Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?
| Divide $$(7x-4)-(7x-5)=(4x-1)-(4x-2)\ \ \ \ \ (1)$$ by the original equation, which is $$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}\ \ \ \ \ (2)$$ You should get:$$\sqrt{7x-4}+\sqrt{7x-5}=\sqrt{4x-1}+\sqrt{4x-2}\ \ \ \ \ (3)$$Now add (2) and (3), you get:
\begin{align}
2\sqrt{7x-4}&=2\sqrt{4x-1}
\\ \\
7x-4&=4x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
| Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:
$$\begin{align}
I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\
= & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\
\implies 4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(... | Notice that
$$(n-2)^3=((n-1)-1)^3=(n-1)^3-3(n-1)^2+3(n-1)-1$$
Anyway, related to the image you posted (which I didn't notice at first sight), the result was really obtained by substitution from the first row ($n\to n-1$), as other people also observed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.
Any hints? Thanks!
| My proposal was that
$$\frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ = \ \frac{40 \sqrt{3}}{45} \ = \ \frac{8}{3 \sqrt{3}} \ = \ (\frac{4}{3})^{3/2} \ . $$
So both logarithms equal 3/2 and the rest follows as Zev Chonoles shows.
EDIT: Just to verify that the other piece checks,
$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need help simplifying complicated rational expression. Studying for my final and I can't figure this out.
Simplify:
$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$
| The shortest route is to note that the least common multiple of all four denominators of the ‘small’ fractions is $x^3y^4$ and to multiply the expression by $1=\frac{x^3y^4}{x^3y^4}$:
$$\large{\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}\cdot\frac{x^3y^4}{x^3y^4}}\;.$$
I’ll let you check that th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{... | we have
$$ \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
replace $x$ with $-x$, divide both sides by $x$ then integrate w.r.t $x$ from $0$ to $1$ , we get:
\begin{align*}
S_1&=\sum_{n=1}^{\infty}(-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ dx=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 2
} |
My son's Sum of Some is beautiful! But what is the proof or explanation? My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same... | The result simply holds because
$$\begin{align}
(100*2^n)+(10*2^{n+1})+2^{n+3} &=(2^6+2^5+2^2)*2^n+(2^3+2)*2^{n+1}+2^{n+3} \\
& = 2^{n+6} +2^{n+5}+2^{n+2} +2^{n+4} + 2^{n+2}+2^{n+3} \\
& =2^{n+6}+2^{n+5}+2^{n+4} + 2^{n+3}+(2^{n+2}+2^{n+2})\\
& =2^{n+6}+2^{n+5}+2^{n+4} +(2^{n+3}+2^{n+3})\\
& =2^{n+6}+2^{n+5}+(2^{n+4}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "443",
"answer_count": 10,
"answer_id": 0
} |
Simple number theory problem I found this question in a textbook on number theory:
For which integer c will $\;\displaystyle{\frac{c^6 - 3}{c^2 + 2}}\;$ also be an integer?
I wonder if there is a solution which is not based on trial and error.
| If $(c^6 - 3)/(c^2 + 2)$ is an integer, then so is $$\frac{c^6 - 3}{c^2 + 2} - (c^4 - 2c^2 + 4) = \frac{c^6 - 3}{c^2 + 2} - \frac{c^6 + 8}{c^2 + 2} = \frac{-11}{c^2 + 2},$$ that is, $c^2 + 2$ divides $11$. The only way this can happen is if $c^2 = 9$, so $c = \pm 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rul... | Two things wrong with what you propose.
*
*Those last two terms could be combined into one, $${B\over x-2}+{C\over x-2}={D\over x-2}$$
*It just doesn't work ---- what would you get for $B$ and $C$ if you tried to do $${1\over(x-2)^2}={B\over x-2}+{C\over x-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For what values of $a$ $\int^2_0 \min(x,a)dx=1$ I want to check for what values of $a$ the integral of this function is equal to $1$
$$\int^2_0 \min(x,a)dx=1$$
What I did is to check 2 cases and I am not sure is enough :
case 1:
$$a<x \rightarrow \int^2_0 a = ax|^{2}_{0}=2a=1 \rightarrow a=\frac{1}{2}$$
now I dont kno... | Note that $0 \le x \le 2$. Thus, you have considered two cases: when $a<0$ and when $a>2$. It remains to check when $0\le a\le2$. Hence, we split the given integral into the following two integrals:
$$ \begin{align*}
1 &= \int_0^a \min(x,a)dx + \int_a^2 \min(x,a)dx \\
1 &= \int_0^a xdx + \int_a^2 adx \\
1 &= \left[\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proving an equality on $\arctan x$ To demonstrate the following equality
\begin{equation}
\arctan x=\pi/2- \arctan(1/x)
\end{equation}
I have proceeded in this way. I know that
\begin{equation}
\arctan x= \int \frac{1}{1+x^2} dx + C
\end{equation}
But:
\begin{equation}
\int \frac{1}{1+x^2} dx=\int \frac{1}{x^2(1+1/x^2)... | Let $f(x) = \arctan x$. Then $f'(x) = \frac{1}{1+x^2}$. Let $\phi(x) = f(\frac{1}{x})$, then $\phi'(x) = f'(\frac{1}{x}) (- \frac{1}{x^2}) = - \frac{1}{1+x^2}$.
Hence $f'(x)+\phi'(x) = 0$, and so $x \mapsto f(x)+\phi(x)$ is constant on $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $f(1) = \phi(1) = \frac{\pi}{4}$, we have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Vector analysis. Del and dot products I am trying to prove that
$$\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{A} \cdot \nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A} + \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$$
I've gotten as far as $\nabla(\mathbf{A} \cdot... | Following your effect of using subscript notation: $\newcommand{\b}{\boldsymbol}$
$$
\nabla (\b{A}\cdot \b{B}) = \nabla_{\b{A}}(\b{A}\cdot \b{B}) + \nabla_{\b{B}}(\b{A}\cdot \b{B}),\tag{1}
$$
where $\nabla_{\b{A}}(\b{A}\cdot \b{B})$ is that $\b{A}$ is differentiated while $\b{B}$ is held constant. We can prove:
$$
\nab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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help understanding step in derivation of correlation coefficient I'm looking to understand the starred step in the derivation below (also, if someone could help with the LaTex alignment, I'd appreciate it).
The regression line is $y= b_0 + b_1 x$, where $b_0$ and $b_1$ can be found by:
1) taking the difference between ... | For the numerator, observe that:
$$
\begin{align}
\frac{1}{n} \left( \sum_{i=1}^n x_iy_i \right) - \bar x \bar y
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i \right) - \dfrac{n}{n}\bar x \bar y & \text{common denominator}\\
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i - n\bar x \bar y \right) & \text{factor out }1/n\\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a b c=1$, $a^2+b^2+c^2\ge 3$. (for $a,b,c\in\mathbb{R}$) I can do this problem using calculus minimization techniques, using Lagrange multipliers to find the equations $a^2=b^2=c^2$, so with $abc=1$, $a$, $b$, and $c$, are either $-1$ or $1$ (making sure you don't end up with $abc$ negative). So if the m... | By AM/GM, we have
$$\frac{a^2+b^2+c^2}{3}\ge (a^2b^2c^2)^{1/3}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given that $x = 4\sin \left( {2y + 6} \right)$ find dy/dx in terms of x My attempt:
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \ov... | You are on the right track.
