Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluating $\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$ How can we evaluate:
$$\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$$
I tried to use integration by parts, but the positive sign of the $x^2$ in the square root doesn't cancel with the integral of $\sin^{-1}(x)$, so how can I proceed ?
| $\int\dfrac{x\sin^{-1}x}{\sqrt{1+x^2}}dx$
$=\int\dfrac{\sin^{-1}x}{2\sqrt{1+x^2}}d(x^2)$
$=\int\sin^{-1}x~d(\sqrt{1+x^2})$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sqrt{1+x^2}~d(\sin^{-1}x)$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+\sin^2u}}{\sqrt{1-\sin^2u}}d(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
find general solution to the Differential equation Find the general solution to the differential equation
\begin{equation}
\frac{dy}{dx}= 3x^2 y^2 - y^2
\end{equation}
I get
\begin{equation}
y=6xy^2 + 6x^2 y\frac{dy}{dx} - 2y\frac{dy}{dx}
\end{equation}
rearrange the equation
\begin{equation}
\frac{dy}{dx} = \frac{y-... | You might use separating variable method: you have: $\frac{1}{y^2}dy = (3x^2 -1)dx$, and you integrate both sides to get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c$ are positive integers and $a^2+b^2=c^2$ and $a$ is a prime, what can we conclude about primeness of b and c? Let $a,b,c$ be positive integers and they satisfy $a^2+b^2=c^2$, and if $a$ is prime, can we conclude whether $b$ and $c$, are both prime, composite or neither? If yes, why, if not why not?
I can conc... | $c$ is prime and $b$ isn't :
$$3^2+4^2=5^2$$
both are composite :
$$7^2+24^2=25^2$$
However, $b$ can never be a prime. Proof :
$a=2$ or $b=2$ is impossible (first pythagorean triplet is $3,4,5$)
if both $a$ and $b$ are odd prime numbers, then $a^2\equiv b^2\equiv 1 \mod 4$ (this is true for all odd numbers), and ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$
My try: since
$$\Longleftrightarrow\sum_{sy... | I'll solve the three variable case (the $a,b,c,d,e$ case is similar).
Assume without loss of generality $a \geq b \geq c$. Multiplying by the denominators we obtain
$$
a^3+a^2 c+a b^2-6 a b c+b^3+b c^2+c^3 \geq 0
$$
Now use the rearrangment inequality twice. We can split the inequality in two parts:
$$
a^2c+b^2a+c^2b \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve in $\mathbb Z$ the equation: $x^5 +15xy + y^5=1$ Solve in $\mathbb Z$ the equation: $x^5 +15xy + y^5=1$
I tried: $x(15y+x^4)+y^5=1$
But don't have much ideas on how to continue, thanks!
| Let's find all pairs $(x,y) \in \mathbb{Z} \times \mathbb{Z} $ that satisfies
$$\tag{*}\label{main-poly} x^5 +15xy + y^5=1$$
It's obvious that $(1,0)$ and $(0,1)$ are solutions of $\eqref{main-poly}$ , because if one of $x,y$ equals $0$, then the other one must be equal to $1$.
It is also obvious, that if $x,y \neq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it i... | You can also use that
$$2\sin(x)\Bigl[\cos 2x+\cos 4x + \cos 6x+\cos 8x\Bigr] = \sin 9x-\sin x = 2\cos 5x\sin 4x$$
so that inserting $x=\tfrac\pi5=\pi-4\tfrac\pi5$ yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
Trigonometry Identity Problem Prove that:
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{ \cos A}$$
I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/cos A and then simplifying but it didn't work.
| You may also simplify the expression by using the definition of the tangent and the secant function:
\begin{align}
&\dfrac{\tan A +\sec A-1}{\tan A - \sec A +1}
\\=&\dfrac{\dfrac{\sin A}{\cos A} +\dfrac1{\cos A}-1}{\dfrac{\sin A}{\cos A} - \dfrac1{\cos A} +1}
\\=&\dfrac{\sin A -\cos A+1}{\sin A+\cos A-1}
\\=&\dfrac{(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$ Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$
Solution :
$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$
and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$
$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$
Can we have othe... | As the Rotation of the axes does not change the eccentricity of a curve
set $2y-x-3=X,2x+y−1 =Y$ so that the given equation becomes $$4X^2-9Y^2=80\iff \frac{X^2}{\frac{80}4}-\frac{Y^2}{\frac{80}9}=1$$
Now we know, $$b^2=a^2(e^2-1)$$ where $a,b\le a$ are the semi-major & semi-minor axes and $e$ is the eccentricity of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove the series identity
Prove an identity:
$$\sum_{n=2}^{ \infty } \frac{2}{(n^3-n)3^n}=- \frac{1}{2}+ \frac{4}{3} \cdot \sum_{n=1}^{ \infty } \frac{1}{n \cdot 3^n}$$
I've checked that the left-hand-side of this identity is convergent absolutely, hence I can write it as:
$$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+... | $$\sum_{n=2}^\infty{2\over(n+1)(n-1)n3^n}=\sum_{n=2}^\infty \left[{1\over (n-1)n}-{1\over (n+1)n}\right]{1\over 3^n}=\sum_{n=2}^\infty\left[\color{red}{{1\over (n-1)}}-{2\over n}+\color{blue}{1\over (n+1)}\right]{1\over 3^n}=\sum_{n=2}^\infty\left[\color{red}{{1\over 3(n-1)3^{n-1}}}-{2\over n3^n}+\color{blue}{3\over (n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability the three points on a circle will be on the same semi-circle Three points are chosen at random on a circle. What is the probability that they are on the same semi circle?
If I have two portions $x$ and $y$, then $x+y= \pi r$...if the projected angles are $c_1$ and $c_2$. then it will imply that $c_1+c_2=\pi... | Without loss of generality, we can assume the circumference of the circle to be equal to $1$. Cut the circle at the first point $A$ and spread it out as a line. Let the other two points $B$ and $C$ be located at distances of $x$ and $y$ from $A$. The midpoint of this line is $M$ and $0 \leq x,y \leq 1$.
Points $B$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$ Find $$\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$$ by hand. I'm aware Mathematics gives $\frac{\pi^2}{4}$, but I need to learn this without the aid of Mathematica. I tried using substitutions like $u= \tan \frac{x}{2}$ and some trig identities, but I still can't ... | I prove lemma that says that $$\int_0^\pi xf(\sin x)dx=\frac{\pi}2\int_0^\pi f(\sin x)dx$$
let $t=\pi-x$,so
\begin{align*}
\int_0^\pi xf(\sin x)dx&=\int_\pi^0(\pi-t)f(\sin(\pi-t))(-dt)=\int_0^\pi(\pi-t)f(\sin t)dt\\
&=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi tf(\sin t)dt
\end{align*}
so
$$\int_0^\pi xf(\sin x)dx=\dfrac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Divisibility Of Positve Integers Suppose a,b and c are three positive integers which satisfy the condition that ($a$2+$b$2+$c$2) is divisible by $(a+b+c)$.
Prove that there exists infinitely many positive integers $n$ for which ($a$n+$b$n+$c$n) is also divisible by$(a+b+c)$.
| We have that $(a^{2^n}+b^{2^n}+c^{2^n})$ is divisible by $(a+b+c)$ for each $n\geq 0$.
