Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why the orbit is of dimension $12$? Let $SL_3$ acts on the variety consisting of all nilpotent $3$ by $3$ matrices over $\mathbb{C}$ by conjugation. Let $S_p$ be the orbit of the matrix $$ a=\left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right). $$ Why the orbit $S_p$ over $\mathbb{R}$ is of d... | By orbit-stabilizer, the dimension of the orbit should be the dimension of the group minus the dimension of the stabilizer, so we should compute the dimension of the stabilizer. Suppose
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the ... | Use the fact
$$e^{\sin(x)} \sim_{x\sim 0} 1+x+x^2/2-x^4/8. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Integrating a rational function. How to integrate $$\int_1^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt3x^2+1}{\rm d}x$$
The bottom is not factorizable hence no partial fractions. There seems no other way.
| $$\begin{align}\int_{1}^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt 3x^2+1}dx&=\int_{1}^{\infty}\frac{2x^3-1}{(x^3+1)^2+\sqrt 3x^2}dx\\&=\int_{1}^{\infty}\frac{(2x^3-1)/(x^2)}{\left((x^3+1)^2+\sqrt 3x^2\right)/(x^2)}dx\\&=\int_{1}^{\infty}\frac{2x-\frac{1}{x^2}}{\left(x^2+\frac 1x\right)^2+\sqrt 3}dx\\&=\int_{2}^{\infty}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Prove a relationship involving floor functions I am trying to prove that a particular expression is a lower bound for a very unusually-behaved function. The whole proof will be complete if I can just nail down the details of one technical lemma involving the floor function. I am completely confident that the lemma is... | Fix $n,b$ with $1\le b\le n$.
Let $k\in\mathbb Z$. For $i\in\{1,2\}$ define
$$\tag1\delta_i:=k- \frac{n+\sqrt{n^2+b+i}}{b+i}.$$
Then $k$ is strictly between the bounds iff $\delta_1<0<\delta_2$.
Then for $i\in\{1,2\}$
$$\begin{align}k\sqrt{n^2+b+i}&=\left( \frac{n+\sqrt{n^2+b+i}}{b+i}+\delta_i\right)\sqrt{n^2+b+i}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Summation Proof Dealing With 3s Multiples So the problem is as follows:
Prove that if the sum of digits of a decimal $n$ is three's multiple, then n is three's multiple by direct proof. For example, $11234567$ is 3's multiple because $1+1+2+3+4+5+6+7=24$, and in fact, $11234567$ is $3744856 * 3$. Use $10^n \equiv 1 \pm... | Maybe you could first prove that the sum of the digits subtracted from the number is a multiple of 9.
For example, a 4 digit decimal number such as $1234=1\times1000+2\times100+3\times10+4$
But $1000=999+1$, $100=99+1$ and $10=9+1$.
So $1234=1\times\left(999+1\right)+2\times\left(99+1\right)+3\times\left(9+1\right)+4$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$ Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is:
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$
| $$
\begin{align}
\int_0^1\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=2\int_{1/2}^1\sqrt{\frac{1-x}{x}}\,\mathrm{d}x\tag{1}\\
&=4\int_{1/\sqrt2}^1\sqrt{1-x^2}\,\mathrm{d}x\tag{2}\\
&=4\int_0^{\pi/4}\sin^2(x)\,\mathrm{d}x\tag{3}\\
&=2\int_0^{\pi/4}1-\cos(2x)\,\mathrm{d}x\tag{4}\\
&=\left[2x-\sin(2x)\vphantom{\int}\right]_0^{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to solve this integral by a simple way? I'm given $$\int \frac{x^3}{\sqrt{x^4+x^2+1}}dx$$
My attempt,
Let $u=x^2$, $du=2xdx$
$$=\frac{1}{2}\int \frac{u}{\sqrt{u^2+u+1}}du = \frac{1}{2}\int \frac{u}{\sqrt{(u+\frac{1}{2})^2+\frac{3}{4}}}du$$
Let $s=u+\frac{1}{2}$, $ds=du$
$$=\frac{1}{2}\int\frac{s-\frac{1}{2}}{\sqrt... | Substituting $\frac{2x^2+1}{\sqrt{3}}=u\implies x=\frac{\sqrt{\sqrt{3}u-1}}{\sqrt{2}}$, the integral becomes:
$$\begin{align}
\int\frac{x^3}{\sqrt{x^4+x^2+1}}\,\mathrm{d}x
&=\int\frac{\left(\sqrt{3}u-1\right)^{3/2}}{\sqrt{6}\sqrt{u^2+1}}\cdot\frac{\sqrt{6}\,\mathrm{d}u}{4\sqrt{\sqrt{3}u-1}}\\
&=\frac14\int\frac{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfr... | This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Asymptotic behaviour of sequences Could anyone explain in details how these approximations as $n \to \infty$ are found? ($a$ is a positive real number)
*
*${x_n} = \frac{1}{n}\left( {\frac{a}{3} - \frac{3}{2}} \right) + O\left( {\frac{1}{{{n^3}}}} \right)$ for ${x_n} = \sqrt[3]{{{n^3} + an}} - \sqrt {{n^2} + 3} $
*... | We have $\sqrt[3]{n^3+an}=n\sqrt[3]{1+\frac{a}{n^2}}$. We expand the latter using Newton's Binomial Theorem to get $$\sqrt[3]{n^3+an}=n\sum_{k\ge 0} {1/3 \choose k} \left(\frac{a}{n^2}\right)^k=n(1+\frac{a}{3n^2}-\frac{a^2}{9n^4}+O(n^{-6}))$$
Repeating, we have $\sqrt{n^2+3}=n\sqrt{1+\frac{3}{n^2}}$, so
$$\sqrt{n^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful... | Recall $1+\tan^2(x) = \sec^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$. Hence,
$$1+\tan^2(x) = 8 \sin^2(x) \implies \sec^2(x) = 8\sin^2(x) \implies 8\sin^2(x)\cos^2(x) = 1$$
This gives us
$$2\sin^2(2x) = 1 \implies \sin(2x) = \pm \dfrac1{\sqrt2} \implies 2x = \dfrac{n\pi}2 + \dfrac{\pi}4 \implies x = \dfrac{n\pi}4 + \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $.
What is the integral of: $\int \frac{1}{5+3\sin x}dx$
My attempt:
Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have:
$\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $
I'll expand the denominator: $5t^2+6t+5=5(... | Taking from
$5t^2$+6t+5, instead of expanding, complete the square.
=5(t+$\frac{6}{10})^2$-$\frac{16}{5}$
Rewrite in the form
=$\frac{-1}{5}$$\int$$\frac{1}{16/25-(t+6/10)^2}$
=$\frac{-1}{5}$[$\frac{5}{4}$$tanh^{-1}$($\frac{t+6/10}{4/5})]$+C
=$\frac{-1}{4}$[$tanh^{-1}$($\frac{5t}{4}$+$\frac{3}{2}$)]+C
Still getting an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluation of $\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx$
Evaluate the integral
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$
My Attempt:
Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to
$$
-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt
$$
Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1... | Alternate Solution:
$$
\begin{align}
\int\frac{1}{x^2\cdot \left(x^4+1\right)^{3/4}}\,dx &= \int\frac{1}{x^2\cdot x^3\cdot \left(1+x^{-4}\right)^{3/4}}\,dx\\
&= \int\frac{1}{x^{5}\cdot \left(1+x^{-4}\right)^{3/4}}\,dx
\end{align}
$$
Now Let $(1+x^{-4}) = t$. Then $x^{-5}\,dx = -\frac{1}{4}\,dt$, so we have
$$
-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has ... | There are 5 types of possibilities for the unordered exponents $a_1,\cdots,a_5$ for 2:
$\textbf{1)}$ $6+0+0+0+0,\;\; 2+1+1+1+1$ $\;\;$(4 of a kind)
$\textbf{2)}$ $5+1+0+0+0,\;\; 3+0+1+1+1,\;\;4+2+0+0+0$ $\;\;$(3 of a kind)
$\textbf{3)}$ $4+1+1+0+0,\;\; 0+2+2+1+1$ $\;\;$(2 pairs)
$\textbf{4)}$ $3+3+0+0+0,\;\; 0+0+2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 0
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Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that
*
*$\color{red}{2\le} a\le b\le c$
*$\sqrt{abc}\in\mathbb N$
*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$
Exampl... | (Too long for a comment.)