Notice that
$$\cos^2(2y+6) = 1-\frac{x^2}{16}$$
implies
$$\cos(2y+6) = \pm\sqrt{1-\frac{x^2}{16}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$
Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\... | I'd just use the identity $\dfrac 1{1-\sin x}=\dfrac{1+\sin x}{1-\sin^2x}=\dfrac{1+\sin x}{\cos^2x}=\sec^2x+\sec x\tan x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Indefinite Integration : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$ Problem :
Solve : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$......(i)
I tried :
Let $x =\sec\theta$ therefore , (i) will become after some simplification
$$\int \frac{\sin\theta}{(1+\sin\theta)^2}d\theta$$
but i think its wrong method of approaching please... | HINT: $$\frac1{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}$$
So, $$\frac1{(x+\sqrt{x^2-1})^2}=(x-\sqrt{x^2-1})^2=x^2+x^2-1-2x\sqrt{x^2-1}=2x^2-1-2x\sqrt{x^2-1}$$
Put $x^2-1=y$ for the last part
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $\log _5 7 < \sqrt 2.$
Prove that $\log _5 7 < \sqrt 2.$
Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.
| Want to prove that
*
*$\log_5{7} = \frac{\lg{7}}{\lg{5}} < \sqrt{2}$
Equivalently we can show that
*
*$\lg{7} < \lg{5}\times\sqrt{2}$
*$7 < 5^{\sqrt{2}}$
where $\lg$ is the base 2 logarithm. Notice that
*
*$5\times5^{\frac{2}{5}}= 5^{1.4} <5^{\sqrt{2}}$
So can we show that $\frac{7}{5} < 5^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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Find the area of the portion of the surface of the sphere $x^2+y^2+z^2=4x$ that is cut off by a nappe of the cone $y^2+z^2=x^2$. I'm having trouble with a Calculus question where I am supposed to use the following formula to calculate the area of a surface:
Suppose that $f$ and its first-order partial derivatives are ... | I will solve this problem again using the suggestion by Ted Shifrin in the comments to this question above. Doing the integral in polar coordinates, the solution becomes much simpler.
I will take the polar plane to be the $yz$ plane.
In $yz$ coordinates, the desired surface is the function $x=f(y,z)=\sqrt{4-y^2-z^2}+2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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problem with partial fraction decomposition I want to do partial fraction decomposition on the following rational function:
$$\frac{1}{x^2(1+x^2)^3}$$
So I proceed as follows:
$$\begin{align}
\frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 +... | Not answering your question, but I see $x^2$ everywhere, so it is better to find a partial fractions expression for $\dfrac{1}{u(1+u)^3}$.
So we are looking for $A$, $B$, $C$, $D$ such that
$$\frac{1}{u(1+u)^3}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{(1+u)^2}+\frac{D}{(1+u)^3}.$$
When we bring the right-hand side to a commo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here? I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was... | $\sin^6x + \cos^6x = (\sin^2x)^3 + (\cos^2x)^3 =(\sin^2x + \cos^2x)(\sin^4x + \cos^4x -\sin^2x\cos^2x)$
$\sin^4x+\cos^4x -\sin^2x\cos^2x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x -\sin^2x\cos^2x$
or $1-3\sin^2x\cos^2x = 1-3\left(\dfrac13\right)^2 = \dfrac23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Fractional overlap of 1/2 and 1/3 Given a subset of the natural number sequence (positive integers starting from 1) we could say that $\frac12$ of the numbers in the set are divisible by 2.
e.g if the set were ${[1,2,3,4,5,6,7]}$ we could say that $3\frac12$ of the numbers in it are divisible by 2.
If we now wanted to ... | Of the numbers $1$ to $n$, $\lfloor n/m \rfloor$ (i.e. the greatest integer $\le n/m$) are divisible by $m$. Now $x$ is divisible by both $m_1$ and $m_2$ if and only if $x$ is divisible by $\text{lcm}(m_1, m_2)$ (the least common multiple of $m_1$ and $m_2$: this is $m_1 m_2$ if $m_1$ and $m_2$ have no common factor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Calculate the value of x, y and z coordinates System of the equations:
Equation 1:
$$-360 = -6x + x^2 - 40y + y^2 - 20z + z^2$$
Equation 2:
$$-600 = -40x + x^2 - 30y + y^2 - 100z + z^2$$
Equation 3:
$$59.85 = -10x + x^2 - 4y + y^2 - 10z + z^2$$
| It looks like you are trying
to find the intersection of three spheres
(with equation $r^2 = (x-a)^2+(y-b)^2+(z-c)^2$
or $x^2-2ax+y^2-2by+z^2-2cz = r^2-a^2-b^2-c^2$).
Anyway, the equations are
$\begin{align}
a &=-360 = -6x + x^2 - 40y + y^2 - 20z + z^2\\
b&= -600 = -40x + x^2 - 30y + y^2 - 100z + z^2\\
c&= 59.85 = -10x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$.
Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$?
Has $f$ a zero in $\mathbb F_p$ ?
| We can use three different factorizations:
If $2$ is a square mod $p$. (which occurs for $p=\pm1$ mod $8$)
$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$ Assume $q^2=2$ mod $p$. Then $x^4+1=(x^2+1+qx)(x^2+1-qx)$.
If $-2$ is a square mod $8$. (which occurs for $p=1$ or $3$ mod $8$)
$x^4+1=x^4-2x^2+1+2x^2=(x^2-1)^2-(-2)x^2=(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$
For any odd positive integer $k\geq1$, the sum $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$.
I used induction principle for the solution but cannot prove it.
I took $P(k) = 1^k+2^k+\dots+n^k$.
For $P(1)$ it is true.
For $P(n)$ let it be t... | As $k$ is odd,
$r^k+(n+1-r)^k$ is divisible by $r+(n+1-r)=n+1$
and $s^k+(n-s)^k$ is divisible by $s+(n-s)=n$
Putting $r=1,2,\cdots,n-1,n$ and adding them we get
$$(n+1)|\sum_{1\le r\le n}\{r^k+(n+1-r)^k\} \implies (n+1)|2\sum_{1\le r\le n}r^k$$
Putting $s=1,2,\cdots,n-1,n$ and adding them we get
$$n|\sum_{1\le s\le n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 1
} |
Find number of solutions of $|2x^2-5x+3|+x-1=0$
Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$
Solution:
Case 1: When $2x^2-5x+3 \geq 0$
Then we get, $2x^2-5x+3+x-1=0$
x=1,1
Case 2: When $2x^2-5x+3 < 0$
Then we get, $-2x^2+5x-3+x-1=0$
x=1,2
In both cases, common value of x is 1
Hence solution is x=1
Am I do... | Nope, if you have $|f(x)| + g(x) = 0$ $(1)$ you solve it as follows:
*
*assume that $f(x)\geq 0$, find solutions of $f+g = 0$ - say $x_1$. Check whether $f(x_1)\geq 0$. If it holds, then $x_1$ solves $(1)$ as well. If it does not hold, it's not a solution of the original equation.
*assume $f(x)<0$, find solutions o... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove a Trigonometric Series Question:
$$\cot^{2}\frac{\pi }{2m+1}+\cot^{2}\frac{2\pi }{2m+1}+\cdots+\cot^{2}\frac{m\pi }{2m+1}=\frac{m(2m-1)}{3}$$
$m$ is a positive integer.
Attempt:
I started by showing that
$$\sin(2m+1)\theta =\binom{2m+1}{1}\cos^{2m}\theta \sin\theta -\binom{2m+1}{3}\cos^{2m-2}\theta \sin^{3}\theta... | Divide this equality by $\sin^{2m+1}\theta$ to get
$$
\frac{\sin(2m+1)\theta}{\sin^{2m+1}\theta}=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1}\cot^{2k} \theta\tag{1}
$$
Denote
$$
P_m(x)=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1} x^m
$$
This is polynomial of degree $m$. From $(1)$ we see that $P_m$ have $m$ roots $x_... | {
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"url": "https://math.stackexchange.com/questions/429877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (exa... | \begin{align*}
x^4 - 11x^2y^2 + y^4
&= x^4 - 2x^2y^2 + y^4-9x^2y^2 \\
&= (x^2-y^2)^2-(3xy)^2 \\
&= (x^2-y^2-3xy)(x^2-y^2+3xy).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
Limit on the expression containing sides of a triangle To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$ and in case of equilateral triangle, the limit is 3.
