Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $36|n^{12k-6}-n^6$ Prove that if gcd(n,6)=1 and k>0, then $36|n^{12k-6}-n^6$.
Idea: I want to show that $2|n^{12k-6}-n^6$ and $3|n^{12k-6}-n^6$.
For the first part, since gcd(n,6)=1, n must be an odd number, so it is easy to show that $n^{12k-6}-n^6$ is divisible by 2. I am stuck by the second part and i ... | We show that $4$ divides our difference, and that $9$ does.
Recall (or prove) that if $n$ is odd then $n^2\equiv 1\pmod{8}$. It follows that $n^{12k-6}\equiv 1\pmod{8}$, and $n^6\equiv 1\pmod{8}$, so in fact our difference is divisible by $8$, and therefore by $4$.
For divisibility by $9$, note that $n^2\equiv 1\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how does row vectors span a row space? the column space is very easy to understand, for example, we have:
$$\begin{pmatrix}
1& 0& 0 \\
0& 1& 0\\
0& 0& 1
\end{pmatrix}\times \begin{pmatrix}
a \\ b\\ c
\end{pmatrix}$$
if we let $V_1=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, V_2=\begin{pmatrix} 0\\ 1\\ 0 \end{pma... | I cannot comment since I don't have enough reps for that.
What about $(a\quad b\quad c)\times \begin{pmatrix} V_1\\ V_2\\ V_3 \end{pmatrix}$?
This is 1 x 3 matrix times 3 x 3 which gives you a 1 x 3 matrix.
Post multiplication of a matrix by a vector gives a linear combination of the columns of the matrix.
Pre multipl... | {
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"url": "https://math.stackexchange.com/questions/1737111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| $\bf{Solution:}$ Using $$\displaystyle (x\cdot \sin x+5\cdot \cos x) = \sqrt{x^2+5^2}\left\{\frac{x}{\sqrt{x^2+5^2}}\cdot \sin x+\frac{5}{\sqrt{x^2+5^2}}\cdot \cos x\right\}$$
$\displaystyle = \sqrt{x^2+25}\cdot \cos\left(x-\phi\right)\;,$ where
$\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+25}}$ and $\displaystyle \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$ A problem in my homework had asked me:
When $x+y+z=0$, evaluate$$\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}$$
Without too much difficulty, one can see that the value should be $1$ using $(x,y,z)=(1,0,-1)$.
I dec... | The first term is $$\frac{x^2}{2x^2-xy-y^2}=\frac{x^2}{(2x+y)(x-y)}=\frac{x^2}{(x-z)(x-y)}$$
Do the same transformation on each of the fractions and add them up.
So the whole expression is $$\frac{x^2}{(x-z)(x-y)}+\frac{y^2}{(y-z)(y-x)}+\frac{z^2}{(z-y)(z-x)}=...=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\f... | The function $\sin^{-1}:[-1,1]\to[-\frac{\pi}2,\frac{\pi}2]$ is increasing. So
$$\frac{\sqrt3}{2}<\frac2{\sqrt5}<1\quad\text{and}\quad\frac{\sqrt3}{2}<\frac3{\sqrt{10}}<1\qquad\implies \qquad\frac{2\pi}{3}<\theta<\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding the coefficient of expansion Question:
Find the coefficient of $x^{11}$ in the expansion of:$$(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$$
The traditional way of doing this, as far as I know, is to first find the coefficient of each term that has $x^{11}$, and then sum it. However, with three individual terms in multiplica... | To calculate the coefficient by hand is somewhat cumbersome. Here is a notation which helps to make it better manageable. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain by successively applying the binomial theorem
\begin{align*}
[x^{11}]&(1+x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}} < 2$
If ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ are distinct odd natural
numbers not divisible by any prime greater than 5, then show that
$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}}
< 2$.
First I have noted that $a_i... | $$\sum_{i=1}^n \frac{1}{a_n}<\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^i5^j}=\sum_{i=0}^\infty\frac{1}{3^i\left(1-\frac{1}{5}\right)}$$
$$=\frac{5}{4}\sum_{i=0}^{\infty}\frac{1}{3^i}=\frac{5}{4}\cdot \frac{1}{1-\frac{1}{3}}=\frac{15}{8}<2$$
So, in fact, you have a stronger statement: $\sum_{i=1}^n \frac{1}{a_n}<\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744755",
"timestamp": "2023-03-29T00:00:00",
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Where can I read about this 'rule'? I was trying to solve an equation, I got nowhere, and the solution used a 'rule' that I have never seen before, $a^TX\,b = b^TX\,a$.
What is this rule?
Where can I read about it?
Note: $a$ and $b$ are vectors and $X$ is a matrix.
| a^t is for transpose, right?
$
\begin{pmatrix}
a_1 & a_2 \\
\end{pmatrix}
$$
\begin{pmatrix}
x_{11} & x_{12} \\
x_{21} & x_{22} \\
\end{pmatrix}
$$
\begin{pmatrix}
b_1 \\ b_2 \\
\end{pmatrix}
$ = $a_1b_1x_{11} + a_2b_1x_{21} + a_1b_2x_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$.
Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$.
Firstly,I've found the zeros of $f(x)$,just by simply substituting the elements of $\Bbb Z_5=\{0,1,2,3,4\}$ in $f(x)$.I ge... | Start by rewriting $\bar{f}(x)=x^5-x^4-x^3-x^2+x+1$. Divide by $x-1$ to get
$$\bar{f}(x)=(x-1)(x^4-x^2-2x-1)$$
Now divide $x^4-x^2-2x-1$ by $x-3$ to get
$$\bar{f}(x)=(x-1)(x-3)(x^3+3x^2+3x+2)$$
If we consider now $g(x)=x^3+3x^2+3x+2$ we notice that $g(3)=0$. So $3$ is a double root of $f$. Let's divide $g$ by $x-3$ we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| The nested radical:
$$
\sqrt{6+2\sqrt{5}}=\sqrt{6+\sqrt{20}}
$$
can be denested using the identity (that you can easely verify).
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
that works if $a^2-b$ is a perfect square. In this case $a^2-b=16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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How to find the projw(x) Sorry about formatting, new to this.
Subspace w =
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} such that $x_1 = x_3$ of R3.
x = \begin{pmatrix}-2\\1\\3\end{pmatrix}
Question is to find $proj_w (x)$.
| We'll assume it's the projection with respect to the standard inner product. The basis for $w$ is $$\left\{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}.$$
Therefore, omitting details, you have a least squares problem for the system $$\begin{bmatrix}1&0\\0&1\\1&0\end{bmatrix}\begin{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747024",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$. ... | The following solution is similar to my answers
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
and An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $x = a + b, \, y = c + d$.
We have $x + y = 1$.
We have
\begin{align*}
M &= ab y + cd x - ab ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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$f$ is an odd, $2\pi$ periodic function. Write explicitly Let $f$ be an odd $2\pi$ periodic function defined on $(0,\pi)$ by $f(x) = \frac12 (\pi - x)$.
I am asked to "write $f$ explicitly". what does this mean? how to do that? thanks for help.
| I assume the question asks for an explicit formula for all the real numbers $(-\infty,\infty)$.
It is an odd function, meaning $f(x) = -f(-x)$ for all $x$, therefore $f(0) = -f(0) = 0$. Therefore we have its values for $(-\pi,\pi)$. Because of the periodicity, this automatically gives us the values for the function for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin ... | So, $\cos(x-\alpha)=\dfrac5{10}=\cos60^\circ$
$\implies x-\alpha=360^\circ n\pm60^\circ$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that m... | From Euler's identity $$e^{i\pi/2} = cos(\pi/2) + i sin(\pi/2) = i
$$
Take the natural log of both sides.
The left $\ln(e^{i\pi/2}) = i\pi/2$ and the right is simply $\ln i$.
Since both are equal, $i\pi/2 = \ln i.$
So, it is not arbitrary log but ln. Otherwise, your identity is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| Here I am giving Solution of yours first question, Next time when you post question,
plz show your try.
Let $x=a,x=b\;,x=c$ be the roots of the equation $(x-a)(x-b)(x-c) =0$
So $$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0.............(1)$$
Now here $a+b+c=2$ and $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
So we get $ab+bc+ca = -1$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Find all integral solutions for the Diophantine Equations $x^4 - x^2y^2 + y^4 = z^2$ and $x^4 + x^2y^2 + y^4 = z^2$.
Find all integral solutions for the Diophantine Equations $$x^4 - x^2y^2 + y^4 = z^2$$ and $$x^4 + x^2y^2 + y^4 = z^2$$
I basically think that to solve these equations we need to use the fact that all ... | Updated
We can try to find solutions ourself.
Equation 1.
$\quad x^4+x^2y^2+y^4 = z^2.$
Let
$$\gcd(x,y) = m,\quad x=mX,\quad y=mY,\quad z=m^2Z,$$
then
$$X^4+X^2Y^2+Y^2=Z^2,\quad\gcd(X,Y)=1.$$
$$(X^2+XY+Y^2)(X^2-XY+Y^2)=Z^2,\quad\gcd(X,Y)=1.$$
Multipliers in the left part are the same parity. If they are even, then b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ ho... | Substitute $$\frac{x}{y} \ \textrm{for}\ \ a,\quad \frac{y}{z} \ \textrm{for}\ \ b\quad \textrm{and} \quad \frac{z}{x} \ \textrm{for} \ \ c$$ This leaves us with the following and the condition $abc = 1$ is eliminated.
$$
\frac{\frac{x}{y}} {2 + \frac{y}{x}} +
\frac{\frac{y}{z}} {2 + \frac{z}{y}} +
\frac{\frac{z}{x}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$ $a, b, c, d$ are positive reals.