The two solutions,
$$a,\,b,\,c = k^2,\;(k+1)^2,\;(k+2)^2$$
$$a,\,b,\,c = 2^2,\;k^2,\;(k+1)^2$$
by users coffeemath and martin, respectively, are special cases of the more general solution,
$$a,\,b,\,c = k^2,\;(km\pm1)^2,\;(km\pm2)^2$$
where coffeemath's had $m=1$, while martin's had $k=2,\, m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 3,
"answer_id": 0
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Show that $\sin\left(\frac\pi3(x-2)\right)$ is equal to $\cos\left(\frac\pi3(x-7/2)\right)$ Show that
$$\sin\left(\frac\pi3(x-2)\right)$$
is equal to
$$\cos\left(\frac\pi3(x-7/2)\right)$$
I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.
| The two angles $\frac{\pi}{3}(x-2)$ and $\frac{\pi}{3}(x-7/2)$ differs by $\pi/2$.
In fact,
$$ \frac{\pi}{3}(x-2) - \frac{\pi}{3}(x-\frac{7}{2}) = \frac{\pi}{3}x- \frac{2}{3}\pi - \frac{\pi}{3}x + \frac{7}{6}\pi = - \frac{7-4}{6}\pi = -\frac{\pi}{2}\, , $$
thus $\sin(\frac{\pi}{3}(x-2)) = \sin(\frac{\pi}{3}(x-7/2) - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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limit of sequnce in real numbers For every tow real numbers $a$ and $b$,with this condition that $0<a<b$ ,define sequence ${x_{n}}$ to the following:
$x_{1}=a$ $x_{2}=b$
${x_{n}}=\frac{ x_{n-1}+ x_{n-2} }{2}$ (for $n>2$)
then which of the following options is limit of ${x_{n}}$ ?
1)b
2)$\frac{a+2b}{3}$
3)$\frac... | The characteristic equation for the recurrence is
$$y^2 - \dfrac{y}2 - \dfrac12 = 0 \implies (y-1)(y+1/2) = 0$$
This gives us
$$x_n = c_0 1^n + c_1 \left(-\dfrac12\right)^n$$
Hence, note that $\lim_{n \to \infty} x_n = c_0$. Setting $n=1$, we have $c_0 - \dfrac{c_1}2 = a$ and setting $n=2$, we have $c_0 + \dfrac{c_1}4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that there is a large value
Suppose not all 4 integers, $a, b, c, d$ are equal. Start with $(a, b, c, d)$ and repeatedly replace $(a, b, c, d)$ by $(a - b, b - c, c - d, d - a)$. Then show that at least one number of the quadruple will become arbitrarily large.
Let $x_k$ be the pair with:
$x_k = (a_k, b_k, c_k, ... | This is problem $E5$ under The invariance principle in Arthur Engel's book titled "Problem Solving Strtegies"
We have
$$(a_{n+1},b_{n+1},c_{n+1},d_{n+1}) = (a_n-b_n,b_n-c_n,c_n-d_n,d_n-a_n)$$
As you have rightly observed, the invariant is $a_n+b_n+c_n+d_n = 0$ for all $n \geq 1$, where $a_0 = a$, $b_0=b$, $c_0 = c$ an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| Notice that $(n^2 - 1)(n^2 + 1) = n^4 - 1$. Fermat's little theorem tells us that $n^4 \equiv 1 \pmod 5$. This means that $n^4 - 1$ is a multiple of $5$ if $n$ is not, and therefore $n^2 (n^4 - 1)$ is also a multiple of $5$. For example, if $n = 2$, then $n^4 - 1 = 15$ and $n^2 (n^2 - 1)(n^2 + 1) = 60$.
This leaves us ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Different Law of Cosines using Sine instead: $c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$ Playing around with Trig and the Law of Cosines (LoC), I came up with this formula given a triangle with sides $a$, $b$, $c$ where we are given $a$, $b$ and angle $\theta$ between them:
$$c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\the... | Really this just boils down to the identity $$\cos^2{x} = 1 - \sin^2{x}$$
So no, your formula is really no different from the Law of cosines. Your proof looks fine however.
Having believed I found a new trigonometric law once (and being incorrect), my advice would be to always exhaust your result for any known identit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Deriving a contradiction How can I derive a contradiction from the following nasty statement:
Assume $\sqrt{5} = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8},$ with $a,b,c,d \in \mathbb{Q}$?
This is the last piece of an effort to prove that the polynomial $x^4-2$ is irreducible over $\mathbb{Q}(\sqrt{5}).$
| Starting from
$$\tag1m\sqrt 5=a+b\alpha+c\alpha^2+d\alpha^3 $$
(where $\alpha^4=22$) with $a,b,c,d\in\mathbb Z$ you find
$$\begin{align}5m^2&=(a+b\alpha+c\alpha^2+d\alpha^3)^2\\&=(a^2+2c^2+4bd)+(2ab+4cd)\alpha+(b^2+2d^2+2ac)\alpha^2+(2ad+2bc)\alpha^3\end{align}$$
If you alredy know that $1,\alpha,\alpha^2,\alpha^3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ Can we possibly compute the following integral in terms of known constants?
$$\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$$
Some progress was already done here http://integralsandseries.prophpbb.com/topic279.html but still we have a hypergeo... | An incomplete answer.
Tackling the inner integral with elementary methods,
\begin{align}
\int^1_0\frac{{\rm d}x}{1-ax+ax^2}
&=\int^1_0\frac{{\rm d}x}{a\left(x-\frac{1}{2}\right)^2+1-\frac{a}{4}}\\&=\int^\frac{1}{2}_{-\frac{1}{2}}\frac{{\rm d}x}{ax^2+1-\frac{a}{4}}\\
&=\frac{1}{\sqrt{a\left(1-\frac{a}{4}\right)}}\int^\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
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Numbers of the form $5 \cdot 2^{n}-1$ divisible by $3^k$ for large values of $k$ Let $n_k$ be the smallest integer such that $5 \cdot 2^{n_k}-1$ is divisble by $3^k$ where $k$ is a positive integer. Can one say something about the growth of $n_k$ with respect to $k$ ? Is it exponential ?
The first values are $n_k=1,1,1... | Here is some progress I made ...
Looking closely at the values given by Lucian, we remark that $$n_k - n_{k-1} = 2 \cdot a_{k} \cdot 3^{k-2} \text{ where } a_{k}=0,1,2 \text{ for } k>1.$$
Example: $n_6 - n_5= 324 = 4 \cdot 3^4$.
My sketch of proof below:
First, by induction, we have
$$ 2^{2 \cdot 3^{k-2}} \equiv 1 + 3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Dynamical System transformation How can the system $$\frac{dx}{dt}=-y+\epsilon x(x^2+y^2)$$$$\frac{dy}{ dt}=x+\epsilon y(x^2+y^2)$$ be transformed into $$\frac{dr}{dt}=\epsilon r^3$$ $$\frac{d\theta}{dt}=1$$ via polar coordinates?
