1\cdot3}\right)^{1\over 4}
\left({2^3\cdot4\over 1\cdot3^3}\right)^{1\over8}\left({2^4\cdot4^4\over1\cdot3^6\cdot5}\right)^{1\over16}\cdots$$ is an accelerate of convergence to
$${\pi\over2}={2\over 1}{2\over3}{4\over3}{4\over5}{6\over5}\cdots$$
As in general using the above idea?
| Take the logarithm of both product representations of $F(p)$ and exchange sums.
$\displaystyle \sum\limits_{k=1}^\infty \ln p_k=\sum\limits_{k=1}^\infty \frac{\ln q_k}{2^k}=\sum\limits_{k=1}^\infty (\ln p_k)\sum\limits_{v=k}^\infty \frac{1}{2^v}\binom{v-1}{k-1}\enspace$ because of $\enspace\displaystyle \ln q_k=\sum\limits_{v=1}^k\binom{k-1}{v-1}\ln p_v$
The reason for the equality is
$$1=[x^k](\frac{1}{1-x})=[x^k](\frac{1}{2}\frac{1}{1-\frac{x+1}{2}})=\sum\limits_{v=k}^\infty \frac{1}{2^v}\binom{v-1}{k-1}$$
with $\enspace k\in\mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{-\infty}^\infty\frac1{1+x^2+x^4+\cdots}\ \text{dx}$ I have calculated that
$$\begin{align}
\int_{-\infty}^\infty\frac1{1+x^2}\ \text{dx}&=\pi \\
\int_{-\infty}^\infty\frac1{1+x^2+x^4}\ \text{dx}&=\frac\pi{\sqrt3} \\
\int_{-\infty}^\infty\frac1{1+x^2+x^4+x^6}\ \text{dx}&=\frac\pi2 \\
\int_{-\infty}^\infty\frac1{1+x^2+x^4+x^6+x^8}\ \text{dx}&=\frac1 5\sqrt{10-2\sqrt5}\ \pi
\end{align}
$$
And thus I am trying to evaluate
$$
\lim_{\mathtt k\rightarrow\infty}\int_{-\infty}^\infty\frac1{1+x^2+\cdots+x^{2\mathtt k}}\ \text{dx}
=
\lim_{\mathtt k\rightarrow\infty}\int_{-\infty}^\infty\frac1{\sum_{n=0}^\mathtt kx^{2n}}\ \text{dx}=\ ?
$$
I intend to post a self-answer to this question, assuming a satisfactory answer is not posted within a few hours.
| This is a bit overkill for solving the question at hand, but for completness I'll add an answer for how one can evaluate the integral for a general value of $n$ using residue calculus giving us the result
$$\int_{-\infty}^\infty\frac{{\rm d}x}{1+x^2+\ldots+x^{2n}} = \frac{2\pi}{n+1}\frac{\cos\left(\frac{\pi}{2n+2}\right)}{\sin\left(\frac{3\pi }{2 n+2}\right)}\tag{1}$$
Taking the limit $n\to\infty$ in $(1)$ we get $\frac{4}{3}$, but this can of course be found by simpler methods as pointed to in the other answers.
By summing the geometrical series in the denominator we get
$$\int_{-\infty}^\infty\frac{{\rm d}x}{1+x^2+\ldots+x^{2n}} = \int_{-\infty}^\infty \frac{x^2-1}{x^{2n+2}-1}{\rm d}x$$
We now integrate the function $f(z) = \frac{z^2-1}{z^{2n+2}-1}$ around a semi-circle contour of radius $R$ in the upper half-plane. By the residue theorem we have
$$\oint f(z)\,{\rm d}z = \int_{-R}^R\frac{x^2-1}{x^{2n+2}-1}{\rm d}x + \int_{\rm semi-circle} \frac{z^2-1}{z^{2n+2}-1}{\rm d}z = 2\pi i \sum \text{Res}[f(z);z_k]$$
where $z_k$ are the poles of $f$ inside the contour. For $R>1$ these are given by the zeros of $z^{2n+2}-1$ in the upper half-plane (the singularities at $z=\pm 1$ are removable) so $z_k = e^{\frac{2\pi i}{2n+2} k}$ for $k=1,2,\ldots,n$. Since the poles of $f$ are simple the residues of $f$ are given by $$\text{Res}[f(z);z_k] = \frac{z_k^2-1}{\frac{d}{dz}(z^{2n+2}-1)_{z=z_k}} = \frac{z_k^2-1}{(2n+2)z_k^{2n+1}} = \frac{1}{2n+2}(z_k^3 - z_k)$$ where we have used $z_k^{2n+2} = 1$ to simplify. Taking $R\to \infty$ we find that the contribution from the semi-circle vanishes by a standard $M-L$ estimate since $|f(z)| \leq \frac{R^2}{R^{2n+2}-1} \sim \frac{1}{R^{2n}}$ and $\lim_{R\to\infty}\pi R \cdot \frac{1}{R^{2n}} = 0$ for all $n\geq 1$. This leads to
$$\int_{-\infty}^\infty\frac{z^2-1}{z^{2n+2}-1}{\rm d}z = \frac{2\pi i}{2n+2}\sum_{k=1}^n e^{3\frac{2\pi i}{2n+2} k} - e^{\frac{2\pi i}{2n+2} k} \\= \frac{2\pi i}{2n+2}\left(\frac{1+e^{\frac{3 i \pi }{n+1}}}{1-e^{\frac{3 i \pi }{n+1}}} - \frac{1+e^{\frac{i \pi }{n+1}}}{1-e^{\frac{i \pi }{n+1}}}\right) = \frac{2\pi}{n+1}\frac{\cos\left(\frac{\pi}{2n+2}\right)}{\sin\left(\frac{3\pi }{2 n+2}\right)}$$
where the sum above is another two geometrical series and I used Euler's formula to simplify the result.
| {
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"url": "https://math.stackexchange.com/questions/2069123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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How to find the maximum value of $|a_{2}|$?
(Tenth grade high school math competition):Let $$f(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4$$
where $a_{i}\in \mathbb R$, and such that for $x\in[-1,1]$ we have
$$|f(x)|\le 1$$
Find the maximum of the value $|a_{2}|$.
I know Chebyshev polynomial $T_{4}(x)=8x^4-8x^2+1$ verifies such condition.
But how to prove that it generates the looked for maximum ?
| Note that we only have to consider $f$ which are even, since if $f$ is a quartic with $|f(x)|\le 1$ for $-1\le x\le 1$, then $e(x) = \frac{f(x)+f(-x)}{2}$ is also a quartic with $|e(x)|\le 1$ for $-1\le x\le 1$, and $e$ is even and has the same coefficient of $x^2$ as $f$.
As such, let $f(x) = a_0+a_2x^2+a_4x^4$. Then $f(x) = g(x^2)$, where $g(x) = a_0+a_2x+a_4x^2$, and $g$ has the property that $|g(x)|\le 1$ for $0\le x\le 1$ (since $-1\le x\le 1\iff 0\le x^2\le 1$). In particular, $|g(x)|\le 1$ for $x=0$, $\frac{1}{2}$, and $1$. Now
$$ a_2 = 4\cdot\left(a_0+\frac{a_2}{2} + \frac{a_4}{4}\right) - (a_0+a_2+a_4) -3a_0 = 4g\left(\frac{1}{2}\right) - g(1) - 3g(0) $$
and so $g\left(\frac{1}{2}\right)\le 1$, $g(1)\ge -1$, and $g(0)\ge -1$ imply that
$$ a_2 \le 4\cdot 1 - (-1) - 3\cdot(-1) = 8.$$
Similarly, $g\left(\frac{1}{2}\right)\ge 1$, $g(1)\le -1$, and $g(0)\le -1$ imply that $ a_2 \ge -8$. Hence, we must have $|a_2|\le 8$.
