Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proof by induction with square root in denominator: $\frac1{2\sqrt1}+\frac1{3\sqrt2}+\dots+ \frac1{(n+1)\sqrt n} < 2-\frac2{\sqrt{(n+1)}}$ I need to prove $\frac{1}{2\sqrt1} + \frac{1}{3\sqrt2} + ... + \frac{1}{(n+1)\sqrt n} < 2 - \frac{2}{\sqrt{(n+1)}}$
by induction for every $n \in \mathbb{N} $.
Please help, I am s... | Great answer by trancelocation, but in case you still want it, here is how to do induction step for an inductive proof. First we note the following general rule of quadratics:
$$(n + \tfrac{a+b}{2})^2 - (n+a)(n+b) = (\tfrac{a-b}{2})^2.$$
Using this rule with $a=1$ and $b=2$, we have:
$$(n+\tfrac{3}{2})^2 > (n+1)(n+2).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Integral $\int_{0}^{1}\frac{\log x}{2 - x} dx$
Calculate: $$\int_{0}^{1}\frac{\log x}{2 - x} dx$$
I've done a lot of research here in the community. I tried varying variants, but I did not get anything. The problem is the $2$ present in the integral.
| \begin{align}J&=\int_{0}^{1}\frac{\log x}{2 - x} dx\\
\end{align}
Perform the change of variable $x=2y$,
\begin{align}J&=\int_{0}^{\frac{1}{2}}\frac{\log(2y)}{1 - y} dy\\
&=\ln 2\left[-\ln(1-x)\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\
&=\ln ^2 2+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Nice power sum inequality $ab(a-b)\leq a^{ab}-b^{ab}$ for $a+b=2$
Let $a\geq b>0$ such that $a+b=2$ then we have :
$$ab(a-b)\leq a^{ab}-b^{ab}$$
My try :
We study the function $f(x)$ on $[0;1]$ such that :
$$f(x)=(2-x)^{((2-x)x)}-x^{((2-x)(x))}-(2-x)x(2-2x)$$
the derivative is equal to :
$$f'(x)= x^{(x (2 - x))} (-... | Let $a=1+x$ and $b=1-x$.
Thus, $x\geq0$ and
$$a^{ab}=(1+x)^{1-x^2}=1+x-x^3-\frac{x^4}{2}+\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7)$$ and
$$b^{ab}=(1-x)^{1-x^2}=1-x+x^3-\frac{x^4}{2}-\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let a,b and c be the side lengths of triangle ABC respectively...find the greatest value of b*c. Let a,b and c be the side lengths of triangle ABC respectively. If the perimeter of $\Delta$ABC is 7, and that $\cos A=-\frac{1}{8}$, find the greatest value of $b*c$.
This is how I start the solution:
$$a+b+c=7 \impli... | Apply a.m.-g.m.-inequality on $\sqrt{bc}$.
$$\frac{(7-a)^2}{4}=\left(\frac{b+c}{2}\right)^2 \ge bc = 4(7-2a)$$
$$49-14a+a^2 \ge 16(7 - 2a)$$
$$a^2+18a-63 = (a+21)(a-3)\ge 0$$
$$a \le -21 \text{(rejected) or } a \ge 3$$
$a\mapsto \dfrac{(7-a)^2}{4}$ is the parabola shifted $7$ units to the right multiplied by $1/4$, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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recursive relation on derangement of objects Let $a_{n}$ represent the number of derangements of $n$ objects . If $a_{n+2}=p a_{n+1}+q a_{n}\;\forall n\in\mathbb{N}$ then what is $\displaystyle \frac{q}{p}$?
What I have tried:
I have used
$$ a_{n}=n!\bigg[1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^n\frac{1}{n!}\bigg],$$... | The question implicitly seems to assume that there exist $p$ and $q$ that satisfy this relation for all $n$. Then in particular they must satisfy it for $n=1$ and $n=2$. This gives you two linear equations in $p$ and $q$ that are easily solved: For $n=1$ and $n=2$ you get
$$2=1\cdot p+0\cdot q\qquad\text{ and }\qquad 9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to solve this two limit tasks Hello i stumbled across this two limits task and i cant find an answer to them:
*
*Find the limit depending on the parameter $A$
$$\lim \limits_{x\to\infty}\left(\left(\sqrt{x+1} - \sqrt[4]{x^2 + x + 1} \right) \cdot x^A\right)$$
I tried by multipliying with
$$\frac{\sqrt{x+1} ... | For 2 we have $\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)=\lim\limits_{x\to\infty}\frac{2x^2+3x+1}{(x+1)^2+(x+1)\sqrt[3]{x^3 + x^2} +\sqrt[3]{(x^3 + x^2)^2}}$.
Now you can factor $x^2$ both in the numerator and the denominator and see that the limit is $\frac{2}{3}$.The first limit can be done si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Am I properly simplifying this geometric progression? I'm studying recurrence relations and am given:
$T(n) = 2 \cdot T(n-1) - 1$
with an initial condition that $T(1) = 3$.
I worked through the first few recurrences:
$T(n-1) = 2^2 \cdot T(n-2) - 2 - 1$
$T(n-2) = 2^3 \cdot T(n-3) - 2^2 - 2 - 1$
and so forth, and came th... | The left sides of your second and third equations should be $T(n)$, not what you have written. Under "Factoring out a $-1$" it would be better to end with $2^{n-3}+2^{n-2}$ to maintain the pattern of the exponents increasing by $1$. Your progression does end with $2^{n-2}$. That is why the numerator of the sum has $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Find maximum value by using AM-GM inequality I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$.
Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.
The ... | Another way based on Michael Rozenberg's solution.
If you prove $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ you can prove the lemma $$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1}\ge 1$$ for $abc=1$ and $a;b;c\in R^+$ $(\text{Prove by C-S and} (a=xy/z^2;b=yz/x^2;c=xz/y^2))$
Note that we have: $$\frac{a^3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ *... | $a\sqrt{bc} = a\sqrt{b}\sqrt c$.
It is FALSE that $a \sqrt{bc} = a\sqrt b\times a\sqrt c$. There is only one $a$; not two.
$6\sqrt{24} = 6\sqrt{4\times 6}= 6\sqrt 4 \times \sqrt 6$.
Your calculation $6\sqrt{4\times 6} = (6\sqrt{4})\times (6\sqrt{6})$ is just plain wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Eisenstein and Weierstrass zeta - series identity Let $\zeta_\Lambda$ be the Weierstrass $\zeta$-function for lattice $\Lambda$ and $G_2$ the Eisenstein series of weight $2$.
The quasiperiod is defined by $\eta_\Lambda(\lambda) := \zeta_\Lambda(z + \lambda) - \zeta_\Lambda(z)$.
Then the following identity holds \begin{... | $$G_2(\tau)=\sum_c\sum_d\frac{1_{(c,d)\ne (0,0)}}{(c\tau+d)^2}$$
It converges and is holomorphic for $\Im(\tau)> 0$ because $\sum_d \frac{1}{(c\tau+d)^2} = \frac{\pi^2}{\sin^2(\pi c\tau)}= O(e^{-2 \pi |c\Im(\tau)|})$. Then $$\zeta_\tau(z) =\sum_c\sum_d\frac{1}{z-c\tau-d}+1_{(c,d)\ne (0,0)}(\frac{1}{c\tau+d}+\frac{z }{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complete the Square with $4x^2 - 4x + 3$ I am trying to complete the square of $4x^2 - 4x + 3$ and using the shortcut approach outlined here to get $d,e$ faster
Here is my work so far,
$$
\begin{align}
4x^2 - 4x + 3 &= 0 \\
x^2 - x + \frac{3}{4} &= 0 \\
x^2 - 2dx + (d^2 + e) &= 0 \\
d = - \frac{1}{2}, e = \frac{1}{2... | You got a good result, just recall that you actually have
$$ax^2+bx+c=a(x-q)^2+ar$$
instead of $(x-q)^2+r$ for the final result. Since $a=4$, you have
$$4\left(x-\frac 12\right)^2+4\cdot\frac 12\\
=(2x-1)^2+2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $ I have a question that:
There is a point $D$ inside the equilateral triangle ABC. If
$$ DB^2 + DC^2 + BC^2 = 100 $$
and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.