Now how to procee... | Let $x=p-a$, $y=p-b$, and $z=p-c$, where $p=(a+b+c)/2$. The main expression is equal to $$F:=\frac{4(x+y+z)^2}{x^2+y^2+z^2+3(xy+yz+zx)}=\frac{4\sum x^2+8\sum yz}{\sum x^2 +3\sum yz}\\=4-\frac{4}{\sum x^2/\sum yz +3}.$$ From Cauchy-Schwarz we know that $\sum x^2 \geq \sum yz$. Thus, $F\geq 3$. Furthermore, it is obvious... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$ 1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$ can be written in the form $\frac{a}{b}$, find $a+b$. The solution to the equation
$$1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$$
can be written in the form $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers. Find $a+b$.
| I assume the question is not a live problem anymore. What you can do to solve this is just solving for $x$. So assume $x$ is a solution, then $\sqrt{1-\sqrt{x^4-x^2}}=x-1$, so $1-\sqrt{x^4-x^2}=(x-1)^2$. Continuing like this you will get to $-4x^3+5x^2=0$. This has two solutions, but one of the solutions is not a solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Which of the following sets is a complete set of representatives modulo 7? Which of the following sets is a complete set of representatives modulo 7?
1) (1, 8, 27, 64, 125, 216, 343)
1 mod 7 = 1
8 mod 7 = 1
27 mod 7 = 6
64 mod 7 = 1
125 mod 7 = 6
216 mod = 6
343 mod = 0
2) (1, -3, 9, -27, 81, -243, 0)
1 mod 7 = 1
-3 mo... | Hint: you want a complete set whose members are equivalent, modulo 7, to one and only one element of the set: $\{0, 1, 2, 3, 4, 5, 6\}$, not necessarily in that order. Each and every one of these numbers must be represented by one and only one of the numbers in such a set. That is, you want representatives of the equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Unintentional Negative Sign in Limit Evaluation I've been working on evaluating the following limit:
$$\lim_{x\to 0} \left(\csc(x^2)\cos(x)-\csc(x^2)\cos(3x) \right)$$
According to my calculator, the limit should end up being 4.
Though I've tried using the following process to find the limit, I continue to get -4:
$$\b... | $$ \bigg(\frac{\cos(x) - \cos(2x+x)}{\sin(x^2)}\bigg)=\bigg(\frac{\cos(x) - \cos(2x)\cos(x) + \sin(2x)\sin(x)}{\sin(x^2)}\bigg)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
For real numbers $x$ and $y$, show that $\frac{x^2 + y^2}{4} < e^{x+y-2} $
Show that for $x$, $y$ real numbers, $0<x$ , $0<y$
$$\left(\frac{x^2 + y^2}{4}\right) < e^{x+y-2}. $$
Someone can help me with this please...
| From the well known inequality: $1+z\leq e^z$, we replace $z$ by $\frac{z}{2}-1$ to get $\frac{z}{2}\leq e^{\frac{z}{2}-1}$. Square both sides to get $\frac{z^2}{4}\leq e^{z-2}$. Now let $z=x+y$, to get: $\frac{(x+y)^2}{4}\leq e^{x+y-2}$. Hence:
$$\frac{x^2}{4}+\frac{y^2}{4}<\frac{(x+y)^2}{4}\leq e^{x+y-2}$$
Note:
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ I encountered the following induction proof on a practice exam for calculus:
$$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$
I have to prove this statement with induction.
Can anyone please help me with this proof?
| If $P(n): \sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6},$
we see $P(1): 1^2=1$ and $\frac{1(1+1)(2\cdot1+1)}{6}=1$ so, $P(1)$ is true
Let $P(m)$ is true, $$\sum_{k=1}^mk^2 = \frac{m(m+1)(2m+1)}{6}$$
For $P(m+1),$
$$ \frac{m(m+1)(2m+1)}{6}+(m+1)^2$$
$$=\frac{m(m+1)(2m+1)+6(m+1)^2}6$$
$$=\frac{(m+1)\{m(2m+1)+6(m+1)\}}6$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
T... | I just read that there is a partial converse to Raabe's test that if $\frac{a_{n+1}}{a_n}\geq 1-p/n$ for $p\leq 1$, then the series diverges. So I think that settles that it diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \se... | Try returning to your integral: $$\int \cos^2\theta = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$$
We get a factor of $\frac 12,$ and not $2$, multiplying the second term in the sum.
Note, more importantly, that your substitutions for $\cos\theta\sin\theta$ in terms of $\theta = \arctan(t/3)$ are also incorr... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here... | Using Titu Andreescu's Lemma, we have
$\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{(1-a-b)^2}{1} \ge \dfrac{(a+b+1-a-b)^2}{3}=\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. Tha... | Your reasoning would be correct if $y_n<\frac{1}{n}$, which is not true. The sequence actually diverges. To prove it, we use the fact that for $k\in[\![1,n-1]\!]$,
\begin{align}
\frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n} &\geq \frac{1}{kn+n}+\frac{1}{kn+n}+...+\frac{1}{kn+n} \\
&\geq \frac{n}{n(k+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Series involving Marcum Q function I would like to have a better form of this series:
$$\sum_{k=0}^{\infty}\,\frac{1}{k!}\,\left(\frac{ab\sin(c)}{\sqrt{2}}\right)^{2k}\,Q_{k+\frac{3}{2}}\left(ab\cos(c),bx\right)$$
where $Q_m(\alpha,\beta)$ is the generalized Marcum Q-function.
Does anyone an idea on how to proceed, if ... | Well, I guess, in some form of. But at the same time you can get a closed form solution.
For simplicity let's set $\alpha=\left(\frac{ab\sin(c)}{\sqrt{2}}\right)^{2}$ and $\beta=ab\cos(c)$
Taking into account the definition of the generalized Marcum Q-function:
$$
\begin{eqnarray}
\sum_{k=0}^{\infty}\,\frac{\alpha^{k... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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What is the limit of $\frac {x^4 +y^4}{x^3 +y^3}$ as $(x,y) \to(0,0)$ What is the limit of $$\lim_{(x,y)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}$$
if it exists?
I have tried to solve it by converting it to polar system $(x,y)=(r\cos a,r\sin a)$ and another settings. However I could not find the limit and not to show that... | The idea to prove that the limit doesn't exist is to choose a direction such that $x^3+y^3$ is convergent to $0$ faster than $x^4+y^4$ so let $y=-x+x^3$ hence we find
$$x^3+y^3=x^3+(-x+x^3)^3=3x^5+o(x^5)\ \text{and}\ x^4+y^4=x^4+(-x+x^3)^4=2x^4+o(x^4)$$
hence
$$\lim_{(x,y=-x+x^3)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are pa... | Here's my elementary proof by induction that
$(1+x/n)^n < 1/(1-x)$.
Let $y = x/n$.
This becomes
$(1+y)^n < 1/(1-ny)$
when $0 < y < 1/n$.
For $n=1$,
this is
$1+y < 1/(1-y)$
or
$1-y^2 < 1$
which is true.
Suppose it it true for $n$,
so that
$(1+y)^n < \frac1{1-ny}$.