Conversely, $g(x) = 8\left(x-\frac{1}{2}\right)^2-1 = 8x^2-8x+1$ satisfies $|g(x)|\le 1$ for $0\le x\le 1$, and hence $f(x) = 8x^4-8x^2+1$ satisfies $|f(x)|\le 1$ for $-1\le x\le 1$. Hence, $8$ is the maximum possible value of $|a_2|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How can we show that $2^3+1^3+3^3+4^3+\cdots+L_n^3={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$ In my recent post I was intended to asked this question
$...,3,-1,2,1,3,4,7,...$ for $n=...,-2,-1,0,1,2,3,4,...$; It is the Lucas numbers
Let the sum of the cube of Lucas series be $S_L$
$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,
show that it has a closed form $$S_L={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$$
I try: I can't think of any simple identities to use.
This is the only one might have some sort of link to it,
$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$
$2^2+1^2+3^2+\cdots+L_n^2=L_nL_{n+1}+2$
Any further hints?
| This answer uses that $$L_n^2-L_{n-1}L_{n+1}=5(-1)^n\tag1$$
whose proof can be seen at the end of this answer.
Using $(1)$, we get
$$\begin{align}&{L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}\\\\&=\frac{L_nL_{n+1}^2+5(-1)^nL_{n-1}+10(-1)^{2n}+9}{2}\\\\&=\frac{L_nL_{n+1}^2+(L_n^2-L_{n-1}L_{n+1})L_{n-1}+10+9}{2}\\\\&=\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}\end{align}$$
So, this answer proves by induction that
$$\sum_{k=0}^{n}L_k^3=\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}$$
It holds for $n=0$.
Supposing that it holds for some $n\ (\ge 0)$ gives that
$$\begin{align}\sum_{k=0}^{n+1}L_k^3&=L_{n+1}^3+\sum_{k=0}^{n}L_k^3\\\\&=L_{n+1}^3+\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}\\\\&=L_{n+1}^3+\frac{L_nL_{n+1}^2-(L_{n+1}-L_n)^2L_{n+1}+(L_{n+1}-L_n)L_n^2+19}{2}\\\\&=\frac{L_{n+1}^3+2L_nL_{n+1}^2+L_n^2L_{n+1}-L_n^3-L_n^2L_{n+1}+L_nL_{n+1}^2+19}{2}\\\\&=\frac{L_{n+1}(L_{n+1}+L_n)^2-L_n^2(L_n+L_{n+1})+L_nL_{n+1}^2+19}{2}\\\\&=\frac{L_{n+1}L_{n+2}^2-L_n^2L_{n+2}+L_nL_{n+1}^2+19}{2}\qquad\blacksquare\end{align}$$
Let us prove $(1)$ by induction.
It holds for $n=0$.
Supposing that it holds for some $n\ (\ge 0)$ gives that
$$\begin{align}5(-1)^{n+1}&=5(-1)^n\cdot (-1)\\\\&=-(L_n^2-L_{n-1}L_{n+1})\\\\&=-L_n^2+L_{n-1}(L_{n-1}+L_n)\\\\&=-L_n^2+L_{n-1}^2-L_nL_{n-1}+2L_nL_{n-1}\\\\&=L_{n-1}^2+2L_nL_{n-1}-L_n(L_n+L_{n-1})\\\\&=L_{n-1}^2+2L_nL_{n-1}-L_nL_{n+1}\\\\&=L_{n-1}^2+2L_nL_{n-1}+L_n^2-L_n^2-L_nL_{n+1}\\\\&=(L_{n-1}+L_n)^2-L_n(L_n+L_{n+1})\\\\&=L_{n+1}^2-L_nL_{n+2}\qquad\blacksquare\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Prove the series $(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}=\pi-4(...)$
Show that for $k\ge1$,
\begin{align}
\\&\quad(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}\\[10pt]&=\pi-4\left(1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots+{1\over 2k-1}\right).
\end{align}
I try:
The RHS sort of the favourite Leibniz $\pi$ series.
We could split into composite of fractions
${A\over 2n+1}+{B\over 2n+3}+{C\over 2n+5}+\cdots$ and then take the sum individually.
We know that
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+1)}=\pi$
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+3)}=3\pi-8$
We don't know the general of
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+2k+1)}=F(k)$
I am sure what to do next? Help please!
| Following Claude Leibovici great comment:
$$
\begin{align}
& S_k= (-1)^k\sum_{n=0}^{\infty}\frac{2^{n+1}(2k-1)!!}{{2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)} = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k+1}{4}\normalsize\right)\right] \\[8mm]
& S_{k} - S_{k-1} = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k+1}{4}\normalsize\right)\right] - (-1)^{k-1}\left[\psi\left(\small\frac{2k+1}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k+3}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] = (-1)^k\left[\psi\left(\small\frac{2k+3\color{red}{-4+4}}{4}\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k-1}{4}+1\normalsize\right) - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] \qquad\qquad \small\left\{\psi(x+1)=\psi(x)+\frac{1}{x}\right\}\normalsize \\[4mm]
& \qquad\qquad\space = (-1)^k\left[\psi\left(\small\frac{2k-1}{4}\normalsize\right) + \frac{4}{2k-1} - \psi\left(\small\frac{2k-1}{4}\normalsize\right)\right] = \color{red}{4\,\frac{(-1)^k}{2k-1}} \\[8mm]
& S_k - S_0 = S_{k} -S_{k-1}+S_{k-1} \cdots -S_{1}+S_{1} -S_{0} = \sum_{n=1}^{k}\left(S_{n} -S_{n-1}\right) = 4\,\sum_{n=1}^{k}\frac{(-1)^n}{2n-1} \\[2mm]
& \small S_0=\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+1)}=\pi \normalsize \Rightarrow \color{red}{S_k = \pi + 4\,\sum_{n=1}^{k}\frac{(-1)^n}{2n-1}} = \pi - 4\,\left(\small 1 - \frac13 + \frac15 -\frac17 + \cdots - \frac{(-1)^k}{2k-1} \normalsize \right)
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Discrepancy in differentiating: $y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$ $$y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
$$\sqrt{1+\sin{x}}=\sin{\frac{x}{2}}+\cos{\frac{x}{2}}$$
$$\sqrt{1-\sin{x}}=\sin{\frac{x}{2}}-\cos{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\tan{\frac{x}{2}}\right)$$Differentiating, we get: $$\frac{dy}{dx}=\frac{1}{2}$$
But taking: $$\sqrt{1-\sin{x}}=\cos{\frac{x}{2}}-\sin{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\cot{\frac{x}{2}}\right)$$
Differentiating, we get: $$\frac{dy}{dx}=\frac{-1}{2}$$
I figured that I needed to use absolute value for the simplification of $\sqrt{1-\sin{x}}$, i.e. $\sqrt{1-\sin{x}}=\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|$.
Subsequently, I put the below functions into Wolfram Alpha's input box to differentiate: $$y_1=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
And
$$y_2=\tan^{-1}\left(\frac{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|+\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|-\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}\right)$$
The answers should ideally match, but they dont. $\frac{dy_1}{dx}=\frac{-1}{2}\neq\frac{dy_2}{dx}$
Why don't they?