This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 ... |
Let $|AB|=|BC|=|CA|=a$,
$|DB|=m$, $|DC|=n$, $|AD|=q$,
$|OB|=\tfrac12a$,
$|OH|=x$, $|DH|=y$
We have two constraints
\begin{align}
m^2+n^2+a^2&=100
\tag{1}\label{1}
,\\
S_{DBC}&=5\sqrt3
\tag{2}\label{2}
.
\end{align}
From \eqref{1},
\begin{align}
(\tfrac12a+x)^2+y^2+
(\tfrac12a-x)^2+y^2+a^2=100
,\\
x^2+y^2 = 50-\tfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F... | As you note, the function $f$ with primitive $F$ satisfies $F(x-1)+f(1-x)=x^3$ if and only if
$$F(x)+f(-x)=(x+1)^3.$$
The primitive $F(x)$ uniquely determines $f(x)$, its derivative, so we'll focus solely on $F(x)$. If $F(x)$ is twice differentiable, or equivalently, if $f$ is differentiable, then differentiating the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the integral $\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$ I have to calculate the integral $\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$. I tried the following substitutions: $x \rightarrow \frac{1+t}{1-t}, x \rightarrow \frac{1-t}{1+t}, x \rightarrow \frac{t^{2}+1}{t^{2}-1}$ but with no good result.... | Hint The standard substitution for integrals like this, where the integrand includes $\sqrt{\frac{1 + x}{1 - x}}$, is simply $u = \sqrt{\frac{1 + x}{1 - x}}$. Rearranging and differentiating gives $$x = \frac{u^2 - 1}{u^2 + 1}, \qquad dx = \frac{4 u \,du}{(u^2 + 1)^2}.$$
So in our case, where the integrand is a product... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For any positive integer $n$, let $f(n) = 70 + n^2$ and $g(n)$ be the HCF of $f(n)$ and $f(n+1)$ find the highest possible value of $g(n)$. For any positive integer $n$, let $f(n) = 70 + n^2$ and $g(n)$ be the HCF of $f(n)$ and $f(n+1)$ find the highest possible value of $g(n)$.
This question is from HKIMO prelims 200... | Using your
$$g(n) = (70+n^2, 2n+1) \tag{1}\label{eq1}$$
note $2n + 1$ is odd, so can multiply $70 + n^2$ by $4$ to $280 + 4n^2$, but with it still giving the same HCF. However, note that $2n \equiv -1 \pmod{2n + 1}$, so $4n^2 \equiv 1 \pmod{2n + 1}$. Thus,
$$280 + 4n^2 \equiv 281 \pmod{2n + 1} \tag{2}\label{eq2}$$
I tr... | {
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"url": "https://math.stackexchange.com/questions/3140817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum $\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)$ based on $a^2-b^2 $? Find the sum of the given series
$$\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right).$$
My attempt : I used the formula $a^2- b^2 = (a+b)(a-b)$, then I get
$$\sum_{n=1}^{\infty} \left((-1)^{n-... | Based on the link that Jack provided,
$${\begin{aligned}
\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\\eta (s)&=(1-2^{1-s})\zeta (s)\\
&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0
\end{aligned}}$$
where $\zeta (s)$ is the zeta function, and $\eta (s)$ is the eta function. Then we can e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^{(10)}(0)$ I have to find Maclaurin series for function $f(x)$ = $2x^2\over{16+x^4}$, it's interval of convergence and $f^{(10)}(0)$. I managed to calculate Maclaurin series and $10^{th}$ derivative, but I'm not sure if it's done... | Your computation of the MacLaurin series of $f$ looks right, but remember that $f^{(10)}(0)$ is a number. You know that$$\frac{f^{(10)}(0)}{10!}=\frac{(-1)^2}{2^{11}}.$$Therefore,$$f^{(10)}(0)=\frac{10!}{2^{11}}.$$
Also, $\left\lvert\frac{-x^4}{16}\right\rvert<1$ doesn't just imply that $x\in(-2,2)$; it is actually equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ $a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(... | By C-S $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=\frac{\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)(a+b+c)}{a+b+c}\geq$$
$$\geq\frac{\left(\frac{x}{\sqrt{a}}\cdot\sqrt{a}+\frac{y}{\sqrt{b}}\cdot\sqrt{b}+\frac{z}{\sqrt{c}}\cdot\sqrt{c}\right)^2}{a+b+c}=\frac{(x+y+z)^2}{a+b+c}.$$
The equality occurs for $$\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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closed form of $\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$ I am looking for the closed form of this product.
$$\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone ... | Let $$a_n=\left(\frac{n}{n+1}\right)^{(-1)^{n-1} n}$$ then
$$a_{2p}= \left(\frac{2p}{2 p+1}\right)^{-2 p}\qquad \text{and}\qquad a_{2p+1}=\left(\frac{2 p+1}{2 p+2}\right)^{2 p+1}$$
Now, using a CAS,
$$\prod_{p=1}^m a_{2p}=\frac{\sqrt[12]{2} \sqrt{\pi } \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+2 \zeta
^{(1,0)}\left(-1,m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Can I find all solutions of $2^{n-1}\equiv k\mod n$? Suppose$\ k\ge 2\ $ is a positive integer.
Can I find all positive integers $\ n>1\ $ with $$2^{n-1}\equiv k\mod n$$ ?
I only found out yet that there is always a solution if $\ k>2\ $ and $\ k-1\ $ is not a power of $\ 2\ $. In this case, $\ k\ $ has an odd prime ... | Clearly n can not be prime. I had following experiment:
$2^{4-1}=8=2\times 4 +0$
$2^{6-1}=32=5\times 6 +2$
$2^{8-1}=128=16\times 8+0$
$2^{9-1}=256=28\times 9 +4$
$2^{10-1}=512=51\times 10+2$
$2^{12-1}=2042=170\times 12 +8$
$2^{14-1}=8192=585\times 14 +2$
$2^{15-1}=16384=1092\times 15 +4$
$2^{16-1}=32768=2048\times 16 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i so... | thanks friends got the result
$\displaystyle \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\int\frac{2x+3x^{-4}}{(x^2-x^{-3})^2}dx$
put $x^2-x^{-3}=t$ and $(2x+3x^{-4})dx=dt$
Integration is $\displaystyle \int t^{-2}dt=-\frac{1}{t}+C=-\frac{x^3}{x^5-1}+C$
$$\int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{x^3-x^8}{(x^5-1)^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Generating function with tough restrictions
In how many ways can a coin be flipped $25$ times in a row so that exactly $5$ heads occur and no more than $7$ tails occur consecutively?
For the heads, I think that it is $\binom{25}{5}$, but I do not know what to do with the tails restriction.
Not sure how to approach t... | This answer is based upon the Goulden-Jackson Cluster Method.
We consider the set words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the set $B=\{TTTTTTTT\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coeffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Unexpected (incorrect) solution to Lagrange Inversion solution to $x^4 - x^3 - x^2 - x - 1 = 0$ about the solution near $x = 2$ I am developing generalized hypergeometric solutions for a set of such polynomials. With this example we can write $x^4 - x^3 - x^2 - x - 1 = \frac{x^5 - 2 x^4 + 1}{x - 1}$.
