Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Critique on a proof by induction that $\sum\limits_{i=1}^n i^2= n(n+1)(2n+1)/6$? I need to make the proof for this
1:$$1^2 + 2^2 + 3^2 + ... + n^2=\frac{(n(n+1)(2n+1))}{6}$$
By mathematical induction I know that,
If P(n) is true for $n>3^2$ then P(k) is also true for k=N and also P(k+1) must be true.
Then,
2:$$1^2 + 2... | For the third step, notice that
$$ 1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2 = \frac{(n(n+1)(2n+1))}{6}+(n+1)^2 $$
$$ = \frac{(n(n+1)(2n+1))}{6}+\frac{6(n+1)^2}{6}$$
$$ = \frac{(n+1)(n(2n+1)+6(n+1))}{6}=\dots $$
Can you finish it?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^... | Hint: Your quartic is palindromic. Divide through by $x^2$. We get the equation
$$x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}=0.$$
Make the substitution $x+\frac{1}{x}=t$. Then $x^2+\frac{1}{x^2}=t^2-2$, and we arrive at the equation
$$t^2+at+b-2=0.$$
Imagine solving the quadratic. We will get two not necessarily real values o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$ Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$
| Remark: @chenbai gave a nice solution using the Buffalo Way (BW).
Here, I give an alternative solution using BW:
By Holder, we have
\begin{align}
\left(\sum_{\mathrm{cyc}} \sqrt{3x + 1/y}\right)^2 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}
\ge \left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3.
\end{align}
It s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the following system of simultaneous congruences: \begin{gather}
3x\equiv1 \pmod 7 \tag 1\\
2x\equiv10 \pmod {16} \tag 2\\
5x\equiv1 \pmod {18} \tag 3
\end{gather}
Hi everyone, just a little bit stuck on this one. I think I am close, but I must be getting tripped up somewhere. Here is what I have so far:
from (2)... | This is not an answer, but a guide to a simplification. The first congruence is fine. Note that the second congruence is equivalent to $x\equiv 5\pmod{8}$. Any solution of this congruence must be odd.
Now look at the third congruence. Note that as long as we know that $x$ is odd, we automatically have $5x\equiv 1\pmod{... | {
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Solve $(a^2-1)(b^2-1)=\frac{1}4 ,a,b\in \mathbb Q$ Does the equation $(a^2-1)(b^2-1)=\dfrac{1}4$ have solutions $a,b\in \mathbb Q$?
I search $0<p<1000,0<q<1000$, where $a=\dfrac{p}q$, but no solutions exist. I wonder is this equation solvable?
| Here are some observations which are too extensive really for a comment, and which reduce the search space somewhat, and which may assist in establishing a contradiction.
You can rewrite your equation as $$(ab+1)^2=\frac 14+(a+b)^2$$
Since this is a rational equation, clearing denominators gives a primitive pythagorean... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \... | It s easier to deal with this form
$$ f(x) = \sin^2(x)\cos^2(x)=\frac{\sin^2(2x)}{4} $$
$$ \implies f'(x) = \sin(2x)\cos(2x)=\frac{\sin(4x)}{2}. $$
Now, you should be able to finish the problem.
| {
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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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Let $X$ be the number on the first ball drawn and $Y$ the larger of the two numbers draw. Find the joint discrete density function of $X$ and $Y$. Consider a sample of size 2 drawn without replacement from an urn containing three balls, numbered 1,2, and 3. Let $X$ be the number on the first ball drawn and $Y$ the larg... | The solution is correct. Congratulations.
| {
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After what interval in degrees or radians do sine, cosine and tangent values repeat? Between $0$ to $2π$, I have noticed that $\sin x$, $\cos x$ and $\tan x$ values repeat for different values of $x$.
For example, $\sin 30 = \sin 150$
What exactly is the interval between two successive values of $x$ such that the value... | $1:$
Let $\sin x=\sin y$
Applying $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2,$
we have $2\sin\frac{x-y}2\cos\frac{x+y}2=0$
If $\sin\frac{x-y}2=0, \frac{x-y}2=n180^\circ\implies x-y=n180^\circ$ where $n$ is any integer
If $\cos\frac{x+y}2=0, \frac{x+y}2=(2m+1)90^\circ\implies x+y=(2m+1)180^\circ$ where $m$ is any i... | {
"language": "en",
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"question_score": "3",
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Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a... | Hint: Note that
$$a_n\le 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{n-1}}\lt 2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$. Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.
I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x... | Let $P(x,y)$ denote $g(x+y) + g(x) g(y) =g(xy) + g(x) + g(y) $. The important statements that we'd consider are
$$ \begin{array} { l l l }
P(x,y) & : g(x+y) + g(x) g(y) =g(xy) + g(x) + g(y) & (1) \\
P(x,1) & : g(x+1) + g(x) g(1) = 2 g(x) + g(y) & (2) \\
P(x, y+1) & : g(x+y+1) + g(x) g(y+1) = g(xy+x) + g(x) +g(y+1) & (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | $$\frac{\pi}{t}=\int_{-\infty}^{\infty} \frac{dx}{t^2x^2+1}\Rightarrow\frac{\pi}{t^2}=\int_{-\infty}^{\infty}\frac{2tx^2}{(t^2x^2+1)^2}\,dx\Rightarrow \frac{\pi}{2}=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2}\,dx$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how t... | The denominators alternate signs, but they are of the form
$$\sum_{k=0}^m (-1)^k x^{2 k} = \frac{(-1)^{m+1} x^{2(m+1)}-1}{x^2+1}$$
The sum may then be written as
$$(x^2+1) \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(-1)^{n+1} x^{2(n+1)}-1} = -(x^2+1) \sum_{n=0}^{\infty} \frac{x^n}{x^{2(n+1)}+(-1)^n}$$
I should note that con... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$ Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$
I'm I want to say that you cross multiply to get the same denominator, but I could be wrong.
Please Help!!
| You are correct that a common denominator is necessary:
\begin{align}
\frac 6 x - \frac {42} {x^2 + 7x} &= \frac 6 x - \frac {42} {x(x + 7)}\\
&= \frac{6(x + 7) - 42}{x(x + 7)} \\
&= \frac{6x}{x(x + 7)} \\
&= \frac{6}{x + 7}
\end{align}
for $x \ne 0$. Do you see how to compute the limit now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove $\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\frac{n}{\ln{n}}\right)=-1$? Let equation $x^n+x=1$ have positive root $a_{n}$.
Show that
$$\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\dfrac{n}{\ln{n}}\right)=-1$$
some hours ago,it prove that $$\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$... | I'd like to post a separate answer here in which I'll use a very different method to obtain the result. The search for this method was prompted by this comment on the previous question where the OP asks for a bound of the form $f(n) \leq a_n \leq g(n)$ with $f(n),g(n) \to 1$ as $n \to \infty$. The bound I obtained tu... | {
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Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive... | $p$ is the greatest integer such that $2^p$ divides $m$.
This means that :
*
*there exists an integer $k$ such that $m = 2^p k$
*for any integer $q$, if there is an integer $l$ such that $m = 2^q l$, then $p \ge q$.
Suppose the $k$ you get by 1. is even. Then $k = 2l$ for some integer $l$. Hence $m = 2^p(2l) = 2^... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.