| $$\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\dfrac{1+\sqrt{\dfrac{1-\sin x}{1+\sin x}}}{1-\sqrt{\dfrac{1-\sin x}{1+\sin x}}}$$
Now $\dfrac{1-\sin x}{1+\sin x}=\left(\dfrac{\cos\dfrac x2-\sin\dfrac x2}{\cos\dfrac x2+\sin\dfrac x2}\right)^2=\left(\dfrac{1-\tan\dfrac x2}{1+\tan\dfrac x2}\right)^2=\tan^2\left(\dfrac\pi4-\dfrac x2\right)$
We know for real $a,\sqrt{a^2}=|a|=\begin{cases}a &\mbox{if }a\ge0\\
-a & \mbox{if }a<0\end{cases}$
For $\tan\left(\dfrac\pi4-\dfrac x2\right)\ge0,$
$$\tan^{-1}\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\tan^{-1}\dfrac{1+\tan\left(\dfrac\pi4-\dfrac x2\right)}{1-\tan\left(\dfrac\pi4-\dfrac x2\right)}=\tan^{-1}\tan\left(\dfrac\pi4+\dfrac\pi4-\dfrac x2\right)=n\pi+\dfrac\pi4+\dfrac\pi4-\dfrac x2$$
where $n$ is an arbitrary constant such that $-\dfrac\pi2\le y\le\dfrac\pi2$
What if $\tan\left(\dfrac\pi4-\dfrac x2\right)<0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Simple algebra question concernig the equation of a circle for some reason, I can't do the math. I have worked out that $(x^2+y^2-1)^2+4y^2=r^2((x+1)^2 + y^2)^2$ should be the circle $$\left(x-\dfrac {1+r^2}{1-r^2}\right)^2 + y^2= \left(\dfrac{2r}{1-r^2}\right)^2,$$ using some prior knowledge, however I can't prove it. Can someone help me on my way? Many thanks in advance.
| Consider the left-hand side as a quadratic equation in $y^2$:
$$ \begin{aligned}
(x^2+y^2-1)^2+4y^2 &= y^4 + 2y^2(x^2-1) + (x^2-1)^2 + 4 y^2 \\
&= y^4 + 2y^2(x^2+1) + (x^2-1)^2 \\
&= \left(y^2 + (x^2 + 1)\right)^2 + (x^2-1)^2 - (x^2+1)^2 \\
&= \left(y^2 + (x^2 + 1)\right)^2 - 4 x^2 \\
&= \left(y^2 + (x + 1)^2\right)\left(y^2 + (x - 1)^2\right)
\end{aligned}$$
Then your equation becomes
$$
\left(y^2 + (x + 1)^2)\right)\left(y^2 + (x - 1)^2)\right)
= r^2\left((x+1)^2 + y^2 \right)^2
$$
and it follows that
$$
y^2 + (x + 1)^2 = 0
$$
which implies that $x=-1, y=0$,
or the common factor can be cancelled:
$$
y^2 + (x - 1)^2 = r^2\left((x+1)^2 + y^2 \right) \\
\Longleftrightarrow
(x^2+y^2+1)(1-r^2)-2x(1+r^2) = 0
$$
If $r=1$ then the solution is the line $x=0$ (i.e. the y-axis), otherwise
the circle equation can now easily be derived.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that the sequence $a_n = -n+2$ is equal to $a_n = a_{n-1} + 2a_{n-2} + 2n - 9$ My Working:
\begin{align}
a_0=&-0+2=2\\
a_1=&-1+2=1\\
a_2=&-2+2=0.
\end{align}
and $$a_2=a_{2-1}+2a_{2-2}+2(2)-9=-4.$$
I don't know what's wrong with my solution. I used this method on various questions but now I'm just ending up with the wrong answer.
| Solving it further we have -
$a_{2} = a_{2-1} + 2a_{2-2} + 2(2) - 9$
$a_{2} = a_{1} + 2a_{0} + 2(2) - 9$
$a_{2}$ = 1 + 2(2) + 2(2) - 9$
$a_{2}$ = 1 + 4 + 4 - 9$
= 0
Your mistake is you are using the value of $a_{0}$ = 0 but it is 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that :$\sum_{j=0}^{n}j{n\choose j}^2H_{n-j}={1\over 4}{2n\choose n}(4nH_{n}-2nH_{2n}-1)$ Where $H_0=0$ and $H_n$ is the n-th harmonic number.
Show that this sum has a closed form
$$\sum_{j=0}^{n}j{n\choose j}^2H_{n-j}={1\over 4}{2n\choose n}(4nH_{n}-2nH_{2n}-1)$$
I try:
$j{2n\choose n}={(2n)!\over (j-1)!(2n-j)!}$ look more messier!
I found two sum on Wikipedia site but it doesn't see more helpful at all!
$$\sum_{j=1}^{n}H_j=(n+1)(H_{n+1}-1)$$
$$\sum_{j=0}^{n}{n\choose j}^2=2^n$$
Any help? Thank you!
| Using the Vandermonde Identity we have, $$\sum\limits_{j=0}^{n-1} \binom{n+x-1}{j}\binom{n}{n-j-1} = \binom{2n+x-1}{n-1}$$ i.e., after reindexing the summation and using the identity $\displaystyle \binom{n+x-1}{j-1} = \frac{j}{n+x}\binom{n+x}{j}$ we have,
$$\sum\limits_{j=0}^{n} j\binom{n+x}{j}\binom{n}{n-j} = (n+x)\binom{2n+x-1}{n-1}$$
Differentiating both sides w.r.t $x$ using $\displaystyle \frac{d}{dx} \binom{n+x}{j} = \binom{n+x}{j}(H_n(x) - H_{n-j}(x))$ where $\displaystyle H_n(x) = \sum\limits_{j=1}^{n} \frac{1}{j+x}$,
\begin{align*}&\sum\limits_{j=0}^{n} j\binom{n+x}{j}\binom{n}{n-j}(H_n(x) - H_{n-j}(x)) \\&= \binom{2n+x-1}{n-1}((n+x)(H_{2n-1}(x) - H_{n}(x)) + 1)\end{align*}
i.e., at $x = 0$ we have, \begin{align*} \sum\limits_{j=0}^{n} j\binom{n}{j}^2(H_n - H_{n-j}) &= \binom{2n-1}{n-1}(n(H_{2n-1} - H_{n}) + 1) \\ \implies \sum\limits_{j=0}^{n} j\binom{n}{j}^2H_{n-j} &= H_n\sum\limits_{j=0}^{n} j\binom{n}{j}^2 - \binom{2n-1}{n-1}(n(H_{2n-1} - H_{n}) + 1)\\&= \frac{1}{2}\binom{2n}{n}\left(2nH_n - nH_{2n} - \frac{1}{2}\right)\end{align*}
Alternative:
One may use the identity $\displaystyle \sum\limits_{j=1}^{k} \frac{(-1)^{j-1}}{j}\binom{n}{k-j} = \binom{n}{k}(H_n - H_{n-k}) \tag{*}$ to write:
\begin{align*} \sum\limits_{k=1}^{n} k\binom{n}{k}^2(H_n-H_{n-k}) &= \sum\limits_{k=1}^{n} \sum\limits_{j=1}^{k}\frac{(-1)^{j-1}k}{j}\binom{n}{k}\binom{n}{k-j} \tag{1}\\&= \sum\limits_{j=1}^{n}\sum\limits_{k=j}^{n}\frac{(-1)^{j-1}k}{j}\binom{n}{k}\binom{n}{k-j}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\sum\limits_{k=j}^{n}\binom{n-1}{k-1}\binom{n}{k-j}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\sum\limits_{k=0}^{n-j}\binom{n-1}{n-k-j}\binom{n}{k}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\binom{2n-1}{n-j}\\&= n\binom{2n-1}{n}(H_{2n-1} - H_{n-1}) \tag {2}\end{align*} where, in steps $(1)$ and $(2)$ we used identity $(*)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Complex numbers: proof of equality Can anyone help me prove following equality?
$$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$ $$=2 \iff \frac{n}3\in \mathbb{N}$$
$$=-1 \iff \frac{n}3\not\in \mathbb{N}$$
This is what I've got:
$$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$
$$= \left(\frac{2(cos(-\pi/3)+i sin(-\pi/3))}2\right)^n + \left(\frac{2(cos(\pi/3)+i sin(\pi/3))}2\right)^n$$
$$=[cos(-\pi/3)+isin(-\pi/3)]^n-[cos(\pi/3)+isin(\pi/3)]^n$$
$$=cos(n\pi/3)-isin(n\pi/3)-cos(n\pi/3)-isin(n\pi/3)$$
$$=-2isin(n\pi/3)$$
and if $n/3 \in \mathbb{N}$, I get $n'\pi$ and $sin(n'\pi)=0$ which isn't the result I needed...
