Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $p$ such that $\sum\limits_{n=1}^\infty n(1+n^2)^p$ converges. Check my work.
Given series
$$
\sum\limits_{n=1}^\infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.
To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,
\begin{eqnarray}
\int\limits_1^\infty x(1+x^2)^pdx &=& \l... | Alternative:
If $p\ge -1$, then $(1+n^2)^p \ge (1+n^2)^{-1}$
$$\sum_{n=1}^\infty n(1+n^2)^p \ge\sum_{n=1}^\infty \frac{n}{1+n^2} \ge \sum_{n=1}^\infty \frac{n}{2n^2}= \sum_{n=1}^\infty \frac{1}{2n}$$
Hence it diverges.
If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} \ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$
$$\sum_{n=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Quotient $\mathbf{F}_3[X]/(X^5+1)$
Factor $X^5+1\in\mathbf{F}_3[X]$ into irreducibles. What does the quotient $\mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has ... | I think the mistake you made is that the ideal $(X^5 + 1) \neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) \neq (X + 1,5)$. Long division only gives us "$\subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) \cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal gene... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:
\begin{equat... | Once again I will offer up a method that first converts the integral to a double integral.
For $r > 0$, we begin by enforcing a substitution of $x \mapsto x^{1/r}$. Doing so yields
$$I = \frac{1}{r} \int_0^\infty \frac{x^{1/r - 1}}{1 + x} \, dx.$$
Now noting that
$$\frac{1}{1 + x} = \int_0^\infty e^{-u(1 + x)} \, du,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\big... | HINT: Notice that
$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=\frac{1}{4}\frac{1}{x^2+x\sqrt{2}+1}+\frac{1}{4}\frac{1}{x^2-x\sqrt{2}+1}+\frac{1}{2}\frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=\frac{1}{4}\int_0^1 \frac{dx}{x^2+x\sqrt{2}+1}$$
$$I_2=\frac{1}{4}\int_0^1 \frac{dx}{x^2-x\sqrt{2}+1}$$
$$I_3=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bma... | Rather than applying row operations, expand the minors.
$\det A = -4\begin{vmatrix} 3&0&1\\0&2&0\\1&0&3\end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$\begin{vmatrix} 1\\&1\\-\frac 13&&1\\&&&1\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046691",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^... | Here I give you an alternative approach. If we make $x = \sinh(u)$, we get
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x & = \int\sinh^{3}(u)\mathrm{d}u = \int(\cosh^{2}(u)-1)\sinh(u)\mathrm{d}u\\
& = \frac{\cosh^{3}(u)}{3} - \cosh(u) + K
\end{align*}
Since $x = \sinh(u) = \pm\sqrt{\cosh^{2}(u)-1}$, we con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | We can maximize $g(a,b) = a^2+b^2$ subject to $(a^2+b^2)^2 =3b^2-1$. Solve for $a^2$ we have: $a^2= \sqrt{3b^2-1} -b^2\implies g(a,b) = \sqrt{3b^2-1}$. Thus the problem now is the find the maximum value of $b^2$. To this end, let $c = b^2 \ge 0$, we have: $a^4+2a^2c+c^2 - 3c+ 1 = 0\implies c^2-3c+1 \le 0\implies c \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of a series of a number raised to incrementing powers-- for any number Sum of a series of a number raised to incrementing powers. I have a sub question on this. As I dont have reputation to ask for in comment. The formula being derived for series of summation of powers to 2 i.e $2^{n+1}-1$ doesnt apply for other nu... | This is a geometric series
$$S=5^0+5^1+5^2+5^3+5^4+\cdots+5^n$$
$$5S=5^1+5^2+5^3+5^4+5^5+\cdots+5^{n+1}$$
$$4S=5^{n+1}-1$$
$$S=\frac{5^{n+1}-1}{4}$$
In general to find $1+x+x^2+\cdots+x^n$,
$$\sum_{j=0}^{n}x^j=\frac{x^{n+1}-1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sum which have binomial coefficients
Finding value of
$\displaystyle \binom{50}{6}-\binom{5}{1}\binom{40}{6}+\binom{5}{2}\binom{30}{6}-\binom{5}{3}\binom{20}{6}+\binom{5}{4}\binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$\bigg[(1+x)^{10}-1\bigg]^5=(1+x)^{50}-\binom{5}{1}(1+x)^{40}+\binom{5}{2}(1+x)^{... | Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x... | To end your computation, you can split your expression in two:
$$\frac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}} =-\frac{\sin t}{ \sin \frac{3t}{2}}+\sqrt 3\,\frac{1- \cos t}{\sin \frac{3t}{2}} $$
Now it is standard that $\;\lim_{t\to 0}\dfrac{\sin at}{\sin bt}=\dfrac ab$. On the other hand
$$\frac{1-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
| If you can prove $x^4 \equiv 1 \pmod {10}$ that's enough as $x^{n+4}\equiv x^n\cdot x^4 \equiv x^n\cdot 1\equiv x^n\pmod {10}$
If you google Euler's theorem that will be true for all numbers relatively prime to $10$. Intuitively (but hand wavy) if $x$ and $10$ are relatively prime then $x^n$ will be relatively prime. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Show that $\frac{x}{\sqrt{1-a^2}}=\frac{y}{\sqrt{1-b^2}}=\frac{z}{\sqrt{1-c^2}}$
Given $3$ equations:
*
*$ x-cy-bz=0$
*
*$-cx+y-az=0$
*$-bx-ay+z=0$
Show that $\dfrac{x}{\sqrt{1-a^2}}=\dfrac{y}{\sqrt{1-b^2}}=\dfrac{z}{\sqrt{1-c^2}}$
Now solving the 3 equations I got:
$\dfrac{x}{{1-a^2}}=\dfrac{y}{... | Hint:
By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$
By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$
$$\dfrac{x(1-c^2)}{z(1-a^2)}=\dfrac{z(b+ca)}{x(ca+b)}$$
What if $b+ca\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,c\in R,a\ne0$)
$(1)\;5a$
$(2)\;3b$
$(3)\;2c$
$(4)\;0$
As $x=1$ is the root of t... | Note that
$$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
and