Lagrange Invers... | You've chosen $w = 2 x^4 - x^5$, so either the inverse function is given by a Puiseux series in powers of $w^{1/4}$ or the center of the expansion is not $(0, 0)$. In the first case,
$$a_k = \lim_{x \to 0^+} \frac 1 {k!} \frac {d^{k - 1}} {d x^{k - 1}}
\left( \frac x {(2 x^4 - x^5)^{1/4}} \right)^{\!k} =
\frac {(-1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt$ Recently I came across with respect to this post of mine hyperbolic solution to the cubic equation for one real root given by
$$
t=-2\sqrt \frac {p}{3} \sinh \left( \frac {1}{3} \sinh^{-1} \left( \frac {3q}{2p} \sqrt \... | We have from symmetry that $$I=\frac23\int_0^1\sinh^{-1}\left[\frac{3\sqrt3}2(1-x^2)\right]dx$$
So we define
$$f(a)=\int_0^1\sinh^{-1}[a(1-x^2)]dx$$
Then we recall that
$$\sinh^{-1}(x)=x\,_2F_1\left(\frac12,\frac12;\frac32;-x^2\right)=\sum_{n\geq0}(-1)^n\frac{(1/2)_n^2}{(3/2)_n}\frac{x^{2n+1}}{n!}$$
so
$$\sinh^{-1}[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a ... | The homogenisation gives
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$
which is true by Schur.
Indeed, we need to prove that
$$\sum_{cyc}(a-1)^3+\frac{3}{4}\geq0$$ or
$$\sum_{cyc}\left(a^3-3a^2+3-1+\frac{1}{4}\right)\geq0$$ or
$$\sum_{cyc}(a^3-(a+b+c)a^2)+\frac{27}{4}$$ or
$$4\sum_{cyc}(-a^2b-a^2c)+(a+b+c)^3\geq0$$ or
$$\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$ Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$
I'm pretty sure this has something to do with Holder's inequality, but I don't know how to solve this. By guessing I found $a=2,b=2,c=d=0$ the smallest solution with $\Si... | Does not exist. Try $a>0$,$c>0$, $d>0$ and $b\rightarrow-\sqrt[3]4^-.$
For positive variables by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b^3+4}=1+\sum_{cyc}\left(\frac{a}{b^3+4}-\frac{a}{4}\right)=1+\sum_{cyc}\frac{-ab^3}{4(b^3+4)}=$$
$$=1+\sum_{cyc}\frac{-ab^3}{2(b^3+b^3+8)}\geq1+\sum_{cyc}\frac{-ab^3}{2\cdot3\sqrt[3]{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$
If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$
Here's what I did.
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \fr... | We can use rearrangement inequality to obtain$$
b^4+c^4\ge b^3c+bc^3=bc(b^2+c^2)=\frac{b^2+c^2}a.
$$ This implies
$$\begin{align*}
\sum_{cyc}\frac{a}{a+b^4+c^4}&\le\sum_{cyc}\frac{a}{a+\frac{b^2+c^2}a}\\&=\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}\\&=\frac{\sum_{cyc}a^2}{a^2+b^2+c^2}\\&=1,
\end{align*}$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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inequality related to square the sum of any two sides of a triangle with respect to square of other side Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.
My approach
As we know that the sum of any two sides ... | Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,\quad |b-c|<a,\quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,\quad (b-c)^2<a^2,\quad (c-a)^2<b^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow ... | Hint: by the binomial theorem, $$\sum_{n\ge 0}\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(\frac{1}{2}-n)}{n!}(-k^2)^n=(1-k^2)^{-1/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $
$$
\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$
RHS-
$$\frac{1}{2} \ - \fr... | Using a telescoping sum, we get
$$
\begin{align}
\sum_{k=1}^n\frac{k+2}{k(k+1)2^{k+1}}
&=\sum_{k=1}^n\left(\frac1{k2^k}-\frac1{(k+1)2^{k+1}}\right)\\
&=\sum_{k=1}^n\frac1{k2^k}-\sum_{k=2}^{n+1}\frac1{k2^k}\\
&=\frac12-\frac1{(n+1)2^{n+1}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove the sequence $\sum_{k=0}^n \frac{1}{(n+k)^2}$ is convergent I'm having a hard time trying to prove that the sequence $\{a_n\}$ whose general term $a_n$ is
$$\sum_{k=0}^n \frac{1}{(n+k)^2}$$
is convergent. I'm trying to prove it by definition, that is to say, by finding a lower/upper bound and by proving that it i... | To prove that this sequence converges, we will use the monotone convergence theorem which states
If $(a_n)$ is either
*
*increasing and bounded above
*decreasing and bounded below
then it is a convergent sequence.
In order to use this result, we first look at the difference $a_{n+1}-a_{n}$ to det... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to perform polynomial long division on 1/(1 - x)? How do I perform polynomial long division on $\frac{1}{1 - x}$ to obtain the sequence $1 + x + x^2 + x^3 + \cdots$?
In this video, the teacher went about it in the following way...
$$
\require{enclose}
\begin{array}{r}
1 + x + x^2 + x^3 + \cdots \\
1 - x \enclose... | Another way to look at it. Start with the known formulas:
$$\begin{align}1-x^2&=(1-x)(1+x)\\
1-x^3&=(1-x)(1+x+x^2)\\
1-x^4&=(1-x)(1+x+x^2+x^3)\\
&\ \ \vdots\\
1-x^n&=(1-x)(1+x+x^2+x^3+\cdots+x^{n-1}) \Rightarrow \\
\frac{1-x^n}{1-x}&=1+x+x^2+x^3+\cdots +x^{n-1}\end{align}$$
If $|x|<1$ and $n\to \infty$, then $x^n\to 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the equation $a^2+d^2+4=b^2+c^2$ have any solutions? Does the equation have $a^2+d^2+4=b^2+c^2$ where $d<c<b<a$ have any integer solutions? This isn't a homework problem, but I need to know for a separate problem I'm doing. Wolfram Alpha isn't very helpful.
| Above equation shown below has parameterization:
$a^2+d^2+4=b^2+c^2$
$(2m)^2+(m-2)^2+(2)^2=(m+2)^2+(2m-2)^2$
For, $m=5$, we get:
$(10,3,2)^2=(7,8)^2$
Hence, the integer $4$ can be represented by sum difference of four squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help verifying proof to proving that for all natural numbers $\sqrt{1} \leq$ than the sum Prove that for all natural numbers $n$,
$$\sqrt{n} \le 1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Solution: We must prove that $1 + \frac 1 {\sqrt{2}}, + \frac 1 {\sqrt{3}} +\,\cdots\, + \... | Multiply both sides by $\sqrt{n}$. Then the assertion is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$
Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$
Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient ... | Hint
Observe that the exponent of $1-x$ is $-\dfrac32$
So, let us find $$\dfrac{d\left(e^x\dfrac{(1+x)^n}{\sqrt{1-x}}\right)}{dx}$$
Compare with the given expression to find the value of $n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x... | LEMMA
Given integers $$ m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with
$$ x^2 - Mxy + y^2 = -m. $$
PROOF
Calculus: $m+2 > \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left( \sqrt{4m+4} \right)^2 = 4m + 4.$ Therefore also
$$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Understanding Mathematical Induction problems I have problem to understand this 3 formulas, I am new in this type of problems.
I have to solve this problems by induction.
\begin{align}
\sum_{j = 1}^{j = n} j^3 &= \left(\frac{n(n + 1)}{2}\right) ^ 2 &\text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j + 1) &= \frac{1}{3... | Here is an example of using induction that doesn't use sigma notation.
The OP can put it in their #Education Reference Folder.
Show that
$$\tag 1 1 + 2 + \dots + n = \frac{n(n+1)}{2}$$
Base Case:
True when $n = 1$ since $1 = \frac{1(1+1)}{2}$.
Inductive Step:
Assume that $1 + 2 + \dots + k = \frac{k(k+1)}{2}$.
Then $1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find $A^{100}$ where $A$ is a matrix with trigonometric entries?