| Theres no point in converting to trig unless you are familiar and comfortable with polar coordinates. If so:
$(\frac {-1 + i\sqrt{3}}2)^n+(\frac {-1 - i\sqrt{3}}2)^n=$
$(\cos \frac {2\pi}{3} + i \sin \frac {2\pi}{3})^n + (\cos \frac {4\pi}{3} + i \sin \frac {4\pi}{3})^n=$
$(e^{i\frac{2\pi}{3}})^n + (e^{i\frac{4\pi}{3}})^n=e^{i\frac{2n\pi}{3}} + e^{i\frac{4n\pi}{3}}=$
$(\cos \frac {2n\pi}{3} + i \sin \frac {2n\pi}{3}) + (\cos \frac {4n\pi}{3} + i \sin \frac {4n\pi}{3})=$
$3|n$ i.e. if $n = 3k$
$= \cos 2k\pi + i \sin 2k \pi + \cos 4k\pi + i\sin 4k\pi = 1 + 0i + 1 + 0i = 2$
if $3\not \mid n$ ie. if $n = 3k \pm 1$
$ = \cos (\pm \frac{2\pi}3) + i \sin (\pm \frac{2\pi}3) + \cos (\pm \frac{4\pi}3) + i \sin (\pm \frac{4\pi}3)$
$= -\frac 12 \pm i\frac{\sqrt{3}}{2} -\frac 12 \mp i \frac{\sqrt{3}}{2}= -1$
But really itd be easier to realize: $(\frac {-1 \pm i\sqrt{3}}2)^3 = 1$ and $(\frac {-1 \pm i\sqrt{3}}2)^2 = \frac {-1 \mp i\sqrt{3}}2$ [just multiply them out]so
$(\frac {-1 + i\sqrt{3}}2)^{3k + i} + (\frac {-1 - i\sqrt{3}}2)^{3k + i} = $
$(\frac {-1 + i\sqrt{3}}2)^i + (\frac {-1 - i\sqrt{3}}2)^i = 1 + 1 = 2$ if $i = 0$
$=(\frac {-1 \pm i\sqrt{3}}2) + (\frac {-1 \mp i\sqrt{3}}2)$
$= -\frac 12 - \frac 12 =-1$ if $i = 1$ or $i = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Finding remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$. I have found a new problem which asks:
Find the remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$.
I am thinking to find the remainder using Fermet's theorem, but I think I am unable to do it.
Please help.
| You are right. You should use Fermat's Little Theorem to solve this problem. Fermat's Little Theorem tells us $10^6 \equiv 1 \pmod{7}$, meaning that we can subtract $6$ from each power without affecting the remainder. This means we can just find the remainder of the powers when divided by $6$ and then use those powers.
$$10 \equiv 4 \pmod 6$$
$$10^2 \equiv 4^2 \equiv 16 \equiv 4 \pmod 6$$
$$10^3 \equiv 4^3 \equiv 64 \equiv 4 \pmod 6$$
Hopefully, you see the pattern. All of the powers are equivalent to $4 \pmod 6$. Thus, we can replace each $10^{10^a}$ with $10^4$, giving us a sum of $10\cdot 10^4=10^5$. Now, we need to compute $10^5 \pmod 7$:
$$10^5 \equiv 3^5 \equiv 243 \equiv 5 \pmod 7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How can I isolate $x$ in this equation? Simultaneous equations - $7^{x-1} = 49 \hspace{1 mm} (7^{y})$ --------- (i)
$\hspace{46 mm} 3^x+3^y = 84$ ---------- (ii)
The first I make x subject of equation in this way:
$\Longrightarrow 7^{x-1} = 49 \hspace{1 mm} (7^{y})$
$\Longrightarrow 7 \times 7^x = 7^2 \times 7^y$
$\Longrightarrow 7^x = 7^{y+1}$
$\Longrightarrow x = y + 1$
The second I tried with logarithms but could not separate x in simple form:
$\Longrightarrow 3^x + 3^y = 84$
$\Longrightarrow 3^x = 84 - 3^y$
$\Longrightarrow \lg3^x = \lg (84 - 3^y)$
$\Longrightarrow x \lg3 = \lg(84 - 3^y)$
$\Longrightarrow x = \dfrac{\lg(84 - 3^y)}{\lg3}$
How can I simplify this further?
| Your solution is incorrect.
It should be as follows:
$$7^{x-1}=49\cdot 7^y$$
$$ \implies 7^{x-1}=7^{y+2}$$
Comparing both sides of the equation, we get
$$x-1=y+2$$
$$ \implies x=y+3\tag1$$
From the second equation, we have
$$3^x+3^y=84$$
$$ \implies 3^{y+3}+3^y=84$$
$$ \implies 3^y(3^3+1)=84$$
$$ \implies 3^y\cdot 28=84$$
$$ \implies 3^y=3=3^1$$
Again comparing both sides of the equation, we get
$$y=1$$
And using $(1)$, we get
$$x=4$$
Hence the required solution is
$$\boxed{x=4,y=1}$$
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form of $ \int_0^\infty x e^{-b \pi x^2}(1+x^{-a})^{-n}\mathrm{d}x$
Is there an exact closed-form expression or a closed-form upper bound for the following integral?
$$I = \int_{0}^{\infty}\frac{x e^{-b \pi x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
where $a > 2$, $b, n > 0$.
If yes, how can we obtain it?
My attempt: If we multiply $I$ in $(1)$ by $2\pi b$, we get the probability density function of Rayleigh random variable, i.e, $f_X(x) = 2\pi b xe^{-b \pi x^2}$. So we can write $I$ as
$$I = \mathbb{E}\left[\left(\frac{1}{1+X^{-a}}\right)^{n}\right].$$
I could not procced in this way.
| WolframAlpha ist able to compute the anti-derivative of your integrand for $a=2$ and small fixed integer $n$. From these solutions I came up with a conjecture for $a=2$ and general integer $n$.
Some notation:
$$ I_{a,n}(c)= \int_{0}^{\infty}\frac{x e^{-c x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
where $a > 2$, $c, n > 0$.
The conjecture for $a=2$:
$$ F_{2,n}(x,c)= \int\frac{x e^{-c x^2}}{(1+x^{-2})^{n}}\mathrm{d}x = p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c(x^2+1)) - \frac{\mathrm{e}^{-cx^2}~p_2(c,x)}{c(x^2+1)^{n-1}}$$
$$ I_{2,n}(c) = -F_{2,n}(0,c) = -p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c) + \frac{p_2(c,0)}{c}$$
Where $p_1(c)$ is a polynomial in $c$ of degree $n-1$ and $p_2(x,c)$ is a polynomial in $c$ of degree $n-1$ and in $x$ of degree $2(n-1)$ with only even terms in $x$.
I don't have a general form for $p_1$ and $p_2$, but using the above restrictions it's possible to solve for their coefficients given some fixed $n$ by differentiating $F_{2,n}$ and comparing with the integrand.
Some solutions:
$$I_{2,1} = \left(\frac{1}{2} \right)\mathrm{e}^c \mathrm{Ei}(-c) +
\frac{1}{c}\left( \frac{1}{2} \right) \text{(already given by JJacquelin)}$$
$$I_{2,2} = \left(\frac{1}{2} \, c + 1 \right)\mathrm{e}^c
\mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{2} \, c + \frac{1}{2}
\right)$$
$$I_{2,3} = \left(\frac{1}{4} \, c^{2} + \frac{3}{2} \, c + \frac{3}{2}
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{4} \,
c^{2} + \frac{5}{4} \, c + \frac{1}{2} \right)$$
$$I_{2,4} = \left(\frac{1}{12} \, c^{3} + c^{2} + 3 \, c + 2
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{12} \,
c^{3} + \frac{11}{12} \, c^{2} + \frac{13}{6} \, c + \frac{1}{2}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.
| Use Cauchy–Schwartz:
$$a+b+c\ \leqslant\ \sqrt{1^2+1^2+1^2}\sqrt{a^2+b^2+c^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Factorization of $(1+2i)z^2+6iz+2i-1$ to find the residues, but it doesn't work I'm doing complex analysis and I have the following $$\int_{C}\frac{2dz}{(2i+1)z^2+6iz+2i-1}$$ where $C$ is the unit circle. I tried factorizing this polynomial and find the residues but I really can't solve this.
