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McLaurin expansion for $f(x)=e^{\sin{x}}$ of the 4:th order. By the 4:th order, they mean using the 4:th derivative. But the differentiation gets a bit ugly quite fast, so instead of computing all the derivatives , I should be able to use the standard expansions:
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+B(t) \\ \sin{x}=x-\frac{x^3}{3!}+B(x),$$
Where $B(t),B(x)$ are functions that are bounded close to $x=0.$ Correct me if I'm wrong, but both of the expansions above are of 4:th order since for the sine function we have two terms vanishing because they become zero?
I'm not sure how I should fuse the separate expansions of $e^t$ and $\sin{x}$ to one expansion. Setting $t=\sin{x}$ seems to give me an expression tedious tedious to work with.
| You already have
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+O(t^5)$$
and
$$\sin{x}=x-\frac{x^3}{3!}+O(x^5),$$
So just substitute:
$$e^{\sin x} = 1 + \left(x-\frac{x^3}{3!}+O(x^5)\right) + \frac{1}{2!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^2 \\+ \frac{1}{3!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^3 +\frac{1}{4!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^4 + O\left(\left(x-\frac{x^3}{3!}+O(x^5)\right)^5\right)\\
=1 + \left(x-\frac{x^3}{3!}\right) + \frac{1}{2!}\left(x^2-2\frac{x^4}{3!}\right) + \frac{1}{3!}\left(x^3\right) +\frac{1}{4!}\left(x^4\right) + O\left(x^5\right).$$
Gather like terms and bob's your uncle.
$$e^{\sin x} = 1 + x + \frac{x^2}{2!} -\frac{x^3}{3!}+\frac{x^3}{3!} -2\frac{1}{2}\frac{x^4}{3!}+\frac{x^4}{4!}+O(x^5)\\=1 + x + \frac{x^2}{2!} - \frac{x^4}{8}+O(x^5).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.
| Infinite Sum
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+n+1}
&=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}-\frac1{n+\frac12+i\frac{\sqrt3}2}\right]\tag1\\
&=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}+\frac1{-n-\frac12-i\frac{\sqrt3}2}\right]\tag2\\
&=\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag3\\
&=\frac1{i\sqrt3}\pi\cot\left(\frac\pi2-i\frac{\pi\sqrt3}2\right)\tag4\\
&=\frac1{i\sqrt3}\pi\tan\left(i\frac{\pi\sqrt3}2\right)\tag5\\
&=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag6
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: rewrite second term
$(3)$: write as a principal value sum
$(4)$: $(7)$ from this answer
$(5)$: $\cot\left(\frac\pi2-x\right)=\tan(x)$
$(6)$: $\tan(ix)=i\tanh(x)$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^2+n+1}=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-1}\tag7
$$
Derivation of Claude Leibovici's Asymptotic Expansion
Since we have
$$
\begin{align}
\frac1{n^2+n+1}
&=\frac1{n^2}\frac{1-\frac1n}{1-\frac1{n^3}}\\
&=\frac1{n^2}-\frac1{n^3}+\frac1{n^5}-\frac1{n^6}+\frac1{n^8}-\frac1{n^9}+\frac1{n^{11}}-\frac1{n^{12}}\cdots\tag8
\end{align}
$$
Applying the Euler-Maclaurin Sum Formula to $(8)$ yields
$$
\begin{align}
&\sum_{k=1}^n\frac1{k^2+k+1}\\
&=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-1\\
&-\frac1n+\frac1{n^2}-\frac2{3n^3}+\frac{11}{15n^5}-\frac1{n^6}+\frac1{3n^7}+\frac1{n^8}-\frac8{5n^9}+\frac{83}{33n^{11}}-\frac1{n^{12}}+O\!\left(\frac1{n^{13}}\right)\tag9
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/2579430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the Integral via Residue Theorem and Laurent Expansion Let $f(z)=\dfrac{z^{15}}{(z^2+1)^2(z^4+2)^3}$ . My goal is to compute $\displaystyle I=\int_{|z|=4}f(z) \,dz$.
Since all the singular points (except $∞$) lie in the circle $|z|<4$, we obtain $I=-2\pi i\operatorname{Res}(f,∞)$. Thus, we just need to find the Laurent coefficient $a_{-1}$ at $∞$.
$f\left(\dfrac 1 w\right) = \dfrac w {(1+w^2)^2(1+2w^4)^3}$. Any convenient ways to find $a_{-1}$?
| You want to find $Res_{w = 0} \left[\frac 1{w^2}f\left(\frac 1w \right)\right] = Res_{w = 0} \left[\dfrac 1{w(1+w^2)^2(1+2w^4)^3}\right]$ . Now we have:
$$a_{-1} = \frac{1}{2\pi i} \int_{C_0} \frac {dw}{w(1+w^2)^2(1+2w^4)^3} = \frac{1}{(1+0^2)^2(1+2\cdot 0^4)^4} = 1$$
by Cauchy Integral Formula, as $\frac{1}{(1+w^2)^2(1+2w^4)^3}$ is analytic at $C_0$, which can be taken as any curve inside a disk $|z| \le \frac 14$, for example.
A direct way of finding the residue is to use partial fraction decomposition. We have:
$$\dfrac 1{w(1+w^2)^2(1+2w^4)^3} = \frac Aw + \frac{F(w)}{(1+w^2)^2(1+2w^4)^3}$$ Now it's not har to see that $A = a_{-1}$, so multiply both sides by $w$ and evaluate it at $w=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| Let $\Lambda =$ the limit we need to find. Then, $ \ \Box\ \Lambda = 1$.
Proof: We will begin our proof using the following Lemma. $$\forall a, b\in\mathbb{R}, \ \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}.\tag1$$
Substitute $a = x$ and $b = x + \sqrt{x}$ into the Lemma. $$(1) = \sqrt{\frac{x + \sqrt{x^2 - x + \sqrt{x}}}{2}} + \sqrt{\frac{x - \sqrt{x^2 - x + \sqrt{x}}}{2}}.$$ Find the limit of the fractions under each root.$$\lim_{x\to\infty}\frac{x \pm \sqrt{x^2 - x + \sqrt{x}}}{2} = \frac 12\lim_{x\to\infty}\bigg(x \pm \sqrt{x^2 - x + \sqrt{x}}\bigg) = \frac12\cdot\infty = \infty$$ $$\therefore \lim_{x\to\infty}\sqrt{x + \sqrt{x + \sqrt{x}}} = \sqrt{\infty} + \sqrt{\infty} = \infty + \infty = \infty.$$ And, $\because \lim_{x\to\infty}\sqrt{x} = \infty$ then we finally have as desired. $$\Lambda = \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x +\sqrt{x}}}} = \frac{\infty}{\infty} = 1$$ $$\therefore \Lambda = 1.\tag*{$\bigcirc$}$$
| {
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"url": "https://math.stackexchange.com/questions/2581135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:
$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$
Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$
| We have: $a_1a_2 = 1 $ , $a_2a_3 = 2$ ,..., $a_na_{n+1}$ = $n$.
Thus , $a_n = \dfrac{a_na_{n-1}}{a_{n-1}}= \dfrac{n-1}{a_{n-1}}= a_{n-2}\dfrac{n-1}{a_{n-1}a_{n-2}}=a_{n-2}\dfrac{n-1}{n-2}= \dfrac{(n-3)(n-1)}{(n-2)a_{n-3}}= a_{n-4}\dfrac{(n-1)(n-3)}{(n-2)(n-4)}= ....=\dfrac{(n-1)(n-3)\cdots 3\cdot 1}{(n-2)(n-4)\cdots 4\cdot 2}a_2= \dfrac{(n-1)(n-3)\cdots 3\cdot 1}{(n-2)(n-4)\cdots 4\cdot 2\cdot a_1} $.
Can you manage to find the limit from this point?
| {
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"url": "https://math.stackexchange.com/questions/2584120",
"timestamp": "2023-03-29T00:00:00",
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Equation of the sphere that passes through 4 points Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing on four points. But I did not succeed
Here is the book answer
$$x^2+y^2+z^2+54x−58y+4z=0$$
| Using the equation for points on spheres:
$\qquad(x-a)^2+(y-b)^2+(z-c)^2=r^2$
Using coordinates of the four points provided, we have four simultaneous equations to solve for $a, b, c, d$.
\begin{cases}
(-5-a)^2+(4-b)^2+(1-c)^2=r^2 \\
(3-a)^2+(4-b)^2+(-5-c)^2=r^2 \\
(0-a)^2+(0-b)^2+(4-c)^2=r^2 \\
(0-a)^2+(0-b)^2+(0-c)^2=r^2
\end{cases}
(Substracting third and fourth equation yield $c$, and the first two yield $a$ then $b$, then you have $r$, easy to solve on paper)
and get a nonnegative solution:
\begin{cases}
a=2 \\
b=\frac{29}{4} \\
c=2 \\
r=\frac{\sqrt{969}}{4}
\end{cases}
(just solved it, maybe you can check the answers :D )
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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} |
Third degree polynomial with zeros $6$ and $-5+2i$ that passes through $(2,-63)$?
Find a third degree polynomial with real coefficients which has $6$ and $-5+2i$ as zeros and $f(2)=-63$
I tried to substitute in the equation
$$ f(x)=ax^3 +bx^2 + cx +d = 0$$
by the two given zeros and use $f(2)= -63$. But I think we need another equation to determine all the coefficients.
| $$\begin{align}f(x)&=a(x-\alpha)(x-\beta)(x-\gamma)\end{align}$$
Using complex conjugate theorem
$$x=-5+2i\text{ is a root}\implies x=-5-2i\text{ is a root}\\(x+5)^2=-4\\x^2+10x+29=0$$
The other factor is $(x-6)$
So far, we have
$$\begin{align}a(x-6)(x^2+10x+29)&=0\end{align}$$
Use the fact that $f(2)=-63$
$$\begin{align}a(-4)(4+20+29)&=-63\\212a&=63\\a&=\dfrac{63}{212}\end{align}$$
$$f(x)=\dfrac{63}{212}(x-6)(x^2+10x+29)$$
is the required polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \frac{(x^2-1)^4-3x^2(x^2-1)^2+x^4}{x[(x^2-1)^3-x^2(x^2-1)]}=1$
how can i solve 8 degree equation
| Let $m-\frac{1}{m}=1$, $t-\frac{1}{t}=m$.
Thus, $x-\frac{1}{x}=t,$ which gives
$$m=\frac{1\pm\sqrt5}{2},$$
$$t^2-\frac{1\pm\sqrt5}{2}t-1=0$$ and
$$x-\frac{1}{x}=\frac{1\pm\sqrt5\pm\sqrt{22\pm2\sqrt5}}{4}.$$
From the last inequality we can get all $8$ distinct real roots.
| {
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"url": "https://math.stackexchange.com/questions/2586733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}$$
Let $x=\frac{P}{\sqrt{a}}$, $y=\frac{P}{\sqrt{b}}$,$z=\frac{P}{\sqrt{c}}$
Then we have $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
By $AM \ge HM$
$$\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Hence
$$x+y+z \ge 9$$
Any way to proceed further?
| I did the following:
Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}$, $y=\sqrt{a}+\sqrt{c}-\sqrt{b}$ and $z=\sqrt{a}+\sqrt{b}-\sqrt{c}$. Then $\frac{x+y}{2}=\sqrt{c}, \frac{x+z}{2}=\sqrt{b}$ and $\frac{y+z}{2}=\sqrt{a}$. Rewriting the original expression, we get $$\frac{1}{2}\left(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)$$
It is easy to see that this expression is greater than or equal to $3$, as $\frac{x}{y}+\frac{y}{x}\geq 2$ by AM-GM inequality.
This minimum value is achieved when $a=b=c$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| By parts: $u=2x+3$ and $\mathrm dv=\sqrt{3x+1}$, then
$$
\int(2x+3)\sqrt{3x+1}\;\mathrm dx=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{4}{9}\int(3x+1)^{3/2}\;\mathrm dx\\
=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{8}{135}(3x+1)^{5/2}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lucas polynomials and primality testing Can you provide a proof or a counterexample for the claim given below ?
Inspired by Agrawal's conjecture in this paper I have formulated the following claim :
Let $n$ be a natural number greater than one . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $L_n(x)$ be Lucas polynomial , then $n$ is a prime number if and only if $L_n(x) \equiv x^n \pmod {x^r-1,n}$ .
You can run this test here or here .
I have tested this claim up to $2 \cdot 10^4$ and there were no counterexamples .
EDIT
Mathematica implementation of test :
n=31;
r=3;
While[Mod[n,r]==0 || PowerMod[n,2,r]==1,r=NextPrime[r]];
If[PolynomialMod[PolynomialRemainder[LucasL[n,x],x^r-1,x],n]-PolynomialRemainder[x^n,x^r-1,x]===0,Print["prime"],Print["composite"]];
| This is a partial answer.
This answer proves that if $n$ is a prime number, then $L_n(x) \equiv x^n \pmod {x^r-1,n}$.
Lemma : For every positive integer $n$, there are integers $a_0,a_1,\cdots, a_{n-1}$ such that
$$L_n(x)=x^n+\sum_{k=0}^{n-1}a_kx^k$$
Proof for lemma :
The claim is true for $n=1$ and $n=2$ :
$$L_1(x)=x^1+0\cdot x^0,\quad L_2(x)=x^2+0\cdot x^1+2\cdot x^0$$
Supposing that the claim is true for $n-2,n-1$ gives
$$\begin{align}L_n(x)&=xL_{n-1}(x)+L_{n-2}(x)
\\\\&=x\left(x^{n-1}+\sum_{k=0}^{n-2}a_kx^k\right)+x^{n-2}+\sum_{k=0}^{n-3}b_kx^k
\\\\&=x^n+\sum_{k=0}^{n-2}a_kx^{k+1}+x^{n-2}+\sum_{k=0}^{n-3}b_kx^k
\\\\&=x^n+a_{n-2}x^{n-1}+(a_{n-3}+1)x^{n-2}+\sum_{k=1}^{n-3}(a_{k-1}+b_k)x^k+b_0\end{align}$$
where $a_0,a_1,\cdots,a_{n-2},b_0,b_1,\cdots,b_{n-3}$ are integers.
So, the claim is true for $n$. $\quad\square$
For $n=2$, we get
$$L_2(x)=x^2+(x^r-1)\times 0+2\times 1\equiv x^2\pmod{x^r-1,2}$$
In the following, $n$ is an odd prime.
By the binomial theorem,$$\begin{align}2^nL_n(x)&=\left(x-\sqrt{x^2+4}\right)^n+\left(x+\sqrt{x^2+4}\right)^n
\\\\&=\sum_{k=0}^{n}\binom nkx^{n-k}\left(-\sqrt{x^2+4}\right)^k+\sum_{k=0}^{n}\binom nkx^{n-k}\left(\sqrt{x^2+4}\right)^k
\\\\&=\sum_{k=0}^{n}\binom nkx^{n-k}\left(\left(-\sqrt{x^2+4}\right)^k+\left(\sqrt{x^2+4}\right)^k\right)
\\\\&=\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}\cdot 2\left(\sqrt{x^2+4}\right)^{2j}
\\\\&=\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}\cdot 2(x^2+4)^{j}
\\\\&=\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}\cdot 2\sum_{k=0}^{j}\binom jk(x^2)^{j-k}\cdot 4^{k}
\\\\&=\sum_{j=0}^{(n-1)/2}\sum_{k=0}^{j}x^{n-2k}\cdot 2^{2k+1}\binom n{2j}\binom jk
\\\\&=2x^{n}+\sum_{j=1}^{(n-1)/2}\sum_{k=0}^{j}x^{n-2k}\cdot 2^{2k+1}\binom n{2j}\binom jk
\\\\&=2x^{n}+\sum_{j=1}^{(n-1)/2}\left(x^{n}\cdot 2\binom n{2j}+\sum_{k=1}^{j}x^{n-2k}\cdot 2^{2k+1}\binom n{2j}\binom jk\right)
\\\\&=2x^{n}+\sum_{j=1}^{(n-1)/2}x^{n}\cdot 2\binom n{2j}+\sum_{j=1}^{(n-1)/2}\sum_{k=1}^{j}x^{n-2k}\cdot 2^{2k+1}\binom n{2j}\binom jk
\\\\&=2x^n\sum_{j=0}^{(n-1)/2}\binom n{2j}+\sum_{j=1}^{(n-1)/2}\sum_{k=1}^{j}x^{n-2k}\cdot 2^{2k+1}\binom n{2j}\binom jk
\\\\&=2x^n\cdot 2^{n-1}+\sum_{j=1}^{(n-1)/2}\binom n{2j}\sum_{k=1}^{j}x^{n-2k}\cdot 2^{2k+1}\binom jk
\end{align}$$
from which
$$L_n(x)=x^n+\frac{1}{2^n}\sum_{j=1}^{(n-1)/2}\binom n{2j}\sum_{k=1}^{j}x^{n-2k}\cdot 2^{2k+1}\binom jk$$
follows.
From the lemma and the fact that $\binom nm\equiv 0\pmod n$ for $1\le m\le n-1$, there is a polynomial $f$ with integer coefficients such that
$$L_n(x)=x^n+(x^r-1)\times 0+nf$$
from which
$$L_n(x)\equiv x^n\pmod{x^r-1,n}$$
follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with $xyz=1$ Given $x,y,z>0$ such that $xyz=1$
Prove that: $\frac{x^7}{x^8+1}+\frac{y^7}{y^8+1}+\frac{z^7}{z^8+1}\leq \frac{3}{2}$
P/s: I tried to solve it by AM-HM but I failed
| since
\begin{align*}
&\frac34-\frac34\cdot\frac1{x^{12}+x^6+1}-\frac{x^7}{x^8+1}\\
={}&\frac{x^6(x-1)^2(3x^{12}+2x^{11}+x^{10}-x^8-2x^7-2x^5-x^4+x^2+2x+3)}{4(x^{12}+x^6+1)(x^8+1)}
\end{align*}
and
\begin{align*}
2x^{11}+2x&\geqslant2x^7+2x^5,\\
x^{10}+x^2&\geqslant x^8+x^4,
\end{align*}
we get
$$\frac{x^7}{x^8+1}\leqslant\frac34-\frac34\cdot\frac1{x^{12}+x^6+1},$$
so, just need to prove
$$\frac94-\frac34\sum\frac1{x^{12}+x^6+1}\leqslant\frac32,$$
or
$$\sum\frac1{x^{12}+x^6+1}\geqslant1,$$
or equivalent to: $a$, $b$, $c>0$, $abc=1$, prove that
$$\sum\frac1{a^2+a+1}\geqslant1,$$
which is well-known.
| {
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"answer_id": 0
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IAS math problem 2017-paper 1 4a Reduce the following equation into standard form and hence determine the nature of coincoid $x^2+y^2+z^2-zx-xy-yz-3x-6y-9z+21=0$
I have reduced the equation into $(x-3/2)^2 +(y-3)^2 + (z-9/2)^2 -zx -xy-yz = 21/2$ from here on I can sense to reduce it to any of the standard 3D form that I am aware of(like ellipse, ellipsoid, hyperboloid, etc.,). Please help me out in the same.
| just to find the nature of the surface does not require an orthonormal coordinate change. Take
$$ u = x+y+z, \; \; \; v = -x + y, \; \; \; w = -x - y + 2z \; \; . $$
$$ 6u = \frac{3}{4} (v-1)^2 + \frac{1}{4} (w-3)^2 + 18 $$
which is a paraboloid. The $6u$ stuff does not show this, but it is a paraboloid of revolution. One could correct $u,v,w,$ in particular dividing by $\sqrt 3, \sqrt 2, \sqrt 6,$ to get a strictly rotated version. The axis of symmetry (actually of revolution, as I said) is parallel to $(1,1,1)$ in the original $(x,y,z)$ coordinates, and the vertex is at $(0,1,2).$
Having done that much, we get an orthonormal change taking $u = p \sqrt 3,$ $v = q \sqrt 2,$ $w = r \sqrt 6.$
$$ 6 p \sqrt 3 = \frac{3}{4} (q \sqrt 2-1)^2 + \frac{1}{4} (r \sqrt 6-3)^2 + 18, $$
$$ 6 p \sqrt 3 = \frac{3}{2} (q -\frac{1}{\sqrt 2})^2 + \frac{3}{2} (r - \frac{3}{\sqrt 6})^2 + 18, $$
$$ 4 p \sqrt 3 =(q -\frac{1}{\sqrt 2})^2 + (r - \frac{3}{\sqrt 6})^2 + 12 \; \; , $$
$$ p = \frac{1}{4 \sqrt 3} \left(\left(q -\frac{1}{\sqrt 2}\right)^2 + \left(r - \frac{3}{\sqrt 6}\right)^2 + 12 \right) \; \; . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of convergence $R$ to be $2$, but I struggle to show that it converges on $x = \pm 2$. Can you give me a hint?