$$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\fra... | An alternative strategy uses $1+\cos x=2\cos^2\tfrac{x}{2}$ to obtain $$\sin a+\sin b+\sin (a+b)=2\left(\sin a\cos^2\tfrac{b}{2}+\sin b\cos^2\tfrac{a}{2}\right).$$To finish, use $\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ to obtain$$\sin a+\sin b+\sin (a+b)=4\cos\tfrac{a}{2}\cos\tfrac{b}{2}\left(\sin\tfrac{a}{2}\cos\tfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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How to prove $\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos(A)^2}$ I am unable to prove this trigonometric identity
$$\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos^2(A)}$$
I have tried to transform the left-hand side and stuck with this
$$\frac{2\sin(A)\cos(A)}{\sin(A)\cos(... | One way we can prove the identity false is as follows:
$$\begin {align}
\dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\
2 = \dfrac {1}{1-2\cos^2 2A} \\
2 (1-2\cos^2 2A) = 1 \\
2 - 4\cos^2 2A = 1 \\
- 4\cos^2 2A = \dfrac {1}{2} \\
\cos^2 2A = -\dfrac {1}{8}
\end {align}$$
Since the last l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\rig... | Observe that
$$a_n=n+na_{n-1}$$
$$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
and so on, giving us eventually
$$a_n=\sum_{k=1}^n\frac{n!}{(k-1)!}$$
Now we have
$$\lim_{n\to\infty}\frac{1}{n!}(a_n+1)=\lim_{n\to\infty}\frac{a_n}{n!}$$
$$=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(k-1)!}$$
$$=\sum_{k=0}^\infty\frac{1}{k!}=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Magnitude of Complex Number And Other Equation Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5\sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $\left| \frac{1}{z} -... | Note that if you replace $\frac 1z$ with $u$ in your second equation $\left| \frac{1}{z} - \frac{1}{a \sqrt{b}} + \frac{i}{c} \right| = \frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $\frac 1d$ and center is $\frac{1}{z} - \frac{1}{a \sqrt{b}}$. Thus the desired circle is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The sum of the series $\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots$ Suppose $S=\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots \dots$
Does it converge? If so find the sum.
What I attempte... | Let
$$f(x)=\sum_{n=1}^\infty t_nx^n=\sum_{n=1}^\infty\frac{5\times7\times\cdots
\times(2n+3)}{4\times6\times\cdots
\times(2n+2)}x^n.$$
Then
$$t_n=\frac23\frac1{(n+1)!}\left(\frac32\right)\left(\frac52\right)\cdots
\left(\frac{2n+3}2\right)=\frac23u_{n+1}$$
where
$$u_n=\frac{(3/2)(5/2)\cdots((2n+1)/2)}{n!}.$$
Then, for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?
HERE IS ONE
Let $n\in\Bbb{N}$ be fixed su... | For $n\ge2$,
$$
\begin{align}
\left(1-\frac1{\sqrt{n}}\right)^n
&=\left(\left(1+\frac1{\sqrt{n}-1}\right)^{\sqrt{n}}\right)^{-\sqrt{n}}\tag{1a}\\
&\le\left(1+\frac{\sqrt{n}}{\sqrt{n}-1}\right)^{-\sqrt{n}}\tag{1b}\\[9pt]
&\le2^{-\sqrt{n}}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $1-\frac1{\sqrt{n}}=\left(1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}... | $$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$
Even quicker, $\sqrt[4]{25}$ means $\sqrt{\sqrt{25}}$, which becomes $\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\... | Let us start with a step of integration by parts:
$$ \int_{0}^{+\infty}\frac{1}{4x}\cdot\frac{4x^3}{(x^4+1)^2}\,dx =\int_{0}^{+\infty}\frac{1}{4x^2}\left(1-\frac{1}{1+x^4}\right)\,dx=\frac{1}{4}\int_{0}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ \frac{1}{8}\int_{-\infty}^{+\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\le... | Beside telescoping, all you need is to reverse the order of the double sum: $$\sum_{n\ge 1}\frac{\sum_{k=1}^n k^{-1}}{n(n+1)}=\sum_{k\ge 1}k^{-1}\sum_{n\ge k}\frac{1}{n(n+1)}=\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $\frac{k^{-1}}{n(n+1)}$ with $n,\,k$ inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
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Relation of Roots of Bi-quadratic equation without using Vieta's formula
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has n... | Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of the following matrix: $A$ Find the inverseof the following matrix:
$$A= \begin{bmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&1 \\
0 & 0 & 1 & 0 &0 \\0&0&0&1&0
\end{bmatrix}$$
My attempt : I think the inverse will not exist because $\det A=0 $
Is it True ?
Any hints/solution will... | Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
$$\det(A)= \begin{vmatrix}
0 & \color{blue}1 & 0 & 0 &0 \\
\color{red}1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&\color{green}1 \\
0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0
\end{vmatrix}=(-1)^3\begin{vmatrix}
\color{red}1 & 0 & 0 & 0 &0 \\
0 & \color{blu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\c... | You have achieved the $cos 3x$ and $sin 3x$ answers by matching the Real and Imaginary components to show them in terms of only $cos x$ and $sin x$ I presume, leading on from an earlier question.
It should be a lot easier if you simply do exactly what you've done for the answers before matching the components
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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What is the radius and center of the image of $|z|=1$ under $ f(z) = \frac{3z+2}{4z+3}$?
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = \frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be tur... | Here's another solution I was able to find. Notice the matrix factorization:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right]
=
\left[ \begin{array}{cc} \frac{1}{5} & 0\\ 0 & 5 \end{array} \right] \times
\left[ \begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array} \right] \times
\left[ \begin{array}{cr} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, on... | Here is an approach that first converts the integral to a double integral.