This is my work:
$$z_{1,2} = \frac{-3i\pm \sqrt{-9-(-4-1)} }{2i+1}=\frac{-3i\pm 2i}{1+2i} \cdot \frac{1-2i}{1-2i}= \frac{-3i-6\pm 2i\pm 4}{5}$$ which gives $z_1 = -\frac{2}{5}-\frac{i}{5}$ and $z_2 = -2-i$. Now clearly $z_2$ is outside the unit circle so I just have to consider the residue of $z_1$.
However I think I'm either making a factorisation mistake or a mistake in the roots. Indeed if we check the roots, I don't get the same polynomial! $$(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right) = z^2 +z\left(\frac{12}{5}+\frac{6i}{5}\right)+\frac{4i}{5}+\frac{3}{5}$$ which is clearly not the same polynomial I started with. Also if I try to calculate the residue using the limit, I don't get the correct one. Where's my mistake?
EDIT
As requested, I'll add some more calculations. To calculate the residue I did the following: $$\lim_{z\to z_1}(z+\frac{2}{5}+\frac{i}{5})\frac{2}{(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right)} = \frac{2}{\frac{8}{5}+\frac{4i}{5}} = \frac{5}{4+2i} = \frac{20-10i}{20}=1-\frac{i}{2}$$ which is clearly wrong as this integral is the mapping of a real integral of the first type (rational function in sine and cosine). Indeed by Cauchy's Residue Theorem this should give that the value of the initial real integral is $2\pi i (1-\frac{i}{2}) = 2\pi i+\pi$ which is a complex value, so this is wrong
| If you were only interested in the value of your integral $f$ you could let $z=\exp (i \theta )$ and get
$$fr=\int_0^{2 \pi } \frac{1}{\sin (\theta )+2 \cos (\theta )+3} \, d\theta$$
In order to calculate the integral we attempt to use the fundamental theorem of calculus, i.e. taking the difference of the anitiderivative at both ends of the integration interval. But this holds only for continuous antiderivatives. If the antiderivative contains jumps these must be taken into account.
An antiderivative of the integrand is
$$fa=-\tan ^{-1}\left(\frac{2}{\tan \left(\frac{\theta }{2}\right)+1}\right)$$
In the region of integration the denominator vanishes at $\theta =\frac{3 \pi }{2}$. Here the antiderivative makes a jump from
$$\lim_{\theta \to \left(\frac{3 \pi }{2}\right)^-} \, fa=\frac{\pi }{2}$$
to
$$\lim_{\theta \to \left(\frac{3 \pi }{2}\right)^+} \, fa=-\frac{\pi }{2}$$
i.e. it jumps by an amount of $\pi$.
Now the antiderivative at both ends of the integration interval is equal to $-\tan ^{-1}(2)$. Hence the difference is zero, and we are left with the jump which gives $f=fr=\pi$.
EDIT
I'm afraid I have made it more complicated than necessary.
Let us therefore apply a "standard" procedure and do this for the more general integral (displayed here in original form and reduced form)
$$\int_0^{2 \pi } \frac{1}{a+b \cos (\theta )+c \sin (\theta )} \, d\theta =\int_0^{2 \pi } \frac{1}{a+\sqrt{b^2+c^2} \cos (\phi )} \, d\phi =\frac{2 \pi }{\sqrt{a^2-b^2-c^2}}$$
which holds for $a^2-b^2-c^2>0$
Making the "standard" substitution $\theta \to 2 \tan ^{-1}(t)$, $\text{d$\theta $}=\frac{2 \text{dt}}{t^2+1}$ and splitting the integral into the two parts from $0$ to $\pi$ and form $\pi$ to $2 \pi$ gives the complete integration range in to from $-\infty$ to $\infty$.
We have
$$\cos (\theta )=\cos ^2\left(\frac{\theta }{2}\right)-\sin ^2\left(\frac{\theta }{2}\right)=\left(1-t^2\right) \cos ^2\left(\frac{\theta }{2}\right)$$
$$\sin (\theta )=2 \sin \left(\frac{\theta }{2}\right) \cos \left(\frac{\theta }{2}\right)=2 t \cos ^2\left(\frac{\theta }{2}\right)$$
$$\frac{1}{\cos ^2\left(\frac{\theta }{2}\right)}=t^2+1$$
The integral becomes
$$2 \int_{-\infty }^{\infty } \frac{1}{a \left(t^2+1\right)+b \left(1-t^2\right)+2 c t} \, dt$$
Supplementing the square, shifting the integration varibale (this makes no dfference in our infnte interval) and extracting factors gives finally
$$\frac{\int_{-\infty }^{\infty } \frac{2}{v^2+1} \, dv}{\sqrt{a^2-b^2-c^2}}=\frac{2 \pi }{\sqrt{a^2-b^2-c^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that : $\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$, for $n\ge0$ Show that
$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$$
$H_n$; is the n-th harmonic number
My try:
We know that
$$\sum_{k=0}^{\infty}{(-1)^k\over (k+1)^2}=\int_{0}^{1}{-\ln{x}\over 1+x}dx$$
But
$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}=\int_{0}^{a}???$$
From wikipedia I found
$$H_n=\int_{0}^{1}{1-x^n\over 1-x}dx=\sum_{k=1}^{n}(-1)^{k-1}{{n\choose k}\over k}$$ may be of useful
Onward I am clueless as what to do next. Please help!
| Start by the binomial expansion
$$\sum^n_{k=0} {n \choose k} x^k = (1+x)^n$$
By integration
$$\sum^n_{k=0} {n \choose k} \frac{x^{k+1}}{k+1} =\frac{(1+x)^{n+1}-1}{n+1}$$
Integrating again
$$\sum^n_{k=0} {n \choose k} \frac{(-1)^{k+2}}{(k+1)^2} =\frac{1}{n+1}\int^{-1}_0\frac{(1+x)^{n+1}-1}{x}\,dx = \frac{1}{n+1}\int^{1}_0\frac{(1-x)^{n+1}-1}{x}\,dx$$
By substitution
$$\sum^n_{k=0} {n \choose k} \frac{(-1)^{k}}{(k+1)^2} =\frac{1}{n+1}\int^{1}_0\frac{x^{n+1}-1}{x-1}\,dx = \frac{H_{n+1}}{n+1} $$
Note that
\begin{align}
\frac{1+\sqrt{1+4(n+1)^2H_nH_{n+1}}}{n+1} &= \frac{1+\sqrt{1+4(n+1)^2H_n^2+4(n+1)H_n}}{n+1}\\&=\frac{ 1+\sqrt{(1+2(n+1)H_n)^2}}{n+1}
\\&=\frac{2+2(n+1)H_n}{2(n+1)^2} \\&= \frac{H_{n+1}}{n+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$
$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$
| $\frac{\log x}{2}=\log{\sqrt{x}}$
Using this,
$( \log{\sqrt{x}})^{(\log x)^2+2\log x-2}= \log{\sqrt{x}}$
$\implies (\log x)^2+2\log x-2=1$
$\implies (\log x+3)(\log x-1)=0$
$\implies x=e^{-3}$ or $e$
Again,
$\log{\sqrt{x}}=1$ gives a solution.
$\implies x=e^2$ is also a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find monic irreducible polynomials in $Z_3[x]$ which divide both f(x) and g(x) The Problem
Find all monic irreducible polynomials in $Z_3[x]$ which divide both $f(x)$ and $g(x)$.
Where
$f(x) = x^5+x^3+x^2+x+2$, $g(x) = x^6+x^5+x^3+1$
This problem is part of an early exam so it comes with a solution.