I sub in $x=r\cos(\theta)$ and $y=r\sin(\theta)$ into both of the original equations, but... | The given equations are
$\dfrac{dx}{dt} = - y + \epsilon x (x^2 + y^2), \tag{0A}$
$\dfrac{dx}{dt} = x + \epsilon y (x^2 + y^2); \tag{0B}$
we have
$r^2 = x^2 + y^2; \tag{1}$
thus,
$2r\dfrac{dr}{dt} = 2x\dfrac{dx}{dt} + 2y \dfrac{dy}{dt}, \tag{2}$
whence
$r \dfrac{dr}{dt} = x\dfrac{dx}{dt} + y \dfrac{dy}{dt}; \tag{3}$
us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Express a quadratic form as a sum of squares using Schur complements I was able to figure out the first part of this problem, but I have no concept of how it relates to Schur complements, so I'm not sure (no pun intended) how to proceed. The question is as follows:
Consider $2x^2 + 2xy + 2y^2 + z^2 + 2xz$. Write the s... | I think what the question means is to apply the $L^TDL$ decomposition that produces Schur complements, rather than to use Schur complements themselves. In other words, you are expected to apply recursively the decomposition
$$
\pmatrix{A&B\\ B^T&C}=
\pmatrix{I&BC^{-1}\\ 0&I}
\pmatrix{A-BC^{-1}B^T&0\\ 0&C}
\pmatrix{I&0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the roots of the polynomial? (Cardano's Method) $y^3-\frac7{12}y-\frac7{216}$
This is part of Cardano's method, so I've gotten my first root to be:
$y_1=\sqrt[3]{\frac7{432}+i\sqrt{\frac{49}{6912}}}+\sqrt[3]{\frac7{432}-i\sqrt{\frac{49}{6912}}}$
I am at a loss on how to evaluate this. I know this is one of three r... | One can proceed as described above and find the three roots:
$\left\{\frac{1}{6} \left(\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i
\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i
\sqrt{3}\right)}\right),-\frac{7^{2/3} \left(1+i \sqrt{3}\right)}{6\ 2^{2/3}
\sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Compute $A^{2013}$.
QUESTION 6
Let $$A=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}$$
Compute $A^{2013}$ (10 marks)
Solution First, observe that
$$A^{2013}=\begi... | Notice that we can write $A$ on this form
$$A=P\operatorname{diag}(-1,R(\pi/3))P^{-1}$$
where
$$R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
is the matrix of rotation of angle $\theta$. Now using the bloc multiplication we get
$$A^{2013}=P\operatorname{diag}(-1,R^{2013}(\pi/3))P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Common tangent to a circle and ellipse Hey guys i am noy able to solve this problem.So please do help me in solving this.The equation of common tangent to ellipse
\begin{equation*}
x^2 +2y^2=1
\end{equation*}
and circle
\begin{equation*}
x^2 +y^2=\frac{2}{3}
\end{equation*}
is?
| A method that would also work for other pairs of conics: $$x^2 +2y^2-1=0 , x^2 +y^2-\frac23=0$$ have dual conics $$X^2+\frac12Y^2-1=0, \frac23X^2+\frac23Y^2-1=0$$ which intersect in $$(X,Y)=(-\frac1{\sqrt2},-1),(-\frac1{\sqrt2},1),(\frac1{\sqrt2},-1),(\frac1{\sqrt2},1)$$
and by substituting into the equation of the un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving Recurrence Relations using Iteration $$a_0 = 2; \qquad a_k=4a_{k-1}+5 ~ \forall\ k\ge 1$$
I have already tried solving for $a_1$ through $a_5$.
| $$\begin{align*}
a_k - a_{k-1} & = 4a_{k-1}+5 -(4a_{k-2}+5) \\
&= 4(a_{k-1} - a_{k-2}) \\
&= 4^2(a_{k-2}-a_{k-3}) \\
& =4^{k-1}(a_1-a_0) \\
& =11\cdot 4^{k-1} \\
\Rightarrow a_k & = (a_k-a_{k-1})+(a_{k-1} - a_{k-2})+\cdots +(a_1-a_0)+a_0 \\
& = 11\cdot 4^{k-1} +11\cdot 4^{k-2} +\cdots + 11\cdot 4 + 11 + 2 \\
& = \ldots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Simplify $\left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$ Question:
Simplify
$$ \left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$$
Attempted solution:
Rewriting it to look a bit better:
$$\left(ab \sqrt[4]{\frac{a^{3}}{\sqrt{b\sqrt{b}}}}\right)^{2}$$
Rewriting the denominator by replacing square roots with ... | we have $a^2b^2\sqrt{\frac{a^3}{b^{3/4}}}=\sqrt{\frac{a^4b^4a^3}{b^{3/4}}}=a^{7/2}(b^{13/4})^{1/2}=a^{7/2}b^{13/8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the eigenvectors. $$\begin{vmatrix}
1&1&1&1\\
1&1&0&0\\
1&0&1&0\\
1&0&0&1\\
\end{vmatrix}$$
for this matrix the eigenvalues are $(1,1,1-\sqrt3,1+\sqrt 3)$
I if I try find the eigenvectors in the cases where the eigenvalue is $1$ I end up with two equations and $4$ unknowns.
I'm not sure not to take the equ... | For $\lambda =1$
$$\begin{align} (A - I)v = 0 &\implies \begin{pmatrix}0&1&1&1\\1&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\\v_4\end{pmatrix}= 0\\&\implies \begin{cases}v_1 = 0\\v_2 + v_3 + v_4 =0\end{cases}\end{align}$$
Then the eigenvectors associated to $\lambda =1 $ are of the form $$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Why does my solutions manual take away or add 180 when finding $\theta$? These are the provided notes:
These are the provided questions:
I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of th... | Consider the following diagram:
The range of the arccosine function is $[0, \pi]$. Since you are working in degrees, this corresponds to $[0^\circ, 180^\circ]$. Thus, you can calculate the measure of the obtuse angle $\theta$ whose cosine is $-3/5$ directly since
$$\cos\theta = -\frac{3}{5} \Longrightarrow \theta =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Prove/disprove an inequality $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ Is $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ TRUE in $(0, \frac{π}{2})$ ?
I proceed by taking two functions $f(x) = \sqrt{1 + x}-1 $ and $g(x) = \frac{x}{2} - \frac{x^2}{8} $.
Then $f(0) = 0 = g(0),$ while $f(\frac{π}{2}) > g (\frac{... | Integration by parts gives
$$ \sqrt{1+x} = 1 +\frac{x}{2} - \frac{x^2}{8} + \int_0^x \frac{(x-t)^2}{2} \frac{3}{8(1+t)^{5/2}} \, dt, $$
and it is easy to see that the integral on the right-hand side is positive for any $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
partial fraction decomposition help now very quick way to solve 2x/x^2-4 as a partial fraction. I have tried the long way and it took over 30 minutes, I got it right but is it easier another way?
| Write $$\frac{2x}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$
Then note that if we multiply both sides by $x-2$ and let $x \to 2$ we obtain
Write $$\left.\frac{2x}{x+2}\right|_{x=2}=B+\left.\frac{A(x-2)}{x+2}\right|_{x=2}$$
which gives $B=1$.
Similarly, note that if we multiply both sides by $x+2$ and let $x \to -2$ we obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
When is $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ solvable?
Exercise: For which prime numbers does the equation $x^2 - 75 y^2 = 0$ have non-trivial solution in the $p$-adic integers $\mathbb{Z}_p$?
For $p\neq 5$, the non-trivial solvability of the equation $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ is equivalent to the solvabil... | Proof looks OK but I would streamline it quite a bit.
$x^2-75y^2=0$ is just the same as $x^2=3(5y)^2$ so for $x,y <>0$ and $p<>5$ (which makes sure that 5 has an inverse) we get $3=[x(5y)^{-1}]^2$ which means that 3 is a quadratic residue mod p or, in Legendre symbol notation,
$$(\frac{3}{p})=+1$$
Using the quadratic r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to establish the identity How do I establish the identity in this problem? Struggling with this one at the moment.
$$\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\tan\theta}{1-\tan^2\theta}$$
| We have
$$\frac{\tan \theta}{1-\tan^2 \theta}=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
a... | Oh, lol!
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n-1)! + (n-2)!}{3!(n-2)!} = \frac{(n-2)!((n-1) + 1)}{3!(n-2)!}=\frac{n}{3!} = \frac{n}{6}$
Correct? :)
How come Wolfram Alpha gives t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Prove that the following formula is true for $n \geq 1$ by induction Prove that the following formula is true for $n \geq 1$ by induction.
$a_{n} = a_{n-1} + 4n - 3 \\ a_{n} = 2n^{2} - n + 1 \\ a_{1} = 2$
My attempt follows below. I almost succeed in proving the formula for $n \geq 2$, but I fail (more specifically, I ... | I find working with $n$ is better with $k$. So: $$a_n = a_{n-1} + 4n-3=2(n-1)^2-(n-1)+1+4n-3=2n^2-4n+2-n+2+4n-3=2n^2-n+1$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Ramanujan's sum related to $\tan^{-1}(e^{-\pi x/2})$ While reading Ramanujan's Collected Papers I came across a nice formula which he mentions without proof $$\tan^{-1}(e^{-\pi x/2}) = \frac{\pi}{4} - \left(\tan^{-1}\frac{x}{1} - \tan^{-1}\frac{x}{3} + \tan^{-1}\frac{x}{5} - \cdots\right)$$ where $x$ is any real number... | If we differentiate both terms, we get an identity that can be easily proved by considering the logarithmic derivatives of the Weierstrass products for the $\sinh$ and $\cosh$ functions.