Let
$$I = \int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = 2 \int_0^\infty \frac{x}{(1 + x^2)^2 (e^{2 \pi x} - 1)} \, dx.$$
Observe that
$$\frac{1}{2} \int_0^\infty y e^{-y} \sin (xy) \, dy = \frac{x}{(1 + x^2)^2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Power tower last digits Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:
$6^{6^{6^{6^{6^{6}}}}}$
Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to s... | We're interested in $6^x \mod 10^6$; this is determined by $6^i \mod 2^6$ and $6^i \mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x \ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trying to solve this triple integral: $\iiint (x-1)(y-1) \,dx\,dy\,dz$ Here's the question
$$\iiint (x-1)(y-1) \,dx\,dy\,dz.$$
I am asked to evaluate this integral over the region $$D:=\left \{ (x,y,z) \in\mathbb{R}^3 :x^2+y^2 \leq z \leq 2x+2y+2 \right \}.$$
There are the bounds of integration in set D (the variable $... | Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
$$(x-1)^2+(y-1)^2=2^2.$$
Hence the given triple integral becomes
$$\iint_{(x-1)^2+(y-1)^2\leq 2^2}\int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
and after integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \d... | Hint
First of all, let $x=\frac 1y$ to make the expression
$$A=\frac{(3 y+1) \log \left(\frac{1+a y}{1+b y}\right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=\frac{a-b+2}{y}+\frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)+\frac{1}{6} y \left(2
a^3-9 a^2-2 b^3+9 b^2\right)+O\left(y^2\right)$$ So
$$a-b+2=0 \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+d\ne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{\color{red}{?}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Prove that for all positive real numbers $x$ and $y$, $(x+y)(\frac1x+\frac1y)\ge4$. Prove that for all positive real numbers $x$ and $y$,
$(x+y)(\frac1x+\frac1y)\ge4$.
Any help would be appreciated thanks.
| To begin with, notice that
\begin{align*}
(x+y)\left(\frac{1}{x} + \frac{1}{y}\right) = 2 + \frac{x}{y} + \frac{y}{x}
\end{align*}
According to the inequality $AM\geq GM$, we have
\begin{align*}
\frac{x}{y} + \frac{y}{x} \geq 2\sqrt{\frac{x}{y}\cdot\frac{y}{x}} = 2
\end{align*}
from whence we obtain the claimed result.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determine Fourier series expansion for $f(\theta)=\cos^4\!\theta$ Q: The function $f(\theta) = \cos^4\! \theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $\cos j \theta$ and $\sin j \theta$ with real coefficients. Determine what the expa... | Using $$\cos^2(\theta )=\frac{1+\cos(2\theta )}{2},$$ one get $$\cos^4(\theta )=\frac{1+2\cos(2\theta )+\cos^2(2\theta )}{4}=\frac{1+2\cos(2\theta )+\frac{1+\cos(4\theta )}{2}}{4}$$
$$=\frac{3}{8}+\frac{1}{2}\cos(2\theta )+\frac{1}{8}\cos(4\theta ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\me... | We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$\frac{a}{c}+\frac{c}{a}=\frac{169}{336}$$ from here you will get the quotient of $$\frac {a}{c}$$ and thus we can calculate the angles of triangle $$\Delta ABC$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 1... | You're on the right track.
When $x=13i+3$, $y=4(13i+3)^2+8(13i+3)+5= 4(169i^2+78i+9)+8(13i+3)+5$.
Leaving out some multiples of $13$, this is $4\times 9+8\times3+5=36+24+5=65=5 \times 13$.
This shows that, when $x=13i+3$, $y$ is a multiple of $13.$
(I'll leave $x=13i+8$ as an exercise.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides ... | It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; \tag 0$
then, note that
$b = a + (b - a) < a + 2\sqrt a \sqrt{b - a} + (b - a) = (\sqrt a + \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| You could compare equation (1) with a standard ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $(x, y)$ has parametric co-ordinates $(acos\theta, bsin\theta)$.
So in your case $(x, y^2)$ could be represented by $(4+cos\theta, 3+sin\theta)$.
Then $x^2 + y^4 = 25 + 1 + 8cos\theta+6sin\theta$ which has a maximum o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding all of the solutions of the functional equation $P(x+2)+P(x-2) = 8x+6$
Let $P(x+2)+P(x-2) = 8x+6$. Find all of the solutions for $P(x)$.
If we put $P(x) = ax + b $ the answer is obvious but how to determine all of the solutions?
| This difference equation is linear so it's solution can be reprasented as
$$
P(x) = P_h(x) + P_p(x)
$$
with
$$
P_h(x+2)+P_h(x-2) = 0\\
P_p(x+2)+P_p(x-2) = 8x+6
$$
for the homogeneous we propose $P_h(x) = \gamma^x$ and after substituting
$$
\gamma^x\left(\gamma^2+\frac{1}{\gamma^2}\right)=0
$$
and the solutions are
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$ How should we find two rational numbers $\alpha$ and $\beta$ such that
$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$
The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you
| You are right.
We can get your answer by the following way.
Let $\sqrt[3]{7+5\sqrt2}=x$.
Thus, $x>2$ and $$x^3=7+5\sqrt2,$$ which gives
$$(x^3-7)^2=50$$ or
$$x^6-14x^3-1=0$$ or
$$x^3-\frac{1}{x^3}-14=0.$$
Now,let $x-\frac{1}{x}=t$.
Thus, $t>0$ and we obtain:
$$t^3+3t-14=0$$ or
$$t^3-2t^2+2t^2-4t+7t-14=0$$ or
$$(t-2)(t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force appro... | Let $$(a;b;c)\rightarrow (\frac{kx}{y};\frac{ky}{z};\frac{kz}{x})\Rightarrow abc=k^3$$
Then we will need to prove $$\frac{yz}{kx\left(ky+z\right)}+\frac{xz}{ky\left(kz+x\right)}+\frac{xy}{kz\left(kx+y\right)}\ge \frac{3}{k^3+1}$$
By Cauchy-Schwarz:
$$LHS=\sum_{cyc}\frac{y^2z^2}{kxyz\left(ky+z\right)}\ge \frac{\left(xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$ After seeing this integral I've decided to give a try to calculate:
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx$$
That is because it's common for many integrals to have a combination of a polynomial in the denominator and a logarithm or an inverse trig function in the numera... | Here is one approach. As a warning, the final answer I find is not pretty.
Let
$$I = \int_0^1 \frac{\tan^{-1} x}{x^2 -x - 1} \, dx.$$
Start by using a self-similar substitution of
$$x = \frac{1 - u}{1 + u}, \,\, dx = -\frac{2}{(1 + u)^2} \, du.$$
So , after having reverted the dummy variable $u$ back to $x$, we have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Integration: Integrating then substituting a number has a different answer to substituting a number then integrating; for trigonometric functions This might be a very dumb question but I am really stuck as to why this happens. Now, the answer to the integral
$$B_n = 2 \int_0^1 k\sin(3\pi x)\sin(n\pi x) dx \qquad (k = 0... | Note that $\dfrac{1}{3\pi - n\pi}$ is not defined at $n=3$. So you need to solve the integral for $n=3$ separately.