How would I prove the inequality
$$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4$$
I have tried using the rearrangement inequality with $a\leq b\leq c\leq d$
But it does... | $$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}$$
$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Show that $\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$ Show that
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}\\=\frac{\cos A}{1-\sin A}+\frac{\cos B}{1-\sin B} \... | The $LHS$ can be written as $\dfrac {1+\sin A}{\cos A} +\dfrac{1+\sin B}{\cos B}$
$=\dfrac{\cos A+\cos B+\sin A\cos B+\cos A\sin B}{\cos A\cos B}$
Multiply Neumaretor and Denominator by $\sin(A-B)+\cos A-\cos B=(\cos A+\sin A\cos B)-(\cos B+\cos A\sin B)$
$=\dfrac{(\cos A+\sin A\cos B)^2-(\cos B+\cos A\sin B)^2}{\cos A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$ $a,b,c >0$, and $a+b+c=3$, prove
$$ a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$$
I try to substitute $c=3-a-b$ to reduce the number of variables, but cannot further proceed to solve the problem. I made an Excel spreadsheet and test 100 pairs of $(a,b,c)$, it seem... | Hint: for $x>0$ and $y>0$ and $z>0$ we have :
$$x+y+z\geq \left(\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1\right)^{\frac{1}{6}}\geq (5)^{\frac{1}{6}}$$
Method:
If we put $x+y+z=\lambda$
with $x=a^{ab}a$
$y=b^{bc}b$
$z=c^{ac}c$
And $a+b+c=3$
The question is : when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a ... | I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{a+b}{13}$$
Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^4}{8x^3+5}+\frac{1}{8+5x^3}\geq \frac{1+x}{13}$$
Or
$$\frac{5}{13}(x - 1)^2 (x + 1) (x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
"answer_count": 8,
"answer_id": 6
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Below is a nice general way of computing the partial fraction that is a generalization of Heaviside's cover up method to nonlinear denominator factors.
$$\begin{eqnarray} \frac{x^2+3}{(x^2+1)x^6} &=&\ \ \frac{ax+b}{x^2+1}+ \frac{f(x)}{x^6}\quad\text{so clearing denominators yields}\\
x^2+3\ &=&\ \ (ax+b)x^6\! + (x^2+1... | {
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"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{... | Your function will be maximum when the denominator will be minimum.
So, your work will be to calculate $$\frac{d}{dx}\left[\left\{(1-x)-\frac{4}{1-x}+1\right\}^2+1\right]=0$$
It will come down to $$2\left\{(1-x)-\frac{4}{1-x}+1\right\}\left\{-1-\frac{4}{(1-x)^2}\right\}=0$$
This will give rise to $2$ cases.
Case-$1$:
$... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 20... | I am not a reputable source, but I think I can prove the following theorem: $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
PROOF
We will need firstly the following
Lemma 1.
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Proo... | {
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Suppose that $E[X^n] = 3n$. Find $E[e^X]$... Suppose that $E[X^n] = 3n$. Find $E[e^X]$.
Hint from my professor: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +···$
Not quite sure how to solve this problem, wouldn't $e^x$ go on exponentially. Any help is really appreciated.
| Warning: What appears below is at best vacuously true --- i.e. true but uninformatitve, since, as several people have pointed out, there is not actually any probability distribution whose $n$th moment, for every positive integer $n$, is $3n$.
\begin{align}
\operatorname{E}\left(e^X\right) & = \operatorname{E}\left( 1 +... | {
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"source": "stackexchange",
"question_score": "3",
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How to prove this geometry inequality (1) with $2(DF+EF)\ge BC$ There is a picture which hope to illustrate the configuration:
$\triangle ABC$ is such that $\angle A=\dfrac{2\pi}{3}$, and $F$ is the midpoint of $BC$, and $D,E$ lie on $AB,AC$ respectively, such that $DE ||BC$.
Show that:
$$2(DF+EF)\ge BC$$
maybe use cos... | This is not a complete solution, just some simplification.
Square both sides of (2) and using (1) to get rid of $a^2$, we obtain:
$$2\sqrt{c^2+b^2(1-2k)^2+bc(1-2k)}\sqrt{b^2+c^2(1-2k)^2+bc(1-2k)}+((b^2+c^2)(1-2k)^2+2bc(1-2k))\ge bc\tag{3}$$
Let $\kappa=1-2k$, thus $-1\le \kappa \le 1$. Then (3) becomes
$$2\sqrt{c^2+b^2... | {
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Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$ The question
Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$
What I've tried
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac... | Note that $$\prod_{n=2}^{\infty} \left(1-\frac{1}{n^{2}}\right) \to \frac{1}{2}$$
This is because $$A_{n} =\prod_{k=2}^{n}\left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \frac{n+1}{2n} \to \frac{1}{2}$$
We have used $\displaystyle \left(1-\frac{1}{n^4}\right) = \left(1+\frac{1}{n^2}\right) \cdot... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$
Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$.
I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty mu... | Suppose that $x$ is a solution, then:
$ 2^{x} = 3.9^{m} + 5 = 3.9^{m} + 3 + 2$, so:
$$2(2^{x-1} - 1) = 3( 9^{m} + 1) \Rightarrow (3.(9^{m} + 1)) | (2^{x-1} - 1)$$
For $m\geq 2$. But $2^{x-1} - 1$ is odd and $3.(9^{m} + 1)$ is pair.
| {
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"source": "stackexchange",
"question_score": "5",
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Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$
Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get
$$I = \int_{0}^{\frac{... | I'm not sure this is the most straight forward way, but here it is anyway.