What I have so far: $R = \frac{1}{\limsup_{n \to \infty} (x_n)} = \frac{1}{\frac{1}{1+1}} = \frac{2}{1} = 2$, where
$$\begin{align} x_n := \sqrt[n]{|a_n|} &= \sqrt[n]{| \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2}|} = \frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{1} \\ &= \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &= \frac{n - 1}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{1 - \frac{1}{n}}{\sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}}} \end{align}$$
How do I determine if $\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} (\pm 2)^n$ converges? The root test doesn't work and the ratio test doesn't seem to be a smart move. What can I do?
| Let's try to bound the sequence here after we call it $a_n$. Note that the following inequalities hold:
$$a_n={{(\sqrt{n^2+n}-\sqrt{n^2+1})^n}\over{n^2}}x^n$$$$={{(n-1)^n}\over{n^2}(\sqrt{n^2+n}+\sqrt{n^2+1})^n}x^n$$$$\le {{(n-1)^n}\over{n^2}(2n)^n}x^n$$$$\le{{n^n}\over{n^2}(2n)^n}={{({x\over 2})^n}\over{n^2}}=b_n$$
where $b_n$ obviously converges when $|{x\over 2}|\le 1$ or $|x|\le 2$.
Now we try to bound $a_n$ from bottom. We have :
$$a_n\ge {{(n-1)^n}\over{n^2}(\sqrt{n^2+n+{1\over 4}}+\sqrt{n^2+n+{1\over 4}})^n}x^n={{(n-1)^n}\over{n^2}(2n+1)^n}x^n={1\over {n^2}}({{n-1}\over{2n+1}})^nx^n={1\over {n^22^n}}(1-{3\over{2n+1}})^{{{2n}\over 3}{3\over 2}}x^n$$
Now note that the term $(1-{3\over{2n+1}})^{{{2n}\over 3}{3\over 2}}$ tends to $e^{3\over2}$ which is a constant so
$$a_n=\Theta({1\over {n^2}}({x\over 2})^n)$$
which diverges when $|x|>2$ so is $a_n$. Therefore we finally deduce that the radius of convergence is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality $a_{n}>\frac{2n}{n+2}$ Let sequence $\{a_{n}\}$ such $a_{1}=1$,and such
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{(n+1)^2}$$\show that
$$a_{n}>\dfrac{2n}{n+2}\tag{1}$$
I try use induction,then It's a pity that failed, becase
$$a_{n+1}>\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}$$
it need to show
$$\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}>\dfrac{2(n+1)}{n+3}$$
since
$$\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}-\dfrac{2(n+1)}{n+3}=-\dfrac{4(n^2+5n+2)}{(n+1)^2(n+2)^2(n+3)}$$
so how prove this (1)? Thanks,maybe $(1)$ is easy to prove it.
| Note that for any $n \in \mathbb{N}_+$,$$
\frac{2n + 2}{n + 3} > \frac{2n}{n + 2} \Longleftrightarrow (n + 1)(n + 2) > n(n + 3),
$$
thus it suffices to prove that$$
a_n \geqslant \frac{2n + 2}{n + 3}. \quad \forall n \in \mathbb{N}_+ \tag{1}
$$
For $n = 1$, $\displaystyle a_1 = 1 = \frac{2 + 2}{1 + 3}$, so (1) holds. Now suppose (1) holds for $n$, then\begin{align*}
a_{n+ 1} &= a_n + \frac{a_n^2}{(n + 1)^2} \geqslant \frac{2n + 2}{n + 3} + \frac{1}{(n + 1)^2} \left(\frac{2n + 2}{n + 3}\right)^2\\
&= \frac{2n + 2}{n + 3} + \frac{4}{(n + 3)^2} = \frac{2n^2 + 8n + 10}{(n + 3)^2}.
\end{align*}
Because\begin{align*}
\frac{2n^2 + 8n + 10}{(n + 3)^2} \geqslant \frac{2n + 4}{n + 4} &\Longleftrightarrow (n + 4)(n^2 + 4n + 5) \geqslant (n + 2)(n + 3)^2\\
&\Longleftrightarrow n^3 + 8n^2 + 21n + 20 \geqslant n^3 + 8n^2 + 21n + 18,
\end{align*}
then$$
a_{n+ 1} \geqslant \frac{2n^2 + 8n + 10}{(n + 3)^2} \geqslant \frac{2n + 4}{n + 4} = \frac{2(n + 1) + 2}{(n + 1) + 3}.
$$
By induction, (1) holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up the following equation (for the purpose of this question I will refer to the leftmost matrix here as A).
$$
\begin{bmatrix}
1 - \lambda & 0 & 0 \\
3 & 3 - \lambda & -4\\
-2 & 1 & -\lambda -2 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
I) Substitute $\lambda_3 = 1$
$$
\begin{bmatrix}
0 & 0 & 0 \\
3 & 2 & -4\\
-2 & 1 & -3 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
II) Reduce the matrix with elementary row operations.
$R_2 \leftarrow R_2 - 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
-2 & 1 & -3 \\
\end{bmatrix}
$$
$R_3 \leftarrow 3R_2 + 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
17 & 2 & 0 \\
\end{bmatrix}
$$
$R_2 \leftarrow \frac{1}{7} R_2$
$R_3 \leftarrow \frac{1}{17} R_3$$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
1 & 0 & 2/7\\
1 & 2/17 & 0 \\
\end{bmatrix}
$$
III) multiply matrices to get a series of equations equal to 0 and rearrange them in terms of a common element.
$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$
$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$
IV) Substitute a value into the vector to get an eigenvector.
Let $\ x_1 = 1 \rightarrow X_3 = \left(\begin{smallmatrix}1\\-2/17\\-2/7\end{smallmatrix}\right)
$
Which at this point we can see is not a multiple of the expected $X_3$. Can anyone highlight my error for me?
Many thanks in advance.
| The error is right in the end after you obtain the system of equations. You have
$$
x_1 + \frac{2}{7}x_3 = 0 \\ x_1 + \frac{2}{17}x_2 = 0
$$
You want to write $x_2$ and $x_3$ in terms of the common element $x_1$ so
$$
x_2 = -\frac{17}{2}x_1 \\ x_3 = -\frac{7}{2}x_1
$$
This means that the vector you are looking for is
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = x_1\begin{pmatrix} 1 \\ -17/2 \\ -7/2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^1 \frac {1- \tan(x)}{\sqrt{x}+\tan(x)}dx$ I am having an extremely hard time finding the anti-derivative of this function:
$$f(x)=\dfrac{1-\tan(x)}{\sqrt{x}+\tan(x)}$$
while the lower bound is 0 and the upper bound is 1.
I tried to expand by $\cos(x)$ so it would look like
$$f(x)=\dfrac{\cos(x)-\sin(x)}{\sqrt{x}\cos(x)+\sin(x)}$$
and then linearly separate it into 2 integrals but it didn't help much at all. Any help would be appreciated, thank you very much.
| This is not an answer but it is too long for a comment.
If you had "extremely hard time finding the anti-derivative of this function", now we are two !
I do not think that we could find the antiderivative
$$I=\int\frac {1- \tan(x)}{\sqrt{x}+\tan(x)}\,dx$$ even using special functions.
What I first did is to let $x=y^2$ to work
$$J=\int\frac{2 y \left(1-\tan \left(y^2\right)\right)}{y+\tan \left(y^2\right)}\,dy$$ If you plot the integrand for $0\leq y\leq 1$, you could notice that it is "close" to $2-2y$ and then, over this range, the definite integral should be "close" to $1$ (smaller that $1$ because or the curvature of the integrand).
What I tried was to approximate the integrand looking for $[n,m]$ Padé approximants (built around $y=0$) which could be workable for integration. However, the problem I faced is that, until $m=4$, I just obtain the linear approximation mentioned above. For $n=2$, the denominator cancels in the range. So, I kept the simplest $(n=3,m=4)$ which gives, as an approximation,
$$\frac{2 y \left(1-\tan \left(y^2\right)\right)}{y+\tan \left(y^2\right)}=\frac {2-2 y^2-\frac{2 }{3}y^3 } {1+y-\frac{1}{3}y^3-\frac{1}{3}y^4 }=6\frac {1- y^2-\frac{1 }{3}y^3 } {(1+y)(3-y^3)}$$ The last expression can be decomposed using partial fraction since,$a,b,c$ being the roots of $y^3=3$,
$$\frac{1- y^2-\frac{1 }{3}y^3 }{(y+1) (y-a) (y-b) (y-c)}=\frac{-a^3-3 a^2+3}{3 (a+1) (a-b) (a-c) (y-a)}+\frac{-b^3-3 b^2+3}{3 (b+1) (b-a)
(b-c) (y-b)}+\frac{-c^3-3 c^2+3}{3 (c+1) (c-a) (c-b) (y-c)}-\frac{1}{3 (a+1)
(b+1) (c+1) (y+1)}$$ Integrating and recombining everything leads to the "small monster"
$$\frac{1}{12} \left(3 \left(\log \left(\frac{16}{9}\right)+\sqrt[3]{3}
\left(\sqrt[3]{3}-1\right) \log \left(\frac{1}{2} \left(2+3 \sqrt[3]{3}-3\
3^{2/3}\right)\right)+2 \sqrt[6]{3} \left(3+3^{2/3}\right) \tan
^{-1}\left(\frac{2+\sqrt[3]{3}}{3^{5/6}}\right)\right)-\sqrt[6]{3}
\left(3+3^{2/3}\right) \pi \right)\approx 0.922454$$ while a numerical integration would lead to $\approx {0.929600}$.
Not very fantastic, I agree.
Edit
Just for the fun of it, I built the $[2,2]$ Padé approximant around $y=\sqrt{\frac{\pi }{4}}$ (I shall not type the the value of the coefficients). So, $J$ can be computed and the result is $\approx 0.928856$ (much better than the previous one).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How to find $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$? The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?
| \begin{align}
\int \sqrt{x^8 + 2 + x^{-8}} dx &= \int \sqrt{\left(x^4+\frac1{x^4}\right)^2} dx \\
&= \int \left|\underbrace{x^4+\frac1{x^4}}_{\ge0}\right| dx \\
&= \int \left(x^4+\frac1{x^4}\right) dx \\
&= \frac{x^5}5-\frac1{3x^3}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Want to use Herbert Wilf's snake oil method to show $\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$ I was trying to use the snake oil method to show that $$\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$$
tl;dr I wasn't able to and desperately need help
My approach was to try to establish a more general identity regarding $$\sum_k \binom{2n+1}{2k}\binom{m+k}{s}$$
So I let $$F(x) = \sum_s \sum_k \binom{2n+1}{2k}\binom{m+k}{s} x^s$$
By changing the order of summation, I obtained
$$F(x) = \sum_k \sum_s \binom{2n+1}{2k}\binom{m+k}{s} x^s = \sum_k \binom{2n+1}{2k}(1+x)^{m+k}$$
After this, I noted that $$\sum_{k \geq 0} \binom{n}{2k}x^{2k} = \frac{(1+x)^n + (1-x)^n}{2}$$
Now, $$F(x) = \frac{(1+x)^m}{2}((1+\sqrt{1+x})^{2n+1} + (1-\sqrt{1+x})^{2n+1})$$
I'm having trouble finding the coefficient of $x^{2n}$ in this generating function.
Alternate approach: I could have created a generating function where the summation was on $m$ but that proved to be fruitless.
| Following request by OP for snake oil method we evaluate
$$\sum_{0\le 2k\le q+1} {q+1\choose 2k} {m+k\choose q}$$
and introduce
$$F(z, u) = \sum_{q\ge 0} \sum_{m\ge 0}
z^q u^m \sum_{0\le k, 2k\le q+1}
{q+1\choose 2k} {m+k\choose q}
\\ = \sum_{q\ge 0} \sum_{m\ge 0}
{m\choose q} z^q u^m
+ \sum_{q\ge 0} \sum_{m\ge 0}
z^q u^m \sum_{1\le k, 2k\le q+1}
{q+1\choose 2k} {m+k\choose q}
\\ = \sum_{m\ge 0} u^m (1+z)^m
+ \sum_{m\ge 0} u^m \sum_{k\ge 1}
\sum_{q\ge 2k-1} z^q
{q+1\choose 2k} {m+k\choose q}
\\ = \frac{1}{1-u(1+z)}
+ \sum_{m\ge 0} u^m \sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} {m+k\choose q+2k-1}
\\ = \frac{1}{1-u(1+z)}
+ \sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} \sum_{m\ge 0} u^m {m+k\choose q+2k-1}$$
Continuing with the sum we find
$$\sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} \sum_{m\ge q+k-1} u^m {m+k\choose q+2k-1}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1}
\sum_{q\ge 0} z^q u^q
{q+2k\choose 2k} \sum_{m\ge 0} u^m {m+q+2k-1\choose q+2k-1}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1}
\sum_{q\ge 0} {q+2k\choose 2k} z^q u^q \frac{1}{(1-u)^{q+2k}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1} \frac{1}{(1-u)^{2k}}
\sum_{q\ge 0} {q+2k\choose 2k} z^q u^q \frac{1}{(1-u)^{q}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1} \frac{1}{(1-u)^{2k}}
\frac{1}{(1-uz/(1-u))^{2k+1}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1}
\frac{1-u}{(1-u(1+z))^{2k+1}}
\\ = (1-u) \sum_{k\ge 0} z^{2k+1} u^{k}
\frac{1}{(1-u(1+z))^{2k+3}}
\\ = \frac{(1-u)z}{(1-u(1+z))^3} \sum_{k\ge 0} z^{2k} u^{k}
\frac{1}{(1-u(1+z))^{2k}}
\\ = \frac{(1-u)z}{(1-u(1+z))^3}
\frac{1}{1-uz^2/(1-u(1+z))^2}
\\ = \frac{(1-u)z}{1-u(1+z)}
\frac{1}{(1-u(1+z))^2-uz^2}$$
Restoring the term in front now yields
$$\frac{1}{1-u(1+z)}
+ \frac{z}{1-u(1+z)}
\frac{1}{1-u(1+z)^2}
\\ = \frac{1}{1-u(1+z)}
\left(1+ \frac{z}{1-u(1+z)^2}\right)
\\ = \frac{1+z}{1-u(1+z)^2}.$$
Extracting coefficients we get
$$[z^q] [u^m] \frac{1+z}{1-u(1+z)^2}
= [z^q] (1+z) (1+z)^{2m} = [z^q] (1+z)^{2m+1} = {2m+1\choose q}.$$
Now put $q=2n$ to obtain
$$\bbox[5px,border:2px solid #00A000]{
{2m+1\choose 2n}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very easily using Muirhead. But my friend has no idea what Muirhead inequality is. So I want to know whether there is any other method to solve this problem except for Muirhead's inequality.
| $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\sqrt{\frac{a^3b^3}{abc^2}}+\sqrt{\frac{b^3c^3}{bca^2}}+\sqrt{\frac{c^3a^3}{cab^2}}=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{\frac{ab^2c}{ac}}+\sqrt{\frac{bc^2a}{ba}}+\sqrt{\frac{ca^2b}{cb}}=a+b+c$$
Above, we have twice used the known inequality $x^2+y^2+z^2\ge xy+yz+xz$, or (equivalently)$x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$
I'm studying the convergence of the series
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$$
*
*$\frac {\sqrt{n^2+1}-n}{\sqrt{n}}>0, \forall n \ge 1$
*$\lim_{n \to +\infty}\frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\lim_{n \to +\infty}\frac {n^2+1-n^2}{\sqrt{n}*(\sqrt{n^2+1}+n)}=0$
*The suggested solution in my book says that the series converges but:
$\frac {\frac {\sqrt{n^2+1}-n}{\sqrt{n}}}{(\frac {1}{n})^{- \frac {1}{2}}}= \frac {n* \sqrt {1+ \frac {1}{n^2}}}{ \sqrt {n}}* \frac {1}{\sqrt {n}} \rightarrow 1 $ for $n \rightarrow \infty $
| Note that
$$ \frac {\sqrt{n^2+1}-n}{\sqrt{n}} \cdot \frac { \sqrt{n^2+1}+n }{\sqrt{n^2+1}+n}= \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}\sim \frac{1}{2\sqrt{n^3}}$$
thus
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\sum_{n=1}^\infty \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}$$
converges by comparison test with $$\frac{1}{\sqrt{n^3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I simplified the equation as $$ \int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\tan^2(x))dx$$
Then I simplified $\tan^2(x)\equiv\frac{\sin^2(x)}{\cos^2(x)}, \sin^2(x)\equiv1-\cos^2(x)$ so it becomes, $\tan^2(x)\equiv\frac{1-\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}-1$ making the overall integral
$$\int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\frac{1}{\cos^2(x)}-1)dx$$
$$=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx-\int_\frac{-π}{3}^{\frac{π}{3}}1dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx$$
I know that $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$ but that's off by heart and not because I can work it out. Since $x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}=0$, the final equation becomes
$$\int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx=\tan(x)|_{\frac{-π}{3}}^{\frac{π}{3}}=2 \sqrt3$$
Is what I did correct because I feel like I've made a mistake somewhere but can't find it. Also why does $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
EDIT - Made an error in $\int\frac{1}{\cos^2(x)}dx=\tan^2(x)+c$, it's actually $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
| Well you know that $(\tan(x))'= \frac 1 {\cos^2(x)}$
$$(\tan(x))'= \frac {\sin^2(x)+\cos^2(x)} {\cos^2(x)}$$
$$(\tan(x))'= {1+\tan^2(x)}$$
Integrate
$$\tan(x)= \int {1+\tan^2(x)}dx$$
$$\tan(x)-x=\int {\tan^2(x)}dx$$
$$\tan(x)-2x= \int -1+ {\tan^2(x)}dx$$
$$2x-\tan(x)= \int 1- {\tan^2(x)}dx=\int 1- {\frac {\sin^2(x)}{\cos^2(x)}}dx=\int {\frac {\cos^2(x)-\sin^2(x)}{\cos^2(x)}}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
| a nice problem! squaring the first equation we obtain
$$xy+xz+yz=0$$
raise to the power $3$ the last equation we have
$$x^2(y+z)+y^2(x+z)+z^2(x+y)+xyz=0$$
and this can be written as
$$x^2(1-x)+y^2(1-y)+z^2(1-z)+xyz=0$$
and this is
$$x^2+y^2+z^2-x^3-y^3-z^3+xyz=0$$ therefore $xyz=0$
| {
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What is the number of ways to put 40 identical balls in 6 boxes while no box contains more than 14 balls. I believe that I have to use the Inclusion–exclusion principle but I am not sure how. I tried to use this formula: $\binom{n+k-1}{n-1}$ for every step of the inclusion–exclusion solution but it didn't work out.
