Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Color integers of finite set $M\subseteq \mathbb{N}$ so that if different $a,b$ are the same color then $a+b$ is different color. $|M|_{\max}=?$
The integers $1, 2, \ldots, n$ are each colored with one of three colors, following the rule that if $a$, $b$ are distinct integers that both have the same color, $a+b$ must ... | An unsatisfactory method, but:
By exhaustive search, there are only three solutions for $n=23$ :
$$\begin{align}
A &= \{1,2,4,8,11,22\},\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,16,17,18,20\}\end{align}$$
$$\begin{align}
A &= \{1,2,4,8,11,16,22\}\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,17,18,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Congruent number problem- one elementary statement We call a positive integer $n$ a congruent number if there is a rational right triangle with area $n$.
Equivalently, $n$ is a congruent number if there exists positive integers $x$ and $y$ such that
$$x^2\pm n=y^2.$$
Somewhere, I came across a statement:
If we take $... | I'll prove that if three of the four integers $\{\,a,b,a+b,a-b\,\}$ are squares, then the fourth is a congruent number. By the way, this result was first mentioned by Leonardo of Pisa, better known as Fibonacci.
First, let's get a correct definition of congruent: a positive integer $n$ is said to be congruent if there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$ How do I evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$?
I don't know what to do. Should I use a contour integration?
| By the cosine addition formulas and symmetry we have
$$ \int_{0}^{+\infty}\cos(x)\cos(x^2)\,dx = \int_{0}^{+\infty}\cos\left(x^2-\tfrac{1}{4}\right)\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\cos\left(x-\tfrac{1}{4}\right)}{\sqrt{x}}\,dx $$
where the last integral is clearly converging by Dirichlet's test. Since the Lap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Cauchy Product starting from $1$ The definition of the Cauchy product from Wikipedia is defined as
$$\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
=\sum_{k=0}^\infty\sum_{\ell=0}^ka_\ell b_{k-\ell}$$
Question: Does this apply if the summation runs from $1$ instead of $0$ for both $i,j$?
In part... | What you need is $\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty a_{i,j} = \sum_{k=1}^\infty \sum_{\ell=1}^k a_{\ell,\,k+1-\ell}.$
$$
\begin{array}{c|ccccccccc}
& 1 & 2 & 3 & 4 & 5 & \longrightarrow j \\
\hline
1 & a_{11} & a_{12} & a_{13} & a_{14} & \cdots \\
2 & a_{21} & a_{22} & a_{23} \\
3 & a_{31} & a_{32} \\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove only one solution exists for $f(\zeta)$ and $g(\zeta)$ using a convex function argument I want to prove that the following only have one solution, for $\zeta\in[0,1]$, at $\zeta =1$.
$$f(\zeta)=\frac{1}{1+\zeta}$$
$$g(\zeta) =
\frac{
(1-\zeta)(2-\zeta)\zeta
-
(2-\zeta)^2\log\left(2 - \zeta \right)
}
{\zeta(\zeta-... | we want to solve g(x)=f(x) for x in [0,1]
by doing the calculations we end up in this equation
$$ \ log(2-x)= \frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
$$h(x)=log(2-x)$$ and
$$ k(x)=\frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
h(x) is convex and k(x) is concave in [0,1] (1)
$$ k'(x)=\frac{-2(x^3+8x^2-17x+6)}{(x-2)^3(x+1)^2} $$
so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$ I want to find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$
To calculate the critical points of $f$ I use the partial derivatives $f_x=2xy+2y$ and $f_y=x^2+2x+24y$ an... | If $x=-1$ and $y=-\frac{3}{4}$ then we get a value $\frac{15}{2}$.
We'll prove that it's a maximal value.
Indeed, by the given $$(x+1)^2+16y^2\leq9.$$
Thus, $$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\leq$$
$$\leq|y|(x+1)^2+12y^2-y\leq|y|(9-16y^2)+12y^2-y.$$
Hence, it remains to prove that $$|y|(9-16y^2)+12y^2-y\leq\frac{15}{2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine $ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $ Determine $ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $
Answer:
$ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5} \\ = \ \large \lim_{x \to \infty} \frac{40}{5-\frac{1}{x}}- \ \la... | Yes, you are right:
$$ \lim_{x \to +\infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}}=\frac{40-\lim\limits_{x\rightarrow+\infty}\frac{32\sin{x}}{2^{0.1x}}}{5-\lim\limits_{x\rightarrow+\infty}\frac{1}{x}}=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove I... | By Holder and AM-GM we obtain:
$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=\frac{1}{4}\cdot(1+1)^2\left(\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3\right)\geq$$
$$\geq\frac{1}{4}\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^3=\frac{1}{4}\left(1+\frac{1}{ab}\right)^3\geq\frac{1}{4}\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
| It's $$\left(\frac{x^2+2}{x}\right)^2-6\left(\frac{x^2+2}{x}\right)+8=0,$$
which gives
$$\frac{x^2+2}{x}=2$$ and $x\in\{1+i,1-i\}$ or
$$\frac{x^2+2}{x}=4,$$ which gives $x\in\{2+\sqrt2,2-\sqrt2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
How to prove that this diophantine equation has only two solutions? Consider an equation of the form.
$$z^3=y^2+4$$
where z ,y $\in \Bbb N$
Hence, the solution of form (z,y) are (2,2) and (5,11).
So How to prove that this diophantine equation only has these two solutions.
| Suppose $z^3=y^2+4 = (2-yi)(2+yi)$, if the two factors $2-yi$ and $2+yi$ are coprime, then $(u+vi)^3 = 2+yi$, and the rest is easy. However, they need not be coprime, a more comprehensive analysis is necessary.
Let us first assume $z$ is even, letting $z=2z_1,y=2y_1$ gives $2z_1^3 = y_1^2+1$. Modulo $4$ shows that $z_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that $\sqrt{(1+1/n)}$ is irrational for all $n\in \mathbb{N}$ I have trouble solving this probably simple problem:
Prove that $\sqrt{1+1/n}$ is irrational for all $n\in \mathbb{N}$.
| Here's another way.
$\sqrt{1 + \frac 1n} = \sqrt{\frac {n+1}{n}} = \sqrt{\frac {n(n+1)}{n^2}}= \frac 1n \sqrt{n(n+1)}$ which rational if and only if $\sqrt{n(n+1)}$ is rational.
Square roots of integers are rational only when that are integer square roots of perfect squares. So $\sqrt{n(n+1)}$ is rational only if $n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Proof: $2n! < 2^{2n} (n!)^{2}$ Could someone help with finishing this inequality induction proof.
Hypothesis: $2n! < 2^{2n}(n!)^2$
Step: $2(n+1)! < 2^{2(n+1)}((n+1)!)^2$
(2n+2)! < 2ˆ(2n + 1) * ((n + 1)!)ˆ2
| Note that $4n^2+6n+2 < 4n^2+8n+4$ so $\color{blue}{2(n+1)(2n+1)< 2^2(n+1)^2}$.
\begin{eqnarray*}
(2(n+1))!=2(n+1)(2n+1) \underbrace{\color{red}{(2n)!} < \color{blue}{2(n+1)}}_{ \text{by the } {\color{red}{ \text{inductive hypothesis}}}}\color{blue}{(2n+1)} \color{red}{2^{2n}}\underbrace { \color{red}{(n!)^2} < \color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve a trigonometric equation involving double and triple angles? The problem is:
$\textrm{From the following equation}$
$$10\cos{2\omega}-13\cos{3\omega}+2\sin{\frac{5\omega}{2}}\sin{\frac{\omega}{2}}=0$$
$\textrm{find the value of:}$
$$\sec{\omega}+\sec{3\omega}$$
$\textrm{Consider:}$
$$\omega\neq (2K+1)\frac... | Consider the equation
$$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3\cos\omega)=0$$
and just multiply it by $2$:
$$11(4\cos^{2}\omega-2)-28(4\cos^{3}\omega-3\cos\omega)=0$$
From this equation and the assumption that $\omega \neq (2k+1)\frac{\pi}{6}$, we conclude that
$$\frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\ome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it actually possible to solve $\frac{xy}{\ln(x^p)(x+y)^2}=c$ for $x$? Is it possible to solve the following for $x$
$$\frac{xy}{\ln(x^p)(x+y)^2}=c$$
where $p\in[0,1]$ and all other variables are positive real numbers ?