The proof is by induction on $n$. The base cases $n=0$ and $n=1$ are given to us; suppose that it is true for $n$ and $n-1$. Then $$2(a^{2^{n-1}}b^{2^{n-1}}+a^{2^{n-1}}c^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}) = (a^{2^{n-1}}+b^{2^{n-1}}+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding relatives of the series $\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}$. Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$
$$
\begin{align*}
\varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\
\varphi &=\frac{... | Yes, you are right. There is a family. A little experimentation shows that,
$$\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{C_{k}}{2^{4k+3}}$$
$$\varphi =\frac{13}{2^3}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}$$
$$\varphi =\frac{207}{2^7}+\sum_{k=0}^{\infty}(-1)^{k+2}\frac{C_{k+2}}{2^{4k+11}}$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $ I can't figure out how to prove the following inequality:
$$
1/2 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2}
$$
Thanks
| Since $\cos\frac{\pi}{7}$ is a root of the Chebyshev polynomial $U_6(x)$
$$U_6(x) = 64x^6-80x^4+24x^2-1$$
we have
$$\frac{1}{4\cos^2\frac{\pi}{7}}=16\cos^4\frac{\pi}{7}-20\cos^2\frac{\pi}{7}+6,$$
so:
$$4\sin^2\frac{\pi}{14}+\frac{1}{\cos^2\frac{\pi}{7}}=4-2\cos\frac{\pi}{7}-2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7},$$
or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of the series. I need to find the following sum:
$$\sum_{s=0}^{n+1}{(-1)}^{n-s}4^s\binom{n+s+1}{2s}$$
First I tried to simplify this:
$$\begin{split}
\sum_{s=0}^{n+1}{(-1)}^{n-s}4^s\binom{n+s+1}{2s}
&= {(-1)}^n\sum_{s=0}^{n+1}{(-1)}^{s}2^{2s}\binom{n+s+1}{2s} \\
&= \left[{(-1)}^{m-1}\sum_{s=0}^m{(-1)}^{s... | Suppose we seek to evaluate
$$\sum_{q=0}^{n+1} (-1)^{n-q} 4^q {n+q+1\choose 2q}
= (-1)^n \sum_{q=0}^{n+1} (-1)^{q} 4^q {n+q+1\choose n+1-q}.$$
We use the integral
$${n+q+1\choose n+1-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+q+1}}{z^{n-q+2}} \; dz.$$
This has the property that it is zero when $q\gt n+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Help with limit calculation Can anyone help me with this limit please:
I have been trying to solve this for 2 hours with no success:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$
| We don't need to make all the detailed calculations as the highest exponent dominates when $n\to\infty$
$$1+4+7+...+(3n-2)=\frac n2\left(1+3n-2\right)=\frac{3n^2-n}2$$
$$\implies \left(1+4+7+...+(3n-2)\right)^2=\left(\frac{3n^2-n}2\right)^2=\frac{9n^4}4+O(n^3)$$
$$1^3+4^3+7^3+...+(3n-2)^3=\sum_{1\le r\le n}(3r-2)^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{c_n(a_1,\ldots,a_n)}{c_{n-1}(a_2,\ldots,a_n)}=a_1 + \frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_{n-1}+\frac{1}{a_n}}}}$ For $n>0$ and $a_1,...,a_n \in K$ let $c_n(a_1,...,a_n)$ be the determinant of the matrix
$$
\begin{pmatrix}
a_1 & 1 & 0 & \cdots & 0 \\
-1 & a_2 & \ddots & \ddots & \vdots \\
0 &... |
but I already fail at the initial step
Not really. You have
$$\frac{c_2(a_1,a_2)}{c_1(a_2)} = \frac{a_1a_2+1}{a_2} = \frac{a_1a_2}{a_2} + \frac{1}{a_2} = a_1 + \frac{1}{a_2}.$$
In your induction step, you also have the necessary ingredients,
$$c_{n+1}(a_1,\dotsc,a_{n+1}) = a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How find this postive integer $a$ such $a(x^2+y^2)=x^2y^2$ always have roots Find all postive integer numbers of $a$,such this equation
$$a(x^2+y^2)=x^2y^2,xy\neq0$$
always have integer roots $(x,y)$
my try: since
$$\dfrac{x^2y^2}{x^2+y^2}\in N$$
and I can't
Thank you
| Take any two positive coprime integers $u$ and $v$, and a third positive integer $t$.
Let $q$ be the square-free part of $u^2 + v^2$: thus $u^2 + v^2 = qs^2$ for some positive integer $s$.
Let $a = u^2 v^2 t^2 q$.
Then the equation $a(x^2+y^2) = x^2y^2$ has the non-trivial solution $x = tsqu$ and $y = tsqv$ because
$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Singular Value Decomposition: Nonsquare matrix, third column of U My textbook says the singular value decomposition of the below matrix:
$A = \left( \begin{matrix} -2&1\\ 1&2 \\ 1 & 2 \end{matrix} \right)$
is given by:
$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-... | It seems like there are some typos in $V^T$, the result should be:
$$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \sqrt{10}&0\\0&\sqrt{5}\\0&0\end{matrix}\right)\left( \begin{matrix} \frac{1}{\sqrt{5}}&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
S... | You want to prove that
$$P(k):\qquad \sum_{i=1}^k{1\over i^2}\leq 2-{1\over k}$$
implies
$$P(k+1):\qquad \sum_{i=1}^{k+1}{1\over i^2}\leq 2-{1\over k+1}\ .$$
Therefore we have to prove that
$$\left(2-{1\over k}\right)+{1\over (k+1)^2}\leq 2-{1\over k+1}\ ,$$
which is the same as
$${1\over (k+1)^2}\leq {1\over k}-{1\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/638418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is completing the square? Why is it called, completing the square? Is square metaphorical in this sense? How do you complete the square and what is it used for?
Thank you, regards.
| Consider the quadratic term $x^2+2x-11$. The goal of completing the square is to write this term in a form where $x$ appears only once and the idea behind this is to use the binomial formula $(a+b)^2=a^2+2ab+b^2$ in opposite direction. So we set $a=x$ and then it looks like $2ab$ should be $2x$ and since $a=x$ we get $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the signature of the quadratic form Very simple question but something doesn't make sense to me.
We are given a quadratic form (bilinear map but on the same vector twice):
$Q(v) = v^t *\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}*v$
and we are asked to find the signature....But this matrix isn'... | Think I got it.
Suppose $u=\begin{pmatrix} x \\ y \\ z\end{pmatrix}$.
$Q(u)=\begin{pmatrix} x & y & z\end{pmatrix} \begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\1 & 1 & 0\end{pmatrix} \begin{pmatrix} x\\y\\z\end{pmatrix} = \begin{pmatrix}x+z \\ x+y+z \\y\end{pmatrix} \begin{pmatrix} x \\y\\z\end{pmatrix}=x^2+xz+xy+y^2+2yz$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic congruence relation Is there a general formula for solving quadratic congruence relation like $ax^2-bx+c = 0$ in $\mathbb{Z}_{n}$ where $n \in \mathbb{N}$?
For example, I am trying to solve $x^2-3x+2 = 0$ in $\mathbb{Z}_{200}$. What kind of tools I need to solve it?
| There are techniques, but no general formula. In this case there is a method, based on the factorization $(x-1)(x-2)$.
For any $x$, the numbers $x-1$ and $x-2$ are relatively prime. Note that $200=2^3\cdot 5^2$, and work modulo $8$ and $25$ separately.
Since $x-1$ and $x-2$ are relatively prime, we have $(x-1)(x-2)\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
conditions on $\{a_n\}$ that imply convergence of $\sum_{n=1}^{\infty} a_n$ (NBHM 2011) Question is :
For a sequence $\{a_n\}$ of positive terms, Pick out the cases which imply convergence of $\sum_{n=1}^{\infty} a_n$.
*
*$\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$
*$\sum_{n=1}^{\infty} n^2a_n^2<\... | For (a) , if we take $$\{b_n\}=\frac{1}{n^{\frac{3}{2}}}$$, then $$\lim_{
n\to \infty}\frac{a_n}{b_n}=lim_{n\to \infty}n^{\frac{3}{2}}a_n=\frac{3}{2}$$ which is finite . Hence by limit comparison test $\{a_n\}$ converges.
For (b) since $$\sum({na_n})(\frac{1}{n})\le \{\sum n^2a_n^2\}^{\frac{1}{2}}\{ \sum\frac{1}{n^2}\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? Is the series $\sum_{n=1}^\infty \frac{1}{n^3+n}$ convergent or divergent? And how? What method is required? Thanks.
| $\int_1^\infty \frac{1}{x^3+x}dx$
$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}$
Therfore, $A+B=0, C=0, A=1$, then $B=-1$.
$\int \frac{1}{x^3+x}dx=\int \frac{dx}{x}-\int \frac{x dx}{x^2+1}=\ln x -\frac{1}{2}\ln (x^2+1)=\ln\frac{x}{\sqrt{x^2+1}}$
For $\int_1^\infty \frac{1}{x^3+x}dx=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/641281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How prove this $\prod\limits_{cyc}(a^3+2b+\frac{2}{a^2+1})\ge 64$ let $a,b,c>0$ and such
$$abc\ge 1$$
show that
$$\left(a^3+2b+\dfrac{2}{a^2+1}\right)\left(b^3+2c+\dfrac{2}{b^2+1}\right)\left(c^3+2a+\dfrac{2}{c^2+1}\right)\ge 64$$
my try:
$$\sum_{cyc}\ln{\left(a^3+2b+\dfrac{2}{a^2+1}\right)}\ge 6\ln{2}$$
Then I can't... | Hint: first prove that $a^3+\frac{2}{a^2+1} \geq 2a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/642523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the given inequality $$\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $$
$$\implies \frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
Given if $A>B>C$ and $A+B+C=180^\circ$
Is that implication correct if not then please correct it otherwise try to solve the first inequality.