Thank you!!
| Jack D'Aurizio gives the clue, it can be solved by using the generating functions. In this case, generating function is
$$G(x) = (1+x+x^2+\ldots+x^{14})^6$$
and we are seeking the coefficient of $x^{40}$. Now, if we manipulate the expression by using:
$$1+x+x^2+\ldots+x^{14} = \frac{1-x^{15}}{1-x}\ and\ \frac{1}{1-x} = 1+x+x^2+...$$
we get
$$G(x) = (1-x^{15})^6 \cdot (1+x+x^2+...)^6$$
Now, we also know that we have
$$(1+x+x^2+...)^6 = \sum_{m = 0}^{\infty}\binom{6+m-1}{6-1}x^m$$
So generating function becomes
$$G(x) = (1-x^{15})^6 \cdot \sum_{m = 0}^{\infty}\binom{5+m}{5}x^m$$
So the coefficient of the term $x^{40}$ is:
$$\binom{6}{0} \binom{45}{5}-\binom{6}{1}\binom{30}{5}+\binom{6}{2}\binom{15}{5}$$
| {
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Calculating the integral $\int \sqrt{1+\sin x}\, dx$. I want to calculate the integral $\int \sqrt{1+\sin x}\, dx$.
I have done the following:
\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin x}}\, dx=\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx\end{equation*}
We substitute $$u=\sqrt{1-\sin x} \Rightarrow du=\frac{1}{2\sqrt{1-\sin x}}\cdot (1-\sin x)'\, dx \Rightarrow du=-\frac{\cos x}{2\sqrt{1-\sin x}}\, dx \\ \Rightarrow -2\, du=\frac{\cos x}{\sqrt{1-\sin x}}\, dx $$
We get the following:
\begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=\int(-2)\, du=-2\cdot \int 1\, du=-2u+c\end{equation*}
Therefore \begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=-2\sqrt{1-\sin x}+c\end{equation*}
In Wolfram the answer is a different one. What have I done wrong?
| You have $$\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin x}}\, dx=\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx\end{equation*}$$ which is true up to your last equality where you forgot your $\sqrt {cos^2x}=|cos(x)|$ and replaced it with $\sqrt {cos^2x}=cos(x)$
| {
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Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough?
| $$
\sqrt{n-2\sqrt{n}}-\sqrt{n}=\sqrt{n\left(1-\frac{2}{\sqrt{n}}\right)}-\sqrt{n}=\sqrt{n}\left(\sqrt{1-\frac{2}{\sqrt{n}}}-1\right)
$$
and
$$
\sqrt{1-\frac{2}{\sqrt{n}}}\underset{(+\infty)}{=}1-\frac{1}{\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)
$$
Finally
$$
\sqrt{n-2\sqrt{n}}-\sqrt{n}\underset{(+\infty)}{=}1+o\left(1\right)
$$
Meaning that
$$
\sqrt{n-2\sqrt{n}}-\sqrt{n}\underset{n \rightarrow +\infty}{\rightarrow}1
$$
| {
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Proving that a limit does not exist with absolute values I'm to prove that the following limit does not exist
$\lim_{x\to -2} \frac{\vert 2x +4\vert -\vert x^3 +8 \vert}{x+2}$
From here, I have taken the method of finding $\lim_{x\to -2^+}$ and $\lim_{x\to -2^-}$ to show that they're not equal
However my problem is that both are simplified to become $\frac{x^3 - 2x +4}{x+2}$
Is there something I'm doing wrong or did i make a mistake when opening the absolutes?
| Your simplification is only correct for one side. If you include more steps, we can pinpoint what went wrong
For $x > -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{(2x+4)-(x^3+8)}{x+2} = \frac{(x+2)(2 - (x^2+2x+4))}{x+2} \\ = -(x^2+2x+2) $$
For $x < -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{-(2x+4)+(x^3+8)}{x+2} = \frac{(x+2)(-2 + (x^2+2x+4))}{x+2} \\ = x^2+2x+2 $$
You can work out the two limits from here.
| {
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A and B are two matrices such that $(A+B)^3=A^3+3A^2B+3AB^2+B^3$ then $ AB=BA$ Let $A$ and $B$ be two invertible matrices in $M_2(\mathbb{R})$such that $(A+B)^3=A^3+3A^2B+3AB^2+B^3$ then prove or disprove that $ AB=BA$
My working:
$$(A+B)^3=A^3+3A^2B+3AB^2+B^3$$
$$\implies BA^2+B^2A+ABA+BAB =2A^2B+2AB^2$$
Now what should I do?
| Counter-example :
$$
A=
\left(\begin{array}{cc}
2 & 0 \\
& \\
0 & 3 \\
\end{array}\right),
B=
\left(\begin{array}{cc}
1 & 0 \\
& \\
1 & -10 \\
\end{array}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)^2 - (b/2)^2$
| Yes it is correct, indeed
$$(x + b/2)^2 - (b/2)^2=x^2+bx+\frac{b^2}4-\frac{b^2}4-=x^2+bx$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?
How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$
I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
|
Since $$\ln(x+1)= x-\frac{x^2}{2}+O(x^3)~~~and ~~~e^x= 1+x+\frac{x^2}{2}+O(x^3)$$
we get
$$(1+x)^{\frac{1}{x}}= \exp\left(\frac{1}{x}\ln(1+x)\right) = \exp\left(\frac{1}{x}(x-\frac{x^2}{2} +O(x^3))\right) \\=\exp\left(1-\frac{x}{2} +O(x^2)\right) =e\exp\left(-\frac{x}{2}+O(x^2)\right) = e(1- \frac{x}{2}+O(x^2)) $$
Hence $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} = \lim_{x\rightarrow 0} \frac{e(1- \frac{x}{2}+O(x^2))-e}{x} =\lim_{x\rightarrow 0} -\frac{e}{2}+O(x) = \color{blue}{-\frac{e}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f(x)$ if $f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$ $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function satisfying $$f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$$
if $f'(0)=2$, find the function
My Try:
we have $$f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}=\frac{2+f(x)}{3}$$ $\implies$
$$\frac{f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}}{\frac{x}{3}}=\frac{2+f(x)}{x}$$
Now taking Limit $x \to 0$ we have
$$\lim_{x \to 0}\frac{f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}}{\frac{x}{3}}=\lim_{x \to 0}\frac{2+f(x)}{x}$$
$\implies$
$$f'\left(\frac{y}{3}\right)=\lim_{x \to 0}\frac{2+f(x)}{x}$$
Now since LHS to be finite , we need $0/0$ form in RHS, hence $f(0)=-2$
Now by L'Hopital's Rule we get
$$f'\left(\frac{y}{3}\right)=f'(0)=2$$
Integrating we get
$$3f\left(\frac{y}{3}\right)=2y+c$$
Putting $y=0$ we get $c=-6$
So
$$3f\left(\frac{y}{3}\right)=2y-6$$
So $$f(y)=2y-2$$
Hence $$f(x)=2x-2$$
But this function is not satisfying given functional equation.
What went wrong?
| Using $x=y=0$ in the functional equation we can see that $f(0)=2$. Further note that $$f(x+h)=\frac{2+f(3x)+f(3h)}{3}, f(x) =\frac{2+f(3x)+f(0)}{3}$$ so that $$\frac{f(x+h) - f(x)} {h} =\frac{f(3h)-f(0)}{3h}$$ Taking limits as $h\to 0$ we get $f'(x) =f'(0)=2$ and thus $f(x) =2x+c$ and from $f(0)=2$ we get $c=2$ so that $f(x) =2x+2$. You can check that it satisfies the functional equation.
| {
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
| Your proof doesn't use any properties of the sequence: you saying that the sequence is Cauchy is no proof. Also, your first sentence is wrong, as the sequence is increasing.
What you want to do is
$$
\frac {3n+5}{2n+6}-\frac {3m+5}{2m+6}=\frac{8 (n-m)}{(2m+6)(2n+6)}
\leq \frac{8 (n-m)}{4mn}=\frac2m-\frac2n.
$$
Now you can show that, the bigger $n,m $ are, the smaller the difference. Usually one needs absolute value, but in this particular case the condition $n>m $ guarantees that everything above is positive.
| {
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Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
| You could try the generating form
$$F(x)=\frac{x^4}{2.3.4}+\frac{x^6}{4.5.6}+...\\
\frac{d^3F}{dx^3}=x+x^3+x^5+...=\frac x{1-x^2}$$
Try to integrate the last expression three times, then take the limit as $x\to1$
| {
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Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$
My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive numbers to cancel the negative. One integer would have to cancel out the second, and the third wouldn't make the sum zero. Am I right?
| Alternatively, denote: $y=ax,z=abx$, then:
$$\begin{cases}x+ax+abx=0 \\ \frac{1}{x}+\frac{1}{ax}+\frac{1}{abx}=0\end{cases} \Rightarrow \begin{cases}1+a+ab=0 \\ 1+\frac{1}{a}+\frac{1}{ab}=0\end{cases} \Rightarrow \begin{cases}a=-\frac{1}{b+1} \\ b^2+b+1=0\end{cases} \Rightarrow \emptyset.$$
| {
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What is the number of integral values of $k$ for which the equation $|x^2 - 5|x| + 6|=k$ has four solutions is? I have done the sum by first plotting the graph of the function in the Left Hand Side of the equation and then plotted the line $y=k$. For the equation to have $4$ solutions, both these two curves must intersect at $4$ different points, and from the two graphs, I could see that for the above to occur, the value of $k$ must lie between $\cfrac{1}{4}$ and $6$, that is k belongs to $( 0.25,6)$. Thus , the integral values of $k$ would be $1,2,3,4,5$; and so the number of integral values of $k = 5$. This was the answer given in the book. But as I was looking at the graph, I realized that $k=0$ also gives $4$ solutions namely $x=-2,-3,2,3$. So, should the number of integral values of $k$ be $6$ $(0,1,2,3,4,5)$?
| Let $f(x)=|x^2-5|x|+6|$. Note that $f(x)=f(-x)$.
$f(0)=6$. So, when $k=6$, the number of solutions of $f(x)=k$ is odd.
When $k\ne6$, the number of solution is double the number of solution of $f(x)=k$ for $x>0$. It suffices to find $k$ such that $f(x)=k$ has two solutions in $(0,\infty)$.
For $x>0$, $f(x)=|x^2-5x+6|$.
Obviously, $f(x)=k$ has solutions only when $k\ge0$.
When $k=0$, the solutions are $2$ and $3$.
$x^2-5x+6=k$ has two distinct positive roots if $0\le k<6$ and one positive root if $k>6$.
Note that $x^2-5x+6=-k$ has discriminant $5^2-4(6+k)=1-4k$. So $x^2-5x+6=-k$ has one positive root when $k=0.25$ and two distinct positive roots when $0\le k<0.25$.
So, $f(x)=k$ has two distinct positive roots if and only if $k=0$ or $0.25<k<6$.
Graphical Method:
I think you are correct. $k=0,1,2,3,4$ or $5$.
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the tangent is $2x$ so the gradient of the normal is $-\frac{1}{2}x$ and the equation of the normal will be $-\frac{1}{2}x + v$
Where $x=\sqrt{c}$
then $y = -\frac{1}{2}\sqrt{c} + v$
We know $y = 2c$ therefore $2c = -\frac{1}{2}\sqrt{c} + v$ and so
$v = 2c + \frac{1}{2}\sqrt{c}$
So the equation of the normal is $y = -\frac{1}{2}x +2c +\frac{1}{2}\sqrt{c}$
We can tidy this up to get $2y = -x + 4c + \sqrt{c}$
Unfortunately the book tells me the answer is $2y\sqrt{c} = -x+\sqrt{c}(4c+1)$
It looks like I lost a $\sqrt{c}$ somewhere? Where did I go wrong?
Thank you
Gary
| Equation of normal is
$$y-y_0=-\frac{1}{f'(x_0)}(x-x_0)$$
In your case as you said $x_0=\sqrt c$ and $y_0=2c$.
$$f'(x_0)=2x_0=2\sqrt c$$
When we put all this back to the equation of normal we get:
$$y-2c=-\frac{1}{2\sqrt c}(x-\sqrt c)$$
$$y=-\frac{1}{2\sqrt c}(x-\sqrt c)+2c$$
$$2y\sqrt c=-x+\sqrt c+4c\sqrt c$$
$$2y\sqrt c=-x+\sqrt c(4c+1)$$
| {
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Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac{3}{4}\big)\implies\cos\big(\tan^{-1}x\big)=\cos\big(\tan^{-1}\frac{3}{4}\big)\\\implies\tan^{-1}x=2n\pi\pm\tan^{-1}\frac{3}{4}\\
\implies \tan^{-1}x=\tan^{-1}\frac{3}{4}\quad\text{ as }0\leq\tan^{-1}\frac{3}{4}\leq\frac{\pi}{2}\\
\implies x=\frac{3}{4}
$$
Is it correct or $\frac{-3}{4}$ also is a solutions ?
What about the condition $0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}$, does this affect the solutions ?
| Well, we can use that:
$$\cos\left(\arctan\left(x\right)\right)=\frac{1}{\sqrt{1+x^2}}\tag1$$
And:
$$\sin\left(\text{arccot}\left(x\right)\right)=\frac{1}{x\cdot\sqrt{1+\frac{1}{x^2}}}\tag2$$
So in your case, we need to solve:
$$\frac{1}{\sqrt{1+x^2}}=\frac{1}{\frac{3}{4}\cdot\sqrt{1+\frac{1}{\left(\frac{3}{4}\right)^2}}}=\frac{4}{5}\space\Longleftrightarrow\space x=\pm\frac{3}{4}\tag3$$
| {
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Dividing one polynomial by another How is this done? For example, how would one simplify the following?
$$\frac{x^3-12x^2+0x-42}{x^2-2x+1}$$
I can do it with long division, but it never makes intuitive sense to me. Either an explanation of the long division algorithm or a new way of solving this would be much appreciated.
| The long division algorithm is a way of taking the numerator, and split it into multiples of the denominator, separate those out into their own fractions, and simplify away the denominator. For instance, we have $x(x^2 - 2x+1) = x^3-2x^2+x$, so
$$
\frac{x^3 - 12x^2 + 0x-42}{x^2-2x+1} = \frac{(x^3-2x+x) - 10x^2 - 1x-42}{x^2-2x+1}\\
= \frac{x(x^2-2x+1)}{x^2-2x+1} + \frac{-10x^2-1x-42}{x^2-2x+1}\\
= x + \frac{-10x^2-1x-42}{x^2-2x+1}
$$
So now we've gotten an $x$, and we're left with a new fraction where the degree of the numerator has gone down by 1 compared to where we started. We can keep going, as long as the degree of the numerator is larger than or equal to the degree of the denominator, which in this case means one more step. This time, we see that $-10(x^2-2x+1) = -10x^2+20x-10$ so
$$
\frac{-10x^2-1x-42}{x^2-2x+1} = \frac{(-10x^2+20x - 10) -21x-32}{x^2-2x+1}\\
= -10 + \frac{-21x-32}{x^2-2x+1}
$$
At this point the algorithm stops, and we're left with
$$
\frac{x^3 - 12x^2 + 0x-42}{x^2-2x+1} = x + \frac{-10x^2-1x-42}{x^2-2x+1} = x -10 + \frac{-21x-32}{x^2-2x+1}
$$
This is exactly the long division algorithm, written out with regular fraction notation. If you do the long division algorithm the normal way, you will recognize the different numerators all round, but that technique is designed for ease of calculation, not transparency. So it's difficult to tell what is really going on unless you already know.
| {
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Alternate proof for Viète's infinite product of nested radicals I am looking for alternate proof for Viete's infinite product of nested radicals. (Reference - Wikipedia)
Basically we need to find $\lim_{n\to \infty}\prod_{k=1}^{n} T_k$ where $$T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}$$ and $T_1 = \sqrt{\frac{1}{2}}$. Series looks like
$$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}...$$
Miss gave a solution treating $\cos(\theta) = \frac{1}{\sqrt2}$ that is $\theta = 45^\circ$. The series result is given easily using the identity $\cos(\theta) + 1 = 2 \cos^2(\theta/2)$ and using $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. The final result is $\frac{\sin(2\theta)}{2\theta} = \frac{2}{\pi}$.
I look for alternate ways to get to this! I am open to calculus methods.
|
Answer: we have $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\sqrt{2}}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$
*
*First check that for all $k$ we have $0\le T_k\le 1$ this is obvious by induction since $$0\le T_1= \frac{\sqrt{2}}{2}\le 1$$
*Hence there exists $a_k \in [0,\frac{\pi}{2} ]$ such that $$T_k = \cos a_k, $$
*Easily, $T_1 =\frac{\sqrt{2}}{2}\implies a_1 =\frac{\pi}{4}$ and $$\cos (a_{k+1})= T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}= \sqrt{\left(\frac{\cos a_k + 1}{2}\right)} = \cos\left(\frac{ a_k }{2}\right)$$
*Since $x\mapsto \cos x$ realize a bijection in $[0,\frac{\pi}{2} ],$ It turn out that $a_k$ is geometric sequence with ratio $1/2$
that is we have $$\color{red}{a_{k+1} = \frac{ a_k }{2}\implies a_k = \frac{a_1}{2^{k-1}}=\frac{\pi}{2^{k+1}}}$$
*Therefore, we have
$$\color{red}{T_{k} = \cos\left(\frac{\pi}{2^{k+1}}\right)}$$
6.By double angle formula we have $$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=4\sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{2}\right)\\=\dots=2^{n}\sin\left(\frac{x}{2^{n}}\right)\prod_{k\leq n}\cos\left(\frac{x}{2^{k}}\right)$$ now remains to note that $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)= \lim_{n\rightarrow\infty} x\frac{\sin\left(\frac{x}{2^{n}}\right)}{\frac{x}{2^{n}}}=\lim_{h\rightarrow 0} x\frac{\sin h}{h}= x .$$
*Thus $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\pi}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$
| {
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Does the constant 1 drop out in this very simple algebraic simplification I was integrating $\int \frac{x^3}{x^2+1}\, dx$, and my answer was
\begin{align}
&\frac{x^2+1}{2}- \frac{\ln(x^2+1)}{2} + C\\&
\text{ Rewrite/Simplify:}\\&=\frac{x^2-\ln(x^2+1)}{2} + C
\end{align}
How exactly does the top simplify into the bottom? Shouldn't it be $x^2+1-ln(x^2+1)$? Why is there no $+1$?
| Let $D= C+\frac12$ which is a constant.
\begin{align}
\frac{x^2+1}{2}- \frac{\ln(x^2+1)}{2} + C=\frac{x^2-\ln(x^2+1)}{2} + D
\end{align}
Here, they just reuse the notation and write $D$ as $C$. It is just an arbitrary constant.
| {
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Finding the maximum value without using derivatives
Find the maximum value of $$f(x)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$$ without using derivatives.