Without the $\ln(x^p)$ in the denominator, the solution simply is
$$x=-\frac{-y+2cy+y\sqrt{-4c+1}... | Probably, the best you can do (without numerical method) is to express $y$ in terms of $x$:
$$
\frac{(x+y)^2}{xy} = \frac{1}{c\ln(x^p)}
$$
$$
\frac{x}{y} + \frac{y}{x} + 2 = \frac{1}{c\ln(x^p)}
$$
Let $z=\frac{y}{x}$, then
$$
z+\frac{1}{z}=\frac{1}{c\ln(x^p)}-2
$$
$$
z=\frac{\frac{1}{c\ln(x^p)}-2\pm\sqrt{\frac{1}{c^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it okay to "ignore" small numbers in limits where $x$ approaches infinity? I got a limit:
$$\lim_{x\to\infty}\frac {(2x+3)^3(3x-2)^2} {(x^5 + 5)}$$
As far as $x$ approaches infinity, can I just forget about 'small' numbers (like $3$, $-2$ and $5$ in this example)? I mean is it legal to make a transition to:
$$\lim_{... | We have $$\lim_{x\to\infty}\frac{2x+3}{2x} = 1,\ \lim_{x\to\infty}\frac{3x-2}{3x} = 1,\ \lim_{x\to\infty}\frac{x^5+5}{x^5} = 1$$
and so $$\lim_{x\to\infty}\frac{(2x+3)^3(3x-2)^2}{x^5+5}= \lim_{x\to\infty}\left(\frac{\frac{(2x+3)^3}{(2x)^3}\cdot\frac{(3x-2)^2}{(3x)^2}}{\frac{x^5+5}{x^5}}\cdot\frac{(2x)^3(3x)^2}{x^5}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 6,
"answer_id": 4
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Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\the... | The trick here is to find $2$ representations of the complex exponential $$e^{ix} = \cos(x)+i\sin(x).$$
With this in mind, we have
$$e^{2ix} = \cos(2x)+i\sin(2x)$$ and
$$e^{3ix} = \cos(3x)+i\sin(3x) $$ along with
$$(e^{ix})^2 = (\cos(x)+i\sin(x))^2 = (\cos(x)^2 - \sin(x)^2) + i(2\sin(x)\cos(x))$$ and
$$(e^{ix})^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Integrate the given expression to find $f(1)$ If $$f(x)=\int \frac {x^2}{(x^2+2)(1+\sqrt{x^2+1})}dx$$ and $f(0)=0$ , then find $f(1)$
I multiplied and divided by $\sqrt{x^2+1}-1$ but later got stuck at $\int \frac{\sqrt{x^2+1}}{x^2+2}.dx$ How to solve this integral. Could someone help me with this?
| $$\int_{0}^{1}\frac{x^2\,dx}{(x^2+2)(1+\sqrt{1+x^2})}\stackrel{x\mapsto\sinh z}{=}\int_{0}^{\log(1+\sqrt{2})}\frac{\sinh^2(z)\cosh(z)\,dz}{(\sinh^2(z)+2)(1+\cosh z)} $$
equals, by the substitution $z=\log t$,
$$ \int_{1}^{1+\sqrt{2}}\frac{(1-t)^2 (1+t^2)}{t(1+6t^2+t^4)}\,dt $$
which can be computed by partial fraction ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove $ \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40} $ Prove:
$$
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40}
$$
i have tried to write $1/40$ as $(1/40^{1/2007})^{2007}$ and prove $(1... | $$\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}=\sqrt{\frac{3\cdot (3 \cdot 5)\cdot(5 \cdot 7)\cdot... \cdot (2005\cdot 2007)\cdot2007}{4\cdot4^2 \cdot 6^2 \cdot ... \cdot2006^2\cdot 2008^2}}<\sqrt{\frac{3\cdot2007}{4\cdot2008^2}}<\frac{1}{40}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How many solutions are there to $x+y+z=14$ where $x,y,z$ are all non-negative integers, $x \leq 5, y \leq 6, z \leq 7$? Through using brute force, I have got 15 triplets of solutions. Have I reached the right answer, and how can I not use brute force?
| For the given conditions, the maximum values of x, y and z are:
$$\begin{align}
x={} & 5 \\
y= {} & 6\\
z = {} & 7 \\
\sup(x+y+z) = {} & 18
\end{align}$$
Hence we must subtract 4 from $x$, $y$ and $z$ collectively such that $x+y+z=14$.
Of the 5 partitions of 4, all except 1+1+1+1 are possible allocations.
*
*4+0+0 -... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$
Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$
My proof :
$$(a+x)^{n+1} = \bigg(n\frac{a}{n} + x\bigg)^{n+1} = \bigg( \frac{a}{n} + \frac{a}{n} + \dots +\frac{a}{n} +x \bigg)^{n+1} ... | use the AM-GM inequality, then
$$\frac{\frac{a}{n}+\frac{a}{n}+...+\frac{a}{n}+x}{n+1}\geq \sqrt[n+1]{\left(\frac{a}{n}\right)^n\cdot x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find value of $b$ and $c$ Suppose $x^{12} - 1$ is divisible by $x^4 + bx^2 + c$ . Find all possible values for $b$ and $c$ . I've tried to use $x^4 + bx^2 + c = 0$ for finding remainder and equaling it with zero but didn't help for finding $b$ and $c$.
| $$x^{12}-1=(x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1)=(x^4-1)(x^4+x^2+1)(x^4-x^2+1)$$
The situation is a bit trickier since those $3$ don't need to be the only possible solutions. Using the substitution suggested in comments by MathematicianByMistake $t=x^2$ we get $t^6-1=(t^3-1)(t^3+1)=(t-1)(t+1)(t^2+t+1)(t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\f... | You brought the wrong thing to the right hand side. Your inequality is equivalent to
$$1 + \frac{1}{n} < \biggl(1 + \frac{1}{n^2-1}\biggr)^n\,.$$
Now applying Bernoulli's inequality to the right hand side we get
$$\biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1} > 1 + \frac{1}{n}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int \sin^3(x)\sqrt{\cos(x)} dx$ I tried to solve $\int \sin^3(x)\sqrt{\cos(x)}\,dx$ by setting it equal to
$$\int \cos^{1/2}(x)\left(1-\cos^2x\right)\sin(x)\,dx $$
and then making $u=\cos(x)$ and $du=-\sin(x)$. I ended up with
$$\frac{-2\cos^{3/2}(x)}{3} + \frac{\cos^2(x)}{2} + C$$
but the book's answer is... | $$\begin{align*}
\int \sin^3 x \sqrt{\cos x} \, dx
&= \int (1 - \cos^2 x) \cos^{1/2} x \sin x \, dx \\
&= \int (\cos^{1/2} x - \cos^{5/2} x) \sin x \, dx \\
&= \int -(u^{1/2} - u^{5/2}) \, du \\
&= \frac{2}{7} u^{7/2} - \frac{2}{3} u^{3/2} + C \\
&= \frac{2}{7} \cos^{7/2} x - \frac{2}{3} \cos^{3/2} x + C. \end{align*}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2... | the trick in c) to transform $2^{15}$ in $(2^5)^3$ and use $a^3-b^3$ formula
| {
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"url": "https://math.stackexchange.com/questions/2521465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area inside a rectangle
SEAMO 2016 paper E
I do not know how to start with this question and I have tried finding heights and bases of the triangle in terms of the sides of rectangle but i could not find the ideal pair
| Let $AC\cap EG=\{H\}$ and $AC\cap GF=\{I\}$.