This is not the original problem, b... | The original inequality is $\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $
And we have to prove this $\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$.
Now we have to interchange only the positions of $tan(B-C)$ and $sin^{2}B$. $sin^{2}B$ is always positive since it is a square. We need to care only about $tan(B-C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to solve this trigonometric identity? I have tried it several times and I do not know what I am doing wrong. This is the identity:
$$\csc^2x - \csc x = \frac{\cot^2 x}{1 + \sin x}$$
Trying with RHS, I get:
$$\frac{\cos^2 x }{ \sin^2 x} + \frac{\cos^2 x }{ \sin x}$$
With LHS:
$$\frac{1 }{ \sin^2 x} - \frac{\sin x ... | $$\frac{1}{sin^2x}-\frac{1}{sinx}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}-\frac{sinx}{sin^2x}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}(1-{sinx})=\frac{1}{sin^2x}(\frac{cos^2x}{1+sinx})$$
$$\frac{1}{sin^2x}(1-{sinx})(1+sinx)=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(1-{sin^2x})=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determine appropriate $c$ and $x_0$ for Big-O proofs. "Prove that $f(x)$ is $O(x^2)$:"
$$f(x) = \frac{x^4+2x-7}{2x^2-x-1}$$
Let $c=10$ (addition of coefficients of the numerator less the addition of coefficients of the denominator), and $x_0 = 1$ (the lowest coefficient among all elements in the function).
$$\frac{x^4+... | Note that
$$\frac{x^4 + 2x -7}{2x^2 - x - 1} \leq \frac{x^4 + 2x^4 -7}{2x^2 - x - 1} \leq \frac{3x^4}{2x^2 - x - 1} \leq \frac{3x^4}{x^2} = 3x^2.$$
The last inequality holds true when
$$2x^2 - x - 1 \geq x^2, $$
that is, for all $x \geq \frac{1 + \sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/646999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of a power series $n x^n$ I would like to know:
How come that
$$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$
Why isn't it infinity?
| Using the ratio test, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)x^{n+1}}{nx^n} = \dfrac{(n+1)x}{n}$. So, the series converges when $|x|<1$.
$$F(x) = \sum_{n=1}^\infty n x^n = x + 2x^2 + 3x^3 + ...$$
$$xF(x) = \sum_{n=1}^\infty n x^{n+1} = x^2 + 2x^3 + 3x^4 + ...$$
$$F(x) - xF(x) = x + x^2 + x^3 + x^4... = \sum_{n=1}^\infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/647587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Determine $2$ missing digits for modulo $11$ An account number verification system works as follows:
All digits in a 10-digit account number all multiplied by following weights:
$$(6 \ 3 \ 7 \ 9 \ 10 \ 5 \ 8 \ 4 \ 2 \ 1)$$
Resulting numbers are summed and divided by $11$. The result must be $0$.
Example:
Account number... | It amounts to solving $\ \color{#c00}8y \equiv -k - 5x\!\pmod{11}\,$ for $\,x =8\,$ or $\,9.\,$ and $\,k=$ sum from other digits.
Notice that $\,{\rm mod}\ 11\!:\,\ 1/\color{#c00}8 \equiv 12/{-}3 = -4,\,$ so $\, y \equiv 4k+20x\equiv 4k-2x \equiv \smash[b]{\underbrace{4k+6}_{\large x=8},\ \underbrace{4k+4}_{\large x=9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$ $a$,$b$,$c$ are non-negative real numbers such that $~x^{2}+y^{2}+z^{2}=1$
show that $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
| By C-S we obtain:
$$\sum_{cyc}\left(\frac{1}{y^2+z^2}+\frac{1}{1-yz}\right)=\sum_{cyc}\left(\frac{(3x+y+z)^2}{(3x+y+z)^2(1-x^2)}+\frac{(y+z)^2}{(y+z)^2(1-yz)}\right)\geq$$
$$\geq\frac{25(x+y+z)^2}{\sum\limits_{cyc}(3x+y+z)^2(1-x^2)}+\frac{4(x+y+z)^2}{\sum\limits_{cyc}(y+z)^2(1-yz)}.$$
Thus, it remains to prove that
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given $x^2 + y^2 + z^2 = 3$ prove that $x/\sqrt{x^2+y+z} + y/\sqrt{y^2+x+z} + z/\sqrt{z^2+x+z} \le \sqrt3$ Given $x^2 + y^2 + z^2 = 3$
Then prove that $${x\over\sqrt{x^2+y+z}} + {y\over\sqrt{y^2+x+z}} + {z\over\sqrt{z^2+x+y}} \le \sqrt 3$$
I tried using the Cauchy-Schwarz inequality but the inequality is coming in oppo... | Using the Cauchy-Schwarz inequality, we have
$$(1+1+1)(x^2+y^2+z^2)\ge (x+y+z)^2\Longrightarrow x^2+y^2+z^2\ge x+y+z$$
and using the Cauchy-Schwarz inequality, we have
$$(x^2+y+z)(1+y+z)\ge (x+y+z)^2\Longrightarrow \dfrac{\sqrt{1+y+z}}{x+y+z}\ge\dfrac{1}{\sqrt{x^2+y+z}}$$
so
$$\Longleftrightarrow\dfrac{x\sqrt{1+y+z}+y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalues in orthogonal matrices Let $A \in M_n(\Bbb R)$.
How can I prove, that
1) if $ \forall {b \in \Bbb R^n}, b^{t}Ab>0$, then all eigenvalues $>0$.
2) if $A$ is orthogonal, then all eigenvalues are equal to $-1$ or $1$
| The first part of the problem is well solved above, so I want to emphasize on the second part, which was partially solved.
An orthogonal transformation is either a rotation or a reflection. I will focus on 3D which has lots of practical use. Let us then assume that $A$ is an orthonormal matrix in $\mathbb{R}^3 \times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 3
} |
Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere.
Hints are also welcome.
| Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=a^2b+abc+c^2b\leq a^2b+2abc+c^2b=b(a+c)^2=4\cdot b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also ... | Square both sides:
$$x + 2 \sqrt{xy} + y = 1.$$
Isolate the radical and square:
$$xy = \left(\frac{1-x-y}{2}\right)^2.$$
Rotate the axes by $\pi/4$ to new axes ${X,Y}$:
$$\frac{Y^2-X^2}{2} = \left(\frac{1 - \sqrt{2}Y}{2}\right)^2.$$
Simplfying gives:
$$X^2 = 2\sqrt{2}Y - 1,$$
which is the equation for a parabola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the... | There is a nice point that tells $$D^n\{e^{kx}f(x)\}=e^{kx}(D+k)^nf(x)$$ It can be proved by an inductive approach on $n$. By using it we get: $$D\left(e^x\sin x\right)=e^x(D+1)\sin x=e^x(\cos x+\sin x)=\sqrt{2}e^x\sin(x+\pi/4)\\D^2\left(e^x\sin x\right)=D(\sqrt{2}e^x\sin(x+\pi/4))=\sqrt{2}e^x(D+1)\sin(x+\pi/4)=\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$. I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is?
Pr... | Hint $\ $ You cannot divide by $\,2\,$ since $\,2 = 0\,$ in $\,\Bbb F_{2^n}.\,$ Instead, notice that
$\, 0 = (a-b)(a^2+ab+b^2) = a^3-b^3.\,$ If $\,a\,$ or $\,b\neq 0,\,$ wlog. $\,b\ne 0,\,$ then $\,c^3 = 1\,$ for $\,c = a/b.\,$ If $\,c \ne 1\,$ then $\,c\,$ has order $= 3,\,$ so Lagrange's Theorem $\Rightarrow 3\,$ di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there a shortcut for taking this matrix to a certain power? One can take a diagonal matrix to a certain power by just taking diagonal elements that power. There is a similar polynomial-time (in respect to the matrix dimensions) shortcut for triangular matrices.
Is there a shortcut for exponentiating a square matrix ... | By using Cayley Hamilton Theorem. If you are doing the calculations by hand, then the problem is simple if the characteristic polynomial has simple roots.