The domain of $f(x)$ is $x \in [1,\infty)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D(f)$ and the maximum value is $f(1)= 2 - \sqrt{2}$. However, this uses derivatives.
| Note that
\begin{align}
f(x) & = (\sqrt{x}-\sqrt{x-1})-(\sqrt{x+1}-\sqrt{x})\\
& = \frac{1}{\sqrt{x}+\sqrt{x-1}} - \frac{1}{\sqrt{x+1}+\sqrt{x}}\\
& = \frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})} \\
& = \frac{2}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}+\sqrt{x-1})}
\end{align}
which is a decreasing function of $x$.
| {
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How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by 11 How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by $11$
I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is $5! = 120$.
| Let the base-$10$ number $abcde$ be divisible by $11,$ where $\{a,b,c,d,e\}=\{2,3,4,5,6\}.$ The base-$10$ number $abcde$ is divisible by $11$ iff $11$ divides $(a+c+e)-(b+d).$
We cannot have $a+c+e\geq 11 +b+d$ because $a+c+e\leq 4+5+6=15$ and $11+b+d\geq 11+2+3=16.$
We cannot have $b+d\geq 11+a+c+e$ because $b+d\leq 5+6=11$ and $11+a+c+e\geq 11+2+3+4=20.$
So the only multiple of $11$ that can be equal to $(a+c+e)-(b+d)$ is $0.$ So $a+c+e=b+d.$ Hence $2(b+d)=(a+c+e)+(b+d)=20.$
So $b+d=10.$
So $6\in \{b,d\}.$ Otherwise $b+d\leq 5+4=9$, but $b+d=10$. Since $6\in \{b,d\}$ and $b+d=10$ we have $\{b,d\}=\{6,4\}.$ Hence $\{a,c,e\}=\{2,3,5\}.$
There are $3!$ permutations of $(2,3,5)$ and $2!$ permutations of $(4,6)$ so there are $3!\cdot 2!=12$ ways.
| {
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Conic section(parabola and ellipse) An ellipse and a parabola have a common focus $S$ and intersect in two real points $P$ and $Q$, of which $P$ is the vertex of the parabola. If $e$ be the eccentricity of the ellipse and $x$, the angle which $SP$ makes with the major axis, prove that
$$\frac{SQ}{SP}=1+\frac{4e^2\sin^2 x}{(1-e\cos x)^2}$$
| Let the parabola be $r=\dfrac{a}{1+\cos (\theta-x)}$, then
$$\theta=x \implies r=\frac{a}{2}=SP$$
Now the the ellipse is
$$r=\frac{a(1+e\cos x)}{2(1+e\cos \theta)}$$
Take $Q$ as $\theta=y$,
$$SQ=\frac{a(1+e\cos x)}{2(1+e\cos y)}=\frac{a}{1+\cos (y-x)}$$
Let $(X,Y)=(\cos x,\cos y)$, then
$$\frac{1+eX}{1+eY} = \frac{2}{1+XY \pm \sqrt{1-X^2}\sqrt{1-Y^2}}$$
Using Mathematica,
\begin{align}
\frac{1+eX}{1+eY} &= \frac{1+4e^2-2eX-3e^2X^2}{(1-eX)^2} \\
&= \frac{(1-eX)^2+4e^2(1-X^2)}{(1-eX)^2} \\
\frac{SQ}{SP} &= 1+\frac{4e^2\sin^2 x}{(1-e\cos x)^2}
\end{align}
| {
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Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.
| we have $$x=\frac{9}{6-y}$$ and $$z=\frac{9}{6-x}$$ putting things together we have
$$6-\frac{9}{y}=\frac{9}{6-\frac{9}{6-y}}$$
simplifying we get
$$9\,{\frac { \left( y-3 \right) ^{2}}{y \left( -9+2\,y \right) }}=0$$
| {
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How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$
| Ok, so building off of what lab bhattacharjee said:
$$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
$$=-\int\sqrt{1-(x-1)^2}dx-\int\dfrac{2-2x}{\sqrt{2x-x^2}}dx+2\int\dfrac1{\sqrt{1-(x-1)^2}}dx$$
Ok, so I use #8 on the 1st integral, u-substitution on the 2nd, and #1 on the 3rd.
$$=-(\frac{(x-1)\sqrt{1-(x-1)^2}}{2}+\frac{1}{2}\arcsin(x-1))-2\sqrt{2x-x^2}+2\arcsin(x-1)+C$$
Simplify.
$$=-\frac{(x-1)\sqrt{1-(x-1)^2}}{2}-2\sqrt{2x-x^2}+\frac{3}{2}\arcsin(x-1)+C$$
| {
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Sequence that converges to a square root Let there be any two numbers $x$ and $y$, where $x,y \in \mathbb R$ and $y \neq -1$.
Consider a sequence where $$a_1 = \frac{y+x}{y+1}$$ and $$a_{n} = \frac{a_{n-1}+x}{a_{n-1} + 1}.$$
For example, when $x=5$ and $y=50$: $$a_1 = \frac{50+5}{50+1} = \frac{55}{51},$$$$a_{2} = \frac{\frac{55}{51}+5}{\frac{55}{51}+1} = \frac{310}{106},$$$$a_3 = \frac{\frac{310}{106}+5}{\frac{310}{106}+1} = \frac{840}{416},$$$$a_4 = \frac{\frac{840}{416}+5}{\frac{840}{416}+1} = \frac{2920}{1256},$$
et cetera. The result approaches very quickly $\sqrt 5$.
My question is: does this sequence always converge to $\sqrt x$?
| Write the recurrence relation as:
$$a_{n}=1+\frac{x-1}{1+a_{n-1}}=1+\cfrac{x-1}{2+\cfrac{x-1}{2+\cfrac{x-1}{1+a_{n-3}}}}$$
Writing the 'infinite' representation (formally so far) for the limit we have:
$$a_{ \infty} =1+\cfrac{x-1}{2+\cfrac{x-1}{2+\cfrac{x-1}{2+\dots}}}$$
For $x=2$ we have:
$$a_{ \infty} =1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\dots}}}=\sqrt{2}$$
This is a simple continued fraction, which is known to converge for sure if the partial denominators $b_k>0$ and their value doesn't decrease faster than $1/k$. In this case they are constant, so it converges.
The limit of $\sqrt{2}$ is well known too, and can be shown by the same method wilkersmon suggested in the comments.
For a general $x>1$ we also have:
$$\sqrt{x} =1+\cfrac{x-1}{2+\cfrac{x-1}{2+\cfrac{x-1}{2+\dots}}}$$
The convergence can be shown by transforming the continued fraction into the form of a 'simple' continued fraction:
$$\sqrt{x} =1+\cfrac{1}{\frac{2}{x-1}+\cfrac{1}{2+\cfrac{1}{\frac{2}{x-1}\cfrac{1}{2+\cfrac{1}{\frac{2}{x-1}+\dots}}}}}$$
Since the partial denominators do not decrease on average, the CF converges.
For $x < 1$ the case for convergence is a little more complicated (because the partial denominators become negative), but we can just consider $1/\sqrt{x}$ instead.
This continued fraction for square roots is referenced at Wikipedia https://en.wikipedia.org/wiki/Continued_fraction.
| {
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Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$ Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that:
$$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$
| By C-S and Schur we obtain:
$$\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}\frac{a^2}{ab^2+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2b+a^2c)}\geq\frac{3(a+b+c)}{\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{a+b+c}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{1}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}.$$
Now, let $a^2+b^2+c^2=x(ab+ac+bc).$
Hence, $x\geq1$ and we need to prove that
$$4(a+b+c)^2(a^2+b^2+c^2)\geq3\left(\sum_{cyc}(a^2+ab)\right)^2$$ or
$$4(x+2)x\geq3(x+1)^2$$ or
$$(x-1)(x+3)\geq0,$$ which is obvious.
| {
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What is the value of $\delta $ if $\epsilon=0.01$?
Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2}, & \text{if
$(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$
Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0)
\vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-
1.$\delta <0.01$
2.$\delta <0.001$
3.$\delta <0.02$
4.no such $\delta $ exists
solution:since $f(0,0)=0$,then consider $\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{xy(2x+3y)}{x^2+y^2}\right\vert\le \frac{(2x+3y)}{2}$ as $xy\le \frac{x^2+y^2}{2}$
From here, how to proceed further...
| Note that we have
$$\frac12|2x+3y|\le |x|+\frac32|y|\le \frac52\sqrt{x^2+y^2}<0.01$$
whenever $\sqrt{x^2+y^2}<\delta=0.004$
| {
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maximum value of $a^2b$ condition is given the maximum value of $a^2b$ subjected to the condition $a+b+\sqrt{2a^2+2ab+3b^2}=10$ given $a,b\geq 0$
solution i try
$\displaystyle \frac{a}{2}+\frac{a}{2}+b\geq \left(\frac{a^2}{4}b\right)^{\frac{1}{3}}$
for $2a^2+2ab+3b^2=2(a+b)^2-ab+b^2$
$a+b+\sqrt{2a^2+2ab+3b^2}=(a+b)+\sqrt{(a+b)^2-ab+b^2}\geq \left(\frac{a^2}{4}b\right)^{\frac{1}{3}}$
How do i solve it from here
| The condition gives
$$2a^2+2ab+3b^2=(10-a-b)^2$$ or
$$a^2+2b^2+20(a+b)=100$$ or
$$(a+10)^2+2(b+5)^2=250$$ or
$$2\left(\frac{a}{2}+5\right)^2+(b+5)^2=125.$$
Now, by AM-GM and Holder we obtain:
$$125=2\left(\frac{a}{2}+5\right)^2+(b+5)^2\geq3\sqrt[3]{\left(\frac{a}{2}+5\right)^4(b+5)^2}=$$
$$=3\left(\sqrt[3]{\left(\frac{a}{2}+5\right)^2(b+5)}\right)^2\geq3\left(\sqrt[3]{\left(\sqrt[3]{\frac{a^2b}{4}}+5\right)^3}\right)^2=3\left(\sqrt[3]{\frac{a^2b}{4}}+5\right)^2.$$
Id est, $$a^2b\leq500\left(\sqrt{\frac{5}{3}}-1\right)^3.$$
The equality occurs for $b=\frac{a}{2}$, which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.
Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.
Let the image be $(a,b,c)$.
Equation of the line joining $(1,3,4)$ and $(a,b,c)$ is $(a-1,b-3,c-4)$ and it will be parallel to the plane given.
Now we have $\dfrac{a-1}{2}=\dfrac{b-3}{-1}=\dfrac{c-4}{1}=k\implies a=2k+1,b=-k+3,c=k+4$.
Since $(a,b,c)$ lies on the plane so we get $k=-1$.
Hence the point will be $(-1,4,3)$.
But the answer is not coming .Where am I wrong??
| Let $P=a(1,3,4)$ then
$$2a-3a+4a+3=0\implies a=-1$$
thus $$P=(-1,-3,-4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double integral over region bounded by nonparallel lines Problem
I'm having trouble with the following integral:
$\int \int_D \frac{x}{y} dx dy $ where D is the area bounded by $1 \leq 2x+y \leq 5, \quad 4x \leq y \leq 8x$
My attempt at a solution
Substitution:
$ u = 2x + y, \quad v= y \ \implies x = \frac{1}{2}(u-v)$
Jacobian:
$ \begin{vmatrix} \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 \end{vmatrix} = \frac{1}{2}$
Limits for u:
$u: 1\to 5$
Limits for y:
$4x \leq y \implies 2(u-v) \leq v \implies \frac{2}{3}u \leq v \\
y \leq 8x \implies v \leq 4(u-v) \implies v \leq \frac{4}{5}u$
$v: \frac{2}{3}u \to \frac{4}{5}u$
Everything put together:
$\int \int_D \frac{x}{y} dx dy = \int \int_D \frac{\frac{1}{2}(u-v)}{v} \cdot \frac{1}{2} dudv = \int_1^5 \int_{\frac{2}{3}u}^{\frac{4}{5}u} \Big(\frac{u}{4v}-\frac{1}{4} \Big)dv du$
I put this into wolframs double integral calculator and get:
$3 \log(\frac{6}{5})-\frac{2}{5}$ which is wrong. Am I doing something wrong?
| Hint. Integrate w.r.t. $y$ immediately:
$$\int_{4x}^{8x} \frac{dy}{y}=\log 2$$
Now the only thing you have left is finding the limits for $x$ from the two inequalities and integrating:
$$\log 2\int_{x_1}^{x_2} xdx$$
Note that the first integration is equivalent to a change of variable:
$$y=x t$$
$$\int_{4x}^{8x} \frac{dy}{y}=\int_{4}^{8} \frac{dt}{t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$ Find period and Fourier expansion Given the following function:
$$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$$
Find period and Fourier expansion.
I think the period is $T=4 \pi$, observing the functions.
As for the Fourier expansion I have no clue on how to proceed. I know that
$$f(t)= \pi a_0 + \sum_{k=1}^\infty a_k \cos(\frac{kt}{2})$$
$$a_k=\frac{1}{2\pi}\int_{-2\pi}^{2\pi} dt f(t)\cos(kt/2)=\frac{1}{2\pi}\int_{-2\pi}^{2\pi} dt \frac{\cos(\frac{3}{2}t) \cos(kt/2)}{1+\cos^2(\frac{t}{4})}=\frac{1}{\pi}\int_{0}^{2\pi} dt \frac{\cos(\frac{3}{2}t) \cos(kt/2)}{1+\cos^2(\frac{t}{4})}$$
Even with the residue theorem the integral looks really bad since there is that $k \in \mathbb{Z}$... Do I really have to compute this thing?
| I will still show the evaluation of the integrals, since it's not always possible to use tricks, sometimes we need the general methods as well. (For what it's worth, in this case the integrals are very simple).
Edited because the period of the function is $4 \pi$ the correct expression should be:
$$a_k=\frac{1}{\pi}\int_{0}^{2\pi} \frac{\cos(\frac{3}{2}t) \cos \left( \frac{kt}{2} \right)}{1+\cos^2(\frac{t}{4})}dt=\frac{4}{\pi }\int_{0}^{\pi} \frac{\cos(3x) \cos(kx)}{3+\cos x}dx=$$
Now the residue theorem is the most convenient way here, in my opinion, so we transform the integral the following way:
$$=\frac{2}{\pi } \Re \int_{-\pi}^{\pi} \frac{\cos(3x) e^{ikx}}{3+\cos x}dx=\frac{2}{\pi } \Re \int_{-\pi}^{\pi} \frac{\left(e^{3ix}+e^{-3ix} \right) e^{ikx}}{6+e^{ix}+e^{-ix}}dx=$$
Setting $z=e^{ix}$ and choosing the unit circle as the contour, we have:
$$=\frac{2}{\pi } \Re \left( -i \oint \frac{\left(z^3+z^{-3} \right) z^{k-1}}{6+z+z^{-1}}dz \right)=\frac{2}{\pi } \Re \left( -i \oint \frac{\left(z^6+1 \right) z^{k-3}}{6z+z^2+1}dz \right)=$$
We assume $k>2$, the cases $k=0,1,2$ can be dealt with separately (there will be an additional pole at $z=0$ of order $1,2,3$). For other cases, we need to consider only the poles coming from the denominator:
$$z^2+6z+1=0$$
$$z= \pm 2 \sqrt{2}-3$$
Of the roots, only one lies inside the unit circle ($z= 2 \sqrt{2}-3$), so we only need to compute one residue:
$$=\frac{2}{\pi } \Re \left( 2 \pi \frac{\left((2 \sqrt{2}-3)^6+1 \right) (2 \sqrt{2}-3)^{k-3}}{2 \sqrt{2}-3+2 \sqrt{2}+3} \right)=\frac{(2 \sqrt{2}-3)^{k-3}+(2 \sqrt{2}-3)^{k+3}}{\sqrt{2}} $$
So, finally, for $k>2$, $k \in \mathbb{Z}$ we have:
$$\frac{1}{\pi}\int_{0}^{2\pi} \frac{\cos(\frac{3}{2}t) \cos \left( \frac{kt}{2} \right)}{1+\cos^2(\frac{t}{4})}dt=\frac{(2 \sqrt{2}-3)^{k-3}+(2 \sqrt{2}-3)^{k+3}}{\sqrt{2}} $$
Note that:
$$2 \sqrt{2}-3=\frac{1-\sqrt{2}}{1+\sqrt{2}}=-\frac{1}{(1+\sqrt{2})^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebra Smallest Possible Value What is the smallest possible value of
$x^2+y^2-x-y-xy$?
Is this even possible to solve? Please help.
| $$2(x^2+y^2-x-y-xy)=(x-y)^2+(x-1)^2+(y-1)^2-1-1\ge-2$$
Alternatively, let $$x^2+y^2-x-y-xy=k\iff x^2-x(1+y)+y^2-y-k=0$$
As $x$ is real, the discriminant must be $\ge0$
$$\implies(1+y)^2\ge4(y^2-y-k)$$
$$\iff4k\ge3y^2-6y-1=3(y-1)^2-3-1\ge-3-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove determinant is negative Prove that the determinant $\Delta$ is negative
$$
\Delta=\begin{vmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{vmatrix}<0
$$
where $a,b,c$ are positive and $a\neq b\neq c$.
My Attempt:
Applying Sarrus' rule,
$$
\begin{matrix}
a&b&c&a&b\\
b&c&a&b&c\\
c&a&b&c&a
\end{matrix}
$$
$$
\Delta=acb+bac+cba-c^3-a^3-b^3=3abc-(a^3+b^3+c^3)\\
=-\Big[(a^3+b^3+c^3)-3abc\Big]
$$
How do I prove that $(a^3+b^3+c^3)-3abc>0$ thus prove $\Delta<0$ ?
| You can factor $$a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca),$$ where the first term is certainly positive. For the second term, observe $$a^2 + b^2 + c^2 - ab-bc-ca = \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2),$$ which is also positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inverse of a $2\times2$ matrix, Gaussian elimination with unknown $x$ $C=AB$ I need to find A so $CB^{-1}=A$ working in modulo 10
and C=$ \begin{bmatrix}2 & 5 \\3 & 1 \end{bmatrix}$ mod 10
with B=$ \begin{bmatrix}3 & 4 \\2 & x \end{bmatrix}$ mod10
calculating $B^{-1}$
I wrote$ \begin{bmatrix}3 & 4 \\2 & x \end{bmatrix}$ $\begin{bmatrix}1&0 \\0 & 1 \\ \end{bmatrix}$
calculating the inverse of $3 (\mod10)$ and =$7$ and multiplying the first row
gives $ \begin{bmatrix}1 & 8 \\2 & x \end{bmatrix}$ $\begin{bmatrix}7&0 \\0 & 1 \\ \end{bmatrix}$
then R2-2R1
$ \begin{bmatrix}1 & 8 \\0 & x-16 \end{bmatrix}$ $\begin{bmatrix}7&0 \\6 & 1 \\ \end{bmatrix}$
but im not sure where to go from this, because I dont know how to calculate the inverse of $x-16(\mod10)$. I guess you could calculate the inverse of $x(\mod10)$ and multiply it through from the start, how would I go about finding A, namely $B^{-1}$
| To find $B^{-1}$, you can start by analogy with inverses over $\mathbb{Q}$: $$B^{-1}=\frac{1}{3x-8} \begin{pmatrix} x&-4\\-2&3\end{pmatrix}$$
Note that if $x$ is even, then $det(B)$ is even, and therefore $B$ has no inverse modulo $10$. Further, if $x=1$, then $det(B)=5$, and therefore $B$ has no inverse modulo $10$. This leaves $x=3,5,-3,-1$.
For $x=5$, $det(B)=7$, and $7^{-1}\equiv 3\pmod{10}$. Hence $$B^{-1}=3\begin{pmatrix}7&-4\\-2&3\end{pmatrix}=\begin{pmatrix}1&-2\\4&-1\end{pmatrix}$$
The other three cases are solved similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\det (A+B)=\det (A-B)$, prove that $B^{-1}$ exists iff $b_{11}\neq b_{21}$
Let $A=\begin{pmatrix}
0 & 1 & 2 \\
0 & 1 & 2 \\
0 & 2 & 3
\end{pmatrix}\in \mathbb{R}^{3\times 3}$ and let $B=\left [b_{ij}\right ]\in \mathbb{R}^{3\times 3}$ such that $\det \left (A+B\right )=\det \left (A-B\right )$. Prove that $B$ is invertible if and only if $b_{11}\neq b_{21}$.