Thus, $$\frac{AH}{HC}=\frac{AE}{GC}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3},$$
Which says that $$AH=\frac{2}{5}AC.$$
Also,
$$\frac{CI}{IA}=\frac{CG}{AF}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4},$$
Which says that $$CI=\frac{3}{7}AC.$$
Thus, $$HI=AC\left(1-\frac{3}{7}-... | {
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"source": "stackexchange",
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Maximising the area of a triangle with its vertices on a parabola.
Let $A\equiv (4,-4)$ and $B\equiv(9,6)$ be tw points on the parabola $y^2 = 4x$. Let $C$ be a point on the parabola between $A$ and $B$ such that the area of the triangle $\triangle ABC$ is maximal. What are the coordinates of $C$?
The parametric co... | The shoelace formula for the points $(4, -4), (9,6), (t^2, 2t)$ gives
$$
A = \frac12\left|4\cdot 6 + 9\cdot 2t + t^2\cdot (-4) - (-4)\cdot 9 - 6\cdot t^2 -
2t\cdot 4 \right|\\
= \frac12|24 + 18t - 4t^2 + 36 - 6t^2 - 8t|\\
= |30 + 5t - 5t^2|
$$
Since we "[l]et $C$ be a moving point on the parabola between $A$ and $B$"... | {
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"source": "stackexchange",
"question_score": "3",
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Find the domain and range of $y=-x^2+4x-3$ Find the domain and range of $y=-x^2+4x-3$
My Attempt:
$$y=f(x)=-x^2+4x-3$$
The given function is a polynomial of degree $2$ in $x$. $f(x)$ is defined for all $x\in R$, so the domain of $f=R$.
Again,
$$y=-x^2+4x-3$$
$$y=-(x^2-4x+3)$$
$$y=-(x-1)(x-3)$$
$$y=(x+1)(x-3)$$.
| Write $f(x)=-x^2+4x-3=1-(x-2)^2$, then
$f(x)=1-(x-2)^2\leq -1$
and the range is $(-\infty,1]$ and the domain is all the set $\mathbb{R}=(-\infty,\infty)$ because $f$ is a polynomial function. It's minimum will be always at the middle point of the two roots: $x_1=-1$ and $x_2=3$ (because $f(x)$ is a polynomial function ... | {
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"source": "stackexchange",
"question_score": "2",
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Values of $a,b,c$ such that $\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$ Find the values of $a,b,c$ such that $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$
Here's what I have got so far
Using L'Hospital's rule,
$$\lim_{x \to 0}\frac{a+b-\cos x+x\sin x-c\cos x}{5x^4}=1$$
So,$a+b-1=0$
Again,
$$\lim_{x \to ... | Expand with Taylor:
$$
x(a+b-\cos x)-c\sin x=
x\left(a+b-1+\frac{x^2}{2!}-\frac{x^4}{4!}\right)
-c\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)+o(x^5)
$$
Thus you want
$$
\begin{cases}
a+b-1-c=0 \\[4px]
1/2+c/6=0 \\[4px]
-1/24-c/120=1
\end{cases}
$$
The second equation becomes $c=-3$, the third becomes $-5-c=120$.
No so... | {
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"url": "https://math.stackexchange.com/questions/2528720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integration by parts of $x^5\ln(x)$ I have to integrate the function: $x^5\ln(x)$
My Attempt
$$\int(x^5\ln(x))dx$$
$u=\ln(x)$ and $du=\frac{1}{x}dx$
$dv=x^5dx$ and
$v=\frac{x^6}{6}$
Using this I can then integrate the function using the method
$$uv-\int v du$$
Which then substituting I get:
$$\frac{x^6\ln(x)}{6}-\int\f... | how about
$u = x^3$
$du=3x^2 dx$
$\int(x^5\ln(x))dx = \int \frac{ux^2 \ln(u^\frac{1}{3})}{3x^2} du $
properties of logs to eliminate cube root
=$\frac{1}{3} \int \frac{u \ln(u)}{3} du $
=$ \frac{1}{9}\int u \ln(u) du $
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \int \frac{1}{2}u^2 / udu )$
= $ \frac{1}{9}(\frac{1}{2}u^2\ln... | {
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Finding the set of values of the real number $a$ for which $\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges
Find the set of values of the real number $a$ for which $\displaystyle\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges.
I had been stuck on this problem for a while but now I have managed... | For the divergence:
\begin{align*}
\sum_{n}\left(\frac{1}{n}-\sin\frac{1}{n}\right)^{a}&\geq\sum_{n}\left(\frac{1}{3!}\frac{1}{n^{3}}-\frac{1}{5!}\frac{1}{n^{5}}\right)^{a}\\
&=\frac{1}{(3!)^{a}}\sum_{n}\frac{1}{n^{3a}}\left(1-\frac{3!}{5!}\frac{1}{n^{2}}\right)^{a}\\
&\geq\frac{1}{(3!)^{a}}\frac{1}{2^{a}}\sum_{n\geq N... | {
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"source": "stackexchange",
"question_score": "2",
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Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ , where $ \ l=constant \ $
Answer
The given series is convergent clearly .
$ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \\ = \frac{1}{1^2+l^2}+\frac{1}{2^2+l^2}+......+\frac{1... | Hint: Use the fact that
\begin{align}
\sum^\infty_{n=-\infty} \frac{1}{x^2+n^2} = \frac{\pi\coth(\pi x) }{x}.
\end{align}
| {
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"question_score": "4",
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Using the factorization of a quadratic equation If $a,b$ are the roots of the equation $x^2 -px +q=0$ and $a_1, b_1$ are the roots of the equation $x^2-qx +p=0$ then find the equation of quadratic whose roots are $\frac{1}{a_1b} + \frac{1}{ab_1}$ and $\frac{1}{aa_1} + \frac{1}{bb_1}$.
I know that the equation $Qx^2 + b... | $$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)+\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(a_1+b_1)(a+b)}{a_1b_1ab}=\frac{pq}{pq}=1.$$
$$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(ab_1+a_1b)(aa_1+bb_1)}{p^2q^2}=$$
$$=\frac{a^2q+b_1^2p+a_1^2p+b^2q}{p^2q^2}=\fr... | {
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$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$ proof by induction
Prove with mathematical induction that for every $n \in \Bbb N$ it holds:
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$
*
*Basis (P(1))
$\frac21=2^1$
*Inductive step - If it holds for P(n), then it also holds for... | In the induction step, you assume
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$
and wish to show that
$$\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}=2^{n+1}.$$
You have
\begin{align}\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}&=\frac{n+2}1 \cdot \frac{n+3}3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ Prove that $\displaystyle 1 < \cos A+\cos B+\cos C \leq \frac{3}{2}$ where $A+B+C = \pi$
Attempt: $\cos A+\cos B+\cos C= 1+4\cos A\cdot \cos B\cdot \cos C$
Now $\displaystyle \cos A\cdot \cos B\cdot \cos C=\frac{1}{2}\cos A\left[\cos (B-C)-\cos A\right]\leq \frac{1}{2}... | This question is as old as the OP's age ( probably )...Let's do the left inequality. $\cos A +\cos B+\cos C > 1\iff \cos A+\cos B -\left(1-\cos C\right) > 0\iff2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^2\left(\frac{C}{2}\right)>0\iff 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Computing: $\lim\limits_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$ $$\lim_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$$
If the limit is a finite non zero number, find n.
I tried using L'Hôpital's rule but failed. I differentiated it but not understanding what exactly should be done. Please help.
| Given that
$$ \cos'x|_{x=0} =\lim_{x\to 0}\frac{\cos x-1}{x} =\sin 0=0~~~and ~~~\lim_{x\to 0}\frac{e^x-1}{x} =(e^x)'|_{x=0}= 1$$
or by taylor series we have, $$\frac{\cos x-1}{x} \sim \frac{x}{2}~~~and ~~~\lim_{x\to 0}\frac{e^x-1}{x} \sim 1$$
and
$$ \frac{(\cos x-1)(\cos x-e^x)}{x^n} = \frac{(\cos x-1)(\color{red}{\c... | {
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Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so... | $I =-\int_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}$
You are mixing $x's$ and $u's$
$I =\int_0^{\infty}\frac {\ln^2 x}{x^2+a^2}\ dx\\
x = \frac {a^2}{u}, dx = -\frac {a^2}{u^2} du\\
\int_{\infty}^0\frac {(\ln \frac {a^2}{u})^2}{\frac {a^4}{u^2}+a^2} (-\frac {a^2}{u^2})\ du\\
\int_0^{\infty}\frac {(2\ln a - \ln... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
If $a=2$,what is the solution set of the equation?