*
*$\int \cos^3 x \, dx$
I did
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x... | You can even put the formula of $$\cos^{3}x = \frac{\cos3x + 3 \cos x}{4}$$
And then integrate it where you get the answer as
$$ \frac{\sin 3x}{12} + \frac{3 \sin x}{4} + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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integral part of surd
$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$
$(\sqrt{a}+b)^{n+2} =M+g$ where $g \in (0,1)$
Given that $0<\sqrt{a}-b<1$ and $(a,b)$
belongs to integers, then
*
*If $n$ is odd, $f>g$
*If $n$ is even, $f<g$
How to prove/disprove it?
| Since $a$ and $b$ are integers (and $a > 0$ is not a square, otherwise $0 < \sqrt{a}-b < 1$ would not hold), we have
$$\begin{align}
(b+\sqrt{a})^n + (b-\sqrt{a})^n &= \sum_{k=0}^n \binom{n}{k}b^{n-k}a^{k/2} + \sum_{k=0}^n (-1)^k\binom{n}{k}b^{n-k}a^{k/2}\\
&= \sum_{k=0}^n \bigl(1 + (-1)^k\bigr)\binom{n}{k}b^{n-k}a^{k/... | {
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"timestamp": "2023-03-29T00:00:00",
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How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
| As noted by Mher Safaryan, your inequality is equivalent to
$$
(1+a)\sqrt{b^2+1} \geq 5\sqrt{5} \tag{1}
$$
or in other words,
$$
(1+a)^2(1+b^2) \geq 125 \tag{2}
$$
or equivalently,
$$
(1+\frac{8}{b})^2(1+b^2) \geq 125 \tag{3}
$$
We are now done, because of the identity
$$
(1+\frac{8}{b})^2(1+b^2)-125=\frac{(b-2)^2\bigg... | {
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"timestamp": "2023-03-29T00:00:00",
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Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
| Let $u=\frac{2x}{3}$ and notice that $u^{4}=\big(\frac{2x}{3}\big)^{4}=\frac{2^{4}x^{4}}{3^{4}}=\frac{16x^{4}}{81}$.
So we get:
$\frac{\frac{16x^{4}}{81}-y^{4}}{\frac{2x}{3}+y}=\frac{u^{4}-y^{4}}{u+y}=\frac{(u^{2}-y^{2})(u^{2}+y^{2})}{u+y}=\frac{(u-y)(u+y)(u^{2}+y^{2})}{u+y}=(u-y)(u^{2}+y^{2})=(\frac{2x}{3}-y)(\frac{4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to get all solutions to equations with square roots I would like to find all solutions to
$$b-a\sqrt{1+a^2+b^2}=a^2(ab-\sqrt{1+a^2+b^2})$$
$$a-b\sqrt{1+a^2+b^2}=b^2(ab-\sqrt{1+a^2+b^2})$$
I found some solutions. For example, $a = 1, b = \pm i$ and $b = 1 , a = \pm i$. How can I find all solutions?
| The first equation becomes
$$b(1-a^3)=a(1-a)\sqrt{1+a^2+b^2},$$
the second
$$a(1-b^3)=b(1-b)\sqrt{1+a^2+b^2}.$$
If $a=0$, then $b=0$.
Clearly $a=b=1$ is a solution. Let $a=1$ and $b\neq1$. Then from the second equation
$$1+b+b^2=b\sqrt{2+b^2}$$
we derive the solutions $b=\pm i$.
Let $a$ and $b$ not equal 0 or 1. Th... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers
| Here's a solution for positive $x$ and $y$.
I will show that
the only solutions
for positive $x$ and $y$ are
$(x, y) = (2, 1)$ and $(11, 3)$.
$x(x+1) = y(y+1)(y^2+2)
=y(y^3+y^2+2y+2)
=y^4+y^3+2y^2+2y
$
Multiplying by 4,
$(2x+1)^2-1
=4y^4+4y^3+8y^2+8y
$
or
$(2x+1)^2
=4y^4+4y^3+8y^2+8y+1
$
My goal is to show algebrai... | {
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"url": "https://math.stackexchange.com/questions/492581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider the funcition and find an expression Consider the function $f(x) = \frac{2}{x-1}$ and find an expression for $\frac{f(x+h)-f(x)}{h}$ and simplify where he cannot be equal to $0$.
| $\require{cancel}$
$$f(x) = \frac 2{x - 1}$$
$$\frac{f(x + h) - f(x)}{h} = \frac{\dfrac{2}{(x + h) - 1} - \dfrac {2}{x - 1}}{h}$$
Multipy numerator and denominator by $(x + h - 1)(x - 1)$
$$\begin {align}\frac{\dfrac{2}{(x + h) - 1} - \dfrac {2}{x - 1}}{h} \cdot \frac{ \dfrac{(x + h - 1)(x - 1)}{1}}{(x + h - 1)(x - 1)... | {
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Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Simila... | Would a=b=4 work? In that case the (double) solution in both cases is 2 and that's positive
| {
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"url": "https://math.stackexchange.com/questions/494679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Recurrence relation problem, need help:) I´m stuck on a problem. Can anyone help me?
The problem: Find the recurrence relation to
$$a_n=a_{n-1}+2a_{n-2}+\cdots+(n-1)a_1+na_0\;(\text{for }n\ge 1),\\a_0=1$$
I guess I have to compare $a_n-a_{n-1}$ with $a_{n-1}-a_{n-2}$?
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write your recurrence as:
$\begin{equation*}
a_{n + 1}
= \sum_{0 \le k \le n} (k + 1) a_{n - k}
\end{equation*}$
Multiply the recurrence by $z^n$ and sum over $n \ge 0$, recognize some sums and use $a_0 = 1$:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ I found the following two relational expressions in a book without any additional information:
$$\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac13(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$$
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\f... | let
$$\sqrt[3]{\dfrac{1}{3}}=x,\sqrt[3]{\dfrac{2}{3}}=y$$
then
$$\sqrt[3]{\dfrac{1}{9}}-\sqrt[3]{\dfrac{2}{9}}+\sqrt[3]{\dfrac{4}{9}}=x^2-xy+y^2$$
then
$$(x+y)(x^2-xy+y^2)=x^3+y^3=\dfrac{1}{3}+\dfrac{2}{3}=1$$
$$\Longrightarrow x^2-xy+y^2=\dfrac{1}{x+y}=\sqrt[3]{3}(\sqrt[3]{1}+\sqrt[3]{2})^{-1}=\sqrt[3]{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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question on separable differential equations I'm trying to solve this equation but always got stuck in a term. How can I separate the $x$,$y$ variables in the equation?
$$(y^2 -yx^2)dy + (y^2 + xy^2)dx =0$$
| For the ODE $(y^2-yx^2)~dy+(y^2+xy^2)~dx=0$ , I don't think you can separate the $x$ and $y$ variables directly, instead you can only solve it like this:
$(y^2-yx^2)~dy+(y^2+xy^2)~dx=0$
$y^2(x+1)~dx=(yx^2-y^2)~dy$
$(x+1)\dfrac{dx}{dy}=\dfrac{x^2}{y}-1$
Let $u=x+1$ ,
Then $x=u-1$
$\dfrac{dx}{dy}=\dfrac{du}{dy}$
$\theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sq... | Take a logarithm from the both sides and use $\log n\leq n-5$ for $n\geq 7$ and we have:
$$
\log C= \log \sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} = \frac{\log 2}{2}+\frac{\log 3}{2^2}+\frac{\log 4}{2^3}+\frac{\log 5}{2^4}+...\\
\leq \frac{\log 2}{2}+...+\frac{\log 6}{2^5}+\frac{2}{2^6}+\frac{3}{2^7}+...\\
\leq 0.92 +\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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integration by parts !!!! PD: I did a little change in the denominator !!!!