When $n=3$, we have
$$k\int_0^1 \cos(3\pi x-n\pi x)-\cos(3\pi x+n\pi x)dx=k\int_0^1 \cos(0)-cos(6\pi x)dx=k\int_0^1 1-\cos(6\pi x)dx.$$
Can you complete this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$
$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$
What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm i... | Use the binomial expansion or, better, Taylor series to show that
$$\sqrt[3]{n^3+2n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt[3]{n^3+2n-3}=n+\frac{2}{3 n}-\frac{1}{ n^2}+O\left(\frac{1}{n^3}\right)$$
$$\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}=\frac{2}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | $$\begin{align}
x^4-2x^3+x-2&= x^4 + (\color{red}{-2x^3+x-2})\\
&= x^4 + \color{red}{P(x)}\\
&= x^4 + \color{green}{-x^3-2x^2+}(\color{green}{x^3+2x^2}+\color{red}{P(x)})\\
&= \color{blue}{(x^4 + -x^3-2x^2)}+(x^3+2x^2+\color{red}{P(x)})\\
&= (x^2-x-2)(x^2)+(x^3+2x^2+\color{red}{P(x)})\\
&= (x^2-x-2)(x^2)+(x^3+2x^2+\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Relation involving generalized Laguerre polynomials Playing around with different approaches to solve the radial part of the Schrodinger equation for the hydrogen-like atom, I have obtained the following expression ($l$ and $n$ are non-negative integers)
$$ \tilde L^{(2l+1)}_n(x) = \frac{1}{n!} x^{-(l+1)}e^{x/2} \left(... | Denoting
\begin{equation}
h(x)=x^{l+1}e^{-x/2};\quad \varphi(x)=x^{2l+1}e^{-x}; \quad\Theta=x \frac{d}{dx}-x \frac{d^2}{dx^2} + \frac{l(l+1)}{x} -\frac{x}{4}
\end{equation}
both polynomial definitions can be written as
\begin{align}
\tilde{L}^{(2l+1)}_n(x) &= \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]\\
L^{(2l+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} + 5^n\binom{n+2}2$? How can we solve the following recurrence relation using GF?
$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 \choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$
I think that most of it is pretty straightforward. What ... | $\sum_{n=0}^\infty {n+2 \choose 2}5^nx^n = \frac12\sum_{n=0}^\infty(n^2+3n+2)(5x)^n=\frac12[\sum_{n=0}^\infty n^2y^n+3\sum_{n=0}^\infty ny^n+2\sum_{n=0}^\infty y^n]$
where $y=5x$. You will need the following series sums:
$(1)\sum_{n=0}^\infty n^2y^n=\dfrac{y(1+y)}{(1-y)^3}\\(2)\sum_{n=0}^\infty ny^n=\dfrac y{(y-1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
| Most answers use functions or derivatives... I'll use another approach: Inequalities!
It's easy to prove that the minimum will be achieved for positive values of $x$ and $y$. Thus, in virtue of the QM-AM inequality:
$$\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}=5\iff x^2+y^2\geq 50$$
Proof of the Quadratic Mean - Ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 9,
"answer_id": 8
} |
Finding the generating function of a sequence for $n$ pennies Let $a_n$ count the number of different ways you can pay a sum of n pennies with 1p and 2p coins. Find the generating function and closed formula for $a_n$. I have attempted this question, finding $a_n$ to be $(1, 1, 2, 2, 3, 3, \ldots)$, and have started c... | A direct way of thinking about the generating function is $(1+x+x^2+...)(1+x^2+x^4+x^6+...)={1\over1-x}\times{1\over1-x^2}$ where the first sequence represents the $1$ pennies and the second represents the $2$ pennies.
Now to find the closed form formula for $a_n$ basically you use the partial sum method to split ${1\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'hig... | First, you could observe that
\begin{cases}
3x \equiv 2 \pmod{5} \Leftrightarrow 3x \equiv 2+10 \equiv 12=3 \cdot 4 \pmod{5} \Leftrightarrow x \equiv \color{red}{4} \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \Leftrightarrow 3x \equiv 4+14 \equiv 18=3 \cdot 6\pmod{7} \Leftrightarrow x \equiv \color{lime}{6} \pmod{6} \\[4px]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} ... | Although each term approaches zero, there are an infinite number of terms as indicated by there being exactly $v$ terms in the addition. The denominator of each term approaches $v^3$ as $v \to \infty$ as $v^3$ is much greater than $v$ or any value less than $v$. This means that the limit can simply be thought of as the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
| $$a^2+{1\over 9} + b^2+{1\over 9} + c^2+{1\over 9}\geq {2\over 3}(a+b+c)$$ by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find at least $5$ integers $n$ such that $\varphi(n)=16$
Let $\varphi(n)$ denote Euler's totient function. Find all integers such that $\varphi(n)=16$.
Answers given were $17,32,34,40,48.$
I am thinking a generalisation of this problem: is there a way to find all positive integers $n$ such that $\varphi(n)=k$ for a... | If $$n= p_1^{a_1}p_2^{a_2}...p_k^{a_k}$$ where $a_i\geq 1$ and $p_1<p_2<...<p_k$ then we have $$\phi(n) = p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_1-1)...(p_k-1)$$
$$16= p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_1-1)...(p_k-1)$$
If $p_1=2$ and $k>1$ $$2^4= 2^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_2-1)...(p_k-1)$$
then $a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - ... | Use $ln(a) + ln(b) = ln(a\cdot b)$ and $aln(b) = ln(b^a)$
You get $ln(\frac{1}{2} \cdot \frac{3}{1} \cdot \frac{1^2}{1} \cdot \frac{1}{2^2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A more natural solution to finding the general terms of a recurrence relation in $2$ variables A high school contest math problem in a problem book:
Find the general terms of
$$a_{1}=a,\quad b_{1}=b,\quad a_{ n + 1 }=\frac { 2 a _ { n } b _ { n } } { a _ { n } + b _ { n } },\quad b_{ n + 1 }=\sqrt{ a_{n+1} b _ {n}}$$
... | The approach is nice.
Note that
$$\cot\left(\arccos\dfrac ba\right) = \frac b{\sqrt{a^2-b^2}}$$
(see also Wolfram Alpha).