Start with the substitution $\sqrt{1-x}\mapsto x $. The integral transforms into
$$I=\int_0^1 \frac{2 x}{2+x+\sqrt{2-x^2}}dx.$$
Next, rationalize the denominator by multiplying the numerator and the denominator by $2+x-\sqrt{2-x^2}.$
This turns ... | {
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"source": "stackexchange",
"question_score": "3",
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Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $... | If $n^2+n+43$ is a square, so it is
$$4n^2+4n+172 = (2n+1)^2 + 171. $$
If $171=a^2-(2n+1)^2$, then $171=(a-2n-1)(a+2n+1).$ Since
$$ 171 = 3^2\cdot 19 $$
the only ways for writing $171$ as a product of two positive integers with the same parity are:
$$ 1\cdot 171,\qquad 3\cdot 57,\qquad 9\cdot 19. $$
For instance, if $... | {
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"url": "https://math.stackexchange.com/questions/1783946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using $\epsilon-\delta$ proof to prove continuity
Use an $\epsilon-\delta$ proof to show that
$f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ , $$f(x) = \frac{3x^2-2x-5}{2x+3}$$
is continuous at $x = -1$
Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ imp... | Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$.
Note that
$$
|f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|.
$$
We are free top choose $\delta$ as we wish, and, in doing so, we will... | {
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with t... | Compare the absolute value of each homogeneously signed sum with the sum containing an equal number of copies of the first term. For example, the all-negative terms from $-1/7$ to $-1/10$ are compared with four copies $-1/7$. The absolute values of the sums then satisfy
$(1/7+1/8+1/8+1/9+1/10)<4/7$
In general the clu... | {
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Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equ... | If $\gcd(x, 25) \neq 1$, then $x$ is a multiple of $5$, so $x^4$ and $x^8$ are multiples of $25$ and don't contribute to the sum. Thus, we will focus on the cases where $\gcd(x, 25)=1$.
Also, $x \equiv x+25 \pmod{25}$, so we really have the following:
$$4(1^4+2^4+3^4+[...]+24^4) \equiv 4(1^8+2^8+3^8+[...]+24^8) \pmod{2... | {
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Thoughts on this limit ? $\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$ $$\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$$
Normally, one multiplies by $\frac{3x}{3x}$ to eliminate the complex fraction. Here, that will not eliminate the problematic denominator term of $(x-3)$. Neither will subt... | $$\lim_{x\to3}\frac{\frac{1}{3}-\frac{1}{x}}{x-3}=\lim_{x\to3}\frac{\frac{x}{3x}-\frac{3}{3x}}{x-3}=\lim_{x\to3}\frac{\frac{x-3}{3x}}{x-3}=\lim_{x\to3}\frac{x-3}{3x(x-3)}=\lim_{x\to3}\frac{1}{3x}=\frac{1}{3\cdot3}=\frac{1}{9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Orthogonal projection question
Consider the (orthogonal) projection $T: \mathbb{R}^3 \to \mathbb{R}^3$ onto the plane $x - y + z = 0$.
*
*(a) Find the standard matrix $[T]_S$ for $T$.
*(b) Find a new basis $B$ so that $[T]_B$ is rather simple.
How do I do this question?
I created an orthogonal basis and p... | Your matrix is wrong. An easy sanity check is that a matrix representing an orthogonal projection should be symmetric with trace equal to the dimension of the space you are projecting onto (in your case, two).
A unit normal vector to your plane is $v = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and ... | {
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Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justi... | As $54\cdot5\equiv90\pmod{180},$
if $\tan5x=\infty,5x=180^\circ n+90^\circ$ where $n$ is any integer
$x=36^\circ n+18^\circ$ where $n\equiv0,\pm1,\pm2\pmod5$
Using $\tan5x$ expansion, the roots of $$5t^4-10t^2+1=0$$ are $\tan(36^\circ n+18^\circ)$ where $n\equiv0,\pm1,2\pmod5$
as $n=2\implies\tan(36^\circ n+18^\circ)... | {
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"timestamp": "2023-03-29T00:00:00",
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Do you write plus/minus if a variable squares equals the square root of a number? For example, if I have $x^2 = \sqrt{49}$. I know that $7$ is the number, but as my final answer, do I write that $x = +\sqrt{7}$ and $-\sqrt{7}$ or just $x = \sqrt{7}$?
| It depends on which field you expect/require the answer to be in:
If the answer is allowed to be a complex number, then the equation $x^2 = \sqrt{49}$ can be rewritten as $x^2 = 7$ OR $x^2 = -7$, which collectively yield four values: $x = +\sqrt{7}$, $x = -\sqrt{7}$, $x = +i\sqrt{7}$, and $x = -i\sqrt{7}$.
If the answe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)
Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.
My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ an... | First of all, the equation in your link is : $(x^2-5x+5)^{x^2+4x-60}=1$.
Simply use $\ln$ in both sides :
$ \ln(x^2-5x+5)^{x^2+4x-60} = \ln1 \Leftrightarrow (x^2 + 4x - 60)\ln (x^2 - 5x + 5) = 0$.