After simplifying by using properties of the determinant, the condition $\det \left (A+B\right )=\det \left (A-B\right )$ becomes $\det \begin{pmatrix}
b_{11} & b_{12} & 2 \\
b_{21} & b_{22} & 2 \\
b_{31} & b_{23} & 3
\end{pmatrix}=-\det \begin{pmatrix}
b_{11} & 1 & b_{13} \\
b_{21} & 1 & b_{23} \\
b_{31} & 2 & b_{33}
\end{pmatrix}$, providing that I did not commit any mistake. This does not seem concluding and I don't think it's the right approach to the solution. Any help?
| $$det(A+B) = det(B) +
det
\begin{pmatrix}
b_{11} & 1 & b_{13}\\
b_{21} & 1 & b_{23}\\
b_{31} & 1 & b_{33}
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & b_{12} & 2\\
b_{21} & b_{22} & 2\\
b_{31} & b_{32} & 3
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} & 1 & 2\\
b_{31} & 2 & 3
\end{pmatrix}
$$
$$det(A-B) = -det(B) +
det
\begin{pmatrix}
b_{11} & 1 & b_{13}\\
b_{21} & 1 & b_{23}\\
b_{31} & 1 & b_{33}
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & b_{12} & 2\\
b_{21} & b_{22} & 2\\
b_{31} & b_{32} & 3
\end{pmatrix}
-
det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} & 1 & 2\\
b_{31} & 2 & 3
\end{pmatrix}
$$
so you get
$$
det(B) = -det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} & 1 & 2\\
b_{31} & 2 & 3
\end{pmatrix} = b_{11} - b_{21}
$$
and it is zero if and only if $b_{11}=b_{21}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_x^{px}\frac{\sin^2 t}{t^2}\,\mathrm dt\leqslant\fracπ2\left(1-\frac1p\right)$ for $x\geqslant0,\ p>1$
Given $x\geqslant 0$ and $p>1$, define $$I (p,x) = \int_x^{px}\frac{\sin^2 t}{t^2} \,\mathrm{d}t.$$ Prove that
$$ I (p,x) \leqslant \frac{\pi}2\left(1-\frac1p\right).$$
So far I only figured out that $I(p,x)$ as a function of $x$, it increases in interval $[0, x_p]$, where $x_p=\dfrac{1}{p^2+p+1}$.
But then I'm lost on how to prove $I(p,x_p) \leqslant \dfrac{\pi}2\left(1-\dfrac1p\right)$…
| $\def\d{\mathrm{d}}$It will be proved that$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) \tag{1}
$$
for any $0 < a < b$.
Case 1: $a \geqslant \dfrac{2}{π}$. Then$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^b \frac{1}{x^2} \,\d x = \frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab}.
$$
Because$$
\frac{b - a}{ab} \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) = \frac{π}{2} \cdot \frac{b - a}{b} \Longleftrightarrow a \geqslant \frac{2}{π},
$$
then (1) holds.
Case 2: $a < \dfrac{2}{π}$ and $b \leqslant \dfrac{π}{2} \cdot \dfrac{a^2}{\sin^2 a}$.
Lemma: $\dfrac{\sin^2 x}{x^2} \leqslant \dfrac{\sin^2 a}{a^2}$ for any $x \geqslant a$.
Proof of lemma: Define $f(x) = \dfrac{\sin x}{x}$, then for any $a < x < \dfrac{π}{2}$,$$
f'(x) = \frac{1}{x^2} (x \cos x - \sin x) = \frac{\cos x}{x^2} (x - \tan x) < 0.
$$
Thus for $a < x < \dfrac{π}{2}$,$$
f(a) \geqslant f(x) > 0 \Longrightarrow \frac{\sin^2 x}{x^2} \leqslant \frac{\sin^2 a}{a^2}.
$$
For $x \geqslant \dfrac{π}{2}$, because $a < \dfrac{2}{π} < \dfrac{π}{2}$, then$$
\frac{\sin^2 a}{a^2} = (f(a))^2 \geqslant \left(f\left(\frac{π}{2}\right)\right)^2 = \frac{4}{π^2} \geqslant \frac{1}{x^2} \geqslant \frac{\sin^2 x}{x^2}.
$$
Now back to Case 2. By the lemma,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^b \frac{\sin^2 a}{a^2} \,\d x = \frac{\sin^2 a}{a^2} (b - a).
$$
Because$$
\frac{\sin^2 a}{a^2} (b - a) \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) = \frac{π}{2} \cdot \frac{b - a}{b} \Longleftrightarrow b \leqslant \frac{π}{2} \cdot \frac{a^2}{\sin^2 a},
$$
then (1) holds.
Case 3: $a < \dfrac{2}{π}$ and $b > \dfrac{π}{2} \cdot \dfrac{a^2}{\sin^2 a}$. Now,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^{+\infty} \frac{\sin^2 x}{x^2} \,\d x = \frac{π}{2} - \int_0^a \frac{\sin^2 x}{x^2} \,\d x,
$$
and$$
\frac{π}{2} \left( 1 - \frac{a}{b} \right) \geqslant \frac{π}{2} \left( 1 - a \cdot \left(\frac{π}{2} \cdot \frac{a^2}{\sin^2 a}\right)^{-1} \right) = \frac{π}{2} - \frac{\sin^2 a}{a},
$$
thus it suffices to prove that$$
g(a) = \int_0^a \frac{\sin^2 x}{x^2} \,\d x - \frac{\sin^2 a}{a} \geqslant 0
$$
for $a < \dfrac{2}{π}$.
Because for $0 < a < \dfrac{2}{π}$,\begin{align*}
g'(a) &= \frac{\sin^2 a}{a^2} - \frac{1}{a^2} (2a \sin a \cos a - \sin^2 a)\\
&= \frac{1}{a^2} (2 \sin^2 a - 2a \sin a \cos a)\\
&= \frac{\sin 2a}{a^2} (\tan a - a) > 0,
\end{align*}
then for any $0 < a_0 < \dfrac{2}{π}$,$$
g(a_0) \geqslant \lim_{a \to 0^+} g(a) = 0 - \lim_{a \to 0^+} \frac{\sin^2 a}{a} = 0.
$$
Thus,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} - \int_0^a \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} - \frac{\sin^2 a}{a} \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right),
$$
i.e. (1) holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Third Moment of Geometric Series? $\sum_{x=1}^{\infty}x(1-p)^{x-1} = \frac{1}{p^2}$
$\sum_{x=1}^{\infty}x^2(1-p)^{x-1} = \frac{2-p}{p^3}$
$\sum_{x=1}^{\infty}x^3(1-p)^{x-1} =$ ... ?
The textbook gives the first 2 moments, but not the third one. I could not find the forumla on Google as well. I think it'll be useful in case I need to find $E(X^3)$ where $X$ is a geometric distribution.
| First, note that
$$\sum_{k=0}^\infty k(1-p)^{k-1} = \sum_{k=1}^\infty k(1-p)^{k-1}$$
This allows us to perform a little trick to calculate the moments. For the first moment, we have:
\begin{align}
\sum_{k=1}^\infty k(1-p)^{k-1} &= \sum_{k=0}^\infty k(1-p)^{k-1}\\
&=\sum_{k=1}^\infty (k-1)(1-p)^{k-2} \\
&=\sum_{k=1}^\infty k(1-p)^{k-2} - \sum_{k=1}^\infty(1-p)^{k-2}\\
&= \frac{1}{1-p}\sum_{k=1}^\infty k(1-p)^{k-1}-\frac{1}{p(1-p)}
\end{align}
Rearranging, we obtain:
$$-\frac{p}{1-p}\sum_{k=1}^\infty k(1-p)^{k-1} = -\frac{1}{p(1-p)}$$
$$\sum_{k=1}^\infty k(1-p)^{k-1}=\frac{1}{p^2}$$
We can use the same trick to calculate the third moment:
\begin{align}
\sum_{k=1}^\infty k^3(1-p)^{k-1} &= \sum_{k=0}^\infty k^3(1-p)^{k-1}\\
&= \sum_{k=1}^\infty (k-1)^3(1-p)^{k-2}\\
&= \sum_{k=1}^\infty k^3 (1-p)^{k-2} - 3\sum_{k=1}^\infty k^2(1-p)^{k-2} + 3\sum_{k=1}^\infty k(1-p)^{k-2} - \sum_{k=1}^\infty(1-p)^{k-2}\\
&=\frac{1}{1-p}\sum_{k=1}^\infty k^3(1-p)^{k-1} - \frac{3(2-p)}{p^3(1-p)}+\frac{3}{p^2(1-p)}-\frac{1}{p(1-p)}
\end{align}
Rearranging, we have:
$$-\frac{p}{1-p}\sum_{k=1}^\infty k^3(1-p)^{k-1} = -\frac{p^2-6p+6}{p^3(1-p)}$$
$$\boxed{\sum_{k=1}^\infty k^3(1-p)^{k-1} = \frac{p^2-6p+6}{p^4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$
$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$
$y\geq 0.5$
Could some help me where I have wrong, thanks
| Let
$$g(t)=\frac{t^2+4t+5}{2t^2+8t+8}=\frac{1}{2}+\frac{1}{2t^2+8t+8},$$
then
$$
g'(t)=-\frac{2(2t^2+8t+8)(4t+8)}{(2t^2+8t+8)^2}=-\frac{1}{(t+2)^3}.
$$
Thus $g$ strictly decreases in the interval $[-1,1]$, and $g(-1)$ and $g(1)$ are maximum and minimum respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Evaluate $ \int_{0}^{\pi/2} \frac{\sin(nx)}{\sin(x)}\,dx $ For every $odd$ $n \geq 1$ the answer should be $\pi/2$
For every $even$ $n \geq 2$ the possible answers are :
$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $
$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$C )$ $ 2(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$D )$ $ \frac{1}{2}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n+1}\frac{1}{n-1} )$
$E )$ $ 1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\cdots - \frac{1}{n-1} $
Does the recurrence $ (n-1)(I_{n}-I_{n-2})=2sin(n-1)x $ help?
| first we note that $I_0=0$ and $I_1 = \frac{\pi}2$.
now, using the trig identity
$$
\sin A - \sin B = 2\sin \frac{A-B}2 \cos \frac{A+B}2
$$
we obtain
$$
I_{n+2} = I_n + \frac2{n+1} \sin \frac{(n+1)\pi}2
$$
if $N$ is odd, therefore, we have $I_N =I_{N-2}=\dots = I_1 = \frac{\pi}2$.
for even values of $N$, say $N=2M$, we have
$$
I_{2M}= I_0+2\bigg(\frac11\sin \frac{\pi}2 + \frac13\sin \frac{3\pi}2+\dots +\frac1{2M-1}\sin \frac{(2M-1)\pi}2 \bigg) = 2\sum_{k=1}^{M} \frac{(-1)^{k-1}}{2k-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Infinite series involving binomial and geometric How to find the following sum:
$$\sum_{x=0}^{\infty}\left(-\frac{1}{16}\right)^x \binom{2x}{x}$$
| \begin{eqnarray*}
\binom{2n}{n} = \binom{-1/2}{n}(-4)^{n}.
\end{eqnarray*}
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \left( \frac{-1}{16} \right)^n = \sum_{n=0}^{\infty} \binom{-1/2}{n} \left( \frac{1}{4} \right)^n =\frac{1}{\sqrt{1+\frac{1}{4}}} =\color{red}{\frac{2}{\sqrt{5}}}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why, my result has the opposite sign to the correct result? I found this exercise of factoring:
$$5a^2 + 3a - 2$$ is a trinomial of the form:
$$ax^2 + bx + c = \frac{(ax + p)(ax + q)}{a}$$
I made this:
$$3 = p + q$$
$$-10 = pq$$
then, I need to get the values of $p$ and $q$.
$$p = -\frac{10}q$$
replace:
$$3 = -\frac{10}q + q$$
$$q^2 -3q - 10 = 0$$
$$(q - 5)(q + 2)$$ That is two possible values of $q$
I need to check the correct value now:
$$3 = p - 5\implies p = -8$$ but $8 \times 5 \neq -10$, so the correct value of $q$ is $2$.
Replace to check if is true:
$$3 = p - 2\implies p = 1$$ but $1 \times -2 \neq -10$
In the end, the correct values impressively were $5$ and $2$, which I had obtained in the equation of $q$, but with a changed sign.
$$p = 5, q = -2$$
So, what am I doing wrong factoring this expression?
Is it a property that the result was that? or was it coincidence
PD: the total factoring was: $(a + 1)(5a - 2)$
|
the total factoring was: $\;(a + 1)(5a - 2)$
You could get that directly by finding the roots using the quadratic formula $\,\dfrac{-3 \pm \sqrt{49}}{10}=\begin{cases}-1 \\ 2/5\end{cases}\,$.
Then the quadratic factors as $5\big(a-(-1)\big)\left(a - \dfrac{2}{5}\right)=(a+1)(5a-2)\,$.
[ EDIT ] In the general case $\,ax^2 + bx + c = \dfrac{(ax + p)(ax + q)}{a}\,$ the factorization follows directly from the quadratic formula as well:
$$
\begin{align}
ax^2 + bx + c = a(x-x_1)(x-x_2) &= a\left(x-\frac{-b - \sqrt{\Delta}}{2a}\right)\left(x-\frac{-b + \sqrt{\Delta}}{2a}\right) \\[10px]
&= \frac{\left(ax + \cfrac{b + \sqrt{\Delta}}{2}\right)\left(ax + \cfrac{b - \sqrt{\Delta}}{2}\right)}{a}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Number of Non negative integral solutions of $x+2y+3z+4w=12$ Find Number of positive integral solutions of $x+2y+3z+4w=12$
I tried to split the problem in to cases :
we have $2z+2y+4w$ is always Even and $12$ is Even number.
Hence $x+z$ should be even number which implies
Case $1.$ $x$ is Even and $z$ is even
Letting $x=2p$ and $z=2q$ we get
$p+y+3q+2w=6$
Now Let $q=0$ we have
$$p+y+2w=6$$
$$p+y=6-2w$$
Number of solutions by stars and bars is given by
$$\sum_{w=0}^{3} 7-2w=16$$
Again letting $q=1$ we have to repeat same process.
This method i feel is so lengthy. Any other way?
| Minimum of $x+2y+3z=6\Rightarrow 4w=4$ since $8+6\gt12$ then $w=1$ necessarily.
Minimum of $x+2y+4w=7\Rightarrow 3z=3$ since $7+6\gt12$ then $z=1$ necessarily.
It follows $$x+2y=12-3-4=5$$ The only solutions of this equation are $(x,y)=(1,2),(3,1)$.
Thus there are only two solutions
$$(x,y,z,w)=(1,2,1,1),(3,1,1,1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of Recursive Sequence $y_{n+1}=y_n+\frac{1}{ny_n}$ with $y_1=1$ I came up with this sequence as I was playing around with another question on this site. And so I decided to ask the users if they can find a solution.
Let $$y_1=1$$
$$y_{n+1}=y_n+\frac{1}{ny_n}.$$
Evaluate $$\lim\limits_{n\to \infty} y_n \leq \infty.$$
| The sequence diverges.
Suppose the sequence was bounded by some constant, i.e. $y_n < C$ for all $n$. Then in particular $\frac{1}{y_n} > \frac{1}{C}$ for all $n$ so:
$y_{2}>y_1 + \frac{1}{C}$
$y_{3} > y_2 + \frac{1}{2C} > y_1 + \frac{1}{C} + \frac{1}{2C}$
$y_{4} > y_3 + \frac{1}{3C} > y_1 + \frac{1}{C} + \frac{1}{2C} + \frac{1}
{3C}$
...
$y_{n+1} > y_1 + \frac{1}{C} \sum_{i=1}^n \frac{1}{i}$
But the harmonic series diverges.
EDIT:
I'd like to expand a bit on the exact asymptotics in response to Did's comment:
Using the same trick as above but with the squares:
$y_2^2 = y_1^2+\frac{2}{1}+\frac{1}{y_1^2}$
$y_3^2 = y_2^2+\frac{2}{2}+\frac{1}{4y_2^2} = y_1^2+(\frac{2}{1}+\frac{2}{2}) +(\frac{1}{y_1^2} + \frac{1}{4y_2^2})$
... (i.e. by induction)
$y_{n+1} = y_1^2 + 2(\sum_{i=1}^n\frac{1}{i}) + (\sum_{i=1}^n \frac{1}{i^2y_i^2})$
$2H_n + 1 < y_{n+1}^2 < 2H_n + 1 + \frac{\pi^2}{6}$
Where in the last step we use that $y_n \ge 1$ for all $n$ and the sum of the reciprocals of squares sums to $\frac{\pi^2}{6}$. So indeed $y_n \approx \sqrt{2H_n}$, as suggested.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How i get the $x$ of the equation I have this:
$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}$ = 1
My main problem is that I want to multiply by something on both sides.
I will move $(5-\frac{5}{x})$ to side of 1, then:
$\frac{32}{20-16} = 1(5-\frac{5}{x}$)
$8 = 5 -\frac{5}{x}$ , now multiply by $x$
$8x = 5x - 5 $
$3x = 5$
$x = 5/3$
But according to symbolab it's $x = 15/19$, could you help me what I'm wrong about? I have done a lot of exercises, but I have a hard time clearing the $ x $ when it is very low.
| This step is uncorrect
$$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=1\implies \color{red}{\frac{32}{20-16} = 1(5-\frac{5}{x})}$$
following this way it should be indeed (for $x\neq 1$)
$$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=1\implies \frac{1}{(5-\frac{5}{x})}\cdot\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=\frac{1}{(5-\frac{5}{x})}\cdot 1\implies \frac{32}{20(5-\frac{5}{x})-16} = \frac{1}{(5-\frac{5}{x})}$$
and then
$$\implies 32(5-\frac{5}{x})={20(5-\frac{5}{x})-16}\\\implies160x-160=100x-100 -16x\implies76x=60\implies x=\frac{15}{19}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?
So this track seems to difficult, another approach. Can I square the area first, find the derivative of that to solve for x?
So the Area = $4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$Area^2 = 16x^2 \cdot (4 - \frac{4x^2}{9}$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} Area^2 = 128x - \frac{256x^3}{9}$$
$$128x(1-\frac{2x^2}{9}$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals 0 when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of x into y we get that $y = \sqrt{2}$ so the Area is 3.
Is this valid? If so why? Does squaring not cause any problems?
| Hint...it would be a lot easier if you wrote $$(x,y)=(3\cos\theta,2\sin\theta)$$ then obtain an expression for the area of the rectangle in terms of $\theta$ and differentiate...
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
| To count the number of factors of $2079000 =2^3\cdot3^3\cdot5^3\cdot7\cdot11$, we would compute
$$
(3+1)(3+1)(3+1)(1+1)(1+1)=256
$$
However, we only want to count the number of factors divisible by $30=2\cdot3\cdot5$. Therefore, we want to compute the number of factors of $\frac{2079000}{30}=2^2\cdot3^2\cdot5^2\cdot7\cdot11$, because each of these factors times $30$ is a factor of $2079000$ which is divisible by $30$, and this would be
$$
(2+1)(2+1)(2+1)(1+1)(1+1)=108
$$
I am not sure exactly why your computation of the number of factors is wrong because you did not explain why you computed
$$
2\cdot2\cdot(3+1)\cdot(1+1)\cdot(1+1)
$$
instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Fraction and simplification
solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$
What are the possible answers ?