If $\underline{\frac ba >1},$ then the hyperbolic functions can be used instead of trigonometric ones, because
$$2 \tanh \frac { \theta } { 2 } = \frac { 2 \tanh \theta \sinh \theta } { \tanh \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^... | $x^2+y^2>1$, and $x^4+y^4 <1$.
Consider $y=x$ in the first quadrant.
$2x^2 >1$, and $2x^4 <1;$
$x ^2 >1/2$ , and $x^2 <1/√2$, i .e.
$1/2=0.5 < x^2 < √2/2=0.7071..$
Points on $y=x$ , line through origin, with
$1<x^2+y^2 < √2 $, satisfy
$x^4+y^4 <1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Can you help me solve this algebra problem? Hi I need to solve this problem and I don’t know how. I’d appreciate your help.
If $x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}$ and $x\neq y\neq z$, then $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$
I think I need... | $$x-\frac{ayz}{x^2}=y-\frac{axz}{y^2}$$ gives
$$x-y+az\left(\frac{x}{y^2}-\frac{y}{x^2}\right)=0$$ or
$$1+\frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
Similarly, $$\frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
$$\frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
$$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Find shortest distance from point on ellipse to focus of ellipse.
Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.
My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and ... | Assuming $a>b$, we have $c=\sqrt{a^2-b^2}$.
Using the parametric equations
$$\begin{cases}x=a\cos\theta,\\y=b\sin\theta\end{cases}$$
we minimize
$$(a\cos\theta-c)^2+(b\sin\theta)^2.$$
The derivative of this expression is
$$-a\sin\theta(a\cos\theta-c)+b\cos\theta(b\sin\theta).$$
There is a root for $$\sin\theta=0$$ (on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
| Calling
$$
AF = a\\
FG=x\\
GC = b\\
EC = c\\
E = (X_0, Y_0)\\
O = \left(a+\frac x2, \frac x2\right)
$$
Considering in the plane $X\times Y$ the square diagonal as the $X$ axis with $A$ at the origin, we have
$$
\frac{\sqrt 2}{2}(a+x+b)= 2c\\
x = \sqrt2 r\\
\left(X_0-\left(a+\frac x2\right)\right)^2+\left(Y_0-\frac x2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 5
} |
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin... | Observe that $\int \sin 2x\ \text {dx} = -\frac {\cos 2x} {2} + C$ and use the fact that $\cos 2x =1-2{\sin}^2 x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$
Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $a,b,c$ are strict positive and dinstinct real numbers.
I know it should be something with arctangent but I don't know how to ge... | Hints
Method 1: Partial fractions Since $a, b, c$ are positive and distinct, a straightforward application of the method of partial fractions gives that the integrand is $$\sum \frac{1}{(a^2 - b^2)(a^2 - c^2)} \frac{1}{x^2 + a^2},$$ where the sum is over the cyclic permutations of the parameters $(a, b, c)$. By lineari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int 2x\sin{(x^2+x)} \,dx$ Evaluate $\int 2x\sin{(x^2+x)} \,dx$ using integration by parts.
Let $u=2x$ and $dv=\sin(x^2+x)dx$ and I got that $du=x^2dx$. I'm stuck when integrating $\sin{(x^2+x)}$. Then I tried the integral calculator and it says:
$$-\cos(x^2+x)-\frac{\sqrt{\pi}\cos{\frac{1}{4}}S\left(\frac{... | Took me a few tries but here goes. First off, we'll integrate $\sin(x^2 + x)$. Then we'll use that to build the original integral in question.
This one requires a trick. You will need to first "complete the square" on $x^2 + x$. Note that (this can be "reverse-engineered" from Wolfram(?)s output by noting the $x + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Factoring a matrix as the product of block triangular and diagonal matrices. How can I check that the matrix
$$K = \left[\begin{array}{c|cc} 1 & 0^{\mathrm T}_m & m \mathbf u^{\mathrm T} \\ \hline 0_m & I_m & I_m \\ m \mathbf u & I_m & O_m \end{array}\right] = \begin{bmatrix}
1 & \textbf{w}^T \\
\textbf... | It works. To multiply
$\begin{equation} K =
\begin{bmatrix}
1 & \textbf{w}^{T}B^{-1} \\
\textbf{0} & I_{2m}\\
\end{bmatrix}
\begin{bmatrix}
s & \textbf{0}^T\\
\textbf{0} & B\\
\end{bmatrix}
\begin{bmatrix}
1 & \textbf{0}^{T}\\
B^{-1}\textbf{w} & I_{2m}
\end{bmatrix}
\end{equation},$
start, e.g.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My ... | For every triplet of real numbers $(x,y,z)$, which satisfies $$xy+yz+zx=1,$$ then there exists an unique triplet of angles $(\alpha, \beta, \gamma)$, such that $0 \le \alpha, \beta, \gamma < \pi $ and $\alpha + \beta + \gamma= \pi $, i.e. they are angles of a triangle, and $$tg(\frac{\alpha}{2}) = x, \text{ } tg(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$ Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$. Is this limit equal to $\lim\limi... | Starting with integration by parts we have
$$\begin{aligned} \displaystyle a_n \displaystyle & = \int_0^1 x^n(1-x)^n\,{dx} \\& = \frac{1}{n+1}\int_0^1 (x^{n+1})'(1-x)^n\,{dx} \\& = \frac{x^{n+1}}{n+1}(1-x)^n\,\bigg|_0^1 -\frac{1}{n+1}\int_0^1 x^{n+1}[(1-x)^n]'\,{dx} \\& = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,{dx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\fra... | Use the Chain Rule:
$\dfrac{\mathrm d}{\mathrm dx}\cos(\sqrt{e^{x^5}\sin x})$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm dx}(e^{x^5}\sin x)$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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solve $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$ $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$
trig sub
$y = \sqrt{3}\sin(u)$
$dy = \sqrt{3}\cos(u)du$
\begin{align}\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}& 4 \sqrt{3-3\sin^2(u)}\sqrt{3}\cos(u)\,du = 12 \int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sq... | Your endpoints are wrong. If $y = \pm \sqrt{3}$, $\sin(u) = \pm 1$. So take $u$ from $-\pi/2$ to $\pi/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral of $x^2 \sin(x^2)$ I was playing around learning the SymPy syntax (python library) e then I saw an example that intrigued me.