How to find $A^{100}$ where $A=\left(\begin{matrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{matrix}\right)$. ? (Given$\theta=\frac{2\pi}{7}$)
I am a highschool student and new in linear algebra and seeking for only problem sol... | From the characteristic equation,
$$\begin{align*}
\lambda^2 - 2\lambda \cos \theta + 1 &= 0\\
\lambda &= \frac{2\cos \theta\pm \sqrt{4\cos^2\theta-4}}{2}\\
&= \cos\theta \pm i\sin\theta
\end{align*}$$
For $\lambda_0 = \cos\theta + i\sin\theta$,
$$\begin{align*}
\pmatrix{\cos \theta-\lambda_0 &-\sin\theta\\\sin\theta &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Elementary way to evaluate $\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}$
Evaluate:
$$
\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x},\ a>0,\ n\in\Bbb N
$$
I've given it several tries but couldn't find an elementary method to find the limit. Two other ways that worked are L'Hospital's rule and generaliz... | Just write
*
*$\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} = 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}$
Now, use $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
So, you get
\begin{eqnarray*} \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}
& = & 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that $\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}<\frac{2}{R'-OO'}$ Let $\Delta ABC$ is acute triangle has $AB>AC$ and $O$ is incircle of $\Delta ABC$ with radius $R$, $O'$ is circumcircle of $\Delta ABC$ with radius $R'$. $OA\cap BC=A';OB\cap AC=B';OC\cap AB=C'$ .Prove that $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{C... | I shall use standard notation in my proof, that is, $R, r, s$ and $\Delta$ are the circumradius, inradius, semi-perimeter and area of the triangle, $AA'$ is the length of $A-\text{internal angle bisector}$ and other similarly.
$$rR(\sin A+\sin B+\sin C)=rs=\Delta=\frac{1}{2}ab\sin C=2R^2 \sin{A}\sin{B}\sin{C}$$$$\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two posit... | Let $y = 1 + (2^k - 1) i$ and $2^k y = 1 + (y^2 - x^2) j$.
Then
1) $2^k y - 1 = (2^k-1) (2^k i+1) = (y^2 - x^2) j$,
2) $(y-1) (y+i) = i j (y^2 - x^2)$,
3) $(-(2^k-1) + j (y^2 - x^2)) ((2^k i+1) + j (y^2 - x^2)) =
i j (y^2 - x^2) 2^{2k}$,
4) $(2y(i j-1)-(i-1))^2 - (i j-1) i j (2x)^2 = (i - 1)^2 - 4 (i j-1) i$,$\quad$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Prove with induction $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$ Prove with induction the identity
$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$
How can I solve this problem?
Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?... | Since $k$ is a dummy variable, you should induct on $n$. In other words, show the claim is right when $n=1$, then show $$\sum_{k=1}^p\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{p(p+2)}{3(2p+1)(2p+3)}\implies\sum_{k=1}^{p+1}\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{(p+1)(p+3)}{3(2p+3)(2p+5)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align... | We have $$\theta=\frac{1}{9}\arctan\frac{3}{4}.$$
Thus,
$$\frac{3}{\sin3\theta}-\frac{4}{\cos3\theta}=\frac{3}{\sin\frac{1}{3}\arctan\frac{3}{4}}-\frac{4}{\cos\frac{1}{3}\arctan\frac{3}{4}}=$$
$$=\frac{3\cos\frac{1}{3}\arctan\frac{3}{4}-4\sin\frac{1}{3}\arctan\frac{3}{4}}{\sin\frac{1}{3}\arctan\frac{3}{4}\cos\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$ Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of
$$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$
To find the minimum of $P$ I rewrite it as
$$P=2\left( ab+... | Let $a=b=0$ and $c=2$.
Thus, $P=4$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$(6-a^2-b^2-c^2)(2-abc)\geq4$$ or
$$\left(\frac{3(a+b+c)^2}{2}-a^2-b^2-c^2\right)\left(\frac{(a+b+c)^3}{4}-abc\right)\geq\frac{(a+b+c)^5}{8}$$ or
$$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2+6a^3bc+6a^2b^2c)\geq0,$$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate by substitution (Not by parts) $\int(x^2-2x)(x^3-3x^2+9)dx$ I have an integration question and I cannot figure out why my answer is incorrect as online calculators will use integration by parts instead, when technically we haven't (and won't) learn that method so I want to be very comfortable with integration... | As per your comment, you can always expand:
$$\int\big(x^2-2x\big)\big(x^3-3x^2+9\big)dx$$
$$= \int x^5-3x^4+9x^2-2x^4+6x^3-18x \ \mathrm{d} x$$
$$= \int x^5-5x^4+6x^3+9x^2-18x \ \mathrm{d} x$$
$$= \frac{1}{6}x^6 - x^5 + \frac{3}{2}x^4 + 3x^3-9x^2 + C$$
$$= \frac{1}{6} \left(x^6 - 6x^5 + 9x^4 + 18x^3 - 54x^2 + C'\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to compute this improper integral?
Let $n\geq1$ be an integer and let
$$I_n=\int\limits_{0}^{\infty}\dfrac{\arctan x}{(1+x^2)^n} \,\mathrm dx$$
Prove that
$$\sum\limits_{n=1}^{\infty}\dfrac{I_n}{n}=\dfrac{\pi^2}{6} \tag{1}$$
$$\int\limits_{0}^{\infty} \arctan x\cdot\ln\left(1+\frac{1}{x^2}\right) \,\mathrm d x=\df... | You can use the monotone convergence theorem to show that the series and the integral are equal:
\begin{align}
\sum \limits_{n=1}^\infty \frac{I_n}{n} &\stackrel{\text{MCT}}{=} \int \limits_0^\infty \arctan(x) \sum \limits_{n=1}^\infty \frac{1}{n (1+x^2)^n} \, \mathrm{d} x = \int \limits_0^\infty \arctan(x) \left[-\l... | {
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Transform $dx/dt$ to $dr/dt$ polar coordinates I've had to screenshot the question and post it as a photo
| The transformations can be written as
$$\tag 1 r^2 = x^2 + y^2~~ \text{and}~~ \theta = \tan^{-1}\left(\frac{y}{x}\right)$$
Differentiating $(1)$
$$r^2 = x^2 + y^2 \implies 2 r r' = 2x x' + 2y y' \implies r ' = \dfrac{x x' + y y'}{r}$$
Substituting $x = r \cos \theta, y = r \sin \theta$, we have
$$\begin{align} r' &= \... | {
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Generating function of binomial coefficients We want to evaluate the sum $$\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L$$
From this set of notes (page 2, equation 8) we find the formula $$\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^n}{(1-y)^{n+1}}$$ which suggests that I can do $$\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2... | Note: The cited formula (8) is not correct. There is a typo since we have
\begin{align*}
\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^{\color{blue}{k}}}{(1-y)^{\color{blue}{k+1}}}\tag{1}
\end{align*}
We therefore expect
\begin{align*}
\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} =\frac{1}{x}\sum_{L=1}^{\inf... | {
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Related to angles in tetrahedron Let $OABC$ be a tetrahedron such that $|OA|=|OB|=|OC|$. Denote by $D$ and $E$ the midpoints of segments $AB$ and $AC$ respectively. If $\alpha=\angle(DOE)$ and $\beta=\angle(BOC)$ what is the ratio $\beta/\alpha$?
It is obvious that $|BC|=2|DE|$ since triangles $\Delta(ABC)$ and $\Delt... | This answer attempts to find a formula for the ratio.
I imagine this becomes finding the distance of two mid-points of a spherical triangle.
Construct a sphere centred at $O$ with radius $r=OA$.