Then,
if $0 < y < \frac1{n+1}$,
$\begin{align}
(1-(n+1)y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$ The question is:
Prove that:
$$ \cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B} $$
I have tried expanding it as $\dfrac{\cos(A+B)}{\sin(A+B)}$ and $\dfrac{1}{\tan(A+B)}$.
| It is probably easier to start from the RHS:
$$\frac{\cot A\cot B-1}{\cot A+\cot B}= \frac{\frac{\cos A}{\sin A}\frac{\cos B}{\sin B} -1}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B} }$$
Now, all the following computations are natural, and pretty authomatic:
$$=\frac{\frac{\cos A \cos B}{\sin A \sin B}-\frac{\sin A\sin ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solve in integers $ (y^3+xy-1)(x^2+x-y)=(x^3-xy+1)(y^2+x-y)$ Solve in integers: $$ (y^3+xy-1)(x^2+x-y)=(x^3-xy+1)(y^2+x-y)$$
My idea:
$$\Longleftrightarrow (y^3+xy-1)(x^2+x-y)-(x^3-xy+1)(y^2+x-y)=0$$
$$\Longleftrightarrow -x^4-x^3y^2+2x^3y+x^2y^3+2x^2y-x^2+2xy^3-2xy^2-2x-y^4-y^2+2y=0$$
$$\Longleftrightarrow 2xy(x... | We first note that if $(x,y)$ is a solution then so is $(-y,-x)$.Consider the term
$$
\frac{x^3-xy+1}{x^2+x-y}
$$
This can be expanded as
$$
x-1+\frac{x+1-y}{x^2+x-y}
$$
For the latter term to be integral implies that $x^2+x-y <= x+1-y$ which implies that $|x|<=1$. This gives us 3 possible values for $x$ namely $0,\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For what $x\in[0,2\pi]$ is $\sin x < \cos 2x$ What's the set of all solutions to the inequality $\sin x < \cos 2x$ for $x \in [0, 2\pi]$? I know the answer is $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi]$, but I'm not quite sure how to get there.
This is what I have so far: $\\
... | You are doing fine, now just need to find the values of $x$ that make $\sin x$ come out to those values. One way to see $\sin \frac \pi6=\frac 12$ is to consider an equilateral triangle cut in half vertically. The base (adjacent) is $\frac 12$ and the hypotenuse is $1$. To see $\sin \frac {3\pi}2=-1$ just consider t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.
Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer.
I've tried proving this by induction, but I've not succeed so far. What I did was:
Let'... | Let $f(n)=\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}$
$$f(n+1)-f(n)=\frac{(n+1)^5}5+\frac{(n+1)^3}3+\frac{7(n+1)}{15}-\left(\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}\right)$$
$$=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac{7(n+1)-7n}{15}$$
$$=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac{7}{15}$$
$$=(n^4+2n^3+2n^2+n)+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Calculating $\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$
Evaluate:$$\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$$
| Rewrite the integral as
$$\int dx \frac{\cot{x} (\csc^2{x}-1)}{\sqrt{1+\csc^4{x}}}$$
Split the integral along the numerator. The first piece is
$$\int dx \frac{\csc^2{x} \cot{x}}{\sqrt{1+\csc^4{x}}} = -\int \frac{d(\cot{x}) \cot{x}}{\sqrt{1+(1+\cot^2{x})^2}} = -\frac12 \int \frac{du}{\sqrt{(u+1)^2+1}}$$
where $u=\cot^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Radius of convergence and interval of convergence of $\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$ $$\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$$
I tried the ratio test on it but got stuck.
| I'm going to throw this in, since it looks more like what is found in calculus texts:
$$\lim_{n \rightarrow \infty} \ \left| \frac{a_{n+1}}{a_n} \right| \ = \ \lim_{n \rightarrow \infty} \ \left| \frac{(-3)^{n+1} x^{n+1}/ (n+1)^{3/2}}{(-3)^n x^n / n^{3/2}} \ \right| $$
$$= \ \lim_{n \rightarrow \infty} \ \left| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with $\int\frac{1}{1+x^8}dx$ I'm not sure how to proceed. I tried factoring like you do to evaluate $\int\frac{1}{1+x^4}dx$, and since it came out nasty I checked out WolframAlpha to see if I was on the right track. In fact, Wolfram doesn't have a step-by-step solution for the integral in question, but the primiti... | Decompose the integrand as
\begin{align}\frac{1}{1+x^8}
&=\frac1{2\sqrt2}\left(\frac{x^2+\sqrt2}{x^4-\sqrt2x^2+1}
- \frac{x^2-\sqrt2}{x^4+\sqrt2x^2+1}\right)\\
&=\frac{2+\sqrt2}8 \left( \frac{1+x^2}{x^4+\sqrt2x^2+1}
+\frac{1-x^2}{x^4-\sqrt2x^2+1} \right)\\
&\hspace{5mm}+\frac{2-\sqrt2}8 \left( \frac{1+x^2}{x^4-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$. If $x,y,z\in\mathbb{R}$ such that $$x+y+z=4,\quad x^2+y^2+z^2=6;$$then show that the each of $x,y,z$ lie in the closed interval $[2/3,2]$.
I have been able to solve using $2(y^2+z^2)\geq(y+z)^2$.
Is there any another met... | We have $(x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=6-8+3=2$, i.e.,
$$(x-1)^2+(y-1)^2+(z-1)^2=1.$$
This implies $(x-1)^2\le1$ and $x\in[0,2]$. Similarly for the other two variables.
Similarly we get
$$\left(x-\frac53\right)^2+\left(y-\frac53\right)^2+\left(z-\frac53\right)^2=1$$
which implies $(x-\frac53)^2\le1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Taylor Polynomial $\;y = e^{\sin x},\,$ $a = \frac{\pi}{2},\;$ $x = 1.5$ $y = e^{\sin x};\;$ $a = \frac{\pi}{2};\;$ $x = 1.5$
So first things get the derivative. I need to calculate the error and the $T_2$
$$f'(x) = \cos x e^{\sin x}$$
$$f''(x) = \cos ^2x e^{\sin x} - \sin x e^{\sin x}$$
So that now gives me
$$f(a) + f... | Your derivatives $f'(x), f''(x)$ are fine. And the following is correct:
$$T_2(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)(x-a)^2}{2}\tag{1}$$
Where the problem seems to be is in your evaluation of $(1)$, given $f(a), f'(a), f''(a)$ with $a = \pi/2, x = 1.5$:
You wrote: $$e^ {\sin \frac{\pi}{2}} + \cos (\frac{\pi}{2}) e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| Straight, you can get the following equation :
$$
\sqrt{(2-\sqrt{x-1})^2} + \sqrt{(3-\sqrt{x-1})^2} =1
$$
which leads to the following equation :
$$
|2-\sqrt{x-1}| + |3-\sqrt{x-1}| =1
$$
Then you will have three cases to discuss :
*
*case : $\sqrt{x-1} \leq2$ (equivalent to $x\leq5$) :
$\sqrt{x-1} = 2$ then $x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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$p\nmid 2n-1,$ then $\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3} \Leftrightarrow \sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0 \pmod{p^2} $ Is it true that if $p$ is a prime and $p\nmid 2n-1,$ then
$$\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3}
\hspace{12pt}\Leftrightarrow \hspace{12pt}
\sum_{k=1}^{p-... | We show something stronger, namely that
$$
2\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}+(2n-1)\cdot p\cdot\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0\mod{p^3}.
$$
Let $n\geq 1$ be a fixed integer. If we compute the difference
$$
\frac{1}{(1+x)^{2n-1}}-\left(1-(2n-1)x+n(2n-1)x^2\right),
$$
we see that it is of the form
$$
\frac{f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ al... | There are other ways to represent the harmonic series, for example, recall the geometric sum:
$$\frac{1-r^n}{1-r}=1+r+r^2+\dots+r^{n-1}$$
And integrate both sides with respect to $r$ from $0$ to $1$:
$$\begin{align}\int_0^1\frac{1-r^n}{1-r}dr&=\int_0^11+r+r^2+\dots+r^{n-1}dr\\&=\left.\frac11r+\frac12r^2+\frac13r^3+\dot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 5,
"answer_id": 3
} |
show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$ show that:
$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$
where $n=0,1,2,3,\ldots$.
is there any help?
thanks for all
| It is easier to find the integral in a more general form below:
$$I_{n}(a)=\int_{-\infty}^{+\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}} \quad \text {, where } n=0,1,2,3, \cdots \textrm{ and }a>0.$$
We first start with the easiest one $I_0(a)$, $$\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\frac{1}{\sqrt{a}} \left[ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
If $(a+1)(b+1)(c+1)=8,$ then prove $abc \le 1$? My proof:
$\sqrt[3]{(a+1)(b+1)(c+1)} = 2 \le \frac{a+b+c}{3} + 1 $according to AM-GM.