$$\int x^2\sin(x^2)dx = \frac{5x\cos(x^2)\Gamma(\frac{5}{2})}{8\Gamma({\frac{9}{4})}} + \frac{5\sqrt{2\pi}C(\frac{x\sqrt{2}}{\sqrt{\pi}})\Gamma{(\frac{5}{4})}}{16\Gamma{(\frac{9}{4})}}$... | With change of variable $X=x^2\quad;\quad dx=\frac{dX}{2\sqrt{X}}$ :
$$I=\int x^2\sin(x^2)dx=\int X\sin(X)\frac{dX}{2\sqrt{X}}=\frac12\int \sqrt{X}\sin(X)dX$$
Integration by part :
$$I=-\frac12\sqrt{X}\cos(X)+\frac14\int\frac{\cos(X)}{\sqrt{X}}dX$$
$$I=-\frac12x\cos(x^2)+\frac14\int\frac{\cos(x^2)}{x}2xdx$$
$$I=-\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.
The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All
I have been able to find ... | Be careful that your final equation has more potential solutions. The equation
$$ \sin \left(x + \frac{\pi}{12}\right) = \sin 2x$$
implies in fact
$$ x + \frac{\pi}{12} = 2x + 2k \pi$$
or
$$ x + \frac{\pi}{12} = \pi - 2x + 2k \pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0... | $$\log_{b}{a}\log_{c}{b}\log_{a}{c}=e^{\ln\left(\log_{b}{a}\log_{c}{b}\log_{a}{c}\right)}=$$
$$e^{\ln\ln{a}-\ln\ln{b}+\ln\ln{b}-\ln\ln{c}+\ln\ln{c}-\ln\ln{a}}=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If complex $z$ satisfies $z+\frac{1}{z}=\cos x$, then evaluate $z^n+\frac{1}{z^n}$ for integer $n$
Let $z$ be a complex number such that
$$
z + \frac{1}{z} = \cos x
$$
Then what is the value of the expression
$$
z^n + \frac{1}{z^n}
$$
where $n$ is an integer?
Please help me. I have tried somehow using the trigonometr... | \begin{cases}
z+\frac1z=\cos x\\
z\cdot\frac1z=1
\end{cases}
Considering $t^2-\cos x\cdot t+1=0$, and we get
\begin{equation}
\{z,\frac1z\}=\{\frac{\cos x+\sqrt{\cos^2x-4}}{2},\frac{\cos x-\sqrt{\cos^2x-4}}{2}\}
\end{equation}
So
\begin{align}
z^n+\frac{1}{z^n}
&= (\frac{\cos x+\sqrt{\cos^2x-4}}{2})^n+(\frac{\cos x-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the midd... | Following the notations of the figure by @Dr. Mathva, knowing that $AD=3, BD=2, CD=1,$ and setting $x=AB=AC=BC$, let us express that the Cayley-Menger determinant of the four points $A,B,C,D$ is zero (see here):
$$\begin{vmatrix}0&x^2&x^2&3^2&1\\
x^2&0&x^2&2^2&1\\
x^2&x^2&0&1^2&1\\
3^2&2^2&1^2&0&1\\
1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Convergence of $a_n = \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}$ I am not finding any path to solve it. Online calculators are not helping either... The answer is it converges and converges to 0.
I believe it is the same as the result of
$$\lim_{x\to\infty}\frac1x\times\frac{(2... | If
$a_n
= \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}
= \frac{\prod_{k=1}^n(2k-1)}{n\prod_{k=1}^n(2k)}
$
then
$a_{n+1}
= \frac{\prod_{k=1}^{n+1}(2k-1)}{(n+1)\prod_{k=1}^{n+1}(2k)}
$
so
$\begin{array}\\
\frac{a_n}{a_{n+1}}
&=\frac{(n+1)2(n+1)}{n(2n+1)}\\
&=\frac{2(n^2+2n+1)}{2n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,
i.e, how to derive this sequence?
The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=1665... | The $n$th such equation is $$\left(\tfrac{5\times 10^{n-1}-2}{3}\right)^3+(5\times 10^{n-1})^3+\left(\tfrac{10^n-1}{3}\right)^3=10^{3n-1}+\tfrac{2}{3}(10^{n-1}-1)10^{2n}+5\times 10^{2n-1}+\tfrac{10^n-1}{3},$$or with $x=10^n$ we can write it as $$\left(\tfrac{x-4}{6}\right)^3+\left(\tfrac{x}{2}\right)^3+\left(\tfrac{x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$)... | The problem is a conflict between these two lines:
$2\arcsin x= \arcsin(2x\sqrt{1-x^2}) \tag1$
and
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
The problem is that if $\frac1{\sqrt2} < \lvert x \rvert < 1,$
then $\frac\pi4 < \lvert\arcsin x\rvert < \frac\pi2,$ and therefo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that for $a,p,q \in \Bbb R$ the solutions of: $\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$ are real numbers. Prove that for $a,p,q \in \Bbb R$ the solutions of: $$\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$$
are real numbers.
I tried manipulating the expression, getting rid of the denominators, but i can'... | We need $a\ne0$. The equation can be written as $(x-p)(x-q)=a^2(2x-p-q)$, or $x^2-(2a^2+p+q)x+a^2(p+q)+pq=0$.
The discriminant is
$(2a^2+p+q)^2-4a^2(p+q)-4pq=(p-q)^2+4a^4>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $f(a)=f(b)=0$ and $|f''(x)|\le M$ prove $|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$ If $f(a)=f(b)=0$ and $|f''(x)|\le M$. Prove
$$|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$$
I have thought about that since $f(a) = f(b) = 0 $ there is $\xi$ such that $f'(\xi) = 0$. Then when $x \le \xi$, $|f'(x)| \le... | Show that $f(a)=f(b)=0$ and $|f''(x)|\le M$ implies
$$
\tag{*}
|f(x)| \le \frac M2 (x-a)(b-x)
$$
for $a \le x \le b$, i.e. $f$ is bounded by the parabola with constant second derivative $-M$ and zeros at $x=a$ and $x=b$.
Then
$$
\left|\int_a^b f(x)\, dx \right| \le \frac M2\int_a^b (x-a)(b-x) \, dx = \frac{M}{12}(b-a)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
GCD of cubic polynomials I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.
Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \b... | As you've already shown, factoring out the cube of the GCD of $a$ and $b$ gives a new equation
$$e = \gcd\left(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2 \right) \tag{1}\label{eq1}$$
where
$$\gcd\left(\alpha,\beta \right) = 1 \tag{2}\label{eq2}$$
Update: Here is a simpler solution than what I originally wrote. F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving combinatorial problems with symbolic method and generating functions I am trying to solve the following problems:
a) Let $\mathcal{F}$ be the family of all finite 0-1-sequences that have no 1s directly behind each other. Let the weight of each sequence be its length.
How can $\mathcal{F}$ be constructed with ... |
Ad c.)
Your approach is fine. With ${\mathcal{K}} = \{1,2\}^{\ast}$ we obtain
\begin{align*}
K(z) &= \frac{1}{1-\left(z+z^2\right)}\\
&=\frac{1}{1-z-z^2}\tag{1}
\end{align*}
and with $\tilde{\mathcal{K}} = \{2,3,4,5,6,...\}^{\ast}$ we obtain
\begin{align*}
\tilde{K}(z) &= \frac{1}{1-(z^2+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $f(x, y) = \frac{x^2}{y}$ is a convex function on the set $\{(x, y) \in \mathbb{R}^2 : y > 0\}$
Prove $f(x, y) = \frac{x^2}{y}$ is a convex function on the set $\{(x,y) \in \mathbb{R}^2 : y > 0\}$.
Attempt:
I start with the basic convexity, i.e.,
\begin{align}
f( \alpha_1 x_1 + \alpha_2 x_2) \leq \alpha_1 f(x... | A twice continuously differentiable function of several variables is convex on a convex set if and only if its Hessian matrix of second partial derivatives is positive semidefinite on the interior of the convex set.
$$f(x,y) = \frac{x^2}{y}$$
Partial derivatives:
$$\frac{\partial f(x,y)}{\partial x} = \frac{2x}{y}, \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating expression with Integer part and Fraction part of a nested radical
Let
$$A= \sqrt{93+28\sqrt{11}}$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I n... | Let's find intgers $x,y$ such that $$(x+y\sqrt{11})^2= 93+28\sqrt{11}$$
Then $$x^2+11y^2 =93 \;\;\;\wedge \;\;\; 2xy = 28$$
Since $11y^2<99\implies y^2 <9 \implies |y|\leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$\boxed{\sqrt{93+28\sqrt{11}} = 7+2\sqrt{11}}$$
Since $$13= 7+2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Need to compute $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x
^3}{3}+\frac{x^5}{5}+...$$
so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
... | $n=k$ $$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$ so
$$\underbrace{2^0+2^1+2^2+...+2^{k-2}}=\overbrace{2^{k-1} -1}$$
Now show for k+1
$$\underbrace{2^0+2^1+2^2+...+2^{k-2}}+(2^{k-1}+1)=\overbrace{2^{k-1} -1}+(2^{k-1}+1)$$
$$=2^{k-1}+2^{k-1}=2*2^{k-1}=2^{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find minimum perimeter of the triangle circumscribing semicircle The following diagram shows triangle circumscribing a semi circle of unit radius. Find minimum perimeter of triangle
My try:
Letting $$AP=AQ=x$$
By power of a point we have:
$$BP^2=OB^2-1$$ where $O$ is center of the circle.
Also $$CQ^2=OC^2-1$$
Let $$OB=... | The problem with that is that there are only really two degrees of freedom; the variables $x,y,z$ are related in some way. We could try to find what that relation is - or we could find another way to express things, one with easier relations to spot.
I'd go with the angles of the triangle. Labeling $O$ as you did, $OP$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| Using AM-GM inequality:$$\frac{x_1+\dots+x_n}{n}\geq (x_1\dots x_n)^{1/n}$$
let $n=3$ and $x_1=a+b,\; x_2=b+c,\; x_3=c+a$:
$$\frac{2(a+b+c)}{3}\geq [(a+b)(b+c)(c+a)]^{1/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| If $-1<x<0$ then $x^{5}+x^{4}=x^{4}(1+x) <1$ because $x^{4}$ and $1+x$ both belong to $(0,1)$. If $ x \leq -1$ then $x^{5}+x^{4}=x^{4}(1+x) \leq 0<1$. Hence there is no root with $x <0$. For $x>0$ the function is strictly increasing so it cannot have a second root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve $4x^3 - 9x - 1 = 0$, correct to 3 decimal places. I am trying to solve the following problem:
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve the equation $4x^3 - 9x - 1 = 0$, correct to 3 decimal places.
Assuming that $x = \cos 3\theta$, you ca... | Render $x=a\cos\theta$. Thereby
$4x^3-9x=4a^3\cos^3\theta-9a\cos\theta$
$=a^3(4\cos^3\theta-(9/a^2)\cos\theta)$
If we put $a=\sqrt{3}$ then $9/a^2=3$ and then
$4x^3-9x=3\sqrt{3}\cos(3\theta)=1$
where the "=1" cones from the constant term of the original equation. So $\cos(3\theta)=\sqrt{3}/9$, from which we can get v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve $ \int_0^a x^4 \sqrt{a^2-x^2}dx$
Solve $$ \int_0^a x^4 \sqrt{a^2-x^2}dx.$$
Substitution: $x = a\sin t \Rightarrow dx = a\cos t dt$
$$ \int_0^a x^4 \sqrt{a^2-x^2}dx = a^6 \int_0^{\pi/2}\sin^4t\cos^2 tdt = \frac{a^6}6 \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(3)} = \frac{a^6 \pi}{288}.$$
When evaluating this integral ... | A direct path to the Beta function can be made by considering the substitution $x = a \, \sqrt{u}$ in the integral:
\begin{align}
I &= \int_{0}^{a} x^{4} \, (a^2 - x^2)^{1/2} \, dx \\
&= \int_{0}^{1} a^4 \, u^2 \, (a^2 - a^2 u)^{1/2} \, \frac{a \, du}{2 \, \sqrt{u}} \\
&= \frac{a^6}{2} \, \int_{0}^{1} u^{3/2} \, (1-u)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3178366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the maximum values of the function $f(x,y,z)=x^2y^2z^2$ subject to the constraint $x^2+y^2+z^2=289$. I have already figured out the majority of the problem. I took the derivative of $x$ and got: $x=0,y=0,z=0$. Initially I put that there were infinitely many solutions, but that is not true. I do not know where I we... | With AM-GM$$x^2+y^2+z^2=289\ge3\cdot \sqrt[3]{x^2y^2z^2}$$ Thus $$\bigg(\frac{289}{3}\bigg)^3\geq x^2y^2z^2$$
Equality holds iff $\pm x=\pm y=\pm z$
Detailed Explanation
Theorem 1: Arithmetic Mean-Geometric Mean (or simply AM-GM) inequality for three variables
Let $a,b,c\in \mathbb R^+$, then $$\frac{a+b+c}{3}\ge\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this with mathematical induction? $$\sum_{i=0}^n \frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
Let's skip the check, since when n = 1, I have $\frac{1}{2} = \frac{1}{2}$
What i will next do ? What for expression i may receive ?