I need to solve this integral using integration by parts.
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
Thanks!
PS: I know that I can to do:
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}=\int\frac{(x-c)\,dx}{\sqrt{(a^2+... | Let $x-c=\sqrt{a^2+b^2} \tan \theta,$ then $d x=\sqrt{a^2+b^2} \sec ^2 \theta d \theta$ and
$$
\begin{aligned}
\int \frac{x d x}{\sqrt{\left(a^2+b^2\right)+(x-c)^2}} &=\int \frac{c+\sqrt{a^2+b^2 \tan \theta}}{\sqrt{a^2+b^2 \sec \theta}} \sqrt{a^2+b^2 \sec ^2 \theta} d \theta\\
&=\int\left(c+\sqrt{a^2+b^2} \tan \theta\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\cdots+\frac{n^2}{\sqrt{n^{6}+n}}$ Let $$x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}.$$ Then $(x_{n})$ converges to
(A)$1$
(B)$0$
(C)$\frac{1}{2}$
(D)$\frac{3}{2}$
My Tr... | $$
x_{n}=\frac{n^2}{\sqrt{n^{6}}}+\frac{n^2}{\sqrt{n^{6}}}+\cdots+\frac{n^2}{\sqrt{n^{6}}} = n(\frac{n^2}{\sqrt{n^{6}}}) = \frac{n^3}{n^3} = 1
$$
This is how I solve this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the equation of an ellipse centered at (-1,3) and through the points (1,3) and (-1,4) I've been taking a computer graphics class and as a review we had some questions about some geometric problems. This one I can not seem to figure out though it seems that it should be straight forward.
I know that the standard... | Substitute into your second equation (corrected with $b^2$ in the second denominator) the $x, y$ values for each of the points on the ellipse $(1, 3), (-1, 4)$, and you'll obtain two equations in two unknowns, from which you can solve for $a^2$ and $b^2$.
That is, evaluate $$\dfrac{(x+1)^2}{a^2} + \dfrac{(y-3)^2}{b^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show $\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{k+1} $ I have been attempting to show $$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1} $$ and my work is
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{r!((k+1)-r)!} + \frac{(k+1)!}{(r+1)!((k+1)-(r+1))!}$$
$$=\frac{(k+1)!(r+1)}{(r+1)!((k+1)-r)!} + \frac{(k+1)!
... | We have $\binom{k+1}{r} = \frac{(k+1)!}{(k+1-r)!r!}$ and $\binom{k+1}{r+1} = \frac{(k+1)!}{(k+1-(r+1))!(r+1)!} = \frac{(k+1)!}{(k-r)!(r+1)!}$. Both have a common factor of $\frac{(k+1)!}{(k-r)!r!}$ so let's factor that out to get
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\left(\frac{1}{k+1-r}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many eight-digit numbers can be formed with the numbers 2, 2, 2, 3, 3, 3, 3, 4, 4? I know how to determine the problem when we form nine-digit numbers that is $\frac{9!}{(3!\cdot 4!\cdot 2!)}$. But what about eight-digit numbers?
| We need to omit exactly one of the numbers. So any $8$-digit number induces the multiset of numbers:
\begin{align*}
& \{2,2,3,3,3,3,4,4\}, \\
& \{2,2,2,3,3,3,4,4\}, \text{or } \\
& \{2,2,2,3,3,3,3,4\}.
\end{align*}
These give rise to
\begin{align*}
\binom{8}{2,4,2}+\binom{8}{3,3,2}+\binom{8}{3,4,1} &= 420+560+280\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I reach this form I just want to check with you guys how to do this correctly:
I have to reduce:
$$ \frac{1}{N}+3=\frac{2R+2}{5R+2}$$
...to the following form:
$$ N=\frac{aR+b}{cR+d}$$
Any leads on how I reach this form? Should I flip all terms, to get N=, then bring 1/3 to the other side and subtract it there?
... | First, subtract $3$ from each side of the equation. Then find a common denominator: $$\begin{align} \frac{1}{N}+3=\frac{2R+2}{5R+2} &\iff \frac{1}{N} =\frac{2R+2}{5R+2} -3 \\ \\ & \iff \dfrac 1N = \dfrac{(2R+2) - 3(5R+2)}{5R + 2} \\ \\ & \iff \frac 1N = \frac{-13 R- 4}{5R + 2}\end{align}$$
Simplify, and "flip"/invert... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all natural numbers n such that $7n^3 < 5^n$ Find all natural numbers $n$ such that $7n^3 < 5^n$.
I drew a graph which showed that $n \geq 4$
wolfram
How can I prove that? I guess I need to use induction with the base case $n=4$?
But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not subst... | First, check for $n = 4$:
$$7n^3 = 448 < 625 = 5^n.$$
Now, assume that it is true for all numbers strictly less than $n$. Let us check for $n$:
\begin{align*}
7n^3 &= 7((n-1)+1)^3 = 7(n-1)^3 + 7 \cdot 3 \cdot (n-1)^2 + 7 \cdot 3 \cdot (n-1) + 7. %\\
%&< 4 \cdot 7 (n-1)^3 < 4 \cdot 5^{n-1} < 5 \cdot 5^{n-1} = 5^n.
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove: $\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left\vert\frac{\sin{(n+1)x}}{\sin{x}}\right\vert dx-\frac{2\ln{n}}{\pi}\right)$ show that
$$\mathop {\lim }\limits_{n \to \infty } \left( {\int\limits_0^{\frac{\pi }{2}} {\left\vert\frac{{\sin \left( {2n + 1} \right)x}}{{\sin x}}\right\vert\,dx - \frac{{2... | Notice for any continuous function $f(x)$ on $[0,\frac{\pi}{2}]$, we have:
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| f(x) dx = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} f(x) dx$$
Apply this to $\frac{1}{\sin x} - \frac{1}{x}$, we get
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and co... | You need to do $$\frac 1{x^2-x-2}=\frac A{x-2}+\frac B{x+1}$$
Clear fractions (multiply both sides by $x^2-x-2$) to obtain $$1=A(x+1)+B(x-2)$$
You should be able to do it from there. Easy way - set $x=2$, $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$
So $\displaystyle \int\frac{t+... | Addendum:
Evaluating $ \displaystyle{\int\frac{t}{t^3 + 1}\,\mathrm{d}t} $ without partial fractions:
$$
\begin{aligned}
\int\frac{t}{t^3 + 1}\,\mathrm{d}t&=\frac{1}{2}\int\frac{t+1+t-1}{t^3+1}\,\mathrm{d}t\\
&=\frac{1}{2}\int\frac{t+1}{(t+1)(t^2-t+1)}\,\mathrm{d}t-\frac{1}{2}\int\frac{t^2 - t + 1 - t^2}{t^3 + 1}\,\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result. Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result.