That means that : $x^2 + 4x - 60 = 0$ or $(x^2 - 5x + 5) = 1$.
Of course, the functions inside of $\ln$ must be bigger than... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to show that any rectangle in ellipse must be oriented parallel to axes? A problem which is often given as an exercise for students learning about calculus and finding extrema, is to find maximal possible area of a rectangle inside an ellipse. Such question was asked, for example, here: Find the area of largest rec... | Suppose we have a rectangle which has all four vertices on the ellipse. If $C$ is the center of the rectangle, then all vertices of the rectangle are in the same distance from $C$. If we choose coordinate system with origin in the point $C$, we get the equation
$$x^2+y^2=r^2.$$
We also want to express the condition tha... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding Modular Multiplicative Inverses (Quickly!) as part of an upcoming number theory exam I will need to find the modular multiplicative inverse of every element of ${Z_n}$ (the ones that exist anyway) very quickly. The only way I know is using the Extended Euclidean Algorithm. Is there no way that is faster? The qu... | Solve $ax \equiv 1 \pmod{n}$
I would tackle these problems keeping this trick in mind and applying the basic tactic of making $a$ smaller by 'multiplying to the modulus'.
Example 1: Solve $12x \equiv 1 \pmod{19}$.
$\; 12x \equiv 1 \pmod{19} \; \text{ iff}$
$\; -7x \equiv 1 \pmod{19}$
and
$\;-7 \cdot (3) \equiv -21 \equ... | {
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For a,b in a group G, $|a| = 2$, $b ≠ e$, and $aba = b^2$. What is $|b|$? If it was $aba^{-1}$ = $b^2$ I could use the theorem that if $aba^{-1}$ = $b^n$ then $a^kba^{-k}$ = $b^{n^k}$ so that $b$ = $ebe$ = $a^2ba^{-2}$ = $b^4$ so that $|b|$ = 3. Is there a similar method here?
| Since $a^2=e$, we have $a=a^{-1}$. From the equation $aba=b^2$
\begin{align*}
ba &= ab^2 \text{(composing a on the left)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
ab &= b^2a \text{(composing a on the right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\
ab^2&=b^2ab \text{(composing b on the right... | {
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Prove $\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43$ for $a^2 + b^2 + c^2 + d^2 = 4$
Let $a,b,c,d \geqslant 0$ and $a^2+b^2+c^2+d^2=4$. Prove that
$$\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43.$$
I try some reverse AM-GM techniques but fail. I don'... | Note: in this solution, $\sum$ denotes the cyclic sum, so that $\sum f(a, b, c, d)=f(a, b, c, d)+f(b, c, d, a)+f(c, d, a, b)+f(d, a, b, c)$.
Lemma: $\sum c^4+2\sum abc^2+6\sum ab\le 36$.
Proof: Expanding the inequality $\sum[2(a-b)^2+(ab-1)^2]\ge 0$ gives:
$$6\sum ab\le 20+\sum a^2b^2$$.
Expanding $\sum (ab-ac)^2\ge... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Compa... | Let $x_1=a, x_2=a+d, x_3=a+2d$.
Use Vieta's Formulas
Then $$\begin{cases}a+a+d+a+2d=-3p
\\
a(a+d)+a(a+2d)+(a+d)(a+2d)=3q
\\
a(a+d)(a+2d)=-r
\end{cases}$$
$$\begin{cases}a+d=-p
\\
3a^2+6ad+2d^2=3q
\\
a(a+d)(a+2d)=-r
\end{cases}$$
Then $-2p^2-3pq+r=-2(a+d)^2+(a+d)(3a^2+6ad+2d^2)-a(a+d)(a+2d)=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1... | You can do in this way:
If $\sum_{n=1}^{\infty} \frac{1}{n+2} $ converges, say $\sum_{n=1}^{\infty} \frac{1}{n+2} = S$.
$\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac12+\sum_{n=1}^{\infty} \frac{1}{n+2}=\frac 32+S$,
Which will contradics with $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$
Prove that if $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$.
I am not sure how to prove this statement, but it seems that from $9 \mid 2^b-2^a$ we have $b-a = 6n$. Then what should I do from here to prove the statement?
| If $9\mid(2^b-2^a),$ then, taking $2$ out, we may suppose that $a=0.$ So $9\mid 2^b-1.$
But we see the powers of $2$ are $2,4,-1,-2,-4,1.$ Thus the order of $2$ modulo $9$ is $6.$ Hence $6\mid b.$
Also $2^6\equiv1\pmod7$ by Fermat's little theorem. Therefore $2^b-1\equiv1-1\equiv0\pmod7.$
Hope this helps.
| {
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Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such th... | Let $\gamma$ be the function that takes a number in the cycle of $\beta$ to the number in $\alpha$ in the corresponding position. That is, send $2 \rightarrow 1$, and $3 \rightarrow 2$, and $4 \rightarrow 5$ et cetera.