Options
*
*{$3,6$}
*{$2$}
*{$3$}
*{$6$} Correct Answer
I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equatio... | By definition $\sqrt{}$ is always the non-negative root. Every positive number has exactly two square roots equal in magnitude, on positive and one negative.
$\sqrt{25} =5$ and $\sqrt {25} \ne -5$. But both $5$ and $-5$ are solutions to $x^2 = 25$.
To solve an equation $x^2 = k$ there will be two answers. One is $\s... | {
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be ... | Since $\alpha $ and $\beta $ are roots of $x^2+3x+1$ we know that $\alpha +\beta = -3$ and $\alpha \beta = 1$
Indeed $$ x^2+\color{red}{3}x+\color{blue}{1} =(x-\alpha)(x-\beta) = x^2 -\color{red}{(\alpha +\beta)}x+\color{blue}{\alpha \beta}$$
Hence,
$$\frac {\alpha}{\beta} + \frac {\beta}{\alpha} = \frac {\alpha^2+\... | {
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"source": "stackexchange",
"question_score": "2",
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Find all prime numbers $p$ such that $16p+1$ is a perfect cube What I have attempted:
Suppose $16p+1=k^3$ where $k \in Z$ then $16p=k^3-1=(k-1)(k^2+k+1)$ so we can say that $k=17$ and thus $p=17^3+17+1=4931$ which is prime.
How would I find the remaining numbers?
| Since $k^3\equiv1\pmod{16}\implies k\equiv1\pmod{16}$, if $16p+1$ is a perfect cube, we must have
$$
\begin{align}
16p+1
&=(16k+1)^3\\
&=4096k^3+3\cdot256k^2+3\cdot16k+1
\end{align}
$$
Thus, we get $p=256k^3+48k^2+3k=(256k^2+48k+3)k$, which can only be prime if $k=1$, that is $p=307$ and thus
$$
17^3=16\cdot307+1
$$
is... | {
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"answer_count": 3,
"answer_id": 1
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Simultaneous recurrence relations: $a_n = a_{n-1} (1 - b_n), b_n = b_{n-1} (1-a_n)$ I want to solve the following simultaneous recurrence relations:
\begin{align*}
a_n &= a_{n-1} (1-b_n) \\
b_n &= b_{n-1} (1-a_n).
\end{align*}
Initial conditions are $a_1 = b_1 = 1$ and $a_2 + b_2 = 1$.
The symmetric case, $a_2 = b_2 = ... | Since$$b_n=1-\frac{a_n}{a_{n-1}}$$
we get
$$1-\frac{a_n}{a_{n-1}}=\left(1-\frac{a_{n-1}}{a_{n-2}}\right)(1-a_n)$$
Multiplying the both sides by $a_{n-1}a_{n-2}$ gives
$$a_{n-1}a_{n-2}-a_na_{n-2}=a_{n-1}a_{n-2}-a_na_{n-1}a_{n-2}-a_{n-1}^2+a_na_{n-1}^2,$$
i.e.
$$a_{n-1}^2(a_n-1)=a_na_{n-2}(a_{n-1}-1)$$
Dividing the both ... | {
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"question_score": "2",
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"answer_id": 3
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$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'s General Solution I was looking for the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$
My work
We need to get the auxillary equation of the given differential equation above...
The auxillary equation would be:
$$r^5 + r^4 - 7r^3 - 1... | $(r+2)^2(r-3)(r^2+1)$
We have one double root and one pair of complex roots.
With complex roots:
$Ae^{it} + Be^{-it} = C_1 \cos t+ C_2\sin t$
are solutions
With roots of multiplicity.
$C_3^{-2t}+ C_4te^{-2t}$ will be solutions.
| {
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"source": "stackexchange",
"question_score": "1",
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How to solve this system of linear equations over $\mathbb{F}_{11}$? Given
$$\begin{pmatrix}0&1&2\\ 4&3&a\\ 1&2&4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1\\ b\\ 4\end{pmatrix}$$
I need to find for what values $a,b \in \mathbb{F}_{11}$ there are solutions and what the general solution is (witho... | Writing $Ax=b$ for your system of linear equations, the determinant of $A$ is given by
$$
\det(A)=2(2a-27).
$$
Over the field $\mathbb{F}_{11}$ the determinant is non-zero if and only if $a\neq 8$ (we could write $a\neq -3=8$ as well). Then the unique solution is given by
$$
x=A^{-1}b=\frac{1}{a+3}\begin{pmatrix} 9(a... | {
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"source": "stackexchange",
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How to solve a system of linear equations modulo n? For example,
$4x - 10y \equiv 8\pmod {20}$
$7x + 2y \equiv 5\pmod {20}$
It resembles linear diophantine equations and the Chinese Remainder Theorem, but I don't know how to actually solve it..
| \begin{eqnarray}
4x - 10y \equiv 8\pmod {20} \\
7x + 2y \equiv 5\pmod {20}
\end{eqnarray}
Method 1 - Take advantage of a unit coefficient (7)
\begin{align}
7x + 2y &\equiv 5\pmod {20} \\
21x + 6y &\equiv 15\pmod {20} \\
x + 6y &\equiv 15\pmod {20} \\
x &\equiv 14y + 15 \pmod{20} \\
\hline
4(14y ... | {
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"question_score": "7",
"answer_count": 2,
"answer_id": 1
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The order of a modulo p.
If $p$is prime and $\operatorname{ord}_p(a)=4$, then $1+a+a^2+a^3≡ 0\bmod p$, where $\operatorname{ord}_p(a)$ is the order of $a$ modulo $p$.
I think it is true statement
$\operatorname{ord}_p(a)=4$, then $a^4≡1\bmod p$ so $a^4-1≡0\bmod p$
since $a^4-1=1+a+a^2+a^3$, then $a^4-1=1+a+a^2+a^3\e... | You can also find this directly since $4$ is conveniently small.
Since $p$ is prime, there are two roots of $1$: $\{1,-1\}$. Since $(a^2)^2\equiv 1$ but $a^2\not\equiv 1 \bmod p$ we must have $a^2\equiv \color{red}{-1} \bmod p$. Then $a^3\equiv a^2\cdot a \equiv \color{blue}{-a} \bmod p$.
Thus $1+a+a^2+a^3\equiv 1+a+ \... | {
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prove that $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ are positive real numbers such that $abc=1,$
prove that $$a^2+b^2+c^2 \geq a+b+c$$
i was thinking of using AM GM inequlaity.but donot have idea on which pairs to apply it.
any hint....
| By QM-AM for the first inequality and AM-GM for the second,
$$\frac{a^2+b^2+c^2}{3}\geq\left( \frac{a+b+c}{3} \right)^2\geq\frac{a+b+c}{3}\left( \sqrt[3]{abc} \right)$$
so since $abc=1, a^2+b^2+c^2\geq a+b+c$
| {
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"question_score": "1",
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Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
| The denominator can be simplified to $$4(\cos(x/2)\cos(\pi/6)-\sin(x/2)\sin(\pi/6))=2(\sqrt3\cos(x/2)-\sin(x/2))$$
The numerator is $$2\cos(x)+1=2(2\cos^2(x/2)-1)+1=4\cos^2(x/2)-1$$
So you have $$\require{cancel}\frac{4\cos^2(x/2)-1}{2\sqrt3\cos(x/2)-2\sin(x/2)}=\frac{4\cos^2(x/2)-\cos^2(x/2)-\sin^2(x/2)}{2\sqrt3\cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Finding $x$ values of $\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $
Solve $$\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $$ for $0^\circ<x<120^\circ$
I simplify it to
$$2 \cos 6x + 2 = 3 + \cos 3x $$
There is $\cos 6x $ and $\cos 3x$ how do I merge them together? To solve the equation from $0^\circ$ to $12... | Here's what I got...(EDIT: $3 \cos 3x$ is incorrect in the third line; the correct quantity is $\cos 3x$.).
$$\cos 6x + 1 = \frac {3}{2} + \frac {1}{2} \cos 3x$$
Multiplying by $2$ we get
$$2 \cos 6x + 2 = 3 + \cos 3x$$
Swapping constants and bringing over $\cos 3x$ we then get
$$2 \cos 6x - \cos 3x = 1$$
Using the dou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac... | Not an answer but too long for a comment.