I've attempted to apply the Squeeze Theorem as such:
$\frac{-1}{x^2 + y^4} \leq \frac{\sin{(x^3 + y^5)}}{x^2 + y^4} \leq \frac{1}{x^2 + y^4}$. Cl... | $$
\left|\frac{\sin(x^{2+\color{red}{n}} + y^{4+\color{blue}{m}})}{x^2 + y^4}\right|
\leqslant\left|\frac{x^{2+\color{red}{n}} + y^{4+\color{blue}{m}}}{x^2 + y^4}\right|
\leqslant\frac{|x|^\color{red}{n}\cdot x^2+|y|^\color{blue}{m}\cdot y^4}{x^2+y^4}
\leqslant|x|^\color{red}{n}+|y|^\color{blue}{m}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Polynomial Question Find polynomials $A(x)$ and $B(x)$ such that $A(x)P(x) + B(x)Q(x) = x + 1$ for all $x$ where $P(x) = x^4 - 1$ and $Q(x) = x^3 + x^2$. I'm stumped on this question. I know that I'm supposed to apply the extended version of Euclid's algorithm for polynomials but I'm unsure of how to do that. I thought... | By using Extended Euclidean Algorithm,
$$\begin{align}
x^4-1 =& \left(x^3+x^2\right)(x-1) +x^2-1\\
x^3+x^2 =& \left(x^2-1\right)(x+1) + x+1\\
x^2-1 =& \left(x+1\right)(x-1) + 0
\end{align}$$
And so $x+1$ is the GCD of $P(x)$ and $Q(x)$. Now, we substitute previous remainders to the second-to-last line:
$$\begin{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real... | A comment to complete the other proof.
We try to show that it is enough to consider the inequality for only positive $a,b,c$. Assume $a\leq b\leq c$.
$$
\sum_{cyc}\dfrac{1}{2a^2-6a+9}=
\sum_{cyc}\dfrac{1}{(b+c)^2+a^2}=
\sum_{cyc}\dfrac{1}{9-2ab-2ac}
$$
Now it can be seen that if $a\leq 0$ and $b\leq 0$, one can choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Partial Fractions-Hows I seem to have serious problem understanding entry points of Partial fractions. I would like to decompose the following:
$$\dfrac{x^4-8}{x^2+2x}$$.
My workings. Please help me judge if I am getting the concept or completely lost:
I first simply the denominator ${x^2+2x}$ to become $x(x+2)$.
Using... | You missed the first step: long division on polynomials to make the numerator of smaller degree than the denominator.
After long division, you should get $$x^2-2x+4+\frac{-8x-8}{x^2+2x}$$ and then you proceed as in your initial attempt, i.e. $$\frac{-8x-8}{x^2+2x}=\frac{A}{x}+\frac{B}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p$. Let $p$ be an odd prime number. Prove that $$1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p.$$ I know I can use Wilson's Theorem somehow. It would make sense if I could show that all the factors $(p-2k)^2$ is congrunt to $-... | I will give you an answer that does not use the 'trick' $a^2\equiv(-1)\cdot a\cdot(p-a)$, which was given in a comment above. (Ok, this is less beautiful but, honestly, this was the first thing I thought and I think I would not have found the 'trick'.)
I will use the fact that $2\cdot4\cdot6\cdots2n=2^n\cdot n!$, hopin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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$n!+k$ is never any power of any prime number if $n\ge 6$ and $2\le k\le n$? Question : Is the following true?
"If $n!+k$ is a power of a prime number, then it is one of $2!+2, 3!+2, 3!+3, 4!+3, 5!+5$ where $n,k\in\mathbb Z$ satisfy $n\ge 2$ and $2\le k\le n$."
Motivation : The following is well known :
1. A sequence ... | Suppose that
$$n!+k=p^r\ \ (r\ge 2).$$
Since it is obvious that $k$ is a power of $p$, let $k=p^s\ (1\le s\lt r).$ However, if $s\gt 1,$ then $n\ge k=p^s\gt p$ leads that $p^{s+1}$ is a divisor of $n!$. Noting that $s+1\le r$ and that $p^{s+1}$ is a divisor of $p^r$, this leads that $p^{s+1}$ is a divisor of $p^r-n!=p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assum... | You assume $\;9\mid\left[n^3+(n+1)^3+(n+2)^3\right]\;$ , and now you want to prove that also
$$9\mid\left[(n+1)^3+(n+2)^3+(n+3)^3\right]\;\;,\;\;\text{but}$$
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36=$$
$$=\left(3n^3+9n^2+15n+9\right)+9n^2+27n+27$$
and you can see the left part between the parentheses is divisible by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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About the area of integer-edge-length triangles Let $a,b,c$ be three edge lengths of a triangle whose area is $S$.
Then, here is my question.
Question : Supposing that $a,b,c$ are natural numbers, then does there exists $(a,b,c)$ such that $S=6k$ for any $k\in\mathbb N$?
Motivation : I've known the following fact:
... | Theorem: There is no integer-sided triangle whose area is 18 units.
Proof: Assume such a triangle exists. Then there are integers $x, y, z$ such that $18^2 = xyz(x + y + z)$. Assume without loss of generality that $x \ge y \ge z$. Since $y \ge 1$ and $z \ge 1$, we have $x(x + 2) \le 18^2$ which certainly forces $x < 18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/513379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Critique of Solution "Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.
In a quadratic in the form of $ax^2+bx+c$, the product of the roots of the quadratic equal the constant ($c$). Since there can only be integer coefficien... | I think it is correct, but it would be easier to use that the discriminant must be a square number, i.e. $n^2+308$ must be a square number. This has the solutions $n=\pm 4, \pm 76$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Functional equation $f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$ Find all of the functions defined on the set of integers and receiving the integers value, satisfying the condition:
$$f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$$
for each pair of integers $(a,b)$.
| Not cleaned. I just wrote it as I thought about it.
$A^3+B^3+C^3-3ABC=(A+B+C)(A+wB+w^2C)(A+w^2B+wC)$. So in our case
$$\left(f(a+b)-f(a)-f(b)\right)\left(f(a+b)-wf(a)-w^2f(b)\right)\left(f(a+b)-w^2f(a)-wf(b)\right)=0.$$
If for some $a,b$ we have $f(a+b)-wf(a)-w^2f(b)=0$ then $f(a+b)-wf(a)-w^2f(b)$ is divisible by $w^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/517749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem. I want to prove:
$$\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt=\frac{2\pi}{ab}$$
by the deformation theorem of complex variable.
Then I consider a parameterization $\gamma:[0,2\pi]\rightarrow A$, traveled... | Let the ellipse $\Gamma$
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
be parameterized by
$$ z = a\cos\theta + ib\sin\theta$$
with $\theta\in[0,2\pi].$
Consider the integral
$$ \int_\Gamma \frac{1}{z} dz.$$
This is equal to
$$ \int_0^{2\pi} \frac{1}{a\cos\theta + ib\sin\theta}
(-a\sin\theta + ib\cos\theta) \; d\theta\\ =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
$z$-transform of $1/n$ How can one calculate the $z$-transform of:
$x(n) = \frac{1}{n}$ , where $n \geq 1$? I have searched for table entries, then got stuck while trying to do it with the definition of $z$-transform (summation).
| Good Question
$$\frac{u(n-1)}{n} \rightleftharpoons ??$$
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
finding range of function of three variables Three real numbers $x$, $y$, $z$ satisfy the following conditions.
$x^{2}+y^{2}+z^{2}=1~$, $~y+z=1$
Find the range of $~x^{3}+y^{3}+z^{3}~$ without calculus.
I solved this problem only with Lagrange-Multiplier and wonder if there exist other methods.
| to find max value, we only consider $f(u)=1-3u^2+2\sqrt{2}u^3, u=\sqrt{y(1-y)}, 0 \le u \le \dfrac{1}{2}$.we try first to see if it's increasing or decreasing function. $f(0)=1, f(\dfrac{1}{2})=1-\dfrac{3-\sqrt{2}}{4} < 1$, so we guess it is a decreasing function. to prove this, let $ u_1>u_2$
$f(u_1)-f(u_2)=2\sqrt{2}u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/519218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$
Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2). Not sure when to apply the induction hypothesis and how to ge... | You can also prove this by power series manipulation (generating functions). Note first that changing the sum by starting at $i=0$ instead of $i=1$ doesn't change its value. Compute as follows:
\begin{align}\sum_{n=0}^\infty \sum_{i=1}^n \frac{i}{(i+1)!}x^{n+1} &= \sum_{n=0}^\infty \sum_{i=0}^n \frac{i}{(i+1)!}x^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$ Intuitively it's true, but I just can't think of how to say it "properly".