My attempt at a solution
What I did was doing $gcd(f(x),g(x))$ and that went alright, the first 2 iterations gave me these results:
$g(x) = (x+1)f(x) + 2x^4+2x^3+x^2+2$
$f(x) = (2x+1)(2x^4+2x^3+x^2+2) + 2x^3$
After finding the gcd you simply find the irreducible factors of the gcd, and then you divide f(x) and g(x) by the gcd and start over. That's what I would do anyway.
The solution given
They start out with the gcd:
$g(x) = (x+1)f(x) + 2x^4+2x^3+x^2+2$
then let $2h(x) = 2x^4+2x^3+x^2+2$
$f(x) = (x+2)h(x) + 0$
Therefore $gcd(f(x),g(x)) = h(x)$
We now look for linear factors in $h(x)$ and find that $h(2)=0$ therefore $x-2=x+1$ is a factor of $h(x)$, polynomial division gives us:
$h(x)=(x+1)(x^3+2x+1)$, and $x^3+2x+1$ has a degree beneath or equal to 3 and no linear factors, therefore irreducible.
Answer: the factors are $x+1$ and $x^3+2x+1$
My questions
Why can we 'take out' the 2 from $2h(x)$ when we continue in our gcd search?
Why do we not divide $f(x)$ and $g(x)$ to search for further gcd:s to find new factors?
| We have $\ \gcd(g,f) = \gcd(g\bmod f,f) = \gcd(2h,f) = \gcd(h,f) = h\ $ since $\ \gcd(2,f) = 1$
Alternatively we can use $\, 2 = -1\ \in\Bbb Z_3\,$ so $\,\gcd(2h,f) = \gcd(-h,f) = \gcd(h,f) = h$
There is an error in your calculation: $\ f\bmod 2h = 0,\, $ not $\, 2x^3 $ (so there's no need to go further)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\int_0^\infty \frac{x\log(1+x^2)}{e^{2\pi x}+1}dx=\frac{19}{24} - \frac{23}{24}\log 2 - \frac12\log A$ Any idea on how to prove the following integral
$$\int_{0}^{\infty} {x\log(1+x^2)\over e^{2\pi x}+1}dx =\require{cancel} \cancel{\frac{17}{24} - \frac{23}{24}\log 2 + \frac{1}{2}\log A}={\frac{19}{24} - \frac{23}{24}\log 2 - \frac{1}{2}\log A }$$
Where $A$ is the Glaisher–Kinkelin constant.
We define
$$A= \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$
Where $$H(n) = \prod^{n}_{k=1} k^k $$
I would start by
$$F(z) = \int^\infty_0 \frac{2xz}{(x^2+z^2)(e^{2\pi x}+1)} \, dx$$
I know that
$$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2} $$
But I can't find a similar one for
$$\frac{1}{e^{2\pi t}+1}$$
| The following approach uses the Abel-Plana formula, which was mentioned in a comment by the user tired.
By applying the Abel-Plana formula to $f(\frac{x}{2})$ and then subtracting the result from the Abel-Plana formula applied to $f(x)$, we get $$\begin{align} &\sum_{n=0}^{\infty} f(n) - \sum_{n=0}^{\infty} f \left(\frac{n}{2} \right) \\ &= \int_{0}^{\infty} f(x) \, dx - \int_{0}^{\infty} f \left(\frac{x}{2} \right) \, dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx - i \int_{0}^{\infty} \frac{f\left(i \frac{x}{2}\right) - f\left(-i\frac{x}{2}\right)}{e^{2 \pi x}-1} \, dx \\ &= \int_{0}^{\infty} f(x) \, dx - 2 \int_{0}^{\infty} f(u) \, du + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx - 2 i \int_{0}^{\infty} \frac{f(iu) - f(-iu)}{e^{4 \pi u}-1} \, du \\ &= -\int_{0}^{\infty} f(x) \, dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}+1} \, dx. \end{align} $$
Let's apply the above formula to the function $$ f(x)= \frac{1}{(1+x)^{s}} , \quad \text{Re}(s) >1.$$
Doing so, we get $$\sum_{n=0}^{\infty} \frac{1}{(1+n)^{s}} - 2^{s} \sum_{n=0}^{\infty}\frac{1}{(2+n)^{s}} = \frac{1}{1-s} + 2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx,$$ which implies that
$$2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx = (1-2^{s}) \zeta(s) + 2^{s} + \frac{1}{s-1}. \tag{1}$$
By analytic continuation, $(1)$ should hold for all complex values of $s$. (For $s=1$, the right side of the equation should be treated as a limit.)
Now if we differentiate under the integral sign and then let $s=-1$, we get
$$\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx + 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx = - \frac{\log 2}{2} \underbrace{\zeta(-1)}_{- \frac{1}{12}}+\frac{\zeta'(-1)}{2} + \frac{\log 2}{2}-\frac{1}{4}. $$
But using Binet's second formula for the log gamma function, we have $$ \begin{align} 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx &= 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx - 4 \int_{0}^{\infty} \frac{\arctan x}{e^{4 \pi x}-1} \, dx \\ &=2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx -2 \int_{0}^{\infty} \frac{\arctan \left(\frac{u}{2}\right)}{e^{2 \pi u}-1} \, du \\ &= 1- \frac{\log (2 \pi)}{2} + \frac{3}{2} \log 2 -2 + \frac{\log(2 \pi)}{2} \\ &= \frac{3 \log 2}{2} -1 .\end{align}$$
Therefore, $$ \begin{align}\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx &= \frac{\log 2}{24} + \frac{\zeta'(-1)}{2} + \frac{\log 2}{2} - \frac{1}{4} + 1 - \frac{3 \log 2}{2} \\ &= \frac{\zeta'(-1)}{2} - \frac{23}{24} \log 2 + \frac{3}{4}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
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} |
$a^4+b^4+c^4+d^4=4\implies\sum\limits_{cyc}\frac{a^3}{bc}\geq4$ Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^4+b^4+c^4+d^4=4$. Prove that:
$$\frac{a^3}{bc}+\frac{b^3}{cd}+\frac{c^3}{da}+\frac{d^3}{ab}\geq4$$
I tried C-S, BW and more, but without success.
| I see a proof by สนอง ห้วยเรไร:
We have:$$\sum_\limits{cyc}a\leq\sum_\limits{cyc}a^2$$and$$abc+bcd+cda+dab$$$$\leq ab+bc+cd+da$$$$\leq a^2+b^2+c^2+d^2\leq 4$$So:$$\sum_\limits{cyc}\frac{a^3}{bc}=\sum_\limits{cyc}\frac{a^4}{abc}$$$$\geq\frac{(\sum_\limits{cyc}a^2)^2}{\sum_\limits{cyc}abc}$$$$\geq\frac{(\sum_\limits{cyc}a^2)^2}{\sum_\limits{cyc}ab}$$$$\geq 4\frac{\sum_\limits{cyc}ab}{\sum_\limits{cyc}ab}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How did people get the inspiration for the sums of cubes formula? I stumbled upon this neat formula for sums of cubes with arbitrary $x,y\in\mathbb{Z}$$$(x^2+9xy-y^2)^3+(12x^2-4xy+2y^2)^3=(9x^2-7xy-y^2)^3+(10x^2+2y^2)^3\tag1$$
With $1729=1^3+12^3=9^3+10^3$ as its first instance. And I believe that this formula was used by Ramanujan to find a formula for$$a^3+b^3=c^3\pm1$$
So my question?
Questions:
*
*What would be someone's thinking process when finding other formulas such as $(1)$?
*Are there any other formulas similar to $(1)$?