The general approach is to divide $x^n$ by $x^4-a x^3 - b x^2 - c x - d$ find the remainder. Thus suppose
$$
x^n = Q(x) (x^4-a x^3 - b x^2 - c x - d) + { \alpha x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/662820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate of number represented by the infinite series $\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$. Evaluate of number represented by the infinite series
$$\sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \sqrt{\tfrac{1}{3} + \cdots }}}$$
| $n > 0:$
Let $x = \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}}$
Assuming convergence:
$$\begin{align*} x = \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}} & \implies x^2 = n + \sqrt{n + \sqrt{n + \sqrt{n + \cdots }}} = n + x \\ & \implies x^2 - x - n = 0\end{align*}$$
This is a simple quadratic to solve (take the positive root as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/663798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove this inequality $\sqrt{\prod\limits_{cyc}(a+\sqrt{\frac{bcd}{a}})}+2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd$ Let $a,b,c,d>0$. Show that
$$\sqrt{\left(a+\sqrt{\dfrac{bcd}{a}}\right)\left(b+\sqrt{\dfrac{acd}{b}}\right)\left(c+\sqrt{\dfrac{abd}{c}}\right)\left(d+\sqrt{\dfrac{abc}{d}}\right)}+2\sqrt{abcd}\ge ab+bc+c... | WLOG, assume $abcd = 1$. The desired inequality is written as
$$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} + 2 \ge ab + bc + cd + da + ac + bd.$$
Since $ab + bc + cd + da + ac + bd \ge 6$, it suffices to prove that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1) - (ab + bc + cd + da + ac + bd - 2)^2 \ge 0.$$
Using $d = \frac{1}{abc}$, we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Why these integrals are evaluated differently? $\cos(x)/(a-b\cos(x))$ In attempt to solve electrostatics problem I came up to this integral that I am trying to integrate:
$$\int_0^{2\pi}\frac{\cos(x)}{a-b\cos(x)} \, dx$$ where $a>b$ and both are real numbers.
For the domain $[0, 2\pi]$ this function is symmetric at $\p... | These integrals are very similar to the ones for which Euler introduced the tangent half-angle substitution, sometimes erroneously called the "Weierstrass substitution" (I hope the quotation marks are intimidating . . . . .)
\begin{align}
t & = \tan\frac x 2 \\[8pt]
2\arctan(t) & = x \\[8pt]
\frac{2\,dt}{1+t^2} & = dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/666275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin... | Alternative solution
$a = |OA| = \sqrt{1^2 + 5^2 + 7^2} = \sqrt{75} = 3 \sqrt{5}$
$b = |AB| = \sqrt{(10-1)^2 + (10--5)^2 + (5--7)^2} = \sqrt{450} = 15\sqrt{2}$
$c = |BO| = \sqrt{10^2+10^2+5^2} = 15.$
Now you can calculate the semiperimeter $s$ which is just $\frac{1}{2} \left(3 \sqrt{5} + 15 \sqrt2 + 15 \right)$, and u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/667560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show that $e^x > 1 + x + x^2/2! + \cdots + x^n/n!$ for $n \geq 0$, $x > 0$ by induction Show that if $n \geq 0$ and $x>0$, then
$$ e^x > 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}.$$
Not sure where to get started with this induction proof.
| Base case: $e^x > 1$ for $x > 0$.
Induction: suppose we are given $k$ such that for all $x > 0$
$$
e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!}
$$
Change variables to $t$ and integrate both sides:
$$
\int_0^x e^t \; dt > \int_0^x \left( 1 + t + \frac{t^2}{2!} + \cdots + \frac{t^k}{k!} \right) \; dt
$$
$$
e^x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/667998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Alternative ways to solve $\frac{\sin{\theta}+1}{\cos{\theta}}\leq 1$? The inequality is:
$$ \frac{\sin{\theta}+1}{\cos{\theta}}\leq 1\quad \text{ with } \cos{\theta}\neq0 \land 0\leq \theta\lt 2\pi$$
I've tried splitting it up into cases of $\theta$ that make $\cos{\theta}$ positive so: $0\leq\theta<\frac{\pi}{2}$ a... | \begin{align}
1 + \sin \theta &= ( \cos (\theta /2) + \sin (\theta /2))^2 \\
\cos \theta &= ( \cos ^2 (\theta /2) - \sin ^2 (\theta /2) \\
\frac{1 + \sin \theta }{\cos \theta } &= \frac{ \cos (\theta /2) + \sin (\theta /2) }{ \cos (\theta /2) - \sin (\theta /2)} = \frac{ 1+ \tan (\theta /2)}{1 - \tan (\theta /2)}
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
solving $\int x^7\sqrt{3+2x^4}dx$ I'm trying to solve $\int x^7\sqrt{3+2x^4}dx$
All I have so far is
Let $u$ = $3+2x^4$
$du$ = $8x^3$ $dx$
$\frac{du}{8x^3}$ = $dx$
Therefore,
$\int x^7\sqrt{u}$ $\frac{du}{8x^3}$
$\frac{1}{8}$$\int x^4\sqrt{u}$ ${du}$
Since there is still a $x$ variable in the integral, I'm not sure wh... | Given $\int x^7 \sqrt{3+2x^4} dx$. Assume that $3+2x^4= u$ then $8x^3 dx=du$. Hence the givenm integral becomes $\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}$ which is \begin{align*}
&\int \sqrt{u}\frac{u-3}{2}\frac{du}{8}\\
=&\frac{1}{16}\int (u^{3/2}-3u^{1/2})du\\
=&\frac{1}{16}(\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2})+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/670447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the surface area obtained by rotating $y=1+3x^2$ from $x=0$ to $x=2$ about the y-axis. Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.
Having trouble evaluating the integral:
Solved for $x$:
*
*$x=0, y=1$
*$x=2, y=13$
$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot... | Note that $((\sqrt{\frac{y-1}{3}})')^2=(\frac{1}{2\sqrt{\frac{y-1}{3}}} (1/3))^2=\frac{1}{4(y-1)}$
$$=\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\frac{1}{4(y-1)}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y-1}3+\frac{1}{12}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y}3-\frac{1}{4}}dy
\\= 6\pi\{(\frac{y}3-\frac{1}{4})^\frac{3}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric identity for $a\sin 2x + b\sin(x+\alpha )$ The following is a known trigonometric identity,
$$a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,$$
where
$$c=\sqrt {a^2+b^2+2ab\cos \alpha },\,$$
and
$$\beta =\arctan \left(\frac {b\sin \alpha }{a+b\cos \alpha }\right) + \begin{cases} 0 & \text{if } a+b \cos \alpha ... | As for the last question:
\begin{align*}
a\sin2x-b\sin(x+\alpha)&=0 \\
2a\sin x\cos x - b(\sin x\cos\alpha + \cos x\sin\alpha)&=0\\
-b\cos\alpha\sin x &= \cos x \cdot(b\sin\alpha-2a\sin x) \\
\frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} &= \cos x \\
\bigg( \frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} \bigg)^2 &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Arc length of parametric equations Find the arc length of the parametrized path $x(t) = t^2/2$ and $y(t) = t^3/3$ for $1 \le t \le 3$.
I try to do the standard integral of $\sqrt{\mathrm dx^2 + \mathrm dy^2}$, but get stuck when I have to integrate $\sqrt{t^2 + t^4}$.
Any ideas?
| $$
\int\mathrm{d}t \sqrt{t^2+t^4} = \int\mathrm{d}t\ \ t\sqrt{1+t^2}
$$
for positive t. Now let $\sqrt{1+t^2}=y-t$:
$$
1+t^2=y^2+t^2-2yt
$$
$$
t=\frac{y^2-1}{2y}
$$
$$
\mathrm{d}t=\mathrm{d}y \frac{y^2+1}{2y^2}
$$
and
$$
\sqrt{1+t^2}=\frac{y^2+1}{2y}.
$$
Substituting yields:
$$
\int \mathrm{d}y \frac{y^2+1}{2y^2} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Adding fractions is not at all obvious Why does $\frac{5}{4} + \frac{2}{3}$ need to be rewritten as $\frac{15}{12} + \frac{8}{12}$ to be added? It's not obvious.
I'm looking towards the fact that any integer can be rewritten as $x=qy$ but these work for rational numbers as well.