Considering
$$f(i)=\sum_{k=0}^p \frac 1 {5k+i}=\frac{1}{5} \left(\psi ^{(0)}\left(\frac{i}{5}+p+1\right)-\psi
^{(0)}\left(\frac{i}{5}\right)\right)$$ then $5\left(f(1)-f(2)-f(3)+f(4)\right)$ write$$\psi ^{(0)}\left(p+\frac{6}{5}\right)-\psi ^{(0)}\left(p+\frac{7}{5}\right)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And bec... | Perhaps,
\begin{align*}
\left|\dfrac{n^{3}-5n^{2}+\pi}{2n^{5}-\sqrt{3}}\right|&\leq\dfrac{n^{3}+5n^{3}+\pi n^{3}}{2n^{5}-\sqrt{3}n^{5}}\\
&=\dfrac{6+\pi}{2-\sqrt{3}}\dfrac{1}{n^{2}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
I think that the only solutions are $n = 0,1,2,4$, but I have no idea on how to prove it.
I tried to write $3^n-1$ as $1+3+3^2+...+3^{n-1}$ and manipulate the sum but found my self at the equall... | The result is clear for $n = 0$. For $ n = 1, 2, 3 \ldots$ let the highest power of $2$ that divides $3^n - 1$ be $2^{p(n)}$. If $n$ is odd, say $n = 2m + 1$, then $3^n - 1 = (3 - 1)(3^{2m} + 3^{2m - 1} + \cdots + 1)$. The sum has an odd number of terms, so $p(n) = 1$. If $n$ is even, say $n = 2m$, then $3^n - 1 = (3^m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find the indefinite integral $\int\frac{dx}{(x^2+2x+5)^2}$ I need help with the indefinite integral
\begin{align}
& \int\frac{dx}{(x^2+2x+5)^2} \\[10pt]
= {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt]
= {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt]
= {} & ... | There is a reduction process. Apply integration by parts.
$$\int \frac{1}{1+x^2}\,dx=\frac{x}{1+x^2}-\int x \,d\frac{1}{1+x^2}$$
$$=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}\,dx-2\int \frac{1}{(1+x^2)^2}\,dx$$
And rearranging,
$$2\int \frac{1}{(1+x^2)^2}\,dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}\,dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do diff... | Hints:
The coefficient of the $n^\text{th}$ term can be written as: $$\frac{1\times 3 \times 5 … (1+4(n-2))}{2\times 4\times 6 …(2+4(n-2))}$$ $$=\frac{1\times 2 \times 3 … (1+4(n-2))\times (2+4(n-2))}{[2\times 4 \times … (2+4(n-2))]^2}$$ $$=\frac{(4n-6)!}{[(2n-3)!(2^{2n-3})]^2}$$ $$=\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Solve $f(x+f(2y))=f(x)+f(y)+y$
Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for each $x$ and $y$ in $\mathbb{R}^+$, $$f(x+f(2y))=f(x)+f(y)+y$$
Note:
$f(x)=x+b$ is a solution for all $b\in\mathbb{R}^+$ but I can not prove it.
| Another solution is
$f(x) =-x/2$.
Here,
at the end of my ramblings,
is how I found it.
Inspired by the comment
that a solution is
$x+b$,
let
$f(x) = x+g(x)$.
Then
$x+2y+g(2y)+g(x+2y+g(2y))
=x+g(x)+y+g(y)+y
$
or
$$g(2y)+g(x+2y+g(2y))
=g(x)+g(y).
$$
Letting $y=0$,
this is
$g(0)+g(x+g(0))
=g(x)+g(0)
$
or
$g(x) = g(x+g(0))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$? How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$?
I tried to find $(a_n-a_{n-1})$ but it didn't help.
| Let's show that for $\forall a_0>0$, the sequence is converging ...
Proposition 1. $3< a_n < 4$, for $\forall n \geq 2$.
If $a_0 > 0 \Rightarrow \frac{1}{a_0}>0 \Rightarrow \color{red}{a_1}=3+\frac{1}{a_0}\color{red}{>3}$. So $a_1 >0 \Rightarrow \frac{1}{a_1}>0 \Rightarrow \color{red}{a_2}=3+\frac{1}{a_1}\color{red}{>3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
H... | $$$$
\begin{align}
\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= x \tag{A}\\
\bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)\bigg(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\bigg)
&= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
(2+\sqrt 3)-(2-\sqrt 3) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
\bigg(\sqrt{2+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 12,
"answer_id": 8
} |
Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then h... | $$\frac{(n+1)^2 -n^2}{n}-2=\frac{n^2+2n+1-n^2}{n}-2=\frac{2n+1}{n}-2=\frac{1}{n}
\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Let $w = \dfrac{1 + i\sqrt{7}}{2}$, then $\dfrac{1 - i\sqrt{7}}{2} = \bar{w}$, its complex conjugate. We are required to find the value of $w^4 + \bar{w}^4$.
Observer that
$$w \bar{w} = 2 \quad {\rm and} \quad w + \bar{w} = 1.$$
Now
$$(w + \bar{w})^2 = w^2 + 2w\bar{w} + \bar{w}^2 \quad \Rightarrow \quad w^2 + \bar{w}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Argue that $f$ is not uniformly continuous Let $f(x)=x^3$. We want to show that $f$ is not uniformly continuous.
Can someone please explain from $|f(x_n)-f(y_n)=\big|(n+\frac{1}{n})^3 - n^3)\big| = 3n+\displaystyle\frac{3}{n}$? How does $\big|(n+\frac{1}{n})^3-n^3\big| $ yield $3n+\displaystyle\frac{3}{n}$? ${}{}{}{}$... | The first part is just the Binomial Theorem:
$$
\left(n + \frac{1}{n}\right)^3 = n^3 + 3 n^2 \cdot \frac{1}{n} + 3 n \frac{1}{n^2} + \frac{1}{n^3} = n^3 + 3n + \frac{3}{n} + \frac{1}{n^3}
$$
So
$$
\left(n + \frac{1}{n}\right)^3 - n^3 = 3n + \frac{3}{n} + \frac{1}{n^3}
$$
The author seems to skip the $\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
| \begin{align}\frac1{1-\cos x}-\frac2{x^2}&=\frac1{2\sin^2\left(\frac x2\right)}-\frac2{x^2}\\&=\frac{\frac{x^2}4}{\sin^2\left(\frac x2\right)}\times\frac2{x^2}-\frac2{x^2}\\&=\left(\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2-1\right)\frac2{x^2}.\end{align}But $\frac x{\sin x}=1+\frac{x^2}6+o(x^3)$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?
| We have: $$x + y = 14\sqrt{xy -48} \implies 196(xy-48)= (x+y)^2 \implies x^2+y^2-194xy+9408 =0$$
Now, rearranging the terms we get: $$ y^2-194xy = -x^2-9408 \implies 9409x^2-194xy +y^2=9408x^2-9408$$ $$ \implies (y-97x)^2 = 9408(x^2-1) \implies y = 97x \pm 56\sqrt{3(x^2-1)}$$
There are many solutions to this: $(1,97), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proof for a prime number formula involving the prime counting function How would I go about proving this? I came across this when I was searching for interesting prime generating function. Or alternatively, could you kindly direct me to a source containing a complete proof for the following formula?
$$p_n=1+\sum^{2^n}_... | Lemma: For any $x \geqslant 0$ and $n \in \mathbb{N}_+$, $\bigl[[x]^{\frac{1}{n}}\bigr] = [x^{\frac{1}{n}}]$.