$$\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
My trying :
$$ \biggl(\sum_... | You should prove that$$\sum_{i=0}^n\frac i{2^i}=2-\frac{n+2}{2^n}\implies\sum_{i=0}^{n+1}\frac i{2^i}=2-\frac{n+3}{2^{n+1}}.$$In order to do that, you do:\begin{align}\sum_{i=0}^{n+1}\frac i{2^i}&=\left(\sum_{i=0}^n\frac i{2^i}\right)+\frac{n+1}{2^{n+1}}\\&=2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}\\&=2-\frac{2n+4}{2^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$. Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$.
I cannot give a proof to this... | After some discussion with my professor, we have solved this problem by using some elementary arguments.
We first prove the following lemma.
Lemma : For $n\geq 4$ we have $$s_n<\bigg(\frac{p_{n+1}-1}{2}\bigg)^2.$$
To show this we first see $s_4=2+3+5+7=17$ and $(p_{5}-1)/2=5$, so obviously $17< 5^2$. Now if ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \su... | Hint: if $G(x)$ is the generating function for $a_n$, what are $x G(x)$ and $x^2 G(x)$ the generating functions for?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$
Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$?
Quick Results
This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{... | First, we prove a lemma on the integral representation of $(H_n)^2$.
$$I_n=\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)=(H_n)^2$$
Let's prove by induction. $\displaystyle I_0=-\int_0^1\frac{\ln x}{1-x}dx=\zeta(2)=\zeta(2)+(H_0)^2$.\
Assume the equation holds for $n-1$,
$$\begin{aligned}
I_n&... | {
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"timestamp": "2023-03-29T00:00:00",
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Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\s... | You must convert the limits to be in terms of $\theta$. When $x=0,1$, what is $\theta$?
Also if you want another method consider partial fractions
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was fro... | *
*You can get the four apparent solutions from $x^{4} - 10x^{2} - x + 20 = (x^2 - x - 5) (x^2 + x - 4)$ and then solving two quadratic equations
*Squaring can produce spurious results:
*
*for example if $x=\frac{-1 - \sqrt{17}}{2}$ then $\sqrt{\sqrt{x + 5} + 5} = -x$ rather than $+x$
*but you lost that distinc... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
| Hint:
Square
$$\sqrt{2+\sqrt{3-3b^2}+b}- \sqrt{2-\sqrt{3-3b^2}+b} = \sqrt{2-2b}.$$
This gives
$$2+\sqrt{3-3b^2}+b+2-\sqrt{3-3b^2}+b-2\sqrt{(2+\sqrt{3-3b^2}+b)(2-\sqrt{3-3b^2}+b)}= 2-2b,$$
$$4+2b-2\sqrt{(2+b)^2-(3-3b^2)}= 2-2b,$$
$$2+4b-2\sqrt{4b^2+4b+1}= 0.$$
Remains to discuss the domain and the signs.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugat... | Hint: It is $$\frac{4+i}{i}+\frac{3-4i}{1-i}=\frac{(4+i)(1-i)+(3-4i)i}{i(1-i)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate problem I've tried u substitution and partial fraction doesn't seem to work either.
$\int \frac{1}{x^2+3}$
Can I take $\int \frac{1}{x^2+\sqrt(3)^2}$ in order to use $x=a\tan\theta$
| We are asked to evaluate $ \int \frac{1}{x^2+3}dx$. We need to know $\int \frac{1}{1+x^2}dx= \arctan(x)+C$ to answer this question.
We need to modify the denominator $x^2+3$. Note $x^2+3 = 3(\frac{x^2}{3}+1)$. So we have
$$ \int \frac{1}{x^2+3}dx = \int \frac{1}{3(\frac{x^2}{3}+1)}dx = \frac{1}{3} \int \frac{1}{\frac... | {
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"timestamp": "2023-03-29T00:00:00",
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Clarification in proof of perpendicular bisectors meeting at a point Context:
My crude drawing from Paint to illustrate triangle OAB:
My working:
$\begin{align*}
(z-\frac{1}{2}(x+y)) \cdot (y-x) &= z\cdot(y-x) + \frac{1}{2}(-x-y) \cdot (y-x)\\
&= z\cdot y -z\cdot x + \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{... | Let $K$ be the point of intersection of the perpendicular bisectors of the sides $OA$ and $OB.$ Let $\overrightarrow {OK} = z.$ Let $M$ be the midpoint of the side $AB.$ Then observe that if we can show that $\overrightarrow {MK} \perp \overrightarrow {AB}$ we are through. So we need only to show that $$\overrightarrow... | {
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"timestamp": "2023-03-29T00:00:00",
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Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum ... | It can be characterized as a sum of non-negative polynomials. This decompostion writes it as follow
$$\!1\!+\!a^{2}\!+\!b^{2}\!+\!c^{2}\!+\!4 abc\!-\!a\!-\!b\!-\!c\!-\!ab\!-\!bc\!-\!ca\!=\\=\!abc\left\{\!\left(\!1\!+\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)\left(\!2\!-\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)^{2}\!+\!\dfrac{3\left... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute a Jacobian using polar coordinates? Consider the transformation $F$ of $\mathbb R^2\setminus\{(0,0)\}$ onto itself defined as
$$
F(x, y):=\left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
Its Jacobian matrix is
$$\tag{1}
\begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\fra... | The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(\phi(\frac{1}{r}, \cos\theta)) × D\tilde{F}(r, \theta)= DF \times D(\phi(r, \theta))$
| {
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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.