Intuitively you can think of this as as you temporarily relabelling the numbers in $\beta$ (applying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric s... | The most straightforward way is of course
$$\frac n2(a+\ell))=\frac n2 \big(1+(4n-3)\big)=\color{red}{2n^2-n}\qquad\blacksquare$$
Another method:
The well-known result of the sum of the first $n$ integers:
$$1+2+3+4+\cdots+n=\frac{n(n+1)}2$$
Subtract $1$ from each term ($n$ in total):
$$0+1+2+3+\cdots+(n-1)=\frac{n(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 3
} |
numbers from $1$ to $2046$ We have randomly taken $21$ integers from $1$ to $2046$.
Show that we can take $a$, $b$ and $c$ from the previous $21$ integers in a way such that the following inequality holds
\begin{equation}
bc<2a^2<4bc \end{equation}
My Attempt:
I have found $3$ triplets of such integers, but I do no... | Note that for any three positive integers $a,b,c$, if $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}$, then
$$bc<ac<a\cdot 2^{k+1}=2a\cdot 2^k\leqslant 2a\cdot a=2a^2,$$
and
$$bc>ba\geqslant 2^ka=\frac{1}{2}\cdot 2^{k+1}a>\frac{1}{2}\cdot a\cdot a=\frac{1}{2}a^2.$$
Thus, $2^k\leqslant b<a<c<2^{k+1}$ for som... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$ I would like to evaluate the following integral:
$$
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}
$$
I tried various methods but without success.
| Substitution $x \rightarrow -x$ gives $$\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^{-x}+1)(x^2+1)}=\int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}$$
Therefore
$$
\int_{-1}^{1} \frac{dx}{(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} + \int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}=2\int_{-1}^{1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Finding $\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}$ using substitution $t = \cosh(x)$ $$\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}$$
Hello,
I have to substitute with $$t = \cosh (x)$$
I just don´t know what to do with the e-function.
Thanks.
| Note that $\cosh(x) = \frac{e^x + e^{-x}}{2}$, so $$\frac{e^{x}}{\cosh(x)} = \frac{2 e^{x}}{e^{x}+e^{-x}} = \frac{2}{1+e^{-2x}}$$
Now subsitute $u=e^{-x}$ (which gives $du = -e^{-x}dx = -u dx$) to find: $$\int \frac{2}{1+e^{-2x}} dx = \int \frac{-2}{u(1+u^2)} du$$
The rest is straightforward.
The remainder of the work... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$ $$I=\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$$
Applying partial fractions
$x^2=A(e^x-x^2)+B$
$$I=\int_{0}^{\infty}{e^x\over (e^x-x^2)^2}-{1\over e^x-x^2}dx$$
Let
$$J=\int_{... | \begin{align}
I = \int_0^\infty {\frac{{x^2 }}{{\left( {e^x + x^2 } \right)^2 }}dx} = \int_0^\infty {\frac{{x^2 e^{ - 2x} }}{{\left( {1 + x^2 e^{ - x} } \right)^2 }}dx}
\end{align}
Using the geometric series we have $$ \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \Rightarrow \frac{1}{{\left( {1 - x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$ Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$
Let one of the $x,y,z$ be even number.Let $x=2p$
$x^2+y^2+z^2=x^2y^2$
This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y... | This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is $x^2+y^2+z^2=t^2$ ,∀ t=xy
all integer positive solution given by
$x=(l^2+m^2-n^2)/n$ , $y=2l$, $z=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then $2l(l^2+m^2-n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Asymptotic expansion of ratio function I want to expand the following function:
$$
f(x)=\frac{1}{(1-e^{-x})}
$$
$f(x)$ can be rewritten as
$$
f(x) \sim \frac{1}{x-x^2/2 + x^3/2/3}
$$
But I want to express big-oh notation such that
$$
f(x) = \frac{1}{x} + .... +O(x^2)
$$
up to $x^2$ order.
How to do it?
| $$\frac{1}{1-e^{-x}}=\frac{1}{x-x^2/2+x^3/6+O(x^4)} \\
=\frac{1}{x} \frac{1}{1-x/2+x^2/6+O(x^3)} \\
=\frac{1}{x} (1+x/2-x^2/6+x^2/4+O(x^3))$$
where in the last step we used the identity $\frac{1}{1-y}=\sum_{m=0}^\infty y^m$ whenever $|y|<1$. Note that in this case $y=(-x/2+x^2/6+O(x^3))$ and so the $x^2$ term has a con... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the hypotenuse ( Trigonometric Identities!) If the sides of a right-angled triangle are $\cos 2a + \cos 2b + 2\cos(a+b)$ and $\sin 2a + \sin 2b + 2\sin(a+b)$, find the hypotenuse-
I can simplify this but I end up having a lot of terms in the end :/
| HINT:
Use Prosthaphaeresis Formulas
$$\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[1+\cos(A-B)]$$
$$=4\cos(A+B)\cdot\sin^2\dfrac{A-B}2$$
Similarly, $$\sin2A+\sin2B+2\sin(A+B)=\cdots=4\sin(A+B)\cdot\sin^2\dfrac{A-B}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's ... | By Fermat's theorem $x^6\equiv1\pmod7$ holds for all $x$ coprime to $7$, so $(x^3)^2\equiv1\pmod7$ hence
$$(x^3)^2-1=(x^3-1)(x^3+1)\equiv0\pmod7$$
and hence $x^3\equiv\pm1\pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + ... | I shall solve the system of modulo equations by Chinese Remainder Theorem. I firstly solve the following system:
$$
\left\{\begin{array}{ll}
5 \times 7 \times 11 \times 13 A \equiv 2 & (\bmod 3) \\
3 \times 7 \times 11 \times 13 B \equiv 3 & (\bmod 5) \\
3 \times 5 \times 11 \times 13 C \equiv 5 & (\bmod 7) \\
3 \times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function. Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function.