Proof: It is easy to see that $\bigl[[x]^{\frac{1}{n}}\bigr] \leqslant [x^{\frac{1}{n}}]$. Now, suppose $a = [x^{\frac{1}{n}}] \in \mathbb{N}$, then$$
x = (x^{\frac{1}{n}})^n \geqslant a^n \Longrightarrow [x] \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that these perpendicular distances are in G.P. Let $BC$ be the chord of contact of the tangents from a point $A$ to the circle $x^2+y^2=1$. $P$ is any point on the arc $BC$. Let $PX, PY$ and $PZ$ be the lengths of the perpendiculars from P on the line $AB, BC$ and $CA $ respectively then prove that $PX, PY$, a... | I suspect there's an elegant synthetic solution to this, but after several hours of scribbling to no avail I resorted to trigonometry. My answer is very similar to mathlove's although I did derive it independently, and I think it's sufficiently different to be worth submitting.
WLOG, we can put $A$ on the +X axis. Let ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5... | You have to insert $(x,y,z)=(1/3,1/3,1/3)$ into the expression $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ to get the value $\dfrac{3}{2}$ because you are looking for the minimum of $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ but not $\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation us... | It's $$\cos2x(\cos2x+\cos4x)=\frac{1}{2}.$$
Now, let $\cos2x=t$.
Thus, $$t(t+2t^2-1)=\frac{1}{2}$$ or
$$4t^3+2t^2-2t-1=0$$ or
$$2t^2(2t+1)-(2t+1)=0$$ or
$$(2t^2-1)(2t+1)=0,$$ which gives
$$\cos4x=0$$ or $$\cos2x=-\frac{1}{2}$$ and we can write the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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On $\sum_{n=0}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}+x$ and solvable Galois groups? In this comment, I hastily assumed that,
$$\sum_{\color{blue}{n=0}}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}$$
In fact, this is... | It looks like the correct thing to look at is
$$\sum_{n=-\infty}^\infty\frac1{n^2+bn+c}.$$
This is surely expressible in terms of the digamma function.
Your rational fraction is surely an artefact of insisting on taking a one-sided sum. As an example,
$$\sum_{n=0}^\infty\frac1{n^2+9n+c}
=\sum_{m=-\infty}^{-9}\frac1{(-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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For $\alpha>1$ find $\lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1}$
For $\alpha>1$ find (w/o lhopital)$$
\lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1}
$$
Progress:
Used this:$$
x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})
$$ to get to this:
$$
\alpha-1=(\sqrt[2^n]{\alpha})^{2^n}-1^{2^... | $\sqrt[2^n]{\alpha}=(1+(\alpha -1))^{2^{-n}}=1+(\alpha -1)2^{-n}+o(2^{-n})$
$\sqrt[n]{\sqrt[2^n]{\alpha}-1}=((\alpha -1)2^{-n}+o(2^{-n}))^{\frac{1}{n}}=(\alpha -1)^{\frac{1}{n}}(2^{-n})^{\frac{1}{n}}+o(1)=1\cdot 2^{-1}=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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For which values of $m$ and $n$ is the probability $P_{(A,A)} = 0.5$? Given are the classes $A$ and $B$.
Suppose there are $n$ instances from class $A$ and $m$ instances from class $B$.
Let $X$ be all instances. Thus, $X=\{~\{a\}_n, \{b\}_m~\}$.
I denote a pair $p$ as a combination of two instances $(i, j$) from $X$, f... | The question, in effect, asks for all possible positive integer pairs $(m, n)$ with$$
\frac{n(n - 1)}{(m + n)(m + n - 1)} = \frac{1}{2}. \tag{1}
$$
Denote $l = n - m$. After some manipulation, (1) becomes$$
(2l - 1)^2 - 8m^2 = 1. \tag{2}
$$
This is a Pell's equation. Because the fundamental solution of $x^2 - 8y^2 = 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of x² from 0 to b - using archimedes sum of squares - Apostol's I'm reading Apostol's Calculus book and in the first chapter is presented the way archimedes found the sum of the square and how it can be used to calculate the integral of $x²$. But I'm not able to follow some steps of the proof.
from the book:
W... | The central idea is that you have to find a unique number (independent of $ n$) which lies between every $S_{\text{small}} $ and $S_{\text{big}} $ both of which depend on $n$. First thing to note here is that there can't be two distinct numbers say $A, A'$ lying between these $S$'s. Without any loss of generality let $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing something is even Show that if $x$ and $y$ are integers and $x^3+6=y^2$, then $x$ is odd.
I tried to do it by contradiction, with something like :
"Suppose for contradiction $x$ is even, then $x=2k$, and $x^3+6=y^2=8k^3+6=2(4k^3+3)$ which means $y=\sqrt{2(4k^3+3)}=\sqrt{2}\sqrt{4k^3+3}$ but I don't know how to... | suppose that $x$ is even, $x=2z$, $x^3+6=8z^3+6=y^2$ if $y$ is odd impossible since$y^2$ is odd and $8z^3+6$ is even, if $y$ is even, then $y=2a$ you have $8z^3+6=4a^2$, this implies that $4$ divides $6$ contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1. Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1.
My approach:
Total number of cases=$9^3$
Considering that a,b,c are 0 to 9 and and for the HCF to be 1 there should be no common factor of 2 or 3 .
So total... | For each leading digit from $1$ to $9$ there are $100$ possible integers. If the second and third digits both share a divisor greater than $1$ with the leading digit then we exclude that case.
So if $n$ digits share a divisor with the leading digit, we exclude $n^2$ cases.
$$
\begin{array}{c|crrr}
\text{Leading digit} ... | {
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Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $ What is $$\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $$
Putting $\sin(x) = t$, then $\cos(x) = \sqrt{1-t^2}$ and $\cos(x)\,dx = dt$
At $x = 0$, $\sin(x) = 0$, $\therefore\ t = 0$
At $x = \pi$, $\sin(x) = 0$, $\therefore\ t = 0$
Integral becomes
$$\int_0^0 t... | $$\int_{0}^{\pi} \sin^4 x\cos^6 x dx =\int_{0}^{\pi} (\frac{1-\cos 2x}{2})^2 (\frac{1+\cos 2x}{2})^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-\cos 2x)^2 (1+\cos 2x)^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-2\cos 2x+\cos^2 2x) (1+3\cos 2x+3\cos^2 2x+\cos^3 2x) dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1+\cos 2x-2\cos^2 2x-2... | {
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"source": "stackexchange",
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Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$ Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$
My attempt
$Mdx + Ndy$ form
$M = (y^2+2x^2y)$, $N =(2x^3-xy) , M_y=2y+2x^2, N_x=6x^2-y$
I am not getting in of standard forms here...
Another attempt:
$2x^2 d(xy)+y(ydx-xdy)=0$,
$2d(xy)=yd(y/x)$ Any clue pls???
| $$(y^2+2x^2y)dx+(2x^3-xy)dy=0....(i)\\
\implies y^2dx+2x^2ydx+2x^3dy-xydy=0\\
\implies y(ydx-xdy)+x^2(2ydx+2xdy)=0$$
Which is the form of :
$$x^ay^b(mydx+nxdy)+x^{a'}y^{b'}(m'ydx+n'xdy)=0$$
So, comparing both equations we get:
$a=0, b=1, m=1, n=-1, a'=2, b'=0, m'=2, n'=2$
Now,
$$\frac{a+h+1}{m}=\frac{b+k+1}{n}\\
\impli... | {
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Finding The Basis Of An Intersection Of Two Subspaces
Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$
I have forgot the algorithm for finding inters... | For the point b, remember that the Null Space of the matrix V = empty set, because they are independent. What you found is the Null Space of the Transpose(V). Same thing about U. But for the intersection you must to swap rows and columns.
| {
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"question_score": "3",
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Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$... | If B=A^-1 and b=<8,8,8,8> then from AV=b you get V=Bb and then BV=(B^2)b. I gess that this is the final result, if there are not other information about A.
| {
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"source": "stackexchange",
"question_score": "2",
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Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| Suppose there exists an $x$ such that $f(x) = x$
$f(x) = \frac {x}{3} + 3 = x\\
x = \frac 92$
It would be a reasonable guess that we get something very close to $\frac {9}{2}$
$f(4) = 4 +\frac 13\\
f(4+ \frac 13) = 4 + \frac 13 + \frac 19\\
f(4+ \frac 13 + \frac 19) = 4 + \frac 13 + \frac 19 + \frac 1{27}$
Now we are o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 ... | This kind or reasoning (let $\sum{a_n}=x$ and so on) is valid iff the series converges. Otherwise is just a mathematical formalism, nice, but meaningless (at least in these context. Probably using Analytic continuation you can give some motivation to your results using generating functions).