Take for example, my answer to the following question:
Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin pri... | A somewhat silly answer:
You probably know that $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$; if not these can be proven easily via induction. Consider then
\begin{align*}
\sum_{i=1}^n (i^2 + 3i + 1) &= \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 14,
"answer_id": 1
} |
Residue at infinity of $f(z)=z^3\cos\big(\frac{1}{z-2}\big)$ I have trouble with the residue of:
$f(z)=z^3\cos\left(\frac{1}{z-2}\right)$ at $z = \infty$.
I tried to solve it at $z=0$ but it turns out that I was wrong while $z=0$ is not a pole.
I must solve it at $z=2$ but I'm stuck.
Any suggestion will be much appreci... | The residue at infinity of a function $f$ holomorphic in a punctured neighbourhood of $\infty$ is the residue in $0$ of the function
$$g(z) = -\frac{1}{z^2}f\left(\frac1z\right).$$
For $f(z) = z^3\cos \frac{1}{z-2}$, that becomes the residue in $0$ of
$$-\frac{1}{z^5}\cos \frac{z}{1-2z}.$$
Expanding $\frac{z}{1-2z}$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $\left|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}\right|<2\sqrt{|x-y|}$? For any real numbers $x,y\neq 0$,show that
$$\left|x\sin{\dfrac{1}{x}}-y\sin{\dfrac{1}{y}}\right|<2\sqrt{|x-y|}$$
I found this problem when I dealt with this problem. But I can't prove it. Maybe the constant $2$ on the righ... | We'll first assume that $0 \le x < y$.
I need three kinds of estimates here.
*
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le x + y$
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2$
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \frac{y}{x} - 1$
Estimates 1 and 2 are tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Prove the following determinant identities without expanding the determinants a)
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos 2x \\
\sin^2 y & \cos^2 y & \cos 2y \\
\sin^2 z & \cos^2 z & \cos 2z \\
\end{vmatrix} = 0;$$
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos^2x-\sin^2x \\
\sin^2 y & \cos^2 y & \cos^2y-\s... | Remember that $\cos 2x = \cos^2 x - \sin^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convert boolean expression into SOP and POS Convert the following expression into SOP (sum of products) and POS (product of sums) canonical forms using boolean algebra method:
$(ac + b)(a + b'c) + ac$
Attempt at solution:
$(ac + b)(a + b'c) + ac$
*
*$(a + b)(c + b)(a + b')(a + c) + ac$
*$...$
*$...$
I'm stuck at ... | One way to get the SoP form starts by multiplying everything out, using the distributive law:
$$\begin{align*}
(ac+b)(a+b'c)+ac&=ac(a+b'c)+b(a+b'c)+ac\\
&=aca+acb'c+ba+bb'c+ac\\
&=ac+ab'c+ab+ac\\
&=ac+ab'c+ab\;.
\end{align*}$$
Then make sure that every term contains each of $a,b$, and $c$ by using the fact that $x+x'=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Factorise $y^2 -3yz -10z^2$ How do I solve this question? I have looked at the problem several times. However, I cannot find a viable solution. I believe that it is a perfect square trinomial problem.
| I made up a method with no guesswork at all, Factoring Quadratics: Asterisk Method
and my earlier answer at How to factor the quadratic polynomial $2x^2-5xy-y^2$?
We require that $b^2 - 4ac$ be a perfect square, and (positive) $\delta = \sqrt {b^2 - 4ac},$ otherwise factoring is impossible anyway. Then: we may simply... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
If $x+y+z=\pi/2$, $\sin{x}\sqrt{1-\sin x}+\sin y\sqrt{1-\sin y}+\sin z\sqrt{1-\sin z} \ge 4\sqrt{2} \sin{x}\sin{y}\sin{z}(\sin{x}+\sin{y}+\sin{z})$
Let: $x,y,z >0$ and $ x+y+z=\dfrac{\pi}{2}$ then prove:
$$\sin{x}\sqrt{1-\sin{x}}+\sin{y}\sqrt{1-\sin{y}}+\sin{z}\sqrt{1-\sin{z}}
\ge 4\sqrt{2} \sin{x}\sin{y}\sin{z}(\s... | Note that $$\sqrt{1-\sin x} = \sqrt{2}\sin\left(\frac{\pi}{4}-\frac{x}{2}\right) = \sqrt{2}\sin\left(\frac{y+z}{2}\right) \geq \frac{1}{2}\sqrt{2}\left( \sin y + \sin z\right).$$ Therefore $$\sin x \sqrt{1-\sin x}+\sin y \sqrt{1-\sin y} + \sin z \sqrt{1-\sin z} \geq \sqrt{2}\left(\sin x \sin y+ \sin x \sin z + \sin y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit in form of 0/0
If I multiply by I get zero/2x anyway.
What manipulation needed to get 2/3?
| $$
\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}=\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}{x(\sqrt[3]{1-x^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=
$$
$$
=\lim_{x\to 0}\frac{1+x-1+x}{x(\sqrt[3]{1-x^2}+\sqrt[3]{(1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/527254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed form for $\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$
| Hint:
$\int_0^\infty\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}e^{-x}~dx$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\sqrt{\sinh^2x+1}}}{\sqrt{\sinh x}\sqrt{\sinh^2x+1}}e^{-\sinh x}~d(\sinh x)$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\cosh x}}{\sqrt{\sinh x}\cosh x}e^{-\sinh x}\cosh x~dx$
$=\int_0^\infty\dfrac{e^{-\sinh x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/527438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}... | This can also be done using a basic complex variables technique.
Suppose we seek to verify that
$$\sum_{k=0}^m {m\choose k} {n+k\choose m}
= \sum_{k=0}^m {m\choose k} {n\choose k} 2^k.$$
Introduce the two integral representations
$${n+k\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n+k} \; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Prove by induction the following inequality for all n∈N $\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}$
I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis.
Please help :... | Let $s_n$ represent the sum up to $n$, and suppose that $s_n \ge \sqrt{n}$ for induction. We want to show that $s_{n + 1} \ge \sqrt{n + 1}$; to this end, note
\begin{align*}
s_{n + 1} - s_n &= \frac{1}{\sqrt{n + 1}}
\end{align*}
Therefore,
\begin{align*}
s_{n + 1} &= \frac{1}{\sqrt{n + 1}} + s_n \\
&\ge \frac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{... | using $maxima$ I get
(%i1) 1/(3*k+2)+1/(3*k+3)+1/(3*k+4)-1/(k+1),ratexpand;
2
(%o1) -------------------------
3 2
27 k + 81 k + 78 k + 24
(%i2) solve(denom(%),[k]);
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework probl... | Let $ \displaystyle f(z) = \frac{\pi}{\sin \pi z} - \frac{1}{z}$.