| $f(x)= \frac{1-\sin x}{1+\sin x} = \frac{(1-\sin x)^2}{(1+\sin x)\times (1-\sin x)} = \frac{1+\sin^2 x-2sinx}{\cos^2 x} = \frac{1}{\cos^2 x}+\tan^2 x -2 \tan x \sec x = sec^2 x+\tan^2 x -2 \tan x \sec x$
Let the even function be $sec^2 x+\tan^2 x $
Let the odd function be $-2 \tan x \sec x $
Because the parity of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?
| Substitute $a=3^x$ and $b=5^x$ to get
$a^2-\frac{34}{15}ab+b^2=0$
Multiply both sides by $15$ to get,
$15a^2-34ab+15b^2=0 \Rightarrow 15a^2-25ab-9ab+15b^2=0 \Rightarrow 5a(3a-5b)-3b(5a-3b)=0 \Rightarrow (3a-5b)(5a-3b)=0$
Now, let $\frac{a}{b}=t$
Dividing both sides by $b^2=5^{2x} \neq 0$, we get
$t=\frac{3}{5}$ or $t=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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$9 \mid a^2 +b^2+ab$. Show that $3$ divides both $a$ and $b$. $a$ and $b$ are integers.
$a^2 +b^2+ab$ is a multiple of $9$.
I have to prove that $3$ divides both $a$ and $b$.
Converse is very easy. Put $a=3k$ and $b=3l$ and that's it.
I was trying factorisation but didn't get anything from it.
| Assuming $a,b$ are distinct,
$9 \mid a^2 +b^2+ab$
$\Rightarrow 9\mid \{(a-b)^2+3ab\} \,\,\,\,\,\,\,\,\, - (1)$
$\Rightarrow 3\mid \{(a-b)^2+3ab\}$
$\Rightarrow 3\mid(a-b)^2$
$\Rightarrow 3\mid(a-b)$
$\Rightarrow 9\mid(a-b)^2 \,\,\,\,\,\,\,\,\, - (2)$
From $(1)$ and $(2)$, we can conclude that
$9\mid3ab \Rightarro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\varepsilon - \delta$ proof that $f(x) = x^2 - 2$ is continuous - question concerning the initial choice of $\delta$ I just realized I did non really internalized $\varepsilon - \delta$ proofs. Here there is an attempt, with some general questions I have.
Proposition: $f(x) := x^2 - 2$ is continuous.
Skratch Work... | Your proof didn't cover the case $c < 0$. I would do it as follows:Let $c \in \mathbb{R}, \epsilon > 0$ be arbitrary, choose $\delta = \min \{1,\frac{\epsilon}{1+2|c|}\}$. Thus if $|x-c| < \delta $, then
$$|x-c||x+c| \leq |x-c|(|x-c| + 2|c|)< |x-c|(1+2|c|)< \dfrac{\epsilon}{1+2|c|}\cdot (1+2|c|) = \epsilon .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$ for $x,y,z \geq 0$ This inequality is wrong - see the accepted answer (it appears there is no general inequality for these two expressions).
On the left we have harmonic mean of pairwise ge... | Your inequality is wrong! For $y=z=1$ and $x\rightarrow+\infty$ we obtain:
$3\leq\frac{2}{3}\left(1+\frac{1}{2}+1\right)$, which is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials I have used the formula
\begin{align}p_0(x)&=1\\
p_1(x)&=x\\
p_2(x)&=\frac12(3x^2−1)\\
p_3(x)&=\frac12(5x^3−3x)
\end{align}
$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$
EDIT-
\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \... | Continuing from your steps, we compare the coefficients of each term on the L.H.S and R.H.S.
$$2x^3-x^2-3x+2=\dfrac A2(5x^3−3x) + \dfrac B2(3x^2−1) + Cx +D$$
So, $$ 2x^3-x^2-3x+2=\dfrac{5}{2}Ax^3+\dfrac{3}{2}Bx^2+(C-\dfrac{3}{2}A)x+(D-\dfrac{B}{2})$$
*
*For the $x^3$ term, we have
$\dfrac{5}{2}A=2 \Rightarrow A=\dfr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
| Hint: Use the fact that ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$. Since $n$ is even, let $n = 2p$ for some $p$. Then we have:
$$\sum_{r\ \text{ odd}}{n \choose r} = \sum_{i=1}^p{2p \choose 2i-1} = \sum_{i=1}^p{2p-1 \choose 2i-2}+{2p-1 \choose 2i-1} = \sum_{i=0}^{2p-1}{2p-1 \choose i}$$
This may be easier t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$ And thanks in advance for your answers. Sorry if the text is badly formatted, I'm new here. Anyway, here is the question:
Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z... | Yes, you can solve your equation by completing the squares.