Extend $OD$ and $OE$ to meet the sphere at $D'$ and $E'$ respectively; $D'$ and $E'$ will be the mid-points of arcs $\overset... | {
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 ... | Mod 11, it reduces to $2\equiv 4y+4$ and $24=2\cdot11+2=4(5)+4$ which then gets $11x+13=189\implies 11x=176\implies x=16$
Note however, this is just one of infinitely many answers. Incrementing x by 35, and y by 11 yields new answers that work. All the x values fall on the line $35z+16$, all the y values fall on the li... | {
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Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force.... | Consider
$$(8+3\sqrt7)^2=127+48\sqrt7,$$
$$(8+3\sqrt7)^3=(8+3\sqrt7)(127+48\sqrt7)=2024+765\sqrt7,$$
$$(8+3\sqrt7)^4=(8+3\sqrt7)(2024+765\sqrt7)=32257+12192\sqrt7$$
etc. Then the first few solutions are $(8,3)$, $(127,48)$, $(2024,765)$,
$(32257,12192)$ etc.
| {
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Laurent series of $f(z)=\frac{4z-z^2}{(z^2-4)(z+1)}$ in different annulus Given $f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$
I need to find the Laurent series in the annulus: $A_{1,2}(0),\;A_{2,\infty}(0),\;A_{0,1}(-1)$
I found the following partial fractions:
$f(z)=\dfrac{-3}{(z+2)}+\dfrac{1}{3(z-2)}+\dfrac{5}{3(z+3)}$,
the p... | There's a typo in your partial fraction decomposition; it should be$$\frac{4z-z^2}{(z^2-4)(z+1)}=-\frac3{z+2}+\frac1{3(z-2)}+\frac5{3(z+1)}.$$If $z\in A_{1,2}(0)$, then:
*
*$\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=-\frac32\sum_{n=0}^\infty\frac{(-1)^n}{2^n}z^n$;
*$\displaystyle\frac1{3(z-2)}=-\frac16\f... | {
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Limit of this sum
Evaluate $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n})$.
I converted the limit of this sum to the sum of limits. Then in each term, I divided the numerator and denominator by $n$. Each limit came out to be zero. Hence I got the answer as $0$. Have I gone wrong so... | Ok: This is inspired on Riemann Sums
Let $S_n=\sum_{k=0}^{n-1}{\frac{1}{n+k+1}}$ be your sum
We first split $[0;1]$ into $n$ intervals of the form $[\frac{k}{n};\frac{k+1}{n}]$, with $k\in\{0,1,2,...,n-1\}$
Note that if $\frac{k}{n}\le x\le \frac{k+1}{n}$ then
$\frac{1}{\frac{k+1}{n}+1}\le\frac{1}{1+x}\le\frac{1}{1+\... | {
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$a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq 2$
*
*Let $a_{n}$ be the sequence satisfying $a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq... | Assume $1\le a_k\le 2$ for all $k<n.$ Then $\frac12(a_{n-1}+\frac2{a_{n-2}})\leq \frac12(2+2/1)=2$ and similarly $\frac12(a_{n-1}+\frac2{a_{n-2}})\ge \frac12(1+2/2)=1$ because $1\leq a_k\leq 2$ for all $ k\le n-1.$
| {
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What can be said about $(\varepsilon-x)y=y'(-x+y^2-2x^2)$ solutions? There was an unanswered question 4 years ago. OP asked for a solution of ODE $(\varepsilon-x)y=y'(-x+y^2-2x^2)$
The comment to the original question proposes an implicit solution, $2\log y + 2\epsilon\log(x + 2 x\epsilon - y^2) - (1+2\epsilon)\log(\ep... | $$(\epsilon-x)y=y'(-x+y^2-2x^2)$$
$$(\epsilon-x)y\:dx-(-x+y^2-2x^2)dy=0 \tag 1$$
The integrating factor is
$$\boxed{\mu=\frac{1}{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}}\tag 2$$
Multiplying Eq.$(1)$ by $\mu$ leads to the total differential of the sought function $F(x,y)$ :
$$\frac{ (\epsilon-x)y\:dx-(-x+y^2... | {
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What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ in terms of $x$?I tried factoring $x$ from both polynomials but I don't know what to do next since there'd be a $1$ in the second ... | $x^{2n}\equiv1 \pmod{x^2-1}$ so $(x^2-1) | (x^{2n}-1)$ so $(x^3-x) | (x^{2n+1}-x)$
so $x^{2n+1}\equiv x \pmod {x^3-x}$
so $x^7+x^{27}+x^{47}+x^{67}+x^{87}\equiv 5x \pmod {x^3-x}$
| {
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How can I show these series combinations are the same algebraically? We have to show:
$$\frac{1}{2}\sum_{r=1}^{n} {\frac{1}{r}} \ + \ \sum_{r=1}^{n} {\frac{1}{2r+1}} \ = \ \sum_{s=1}^{2n}{\frac{1}{s+1}}$$
I understand how this works intuitively and the most formal way I can think to show it is as follows:
$$\frac{1}{2}... | You can use the following trick to only count the even terms:
$$\frac{1}{2}\sum_{r=1}^n\frac{1}{r} = \sum_{r=1}^n\frac{1}{2r} = \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r}+1}{2}$$
And a similar trick to only count the odd terms. Notice that I've manually added two terms $\left[\frac{1}{2n+1}-\frac{1}{1}\right]$ to ... | {
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Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$
By using Mathematical Induction, prove the following equation for all positive integers $n$:
$$\tan^{-1}\frac{1}{2 \cdot 1^{2}} + \tan^{-1}\frac{1}{2 \cdot 2^{2}} +\cdo... | All you need to do is to prove
$$ \tan^{-1}\dfrac{1}{2k+1}-\tan^{-1}\dfrac{1}{2(k+1)^{2}} = \tan^{-1} \dfrac{1}{2k+3}.$$
You can use the formula you mentioned in this way:
$$ \begin{align}\tan^{-1}\dfrac{1}{2k+1} - \tan^{-1} \dfrac{1}{2k+3} &= \tan^{-1} \dfrac{\frac{1}{2k+1}-\frac{1}{2k+3}}{1+\frac{1}{2k+1}\frac{1}{2k+... | {
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How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$). We know that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See here).
However, we want to write the formula
\begin{align}
&\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqr... | Let $f(n)=
\dfrac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \dfrac{3-\sqrt{3}}{6} (2-\sqrt{3})^n$ and $g(n)=\dfrac{\sqrt3} 6(2+\sqrt3)^n-\dfrac{\sqrt3}6(2-\sqrt3)^n$.
Can you show $f(n)^2+2\times g(n)^2=f(2n)$ and $f(n)^2+2\times g(n+1)^2=f(2n+1)?$
| {
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What is $\cos\left(\frac{\arctan(x)}{3}\right)$ and $\sin\left(\frac{\arctan(x)}{3}\right)$? I know that $$\cos(\dfrac{\pi}{3} - \arctan(x))= \dfrac{1}{2\sqrt{(1+x^2)}} + \dfrac{\sqrt{3}x}{2\sqrt{(1+x^2)}}$$
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right)$ = ?
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\... | Triple cosine formula is more well-known: $\cos(3\theta)=4\cos^3\theta-3\cos\theta$. If we let $3\theta=\arctan x$ and $y=\cos\theta$, then
$4y^3-3y=\cos3\theta=\cos\arctan x=\frac{1}{\sqrt{x^2+1}}$. Hence, we have to solve the equation
$$y^3-\frac34y-\frac{1}{4\sqrt{x^2+1}}=0.$$
Its Lagrange resolvent is $z^2--\frac{1... | {
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A geometry problem about triangle angles and perimeter Consider $\Delta ABC$ with three acute angles, we draw its altitudes and make $\Delta MNP$ triangle
if $\frac{PN}{KN}=\frac{3}{2}$ and $\frac{\sin{\alpha}}{\cos{\frac{\alpha}{2}}}+\frac{\sin{\theta}}{\cos{\frac{\theta}{2}}}+\frac{\sin{\gamma}}{\cos{\frac{\gamma}{2... | I will attempt to solve this problem with as little trigonometry as possible.