Thus, $1 \le \frac {a+b+c}{3}$.
Also, $ \sqrt[3]{abc} \le \frac{a+b+c}{3}$.
Then, either $\frac {a+b+c}{3} \le 1$ or $1 \le \frac {a+b+c}{3}$.
Suppose the latter equation is true. Then $... | Your proof doesn't work, as pointed out by vadim
Hint: Apply AM-GM directly to each term, we get that
$$ 8 = (a+1)(b+1)(c+1) \geq 2\sqrt{a} 2 \sqrt{b} 2 \sqrt{c} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $ I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!
Below is how I derive my approximation. I use mainly Stirling Approximat... | The easiest way to get the correct approximation is to use Stirling in the logarithmic form:
$$\ln n!=n(\ln n-1)+\frac12 \ln \left(2\pi n\right)+\frac{1}{12n}+O(n^{-3}),$$
which also implies
$$\ln\frac{n!}{(n-2x)!}=2x \ln n+\frac{x-2x^2}{n}+O(n^{-2}).$$
Now using that
$$\ln (n-1)^{-2x}=-2x\ln(n-1)=-2x\ln n+\frac{2x}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will... | $$
\begin{align}
(a-2)(a-1)a\cdot a(a+1)(a+2)
&=5!\,a\binom{a+2}{5}\\
&=120[(a+3)-3]\binom{a+2}{5}\\
&=120(a+3)\binom{a+2}{5}-360\binom{a+2}{5}\\
&=120\cdot6\binom{a+3}{6}-360\binom{a+2}{5}\\
&=360\left[2\binom{a+3}{6}-\binom{a+2}{5}\right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
A problem of divisibility ... If $7\nmid n$, $7\nmid n-1$ and $7\nmid n^{3}+1$ then $7\mid n^{2}+n+1$ $7\nmid n$, $7\nmid n-1\;\;$ and $\;\;7\nmid n^{3}+1$ $\Longrightarrow 7\mid n^{2}+n+1$$$$$Do not quite understand how to do, but do not want to use modular arithmetic, thought of using consecutive numbers, but did not... | $$n^7-n=n(n^6-1)=n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1).$$
Ask Little Fermat to help.
If you are really into consecutive numbers, then you may also observe that
$$
(n-2)(n-4)=n^2-6n+8=n^2+n+1-7(n-1),
$$
and
$$
(n+2)(n+4)=n^2+6n+8=n^2-n+1+7(n+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inductive demonstration I have this:
$$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$
So I was suggested of doing this:
$$\begin{align}
(1^2+2^2+3^2+\dots+k^2)+(k+1)^2
&= \frac{k(k+1)(2k+1)}6+(k+1)^2 \\
&= \frac{k(k+1)(2k+1)+\color{blue}{6(k+1)^2}}6 \\
&= \frac{(k+1)[k(2k+1)+6(k+1)]}6 \\
&= \frac{(k+1)[2k^2+7k+6]}6 \\
&... | \begin{align*}
\frac{k(k+1)(2k+1)}6+(k+1)^2&=\frac{k(k+1)(2k+1)}6+\frac{6(k+1)^2}{6}\\
&=\frac{k(k+1)(2k+1)+6(k+1)^2}6
\end{align*}
For the last line,
$$\frac{(k+1)(k+2)(2k+3)}6=\frac{(k+1)[(k+1)+1][2(k+1)+1]}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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LU factorization with pivot to solve linear system I read that LUP matrix exist for any square matrix such that it is not a singular one.
But I came across a matrix that when I get the LUP and calculate Lz=Pb -> Ux = z the answer is wrong, I don't know why.
The linear equation was like that
x+y+z=2
z=2
2x+3y+z=0
the m... | As noted in the comments, we have:
$$A = P L U x = b \rightarrow LUx = P^{-1}b \rightarrow Ly = P^{-1}b, Ux = y$$
Note: you can see my $L$ and $U$ below.
$$P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$$
Using this, we have:
$Ly = P^{-1}\cdot b \rightarrow \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0 \\ 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
| $$\dfrac{r^3-a^3}{r^3+a^3}=\dfrac{r-a}{r+a}\cdot\dfrac{r^2+ra+a^2}{r^2-ra+a^2}$$
Let $f(r)=r^2-ra+a^2$
$$f(r+h)=(r+h)^2-(r+h)a+a^2=r^2+r(2h-a)+a^2-ah$$
If we set $r^2+ra+a^2=f(r+h),$
$$r^2+ra+a^2=r^2+r(2h-a)+a^2-ah$$
Comparing the coefficients of $r,$ $$a=2h-a\iff h=a$$
Compare the constants $a^2=a^2-ah+h^2\iff h(h-a)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
$2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $
Problem: $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $
Solution: $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$
$2 \cos^{-1}x =2 \sin^{-1}x$
$ \cos^{-1}x = \sin^{-1}x$
Am I doing right ?
| As the principal value of $\arccos x$ lies in $\in[0,\pi]$
$\implies 2\arccos x$ will lie in $\in[0,2\pi]$
Again as the principal value of $\arcsin x$ $\in[-\frac\pi2,\frac\pi2]$
Observe that the intersection of region is $[0,\frac\pi2]$
So, $\sin^{-1}(2x \sqrt {1-x^2})=2\arccos x$
if $0\le 2\arccos x\le \frac\pi2\if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Closed form of $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\varphi-1)$ I am trying to calculate the dilogarithm of the golden ratio and its conjugate $\Phi = \varphi-1$. Eg the solutions of the equation $u^2 - u = 1$.
From Wikipdia one has the following
\begin{align*}
\operatorname{Li}_2\left( \frac{1 ... | This solution is very similar to @Start wearing purple's solution above.
Using the following identities (check Eq $3$,$4$ and $5$).
$$\text{Li}_2(1-x)+\text{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2\left(\frac{1}{x}\right)\tag1$$
$$\text{Li}_2(x)+\text{Li}_2(-x)-\frac12\text{Li}_2(x^2)=0\tag2$$
$$\text{Li}_2(x)+\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Finding the $n^{\text{th}}$ term of $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots$ I need help on finding the $n^{\text{th}}$ term of this infinite series?
$$
s=\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots
$$
Could you help me in writing th... | The first thing you can do is start with $a_1=\frac{1}{4}$ and then realize that $$a_{n+1} =a_n\frac{2n-1}{2n+2}$$
that doesn't seem to get you anywhere, however.