(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0
The answer from where i've referred this is (B), but when i simplify it I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$$
$$ \frac{x +x(x-1)}{x(x-1)\cdot x} = \frac{1}{x-1} \text{ (took l.c.m on l.h.s)}$$
$$ \frac{x + (x^2 -x)}{(x^2 - x)\cdot x}= \frac{1}{x-1}$$
$$\frac{x^2}{x^3 - x^2} = \frac{1}{x-1}$$
$$ x^2(x-1) = x^3 - x^2$$
$$ x^3 - x^2 = x^3 - x^2$$
Have I simplified it correctly?
| You can simply multiply by the LCD, $x(x-1)$ on both sides of the equation.
$$1+(x-1)=x$$
$$x=x$$
Therefore, there are infinitely many solutions for $x$, where $x\ne0,1$.
If $x=0, 1$, then the denominator(s) of the original expression would equal $0$, so the expression would be undefined.
In interval notation, the solutions can be expressed as $(-\infty,0)\cup(0,1)\cup(1,\infty)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of picking a certain fruit from a tree A tree has 20 fruits. 15 of which have no seeds and the rest do have seeds. A bird eats 5 of these fruits picked at random.
a) If i pick one fruit from whats left on the tree whats the probability it has seeds
b) Given that this one fruit i pick has seeds whats the probability that the bird had consumed at least one with seeds.
What i have so far:
So before the bird eats anything the probability of a fruit with no seeds is 15/20=0.75 and the probability of fruit with seeds is 5/20=0.25.
For part a is the probability that the fruit i pick(i.e. 6th fruit picked) has seeds is given by (15/20)* (14/19)* (13/18)* (12/17) *(11/16) *(5/15)=0.065
Am i on the right track with part a? As for part b im not sure where to begin.
| Since there bird eats only five fruits, it's relatively easy to consider all possible cases. Let's write
*
*$P_0 = P(\text{bird eats zero fruits with seeds}) = \frac{15}{20} \frac{14}{19} \frac{13}{18} \frac{12}{17} \frac{11}{16} = \frac{1001}{15504}$
*$P_1 = P(\text{bird eats exactly one fruit with seeds}) =$ $5\choose 1$$\frac{5}{20} \frac{14}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{5005}{15504}$
*$P_2 = P(\text{bird eats exactly two fruits with seeds}) =$ $5\choose 2$$\frac{5}{20} \frac{4}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{715}{3876}$
*$P_3 = P(\text{bird eats exactly three fruits with seeds}) =$ $5\choose 3$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{12}{17} \frac{11}{16} = \frac{55}{1292}$
*$P_4 = P(\text{bird eats exactly four fruits with seeds}) =$ $5\choose 4$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{11}{16} = \frac{55}{15504}$
*$P_5 = P(\text{bird eats exactly five fruits with seeds}) =$ $5\choose 5$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{1}{16} = \frac{1}{15504}$
So for part a, you can consider the five cases separately.
*
*If the bird ate zero fruits with seeds, there are $5$ out of $15$ fruits with seeds left. So you have to multiply $\frac{1001}{15504}\frac{5}{15}$
*If the bird ate one fruit with seeds, there are $4$ out of $15$ fruits left with seeds. So you have to multiply $\frac{5005}{15504} \frac{4}{15}$
etc ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $B=(x,y)$ so triangle is equilateral Let $O=(0,0)$, $A=(3,4)$ and $B=(x,y)$ be three points in $xOy$. Find real numbers $x$ and $y$, so that $OAB$ is an equilateral triangle.
I'm really struggling with this one, can someone help?
| We need to have $|\overline{OB}|=|\overline{AB}|=|\overline{OA}|=5$. Thus from
$$ |\overline{OB}|^2=x^2+y^2=25$$
and $$|\overline{AB}|^2=(x-3)^2+(y-4)^2=25$$we reach
$$x^2-6x+9+y^2-12y+16=x^2+y^2 $$
or $$6x+12y=25\text{ or }x=2y-25/2.$$
Replace this value to $x^2+y^2=25$:
$$4y^2+50y+625/4+y^2=25$$
or $$y^2+10y-\frac{150}{4}=0 .$$
The solutions of this equation are
$$y=\frac{-1\pm5\sqrt{10}}{2} $$
The rest is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692477",
"timestamp": "2023-03-29T00:00:00",
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Probability of rolling a 2 before a 6 and a 1 in that order
I am rolling a fair die. What is the probability that the first time I see a $2$ is before the first time I see a $6$ , and the first time I see a $6$ is before the first time I see a $1$ ?
So I thought about doing it with an infinite sum:
\begin{align}
\frac{1}{6}\frac{1}{6} + \frac{1}{6}\frac{4}{6}\frac{1}{6} + \frac{1}{6}\big(\frac{4}{6}\big)^2\frac{1}{6} + \dots
\end{align}
where the first term is rolling a $2$ and then rolling a $6$ , the second term is rolling a $2$ , rolling anything but a $1$ or $6$ and then rolling a $6$ , etc. but I don't believe this is correct.
| Your roll will look like this:
$\qquad (\text{1})$: $k$ rolls, each roll being one of $3,4,5$
$\qquad (\text{2})$: a $2$
$\qquad (\text{3})$: $m$ rolls, each roll being one of $2,3,4,5$
$\qquad (\text{4})$: a $6$
$\qquad (\text{5})$: $n$ rolls, each roll being one of $2,3,4,5,6$
$\qquad (\text{6})$: a $1$
For each $k$, $(1)$ has probability $(1/2)^k$.
For each $m$, $(3)$ has probability $(2/3)^m$.
For each $n$, $(5)$ has probability $(5/6)^n$.
Steps $(2)$, $(4)$ and $(6)$ each have probability $1/6$.
All in all, that gives you
\begin{align}
\left[\sum_{k\geq 0} {\left(\frac12\right)}^k\right]
\cdot \frac16\cdot
\left[\sum_{m\geq 0} {\left(\frac23\right)}^m\right]
\cdot \frac16\cdot
\left[\sum_{n\geq 0} {\left(\frac56\right)}^n\right]
\cdot \frac16
&=
\frac{1}{\left(1-\frac12\right)\,\left(1-\frac23\right)\,\left(1-\frac56\right)\,6^3}
\\&=
\frac{1}{\frac12\cdot\frac13\cdot\frac16\cdot 6^3}
\\&=
\frac16
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big(\frac{\frac{2A}2}2\Big)\cos\frac02 =
2\sin\frac{A}2
$$
where $\cos\frac02 = \cos0 = 1$
| First of all $$\sin x\neq \sin (x/2)+\sin (x/2)$$
In fact $$\sin (a+b)=\sin a\cos b+\cos a\sin b$$
Substitute $a=b=x/2$ to get $$\sin x= 2\sin (x/2)\cos (x/2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Formula for the large derivative Is there any formula for the large number derivative?
I need to find $y^{(100)}$ at $x=0$, if $y=(x+1)2^{x+1}$
I tried to find a pattern, but 2nd and 3rd derivatives are already too hairy. I see no pattern, how it evolves.
1st derivative $2^{x+1}+\ln \left(2\right)\cdot \:2^{x+1}\left(x+1\right)$
or $2^{x+1}(1+ln(2)(x+1))$
2nd $\ln ^2\left(2\right)\cdot \:2^{x+1}x+\ln ^2\left(2\right)\cdot \:2^{x+1}+\ln \left(2\right)\cdot \:2^{x+2}$
3rd $\ln ^2\left(2\right)\left(\ln \left(2\right)\cdot \:2^{x+1}x+2^{x+1}\right)+\ln ^3\left(2\right)\cdot \:2^{x+1}+\ln ^2\left(2\right)\cdot \:2^{x+2}$
| Let $u=x+1$ and $v=2^{x+1}$.
$u^{(1)}=1$, $u^{(k)}=0$ for $k\ge 2$.
$v^{(k)}=2^{x+1}(\ln2)^k$ for $k\ge 1$.
By the general Leibniz rule, (https://en.wikipedia.org/wiki/General_Leibniz_rule)
$$y^{(100)}=uv^{(100)}+\binom{100}{1}u^{(1)}v^{(99)}=(x+1)2^{x+1}(\ln 2)^{100}+100\cdot2^{x+1}(\ln 2)^{99}$$
$$y^{(100)}(0)=2(\ln 2)^{100}+200(\ln 2)^{99}$$
| {
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If $P_n$ is $n$th Legendre polynomial and $|x|\leqslant1$ then $\frac{1-x^2}{n^2}(P'_n)^2+P^{2}_{n} \le 1$ I am trying to prove theorem 17 (h) of chapter 7 from Kaplan's Advanced Calculus book. Being $P_{n}$ a legendre polynomial and $P^{\prime}_{n}$ its derivative, I need to prove:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1 , |x| \le 1)$$
I already proved that (see scholia):
$\tag{1}\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$
So I tried to do a "recursive prove":
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le \frac{1-x^2}{(n-1)^2}P^{\prime 2}_{n-1}+P^{2}_{n-1} \le \frac{1-x^2}{(n-2)^2}P^{\prime 2}_{n-2}+P^{2}_{n-2} \le ... \le (1-x^2)P^{\prime 2}_{1}+P^{2}_{1}$$
And as:
$$(1-x^2)P^{\prime 2}_{1}+P^{2}_{1} = (1-x^2) +x^2 = 1$$
It seems to me that:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1)$$
Should be valid for any value of $x$. Not only for $|x| \le 1$. I am afraid I am missing some part of the whole history here. Can anyone help me?
Scholia:
As some people doubted (1) I will show the prove for it.
As cited in Kaplan's book the Legendre polynomials have these two properties:
$\tag{2}P^{\prime}_{n}(x) = xP^{\prime}_{n-1}(x)+nP_{n-1}(x)$
And
$\tag{3}P_{n}(x) = \frac{(x^{2}-1)}{n}P^{\prime}_{n-1}(x)+xP_{n-1}(x)$
So we can substitute (2) and (3) on the LHS of (1) and get:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{(1-x^2)}{n^2}(x^{2}P^{\prime 2}_{n-1} + 2xnP^{\prime}_{n-1}P_{n-1} + n^{2}P^{2}_{n-1}) + \frac{(x^{2}-1)^2}{n^2}P^{\prime 2}_{n-1}+ 2x\frac{(x^{2}-1)}{n}P^{\prime}_{n-1}P_{n-1}+x^{2}P^{2}_{n-1}$$
$$\frac{P^{\prime 2}_{n-1}}{n^2}(-x^{4}+x^{2}+x^{4}-2x^{2}+1)+P^{2}_{n-1}(1-x^{2}+x^{2}) = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$$
And we get (1). The properties (2) and (3) I used in order to demonstrate (1) can be demonstrated using Rodrigues formula and Leibnitz product rule for $n$th order derivatives. I can provide the step-by-step if needed.
| Given the identity
$$\begin{eqnarray*} \frac{1-x^2}{n^2}P_n'^2+P_n^2 &=& \frac{1-x^2}{n^2}P_{n-1}'^2+P_{n-1}^2\\&=&\left(1-\frac{1}{n}\right)^2\left(\frac{1-x^2}{(n-1)^2}P_{n-1}'^2+P_{n-1}^2\right)+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\\\\&\stackrel{\text{Ind.Hyp.}}{\leq}&\left(1-\frac{1}{n}\right)^2+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\end{eqnarray*}$$
($\text{Ind.Hyp}$ stands for Inductive Hypothesis) the claim is straightforward, since $P_{n-1}(x)^2$ attains its maximum value over $[-1,1]$ at the endpoints of such interval, and such maximum value is $1$.
The inequality can be refined for $x$-points belonging to a neighbourhood of the origin, since as shown here we have
$$ P_n(x)^2\leq \frac{4}{\pi(2n+1)}\cdot\frac{1}{\sqrt{1-x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$ I'm stuck here:
Prove that $w=\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$
By using the definition of $z=\sin(w)$, I found that
$$\arcsin(z)=\frac{1}{i}\log(iz+\sqrt{1-z^2})$$
But where that $\frac{\pi}{2}$ comes from? How can I rearrange my expression to have $\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$?
Thanks for your time.
| In order to get to the final expression you need to use the log identity:
$$
\ln(a+b)=\ln(a)+\ln(1+\frac{b}{a})
$$
So in your case:
$$
\frac{1}{i}\left[\ln(iz+\sqrt{1-z^2})\right] = \frac{1}{i}\left[\ln(iz) + \ln\left(1+\frac{\sqrt{1-z^2}}{iz}\right)\right]
$$
We know that when $z>0$,
$$
\ln(iz)=\frac{i\pi}{2}+\ln(z)
$$
Also, by doing some algebra:
$
\ln\left(1+\frac{\sqrt{1-z^2}}{iz}\right) = \ln\left(1+\frac{i\sqrt{1-z^2}}{(-1)z}\right) = \ln\left(\frac{z}{z}-\frac{i\sqrt{1-z^2}}{z}\right)
$
$
= \ln\left(\frac{z-i\sqrt{1-z^2}}{z}\right)=\ln\left(z-i\sqrt{1-z^2}\right) - \ln(z)
$
Putting it all together:
$
\frac{1}{i}\left[\ln(z)+\frac{i\pi}{2}+\ln(z-i\sqrt{1-z^2}) -\ln(z)\right]
$
$
=\frac{\pi}{2}+\frac{i}{i^2}\ln(z-i\sqrt{1-z^2}) =
=\frac{\pi}{2}-i\ln(z-i\sqrt{1-z^2})
$
Doing some more algebra:
$
-i\ln(z-i\sqrt{1-z^2}) = i \ln\left(\left(\frac{1}{z-i\sqrt{1-z^2}}\right)\left(\frac{z+i\sqrt{1-z^2}}{z+i\sqrt{1-z^2}}\right)\right)
$
$
= i\ln\left(\frac{z+i\sqrt{1-z^2}}{z^2-i^2(1-z^2)}\right)=i\ln(z+i\sqrt{1-z^2})
$
Hence the expression is true:
$$
\arcsin(z)=\frac{\pi}{2}+i\ln(z+i\sqrt{1-z^2})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| $$b^2-4c<0\le(2x+b)^2$$
$$\implies0<(2x+b)^2-b^2+4c=4x^2+4bx+4c$$
$$\implies x^2+bx+c>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
Compute the intersection of perpendicular lines Lines Ax + By = C and Bx – Ay = 0 are perpendicular.
I am trying to understand how their intersection is computed below:
$$ x = \frac{AC}{A^2 + B^2}$$
$$ y= \frac{BC}{A^2 + B^2}$$
I tried solving for y first but I am not sure how to continue:
$$ Ax+By-C= Bx-Ay $$
$$By+Ay =-Ax +C +Bx $$
$$y(B+A) = -Ax +C +Bx$$
$$y= \frac{-Ax + Bx +C}{ B +A} $$
$$y= \frac{x(-A +B) +C}{ B +A} $$
| If $B \ne 0$, then
$Bx - Ay = 0 \iff x = \dfrac ABy$
Substituting into $Ax+By=C$, we get
\begin{align}
\dfrac{A^2}{B}y+By=C
&\implies A^2y + B^2y = BC \\
&\implies y = \dfrac{BC}{A^2+B^2} \\
&\implies x = \dfrac ABy = \dfrac{AC}{A^2+B^2}
\end{align}
If $B=0$ and $A \ne 0$, then $Ax+By=C \implies Ax=C \implies x = \dfrac CA$
and $Bx-Ay=0 \implies -Ay=0 \implies y=0$. Which is consistent with
$$(x,y) = \left(\dfrac{AC}{A^2+B^2}, \dfrac{BC}{A^2+B^2} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$ $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$
I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2(1+\frac{2}{n}+\frac{1}{n^2})})=n^{\alpha+\frac{2}{5}} (\sqrt[5]{1+\frac{1}{n}}-\sqrt[5]{1+\frac{2}{n}+\frac{1}{n^2}})$
Then
$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}+ \infty & \alpha>-\frac{2}{5}\\
2 & \alpha=-\frac{2}{5}\\0 & \alpha<-\frac{2}{5}\end{cases}$
Is it right?
| It's simpler if you consider the limit you get by (formally) substituting $n=1/x$:
$$
\lim_{x\to0^+}\frac{1}{x^{\alpha}}\frac{\sqrt[5]{1+x}-\sqrt[5]{1+2x+x^2}}{\sqrt[5]{x^2}}
$$
If this limit exists (finite or infinite), then it's the same as the limit of your sequence. Now use Taylor:
$$
\frac{1+\dfrac{1}{5}x-1-\dfrac{2}{5}x+o(x)}{x^{\alpha+2/5}}=
\frac{-\dfrac{1}{5}+o(1)}{x^{\alpha-3/5}}
$$
Now it should be easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\sqrt{1-\frac{1}{x^3}}+\frac{1}{x^2}}$$
which evaluates to$\frac{1}{1+0}$.
For the second term $\frac{\sqrt[3]{x+1}}{{\sqrt{x^{3}-1}+x}}$ I can't get the manipulation right. Help is much apreciated!
| $$\lim_{x \rightarrow \infty}\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
Using the "divide top and bottom by the highest power" method, the expression simplifies:
Top first term:
$$x\sqrt{x}=x^{3/2};\frac{x^{3/2}}{x^{3/2}}=1$$
Top second term:
$$\sqrt[3]{x+1}=(x+1)^{1/3}$$
$$\frac{(x+1)^{1/3}}{x^{3/2}}=\frac{(x+1)^{2/6}}{x^{9/6}}=\frac{(x^2+2x+1)^{1/6}}{x^{9/6}}=(\frac{x^2}{x^9}+\frac{2x}{x^9}+\frac{1}{x^9})^{1/6}=(\frac{1}{x^7}+\frac{2}{x^8}+\frac{1}{x^9})^{1/6}$$
Bottom first term:
$$\frac{(x^3-1)^{1/2}}{x^{3/2}}=(\frac{x^3-1}{x^3})^{1/2}=(1-\frac{1}{x^3})^{1/2}$$
Bottom second term:
$$\frac{x}{x^{3/2}}=\frac{x^{1/2}x^{1/2}}{x^{3/2}}=\frac{1}{x^{1/2}}$$
Now we have
$$\lim_{x \rightarrow \infty}1=1$$
$$\lim_{x\rightarrow \infty}(\frac{1}{x^7}+\frac{2}{x^8}+\frac{1}{x^9})^{1/6}=0$$
$$\lim_{x\rightarrow \infty}(1-\frac{1}{x^3})^{1/2}=1$$
$$\lim_{x\rightarrow \infty}\frac{1}{x^{1/2}}=0$$
So
$$\frac{1+0}{1+0}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Some details about 'Collatz Conjecture'? Yes, there is no one who doesn't know this problem.My question is only about curiosity.
$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
On this problem, I caught something like this.I'm sure, We all realized that.
For example, $n=19$, we have $6$ odd steps.
We know that, even steps are not important, because each even number is converted to an odd number.
$19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$
Then, for $n=77$, We have also $6$ odd steps.
$77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$
For $n=9$
$9\Longrightarrow 7 \Longrightarrow 11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$
Again we have $k=6$ odd steps.
I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result $1$?
Thank you!
| If you search a single formula for any $k$, here it is:
$$n_k=\frac{2^{l_1+l_2+...+l_k}}{3^k}-\frac{2^{l_2+l_3+...+l_k}}{3^k}-\frac{2^{l_3+l_4+...+l_k}}{3^{k-1}}-\frac{2^{l_4+l_5+...+l_k}}{3^{k-2}}-...-\frac{2^{l_{k-1}+l_k}}{3^3}-\frac{2^{l_k}}{3^2}-\frac{2^0}{3^1}$$
e.g.
$$19=\frac{2^{4+3+2+1+3+1}}{3^6}-\frac{2^{3+2+1+3+1}}{3^6}-\frac{2^{2+1+3+1}}{3^5}-\frac{2^{1+3+1}}{3^4}-\frac{2^{3+1}}{3^3}-\frac{2^{1}}{3^2}-\frac{2^0}{3^1}$$
The dificulty is to find the $l_k$ for which $n_k$ is an integer.