I'm thinking along the lines of starting with $$(x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3=(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3$$
But even Mathematica can't solve the ensuing system that follows. So for the moment, I'm stuck.
| The formula that Ramanujan actually recorded, as well as Euler's solution, are discussed by Ono in https://arxiv.org/abs/1510.00735
The original formula was
$$ \left( 6 a^2 - 4ab + 4 b^2 \right)^3 = \left( 3 a^2 +5ab - 5 b^2 \right)^3 + \left( 4 a^2 - 4ab + 6 b^2 \right)^3 + \left( 5 a^2 - 5ab -3 b^2 \right)^3 $$
The formula can be made more symmetric still. The two classes of forms of discriminant $85$ are represented by
$$ x^2 + 9 xy - y^2, $$
$$ 3 x^2 + 7 xy - 3 y^2. $$ The latter is equivalent to $3 x^2 - 5 xy - 5 y^2, $ off by a single minus sign.
The two classes of (primitive) forms of discriminant $-20$ are represented by
$$ x^2 + 5 y^2, $$
$$ 2 x^2 + 2 xy + 3 y^2. $$
That is, one may pass between Ramanujan's version and yours by using Gauss composition in order to multiply by $3,$ which passes between the principal genus and the other genus, both for discriminant $85$ and $-20.$
On page 2 they give Ramanujan's modern version of Euler's complete solution.
When
$$ \alpha^2 + \alpha \beta + \beta^2 = 3 \lambda \gamma^2, $$
$$ \left( \alpha + \lambda^2 \gamma \right)^3 + \left( \lambda \beta + \gamma \right)^3 = \left(\lambda \alpha + \gamma \right)^3 + \left( \beta + \lambda^2 \gamma \right)^3 $$
http://esciencecommons.blogspot.com/2015/10/mathematicians-find-magic-key-to-drive.html
| {
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"url": "https://math.stackexchange.com/questions/2078792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Finding a singluar cardinality Given an infinite cardinal b such that $b^2=b$ I need to find a cardinal $x$ such that:
$$(x^b = 2^b)\quad \land \quad (\forall c.(c>x)\to (c^b>2^b))$$
Furthermore, I need to show there is only one such $x$.
I think the answer is $ x = 2^b$ but have trouble proving the second part:$$(c>2^b) \to (c^b >2^b) $$
Any thoughts?
| Certainly $x=2^b$ satisfies the given conditions: $x^b=\left(2^b\right)^b=2^{b^2}=2^b=x$, and if $c>x=2^b$, then $c^b\ge c>2^b$. To show uniqueness, suppose that $y$ is an infinite cardinal such that $y^b=2^b$, and $c^b>2^b$ whenever $c>y$. Then on the one hand we must have $y\le x$, since $y^b=2^b$, but on the other hand we must have $x\le y$, since $x^b=2^b$, so $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric functions I need some help with these tasks and would be very grateful if someone shows me the way of solving them.
If $\sin\left(a\right)+\cos\left(a\right)=b$ and $\left|b\right|\le \sqrt{2}$, represent the following expressions with $b$:
$A=\sin\left(a\right)\cdot \cos\left(a\right)$,
$B=\left|\sin\left(a\right)-\cos\left(a\right)\right|$,
$C=\left|\sin^2\left(a\right)-\cos^2\left(a\right)\right|$,
$D=\left|\sin^3\left(a\right)+\cos^3\left(a\right)\right|$,
$E=\sin^4\left(a\right)+\cos^4\left(a\right)$
Answer: $A\:=\:\frac{1}{2}\left(b^2-1\right),\:B\:=\sqrt{2-b^2},C=\left|b\right|\sqrt{2-b^2},\:D\:=\:\frac{\left|b\right|}{2}\left(3-b^2\right),\:E=\frac{1}{2}\left(1+2b^2-b^4\right)$
| For $A$, note that
$$
b^2=(\sin a+\cos a)^2=\overbrace{\sin^2a+\cos^2a}^1+2\sin a\cos a
$$
so that
$$
\sin a\cos a=\frac{b^2-1}{2}
$$
A very similar approach works for the rest of expressions.
If you are given the solutions, the best approach is to work backwards: plug in the value of $b$ into the proposed solutions, and simplify them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
in triangle ABC there satisfies equation: $\cos A \cos B+\sin A\sin B\sin C=1$, determine possible values of $C$ Question
in triangle ABC there satisfies equation: $\cos A\cos B+\sin A\sin B\sin C=1$, determine possible values of C
What I have so far
I've noticed that the given equation looked similar to $\cos(A-B)$ which was $\cos A\cos B+\sin A\sin B$.
This I can extrapolate that $\cos(A-B)-\sin A\sin B+\sin A\sin B\sin C=1$
Thus $\cos(A-B)=1+\sin A\sin B+\sin A\sin B\sin C$
and when i factor out the common terms:
$\cos(A-B)=1+\sin A\sin B(1-\sin C)$
Right here I am not sure how to proceed and I got stuck
| $\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\leq 1$
$\cos A\cos B+\sin A\sin B\geq 1$
$\cos\left(A-B\right)\geq 1$
$\cos\left(A-B\right)=1$
$A-B=0$
$A=B$
$\sin C=\frac{1-\cos^2 A}{\sin^2 A}=1$
$C=90$
$A=B=45$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I find the slope of an angle bisector, given the equations of the two lines that form the angle? The equation for the first line is $y = \frac{1}{2}x - 2$, and the equation for the second line is $y = 2x + 1$. They intersect at $(-2, -3)$.
Someone told me I can just average the slopes of the two lines to find the slope of the bisector, but I'm not sure if it's right.
| Take the unitary vectors parallel to the lines: their sum will be parallel to one of the bisecting line, their difference will be parallel to the other one.
To determine which is the one bisecting the acute /obtuse angle, just take the dot product of the two vectors:
*
*if that is positive, then their sum will be parallel to the acute
bisectrix and their difference to the obtuse one
*viceversa if the dot product is negative.
note: this method is valid also in 3D.
In 2D you can apply the same method to the normal (instead than parallel) unitary vectors.
in your example (using parallel vectors)
Rewrite the line equations in the proportional form
$$
\left\{ \begin{gathered}
y = \frac{1}
{2}x - 2\quad \Rightarrow \quad \frac{{x - 0}}
{2} = \frac{{y - \left( { - 2} \right)}}
{1} \hfill \\
y = 2x + 1\quad \Rightarrow \quad \frac{{x - 0}}
{{1/2}} = \frac{{y - 1}}
{1}\quad \Rightarrow \quad \frac{{x - 0}}
{1} = \frac{{y - 1}}
{2} \hfill \\
\end{gathered} \right.
$$
then the parallel unitary vectors are
$$
\mathbf{u} = \frac{1}
{{\sqrt 5 }}\left( {2,1} \right)\quad \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {1,2} \right)\quad 0 < \mathbf{u} \cdot \mathbf{v} = \frac{4}
{5}
$$
and the angle between them is acute. Their sum and difference is
$$
\mathbf{u} + \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {3,3} \right)\quad \mathbf{u} - \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {1, - 1} \right)
$$
and the equations of the bisecting lines will be:
$$
\left\{ \begin{gathered}
\frac{{x - x_{\,c} }}
{3} = \frac{{y - y_{\,c} }}
{3}\quad \Rightarrow \quad x + 2 = y + 3\quad \Rightarrow \quad y = x - 1\;\;acute \hfill \\
\frac{{x - x_{\,c} }}
{1} = \frac{{y - y_{\,c} }}
{{ - 1}}\quad \Rightarrow \quad x + 2 = - y - 3\quad \quad \Rightarrow \quad y = - x - 5\;\;obtuse \hfill \\
\end{gathered} \right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Irreducible factorization of quadratic function with complex roots? How can we get the irreducible factorization of function
$$f(x)=1+4x+8x^2$$
which is $$f(x)=(1+(2+2i)x)(1+(2-2i)x)?$$
If we find the discriminant, which is $\Delta=-16$, we get
$$\sqrt{\Delta}=4i,x_1=\frac{i-1}{4},x_2=\frac{-1-i}{4}$$
Now we have $$f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$$
My question is how can we get $f(x)=(1+(2+2i)x)(1+(2-2i)x)$ from $f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$?