Then according to the Mittag-Leffler pole expansion theorem, $$ \frac{\pi}{\sin \pi z} - \frac{1}{z} = f(0) + \sum_{n=1}^{\infty} \text{Res}[f,n] \Big( \frac{1}{z-n} + \frac{1}{n} \Big) + \sum_{n=1}^{\infty} \text{Res}[f,-n] \Big( \frac{1}{z+n} - \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/535116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$) It appears that the only two solutions are always $3$ and $p^k-3$, I want to prove this, here has been my approach, I think I am close but just missing something, would really appreciate any he... | Let $p\gt 3$. Note that $x^2\equiv 9\pmod{p^k}$ if and only if $p^k$ divides $(x-3)(x+3)$. But $p$ cannot divide both $x-3$ and $x+3$, else it would divide their difference $6$. So $(x-3)(x+3)$ is divisible by $p^k$ if and only if $p^k$ divides $x-3$ or $p^k$ divides $x+3$. Thus there are two solutions, $x\equiv 3\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find value of $x+y+z+u+v+w$ let $x,y,z,u,v,w$ be positive integer numbers,and such
$$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)=2004(yzvw+yzu+yzw+uvw+y+u+w)$$
Find this value of
$$x+y+z+u+v+w=?$$
My try:
maybe use
$$(x+1)(y+1)(z+1)(u+1)(v+1)(w+1)=(xyz+xy+yz+xz+x+y+z+1)(uvw+uv+uw+vw+u+v+w+1)$$
| This is not a solution but a little progress towards the solution pointed out in Ewan Delanoy's comment to show that $x=u=1$ and $y \ge 36$.
Let each side of original equation be $2004 \times 1949 \times k$. Multiplying the parentheses on right-hand-side with $x$ and subtracting from the parentheses on left-hand-side g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Sketch curve $y = (4x^3-2x^2+5)/(2x^2+x-3)$ I'm trying to sketch the curve
$$
y = (4x^3-2x^2+5)/(2x^2+x-3).
$$
I tried to find the first and second derivative but I don't know how to find the roots of these.
\begin{align}
y' &= \frac{-5-8 x-38 x^2+8 x^3+8 x^4}{(-3+x+2 x^2)^2}
&&\text{Fermat Theorem}\\
0 &= -5-8 x-38 x... | You don't need the second derivative.
*
*Factorise the denominator to find vertical asymptotes.
*Ask yourself what happens when $x \to \pm \infty$ to find the horizontal asymptotes.
*Put $x=0$ to find the $y$-intercept.
*Look for the roots of the numerator to give you the $x$-interecepts.
*Solve $y'=0$ to give y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Problem involving trigonometry and cubics One of my teachers proposed me the following problem:
$$\text{If } (3\sec x+\csc x)\sin x=5\cos^2 x\text{, calculate } z=\tan x+\sec x$$
I started by manipulating
$$3\tan x +1=5\cos^2 x$$
$$\sec^2 x(3\tan x +1)=5$$
$$(1+\tan^2 x)(3\tan x +1)=5$$
$$3\tan^3x +\tan^2 x + 3\tan x -... | Maybe this will be helpful for you. It seems the following.
Put $y=\pi/2-x$. Then $\cos x=\sin y$, $\sin x=\cos y$ and $\frac 1z=\frac {\cos x}{1+\sin x}=\frac {\sin y}{1+\cos y}=\tan\frac y2$. Then $\sin y=\frac {2z}{z^2+1}$,
$\cos y=\frac {z^2-1}{z^2+1}$ and
$$\frac{3(z^2-1)}{z^2+1}+\frac{2z}{z^2+1}=5\left(\frac {2z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$ $(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geq... | One of the best ways in general to handle questions about quadratics is via the process of completing the square: that is, turning a quadratic of the form $P(x) = x^2+bx+c$ into one of the form $Q(x) = (x+r)^2+s$. To do this, note that the latter can be written as $x^2+2rx+r^2+s$, and we can equate terms; this means t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| The three solutions of the equation $a^3+b^3+3ab=1$ are
$$b_1=1-a,$$
$$b_2,b_3 = \frac{1}{2}\left(a-1\pm i\sqrt{3}(1+a)\right).$$
If $a\neq -1$, since $a,b\in\mathbb{R}$, we must have $a+b=a+b_1=1$. If
$a=-1$, then the imaginary part of $b_2,b_3$ vanishes and we find two solutions for $(a,b)$: $(-1,-1)$ and $(-1,2)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
} |
how to solve $x^2 \equiv -1 \pmod{13}$ Knowing that $p$ is prime and if $p \equiv 1 \pmod 4$, then $\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -1 \pmod p$; how do I solve $x^2 \equiv -1 \pmod{13}$?
| Observe that $$p-r\equiv-r\pmod p$$
putting $r=1,2,\cdots,\frac{p-1}2$ and multiplying we get
$$\prod_{\frac{p+1}2\le r\le p-1 }r\equiv(-1)^{\frac{p-1}2} \prod_{1\le s\le \frac{p-1}2 }s$$
$$\prod_{\frac{p+1}2\le r\le p-1 }r\equiv\prod_{1\le s\le \frac{p-1}2 }s\text{ as } \frac{p-1}2 \text{ is even}$$
Multiplying by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)$, Any quick methods? How to prove the following equation by a quick method?
\begin{eqnarray}
\\\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)\\
\end{eqnarray}
If I use so much time to expand it and take extra care of the calculation process, I can find the ans... | To flesh out Thomas's comment into another answer: start with the factorization $x^3-y^3 = (x-y)(x^2+xy+y^2)$; this can be derived easily from the finite geometric series. Replacing $y$ with $-y$ yields $x^3+y^3 = (x+y)(x^2-xy+y^2)$, and then replacing $x$ with $x^2$ and $y$ with $y^2$ yields $x^6+y^6 = (x^2+y^2)(x^4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How can I Prove $\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$ for $x,y>0$ prove that
$\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$
I have tried to develop $(x+y)^2=$ and to get to an expression that must be bigger than those above
Thanks!
| Start by noticing that since $(x-y)^2 \geq 0$ then $x^2 - 2xy +y^2 \geq 0$ which implies that $x^2+y^2 \geq 2xy$. Now we attack these inequalities one at a time.
To show the left inequality, we can show $2xy \leq (x+y)\sqrt{xy}$, and since everything is positive we can square both sides and further reduce the problem ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | Using the partial fractions identity, it's possible to split the summation into two
$$ \frac{2}{\sin^2 \theta} = \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta}$$
Now our sum over the nth roots
$$ (\star) =\sum_{k=1}^{m-1} \frac{1}{\sin^2 \frac{\pi k }{n}}
= \frac{1}{2}\sum_{k=1}^{m-1} \frac{1}{1-\cos \frac{\pi k }{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 6
} |
Inequality $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c$ when $abc = 1$
If $a,b,c > 0$ are such that $abc=1$, then
$$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$
I would be pleased if you give me a hint. Thanks in advance.
| $$a+b+c =$$
$$\{ \mbox{multiply by 1: maybe } abc\mbox{, maybe }\sqrt{abc}, ... \}$$
$$= \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}=$$
$$
=\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}}
+\sqrt[3]{\frac{b}{c} \cdot \frac{b}{c} \cdot \frac{c}{a}}
+\sqrt[3]{\frac{c}{a} \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/546685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours an... | By applying CBS is obtained
$$\frac{a^2}{na^2+kab} + \frac{b^2}{nb^2+kbc} + \frac{c^2}{nc^2+kca} \ge \frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}.$$
But
$$\frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}\ge \frac{3}{n+k}$$
is equivalent to
$$(k-2n)(a^2+b^2+c^2-ab-bc-ca)\ge 0$$
which is true under the additional condition ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Generalizing the sum of consecutive cubes $\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$ to other odd powers We have,
$$\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$$
$$2\sum_{k=1}^n k^5 = -\Big(\sum_{k=1}^n k\Big)^2+3\Big(\sum_{k=1}^n k^2\Big)^2$$
$$2\sum_{k=1}^n k^7 = \Big(\sum_{k=1}^n k\Big)^2-3\Big(\sum_{k=1}^n k... | This is not an answer but does not fit in a comment-box.