Obviously the fibonacci's s... | {
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"source": "stackexchange",
"question_score": "5",
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Finding side length and circumradius in a triangle
in a triangle $ABC$ if $2\sin B= \sin A $ and $a,b,c$ be the length of side $BC,CA$ and $AB$ and length of internal angle bisector through $A$ is $\displaystyle \frac{\sqrt{2}}{3}$ unit and the equation $25\cos^2(A-B)+x^2-40\cos(A-B)-2x+17=0$ has at least one solutio... | We have $$2\sin B=\sin A\tag1$$$$\implies \cos^2B=\frac{3+\cos^2A}{4}\tag2$$$$\implies b=\frac a2\tag3$$
You already have
$$\cos(A-B)=\frac 45,$$
i.e.
$$\cos A\cos B+\sin A\sin B=\frac 45\tag4$$
From $(1)(4)$, we have
$$\cos A\cos B=\frac 45-\frac{\sin^2A}{2}$$
Squaring the both sides and using $(2)$ give
$$\cos^2A\cdo... | {
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"source": "stackexchange",
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$\exp(z^2)=i$ solution check Find all complex solutions to the equation $\exp(z^2)=i$.
Attempt
We have $z^2 = x^2-y^2+2ixy$ where $z=x+iy$. Then
$$exp(x^2-y^2)=1 \text{ and } 2xy=\frac{\pi}{2}$$
So $x^2=y^2$ and $xy = \frac{\pi}{4}$ Thus $x = y$ and so $x=y= \pm \frac{\sqrt{\pi}}{2}$
I just don't feel very confident ab... | Your answer is fine except at the step where you say $2xy=\frac{\pi}{2}.$ In fact, you can have $2xy=\frac{\pi}{2}+2\pi k$ for any integer $k$, or $xy=\frac{\pi}{4}+\pi k.$
If $k\geq 0$ then you get $x=\pm\sqrt{\frac{\pi}{4}+\pi k}=\pm\frac{\sqrt{\pi}}{2}\sqrt{4k+1}$ and $y=x.$
If $k<0$ then you get $x=\pm\sqrt{-\pi k ... | {
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Count the integer solutions of $x_1+x_2-x_3+x_4-x_5=3$ I am asking for help with solving this exercise:
Find the count of possible integer solutions for equation:
$$x_1+x_2-x_3+x_4-x_5=3$$
There are restrictions for possible values of $x$:
$$
\begin{aligned}
&0 < x_1 \le 6\\
-&8 \le x_2 < -2\\
&x_3 \le 1\\
&3 < x_4\\
&... | By translations $x\to x+c$ and symmetries $x\to-x$ you can reformulate your problem as
Find all $x_1,x_2,x_3,x_4,x_5$ integers that satisfy
$$
1+x_3+x_4 = x_1+x_2+x_5$$
with bounds
$$0\le x_1\le 5\qquad 0\le x_2\le 5\qquad 0\le x_5\le 6$$
$$0\le x_3\qquad 0\le x_4$$
Now, call $p(n)$ the number of triples $(x... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove Stirling numbers of the second kind, $S(n,n)=1$ Stirling numbers of the second kind,
$$
S(m,n)={m \brace n}=\frac{1}{n!}\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^m
$$
When $n=m$, I can easily find $S(n,n)=1$ if I set values for $n$. Ex:
$$
S(2,2)=\frac{2^2-2.1^2+0}{2!}=\frac{2}{2}=1
$$
How do i prove it for a gener... | Use the old coefficient trick
\begin{eqnarray*}
[y^n]: e^{yk} = \frac{k^n}{n!}.
\end{eqnarray*}
Now the sum can be written as (reverse the index)
\begin{eqnarray*}
\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \frac{k^n}{n!} &=& [y^n]: \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} (e^{y})^{k} \\
&=& [y^n] : (e^y -1)^n = 1.
\end{eqna... | {
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"source": "stackexchange",
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How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac... | Let $f(x)=\frac{x}{1+x}.$
Then by Jensen inequality
$$
f(a)+f(b)=\frac{a}{1+a}+\frac{b}{1+b} \geq 2 f( \frac{a+b}{2})=\frac{a+b}{1+\frac{a+b}{2}} \geq \frac{a+b}{1+a+b}.
$$
| {
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"answer_id": 1
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What are the last 2 digits of $31^{41}$? By using a slick trick, I found that the last two digits were $31.$ However, I want to verify this using Fermat's little theorem or some alternate. How would I apply Fermat's Little Theorem?
| You changed the problem on me after I answered. So:
Two find the last two digits, just work in $\mathbb Z\bmod100.$
\begin{align}
33^1 & \equiv 31 \\
31^2 & \equiv 31\times 31 \equiv 61 \\
31^3 & \equiv 31\times61 \equiv 91 \\
31^4 & \equiv 31\times91 \equiv 21 \\
31^5 & \equiv 31\times21 \equiv 51 \\
31^6 & \equiv 31\... | {
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Find $ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2} $ I want to find the following limit:
$$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$
I guess it converges to 0 and I tried to prove it using the squeeze theorem b... | Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.
This is a Riemann sum:
\begin{align}
\lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^... | {
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"url": "https://math.stackexchange.com/questions/2596230",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Definite Trigonometric Integration
Calculate the integral : $$\ \int_{0}^{\pi}\frac{x}{a-\sin(x)}dx , \quad a>1 $$
I have tried trig substitutions, and replacing $\ x$ with $\ \pi-y $, along with other commonly used "tricks", but none seem to work.
| As mentioned in the comments, the substitution $x \to \pi - x$ yields :
$$\int_0^{\pi}\frac{xdx}{a-\sin x}=\frac{\pi}{2}\int_0^{\pi}\frac{dx}{a-\sin x}$$
I'll go over the calculation of the integral on the RHS.
We have :
$$ I ={\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x =-{\displaystyle\int}\dfrac... | {
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"source": "stackexchange",
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Is this substitution to solve an integral correct? To solve the integral:
$\int\frac {x}{\sqrt{4-x^2}}\ dx$
I used the substitution:
$u = \sqrt{4-x^2}$
hence:
$\frac{du}{dx} = -\frac {x}{\sqrt{4-x^2}}$
$dx = -\frac {\sqrt{4-x^2}}{x}\ du$
So the integral becomes:
$\int\frac {x}{u}(-\frac {\sqrt{4-x^2}}{x})\ du$
$= \int\... | Let $4-x^2=u$
Then $xdx=-\frac {du} 2$......
| {
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Infinite series summation
$$\sum^{\infty}_{n=0}\frac{1}{(4n+5)(4n+6)\cdots \cdots (4n+11)}$$
Try: $$\sum^{\infty}_{n=0}\frac{(4n+4)!}{(4n+11)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\frac{(4n+4)!\cdot 6!}{(4n+4+6+1)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\int^{1}_{0}x^{4n+4}(1-x)^6dx$$
$$ \frac{1}{6!}\int^{1}_{0}\left(\sum^{\... | Hint:
Another way:
$$\prod_{r=5}^{11}\dfrac{4n+11-(4n+5)}{4n+r}=\prod_{r=6}^{11}\dfrac1{4n+r}-\prod_{r=5}^{10}\dfrac1{4n+r}$$
See Telescoping series
| {
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Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $
I've got in my assignment to show if the following series converges or diverges.