The value of $\frac{144}{100}=(1.2)^2$ is actually a red herring. First we note by $a$, $b$, $c$, $\angle{A}$, $\angle{B}$, $\angle{C}$, $S$, $R$ and $r$ the sides, angles, area, circumradius and inradius of $ABC$. Note that: $$\frac{\sin \a... | {
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Power series differential check $(x-1)y'' + y' = 0$ I'm a bit stuck on how to solve this:
$$(x-1)y'' + y' = 0 $$
so assuming y is a solution in this form:
$$\sum_{n=0}^\infty C_nx^n$$
$$\sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
subbing in and distributing:
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-... | You initially wrote that,
$$ y = \sum_{n=0}^\infty c_n x^n,$$
and then you determined that for $n \geq 1$ we have $c_n = c_1/n$.
$$ y = c_0 + \sum_{n=1}^\infty c_n x^n,$$
$$ y = c_0 + \sum_{n=1}^\infty \frac{c_1}{n} x^n,$$
$$ y = c_0 + c_1 \sum_{n=1}^\infty \frac{x^n}{n},$$
which is consistent with the answer given by... | {
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find the number of perfect squares of the form : $n^6 + n^4+1$ find the number of perfect squares of the form -
$$n^6 + n^4+1$$
where n is a natural number.
I've reached this equation.
$$n^4(n^2+1)=(k+1)(k-1)$$
I thought this might have something to do with prime factorization but I don't seem to be getting anywhere.
A... | We do not have a solution for $n=1$, so $n\geq 2$. Because $(k-1, k+1)\leq 2$ we should have $k+1\geq \frac{n^4}{2}$ and $k-1\leq 2(n^2+1)$. $n^4-2(2(n^2+1)+2)=n^4-4n^2-8=(n^2-2)^2-12>0$ for $n\geq3$. That means we only need to check $n=2: 64+16+1=81=9^2$ - unique solution
Also, there is an alternative way. Considering... | {
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using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&... | Note that
$$\begin{align}\sin^4x +\cos^2x &=
\sin^2x(1-\cos^2x) +\cos^2x\\
&=\sin^2x+\cos^2x(1-\sin^2x)=\sin^2x +\cos^4x.
\end{align}$$
Hence, according to your work,
$$\begin{align}
2(\sin^4x +\cos^2x)&=(\sin^4x +\cos^2x) +(\sin^2x +\cos^4x)\\
&=\sin^4x +\cos^4x+1=\frac{7+\cos 4x}{4}.\end{align}$$
Can you take it fr... | {
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Matrix Normalization From the eigenvectors matrix:
I did normalization
but I think there's an error I could not find.
| Let$$
v_1=\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}
\quad
v_2=
\begin{bmatrix}
1 \\
\frac{-\rho_2 + \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{2\rho_1} \\
1
\end{bmatrix}
\quad
v_3=
\begin{bmatrix}
1 \\
\frac{-\rho_2 - \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{2\rho_1} \\\\
1
\end{bmatrix}
$$
the column vectors of mat... | {
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^... | Use the following fact: when $x\to 0$,
$$(1+x)^{\alpha}-1\sim\alpha x.$$
So
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}=n\left(\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)\right)$$
$$=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}},$$
use the above fact:
$$\sqrt[3]{1+\frac{3}{... | {
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Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$. If $f(x), g(x)$ are defined as following on $[0 , \infty)$,
$$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$
$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + ... | First note that for $x \ge 0$ and $n \in \Bbb N$
$$
\frac{x^4n}{(n^3 + x^3)^2} =\frac{x^3}{n^3+x^3}\cdot\frac{nx}{n^3 + x^3}
< \frac{nx}{n^3 + x^3}
$$
and therefore $g(x) \le f(x)$, so that it suffices to show that the function $f$ is bounded on $[0, \infty)$.
For $0 \le x \le 1$ we have
$$
\frac{nx}{n^3 + x^3} \le ... | {
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In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$? Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.
$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.
Is $a$ divisible by 3?
| Yes, $a$ must be divisible by $3$. If the integer $a=3k+1$ or $3k+2$ for arbitrary $k$, (which are the cases in which $a$ is not divisible by $3$) then the remainder is $9k^2+6k+1\equiv1\pmod3$ or $9k^2+12k+4\equiv1\pmod3$.
Therefore, $a^2|3$ cannot occur if $a$ is not divisible by $3$, and thus, $a$ must be divisible ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Efficient computation of conjugacy classes of a small group. I'm trying to construct a character table for a group of order 54 given by:
$$ \langle a,b : a^9 = b^6 = 1, b^{-1} a b = a^2\rangle $$
To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on ho... | $G = \{a^mb^n:0\le m\le 8,0\le n\le 5\}$. For an element $a^mb^n\in G$, $b^{-1}a^mb^nb = a^{2m}b^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^{-1}a^mb^na = a^{m-1}b^nab^{-n}b^n = a^{m-1}a^{5^n}b^n$$since $ba^2b^{-1} = a = (a^2)^5$ and $a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | Let us first simplify the expression. Assume $a^2=1-x$ $ $ & $ $ $b^2=1+x$
Please note that $a^2+b^2=2$
Now our equation becomes $\sqrt{1+ab}(a^3+b^3)=2+ab$
$\sqrt{\frac{a^2+b^2+2ab}{2}}(a+b)(a^2+b^2-ab)=2+ab$ which further simplifies to
$\frac{(a+b)^2}{\sqrt{2}}(2-ab)=2+ab$
$\frac{(2+2ab)(2-ab)}{\sqrt{2}}=2+ab$
Find ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 4
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Find limit in use of integrals Find limit $\lim_{n \rightarrow \infty} \int_{-1}^{1} x^5 \cdot \arctan{(nx)} dx $
From mean-value-theorem we have
$$ \frac{1}{2} c^5 \cdot \arctan{(nc)} \mbox{ for some c } \in (-1,1) $$
$$ \underbrace{\frac{1}{2} \cdot \arctan{(-n)}}_{\rightarrow - \pi /4} \le \frac{1}{2} c^5 \cdot... | We'll simplify with $y=nx$. Integration by parts gives $$\int y^5\arctan ydy=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\frac{y^{6}}{1+y^{2}}dy\\=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\left(y^{4}-y^{2}+1-\frac{1}{1+y^{2}}\right)dy\\=\frac{y^6+1}{6}\arctan y-\frac{1}{30}y^5+\frac{1}{18}y^3-\frac16 y+C.$$Hence $$\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Find minimize value of $P=(a^2+b^2+c^2)(|a-b|+|b-c|+|c-a|)$
For $a,b,c$ are real number satisfied: $a^3+b^3+c^3=3abc+32$. Find minimize value of $P=(a^2+b^2+c^2)(|a-b|+|b-c|+|c-a|)$
This is my try: WLOG, I suppose that: $a\ge b\ge c$.
Then we have $P=(a^2+b^2+c^2)(a-b+b-c+a-c)=(a^2+b^2+c^2)(2a-2c)$
In the other hand,... | Let $a\geq b\geq c$.
Thus, since $(a-c)^2\geq\sum\limits_{cyc}(a^2-ab)$ it's $(a-b)(b-c)\geq0,$ by AM-GM we obtain:
$$P=\sqrt{(a^2+b^2+c^2)^2(a-b+b-c+a-c)^2}=$$
$$=\frac{1}{3}\sqrt{\left((a+b+c)^2+\sum_{cyc}(a-b)^2\right)^2\cdot4(a-c)^2}\geq$$
$$\geq\frac{1}{3}\sqrt{4(a+b+c)^2\sum_{cyc}(a-b)^2\cdot4\sum_{cyc}(a^2-ab)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exercise about a 2-dimensional random vector I solved the following exercise and want to know, if I did it correct
Consider a 2-dimensional random vector $(X,Y)$ with density
$$f_{X,Y}(x,y)=c\frac{y}{x^3}1_{\{0<x\leq 1\}}1_{\{0<y\leq x^2\}}$$
a) Compute $c$, $f_X$ of $X$, $P[X\leq 1/2]$
b) Compute $E[X/Y]$, $E[Y\m... | Since the joint pdf factorises as $$f_{X,Y}(x,y)=\frac{2y}{x^4}\,\mathbf1_{0<y<x^2}\underbrace{2x\mathbf1_{0<x<1}}_{f_X(x)}\quad,$$
this automatically shows that the conditional density of $Y\mid X$ is
$$f_{Y\mid X=x}(y)=\frac{2y}{x^4}\mathbf1_{0<y<x^2}$$
This distribution depends on $X$, so that rules out independence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx$ Can we find the exact value expressed by elementary functions of
$$\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx?$$
| $$\mathcal I=\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx\overset{x\to \frac{1}{x}}=\int_0^{\infty} \frac{x^2}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx$$
Summing up the two integrals from above and rearranging the logarithm gives:
$$2\mathcal I= -\int_0^{\infty} \frac{1+x^2}{1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How do I solve this log equation for x? $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$ $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$
I only managed to solve up to this step: $\left(2+x\right)^{\frac{3}{2}}\sqrt{2-... | $$
(\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{4-x^2}\\
(\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{2-x}\sqrt{2+x}\\
(\sqrt{2+x})^2=125\\
2+x=125\\
x=123.