Another way to describe the relation between the sums and the sums of squares of various sums-of-like-powers makes use of the Eulerian-numbers, which converts the expressions for the sums-of-like-powers into polynomials. The sums of like powers can be expresse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 4,
"answer_id": 1
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Find the Laurent series for $f(z) = (z^2 - 4)/(z-1)^2 $ for $z=1$ What I understand is that we have to expand $f(z$) in the positive and negative powers of $(z-1)$.
Hence I tried factorizing the numerator $(z^2-4)=(z+2)(z-2)$ , which can then be written in terms of $(z-1)$ as:
$(z-1-1)(z-1+3)/(z-1)^2$ . however i canno... | Use
$$z^2-4 = (z-1+1)^2 -4 = (z-1)^2+ 2 (z-1)-3$$
so that
$$f(z) = -\frac{3}{(z-1)^2} + \frac{2}{z-1}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/550014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$. Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$.
I did
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9+h}-3}{h} \frac{\sqrt(9+h)+3}{\sqrt(9+h)+3} = \frac{9+h-3}{h\sqrt{9+h}+3}= \frac{6}{\sqrt{9+h}+3}= \... | To follow up on Oria Gruber's answer:
If
$y = \sqrt x, \tag{1}$
then
$x = y^2, \tag{2}$
so
$dx/dy = 2y = 2\sqrt x, \tag{3}$
whence
$dy/dx = (dx/dy)^{-1} = \frac{1}{2 \sqrt x}; \tag{4}$
thus when $x = 9$ we obtain
$dy/dx = 1/6, \tag{5}$
so the equation for the desired tangent line is
$y - 3 = \frac{1}{6}(x - 9). \tag{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/553494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Laplace differential equation Can somebody help me work out $2y''+y'-y=\mathrm{e}^{3t}$, y(0)=2 and y'(0)=0 with the method of Laplace? I got
\begin{align*}
Y(s)&=\frac{1}{(s-3)(2s^2+s-1)}+ \frac{2+4s}{(s-3)(2s^2+s-1)}\\
&=-\frac{4}{15}\frac{1}{2s-1}+\frac{1}{12}\frac{1}{s+1}+\frac{1}{20}\frac{1}{s-3}-\frac{1}{6}\frac{... | I think the answer is $$ L(y)=\frac{-\exp(t)}{6}+\frac{\exp(3t)}{4}+\frac{2\exp(-.5t)}{21}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/554187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfra... | $z=\dfrac{1}{xy}$, put in LHS and and clean the denominators, we have:
edit:
LHS-RHS=$ 9y^2x^4-8y^3x^3+2y^2x^3+9y^4x^2+2y^3x^2-9y^2x^2-8yx^2-8y^2x+2yx+9 \ge0 \iff $
$4y^2x^4-8yx^2+4\ge 0,\\4y^4x^2-8y^2x+4\ge0,\\5y^2x^4+5y^4x^2\ge 10x^3y^3,\\2y^2x^3+2y^3x^2\ge 4(xy)^{\frac{5}{2}} \iff\\ LHS \ge 2x^3y^3+4(xy)^{\frac{5}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Is $\int\frac{\cos^5x\sin^3x}{1+\cos2x}dx = \frac{\sin^4x}{8}-\frac{\sin^6x}{12} +C$ or $\frac{\cos^6x}{12}-\frac{\cos^4x}{8} +C$? $\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx = \dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$ or $\dfrac{\cos ^6x}{12}-\dfrac{\sin ^4x}{8} +C$?
$\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx$ = $\d... | You can differentiate and simplify your two answers to see that they are both correct. If you take the first answer, expand it using the identity $\sin^2 x = 1-\cos^2 x$, and simplify, you will get the second answer plus an additional constant. Thus the two expressions (ignoring the $C$'s) differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the first three terms of the Maclaurin Series
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.
I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kin... | $\sec(x)=\frac{1}{\cos x}$. The three first terms are $1,x^2$ and $x^4$. Then we write: $$\cos x=1-\frac{x^2}2 + \frac{x^4}{24} + o(x^4)$$
Putting $$u=-\frac{x^2}2 + \frac{x^4}{24}=-\frac{x^2}{2}\left(1-\frac{x^2}{12}\right)$$ we have:
$$\sec(x)=(1+u)^{-1} =\operatorname{Tronc}_4 (1 -u +u^2-u^3+u^4)=\operatorname{Tro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556423",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Check my solution: Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$ Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$
First I rationalized the numerator,
$$
\begin{align*}
&\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x) \cdot \frac{\sqrt{(x+a_1)(x+a_2)}+x}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{x(a_1+a_2... | The mistake happens when you divide numerator and denominator by $x$. Inside the square root this means division by $x^{2}$ and so you should divide one factor $(x + a_{1})$ by $x$ and another factor $(x + a_{2})$ by another $x$ so that after the division the denominator will be $$\sqrt{\left(1 + \frac{a_{1}}{x}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
last three digits of $7^{100}-3^{100}$ How can I find the las three digits of $7^{100}-3^{100}$ ? I know one way is to use $7^{100}=(10-3)^{100}=\sum_{k=0}^n{100 \choose k}10^{100-k}(-3)^k$ but I'm totally stuck...
Thanks
| Well, note that for $0\le k\le 97,$ we have that $1000$ is readily a factor of $\binom{100}{k}10^{100-k}(-3)^k,$ so the last three digits of all of those numbers will be $0$s. Hence, the last three digits of $7^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}+\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$ for real numbers $a,b,c$ $a,b,c$ reel sayılar için ;
$$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$
Olduğunu gösteriniz.
Translation:1 For real numbers $a,b,c$, show that:
$$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$
1With some help... | let $$u=a+b+c,v=ab+bc+ac,w=abc$$
it suffices to show that
$$2(u^2+v^2+w^2-2wu-2v+1)\ge (u+v-w-1)^2$$
this inequality is equivalent to
$$u^2+v^2+w^2-2uv+2vw-2uw+2u-2v-2w+1\ge 0$$
or
$$(u-v-w+1)^2\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/565300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
find all the values of a and b so that the system has a) no solution b) 1 solution c) exactly 3 solutions and 4) infinitely many solutions $$\begin{cases}
&x &- &y &+ &2z &= 4 \\
&3x &- &2y &+ &9z &= 14 \\
&2x &- &4y &+ &az &= b
\end{cases}
$$
I know that $a$ and $b$ has to either equal to something or not in order to... | Apply Gaussian Elimination to the reduced matrix.
$$
\pmatrix{1 & -1 & 2 & 4\\3 & -2 & 9 & 14\\2 & -4 & a & b} \to
\pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & -2 & a-4 & b-8} \to
\pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & 0 & a+2 & b+4}
$$
So
*
*if you want a unique solution, $\frac{b+4}{a+2}$ must be defin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/567213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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How show $8|[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]+1$ let $n$ be postive integer numbers,and such $n>2$,show that
$$
8\,\,\left.\right\vert\,\,\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor + 1
$$
where $\left\lfloor x\right\rfloor$ is is the largest integer not greater... | Claim For $n >2$ we have
$$\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor =8n+7$$
This is equivalent to
$$8n+7 \leq
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}< 8n+8 \Leftrightarrow \\
8n+7 \leq 2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/567745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\lim\limits_{x\to 0} (\sqrt{4+x}-1)^{1/(e^x-1)}$ I have to find the limit of $(\sqrt{4+x}-1)^{1/(e^x-1)}$ as $x\to0$ without de l'Hopital's rule and only with notable limits
| $\ln \left( \sqrt{4+x}-1 \right)^{\frac{1}{e^x-1}}= \frac{x}{e^x-1} \frac{\ln \left(1+(\sqrt{4+x}-2)) \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{\frac{x}{2+\sqrt{4+x}}} \frac{1}{2+\sqrt{4+x}} \rightarrow 1 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substi... | We will assume that $a>0$ for the following.