Can anyone clarify why exactly fraction... | First, we have to define: what is a fraction? For now, we will just say it is an ordered pair $(a, b)$, where $b \ne 0$. We identify $(a, 1)$ with the integer $a$. (The notation doesn't matter, I just chose this so that we don't accidentally use identities we already know)
Here is what we want:
*
*$(a, 1) + (b, 1) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to solve this trig problem? $\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$ Basic trig problem my brother ask me, but I don't know how to do it:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$
| Since $\sec(x) = 1/\cos(x)$, let's first consider
$$\cos\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right].$$
Using the addition formula for $\cos$, this is equal to
$$\cos(\sin^{-1}(-5/13)) \cos(\tan^{-1}(4/3)) + \sin(\sin^{-1}(-5/13)) \sin(\tan^{-1}(4/3))$$
Now let's work term-by-term.
The key is to recognise that the ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/678609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.
You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$
Now, from, $AM\ge GM$, we have $$\frac {a^4+
a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
modular arithmetic proof Suppose $x$, $y$, and $z$ are integers and $x= 3y^2 -z^2$. Prove that $x\not\equiv1\mod4$.
My thoughts: So I am not sure the route that can prove this. I am trying to just use the simple stuff to set me up. I started with saying if $x\equiv1\mod4$ you would have $4|x-1$ and there will exist a... | The base is small enough for this problem that we can use proof by exhaustion by letting $y,z \equiv 0,1,2,3 \pmod 4$. And in fact, because $0^2 \equiv 2^2 \pmod 4$ and $1^2 \equiv 3^2 \pmod 4$, you only have to check $4$ different cases:
*
*$y \equiv 0, z \equiv 0 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 0^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding intersection points of 2 functions. My method is incomplete. These are the 2 functions :
$y = x^{4}-2x^{2}+1$
$y = 1-x^{2} $
Here's how I solved It :
$x^{4}-2x^{2}+1 = 1-x^{2}$
$x^{4}-x^{2} = 0$
$x^2(x^2-1)=0$
$x^2-1=0$
$x=\pm \sqrt{1} $
Value of $y$ when $x=1$
$y=1-x^2\\y=1-1\\y=0$
Value of $y$ when $x=(-1)$
$... | $x^2(x^2-1)=0\Rightarrow \text{either } x^2=0 \text{ or } x^2-1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How prove this limit $\lim\limits_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a$ given two other limits related to $f_n$ Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}... | I'm going to post a failed approach, just so you can see what I tried. This is here only to show an approach and (maybe) generate new ideas.
Let $$x:=a+b\;\;\text{ and }\;\; y:=ab.$$ Assume for now $0<a<b$. Then
$$\frac{x-\sqrt{x^2-4y}}{2}
=\frac{a+b-\sqrt{(a+b)^2-4ab}}{2}
=\frac{a+b-\sqrt{(a-b)^2}}{2}
=\frac{a+b-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/682025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $(a+b)(a+c)=(b+c)^2$, prove that $(b-c)^2>8(b+c)$.
If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $$(a+b)(a+c)=(b+c)^2$$prove that $$(b-c)^2>8(b+c).$$
The first thing I did after I saw the problem was turning the inequality into this: $$(b+c)^2-8(b+c)-4bc>0$$
After... | $(a+b)(a+c)=(b+c)^2\implies a^2 +a(b+c)+bc-(b+c)^2=0$, which is a quadratic in $a$. Solving for $a=\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}=\frac {-2(b+c)+\sqrt{(b-c)^2+4(b+c)^2}}{2}$, (we take the positive root since $a\ge1$).
or, $\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}\ge1$
or, $(b-c)^2+4(b+c)^2\ge(2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.
I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}... | $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\down}{\downarrow}%
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Finding sums of infinite series as stated in the title the series are infinite and im struggle with how to find the common ratio. and first term (kinda) well I want it to be on the form: ar^k where r is the ratio and a the first term
(3+2^n)/(2^(n+2)) (n=1)
any tips on how to get that series on the form ar^k ?
| It may help to decompose the series:
$$S = \displaystyle \sum_{n = 1}^{k} \frac{3 + 2^n}{2^{n + 2}} = \displaystyle \sum_{n = 1}^{k} \frac{3}{2^{n + 2}} + \displaystyle \sum_{n = 1}^{k} \frac{1}{4}$$
Going a bit further,
$$S = (\frac{3}{8} + \frac{3}{16} + \frac{3}{32} + \dots + \frac{3}{2^{k+2}}) + (\frac{1}{4} k)$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x+y+z=xyz$, find $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ I found this question in a maths worksheet of trigonometry (kinda odd, right?), but I dont know how to figure it out.
If $\displaystyle x+y+z=xyz$, find $\displaystyle\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2... | Case I: At least one of $x^2$, $y^2$ and $z^2$ is equal to $1/3$. This case is possible, e.g., $x=\frac{\sqrt{3}}{3}=-y$ and $z=0$. In this case, the formula $\displaystyle \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ does not make any sense.
Case II: None of $x^2$, $y^2$ and $z^2$ is equal to $1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/691180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Two dice are rolled, what is the probability that the minimum value of the two dice is $3$? Let $D_1$ and $D_2$ represent the values obtained from dice $1$ and dice $2$. We are looking for $P(\min(D_1, D_2) = 3)$
I.e.
$$
\begin{align}
& P([D_1 = 3, D_2 = 3, 4, 5, 6] \cup [D_2 = 3, D_1 = 3, 4, 5, 6])
= 2P([D_1 = 3, D_2... | Be careful: the probability of the union is not equal to the product of the probabilities (that would be the intersection, and it's only true if the events are independent). Nor is the probability of the union equal to the sum of the probabilities (unless the events are disjoint / "mutually exclusive").
\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$
I could not approach the p... | Let
$$
T_1=\sum_{cyc} \frac{a}{b-c}
$$
By hypothesis we have $T_1=0$, but if we put
$T_2=T_1(ab+ac+bc-a^2-b^2-c^2)$, we have
$$
T_2=\sum_{cyc} \frac{a(ab+ac+bc-a^2-b^2-c^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2c}{b-c}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Substitution for $\int\sqrt{9-x^2}\,\mathrm dx$ Question: Use the substitution $x=3\sin(t)$ to evaluate the integral of $\int\sqrt{9-x^2}\,\mathrm dx$.
I started by making a right triangle and solving for $\sin(t)$ and $\cos(t)$.
*
*$\sin(t)=\frac{x}{3}$ and $\cos(t)=\frac{\sqrt{9-x^2}}{3}$
Then, I so... | $$
\int \sqrt{9-x^2} dx
$$
let $x=3\sin\phi$,$dx=3\cos\phi d\phi$, thus we have
$$
3\int d\phi\cos\phi\sqrt{9(1-sin^2\phi)}=9\int d\phi \cos^2\phi =9\int d\phi \frac{1}{2}(1+cos(2\phi))=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C
$$
Thus we see that
$$
\int \sqrt{9-x^2} dx=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C
$$
wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider t... | This is not true. Let $m=2$ and $n=3$. Then $3$ does not divide $1^2+2^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute a linear transformation which carries the circle $|z|=2$ into $|z+1|=1$?
Find the linear transformation which carries the circle $|z|=2$ into $|z+1|=1$, the point $-2$ into the origin, and the origin into $i$.
In order to find the linear transformation will I use the following formula?
$$(z_1−z_3)(z_2... | Not sure if this is correct so please let me know
Let $C_1 = |z| = 2$, and $C_2 = |z + 1| = 1$. Since the origin is in $C_1$ but not in $C_2$, we need to make the circle go inside out. Thus we get $T = \frac{az + b}{cz + d}$.
We know $T = z*(\frac{z + b}{z + d})$ $z*$ be the reflection point of the origin. We let $R$ b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve $6^{2x}-10\cdot 6^x=-21$ using logarithms?
What do I do with $\large 6^{2x}-10\cdot 6^x=-21$?