The $l_k$ are the number of times you divide by 2 to jump from an odd to another odd.
e.g. for $19$, $l_6=1$ because you divide $3*19+1$ only once to get the next odd $29$. $l_5=3$ because you divide $3*29+1$ three times by 2 to get the next odd $11$...
When an $l_k$ is known, any $l_k$ of the same parity will work (e.g. for $19$, $l_6=1$ is odd, so any odd value of $l_6$ will work).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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How do we count solutions of $a^2 + b^2 + c^2\equiv n\pmod p$? Let's count number of solutions of $a^2 + b^2 + c^2 \equiv n\pmod p$ for each integer $n$ and each prime $p$. I'd start by just writing a function:
$$V_n(\mathbb{C}) = \left\{ (a,b,c) \in \mathbb{C}^3: a^2 + b^2 + c^2 = n \right\}$$
Maybe we could arrange this into some kind of Dirichlet series:
$$ f(s) = \prod_p \frac{\# V_n(\mathbb{F}_p)}{p^s}
= \prod_p \frac{\# \{ (a,b,c): a^2 + b^2 + c^2 = n \}}{p^s} $$
This is now a Hasse-Weil zeta function or an Arithmetic zeta function, although I didn't mean to be that advanced.
Here's the sequence of numbers for $n = 3$. The left column is a prime number and the right column is the number of solutions. The result looks somewhat random:
[(2, 4),
(3, 9),
(5, 20),
(7, 56),
(11, 110),
(13, 182),
(17, 272),
(19, 380),
(23, 506),
(29, 812)]
Here is $n=7$
[(2, 4),
(3, 6),
(5, 20),
(7, 49),
(11, 132),
(13, 156),
(17, 272),
(19, 342),
(23, 552),
(29, 870)]
| Partial solution.
You can also rewrite it as: $$\sum_{u+v+w=n}\left(1+\left(\frac u p\right)\right)\left(1+\left(\frac v p\right)\right)\left(1+\left(\frac{w} p\right)\right)$$
This is because $1+\left(\frac mp\right)$ is the number of solutions to $x^2\equiv m\pmod{p}$.
This can then be rewritten as:
$$\sum_{u+v+w=n} \left(1+3\left(\frac{u}{p}\right)+3\left(\frac{uv}{p}\right)+\left(\frac{uvw}{p}\right)\right)$$
(Why?)
Now, $$\sum_{u+v+w=n} 1 = p^2.$$
Then convince yourself that $$\begin{align}\sum_{u+v+w=n} \left(\frac u p\right) &=0\\
\sum_{u+v+w=n}\left(\frac{uv}{p}\right)&=0
\end{align}$$
So the total is $$p^2+\sum_{u+v+w=n}\left(\frac{uvw}{p}\right)$$
Not sure how to evaluate that right-hand sum.
When $p=3,n=5$ the total is $12.$
This corresponds to three permutations each of $1^2+1^2+0^2\equiv 5,\, 2^2+2^2+0^2\equiv 5,$ ands six permutations of $2^2+1^2+0^2\equiv 5.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? I'd like to find the four independent units in (the ring of integers of ) $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{R}$ We also have that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$, as a field extension.
I just want to find the Norm, $\mathfrak{N}(x)$ for $x = a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The conjugates are like this:
$$ \big(a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6}\big)
\big(a - b \sqrt{2} + c \sqrt{3} - d\sqrt{6}\big)
\big(a + b \sqrt{2} - c \sqrt{3} - d\sqrt{6}\big)
\big(a - b \sqrt{2} - c \sqrt{3} + d\sqrt{6}\big)$$
If we multiply all four of these things together, we obtain a mess. I used sympy:
$$
a^4 - 4\,a^2b^2 - 6\,a^2c^2 - 12\,a^2d^2 + 48\,abcd + 4\,b^4 - 12\,b^2c^2 - 24\,b^2d^2 + 9\,c^4 - 36\,c^2d^2 + 36\,d^4
$$
Instead we can rearrange the terms it looks almost manageable:
$$ (a^4 + 4\,b^4 + 9\,c^4 + 36\,d^4)- (4\,a^2 b^2 + 6 \, a^2 c^2 + 12\,a^2 d^2 + 12\,b^2c^2 + 24\,b^2 d^2 + 36\, c^2 d^2 ) + (48\, abcd)$$
and Dirichlet's Unit theorem says we can find integers $a,b,c,d \in \mathbb{Z}$ such that this thing $=1$.
Fortunately, I can find two subfields off the bat: $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{2})] = 2$ and $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{3})] = 2$ and we get that :
$$ 1, 3 + 2\sqrt{2}, 2 + \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3}) $$
are still units in this quartic field (by Pell Eq). There's one left. Which is it?
Related:
*
*Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
| Just a check that Pell unit provided by @Rene Schipperus is indeed independent of the other Pell units. For assume that we have
$$(3+2\sqrt{2})^m(2+\sqrt{3})^n(5+2\sqrt{6})^p=1$$
Applying the Galois maps $\sqrt{2}\mapsto - \sqrt{2}$ ,$\sqrt{3}\mapsto - \sqrt{3}$ we get
$$(3-2\sqrt{2})^m(2-\sqrt{3})^n(5+2\sqrt{6})^p=1$$
so
$$(3+2\sqrt{2})^m(2+\sqrt{3})^n=(3-2\sqrt{2})^m(2-\sqrt{3})^n$$
and therefore
$$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)^n$$
so both must be in $\mathbb{Q}$, since $\mathbb{Q}(\sqrt{2})\cap \mathbb{Q}(\sqrt{3})=\mathbb{Q}$. Now, from $\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m \in \mathbb{Q}$, applying the Galois map $\sqrt{2}\mapsto -\sqrt{2}$ we conclude that
$$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\right) ^m$$
so the common value must be $\pm 1$. We conclude that $m=0$. Similarly we get $n=0$.
ADDED: Using WolframAlpha I got this result. The surprise was
$$2+\sqrt{3}=\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right )^2$$
Moreover, the units
$$1+\sqrt{2}, \frac{\sqrt{2}+\sqrt{6}}{2}, \sqrt{2}+\sqrt{3}$$
form a fundamental system.
Note that $\frac{\sqrt{2}+\sqrt{6}}{2}=\frac{1+\sqrt{3}}{\sqrt{2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to interpret $\left(\begin{smallmatrix}p_n & p_{n-1}\\ q_n & q_{n-1}\end{smallmatrix}\right)$? We know that \begin{equation*}
a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots+\cfrac{1}{a_n}}}}}=[a_0,a_1, \cdots, a_n]
\end{equation*}
If $\frac{p_n}{q_n}=[a_0,a_1, \cdots, a_n]$.
How to prove that $$
\begin{pmatrix}
p_n & p_{n-1} \\
q_n & q_{n-1} \\
\end{pmatrix}=\begin{pmatrix}
a_0 &1 \\
1 & 0 \\
\end{pmatrix}\begin{pmatrix}
a_1 &1 \\
1 & 0 \\
\end{pmatrix}\cdots\begin{pmatrix}
a_n &1 \\
1 & 0 \\
\end{pmatrix}
$$.
I am getting the answer while checking with $n=0,1,2,3$. I think that it could be done by induction but after assuming $k=n-1$ when I am going to prove $k=n$ the calculation is getting messy. Please help me out in proving this.
| Consider the canonical map from $GL_2(\mathbb{R})$, the group of invertible $2 \times 2$ matrices, to the group $PGL_1(\mathbb{R})$, the group of projective linear transformations $\mathbb{P}^1_{\mathbb{R}} \to \mathbb{P}^1_{\mathbb{R}}$:
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \left(x \mapsto \frac{ax+b}{cx+d}\right). $$
Then each matrix $\begin{pmatrix} a_i & 1 \\ 1 & 0 \end{pmatrix}$ corresponds to the transformation $x \mapsto a_i + \frac{1}{x}$. Therefore, their product corresponds to the composition
$$ x \mapsto a_0 + \frac{1}{a_1 + \frac{1}{\ddots + \frac{1}{a_n + \frac{1}{x}}}}. $$
Suppose the matrix product is $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Then the function above would be equal to $\frac{ax+b}{cx+d}$. Now, if we substitute $x := 0$, we get
$$ \frac{b}{d} = a_0 + \frac{1}{a_1 + \frac{1}{\ddots + \frac{1}{a_{n-1}}}} = \frac{p_{n-1}}{q_{n-1}}. $$
On the other hand, if we substitute $x := \infty$, we get
$$ \frac{a}{c} = a_0 + \frac{1}{a_1 + \frac{1}{\ddots + \frac{1}{a_n}}} = \frac{p_n}{q_n}. $$
But also, since determinants are multiplicative, we see that the matrix product has determinant $ad - bc = \prod_{i=0}^n (a_i \cdot 0 - 1 \cdot 1) = (-1)^{n+1}$. It follows that $a$ and $c$ are relatively prime, as are $b$ and $d$. Since we also know that $a,b,c,d$ are all nonnegative integers, the desired result follows.
Another way to view what this argument is doing, in a more elementary way: first calculate that if
$$ \begin{pmatrix} a_i & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} c \\ d \end{pmatrix}, $$
then $\frac{c}{d} = a_i + \frac{1}{\frac{a}{b}}$.
Now, if we start multiplying
$$ \begin{pmatrix} a_0 & 1 \\ 1 & 0 \end{pmatrix} \cdots \begin{pmatrix} a_{n-1} & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a_n & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$
from the right to the left, then the first product results in $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$; the second product results in $\begin{pmatrix} a_{n-1} \\ 1 \end{pmatrix}$; the third product results in a vector whose entries have ratio $a_{n-2} + \frac{1}{a_{n-1}}$; and so on. Therefore, the overall product results in a vector whose entries have ratio $\frac{p_{n-1}}{q_{n-1}}$. Similarly, if we replace $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ above with $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, then the first product results in $\begin{pmatrix} a_n \\ 1 \end{pmatrix}$; the second product results in a vector whose entries have ratio $a_{n-1} + \frac{1}{a_n}$; and so on. Therefore, the overall product results in a vector whose entries have ratio $\frac{p_n}{q_n}$.
However, these are just the columns of the product matrix. Now, as above, we consider determinants to show that the columns have entries which are relatively prime nonnegative integers, so that the product matrix must be of the desired form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is of the form $k^{4}$ with $k$ an integer. Find the number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is not of the form $k^{4}$ with $k$ an integer.
Notice that
$$ n^{4} - 20n^{2} + 100 = (n^{2} - 10)^{2} $$
We can find the number of elements such that $(n^{2}-10)^{2} = k^{4}$.
$$(n^{2}-10)^{2} = k^{4} = (k^{2})^{2} \implies n^{2} - k^{2} = 10$$
$$ (n-k)(n+k) = 10 $$
so the possibilities are
$$ n-k = 1, n+k = 10 \implies n, k \notin \mathbb{Z}$$
$$ n-k = 2, n+k = 5 \implies n, k \notin \mathbb{Z}$$
if we swap the cases, same thing also applies. So there is no $n$ such that
$ n^{4} - 20n^{2} + 100 = k^{4} $
Thus the answer is $100$, all elements in $\{1, ..., 100 \}$.
Is this sufficient? Is there another way to solve this using techniques which are used in contests. Thanks.
Edit :
Also noting Prathyush's answer.
with $(n^{2}-10)^{2} = k^{4}$
we also need to check
$10 - n^{2} = k^{2}$
which means
$$ 10 = n^{2} + k^{2} $$
only solution is $n=1,k=3$ and $n=3, k=1$. So the answer is $98$ elements.
| $$n^{4} - 20n^{2} + 100= (n^2-10)^2$$
Thus the condition reduces to finding all $n$ such that $|n^2-10|$ is of form $k^2$. $n=1,3$ satisfy this. Let $n>3$ so that $|n^2-10|=n^2-10$. Suppose there does exist a $k$ satisfying this, then
$$n^2-10=k^2$$
$$(n-k)(n+k)=10$$
Now the highest power of $2$ in $10$ is $2^1$. Thus if one factor ($n-k$) is even, the other ($n+k$) is odd. But then $n=\frac{(n+k)+(n-k)}{2}$ will not be an integer, which is a contradiction. Thus there doesn't exist any value of $n>3$ which satisfies the equation.
Conclusion: all values of $n$ except $n=1,3$ are such that $n^{4} - 20n^{2} + 100$ is not of the form $k^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$\int \frac {4}{7\sqrt {5x-6}}dx$ Why can't I use the substitution
$$u=\sqrt {5x-6}$$
$$u^2=5x-6$$
$$2udu=5dx$$
$$dx=\frac {2u}{5}du$$
$$\int\frac {4}{7u}\cdot \frac {2u}{5}du=\int \frac {8u}{35u}du=\int \frac {8}{35}du=\frac {8}{35}+C$$
The answer is obviously incorrect, but what have I done wrong?
| $$\int\frac {4}{7u}\cdot \frac {2u}{5}du = \int \frac{8}{35} du =\frac{8}{35} \color{red}{u} + C = \frac{8}{35} \sqrt{5x-6} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Explain how to solve this trigonometric limit without L'Hôpital's rule? In my previous class our professor let us the following limit:
$$
\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)}
$$
He solved it by applying L'Hôpital's rule as follow:
$
\lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^2(x) + \cos(x) + 1}{\cos^2(x)} = \frac{3}{1} = 3
$
He only wrote that in the blackboard and then told us to solve it but without using L'Hôpital's rule. So I proceeded in this way:
$
\lim_{x\to0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{x - \sin(x)} = \lim_{x\to0} \frac{\frac{\sin(x)(1 - \cos(x))}{\cos(x)}}{x - \sin(x)} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))}
$
From there I multiply by its conjugate $(1 - \cos(x))$ but then I get more confused:
$
\lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} \frac{(1 + \cos(x))}{(1 + \cos(x))} = \lim_{x\to0} \frac{\sin^3(x)}{(x - \sin(x))(\cos(x))(1 + \cos(x))}
$ ...
Can someone give me a better advice on how to get the right result.
| $$\dfrac{\tan x-\sin x}{x-\sin x}=\dfrac1{\cos x}\cdot\dfrac{\sin x}x\cdot\dfrac{1-\cos x}{x^2}\cdot\dfrac1{\dfrac{x-\sin x}{x^3}}$$
$\dfrac{1-\cos x}{x^2}=\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{1+\cos x}$
Finally for $$\dfrac{x-\sin x}{x^3}$$ use Are all limits solvable without L'Hôpital Rule or Series Expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Binomial sum involving power of $2$
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$
Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\binom{n}{n}\frac{x^{n+1}}{n+1}$$
Could some help me to solve it , Thanks
|
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}2^{k+2}\binom{n}{k}}
&=\frac{1}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{k+2}2^{k+2}\tag{1}\\
&=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}2^k\tag{2}\\
&=\frac{1}{(n+1)(n+2)}\left(3^{n+2}-1-2(n+2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{(n+1)(n+2)}\left(3^{n+2}-2n-5\right)}
\end{align*}
Comment
*
*In (1) we apply the binomial identity $\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$ twice.
*In (2) we shift the index to start with $k=2$.
*In (3) we apply the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Why do we have $ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $ Show that
$$ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $$
I know $ 1 - \sin^2x = (1 - \sin x) (1 + \sin x) $ but I'm not sure how to apply it.
Or maybe we need $ 2 \sin^2 (x) = 1 - \cos(2x)$?
| Remember that $$ \sin 3x = 3\sin x-4\sin^3 x \implies \frac{\sin 3x}{3\sin x} = 1- \frac 43 \sin^2 x$$
Now you can see the telescoping product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a similar pattern for 5 x 5, 6 x 6, ..., n x n matricies.
How would we prove that we would achieve this pattern for all inverses of n x n matricies?
| To elaborate on the other answer, which is correct, though presented with no concrete examples, note that your matrix $A$ is a sum $A = I + N$, where $N$ is the nilpotent matrix
$$N = \begin{bmatrix} 0 & 1 & & \\ & 0& 1 & \\ & & 0& 1 \\ & & &0 \end{bmatrix}$$
The matrix $N$ has powers of a simple form, for example
$$N^2 = \begin{bmatrix} 0 & 0 &1 & \\ & 0& 0 &1 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^3 = \begin{bmatrix} 0 & 0 &0 & 1\\ & 0& 0 &0 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^4 = 0$$
It is a basic fact in commutative algebra that when you have a nilpotent element $N$, then $1 + N$ is invertible, since (I'll suppose here that $N^4 = 0$) we have
$$I = I^4 + N^4 = (I + N)(I - N + N^2 - N^3)$$
and so $(I - N + N^2 - N^3)$ must be the inverse of $(I + N)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Orthogonal diagonalize this matrix how would one orthogonally diagonalize the matrix $A$ = $\begin{pmatrix}a&0&b\\ 0&a&0\\ b&0&a\end{pmatrix}$
where $b≠0$?
| This is a classic problem. Read up on "Jacobi rotations"
Here is the short answer:
Focus on the matrix $ \left( \begin{array}{c c}
a & b\\
b & a
\end{array}
\right)
$.
A Jacobi rotation is given by
$$ \left( \begin{array}{c c}
c & s\\
-s & c
\end{array}
\right)
$$
where $ c = \cos( \theta ) $ and $ s = \sin( \theta ) $
such that, for this problem,
$$ \left( \begin{array}{c c}
c & s\\
-s & c
\end{array}
\right)
\left( \begin{array}{c c}
a & b\\
b & a
\end{array}
\right)
\left( \begin{array}{c c}
c & s\\
-s & c
\end{array}
\right)^T
$$
is diagonal.
You then take that, and use it as follows:
$$ \left( \begin{array}{c c}
c & 0 & s\\
0 & 1 & 0 \\
-s & 0 & c
\end{array}
\right)
\left( \begin{array}{c c c }
a & 0 & b\\
0 & a & 0 \\
b & 0 & a
\end{array}
\right)
\left( \begin{array}{c c}
c & 0 & s\\
0 & 1 & 0 \\
-s & 0 & c
\end{array}
\right)^T
$$
The exact formula for $ c $ and $ s $ I don't quite remember.
A rotation is an orthogonal matrix, so it what you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity.
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix}
1-\lambda && 7 && -2 \\
0 && 3-\lambda && -1 \\
0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix}
-2 && 7 && -2 \\
0 && 0 && -1 \\
0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && -\frac{7}{2} && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
\frac{7}{2}\\
1 \\
0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix}
-1 && 7 && -2 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && 0 && -5 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
5\\
1 \\
1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally,
$ A - I = \begin{bmatrix}
0 && 7 && -2 \\
0 && 2 && -1 \\
0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting $\begin{bmatrix}
x \\
y \\
z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix}$ ? or is that wrong?
| Yes, it works.
$$\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix} \begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \end{bmatrix} $$
Thus you are correct in choosing the third eigenvector as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding complex roots of fourth degree polynomial $z^4 + 8z^3 + 16z^2 + 9$ I have the equation: $$z^4 + 8z^3 + 16z^2 + 9 = 0$$
I need to find all the complex solutions and I've got no clue how to approach it. I've tried factoring but nothing came out of it. I'm still very new to the world of complex numbers so I'll appreciate any help.
| We can use factorization $x^2+y^2 = (x+iy)(x-iy)$:
\begin{align}
z^4+8z^3+16z^2+9 &= z^2(z^2+8z+16) + 9 \\
&= z^2(z+4)^2+9 \\
&= (z(z+4)+3i)(z(z+4)-3i) \\
&= ((z+2)^2 - (4-3i))((z+2)^2-(4+3i))\\
&= (z+2-z_0)(z+2+z_0)(z+2-\overline {z_0})(z+2+\overline {z_0})
\end{align}
where $z_0^2 = 4-3i$.