$$
Note that you have the following constraints on $x$:
$$
\sqrt{2+x}>0\implies 2+x>0\implies x>-2\\
\sqrt{2-x}>0\implies 2-x>0\implies x<2
$$
If the original equation has a solution, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$... | Some other basic method:
$$\int_0^1 -\log(1+x)\log(1-x) \, {\rm d}x \\
= (1-x)\log(1+x)\log(1-x)\Big|_0^1 - \int_0^1 (1-x) \left[ \frac{\log(1-x)}{1+x} - \frac{\log(1+x)}{1-x}\right] {\rm d}x \\
\stackrel{t=1-x}{=} (1+x)\left[\log(1+x)-1\right] \Big|_0^1 - \int_0^1 \frac{t/2 \cdot \log t}{1-t/2} \, {\rm d}t \\
=2\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Proof of the ring isomorphism: $ \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3) $
I am asked to prove that
$$
\mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3)
$$
First, if we define the following homorphism : $\phi:\mathbb{Z} \rightarrow \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7})$, we can try to ... | Consider the natural homomorphism $$\varphi \colon \mathbb{Z} \hookrightarrow \mathbb{Z}\left[ \sqrt{7} \right] \twoheadrightarrow \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$$ which sends any integer $n \in \mathbb{Z}$ to its class $n +\left\langle 5+ 2 \sqrt{7} \right\rangle \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Odd & Even Combinations If you have a range of numbers from 1-49 with 6 numbers to choose from, how many combinations are there containing all odd, all even, only 1 odd, and only 1 even number?
| There are $25$ odd and $24$ even numbers in the range $1 - 49$.
The number of selections of six numbers from the set $\{1, 2, 3, \ldots, 49\}$ that contain exactly $k$ odd numbers and $6 - k$ even numbers is
$$\binom{25}{k}\binom{24}{6 - k}$$
Therefore, the number of selections that contain just odd numbers is $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lceil z \rceil=z+\frac12-\frac{\tan^{-1}(\tan(\pi(z+0.5)))}{\pi}$ when $z$ is not an integer How can I prove that $\lceil z \rceil=z+\dfrac12-\dfrac{\tan^{-1}(\tan(\pi(z+0.5)))}{\pi}$ for all non-integer real numbers $z$?
Z cannot be an integer because then tan(pi*z + pi/2) would be undefined. I got this equat... | $\tan (x)$ has a period of $\pi$, so $\tan x = \tan (\pi(\frac{x}{\pi}-\lceil \frac{x}{\pi} \rceil + \epsilon))$, where $\epsilon \in \mathbb{Z}$. Consider $x = \pi(z+\frac{1}{2})$, then
$$z + \frac{1}{2} - \frac{\tan^{-1} \tan (\pi(z+\frac{1}{2}))}{\pi} = z + \frac{1}{2} - \frac{\tan^{-1} \tan (\pi(z+\frac{1}{2} - \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expr... | By the double angle formulas, the expression is equivalent to
$$\frac12\left(a+c+(a-c)\cos2\theta+b\sin2\theta\right).$$
Now the expression $(a-c)\cos2\theta+b\sin2\theta$ can be seen as the dot product of a vector with a rotating unit vector, which takes its extreme values when the vectors are parallel or antiparallel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an ... | In M2:
R=QQ[x,y,z,a,b,c]
I=ideal(x^3-a,y^3-b,z^3-c,x+y+z)
gens gb I
toString oo -- note a^3+3*a^2*b+3*a*b^2+b^3+3*a^2*c-21*a*b*c+3*b^2*c+3*a*c^2+3*b*c^2+c^3
-- writing x^3 for a, y^3 for b and z^3 for c:
factor(x^9+3*x^6*y^3+3*x^6*z^3+3*x^3*y^6-21*x^3*y^3*z^3+3*x^3*z^6+y^9+3*y^6*z^3+3*y^3*z^6+z^9)
tex oo
$\left(x+y+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ Various examinations ask students to prove that $$\frac{1}{{n \choose k}}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx ~~~~~(1)$$ by evaluating the integral $\int_{0}^{1}
(tx+1-x)^n~dt$ two ways. When I came across (1), I could prove that $$... | A telescoping approach:
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}}
&=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\\
&=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\cdot\frac{(n-(k-1))+(k+1)}{n+2}\\
&=\frac{n+1}{n+2}\sum_{k=0}^n(-1)^k\frac{k!(n-(k-1))!+(k+1)!(n-k)!}{(n+1)!}\\
&=\frac{n+1}{n+2}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
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Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(... | For any finite $n$,
$$\sum_{k=0}^n \frac{1}{k!} < e < 3.$$
UPDATE
$$1+1+\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n! }< 1 + 1+ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} = 1+2 - \frac{1}{2^{n-1}} <3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometric series in a probability question When $A$ and $B$ flip coins, the one coming closest to a given line wins $1$ penny from the other. If $A$ starts with $3$ and $B$ with $7$ pennies, what is the probability that $A$ winds up with all of the money if both players are equally skilled?
What if A were a better play... | For $1\le n\le 9$, let $p_n$ be the probability that $A$ wins if $A$ has $n$ pennies before starting the next round.
Then we have the following system of $5$ equations in $5$ unknowns . . .
\begin{cases}
p_1=\frac{1}{2}p_2\\[4pt]
p_2=\frac{1}{2}p_3+\frac{1}{2}p_1\\[4pt]
p_3=\frac{1}{2}p_4+\frac{1}{2}p_2\\[4pt]
p_4=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{... | Euler's theorem says that if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$ but Euler's theorem does not say $\phi(n)$ is the smallest such number.[1]
But we can easily verify the smallest such power (called the order of $a$) will have to be a factor of $\phi(n)$ [2]
So test $17^k$ where $k|40$.
The powers twos are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Find the equilibria of the following system of ODEs
Find the equilibria of the following system of ODEs
\begin{align*}
\dot{x} = ax - y + kx(x^2 + y^2)\\
\dot{y} = x - ay + ky(x^2 + y^2)
\end{align*}
where $a$ and $k$ are constants, $a > 1$ and $a^2 \geq 1$.
I want to find the equilibria of this system and say... | Set $\dot x=\dot y=0$. Multiply the first equation with $x$, the second with $y$ and add to get
$$
0=a(x^2-y^2)+k(x^2+y^2)^2.
$$
Now do everything askew, multiply the first equation with $y$, the second with $x$ and subtract to get
$$
0=2axy-(x^2+y^2)
$$
Apart from the origin, there are also solutions for $y=q_\pm x=(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? ... | Clearly if one of them is $0$ in modulo $17$ the other is. We show no other solutions exist.
Assume both $k, l$ are nonzero. Note that since $k^2 + l^2 \equiv 0$, we multiply both sides by $k+l$ and remove the $k^3+l^3$ to get $k^2l + l^2k =0$. Dividing both sides by $lk \not \equiv 0$ gives $k+l \equiv 0$. Squaring g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Finding unknowns in polynomial with two factors and remainder. $x^2-4x-12$ is a factor of $rx^3-sx^2+36$, find $r$ and $s$.