As commentator Tenali notes, you can write the numerator as $$1\cdot 3\cdots (2n-1) = \frac{1\cdot 2 \cdot 3 \cdots (2n)}{2\cdot 4\cdots (2n)}=\frac{(2n)!}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) ... | For typing ease, let $a=\cos^2\theta$ and $b=\sin^2\theta$. Thus $a+b=1$. We are told that $a^2-b^2=x$, or equivalently that $a-b=x$.
We want $a^3-b^3$. It will be enough to find $a^2+ab+b^2$. Note the identity
$$4(a^2+ab+b^2)=3(a+b)^2+(a-b)^2.$$
Remark: For other problems of a similar character, it might be preferab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Show that $19\mid 12^{2013}+7^{2013}$ Show that $$19\mid 12^{2013}+7^{2013}$$$$$$No use of modular arithmetic, have not come this content.
| First we prove two theorems that we will use
If $a,\;b,\;c\;\in \mathbb{N}$ with $a\neq0$ and $x,y\in\mathbb{N} $such that $a\mid b$ and $a\mid c$, then, $a\mid bx+cy$;
Show:$a\mid b$ and $a\mid c$ implies that there $m,n\in\mathbb{N}$ such that $b=a\cdot m$ and $c=a\cdot n$;$$bx+cy=am\cdot x+an\cdot y=a(mx+ny)\Longr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation : $x^2 − 6 |x − 2| − 28 = 0$ The following is an absolute value quadratic equation that I want to solve:
$$x^2 − 6 |x − 2| − 28 = 0$$
Here is what I did :
$x^2 − 6 |x − 2| − 28 = 0$
$x^2 − 6 |x − 2| − 28 = 0$
$-6|x-2|=28-x^2$
$6|x-2|=x^2-28$
$6x-12=x^2-28$ or $28-x^2$
(Is this step correct ?)
Sol... | Once you get to $6|x-2|=x^2-28$, you need to consider two cases.
Case $x \ge 2:$
If $x \ge 2$, then $6x-12 = x^2-28 \implies x^2-6x-16=0 \implies (x+2)(x-8)=0
\implies x \in \{-2, 8\}$ But we assumed $x \ge 2$, so we have to reject $x=-2$ as an extraneous root. Hence if $x \ge 2$, then $x=8$.
Case $x < 2:$
If $x < 2$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Marginal Density - Probability (X,Y) is a two-dimensional abolute continuous random vector with the density function fx,y given by:
(1)
$f_{X,Y}(x,y) = \begin{cases}
\frac{1}{2} & 0 \le x \le 1, 0 \le y \le 4x \\
0 & \text{otherwise} \\
\end{cases}$
Show the density functions fx and fy for X and Y are:
$f_{X}(x) = \b... | What you did for $f_{X}$ is OK ! For $f_{Y}$, if you just write $f_{Y}(y) = \int_{0}^{1} \frac{1}{2} \: dx$, you're missing that the condition on $y$ (which is $0 \leq y \leq 4x$) also depends on $x$.
To do this right, use indicator functions. $f_{(X,Y)}$ can be re-written as follows :
$$ f_{(X,Y)}(x,y) = \frac{1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well ... | WLOG, assume that $a = \max(a, b, c, d)$.
Let $x = a- b, \ y = a-c, \ z = a-d$. Then $x, y, z \ge 0$.
Let $w = b+c+d - a$. Then $w > 0$ since $a, b, c, d$ are the sides of a quadrilateral.
The inequality is written as $$\frac{1}{4}Aw^2 + \frac{1}{4}Bw + \frac{1}{4}C \ge 0$$ where
\begin{align}
A &= 2\, x^2 - x\, y - 2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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How to find the second derivative of $2x^3 + y^3 = 5$? I'm finding it difficult to find the second derivative of the following equation: $2x^3 + y^3 = 5$.
My answer is $\dfrac{-4xy^3 - 8x^4}{y^5}$, but the answer key says $\dfrac{-20x}{y^5}$.
My friends and I have the same answer, but I don't know where I went wrong.
H... | Your answer only requires some re-arrangement:
$$
2x^3 + y^3 = 5
$$
multiply by $x$ on both sides to get:
$$
2x^4 + xy^3 = 5 x
$$
or:
$$
xy^3 = 5x - 2x^4
$$
...(substitute this in your equation to get):
$$
\frac{-4(5x - 2x^4 )- 8x^4}{y^5}\to y''=\frac{-20x}{y^5}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
| If you are familiar with Minkowski's inequality, this is straightforward.
$$\sqrt{x^2+1}+\sqrt{y^2+1} \ge \sqrt{(x+y)^2 + (1+1)^2} = \sqrt5$$
| {
"language": "en",
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"source": "stackexchange",
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equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$ In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
Please prove this identi... | Here's a brute-force approach:
Abbreviating the measures of angles surrounding point $P$ thusly,
$$\alpha := |\angle BPC| \qquad \beta := |\angle CPA| \qquad \gamma := |\angle APB|$$
we have
$$\cos\gamma = \cos\left(2\pi - \alpha - \beta\right) = \cos\left(\alpha+\beta\right) = \cos\alpha \cos \beta - \sin\alpha \sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof for the Ramanujan's formula ? I found this formula in a textbook in which the proof to the formula was not given
Ramanujam's formula
$$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$
Its a great equation andhow do you prove this. its a bit difficult for me and tried differen... | Here is this answer adapted to this question.
Define
$$
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\dots}}}}
$$
then $f(x)^2=1+xf(x+1)$. This indicates we should look at $f(x)=x+1$.
Considering $f(x)=x+1$, we are lead to show inductively that
$$
\hspace{-18pt}x+1=\small\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\dots+\sqrt{1+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding $x^4 + y^4 + z^4$ using geometric series This is a problem from the 2001 Stanford Math Tournament Algebra section.
$$$$Given that $$x+y+z=3$$ $$x^2 + y^2 + z^2 = 5$$$$x^3+y^3+z^3=7$$Find $x^4+y^4+z^4$. $$$$My friend claimed that he was able to solve this question using geometric series, i.e. adding the above t... | This is a rather encumbered solution for a not-as-hard problem, and it is the only way I see to make sense of what your friend is saying. On the other hand, it is cool that it let us know about Newton's identities and their proof using generating functions.
Let us form the formal series \begin{align}F(x,y,z,t)&:=(x+y+z... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of a sequence of integrals Determine
$$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n}$$
Since $e^{x} \geq t+1$ we know that $e^{x^2}-1 \geq 2t +t^2 \geq t^2$ if $t\geq 0$
So due to symmetry, we should be able to use DCT?
But my solution is $$\lim _{n \rightarrow \infty} \int_{-\in... | We start from the the inequality $ e^y \le 1+y +\frac {3y^2} {2}$ if $ y \ge 0,\, y \le 1$. This follows from the Taylor's theorem
in the Lagrange form. Next, we obtain with help of Maple
$$J_n:=\int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \ge \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The Converse of the Pythagoras Theorem Let $ABC$ be a triangle in the plane. Suppose that $AB^2+AC^2=BC^2.$ Prove that:
$\angle BAC$ is right angle.
Remark: I believe this to be true. Now I have the following difficulty. I'm looking for a proof which a 10th grade student can understand, but any effort of mine is via pr... | Well, what always worked for me as a tenth grader was to cut a triangle into right triangles:
Construct a point $D$ on $BC$ so that $DA$ is perpendicular to $BC$ so $ADC$ is a right triangle and $AD^2 + DC^2 = AC^2$.
THere are three possibilities.
i) $\angle ABC$ is acute.
Then $D$ is between $B$ and $C$ and $BC = BD... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How do I factorise this polynomial Please help factorise this:
$$6x^2+x+4=0$$
In my attempts, I assumed $a=6$, $b=1$ and $c=4$.