Since $6$ and $-60$ are not of the same base (nor can they be written as exponents of the same base cleanly) I am having trouble solving for $x$.
| Let $6^x=y$
$$6^{2x}-10\cdot 6^x=-21$$
$$y^2-10y=-21$$
$$y^2-10y+21=0$$
Factor.
$$(y-3)(y-7)=0$$
$$y=3, \ 7$$
Replace $y$ with $6^x$
$$6^x = 3, \ 7$$
I will solve for both equations separately. Let's start with $6^x=3$
$$6^x=3$$
$$\ln(6^x)=\ln(3)$$
$$x\ln(6)=\ln(3)$$
$$x=\dfrac{\ln(3)}{\ln(6)}$$
Now for $6^x=7$
$$6^x=7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric reverse substitution integral question. Evaluate the following integral.
$$ \int \frac{1}{(x^2-4x+13)^2}\ dx $$
My try :
$$ \int \frac{1}{(x^2-4x+13)^2}\ dx $$
$$ \int \frac{1}{[(x-2)^2+9]^2}\ dx $$
$$ (x-2)^2-9 = x^2-a^2\ ,\ a=3,\ x=3\sec\theta\ x^2-a^2=9\tan^2\theta\ dx=3\sec\theta \tan\theta\ d\theta $... | Put $x^2-4x+3=(x-2)^2+3^2$ and then $\tan \theta=\frac{x-2}{3}\Rightarrow x=2+3\tan \theta$, and $dx=3\sec^\theta d\theta$; $\cos \theta = \frac{3}{\sqrt{(x-2)^2+3^2}}$:
$$\int \frac{dx}{[(x-2)^2+3^2]^2}=\int \frac{3\sec^2\theta d\theta}{3^4 \sec^4\theta} =\frac{1}{3^3}\int \frac{d\theta}{\sec^2\theta}=\frac{1}{27}\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int \cos^4(x)dx$ We have: $\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.$
Find $\int \cos^4x\ dx$ by using the formula twice
What I have so far is:
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}\int \cos^{2} x\
dx$
Now we use the formula for $\int co... | You are right.
Look at the alternate forms in WolframAlpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the integral $\int \frac{1}{x^2 \cdot \tan(x)} \ dx$ This problem seems pretty tricky. I need to find the integral of
$$\int \dfrac{1}{x^2 \cdot \tan(x)} \ dx$$
Any help would be greatly appreciated!
| I would use the decomposition of $\frac{1}{\tan x}$ into simple fractions:
$$\frac{1}{\tan x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}$$ So
$$\frac{1}{x^2\tan x}=\frac{1}{x^3}+2\sum_{n=1}^{\infty}\frac{1}{x(x^2-n^2\pi^2)}$$ and
$$\int\frac{dx}{x^2\tan x}=-\frac{1}{2x^2}+\sum_{n=1}^{\infty}\frac{\ln\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/702707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Generating function - technical issue. Which sequence is generated by $\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}}$?
We know that:
$$\frac{1}{{{{(1 - x)}^3}}} = \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}} $$
So we have:
$$(5x - 3{x^2}) \cdot \sum\limits_... | First, note that
$$\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}} = \frac{3}{(x-1)}+\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}$$
And,
$$\frac{3}{(x-1)} = \sum \limits_{k=0}^{\infty}(-3)x^k$$
$$\frac{1}{(x-1)^2}= \sum \limits_{k=0}^{\infty}(k+1)x^k$$
$$\frac{-2}{(x-1)^3}= \sum_{k=0}^{\infty} x^k(k+1)(k+2)$$
Add them up and you get
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How are the Taylor Series derived? I know the Taylor Series are infinite sums that represent some functions like $\sin(x)$. But it has always made me wonder how they were derived? How is something like $$\sin(x)=\sum\limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}\cdot(-1)^n = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7... | This simple derivation of the Taylor series of a function is taken from lecture 37 of the MIT course Single Variable Calculus. More specifically, this derivation is taken from these lecture notes. A similar derivation can also be found here.
Given an infinitely differentiable function $f(x)$, we want to represent it as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/706282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 9,
"answer_id": 8
} |
Prove $1^2+2^2+\cdots+n^2 = {n+1\choose2}+2{n+1\choose3}$ Prove that:
$$
1^2+2^2+\cdots+n^2 = {n+1\choose2}+2{n+1\choose3}
$$
Now, if I simplify the right hand combinatorial expression, it reduces to $\frac{n(n+1)(2n+1)}{6}$ which is well known and can be derived by the method of common differences.
This though is in ... | The identity is an application of Worpitzky's identity involving Eulerian numbers.
Worpitzky's theorem states:
$$x^n=\sum_{k=0}^{n}A(n,k) \binom{x+k}{n}$$
where Eulerian number $A(n, k)$ is defined to be the number of permutations of the numbers 1 to n in which exactly k elements are greater than the previous element (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/710452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$.
Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$.
Here $S_n$ denotes $x_1+x_2+\ldots+x_n$. Also $n\in\mathbb N_{>0}$.
It's easy to see that $x_1=2$.
I'... | Noting that $S_n = x_n + S_{n-1}$, you have
$$
x_n^2 + 4x_n + 4 = (x_n + 2)^2 = 8S_n = 8x_n + 8S_{n-1},
$$
or
$$
x_n^2 - 4x_n + (4 - 8S_{n-1}) = 0.
$$
The solutions are
$$
x_n = \frac{1}{2}\left(4 \pm \sqrt{16 - 4(4-8S_{n-1})}\right)=2\pm\sqrt{8S_{n-1}};
$$
only the positive one is relevant, so you have
$$
x_n=2+2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic Partial Fraction Decomposition I am trying to find the inverse laplace transform of $(s^2+4) \over (s-2)(s+2)$.
The solution is $ {2\over(s-2)} - {2\over(s+2)} + 1 $.
But I can't figure out how to break it up so I can find the solution algebraically.
i.e $ (s^2+4)/((s-2)(s+2)) = As+B/(s-2)+C/(s+2) $ etc. Wh... | $$\begin{align}\frac{s^2+4}{s^2-4}&=\frac{s^2-4+8}{s^2-4}\\&= 1+\frac{8}{s^2-4}\\&=1+2\frac{4}{(s+2)(s-2)}\end{align}$$
Obviously $4=(s+2)-(s-2)$
substituting, we have
$$1+2\frac{(s+2)-(s-2)}{(s+2)(s-2)}$$
$$1+2\frac{(s+2)}{(s+2)(s-2)}-2\frac{(s-2)}{(s+2)(s-2)}$$
Cancelling,
$$\require{cancel}{1+2\frac{\cancel{(s+2)}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try:
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for
$$a... | Given:
$$ab+\frac 1{a^2}+\frac 1{b^2}=ab+\frac 1{a^2}+\frac 1{b^2}+\frac {2}{ab}-\frac {2}{ab}$$
$$\Rightarrow ab+\frac 1{a^2}+\frac 1{b^2}=ab+(\frac 1a-\frac 1b)^2+\frac 2{ab}$$
Let us analyse the RHS:
The minumum value of $(\frac 1a-\frac 1b)^2$ is 0 when a=b and the minumum value of $ab+\frac 2{ab}$ is $2\sqrt 2$(A.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving that the range of linear transformation is a linear subspace I need some help figuring out how to prove this question.
True or false, the set $S = \left \{ A\mathbf{y}: \mathbf{y} \in \mathbb{R}^4\right \}$ is a subspace of $\mathbb{R}^3$ where A is a fixed $3\times4$ real matrix.
Well I will need to show tha... | I think I figured it on, now I just need some confirmation so continuing from what I have above, we have $A\mathbf{u}+cA\mathbf{v} = A\begin{pmatrix}
u_{1}\\
u_{2}\\
u_{3}\\
u_{4} \end{pmatrix} + cA\begin{pmatrix}
v_{1}\\
v_{2}\\
v_{3}\\
v_{4} \end{pmatrix} $
$
= A\begin{pmatrix}u_{1}\\
u_{2}\\
u_{3}\\
u_{4} \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sum\limits_{r=0}^{2n} r \binom{2n}{r}^2 = 2n \binom{4n-1}{2n}$ I expanded
$(1+x)^{2n}$ = $\sum\limits_{r=0}^{2n} \binom{2n}{r} x^r $
Differentiating both sides, we get
$2n(1+x)^{2n-1}$ = $0$ + $\binom{2n}{1}$ + $2\binom{2n}{2}x$ + $3\binom{2n}{3}x^2$ .....
Put $x=1$.
As you see, I'm not able to get the 'squar... | $$
\begin{align}
\sum_{r=0}^{2n}r\binom{2n}{r}^2
&=\sum_{r=1}^{2n}r\binom{2n}{r}\binom{2n}{2n-r}\tag{1}\\
&=\sum_{r=1}^{2n}2n\binom{2n-1}{r-1}\binom{2n}{2n-r}\tag{2}\\
&=2n\binom{4n-1}{2n-1}\tag{3}\\
&=2n\binom{4n-1}{2n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $k\binom{n\vphantom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove if $n^3$ is odd, then $n^2 +1$ is even I'm studying for finals and reviewing this question on my midterm. My question is stated above and I can't quite figure out the proof.