Long division gives $rx+(-s+4r)$ with remainder $12rx+4(-s+4r)x+36+12(-s+4r)$
Where I have difficulty with the logic is that since $x^2-4x-12=0$ then the remainder is 0?
And from that we can say(... |
Factor Theorem
$x-a$ is a factor of $p(x)$ iff $p(a)=0$
Since $x^2-4x-12=(x-6)(x+2)$ is a factor of $p(x)=rx^3-sx^2+36$,
we know that $x-6$ and $x+2$ are factors of $p(x)$.
Thus $p(6)=0$ and $p(-2)=0$, that is $216r-36s+36=0$ and $-8r-4s+36=0$.
Solving the simultaneous linear equations give you $r=1,s=-7$.
To use lon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double integrals involving incomplete beta function
I am trying to solve without success the following double integral
$$I_1^{(p)}(N)\equiv\frac{1}{2^p}\int_0^1\text{d}x\int_0^1\text{d}y(1+y-x)^{N+p}(1+x-y)^{N-2}B\left(\frac{1}{1+y-x};N,p+1\right)\cdot\theta(y-x)\theta(1-x-y),$$
where $N\in\mathbb{N}$, $p>0$, $\th... | a) the Incomplete Beta function is
$$
B \left( {x\;;a,b} \right) = \int_{t\, = \,0}^{\;x} {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} dt}
$$
the integration variable is different from the upper bound
b) It might be useful to change the Step function with the Iverson bracket.
So
$$
\begin{split}
I^{\left( p \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of two variables system this might be a very simple question but it has its importance for me. In my multivariable calculus class, I have to find all $(x,y)$ in $R^2$ verifying the system
I will call $(S)$ :
\begin{cases}
3(x-y)^2+8x-3=0 \\
-3(x-y)^2+3=0
\end{cases}
What I did :
This system is equivalent to... | Note that the first equation gives $3(x-y)^2 = 3-8x$, and the second gives $3(x-y)^2 = 3$. So we equate to get $3-8x = 3 \implies x = 0$. Now $3(x-y)^2 = 3$ becomes $(-y)^2 = 1$, so $y$ is $1$ or $-1$, and it is straightforward to check that both $(x,y) = (0,1), (0,-1)$ satisfy both equations.
| {
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"url": "https://math.stackexchange.com/questions/3257675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2... | $$\tan\frac{2\pi}{29}+4\left(-\sin\frac{2\pi}{29}+\sin\frac{6\pi}{29}+\sin\frac{8\pi}{29}-\sin\frac{20\pi}{29}+\sin\frac{22\pi}{29}\right)=\sqrt{29-2\sqrt{29}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Solve and asymptotic expansion of $\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$ I am solving constrained polynomial systems resulting in constrained sums. I am looking to see if $$\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$$ is expressible in terms of know... | I get
$\frac14 n\ln^2(n)+O(n\ln(n))
$.
If
$\begin{array}\\
s(n)
&=\sum_{a=1}^{n} \sum_{b=a+1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
\text{then}\\
s(n)
&=\sum_{b=1}^{n} \sum_{a=1}^{b-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&=\sum_{a=1}^{n} \sum_{b=1}^{a-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How is Wolfram Alpha and the reduction formula arriving at a different result for the integral of $\int \sec^4 x\,dx$ than naive $u$-substitution? I calculated the following on paper for the value of $\int \sec^4 x\,dx$.
$$\int \sec^4 x\,dx=\int \sec^2 x \sec^2 x\,dx=\int (\tan^2 x + 1)(\sec^2 x)\,dx.$$
Let $u = \tan x... | You made an arithmetic error. It should be
$$
\frac13\tan^3 x+\tan x=\frac13\tan(x)(\tan^2x+3)
$$
not $\frac13\tan(x)(\tan^2x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies ... | OP did most of the work by showing that
*
*$xy^2 + y + 1 \mid x - y^2$.
*$ xy^2 + y+ 1 \mid x^2 + y + 1 $
*Since $x^2 + y + 1 > 0$ hence $x^2 + y + 1 \geq xy^2 + y + 1$ so $x \geq y^2$.
To complete the solution using these ideas, suppose $ x - y^2 > 0 $, then $ x-y^2 \geq xy^2 + y + 1$.
We can verify that this is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving system of linear second order differential equations Solve this system of differential equations:
$$y_1'' + y_2' + 3y_1 = e^{-x}$$
$$y_2''-4y_1' + 3y_2 = \sin(2x)$$
My try: Using $y_1' = p$ and $y_2'= q$ , we get:
$$\begin{pmatrix} p' \\q'\\y_1'\\y_2'\end{pmatrix} =\begin{pmatrix}
0 & -1 & -3 & 0 \\
4 & 0 & ... | I am going to rewrite the system as
$$\tag 1 x'' + y' + 3 x = e^{-t} \\ y'' - 4 x' + 3 y = \sin 2t $$
Using the first equation, we have
$$\tag 2 y' = -x'' - 3 x + e^{-t} \\ y'' = -x''' - 3 x' - e^{-t} \\ y''' = - x'''' - 3 x'' + e^{-t}$$
Taking the derivative of the second equation
$$\tag 3 y''' - 4 x'' + 3 y' = 2 \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}$ Compute $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}.$$
I started by making the substitution $\arccos x = t$. Hence, $-\frac{dx}{\sqrt{1-x^2}}=dt$.
Now I get that $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}=-\int\limits_... | $\arccos(-0.5)=\frac{2\pi}{3},\arccos(0.5)=\frac{\pi}{3}$ thus transformed integral must be $-\int_{\frac{2\pi}{3}}^{\frac{\pi}{3}}(...)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the definite integral $\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx$ Writing:
Integrate[ArcTan[(a Cos[x] + b Sin[x])^2], {x, 0, 2 Pi}, Assumptions -> a^2 + b^2 > 0]
$$\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx,$$
where $a $ and $b $ are real numbers.
I get:
2 Pi ArcTan[Sqrt[1/2 (... | Denote $$C := a^2 + b^2 .$$ Then, we can find an angle $x_0$ such that $a = \sqrt{C} \cos x_0$ and $b = -\sqrt{C} \sin x_0$. The angle sum formula for $\sin$ lets us rewrite the quantity in the inner parentheses of the integrand as $$a \sin x + b \cos x = \sqrt{C} \sin (x - x_0).$$ Then, appealing to the periodicity of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Solving $\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$ Is there a way to compute the solution of the ode:
$\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$
where in $t = 0$ we define the field as $-x^2$.
I need for an application of Guiding the solution of ODE with curves. In... | Hint:
Let $x=\dfrac{1}{u}\dfrac{du}{dt}$ ,
Then $\dfrac{dx}{dt}=\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\therefore\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2=t\sin^2\dfrac{1}{t}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\dfrac{d^2u}{dt^2}-\left(t\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$
Calculate: $$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
The solution of this exercise:
Let $$S_1=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{8}+\cdots$$
$$S_2=\binom{n}{1}-\binom{n}{3}+\binom{n}{5}-\cdots$$
$$S_3=\binom{n}{0}+\binom{n}{4}+\binom{n... | The part with $(1+i)^n$ is explained by De Moivre:
$$(1+i)^n=\sqrt2^ne^{i\frac{n\pi}{4}}=\sqrt2^n(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^n=\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4})$$
Now comparing the $\Re$ parts of the LHS and RHS of the given equation:
$$\Re(\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}))=\Re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the n... | General method, without the formal derivatives.
Suppose
$$
f(x) = g(x) (x-1)^2 + (x+3)= h(x) x^2 + (2x + 4).
$$
Then
$$
(x-1)^2 f(x) = (x-1)^2 x^2 h(x) + (x-1)^2 (2x+4) \tag 1
$$
and
$$
x^2 f(x) = x^2 (x-1)^2 g(x) + x^2(x+3). \tag 2
$$
Now do the Euclidean algorithm to $x^2, (x-1)^2$:
\begin{align*}
x^2 &= (x-1)^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
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.