We just need to prove:
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)+\frac{z^2}{2n+1}} = \frac{\pi}{4\cosh\frac{\pi z}{2}}.\tag{1}$$
From:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Show that $ \int_{-\infty}^{\infty} \frac{x^3}{(x^2+4)(x^2+1)}\, dx$ does not converge I noticed that $\displaystyle \int_{-a}^{b} \frac{x^3}{(x^2+4)(x^2+1)}$ will converge to $0$ whenever $a=b$ and will converge to some value whenever $a,b$ are in the reals (excluding infinity). How would you show that this integral d... | Beside the simple point from Aloizio Macedo which would apply to many cases, may be, going further, you could use the fact that $$\frac{x^3}{(x^2+4)(x^2+1)}=\frac{4 x}{3 \left(x^2+4\right)}-\frac{x}{3 \left(x^2+1\right)}=\frac 23 \frac{2 x}{x^2+4}-\frac 16 \frac{2 x}{x^2+1}$$ So $$\int\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Estimate from below $\int_0^\pi e^{-t}\cos nt dt$ without calculate it. Estimate from below the following integral
$$\int_0^\pi e^{-t}\cos nt dt$$
without calculate it. Here $n\in\mathbb N$. Any suggestions please?
| A simpler way.
Since
$\cos(z) = \Re(e^{iz})$,
$\begin{array}\\
I(n)
&=\int_0^\pi e^{-t}\cos nt dt\\
&=\Re \int_0^\pi e^{-t}e^{int} dt\\
&=\Re \int_0^\pi e^{-t+int} dt\\
&=\Re \frac{e^{t(-1+in)}}{-1+in}\big|_0^\pi\\
\end{array}
$.
$\begin{array}\\
\frac{e^{t(-1+in)}}{-1+in}
=\frac{e^{-t}+e^{int}}{-1+in}\\
=\frac{e^{-t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this... | Perhaps a little bit less tedious then a binomial expansion to the power of $6$, you could use the triple and double angle identities $\cos 3x=4\cos^3x-3\cos x$ and $\cos 2x=2\cos^2x-1$, so that we have
$$\begin{align}\cos (6x)&=\cos(3\cdot 2x)\\&=4\cos^3(2x)-3\cos(2x)\\&=4(2\cos^2x-1)^3-3(2\cos^2x-1)\\&=4(8\cos^6x-12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
How to proceed with Euclidean algorithm for finding greatest common divisor of two polynomials. I am trying to find
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2).
\end{equation*}
I have done the first step of long division and found.
\begin{equation*}
x^4-x^3-4x^2-x+5=(x^2-2x)(x^2+x-2)-5x+5
\end{equation*}
so we ha... | You are working with polynomials with coefficients in $\Bbb Q$ or in a larger field. $\Bbb Z[x]$ is not a principal ideal domain ($\langle x,2\rangle$ is not principal, for example), and the existence of $\gcd$ is not guaranteed.
That said, $5$ is an unit, and hence $-x+1$ and $-5x+5$ are associated. Therefore, you can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
the series: compute $ \sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4} $ Compute
$$
\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4}
$$
the result is $\frac{\pi^4+30\pi^2-384}{768}$, so I'm sure the sums $\sum\frac{1}{n^2}$ and $\sum\frac{1}{n^4}$ should appear in the solution.
The standard method $\frac{1}{4n^2-1}=\frac{1}{2}(\frac{1}{... | Although it is tedious, you can use partial fractions to get:
$\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(4n^2-1)^4}$ $= \dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right)$ $+\dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^2}+\dfrac{1}{(2n-1)^2}\right)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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A tough limit problem involving $1/(\sin x - \sin a)$ and its generalization Long back I had encountered the following problem in Hardy's Pure Mathematics (originally from the infamous Mathematical Tripos 1896):
If $$f(x) = \frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a}$$ then show that $$\frac{d}{da}\{\lim_{x \to... | $$\begin{align} \lim_{x \to a} f(x) &= \lim_{x\to a}\frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a} \\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)}\\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)} \end{align}$$
Let $ y = x - a$; Since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Scale the distance of lattice points I know that for a hexagonal lattice generated by (0,1) and ($\sqrt{3}/2$,1/2) (i.e., when the distance between lattice points is 1), the number of lattice points in a circle of given radius $r$ can be calculated by the equation below:
$$n(r) = \sum_{x=-\lfloor \frac{r}{\sqrt{3}}\rfl... | Yes, if the first one is correct (which I will take your word for!) because you are simply saying how many points there are in a circle of radius $\frac rd$ instead of radius $r$, which is equivalent to what you are looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Product of two integers of the form $x^2+my^2$ is of the same form. Let $x,y,a,b\in \mathbb Z$. Prove that there are integers $c$ and $d$
so that
\begin{equation*}
(x^2+y^2m)(a^2+b^2m)=c^2+d^2m.
\end{equation*}
I'm stuck, I took the product and got $x^2a^2+b^2y^2m^2+(a^2y^2+x^2b^2)m$ but the numbers in the parenthesis ... | Try the following
$$\begin{align}(ax+mby)^2+m(ay-bx)^2
&=a^2x^2+2maxby+m^2b^2y^2+ma^2y^2-2maxby+mb^2x^2\\
&=(a^2+mb^2)(x^2+my^2)\end{align}$$
The easy way to remember it is the identity with $m=1$
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How can I determine the value of $a_1 + \displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2}$ For reals $x \ge 3$, let $f(x)$ denote the function $f(x) = \frac{-x + x\sqrt{4x - 3}}{2}$. Now suppose that $a_1, a_2, \ldots, a_{2013}$ is a sequence of real numbers such that $a_1 > 3, a_{2... | Let $$\displaystyle A= a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2} = a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3-a_i^3+a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$
$$\displaystyle = a_1 +\sum_{i=1}^{2012} (a_{i+1} -a_i) +\sum_{i=1}^{2012} \frac{a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$
The first sum is solved usi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ We want to find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ for an arbitrary "n".
I have tried writing out a few elements of the sequence as $n \to \infty$:
*
*$A^2 = \left( \begi... | In problems such as these one of the most efficient ways to calculate powers of a matrix is through diference equations. For the matrix of this problem the following is obtained.
Given
\begin{align}
A = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)
\end{align}
it is found that
\begin{align}
A^{2} = \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Find the maximum and minimum of the function $f$ Find the maximum and minimum of $f(x, y)=xy-y+x-1$ at the set $x^2+y^2\leq 2$.
I have done the following:
Since the region $x^2+y^2\leq 2$ is closed, $f$ has a maximum and a minimum, which is either at the boundary or at the critical points of the function.
To find t... | Looks good, I have just a few comments:
"So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema.
"$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the argument I used to evaluate the convergence of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$ right?
If $a,b,c$ be real constants, analyze the convergence of $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$
What I tried to:
I compared the general term of my series to $\frac{1}{n}$: $... | Since if $b\neq c$ then there exists $A,B\in\mathbb{R}$ such that $$\frac{n+a}{(n+b) (n+c)}=\frac{A}{n+b} +\frac{B}{n+c} $$
the series is convergent by Leibniz test.
If $b=c$ then $$\frac{n+a}{(n+b) (n+c)}=\frac{1}{n+b} +\frac{a-b}{(n+b)^2}$$
and the series is also convergent by Leibniz test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Joint probability distribution (over unit circle) A couple of two continuous random variables $(X,Y)$ is distributed uniformly over the closed unity circle (so $-1\leq x \leq 1$ , $y$ analog). $U$ is defined as the distance from $O$ to the point $(X,Y)$. Calculate U:
My calculations so far:
$$
f_{XY}(x,y) = \frac{1}{\... | Answer: On the second line of your derivation.