An explanation for the last step: if $z_0$ is a root of $z^2 = w$, then $-z_0$ is a root as well, and $\pm\overline{z_0}$ are the roots of $z^2 = \overline w$. Notice how this is in agreement with the known fact that roots of a polynomial with real coefficients are either real or come in complex conjugate pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Symmetric inequality with three variables including radicals Let $a, b, c$ be positive reals. Show that the following inequality holds:
$$\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{b^2 + c^2}} + \frac{c}{\sqrt{c^2 + a^2}} \le \frac{3}{\sqrt2}$$
I managed to do the following:
The ineq. is equivalent to this:
$$\sqrt{\frac{a^2}{a^2 + b^2}} + \sqrt{\frac{b^2}{b^2 + c^2}} + \sqrt{\frac{c^2}{c^2 + a^2}}\le \frac{3}{\sqrt2}$$
Applying Jensen to the concave $f(x) = \sqrt x$ we get:
$$\sum{\sqrt{\frac{a^2}{a^2 + b^2}}} \le 3 \sqrt{\frac{\sum \frac{a^2}{a^2 + b^2}}{3}}$$ with $\sum$ denoting cyclic sums.
It suffices to prove:
$$\sum\frac{a^2}{a^2 + b^2} \le \frac{3}{2}$$
Or:
$$\sum\frac{b^2}{a^2 + b^2} \ge \frac{3}{2}$$
This is where I am stuck. As Michał Miśkiewicz's comment stated, this can't hold, so we must try something else, but I have no ideas. Can someone help?
| By C-S
$$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+b^2}}\right)^2\leq\sum_{cyc}\frac{a^2}{(a^2+b^2)(a^2+c^2)}\sum_{cyc}(a^2+c^2)\leq\frac{9}{2},$$
where the last inequality it's
$$9\prod_{cyc}(a^2+b^2)\geq8\sum_{cyc}a^2\sum_{cyc}a^2b^2,$$ which is
$$\sum_{cyc}c^2(a^2-b^2)^2\geq0.$$
Also, we can use Jensen here:
$$\sum_{cyc}\frac{a}{\sqrt{a^2+b^2}}=\sum_{cyc}\left(\frac{a^2+c^2}{2(a^2+b^2+c^2)}\cdot\sqrt{\frac{4a^2(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)^2}}\right)\leq$$
$$\leq\sqrt{\sum_{cyc}\left(\frac{a^2+c^2}{2(a^2+b^2+c^2)}\cdot\frac{4a^2(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)^2}\right)}=$$
$$=\sqrt{\frac{4(a^2+b^2+c^2)(a^2b^2+a^2c^2+b^2c^2)}{(a^2+b^2)(a^2+c^2)(b^2+c^2)}}\leq\frac{3}{\sqrt2},$$
where the last inequality it's $\sum\limits_{cyc}c^2(a^2-b^2)^2\geq0$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Compute the surface area of the unit sphere $x^2+y^2+z^2=1$ Compute the surface area of the unit sphere $x^2+y^2+z^2=1$
The following is a solution the book suggests:
The upper hemisphere is the graph of the function $\varphi(x,y)=\sqrt{1-x^2-y^2}.$
A little calculation yields $$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac 1{\sqrt{1-x^2-y^2}} \tag {1}$$
Note that $d A=\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}d x d y \tag {*} $
So, by $(*)$, the area of the hemisphere is obtained by integrating the function in $(1)$ over the unit disc. Switching to polar cordinates yields $$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi \tag{2}$$
Firstly, in $(1)$ how does this "$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}$ " become equal to $\sqrt{1-x^2-y^2}$? Also how do we get $(2)$ , aren't we supposed to compute $d A=\sin\varphi d\theta d\varphi $?
| This is an alternative solution in case one's interested in:
By the spherical coordinates we get $x=\cos\theta\sin\varphi, \; \;y=\sin\theta\sin\varphi, \; \;z=\cos\varphi$
And, since $$d A= \sqrt{\left[ \frac {\partial y \partial z}{\partial \theta \partial \varphi}\right]^2 +\left[ \frac {\partial z \partial x}{\partial \theta \partial \varphi}\right]^2+\left[ \frac {\partial x \partial y}{\partial \theta \partial \varphi}\right]^2 }d \theta d \varphi$$
we get, $$d A=\sqrt{(\sin^2\varphi\cos\theta)^2+(\sin^2\varphi\sin\theta)^2+(\cos\varphi\sin\varphi)^2}d \theta d \varphi=\sin\varphi d \theta d \varphi$$
Thus, the area of the sphere is $$\int_0^{\pi}\int_0^{2\pi}\sin\varphi d \theta d \varphi=-2\pi\cos\varphi|_0^\pi=4\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Number of $6$-digit numbers made up of the digits $1$, $2$, and $3$ with no digit occurring $3$ or more times consecutively? Find the number of 6-digit numbers made up of the digits $1$, $2$, and $3$ that have no digit occur three or more times consecutively. (For example, $123123$ would count, but $123111$ would not.)
We know the total number of possibilities without the occurrence restriction is $3^6 = 729$.
In order to eliminate the possibilities that can't be used, I set a block of 3 numbers as my 3 consecutive digits block. This block would have 3 different combinations. I then split the number of eliminations into 2 cases: one where the block of 3 is in the front, and one where the block of 3 is in the middle.
For case 1: I computed $3\cdot 24$, with $3$ being the ways to change the block of 3, and $24$ being the number of ways to change the other 3 numbers while not having another block of 3 consecutive.
For case 2: I computed $3^4$, ignoring the restriction of case 1.
Summing the 2 cases, I got $72+ 3\cdot 81 = 315$, I then subtracted from $729$ and got $414$, which turns out to be incorrect. What am I doing wrong?
| Discussion of what's missing in your answer
There are four places a block of three can start in the number.
$$\begin{matrix}
(1): && d & d & d & \cdot & \cdot & \cdot \\
(2): && \cdot & d & d & d & \cdot & \cdot \\
(3): && \cdot & \cdot & d & d & d & \cdot \\
(4): && \cdot & \cdot & \cdot & d & d & d \\
\end{matrix}$$
You have only discussed the first two. Additionally, if we just count these four configurations, we overcount configurations where there are four, five, or six consecutive digits. For instance, $111111$ can only be generated one way, but it has been included in the counts for configurations (1), (2), (3), and (4), so has been quadruple-counted.
Corrected version of your answer
There are ten ways to have three (or more) consecutive digits:
$$\begin{matrix}
(1): && d & d & d & \cdot & \cdot & \cdot \\
(2): && \cdot & d & d & d & \cdot & \cdot \\
(3): && \cdot & \cdot & d & d & d & \cdot \\
(4): && \cdot & \cdot & \cdot & d & d & d \\
(5): && d & d & d & d & \cdot & \cdot \\
(6): && \cdot & d & d & d & d & \cdot \\
(7): && \cdot & \cdot & d & d & d & d \\
(8): && d & d & d & d & d & \cdot \\
(9): && \cdot & d & d & d & d & d \\
(10): && d & d & d & d & d & d
\end{matrix} $$
We will proceed by inclusion-exclusion.
And then proceed as shown by N.F.Taussig's answer. (I was about half way through when he saved me having to finish it.)
A solution that avoids explicit inclusion-exclusion
From each string of six digits, $d_1 d_2 d_3 d_4 d_5 d_6$, let us construct a new string of length $5$ having an "s" in position $i$ if $d_i = d_{i+1}$ and an "n" if $d_i \neq d_{i+1}$, as $i$ ranges from $1$ to $5$. A digit occurring three or more times consecutively corresponds to two or more consecutive "s"s. There are only $3$ choices of first pairs of digits which produce an "s" in the first position and $6$ choices of pairs of digits which produce an "n" in the first position. Following an "s" only one choice of digit can produce an "s" (which corresponds to disallowed three consecutive digits) and two choices of digit produce an "n". Following an "n" one choice of digit produces an "s" and two choices of digit produce an "n".
The acceptable 5 letter strings, with their counts and a running sum, are \begin{align*}
snsns && 3 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 12 && 12 \\
snsnn && 3 \cdot 2 \cdot 1 \cdot 2 \cdot 2 = 24 && 36 \\
snnsn && 3 \cdot 2 \cdot 2 \cdot 1 \cdot 2 = 24 && 60 \\
snnns && 3 \cdot 2 \cdot 2 \cdot 2 \cdot 1 = 24 && 84 \\
snnnn && 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 48 && 132 \\
nsnsn && 6 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 24 && 156 \\
nsnns && 6 \cdot 1 \cdot 2 \cdot 2 \cdot 1 = 24 && 180 \\
nsnnn && 6 \cdot 1 \cdot 2 \cdot 2 \cdot 2 = 48 && 228 \\
nnsns && 6 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 24 && 252 \\
nnsnn && 6 \cdot 2 \cdot 1 \cdot 2 \cdot 2 = 48 && 300 \\
nnnsn && 6 \cdot 2 \cdot 2 \cdot 1 \cdot 2 = 48 && 348 \\
nnnns && 6 \cdot 2 \cdot 2 \cdot 2 \cdot 1 = 48 && 396 \\
nnnnn && 6 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 96 && 492 \\
\end{align*}
So there are $492$ allowed configurations.
I generally prefer this method because it more directly corresponds to the tree of consecutive digit choices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
| Here is another proof.
Say $$a_n = \frac{p_n}{q_n}$$
then from the identity we have that
$$p_{n+1} = q_n, q_{n+1} = 2q_n + p_n = 2q_n + q_{n-1}$$
$q_n$ are of the form
$$ A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n $$
($\sqrt{2} + 1$ and $1 - \sqrt{2}$ are roots of $x^2 = 2x + 1$)
Thus $$a_n = \frac{q_{n-1}}{q_n} = \frac{A (\sqrt{2} + 1)^{n-1} + B (1 - \sqrt{2})^{n-1}}{A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n}\to \frac{1}{1 + \sqrt{2}} = \sqrt{2} -1 $$
(Divide both numerator and denominator by $(\sqrt{2} + 1)^n$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\binom{30}{6}$
P = 1 - P(All red + All Green + All Blue + Only red and Green + Only Red and Blue + Only Green and Blue )
$$P = \Large 1 - \frac{\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}}{\binom{30}{6}}$$
According to this, I got $\large 1 - \frac{133099}{593775}$,
which is $0.7758$?
Is my approach correct?
| Let $E_1$ be the event that no red ball is chosen, $E_2$ the event that no green ball is chosen, and $E_3$ the event that no blue ball is chosen.
The probability that at least ball of each color is chosen is
$1 - P(E_1 \cup E_2 \cup E_3).$
By the inclusion-exclusion principle,
$$
P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3) - P(E_1 \cap E_2) - P(E_1 \cap E_3) - P(E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3).
$$
And we can see that
$$
P(E_1) = \frac{\binom{22}{6}}{\binom{30}{6}}, \quad
P(E_2) = \frac{\binom{20}{6}}{\binom{30}{6}}, \quad
P(E_3) = \frac{\binom{18}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_2) = \frac{\binom{12}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_3) = \frac{\binom{10}{6}}{\binom{30}{6}}, \quad
P(E_2 \cap E_3) = \frac{\binom{8}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_2 \cap E_3) = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far
| This can be not the only way to proceed, but it is convenient to avoid working with $e^{(i \pm 1) x}$ terms too much. Don't be afraid by the amount of lines, I simply did almost all the steps.
$$I = \int \sin x \sinh x \ \mathrm{d}x =\\
\int \frac{e^{ix} - e^{-ix}}{2i} \cdot \frac{e^{x} - e^{-x}}{2} \ \mathrm{d}x =\\
\frac{1}{4i} \int \left( e^{ix} e^x - e^{-ix}e^x - e^{ix}e^{-x} + e^{-ix}e^{-x} \right) \ \mathrm{d}x =\\
\frac{1}{4i} \int \left[ e^{(i + 1)x} - e^{(1 - i)x} - e^{(i - 1)x} + e^{-(i + 1)x} \right] \ \mathrm{d}x =\\
\frac{1}{4i} \int \left[ e^{(i + 1)x} - e^{-(i - 1)x} - e^{(i - 1)x} + e^{-(i + 1)x} \right] \ \mathrm{d}x =\\
\frac{1}{4i} \int \left[ e^{(i + 1)x} + e^{-(i + 1)x} - \left( e^{(i - 1)x} + e^{-(i - 1)x} \right) \right] \ \mathrm{d}x
$$
What can be recognized here is a hyperbolic cosine (not a trigonometric cosine, because it would be $\cos \alpha = (e^{i \alpha} + e^{-i \alpha})/2$ with $\alpha$ real: here, instead, the exponents contain complex quantities, $(i \pm 1)x$): simply applying the definition
$$\cosh z = \frac{e^{z} + e^{- z}}{2}$$
The integral becomes:
$$I = \frac{1}{4i} \int \left[ 2 \cosh \left[ (i + 1)x \right] - 2 \cosh \left[ (i - 1)x \right] \right] \ \mathrm{d}x =\\
= \frac{1}{2i} \int \left[ \cosh \left[ (i + 1)x \right] - \cosh \left[ (i - 1)x \right] \right] \ \mathrm{d}x =\\$$
Hyperbolic functions, like trigonometric functions, have sum and subtraction identities:
$$\sinh(\alpha + \beta) = \sinh \alpha \cosh \beta + \cosh \alpha \sinh \beta\\
\sinh(\alpha - \beta) = \sinh \alpha \cosh \beta - \cosh \alpha \sinh \beta\\
\cosh(\alpha + \beta) = \cosh \alpha \cosh \beta + \sinh \alpha \sinh \beta\\
\cosh(\alpha - \beta) = \cosh \alpha \cosh \beta - \sinh \alpha \sinh \beta$$
The integrand function is currently of the type $\cosh(\alpha + \beta) - \cosh(\alpha - \beta)$, whose value would be (by side-by-side subtraction of the above identities):
$$\cosh(\alpha + \beta) - \cosh(\alpha - \beta) = 2 \sinh \alpha \sinh \beta$$
But this would lead you back to the starting point. So, it is not convenient to use those identities now. It is convenient to directly integrate:
$$I = \frac{1}{2i} \left\{ \frac{\sinh \left[ (i + 1)x \right]}{i + 1} - \frac{\sinh \left[ (i - 1)x \right]}{i - 1} \right\} + C = \\
\frac{1}{2i} \left\{ \frac{i - 1}{i - 1} \frac{\sinh \left[ (i + 1)x \right]}{i + 1} - \frac{i + 1}{i + 1} \frac{\sinh \left[ (i - 1)x \right]}{i - 1} \right\} + C = \\
\frac{1}{2i} \left\{ \frac{i \sinh \left[ (i + 1)x \right] - \sinh \left[ (i + 1)x \right]}{-2} - \frac{i \sinh \left[ (i - 1)x \right] + \sinh \left[ (i - 1)x \right]}{-2} \right\} + C = \\
\frac{1}{4i} \left\{ \sinh \left[ (i + 1)x \right] - i \sinh \left[ (i + 1)x \right] + i \sinh \left[ (i - 1)x \right] + \sinh \left[ (i - 1)x \right] \right\} + C = \\
\left\{ \frac{\sinh \left[ (i + 1)x \right]}{4i} - \frac{ \sinh \left[ (i + 1)x \right] }{4} + \frac{ \sinh \left[ (i - 1)x \right] }{4} + \frac{ \sinh \left[ (i - 1)x \right] }{4i} \right\} + C = \\
\left\{ \frac{\sinh (ix + x) + \sinh (ix - x)}{4i} - \frac{ \sinh (ix + x) - \sinh (ix - x) }{4} \right\} + C = \\
$$
Now, the sum/subtraction identities for the hyperbolic sine
$$\sinh(\alpha + \beta) + \sinh(\alpha - \beta) = 2 \sinh \alpha \cosh \beta\\
\sinh(\alpha + \beta) - \sinh(\alpha - \beta) = 2 \cosh \alpha \sinh \beta$$
can be applied.
$$I = \frac{2 \sinh (ix) \cosh (x)}{4i} - \frac{2 \cosh (ix) \sinh (x)}{4} + C$$
Applying the definitions,
$$\sinh(ix) = \frac{e^{ix} + e^{-ix}}{2} = i \sin x$$
and
$$\cosh(ix) = \frac{e^{ix} + e^{-ix}}{2} = \cos x$$
So,
$$I = \frac{2 i \sin (x) \cosh (x)}{4i} - \frac{2 \cos (x) \sinh (x)}{4} + C =\\
\frac{1}{2} \left[ \sin (x) \cosh (x) - \cos (x) \sinh (x) \right] + C$$
which is the expected result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
In regards to a game scenario I am designing In the game there is a 6 sided die which has 5 different faces (i.e. two of the faces are the same.) Considering that the die is fair and has 1/6 chance to roll any one side on each individual roll. What is the average number of rolls until I see the 5 different faces of the 6 sided die.
(I've seen a lot of different answers in regards to if each face of the die is different but not that pertains to if two sides have the same face. I'm sure it would change the formula significantly but I may be wrong.)
The above scenario is in the actual game. I'm trying to make my own scenario that is comparable. Thank you very much with your help.
| In general, if your probability of "success" for an event is $p$, then the expected number of trials until success is $1/p$. If there are $k$ faces of a die you haven't seen, then the probability of seeing a face you haven't seen yet is $k/6$ and you'll have to throw the dice an expected amount $6/k$ before you see one of them.
Consider the situation where you roll the die repeatedly and you see every face. It happens that the last face to come up is the doubled one. In this case, there are six faces you haven't seen at first, so you expect to wait wait 6/6=1 round (i.e. you see a new face immediately). Then there are five faces you haven't seen. You expect to wait 6/5 rounds. Then there are four faces you haven't seen: 6/4 rounds. Then three: 6/3 rounds. Then two: 6/2. The last two faces are identical, so we're finished once we've seen one of them. The expected wait time in this special case is $6/6 + 6/5 + 6/4 + 6/3 + 6/2$. If you had seen the doubled side earlier, you can see that the subsequent wait times would have been different.
We can compute the expected wait time overall. We compute the expected wait time depending on if you see the doubled face first, second, third, and so on. We weight each of those times by the likelihood of seeing the doubled face first, second, third, and so on. Then we add up the weighted expectations.
$$\frac{2}{6}\left(\frac{6}{6}+\frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1}\right) + \frac{4}{6}\cdot\frac{2}{5}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1}\right) +\\ \frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{2} + \frac{6}{1}\right) + \frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\cdot \frac{2}{3}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{1}\right) + \\\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\cdot \frac{1}{3}\cdot \frac{2}{2}\left(\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2}\right)$$
$$= \frac{9}{2} + \frac{88}{25} + \frac{127}{50} + \frac{39}{25} + \frac{29}{50}\\= \frac{225 + 176 + 127 + 78 + 29}{50} = \frac{635}{50}$$
$$\frac{635}{50} = 12.7 \text{ rounds}$$
as opposed to 14.7 if you have six different faces.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find k with little oh notation? For which values of $k$ are the following true (as $x \to 0$)?
(a) $\sqrt{1+x^2} = 1 + o(x^k)$
(b) $\sqrt[3]{1+x^2} = 1 + o(x^k)$
(c) $1 - \cos(x^2) = o(x^k)$
(d) $1 - \cos^2 x = o(x^k)$
How would you find $k$ for problems like these?
| For (a), have a look at $\sqrt{1+x^2}-1$, which we can transform with the standard trick of multiplying and dividing by the conjugate:
$$ \sqrt{1+x^2}-1=\frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{\sqrt{1+x^2}+1}=\frac{x^2}{\sqrt{1+x^2}+1}\sim \frac{x^2}2$$
A similar trick helps with (b). For (c) and (d), have a look at the Taylor-expansion of the cosine )as the tagging suggests)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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