I multiplied $c$ by $a$ and attempted to get the factors that give us the sum of $b$.
The only factor pairs of $24$ are $1\cdot 24$, $2\cdot 12$, $3\cdot 8$ and $4\cdot 6$.
None of the above ... | Let us complete the square:
\begin{align*}
6 x^2 + x + 4 &= 6 \left(x^2 + \frac{1}{6} x + \frac{3}{2}\right) \\
&= 6 \left(x^2 + \frac{1}{6} x + \frac{1}{144} + \frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2 + 6 \left(\frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum of a trigonometric Polynomial
Given
$$x+y+z=\pi$$
$$3\sin(x)+4\sin(y)+18 \sin(z)=A$$
Question：find maximum of $A$.
I spend so many time on this question.
answer is $ 35\sqrt{7} /4$, but why?
| Assuming that $x,y,z>0$, you can write this as:
$$3\sin(x)+4\sin(y)+18 \sin(\pi - x - y)=A$$
Now find when the derivatives:
$$\frac{\partial A}{\partial x}, \frac{\partial A}{\partial y}$$
are equal to zero. You'll get:
$$3 \cos(x) + 18 \cos(x + y)= 0,\ 4 \cos(y) + 18 \cos(x + y) = 0$$
Which leads to:
$$y=\arccos\left(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If this is a telescoping series then how does it collapse? $\frac{3r+1}{r(r-1)(r+1)}$ Express $$\frac{3r+1}{r(r-1)(r+1)}$$ in partial fractions. Hence, or otherwise, show
$$\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}=\frac52-\frac2n-\frac{1}{n+1}$$
So, I have obtained the partial fractions $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}... | HINT:
There are at least two methods for $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
Method $1:$
$$\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}=-\left(\frac1{r+1}-\frac1r\right)-2\left(\frac1r-\frac1{r-1}\right)$$
The survivor of the summation $2\le r\le n$ after cancellation,
for the first part w... | {
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"timestamp": "2023-03-29T00:00:00",
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Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$ Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $$\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$$
I have got rid of $i$ in the denominator, leaving $$\frac{x^2-y^2-2ixy}{x^2+y^2}$$
but now... | You made an error somewhere.
Writing $T = \frac{a}{b}$, the expression you are trying to simplify is
$$\frac{1+\frac{a^2}{b^2}}{2\frac{a}{b}}$$
which after clearing out the denominators is
$$\frac{b^2+a^2}{2ab}.$$
Plugging in $a$ and $b$ and using the fact that they are complex conjugates gives
$$\frac{x^2 +2ixy - y^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Partial Fraction decomposition when denominator is in $x^2 + a$ form Why isn't the expansion of $$ \frac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} $$
in the form of
$$ \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+16} $$ since (as I recall) the denominator is of square power and you should decompose it (in the numerator) until t... | We have:
$$ \dfrac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} = \dfrac{as+b}{s^2+1} + \dfrac{cs+d}{s^2+16} $$
When we multiply things out and equate sides, we have:
$$ s^3 - 2s^2 + 16s - 2 = 16 b + d + 16 a s + c s + b s^2 + d s^2 + a s^3 + c s^3$$
Equating like powers leads to:
$$a+c = 1, b+d = -2, 16a+c = 16, 16b+d=-2$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| As suggested in another answer, one can write:
$$\prod_{k=1}^n \frac{2k+1}{2k+4}=\frac{8}{\pi}\cdot \frac{\Gamma (n+\frac{3}{2})\Gamma (\frac{3}{2})}{\Gamma (n+3)}=\frac{8}{\pi}\text{B}\left(n+\frac{3}{2},\frac{3}{2}\right)$$
Now, using the definition of the beta function, our sum is:
$$1+\frac{8}{\pi}\sum_{n\geq 1}\in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving Inequality with Square Roots and Algebraic Manipulation Question:
Edit: Original question is: Estimate $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10000}$ to nearest hundred.
I used $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\geq \int_{0}^{n}\sqrt{x}dx$ and $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\leq \frac{4n+3}{6}\sqrt{n}$, which I nee... | Your desired inequality
$$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$
is equivalent to
$$(4n+3)\sqrt{n}\le(4n+1)\sqrt{n+1}\;.$$
Since we’re dealing with non-negative numbers here, this is equivalent to
$$n(16n^2+24n+9)\le(n+1)(16n^2+8n+1)\;.$$
If you expand that, you’ll see that it’s true.
| {
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Can't simplify this fraction: $ \frac{1+x^6}{1+x^2}$ I've been having trouble simplifying this fraction :
$$
\frac{1+x^6}{1+x^2}
$$
Can anyone explain step by step on how to solve this?
Thank you.
| $1+x^6=1+({x^2})^3$ which is equivalent to
$(1+x^2)(x^4+1-x^2)$ using $a^3+b^3=(a+b)(a^2+b^2-ab)$
Hence the expression becomes $$\frac{(1+x^2)(x^4+1-x^2)}{1+x^2} = (x^4+1-x^2).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that t... | For given positive integers a and b, and applying simple division of $(a^2 + b^2)$ by $(ab + 1)$ we have quotient as a/b and remainder as $(b^2 - a/b)$.
For the result to be a perfect division, the remainder must be zero. Therefore $(b^2 - a/b) = 0$ which gives the relation $a=b^3$.
Applying $a=b^3$ in the equation $x=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \... | It's the last step, where you're solving for $x$.
$4x + 12 \equiv 0 \Rightarrow 4x\equiv 3 \Rightarrow x\equiv 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of Fibonacci-Numbers Is there a closed formula for the sum of the first $n$ even (or odd) Fibonacci numbers, like there is one for the $n$th number (Moivre-Binet)?
| Let $\varphi=\frac12\left(1+\sqrt5\right)$ and $\widehat\varphi=\frac12\left(1-\sqrt5\right)$, so that $F_n=\frac1{\sqrt5}\left(\varphi^n-\widehat\varphi^n\right)$. Then
$$\varphi^3=\left(\frac{1+\sqrt5}2\right)^3=2+\sqrt5\qquad\text{and}\qquad\widehat\varphi^3=\left(\frac{1-\sqrt5}2\right)^3=2-\sqrt5\;,$$
so the sum ... | {
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Olympiad number theory problem I found this problem in previous problems of the olympiads of my country
If $t^2+n^2=r^2$, where $t$ has $3$ positive divisors, $n$ has $30$ positive divisors and $t,n,r$ are natural numbers, find the sum of all the possible values of $t$
I did gave it a try but only solved it partiall... | We add a little to the analysis by Yury. It could be that $a^2-b^2=p$, the odd leg of our triple is $p(a^2-b^2)$, and the even leg $p(2ab)$, where $(a^2-b^2,2ab,a^2+b^2)$ is a primitive triple.
Then $2ab$ must have $15$ divisors. That leaves very few cases, since the power of $2$ that divides the even one of $a$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Binomial expansion of modulus? How one can perform binomial expansion of modulus of a quantity.
For example, can we expand |-1+x| ? If, yes how?