We have
$$\int x^2\sqrt{a^2-x^2} \,dx.$$
Let
$$
\begin{align*}
x&=a\sin{t} \\
dx &= a\cos{t} \, dt \\
a^2-x^2&=a^2-a^2\sin{t}\\
&=a^2\left( 1-\sin^2{t} \right)\\
&=a^2\cos^2{t}.
\end{align*}
$$
We substitute and integrate,
$$
\begin{align*}
&\int a^2\sin^2t \cdot \sqrt{a^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Interesting determinant: Let $A$ be an $n$ by $n$ matrix with entries $a_{i,j}$ given that $a_{i,j}=2$ if $i=j$ Let $A$ be an $n$ by $n$ matrix with entries $a_{i,j}$ given that $a_{i,j}=2$ if $i=j$, $a_{i,j}=1$ if $i-j\equiv\pm2\pmod n$, and $a_{i,j}=0$ otherwise.
Find $\det A$.
It seems that the determinant is 4 for... | Note that we have the degenerate cases $n=1, 2, 4$, where there are less than $3$ non-zero elements in each row/column.
When $n=1, 2, 4$ respectively, we get $\det(A)=2, 4, 9$.
Let us now consider $n \not =1, 2, 4$, so that each row/column has one $2$ and two $1$s and no other non-zero element.
Observe that $A$ is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Showing that $\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$ Wolfram Alpha tells me that $$\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$$
What are some quick/elegant ways of proving this?
| $$1-\cos(16x) = 2\sin^2(8x) = 8 \sin^2(4x) \cos^2(4x) = 32 \sin^2(2x) \cos^2(2x) \cos^2(4x)$$
Hence,
$$I = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) \cos^2(4x) dx = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) (2\cos^2(2x)-1)^2 dx$$
Now let $\cos(2x) = t$ and compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
| Using twice the result that $\int_0^{\infty} \frac{\ln x}{1+x^2} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^2} d x&=\int_0^{\infty} \frac{\left(1+x^2-x^2\right) \ln x}{\left(1+x^2\right)^2} d x \\
&=\int_0^{\infty} \frac{\ln x}{1+x^2} d x-\int_0^{\infty} \frac{x^2 \ln x}{\left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 12,
"answer_id": 11
} |
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(... | $$0 = \left( \dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2} \right)\left( \dfrac{a}{bc-a^2}+\dfrac{b}{ca-b^2}+\dfrac{c}{ab-c^2} \right) =$$ $$ \dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2} + \frac{a(ca-b^2) +a(ab-c^2) + b(bc-a^2) +b(ab-c^2)+c(bc-a^2)+c(ca-b^2)}{(bc-a^2)(ca-b^2)(ab-c^2)}$$ $$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Solve $3^x + 28^x=8^x+27^x$ The equation $3^x + 28^x=8^x+27^x$ has only the solutions $x=2$ and $x=0$? If yes, how to prove that these are the only ones?
| Since for all sufficiently large $x$ the expression $28^x + 3^x - 8^x - 27^x$ is positive and increasing, there cannot be roots $x$ beyond that point.
The question is how to pin down where the expression becomes positive and increasing.
Now $28^x$ grows fastest, by a factor of $28$ when $x$ is incremented by 1. This s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
$(a^2+b^2)\mid (ax+by)$, implies $\gcd(x^2+y^2,a^2+b^2)>1$ Suppose $a,b,x,y$ are natural number such that $(a^2+b^2)\mid (ax+by)$, prove that $\gcd(x^2+y^2,a^2+b^2)>1$. Can anyone give me a hint to solve the problem? I am rather stuck, thanks!
| First notice that $$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(bx-ay)^2$$
If $a=da_1$ and $b=db_1$ where $(a_1, b_1) = 1$ the equation can be rewritten as $$(a_1^2+b_1^2)(x^2+y^2)=(a_1x+b_1y)^2+(b_1x-a_1y)^2$$
Suppose now that $p$ is a prime such that $p \mid a_1^2+b_1^2$. Since we have $$d^2(a_1^2+b_1^2)=(a^2+b^2)\mid (ax+by)=d(a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Ring theory: Ideals being equal Question: Prove directly, without gcd computations, the following equalities of ideals.
(i) $(5, 7) = (1)$ in $\Bbb{Z}$ (of course (1) = $\Bbb Z$).
(ii) $(15, 9) = (3)$ in $Z$.
(iii) $(X^3 −1,X^2 −1) = (X −1) \text{ in } \Bbb{Q} [X]$
Attempted solution:
Try to show that $(1)$ is a sub... | For (i) you have that $1 = 5x + 7y$ for some $x,y \in \Bbb{Z}$. This can be done using the Euclidean algorithm. One solution is $x = 3$, $y=-2$. Thus any integer $z$ can be written as $z = z\cdot1 = z\cdot(5\cdot3 - 14\cdot2)$, and we have $\Bbb{Z} = (1) = (5,7)$.
For (ii), we use basically the same process, but show ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How solve equation:$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$ Let $x,y\in \mathbb{R}$, solve this follow equation:$$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$$
My try: Since
$$8x^2+4x(y^2+y-10)+y^4-11y^2-8y+52=0$$
then I can't. Maybe this problem have other nice methods.Thank you
| $$8x^2+x\left( 4y^2+4y-40\right) +y^4-11y^2-8y+52=0$$
This is quadratic in $x$, thus
$$x_{1,2}=\frac{-4y^2-4y+40\pm \sqrt{\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)}}{16}$$
Now, use
$$\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)=-16(y-2)^2(y+1)^2$$
to show that the equation has solutions i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why can't I simply use algebra to solve this inequality? Consider the inequality:
$\frac{(x+3)(x-5)}{x(x+2)}\geq 0$
Why can't I simply multiply both sides by $x(x+2)$ and get $(x+3)(x-5)\geq 0$ ?
Which would yield: $x^2-2x-15\geq 0$ and I could then use the quadratic formula to derive the answer..?
This seems algebraic... | if we assume that $x \gt 0$ then $(x+2) \gt 0$ so
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \ge 0$$
but if $x \cdot (x+2) < 0$
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \le 0$$
and if $x = 0$ or $x = -2$ then $\frac{(x+3)(x-5)}{x(x+2)}$ is undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/584433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+1$.
Let the coefficients of
$$a(x) = x^3+x^2+1,$$
$$b(x) = x^2+x+1,$$
$$c(x) = x^3+x^2+x+1$$
be in $\mathrm{GF}(2)$.
Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+... | I assume that Pablo's interpretation of your question is correct, and answer accordingly.
I just added a self-answered question giving a useful discrete logarithm table. I rely heavily on that.
Below $\gamma$ will denote the coset $x+I$ in the quotient ring $GF(2)[x]/I$, where the ideal $I=\langle x^4+x+1\rangle$. That... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.