Let's do it correctly:
$$
\begin{eqnarray}
F_U\left(u\right) &=& \Pr\left(\sqrt{X^2+Y^2} \leqslant u \right) \\
&=& \int_S f_{X,Y}\left(x,y\right) I\left(\sqrt{X^2+Y^2} \leqslant u\right) \mathrm{d}x\mathrm{d}y \\
&=& \left.\int_{-1}^{1} \left(\int_{-\sqrt{1-x^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A recursive sequence is defined by... A sequence is defined recursively by $a_1=1$ and $a_{n+1} = 1 + \frac{1}{1+a_{n}}$.
Find the first eight terms of the sequence $a_n$. What do you notice about the odd terms and the even terms? By considering the odd and even terms separately, show that $a_n$ is convergent and dedu... | We have
$a_{n+1}
= 1 + \frac{1}{1+a_{n}}
= \frac{2+a_n}{1+a_{n}}
$.
Therefore,
$a_n > 1$
for all $n$.
Also,
$\begin{array}\\
a_{n+1}-\sqrt{2}
&= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\
&= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\
&= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\
&= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
prime number and order problem Does anyone can solve problem stated at
Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$?
?
It should have the additional constraint, k < p.
| Let $o$ be the order of $n$ to base $2$.
In other words, $o$ is the smallest positive natural number with $2^o\equiv 1\ \ (\ mod\ n\ )$
Because of $2^{n-1}\equiv 1\ (\ mod\ n\ )$, $o$ must be a divisor of
$n-1=kp^2$.
$o$ cannot be a divisor of $k$, because then we would have $2^k\equiv 1\ (\ mod\ n\ )$
So, $o$ must be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ I want to find the power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ and to find the radius of convergence, what does this take?
Is this just a taylor series with $z=4i$ subbed in? What about finding radius of convergence?
| This is an expansion in powers of $(z-4i)$, so the first step is to locate $z-4i$ in this expression.
$$
\frac 1 {3-((z-4i)+4i)} = \frac 1 {(3-4i) - (z-4i)}.
$$
To get it into the form $\dfrac 1 {1-r}$ we need to divide the numerator and denominator by the constant term in the denominator, which is $3-4i$:
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial deri... | The reciprocal is
$$\frac{x^2+xy+y^2}{(x+y)^2}=1-\frac{xy}{(x+y)^2} $$
and by the AMGM inequality, $\sqrt{xy}\le \frac{x+y}2$ with equality iff $x=y$, hence $\frac{xy}{(x+y)^2}\le \frac14$ with equality iff $x=y$ and from this $\frac{x^2+xy+y^2}{(x+y)^2}\ge \frac 34$ and finally $\frac{(x+y)^2}{x^2+xy+y^2}\le \frac 43... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to distribute $x \times \sqrt{x^2 + 1}$ How do I distribute the x in this problem?
How do I "gain access" so to say. Does it become $x^\frac{1}{2}(x^2 + 1)^\frac{1}{2}$ or $(x^3+x)^\frac{1}{2}$ ?
Or do I need to even do that in order to integrate $\int x(x^2+1)^\frac{1}{2}$ ?
| $x\sqrt{x^2+1} = \dfrac{x(x^2+1)}{\sqrt{x^2+1}}=(x^2+1)\sqrt{\dfrac{x^2}{x^2+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Boolean Algebra fundementals A disjunction A OR B truth table has A , B , and A OR B but mine has A ,B C, with A or B or C could some please explain this
| It's really all the same thing.
$$\begin{array}{c|c|cc}
A & B & A\vee B \\ \hline
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 & \star\end{array}
\qquad
\begin{array}{c|c|c|cc}
A & B & C & A\vee B\vee C \\ \hline
1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 \\
0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, th... | Use this properties: $$\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\tag{1}$$
$$\lim_{x\to\infty}\left[f(x)\pm g(x)\right]=\lim_{x\to\infty}f(x)\pm \lim_{x\to\infty}g(x)\tag{2}$$
if and only if the limit of $f(x)$ and $g(x)$ exists
Therefore:
$$\begin{align}\lim_{x\to\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1 = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But... | Answer for the second question.
We have
\begin{align}
a = k q_{1} + r_{1} \\
a = k q_{2} + r_{2} \\
a = k q_{3} + r_{3} \\
\end{align}
Where $a > b > c \geq k$.
Let suppose that $r_{1} = r_{2} = r_{3}$
We get $a - b = k (q_{1} - q_{2})$; $a - c = k (q_{1} - q_{3})$; and $b - c = k (q_{2} - q_{3})$.
We deduce that $k$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 9
} |
Simple Absolute Value Inequality
$$|\frac{x-1}{x+3}|\leq 2$$
I solve it as follow:
$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{... | $\left|\dfrac{x-1}{x+3}\right| \leq 2 \iff |x-1| \leq 2|x+3| \iff (x-1)^2 \leq 4(x+3)^2 \iff x^2-2x+1 \leq 4(x^2+6x+9) \iff 3x^2 + 26x+35 \geq 0 \iff (x+7)(3x+5) \geq 0 \iff x \leq -7$ or $x \geq -\dfrac{5}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Closed form for the sum $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$ I want a closed form for the sum $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$
where: $a\ne 1<p<n;\quad p\in\mathbb Z$
I know a related identity,
$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1... | Write $n=cp+r$, with $c,r\in\mathbb N\cup\{0\}$, and $r<p$. Then, for $k\leq n$,
$$
\left\lfloor\frac kp\right\rfloor=\begin{cases}0,&\ 0\leq k<p\\ 1,&\ p\leq k<2p,\\
\vdots\\
m,&\ mp\leq k<(m+1)p\\
\vdots\\
c,&\ cp\leq k\leq n
\end{cases}
$$
Then
$$
S=\sum_{k=0}^na^k\,\left\lfloor\frac kp\right\rfloor=\sum_{m=0}^{c-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?
| Considering that the answer is apparently
$$ \frac{7 \left(489857+280140 \sqrt{3}\right)}{8388608}, $$
I doubt there's an easy way.
You know $\sin{\frac{\pi}{6}}=\frac{1}{2}$, $\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}$, so using the half-angle formulae
$$ 2\sin^2{\frac{x}{2}} = 1-\cos{x},\\
2\cos^2{\frac{x}{2}} = 1+\cos{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
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Recursive random draw Let $R(n)$ be a random draw of integers between $0$ and $n − 1$ (inclusive). I
repeatedly apply $R$, starting at $10^{100}$. What’s the expected number of repeated applications until I get zero?
| Just to fill in the algebra in the solution of BruceZ why $E(n) = E(n-1) + \frac{1}{n}, \:\:n > 1:$
Firstly, we re-arrange $E(n) = 1 + \frac{1}{n}\sum_{k=1}^{n-1}E(k)$ to $\sum_{k=1}^{n-1}E(k) = n \cdot [E(n)-1].$ (*)
We also have:
$E(n+1) = 1 + \frac{1}{n+1}\sum_{k=1}^{n}E(k) = 1 + \frac{1}{n+1} \sum_{k=1}^{n-1}E(k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Show that $\cos4x=8\sin^4x-8\sin^2x+1$ using De Moivre's theorem
Use De Moivre's theorem to show that $$\cos4x=8\sin^4x-8\sin^2x+1.$$
Hence show the one of the roots of the equation $8z^4-8z^2+1=0$ is $\sin\frac{\pi}{8}$ and express the other roots in polar form.
Deduce that $\sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqr... | let $\cos x =c $, $\sin x = s$
$\cos4x = (1-s^2)^2-6(1-ss)\cdot s^2+s^4$
$= 1 -2s^2 +s^4 - 6s^2 +6s^4 +s^4$
$= 8s^4 -8s^2 +1$
$\cos 4x = 0 = 8s^4 -8s^2 +1$
$\cos \frac{\pi}{2} =0$
$x= \frac{\pi}{8}$
so $\sin \frac{\pi}{8}$ is a solution
Look at the roots of $8s^4-8s^2 +1$ and use the quadratic equation to solve
we see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}$
How to calculate
$$\lim_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}?$$
I know the result is $1/(6e)$.
| Re-write the logarithm of the original expression as follows
\begin{eqnarray*}
\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}} &=&\frac{1}{x^{2}}\ln \left(
1+\left( \frac{\sin x}{x}-1\right) \right) \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left(
\frac{\sin x}{x}-1\right) }\frac{\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2=1}$ and origin $(1,-i)$? Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2}=1$ and origin $(1,-i)$?