Thanks
| Two cases it depends on (-1+x) value
Case 1: (-1+x) is positive then write it as $(x-1)^n$ and use normal binomial theorem
Case 2: if (-1+x) is negative your answer is $-(x-1)^n$
Again if n is not a positive integer then you have to use
$$(1+x)^n=1+nx+ \frac{n(n-1)}{1 \cdot 2} x^2 + \frac{n(n-1)(n-2)}{1\cdot2\cdot3... | {
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Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not kno... | Let $f(x)=\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$. Note that $f(x)$ is convex on the $(0;+\infty)$ because its second derivative is equal to
$$
f''(x)=2+\frac{6}{x^4}>0,~\text{for all}~x\in(0;+\infty).
$$
Hence, by Jensen's inequality ($p+q=1$)
$$
f(p)+f(q)\geqslant 2f\left(\frac{p+q}{2}\right)=2f\left(\frac... | {
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"url": "https://math.stackexchange.com/questions/487486",
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Proving $\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab} \le\frac{1}{2}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})$
Prove $$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right),$$
where $a,b,c > 0$ and $a,b,c \in \mathbb{R}$
Well, I've been trying for ... | Hint: $$\frac{1}{a^2+bc} \le \frac{1}{2a\sqrt{bc}} = \frac{1}{2\sqrt{ab}\sqrt{ac}}.$$
Now apply AM-GM inequality.
$$\frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \le \frac{1}{2\sqrt{ab}\sqrt{ac}} + \frac{1}{2\sqrt{ab}\sqrt{bc}} + \frac{1}{2\sqrt{ac}\sqrt{bc}} \le $$
$$\le \frac14 \left(\frac{1}{ab}+\frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving that a sequence is between certain values at certain n I'm given that $a_1=1$, and for every $n \gt1, a_{n+1} = a_n + \frac{1}{a_{n}}$. I need to prove that $20 < a_{200} < 24$. I tried finding a limit at infinity setting both limits to $L$ ( for $a_n$ and $a_{n+1}$ but that gave me nothing useful. All I figure... | Lower bound:
Let's prove that $a_n>\sqrt{2n}$ for $n\geqslant 3$, other words:
$$
(a_n)^2>2n, \qquad (n \geqslant 3).
\tag{1}
$$
$a_2 = 1+1=2$.
First, $(a_3)^2 = (2+\frac{1}{2})^2 = 4+2+\frac{1}{4} > 2\cdot 3$.
Now, using math. induction, we will show, that
$$(a_n)^2>2n \implies (a_{n+1})^2>2(n+1).$$
Yes, if $(a_n)^... | {
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Bessel function to $\sin(kr)$ $J_{\frac{1}{2}}(kr)=\frac{\sqrt{\frac{2}{\pi }} \text{Sin}[\text{kr}]}{\sqrt{\text{kr}}})$ This can be easily obtained by Mathematica,
How to do the details?
| Here's the quick and dirty way to do it for $kr\ge0$. The case of $kr<0$ involves a little trick with Bessel functions of negative integers. I will show this below. The following is often taken as the definition for $J_{\nu}$.
$$J_{\nu}(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\nu+1)}\left(\frac{x}{2}\right)^{2n+... | {
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Find all positive integers $n$ for which $1 + 5a_n.a_{n + 1}$ is a perfect square. The sequence $a_1, a_2, \ldots $ is defined by the initial conditions $$a_1 = 20; \quad a_2 = 30$$ and the recursion
$$a_{n+2} = 3a_{n+1} - a_n$$ and
for $n \geq 1$. Find all positive integers $n$ for which $1 + 5a_n * a_{n+1}$ is a pe... | Note:Just a try. If I'm wrong feel free to comment. Maybe this approach will give someone an idea
I'll post a solution, or rather attempt to solve this problem.
Let $1 + 5 \cdot a_n \cdot a_{n+1} = x^2$
First obviously every number in the sequence $a$ is multiple of $10$. So:
$$a_n \cdot a_{n+1} \equiv 0 \pmod {100}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $a_{n+1} = a_n^2$, the last $n$ digits of $a_{n+1}$ are the same as the last $n$ digits of $a_n$. I have been working on this problem for a while. I know that I have to prove it using induction, but I'm unsure of the next step. The formula for the terms is: $a_{n+1} = 5^{2n}$ with $a_1 = 5$. The sequence ... | Assuming that the formula is actually $a_{n+1} = 5^{2^n}$ (which matches up with $a_{n+1} = a_n^2$ and $a_1 = 5$), then we have as follows:
$a_{n+1} - a_n = 5^{2^n} - 5^{2^{n-1}} = (5^{2^{n-1}})^2 - 5^{2^{n-1}} = 5^{2^{n-1}}(5^{2^{n-1}} - 1)$, and we want to show that this number is divisible by $10^n$.
Since $5$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Definite integral involving square root of polynomial How can I solve the integral below?
$$\int_0^1 (x^4+x^3+x^2+x^1+1)^{1/2} \mathrm dx $$
| You can use the horrible result of calculating $\int(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$ by Wolfram Integrator, but I have the following relatively simpler approach:
$\int_0^1(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$
$=\int_0^1\left(\dfrac{1-x^5}{1-x}\right)^{\frac{1}{2}}~dx$
$=\int_0^1\dfrac{(1-x^5)^\frac{1}{2}}{(1-x)^\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int_{3\sqrt{2}}^6 1/\big(t^3\sqrt{t^2-9}\big)\;dt$ Here's the full problem
$$\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt.$$
The problem for me is what to do with the $-9$. I know that $1 - \sec^2(t) = \tan^2(t)$, so I'm trying to substitute $9\sec(x)$ for $t$ to get rid of the square root. So t... | $$
\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt = \frac12 \int_{3\sqrt{2}}^6\frac{2t\,dt}{t^4\sqrt{t^2-9}} = \frac12 \int_{18}^{36} \frac{du}{u^2\sqrt{u-9}}.
$$
Now let $w=\sqrt{u-9}$, so that $w^2 = u-9$ and $2w\,dw=du$, and $u = w^2+9$. Then the integral is
$$
\frac12 \int_3^{3\sqrt{3}} \frac{2w\,dw}{(w^2+9)^2 w}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration to minimize area I have the following problem:
Given a line $f(x) = ax + b$ which passes trough the point $(1, 2)$, find the values for $a$ and $b$ which results in the minimum value for $\int_{-1}^{1}(ax + b)^2\mathrm{d}x$. I already know the result of the area will be $\dfrac{2a^3}{3} + 2b$. But I don't e... | Firstly, the result of the integration is incorrect:
\begin{align*}
\int_{-1}^1 (ax+b)^2\ dx &= \int_{-1}^1 a^2x^2+2abx+b^2 \ dx \\
&=\left[ \frac{a^2}{3}x^3+abx^2+b^2x \right]_{-1}^1 \\
&=\frac{a^2}{3}1^3+ab \cdot 1^2+b^2 1 - \left( \frac{a^2}{3}(-1)^3+ab(-1)^2+b^2 (-1) \right) \\
&=\tfrac{2}{3}(a^2+3b^2).
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Digit Root of $2^{m−1}(2^m−1)$ is $1$ for odd $m$. Why? The Wiki page on Perfect numbers says:
[A]dding the digits of any even perfect number (except $6$), then adding the digits of the resulting number, and repeating this process until a single digit (called the digital root) is obtained, always produces the number $... | Let $e = m-1$ be even. Then $2^e$ can be congruent to one of $1, 4, 7$ modulo $9$. Then $2^{e+1}-1$ is congruent to $2\cdot 1-1 = 1,\, 2\cdot 4 - 1 = 7,\, 2\cdot 7 - 1 = 13 \equiv 4$ modulo $9$, hence $2^e(2^{e+1}-1)$ is congruent to $1\cdot 1$ or $4\cdot 7 = 28 \equiv 1$ modulo $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| Expand $(x+y)^7$, you would get:
\begin{align*}
(x+y)^7-(x^7+y^7)&=x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7-x^7-y^7
\\ &=7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6
\\ &=7xy(x^5+3x^4y+5x^3y^2+5x^2y^3+3xy^4+y^5)
\\ &=7xy(x+y)(x^4+2x^3y+3x^2y^2+2x^3+y^4)
\\ &=7xy(x+y)(x^2+xy+y^2)^2,\text{which is your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
} |
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