On my midterm I used proof by contraposition by stating:
If $n^2 +1$ is odd then $n^3$ is even.
I let $n^2+1 = (2m+1)^2 + 1$
$= (4m^2 + 4m +... | If $n$ is even, then $n=2k$ and $n^3=8k$, and we have that $n^3$ is even.So, if $n^3$ is
odd then $n$ is odd, otherwise if $n$ is even, by the previous statement we would have $n^3$ even given a contradiction. So, $n$ is odd, and then $n=2t+1$ so $n^2+1=(2t+1)^2+1=
4t^2+2t+1+1=2(2t^2+t+1)$ then, $n^2+1$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
limit $\lim_{n \to \infty} \frac{(2n+1)(2n+2) n^n}{(n+1)^{n+2}}$? Wolfram alpha says that this limit (arising from a ratio test to determine the radius of convergence of a series) should be $4/e$. How does it get this result?
| Notice that
$$\frac{2n+1}{n+1} = 2 - \frac{1}{n+1}.$$
Then
\begin{eqnarray*}
\lim_{n \to \infty} \dfrac{(2n+1)(2n+2) n^n}{(n+1)(n+1)^{n+1}} &=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2(n+1)n^n}{(n+1)^{n+1}} \\
&=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2n^n/n^n}{(n+1)^{n}/n^n} \\
&=&\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Number of Unique Sequences with Circular Shifts I'm trying to determine the number of binary sequences containing exactly $a$ $1$'s and $b$ $0$'s if two sequences are considered identical if they can be reached from the each other by a series of circular shifts (moving the first bit to the end). For instance, for $a=b=... | The total number of sequences (without circular shifting) with precisely $a$ $1$'s and $b$ $0$'s is $\tbinom{a+b}{a}=\tbinom{a+b}{b}$. If either $a=0$ or $b=0$ then there is of course only one such sequence.
For any sequence of length $a+b$, the number of cyclic shifts $d$ required to return the original sequence is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Function extremal - calculus of variations Find a curve passing through (1,2) and (2,4) that is an extremal of the function:
$J(x,y')=\int_1^2 xy'(x)+(y'(x))^2dx$
I don't know what methods to use at all.
| This is a very simple problem. You are trying to find the function $y$ which gives a minimum of the following integral:
$$
J = \int\left(xy' + y'^2\right)dx = \int\mathcal{L}(y, y', x)dx
$$
Where $y$ is a function of $x$. The very first thing to notice is that $\mathcal{L}$ has no dependence on $y$! Therefore $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that $7\mid a^2+b^2$ implies $7\mid a$ and $7\mid b$? For my proof I distinguished the two possible cases which derive from $7 \mid a^2+b^2$:
Case one: $7\mid a^2$ and $7 \mid b^2$
Case two (which (I think) is not possible): $7$ does not divide $a^2$ and $7$ does not divide $b^2$, but their sum.
I've shown... | a = 7k + m, and b = 7p + n ==> a^2 + b^2 = m^ + n^2 ( mod 7 ).
Case 1: m = 1, n = 1 => m^2 + n^2 = 2. ( mod 7 )
Case 2: m = 1, n = 2 => m^2 + n^2 = 5.
Case 3: m = 1, n = 3 => m^2 + n^2 = 3.
Case 4: m = 1, n = 4 => m^2 + n^2 = 3.
Case 5: m = 1, n = 6 => m^2 + n^2 = 2.
....
All cases yield non-zero mod 7. So it can happ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Estimate the error of interpolating ,,, Estimate the error of interpolating (${lnx}$) . at ${x=3}$ with an interpolation polynomial with base points ${x=1 , x=2 , x=4 , x=6 }$ .
| Well, the polynomial $p$ that you need has
$$
p(1) = ln(1) = 0\\
p(2) = ln(2)\\
p(4) = ln(4) = 2 ln(2) \\
p(6) = ln(6)
$$
Writing
$$
p(x) =
a_1 \frac{(x-2)(x-4)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_2 \frac{(x-1)(x-4)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_3 \frac{(x-1)(x-2)(x-6)}{(x-1)(x-2)(x-4)(x-6)}+
a_4 \frac{(x-1)(x-2)(x-4)}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$? I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of t... | We have, (x+1)^2 - x^2 = 2x + 1
Using this, we can say,
1^2 = 0^2 + 1 = 0 +1 = 1
2^2 = 1^2 + 3 = 1 + 3 = 4
3^2 = 2^2 + 5 = 1 + 3 + 5 = 9 and so on...
This explains why the sum is n^2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 11,
"answer_id": 5
} |
One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your ... | I don't think your formula exhaust all solutions of the Diophantine equation
$$2(a^2+b^2+c^2+d^2) = (a+b+c+d)^2\tag{*1}$$
Your equation can be rewritten as
$$(a+b+c+d)^2 = (-a+b-c+d)^2 + (-a+b+c-d)^2 + (-a-b+c+d)^2$$
This is the equation for a Pythagorean quadruple. The set of all Pythagorean quadruples can be parametr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve: $\frac{d^2x}{dt^2}+\Bigl(\frac{d^2y}{dt^2}\Bigr)^2=0$ We have the following coupled Diferential equation:
$\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2$
$\dfrac{d^2y}{dt^2}=\dfrac{y}{x}$
Then find the solution $x$ and $y$ in terms of $t$ .
What we have that
$\dfrac{d^2x}{dt^2}+\left(\dfrac{d^2y}{dt^2}\right)^... | $\begin{cases}\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2~......(1)\\\dfrac{d^2y}{dt^2}=\dfrac{y}{x}............(2)\end{cases}$
From $(2)$, $x=\dfrac{y}{\dfrac{d^2y}{dt^2}}......(3)$
Put $(3)$ into $(1)$ :
$\dfrac{d^2}{dt^2}\left(\dfrac{y}{\dfrac{d^2y}{dt^2}}\right)=-\dfrac{y^2}{\dfrac{y^2}{\biggl(\dfrac{d^2y}{dt^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
prove that $\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}$ if $a+b+c=3$ $a,b,c\in[0,2]$
observation
by triangle inequality $|a|+|b|\ge |a+b| $
$|a-1|+|b-1|+|c-1|\ge |a+b+c-3|$
but $a+b+c=3$
hence
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge \sqrt{4-... | Here is a simple solution.
First I would like to emphasize that $a,b,c$ must be non-negative for the inequality to hold. Otherwise a counter-example is $(a,b,c)=(-1,2,2)$.
Thus, we need to prove the following inequality for $0\le a,b,c\le 2$:
$$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Confusing rational numbers Question:
If $$x = \frac{4\sqrt{2}}{\sqrt{2}+1}$$ Then find value of, $$\frac{1}{\sqrt{2}}*(\frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x - 2\sqrt{2}})$$
My approach:
I rationalized the value of $x$ to be $8-4\sqrt{2}$, then substituted values to get:
$$\frac{1}{\sqrt{2}}* (\frac{10 - 4\sqrt{2}}{6 -... | $$\frac x2=\frac{2\sqrt2}{\sqrt2+1}$$
Applying Componendo & Dividendo, $$\frac{x+2}{x-2}=\frac{2\sqrt2+(\sqrt2+1)}{2\sqrt2-(\sqrt2+1)}=\frac{3\sqrt2+1}{\sqrt2-1}=\frac{(3\sqrt2+1)(\sqrt2+1)}{2-1}$$
$$=7+4\sqrt2$$
Similarly, $$\frac x{2\sqrt2}=\frac2{\sqrt2+1}$$
Apply Componendo & Dividendo again
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality doubt with taylor expansion Can I prove that $\forall x>0$ $$e^{x/(1+x)} < 1+x$$
Showing that $e^{x/(1+x)} = 1+x-\frac{x^2}{2}+o(x^2)$ and so $-\frac{x^2}{2}+o(x^2)<0$ for all $x>0$? How i can be sure that $o(x^2)$ doesn't interfere in the inequality also for large values of $x$?
| Or using the Taylor expansion for $e^t$:
$$e^{\frac{x}{x+1}} = 1+ \frac{x}{x+1} + \frac1{2!} \left(\frac{x}{x+1} \right)^2 + \cdots < 1+ \frac{x}{x+1} + \left(\frac{x}{x+1}\right)^2 + \cdots = \frac1{1-\frac{x}{x+1}} = 1+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/737019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Bounding $\sum_{n=n_1}^\infty x^n (n+1)^2$ I need to upperbound the sum
$$\sum_{n=n_1}^\infty x^n (n+1)^2$$
where $0<x<1$ is a parameter.