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$
Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right... | Thank you for the comment. By using the comparison test I easily proved the sum converges.
$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my p... | Note that
$$a_{n}=a_{n-1}^{2}-2\implies a_n^2=a_{n-1}^{4}-4a_{n-1}-4\implies a_n^2-4=a_{n-1}^{2}(a_{n-1}^{2}-4) $$
thus
$$a_{n-1}^2-4=a_{n-2}^{2}(a_{n-2}^{2}-4) \implies a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}(a_{n-2}^{2}-4) $$
and so on, thus it can be easily proved by induction that
$$a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}...a_2^2a_1... | {
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Why is this approximation correct? I came across an interesting approximation today in an indian olympiad paper (page 6). It essentially boils down to:
$$V(x)=\frac{k}{\sqrt{x^2+4/3}}+\frac{k}{\sqrt{(x+1)^2+1/3}}+\frac{k}{\sqrt{(x-1)^2+1/3}}$$
$$\approx k\sqrt{\frac 34}(3+\frac {9}{16}x^2)$$
Even before I got down to a... | Consider $$y(a,b)=\frac{1}{\sqrt{(x+a)^2+b}}$$ Using Taylor series around $x=0$, you get
$$y(a,b)=\frac{1}{\sqrt{a^2+b}}-\frac{a x}{\left(a^2+b\right)^{3/2}}+\frac{x^2 \left(2
a^2-b\right)}{2 \left(a^2+b\right)^{5/2}}+\frac{x^3 \left(3 a b-2 a^3\right)}{2
\left(a^2+b\right)^{7/2}}+O\left(x^4\right)$$ Using this ... | {
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Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate pro... | $a,b$ are the solutions of $x^3+x+1=0$, so we have
$$a^3+a=-1=b^3+b$$
and so,
$$(a^3-b^3)+(a-b)=0\implies (a-b)(a^2+ab+b^2+1)=0$$
Suppose here that $a=b$. Then, since $a+b+c=0\implies c=-2a$, we have
$$abc=-1\implies -2a^3=-1\implies a^3=\frac{1}{2}\implies a=-1-a^3=-\frac 32$$
However, $a=-\frac 32$ is not a solution ... | {
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"question_score": "3",
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Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\be... | To use L'Hospital rule, one needs to check if $\lim_{x\rightarrow 0}\dfrac{f'(x)}{g'(x)}$ exists. So in this case the topologist's sine curve $\sin(1/x)$ does not have limit whenever $x\rightarrow 0$, so L'Hospital rule fails to apply.
Now use the trick like
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{x}{\sin x}=1
\end... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Probability of someone picking the same toy For someone to pick the same toy:
Jon picks car1 and car2
Given 3 people
Hans
Thomas
Jon
And 6 toys:
Car1, Car2, Doll1, Doll2, Ball1, Ball2
Each person is given 2 toys, calculate the probability of someone getting the same toy. (there are two instances of each toy).
T... | Someone picking a toy:
One in left hand one in right.
Left hand first, will chose between 6 options.
Then the right hand will have 5 options.
$5\cdot 6=30$ options
Do we care about what is in which hand? No... So:
In any case all the ways for the first "player" are $\frac{6\cdot 5}{2}=15$.
Second person has to chose b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Calculate length of one side given all other sides in a triangle.
All the lengths in red had been given in the problem statement. The task is to find $BC$. The green ones were calculated from the information in red.
Here's two ways I tried to solve it (and I would like feedback):
*
*
The green line is drawn parall... | Your question doesn't mention $PQ$ is parallel to $BC$ but your answer does.
So assuming that line $PQ$ is parallel to $BC$ then by similar triangles
( Considering the triangles $APQ$ and $ABC$)
$$\frac {AP}{AB}= \frac {AQ}{AC}=\frac {PQ}{BC}$$
Hence we get $$\frac {3}{12}=\frac {4}{BC}$$
$$\Rightarrow BC=16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital
$$\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=\frac{1}{e^2}.$$
But how do I prove this without using de l'Hôspital twice?
| Without change of variables, note that
$$ \left ( \frac{2n-1}{2n+3} \right ) ^n =\left ( \frac{2n+3-4}{2n+3} \right ) ^n=\left ( 1+\frac{-4}{2n+3} \right ) ^n=\sqrt{\left ( 1+\frac{-4}{2n+3} \right ) ^{2n+3}\left ( 1+\frac{-4}{2n+3} \right ) ^{-3}}\to \sqrt{e^{-4}\cdot1}=\frac1{e^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$
Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$
My approach:
I was thinking to use the binomial f... | It follows from Pascal's identity that\begin{align}\binom{n+1}k&=\binom nk+\binom n{k-1}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-1}{k-2}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-2}{k-2}+\binom{n-2}{k-3}\\&=\cdots\end{align}Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving $ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$
Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$
Well I first observe... | Let $$A=\frac{x^2}{x-1},\qquad B=\sqrt{x-1},\qquad C=\frac{\sqrt{x-1}}{x^2}$$
Then, we have
$$A+B+C=\frac 1A+\frac 1B+\frac 1C\qquad\text{and}\qquad ABC=1$$
from which we have
$$A+B+C=AB+BC+CA\qquad\text{and}\qquad ABC=1$$
Let $s:=A+B+C=AB+BC+CA$.
It follows from these that $A,B,C$ are the solutions of
$$t^3-st^2+st-1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How to remove absolute value from inequality So I know that $|x| \le 2 \iff x\le 2 \text{ and } x \ge -2$. But, I was wondering would the same rules apply for $|x+2| + |y| \le 5$. Would it be correct to say that this inequality is equivalent to $x + y \le 3 \text{ and }x + y \ge -7$?
| For a problem involving two variables such as $$ |x+2| + |y| \le 5$$ you need to look at 4 regions in the $xy-plane$
1) $y\le 0$ and $x+2 \le 0$. In this region your inequality is $-(x+2)-y \le 5$
2) $y\le 0$and $x+2 \ge 0$. In this region your inequality is $(x+2)-y \le 5$
3) $y\ge 0$ and $x+2 \ge 0$. In this region y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding minimum value of logarithmic series Find the min value of $$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \cdot \cdot +\log_{x_{2016}} \left ( x_{2017} - \frac {1}{4}\right) +\log_{x_{2017}} ... | Note that
$$\log_{x_i} \left ( x_j - \frac {1}{4}\right)=\frac {\log \left ( x_j - \frac {1}{4}\right)}{\log x_i}$$
thus by Rearrangement inequality
$$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding value of $k$ in trigonometric equation.
$$\prod^{50}_{r=1}\tan\bigg[\frac{\pi}{3}\bigg(1+\frac{3^r}{3^{50}-1}\bigg)\bigg]=k\prod^{50}_{r=1}\cot\bigg[\frac{\pi}{3}\bigg(1-\frac{3^r}{3^{50}-1}\bigg)\bigg]$$
Try:$$\tan (60^\circ+x)=\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}$$
and $$\cot(60^\circ-x)=\frac{\sqrt{3}-... | $$X= \prod^{50}_{r=1}\tan \left(\frac{\pi}{3}\left(1+\frac{3^r}{3^{50}-1}\right)\right)$$ and $$Y = \prod^{50}_{r=1}\cot \left(\frac{\pi}{3}\left(1-\frac{3^r}{3^{50}-1}\right)\right)$$
Assuming $$a_r=\frac{\pi}{3}\cdot \frac{3^r}{3^{50}-1}$$.Then
$$\tan(\frac{\pi}3+a_r)=\frac{\sqrt 3+\tan a_r}{1-\sqrt 3\tan a_r}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question related to beta and gamma function I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?
| Here is a way to evaluate the integral without invoking the Beta function.
Writing the integral as
$$\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx = \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_1^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx.$$
In the second of these integral, if a substitution o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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.