Because I know $|z+i| = 3$ is is a circle with radius $3$ and origin $(0,i)$.
| For finding out the locus of point $z$, let's assume, $z=x+iy$ then substitute it in the relation given as follows
$$|z+i|=|z-1| $$$$ \implies |x+iy+i|=|x+iy-1| $$$$ \implies \sqrt{x^2+(y+1)^2}=\sqrt{(x-1)^2+y^2}$$ $$\implies x^2+(y+1)^2=(x-1)^2+y^2$$
$$\implies x^2+y^2+2y+1=x^2-2x+1+y^2$$ $$\implies \color{blue}{y=-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the locus of midpoint of $AB$ The normal to the ellipse $b^2x^2+a^2y^2=a^2b^2$ is passing through the x-axis in point $A$ and through the Y-axis in point $B$.
Point $P$ is the midpoint of $AB$.
Need to find the locus of $P$.
My attempt :I tried to use the fact of the tangent to the ellipse $n^2=a^2m^2+b^2$ an... | i will parametrize the ellipse by $$x = a \cos t, y = b \sin t $$ the slope of the tangent is $$\frac{dy}{dx} = -\frac{b\cos t}{a \sin t} $$ therefore the slope of the normal is $$\frac{a\sin t}{b \cos t}.$$ the equation of the normal line is $$(y-b\sin t)b \cos t = a \sin t(x -a \cos t) $$ the $x$-intercept is given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation of product of $m+1$ alternate numbers We know that
$$\begin{align}
\sum_{r=1}^n \prod_{j=0}^m (r+j)&=\sum_{r=1}^n r(r+1)(r+2)\cdots(r+m)\\
&=(m+1)!\sum_{r=1}^n\binom {r+m}{m+1}\\
&=(m+1)!\binom {n+m+1}{m+2}\\
&=\frac {(n+m+1)^\underline{m+2}}{m+2}\end{align}$$
but is there a similar closed form for
$$\sum_{... | Simply for my result, I have:
$$\sum_{r=1}^n\prod_{j=0}^m(r+2j)=S(n,m)=\frac{1}{2m+4}\left(\prod_{k=0}^m(2k+1)+\prod_{k=0}^{m+1}(n+2k)+\prod_{k=0}^{m+1}(n+2k-1)\right)$$
That's a polynomial of $n$ with $m+2$ degrees.
The first few values of $m$
\begin{align*}S(n,1)&=\frac{1}{6}\left(\prod_{k=0}^1(2k+1)+\prod_{k=0}^2(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Evaluating the Sum of $ \frac {1} {3}+\frac {1} {15}+\frac {1} {35}+\ldots +\frac {1} {4n^{2}-1} $ How would you evaluate the sum of this sequence ?
$$
\dfrac {1} {3}+\dfrac {1} {15}+\dfrac {1} {35}+\ldots +\dfrac {1} {4n^{2}-1}
$$
I realise the expression can be factorised but I can't really see what this can tell you... | $$\sum_{n=1}^{\infty }\frac{1}{4n^2-1}=\sum_{n=1}^{\infty }\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$$
$$=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....\right]=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluating $\lim\limits_{x\to\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)$ I could transform it to
$$\lim_{x\to\infty}x\left(\dfrac{1}{e}-\left(1+\dfrac{1}{x}\right)^{-x}\right).$$
I think using sandwich theorem would help but I was unable to find the bounds.
| Another way based on Taylor expansions : consider $$A=\left(\dfrac{x}{x+1}\right)^x$$ Taking logarithms $$\log(A)=x\log\left(\dfrac{x}{x+1}\right)=-x\log\left(\dfrac{x+1}{x}\right)=-x\log\left(1+\dfrac{1}{x}\right)$$ Now, remember that, for small $y$, $$\log(1+y)=\frac {y}{1}-\frac {y^2}{2}+\frac {y^3}{3}+\cdots$$ Repl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ For any natural number $m$,how to evaluate $\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ ?
| Let $$\displaystyle I = \int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx\;,$$ Where $x>0$
Now Let $x^{m}=t\;,$ Then $$\displaystyle mx^{m-1}dx = dt\Rightarrow dx = \frac{dt}{mx^{m-1}}=x\frac{dt}{mt} = \frac{t^{\frac{1}{m}}}{mt}dt$$
So $$\displaystyle I = \frac{1}{m}\int (t^3+t^2+t)\cdot (2t^2+3t+6)^{\frac{1}{m}... | {
"language": "en",
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"source": "stackexchange",
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Resolve this system: Im tried to resolve this problem:
$$\max\quad f\left( x,y \right) =xy\quad \text{s.a}\quad \begin{cases} x^2 +y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
Well, i form the lagrangian and the respective gradient, so i had this system to resolve:
$$\begin{cases} y+2\lambda_1 x+\lambda_2=0 \\ x+2\lambda_1 y+... | W/O using calculus,
Let $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$
$\implies \cos A\cos B+\cos A\sin B+\sin A=0\ \ \ \ (1)$
Method $\#1:$
$\iff\dfrac{-\sin A}{\cos B+\sin B}=\dfrac{\cos A}1=\pm\sqrt{\dfrac1{(\cos B+\sin B)^2+1}}$
$\implies\cos^2A=\dfrac1{2+2\cos B\sin B}$
Method $\#2:$
By $(1),\tan A=-(\cos A+\cos B)$
$... | {
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"answer_id": 0
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Prove the inequality $4+xy+yz+zx \ge 7xyz$ Let $x,y,z$ be non negative real numbers such that $x+y+z=3$, prove the following inequality:
$$4+xy+yz+zx \ge 7xyz$$
I tried MV and taking out one variable but I got nothing
| Note that because $x+y+z=3$ we have $1\geq xyz$ and $4\geq 4xyz$.
In addition, $xy+yz+xz \geq 3(xyz)^\frac{2}{3}$ but because $1\geq xyz$ we have $1\geq xyz^\frac{1}{3}$ and $3(xyz)^\frac{2}{3}\geq3xyz$. Hence, $xy+yz+xz \geq 3xyz$.
Adding the two inequalities we have $4+xy+yz+xz \geq 7xyz$.
| {
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Finding a (nonidentity) rational map of the plane with period $7$
Does there exist a nonidentity (which also is not a rotation) rational map $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with period $7$, i.e., for which the seventh iteration $f^7$ is the identity map?
I know such a function, $f:(x,y)\mapsto \left(\frac... | The $2 \times 2$ rotation matrix for an (anticlockwise) rotation by $\frac{\pi}{7}$ is
$$\pmatrix{\cos \frac{\pi}{7} & -\sin \frac{\pi}{7} \\ \sin \frac{\pi}{7} & \cos \frac{\pi}{7}}.$$ Via the usual identification of that group with the group of Mobius transformations (and clearing the cosine factors for readability),... | {
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"source": "stackexchange",
"question_score": "2",
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Solve in positive integers: $5^x 7^y +4=3^z$ Solve in positive integers: $5^x 7^y +4=3^z$.
I tried to solve it with log but I couldn't complete.
| [Based on the work from @Adam Hughes]
We assume $x>1$ and rewrite it:
$$3^z-4=5^x7^y \Rightarrow
3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$
so $z=2j$ is even.
Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j+2)-(3^j-2)=4$--we have that they are coprime (any larger factor w... | {
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Series involving complex roots: $\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = \frac{(n-2)2^{n-1}+1}{2^n - 1}$ $$
\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} =
\frac{(n-2)2^{n-1}+1}{2^n - 1}
$$
Here $1,a_1,a_2,\dots,a_{n-1}$ are $n$-th roots of unity
I know the sum of roots is 0... | Let $y=\dfrac1{2-x}\iff x=\dfrac{2y-1}y$ where $x^n=1$
$\implies\left(\dfrac{2y-1}y\right)^n=1$
$\displaystyle\iff y^n(2^n-1)-\binom n1 2^{n-1}y^{n-1}+\cdots+(-1)^n=0$
Using Vieta's formula, $$\displaystyle\sum_{r=0}^{n-1}y_r=\dfrac{\binom n1 2^{n-1}}{2^n-1}=\dfrac{n 2^{n-1}}{2^n-1}$$
If $x=1,y=1=y_0$(say)
$\displaysty... | {
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"source": "stackexchange",
"question_score": "2",
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Find the fourier series of the function Find the fourier series of the function
$g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)}$ for $x \not= k\pi$, and $g(k\pi) = \lim_{x\to k\pi} g(x)$, $(k \in \mathbb{Z})$
| $g(x)=\sum\limits_{n=1}^\infty \frac{\alpha^n}{sin(x)}(sin(n-2)x cos2x + cos(n-2)x sin2x)$
$ = \sum\limits_{n=1}^\infty \alpha^n\frac{sin(n-2)x}{sin(x)} + 2\sum\limits_{n=1}^\infty \alpha^n cos(n-1)x$
$= -\alpha + \alpha^2\sum\limits_{n=1}^\infty\alpha^n \frac{sin(nx)}{sinx} +
2\sum\limits_{n=1}^\infty\alpha^n cos(n-1... | {
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What is the value of $x*y$?
Given that $$\left(\frac{x}{y}\right)^{-2} + \left(\frac{y}{x}\right)^{-2} = \frac{10}{3}$$
find the value of $x*y$.
My question is, can we calculate the value of $x*y$ or not? If yes, then how? If not, then why?
| Let $u = (\frac{x}{y})^2$.
Then, your equation becomes: $\frac{1}{u} + u = \frac{10}{3}$.
Multiplying both sides by u, we get:
$1 + u^2 = \frac{10}{3} \cdot u$
Now, we have a quadratic equation to solve. Subtract $\frac{10}{3} \cdot u$ from both sides:
$1 + u^2 - \frac{10}{3} \cdot u = 0$
For look's sake, let's put our... | {
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If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$? What I know is that for equations of type $x+y=8$, $xy$ attains its maximum value when $x=y$ and this can be proved by either solving the quadratic equation with completing the squares or finding the first and seco... | You known AM-GM inequality?
$$(a+1)(b+1)(c+1)(d+1)\le\left(\dfrac{a+b+c+d+4}{4}\right)^4$$
if you don't known AM-GM inequality,you also can following
In fact ,if $x,y,z,w\in R^{+}$,then we have
$$x^4+y^4+z^4+w^4-4xyzw\ge 0$$
because
$$x^4+y^4+z^4+w^4-4xyzw=(x^2-y^2)^2+(z^2-w^2)+2(xy-zw)^2$$
so $=$ iff $x=y=z=w$
so
$$4x... | {
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Prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3$ for $x,y,z>0$ By considering that $$\frac{x}{y}+\frac{y}{x} \geq 2$$I can show that $$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y} \geq 6$$
But how would one go from here to prove the required result? It feels like I'm almost there but I... | By AM-GM inequality, since $x,y,z>0$:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3\sqrt[3]{\frac{x}{y}\frac{y}{z}\frac{z}{x}}=3$$
with equality iff $\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$, i.e. $x=y=z$.
Alternatively, rearrangement inequality. $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ is cyclic - two cases:
*
*wlog $x\ge y... | {
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system of modular equations. $x\equiv 2\pmod3$
$x\equiv 3\pmod 5$
$x\equiv 7 \pmod{11}$
How can I solve this system for $x$? I've tried all kinds of things using divisibility but no success. Any hints of solutions are greatly appreciated. What value of $x$ satisfies these three equations?
| When I create new variables, they're integers.
$x\equiv 2\,\pmod{\! 3}\iff x=3k+2$.
$x\equiv 3\,\pmod{\! 5}\iff 3k+2\equiv 3\,\pmod{\! 5}$
$\iff 3k\equiv 1\equiv 6\stackrel{:3}\iff k\equiv 2\,\pmod{\! 5}$.
I.e. $k=5m+2$, then $x=3(5m+2)+2=15m+8$.
Here we could've solved $3k\equiv 1\pmod{\! 5}$ by using Extended Euclide... | {
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$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ From the post Evaluating limit (iterated sine function) and some discussions inside, one can collect the
following three limits
\begin{eqnarray*}
\lim_{x\rightarrow 0 }\frac{\sin _{n}x}{x} &=&... | Continuing in the same spirit as in my answer to the first question, we can probly admit that the limit could be expressed as a cubic polynomial in $n$. Going on with the expansions (tiring !!), I arrived to $${\sin _{n}(x)-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}=-\Big(\frac{5 n^3}{432}-\... | {
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A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:
$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$
What I... | Hint: Simplify the denominator to
$$\frac{\frac{1}{3^{n+1}}-1}{\frac{1}{3}-1}=\frac{3^{n+1}-1}{2\cdot 3^n}$$
| {
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What am I doing wrong? - Change of basis matrix Problem: Let $\alpha$ be the standard basis of $\mathbb{R}^3$ and let $\beta = \left\{(1,0,0), (1,1,0), (1,1,1)\right\}$ be another basis. Consider the linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by \begin{align*} T(x,y,z) = (x+2y+z, -y, x+4z). \end{al... | I'm not sure where exactly $T$ plays a role in this problem. To find $P$, write the vectors in $\alpha$ in terms of $\beta$
$$
\alpha_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}_\beta,
\alpha_2 = \begin{pmatrix}-1\\1\\0\end{pmatrix}_\beta,
\alpha_3 = \begin{pmatrix}0\\-1\\1\end{pmatrix}_\beta
$$
and so
$$
P = \begin{pmatri... | {
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Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.
| Instead of using power series, which feels like a way of hiding the use of derivatives and, ultimately, L'Hôpital, we can use a trigonometric identity and take limits.
Since
$$
\frac{4\sin(x)-2\sin(2x)}{2\sin(2x)-\sin(4x)}
=\frac1{2(1+\cos(x))\cos(x)}\tag{1}
$$
we have
$$
\lim_{x\to0}\frac{2^{k+2}\sin\left(\frac{x}{2^{... | {
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"question_score": "3",
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solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0... | Case $x>3$ :
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \geq 2x(x-3) \Rightarrow x^2-x-12 \geq 2x^2-6x \\ \Rightarrow x^2-5x+12 \leq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So we reject this case. } $
Case $x<3$ :
*... | {
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's wh... | Set $x=1/h$ to get $$\lim_{h\to0^+}\dfrac{\sqrt{1+h}(h-\sqrt{2+3h})}{7h^2-6h+4}=\dfrac{\sqrt{1+0}(0-\sqrt{2+3\cdot0})}{7\cdot0^2-6\cdot0+4}=?$$
| {
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"source": "stackexchange",
"question_score": "2",
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Why am I getting two different answers for this diff. equation? Completing the square vs quad form. With this differential equation, after seperating and integrating, and using the initial condition to solve for C, and then substituting that value of C into the general solution, I must solve for Y to get the particular... | $$\left( y+ \frac{1}{2}\right)^2= x^2-\frac{15}{4} \Rightarrow y+ \frac{1}{2}= \pm \sqrt{\left( x^2-\frac{15}{4}\right) }\Rightarrow y=-\frac{1}{2} \pm \sqrt{\left( x^2-\frac{15}{4}\right)} \\=-\frac{1}{2} \pm \sqrt{ \frac{1}{4}\left( 4x^2-15\right)}= -\frac{1}{2} \pm \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$$
EDIT: B... | {
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Derivative of the matrix exponential with respect to its matrix argument I was trying to find the Frechet derivative of $f = \exp(X)$, where $X \in \mathbb{R}^{n\times n}$ is positive definite. I thought it ought to be $\exp(X)$.