I know it can be done starting from
$$\sum_{n=n_1}^\infty x^n (n+1)^2\le \sum_{n=0}^\infty x^n (n+1)(n+2)-\sum_{n=0}^{n_1} x^n (n+1)(n+2)$$ using nasty calculations (integrating twic... | Write
$$
\sum_{n \in S} x^n(n+1)^2
= \sum_{n \in S} \frac{d}{dx} x^{n+1} (n+1)
= \frac{d}{dx} \left( x \sum_{n \in S} x^n (n+1) \right)
= \frac{d}{dx} \left( x \frac{d}{dx} \sum_{n \in S} x^{n+1} \right)
$$
for any index set $S$. In the case that $S = \{n_1,n_1+1,\ldots\}$, this becomes
\begin{align}
\sum_{n = n_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/737357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What converges this series? What exactly converges the series?
$\sum _{k=3}^{\infty \:}\frac{2}{k^2+2k}$
I tried taking out the constant
$=2\sum _{k=3}^{\infty \:}\frac{1}{k^2}$
then $p=2,\:\quad \:p>1\quad \Rightarrow \sum _{k=3}^{\infty \:}\frac{1}{k^2}$
but I really don't know what i'm doing
| Note that $\dfrac{2}{k^2+2k}=\dfrac{1}{k}-\dfrac{1}{k+2}$. So our sum is
$$\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+ \left(\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{10}\right)+\cdots.$$Remove the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/737816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding limits of diagonalised matrix I diagonalised
$A=$
$\left[\begin{array}[c]{rr} 0.6 & 0.9\\ 0.4 & 0.1\end{array}\right]$
and got
$SAS^-$$^1$$= V = (-1/13)$
$\left[\begin{array}[c]{rr} -1 & -1\\ -4 & 9\end{array}\right]$
$\left[\begin{array}[c]{rr} 1 & 0\\ 0 & -0.3\end{array}\right]$
$\left[\begin{array}[c]{r... | Your solution looks close.
Write the matrix as fractions:
$$A=\left[\begin{array}[c]{rr} \dfrac{6}{10} & \dfrac{9}{10}\\ \dfrac{4}{10} & \dfrac{1}{10}\end{array}\right]$$
I have my eigenvalues and eigenvectors swapped from the order you show yours in.
Diagonalization yields:
$$A = PJP^{-1} =
\left(\begin{array}{cc}
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/738821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given length of two medians and one altitude , find the length of one side. In $\triangle ABC$, altitude $AD = 18$, median $BE = 9\sqrt5$ and median $CF = 15$. Find $BC$.
(Note that I've drawn median AG)
By appolonius theorem ,
$$2(15)^2+ 2x^2=(2y)^2+(2z)^2$$
$$2(9\sqrt5)^2+2y^2=(2x)^2+(2z)^2$$
By herons formula an... | First, note that your diagram is somewhat inaccurate. The (undrawn) $\overline{EF}$ should be parallel to $\overline{BC}$, and half as long. (This is the "Midpoint Theorem for Triangles".) With that in mind ...
Let $P$ and $Q$ be the feet of perpendiculars from $E$ and $F$ to $\overline{BC}$. Then
$$|\overline{EP}| =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{... | By Fermat's little theorem, if $n\equiv m\ \ \text{mod}\ (p-1)$, then $a^m \equiv a^n\ \ \text{mod}\ p$.
Now, since any power of two modulo $18$ is one of the six numbers in $S=\{2,4,8,10,14,16\}$, it remains to check that:
$$2^k+3^k+5^k \equiv 0 \ \ \text{mod}\ 19$$
For all $k\in S$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/744328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the limit of a function $n^3 / 3^n$ $$\lim_{n→\infty} \frac{n^3}{3^n} =0 $$
The answer is 0 but how would i go about proving that?
| Let $a_n = \frac{n^3}{3^n} $. Then
$$ \left| \frac{ a_{n+1}}{a_n} \right| = \frac{(n+1)^3}{3^{n+1}} \frac{3^n}{n^3} = \frac{1}{3} \left( \frac{n+1}{n} \right)^3 \to \frac{1}{3} < 1$$
Therefore $a_n \to 0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Finding volume of convex polyhedron given vertices I am trying to compute the volume of the convex polyhedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,2,0)$, $(0,0,3)$, and $(10,10,10)$. I am supposed to use a triple integral but am struggling with how to set it up.
| First we can find the equation for the plane that goes through $A = (1,0,0)$, $B = (0,2,0)$, and $C = (0,0,3)$. It is: $x + \dfrac{y}{2} + \dfrac{z}{3} = 1$ or $6x + 3y + 2z - 6 = 0$. Now let $O = (0,0,0)$, and $D = (10, 10, 10)$. We now find the volume $V_1$ of the tetrahedron $OABC$. $V_1 = \frac{1}{6}\cdot OA\cdot O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Square root of a Mersenne number is irrational Defining a Mersenne Number like this:
k = $2^n -1$
I have to prove that the square root of a Mersenne number is irrational (has no solution in $\mathbb Q$). I know that it can be proven that the square root of a non-perfect square number is always irrational, but is there ... | In binary, $2^n - 1$ is:
$$
111...1\text{, } n \text{ } 1\text{'s}
$$
We can attempt to construct a number that squared gives this, start with $1\times1 = 1$ (that works). Obviously both numbers must be odd (since $2^n - 1$ is certainly odd). Now let's try to add to this to still get all $1$'s:
\begin{align}
&0011 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Separate Into Real and Imaginary Parts Separate the following trigonometric function into Real and Imaginary Parts
$$\tan^{-1}e^{i\theta} $$
or
$$\tan^{-1}(\cos\theta+i\sin\theta)$$
I Have made till here
Assuming $x+iy$ is the final outcome after separation
$$\therefore \tan^{-1}e^{i\theta}=x+iy$$
$$\therefore e^{i\th... | I have come up with this solution Hope this might be the correct one
Let$$\displaystyle\tan^{-1}e^{i\theta}=x+iy$$
$$e^{i\theta}=\tan(x+iy)$$
$$\cos\theta+i\sin\theta=\tan(x+iy)$$
$$\cos\theta-i\sin\theta=tan(x-iy)$$
Now,$$ \tan2x=tan[(x+iy)+(x-iy)]$$
$$=\frac{\tan(x+iy)+\tan(x-iy)}{1-\tan(x+iy)\tan(x-iy)}$$
$$=\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/750552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that every nonzero integer has balanced ternary expansion? show that every nonzero integer can be uniquely represented in the form
$e_k3^k + e_{k-1}3^{k-1}+ … + e_13+e_0$ where $e_j= -1, 0, 1$ for $j = 0,1,2,…k$ and $e_k \neq 0$
| Proof by Strong Induction.
Let $P(n) : n = 3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b^0 \text{ where each b = -1, 0 or 1 }$.
Base Case: P(1) is true. Since $1 = 3^0$.
Assuming $P(1), P(2), \cdots, P(n-1)$ is true to prove it for $n$.
Considering cases:
Case 1: $n = 3m$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to su... | Express $$x^2+bx+a=(x-c)(x-d)$$
$$x^2+ax+b=(x-c)(x-f)$$
Matching the coefficients of the powers of $x$ for both equations, we have
$$c+d=-b$$
$$c+f=-a$$
$$cd=a$$
$$cf=b$$
Solving for $d$ and $f$ in terms of $a$ and $b$ we have
$$d=a,f=b$$
leading to $c$ having only one possible value, which is $1$.
Thus we have $$c+a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/759780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding all primes $(p,q)$ for perfect squares. Find all prime pairs $(p,q)$ such that $2p-1, 2q-1, 2pq-1$ are all perfect squares.
Source: St.Petersburg Olympiad 2011
I have only found the pair $(5,5)$ so I am thinking that maybe a modulo $5$ approach could work.
| I'd like to offer another answer based on the unique factorization of $\mathbb{Z}[i]$.
Notice that if we let $a^2=2p-1$, $b^2=2q-1$ and $c^2=2pq-1$, we can write $$p=\left(\frac{a+1}{2}\right)^2+\left(\frac{a-1}{2}\right)^2, q=\left(\frac{b+1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2$$ and $pq=\left(\frac{c+1}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.