| $$ 1 + 4x + 8 x^2 = (1 + 2x)^2 + (2x)^2 = (1 + 2x)^2 - (2ix)^2 = $$
$$ (1 + 2x)^2 - (2ix)^2 = \left( 1 + 2x + 2ix \right) \left( 1 + 2x - 2ix \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $ Evaluate
$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$
Several days ago,I found this interesting integral from a paper about generalized log-sine integrals,but I can't remember the title of it. The answer of the integral is
\begin{align*}
-\mathrm{Ls}_{7}^{\left ( 3 \right )}\left ( \pi \right)&=\frac{9}{35}\log^72+\frac{4}{5}\pi ^{2} \log^52+9\zeta \left ( 3 \right )\log^42-\frac{31}{30}\pi ^{4}\log^32\\
&-\left [ 72\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{9}{8}\zeta \left ( 5 \right )-\frac{51}{4}\pi ^{2}\zeta \left ( 3 \right ) \right ]\log^22\\
&+\left [ 72\mathrm{Li}_{5,1}\left ( \frac{1}{2} \right )-216\mathrm{Li}_6\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_4\left ( \frac{1}{2} \right ) \right ]\log2+72\mathrm{Li}_{6,1}\left ( \frac{1}{2} \right )\\
&-216\mathrm{Li}_7\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{1161}{32}\zeta \left ( 7 \right )-\frac{375}{32}\pi ^{2}\zeta \left ( 5 \right )+\frac{1}{10}\pi ^{4}\zeta \left ( 3 \right )
\end{align*}
where
$$\mathrm{Ls}_n^{\left ( k \right )}\left ( \alpha \right ):=-\int_{0}^{\alpha }\theta ^{k}\log^{n-1-k}\left | 2\sin\frac{\theta }{2} \right |\mathrm{d}\theta $$
is the generalized log-sine integral and
$$\mathrm{Li}_{\lambda ,1}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{\lambda }}\sum_{j=1}^{k-1}\frac{1}{j}$$
is the multiple polylogarithm.
I found a beautiful way to solve the integrals below
$$\int_{0}^{\frac{\pi }{2}}t^{2n}\log^{m}\left ( 2\cos t \right )\mathrm{d}t $$
Let's consider
$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( xt \right )\left ( 2\cos t \right )^{y}\mathrm{d}t$$
By using Gamma function,the integral become
$$\mathcal{I}\left ( x,y \right )=\frac{\pi \, \Gamma \left ( y+1 \right )}{2\Gamma \left ( \dfrac{x+y+2}{2} \right )\Gamma \left ( \dfrac{y-x+2}{2} \right )}$$
Then we can get
$$\mathcal{I}\left ( x,y \right )=\frac{\pi }{2}\exp\left ( \sum_{k=2}^{\infty }\frac{\left ( -1 \right )^{k}}{k\cdot 2^{k}}\zeta \left ( k \right )\left [ \left ( 2y \right )^{k}-\left ( y-x \right )^{k}-\left ( x+y \right )^{k} \right ] \right )$$
On the other hand,using taylor series
$$\mathcal{I}\left ( x,y \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\sum_{m=0}^{\infty }\frac{y^{m}}{m!}\int_{0}^{\frac{\pi }{2}}t^{2n}\log^m\left ( 2\cos t \right )\mathrm{d}t$$
So,the comparison of coefficient shows the answer.For example
$$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$
I wonder can we use the same way to prove the integral in the beginning,if not,is there another way to handle it?
| First I want to define with the Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ a special generalization of the Riemann Zeta function :
$$\zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
and
$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
which are convergent for the integer values $\enspace m\geq 2$ .
For $\enspace n=0\enspace$ we have $\enspace\zeta_0(m)=\zeta(m)\enspace$ and $\enspace\eta_0(m)=\eta(m)\enspace$ .
Note: Obviously (because of the other results) these series can be expressed by sums of the polylogarithm function and modifications of that.
Please also see here, part Expansion by harmonic numbers, with $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ and it's recursion formula.
Secondly, an extension of an integral as a series, $n\in\mathbb{N}_0$ and $z\in\mathbb{R}\setminus \{2\mathbb{N}\}$ and $nz>-1$:
$ \displaystyle \int\limits_0^\pi x^n \left(2\sin\frac{x}{2}\right)^z dx=i^{-z} \int\limits_0^\pi x^n e^{i\frac{xz}{2}}(1- e^{-ix})^z dx= e^{-i\frac{\pi z}{2}} \int\limits_0^\pi x^n \sum\limits_{k=0}^\infty\binom{z}{k}(-1)^k e^{-ix(\frac{z}{2}-k)} dx$
$\displaystyle =\int\limits_0^\pi x^n e^{i(x-\pi)\frac{z}{2}} dx+ \sum\limits_{v=0}^n \frac{(-1)^v\pi^{n-v} n!}{i^{v+1}(n-v)!} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}} $
$\displaystyle \hspace{3.5cm} -i^{n-1}n!e^{-i\frac{\pi z}{2}} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{ (-1)^k}{(\frac{z}{2}-k)^{n+1}}$
using the main branch of the logarithm and therefore $\displaystyle i=e^{i\frac{\pi}{2}}$ .
The Stirling numbers of the first kind are usually defined by $\enspace \displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $ .
Because of $\enspace \displaystyle (\sum\limits_{v=0}^\infty x^v \frac{d^k}{dz^k}\binom{z}{v}) |_{z=0} =\frac{d^k}{dz^k}(1+x)^z |_{z=0} =(\ln(1+x))^k=k!\sum\limits_{v=k}^\infty (-1)^{v-k} \left[\begin{array}{c} v \\ k \end{array} \right] \frac{x^v}{v!}$
we get $\enspace \displaystyle \binom{z}{k}|_{z=0}=0^k\enspace$ , $\enspace \displaystyle \frac{d}{dz} \binom{z}{k} |_{z=0} = (-1)^{k-1} \left[\begin{array}{c} k \\ 1 \end{array} \right] \frac{1}{k!}= \frac{(-1)^{k-1}}{k} \enspace$ , $\enspace \displaystyle \frac{d^2}{dz^2} \binom{z}{k} |_{z=0} = (-1)^{k-2} \left[\begin{array}{c} k \\ 2 \end{array} \right] \frac{2!}{k!}= \frac{(-1)^k 2}{k}\sum\limits_{j=1}^{k-1}\frac{1}{j} \enspace$ and $\enspace \displaystyle \frac{d^3}{dz^3} \binom{z}{k} |_{z=0} = (-1)^{k-3} \left[\begin{array}{c} k \\ 3 \end{array} \right] \frac{3!}{k!}= \frac{(-1)^{k-1} 3}{k}( (\sum\limits_{j=1}^{k-1}\frac{1}{j})^2 - \sum\limits_{j=1}^{k-1}\frac{1}{j^2} ) $ .
For $(n;k):=(3;3)$ follows
$\displaystyle \int\limits_0^\pi x^3 \left(\ln\left(2\sin\frac{x}{2} \right)\right)^3 dx =$
$\hspace{2cm}\displaystyle =\frac{9\pi^2}{2}\left(\zeta(5)+3\eta(5)-4\eta_1(4)+2\eta_2(3)\right) $
$\hspace{2.5cm}\displaystyle - 90\left(\zeta(7)+\eta(7)\right) +72\left(\zeta_1(6)+\eta_1(6)\right) - 18\left(\zeta_2(5)+\eta_2(5)\right) $
Note:
For the calculations I have used $\enspace\displaystyle\int\limits_0^\pi x^n e^{iax}dx = \frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{i\pi a}\sum\limits_{v=0}^n\frac{(-1)^v \pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$
with $\enspace\displaystyle a=-(\frac{z}{2}-k)$ .
And it was necessary to calculate $\enspace\displaystyle\frac{d^m}{dz^m} \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}}|_{z=0}\enspace$ and $\enspace\displaystyle\frac{d^m}{dz^m} e^{-i\frac{\pi z}{2}}\binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{n+1}}|_{z=0}\enspace$ for $\enspace m\in\{0,1,2,3\}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 0
} |
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