I see results where the derivative is with respect to a scalar argument, but this question... | Let $f : \mathbb R^{n \times n} \to \mathbb R^{n \times n}$ be defined by $f (X) = \exp(X)$. Hence,
$\begin{array}{rl} f (X + h V) &= \exp(X + h V)\\ &= I_n + (X + h V) + \frac{1}{2!} (X + h V)^2 + \frac{1}{3!} (X + h V)^3 + \cdots\\ &= I_n + X + h V + \frac{1}{2!} X^2 + \frac{h}{2!} (X V + V X) + \frac{h^2}{2!} V^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361636",
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Solve equation: $25x+9\sqrt{9x^2-4}=\frac{2}{x}+\frac{18x}{x^2+1}$ Solve equation: $25x+9\sqrt{9x^2-4}=\dfrac{2}{x}+\dfrac{18x}{x^2+1}$
I used wolframalpha.com and got the only solution $x=-\dfrac{1}{\sqrt2}$
And this is my try:
Domain: $|x|\ge\dfrac{2}{3}$
If $x\ge\dfrac{2}{3}$, we have:
$\dfrac{2}{x}+\dfrac{18x}{x^... | rewrite your equation in the form
$$9\sqrt{9x^2-4}=\left(\frac{2}{x}+\frac{18x}{x^2+1}-25x\right)^2$$
this can be written in the form
$$4(-1+2x^2)(1+78x^2+117x^4+13x^6)=0$$
solving this we get $$x=\pm\frac{1}{\sqrt{2}}$$
only the number with $$x=-\frac{1}{\sqrt{2}}$$ is the searched solution.
| {
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This is equation is giving me issues $x^2 - 6x + 15 = 0 $ I was given this equation $x^2 - 6x + 15 = 0 $
I tried to look for numbers whose sum is big and product of ac and i could not find any. I tried using the quadratic formula
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
and this is what i got
$x = \frac{6\pm\sqrt{-24}}{2}$
... | So you need to find the roots as follow:
Note that:
$$ i^2 = -1$$
$$ x = \frac{6\pm \sqrt{-24}}{2}=\frac{6\pm i\sqrt{24}}{2} $$
$$ x= \frac{6\pm 2i\sqrt{6}}{2} = 3\pm i\sqrt{6}$$
Further explanation ($i^2 = -1$):
let $a = \sqrt{-24}$
$a^2 = -24= 24 \times -1= 24i^2$
$a =\pm \sqrt{24} \times \sqrt{i^2}= \pm i\sqrt{24} =... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof question: Prove that 2^(odd integer) + 5^(odd integer) + 2 is a multiple of 3, and 4^(any integer) + 1 can be expressed as 5n, 5n + 1 or 5n + 2. I had been working on the claim in the above question for sometime now. Statistically speaking, it works. For example:
$$\left(2^1 \right)+\left(5^1\right)+2=9,\\
\left(... | Anyone who has studied the $3x + 1$ problem with any degree of detail is aware of this property of numbers of the form $2^n$: either $2^n - 1$ is a multiple of $3$, or $2^n + 1$ is.
Obviously, $2^n$ itself can't be a multiple of $3$ (as long as $n$ is an integer). If $2^n = 3m - 1$, then $2^{n + 1} = (3 - 1)(3m - 1) = ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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a log inequality Can anyone offer some guidance on proving the following inequality? Define $\Lambda_1(a)=-a\log a$ and $\Lambda_2(a,b)=-(a+b)\log(a+b)$. Then if $a$, $b$, $c$, and $d$ are non-negative numbers summing to one, the following holds:
\begin{align}
\Lambda_2(a,b)+\Lambda_2(b,c)+\Lambda_2(c,d)+\Lambda_2(d,a)... | By weighted AM/GM,
\begin{align*}
&\left(\frac{(a+b)(d+a)}{a}\right)^a
\left(\frac{(b+c)(a+b)}{b}\right)^b
\left(\frac{(c+d)(b+c)}{c}\right)^c
\left(\frac{(d+a)(c+d)}{d}\right)^d \\
&\le
a\cdot\frac{(a+b)(d+a)}{a}
+b\cdot\frac{(b+c)(a+b)}{b}
+c\cdot\frac{(c+d)(b+c)}{c}
+d\cdot\frac{(d+a)(c+d)}{d}
\\
&=
(a+b)(d+a)
+(b+c... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about... | HINT
If $z = a+bi$ note that
$$
z^3 = a^3 - 3ab^2 + 3a^2bi - b^3i = (a^3 - 3ab^2) + (3a^2b-b^3)i
$$
so
$$
0 = \bar{z} + z^3 = (a^3 - 3ab^2+a) + (3a^2b-b^3-b)i,
$$
which implies you have 2 equations in 2 unknowns.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root. Let $W(x)$ be a polynomial of degree> 2 having at least three different real roots. How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root?
Whether the assumption of the theorem can be weakened?
| By assumption (a,b,c being different reals)
$$W(x)=(x-a)(x-b)(x-c)Q(x)$$
then
$$W'(x)=((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b))Q(x)+(x-a)(x-b)(x-c)Q'(x) \\
=(3x^2-2(a+b+c)x+ab+bc+ac) Q(x)+(x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc) Q'(x)$$
and
$$W''(x)=(2(x-c)+2(x-a)+2(x-b))Q(x)+(x-a)(x-b)(x-c)Q''(x)+2((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int \frac{\arctan x}{x^4} dx$? I have written the integral as $\int x^{-4} \arctan x dx$. Then, by applying by parts, I got $-3\dfrac{\arctan x}{x^3} + 3\int \dfrac{1}{x^3(1 + x^2)} dx$. Now, how can I solve the later integral? Is there any other trick to do this?
|
$$\int\frac{\arctan x}{x^4}dx$$
By parts:
$$=-\frac{\arctan x}{3x^3}+\frac{1}{3}\int\frac{1}{x^3(x^2+1)}dx$$
$u=x^2,\;du=2xdx$
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\frac{1}{u^2(u+1)}du$$
Partial fractions
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\bigg(\frac{1}{u^2}+\frac{1}{u+1}-\frac{1}{u}\bigg)du$$
$$=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding Linear Combination of Polynomials I am stuck on a question involving finding the greatest common divisor of polynomials and then solving to find the linear combination of them yielding the greatest common divisor. My work thus far is as follows:
Define: $f(x) = x^3-3x+3$, $g(x) = x^2 - 4$.
Execute the Euclidea... | The first two lines are fine. Then you are finished, though you may want to write $x+3=\frac{x+3}{5}\cdot 5+0$.
Work backwards from the second line. We get $5=x^2-4 -(x-3)(x+3)$. But from the first line we have $x+3=x^3-3x+3-x(x^2-4)$. Substituting we get
$$5=x^2-4-(x-3)(x^3-3x+3-x(x^2-4)).$$
This simplifies to
$$5=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $p_{ij}^{(n)}$ for the transition matrix
Let
$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}$$
find $p_{11}^{(n)},p_{12}^{(n)},p_{13}^{(n)}$
Since the characteristic function is $$-\mu^3+\frac{4}{3}\mu^2-\frac{1}{3}\mu=0$$
the eigenval... | Why not just computing $P^2$, then interpolating? We have:
$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad P^2=\begin{bmatrix}\frac{1}{3}&\frac{2}{9}&\frac{4}{9}\\\frac{1}{3}&\frac{4}{9}&\frac{2}{9}\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Consider $$(1+x+x^2+x^3+\cdots)(1+x+^2+x^3+\cdots)$$ Coefficient of $x^k$ in this expansion is just $k+1.$ Because $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Therefore $$(1+x+x^2+x^3+\cdots)^2=1+2x+3x^2+4x^3+\cdots.$$
Now evaluate (you can use the same procedure) $$(1-2x+x^2)(1+2x+3x^2+4x^3+\cdots)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 12
} |
Prove that $\alpha + \beta=\frac {\pi}{2}$ It is given that-
(1) $0<\alpha,\beta<90$.
(2) $\sin^2\alpha+\sin^ 2\beta=\sin(\alpha+\beta).$
Prove that $\alpha + \beta=\frac {\pi}{2}$
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\sin^2A+\sin^2B=1-(\cos^2A-\sin^2B)=1-\cos(A+B)\cos(A-B)$$
$$\implies\sin(A+B)+\cos(A+B)\cos(A-B)=1$$
Using Weierstrass substitution for $\sin(A+B),\cos(A+B)$ and setting $\tan\dfrac{A+B}2=t,$
$$[1+\cos(A-B)]t^2-2t+1-\cos(A-B)=0$$
Using Weierstras... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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