$$\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v\tag{*1}$$
For each variable, subtract its appearance in LHS from corresponding term in RHS, you have:
$$
\begin{array}{rl}
x - \sqrt{x} &= (\sqrt{x}-\frac12)^2 - \frac14\\
y - \sqrt{y} &= (\sqrt{y}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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New Maths 9-1 GCSE for 2017 Sample Question My teacher gave me some practice questions for my end of year exam which will be like the new GCSE and this question is very tackling to me. Could any with clear working solve the question and show me how they did it:
A(-2,1), B(6,5), and C(4,k) are the vertices of a right an... | Since $\angle ABC$ is the right angle, according to Pythagorean theorem, we have $AB^2+BC^2=AC^2$
$AB^2=(6-(-2))^2+(5-1)^2=80$
$BC^2=(k-5)^2+(4-6)^2=(k-5)^2+4$
$AC^2=(k-1)^2+(4-(-2))^2=(k-1)^2+36$
So we have $80+(k-5)^2+4=(k-1)^2+36 \Rightarrow k=9$
The line passing through A and C has the form $y=\alpha x+\beta$
Calcu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to express $\binom{a+b}{n}$ as a sum of regular coefficients I am trying to prove that $(1 + f(x))^a(1 + f(x))^b = (1 + f(x))^{a+b}$ in the world of formal power series. At a certain point in the prove I get
\begin{align*}
(1 + f(x))^a(1+f(x))^b& = \ldots \\
& = \lim\limits_{N \to \infty} \left(\left[\sum_{n=0}... | The development chain goes like this:
$$
\eqalign{
& \left( {1 + f(x)} \right)^a \left( {1 + f(x)} \right)^b = \sum\limits_k {\left( \matrix{
a \cr
k \cr} \right)f(x)^k } \sum\limits_j {\left( \matrix{
b \cr
j \cr} \right)f(x)^j } \cr
& = \sum\limits_j {\sum\limits_k {\left( \matrix{
a \cr
k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Show $\frac{\pi^2}{8}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$ Knowing $$\frac{\pi^2}{sin^2(\pi z)}=\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^2}$$ how can I prove $$\frac{\pi^2}{8}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\qquad?$$
| We know that,
$\sum_{n=0}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$
We also know,
$\sum_{n=0}^{\infty}\frac{1}{n^2}=\sum_{n=0}^{\infty}\frac{1}{(2n)^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$.
Now,$\sum_{n=0}^{\infty}\frac{1}{(2n)^2}=\sum_{n=0}^{\infty}\frac{1}{4n^2}=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{n^2}=\frac{1}{4}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction decomposition of $\pi\cdot \tan(\pi z)$
Evaluate the partial fraction decomposition of $\pi \tan(\pi z)$
$$2\pi \tan(\pi z)=\cot\left(\frac{\pi}{2}-\pi z\right)-\cot\left(\frac{\pi}{2}+\pi z\right)$$ $$=\frac{2}{1-2z}+\sum_{k=1}^\infty \frac{1-2z}{(\frac{1}{2}-z)^2-k^2}-\frac{2}{1+2z}-\sum_{k=1}^\in... | Hint. If you know that, for $0<z<1/2$,
$$
\pi \cot (\pi z)=\sum_{k=1}^\infty \left(\frac{1}{z-k}+\frac{1}{z+k-1}\right)
$$ then by putting $z \to 1/2-z$ you get
$$
\begin{align}
\pi \cot (\pi (1/2-z))&=\pi \tan(\pi z)
\\\\&=\sum_{k=1}^\infty \left(\frac{1}{1/2-z-k}+\frac{1}{1/2-z+k-1}\right)
\\\\&=\sum_{k=1}^\infty\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
| Alternative solution:
Recall the known inequality $a^2b+b^2c+c^2a \le \frac{4}{27}(a+b+c)^3 - abc$ for $a, b, c \ge 0$.
See: https://artofproblemsolving.com/community/c6h478168p2677448,
https://artofproblemsolving.com/community/c4h1705936p10979968,
https://artofproblemsolving.com/community/c6h479531p2685016
It suffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof ... | We can also proceed as follows:
\begin{align*}
a^2+b^2+c^2 &\ge \frac{(a+b+c)^2}{3}\\
&= (a+b+c)\,\frac{a+b+c}{3}\\
&\ge (a+b+c) \sqrt[3]{abc}\\
&=a+b+c.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern Consider the sequence:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$
$$\lim_{n \to \infty} a_n=\lim_... | Geometrically:
$(a_{n+1}, b_{n+1})= \frac 12 (a_n,b_n) + \frac 12 \sqrt{a_n^2+b_n^2} (\cos \pi/4, \sin \pi/4)$
So we draw a line between $(a_n,b_n)$ and connect it to a point equidistant from the origin on the line $x = y$, and find the mid-point.
$r_n = \sqrt{a_n^2 + b_n^2}\\
\phi_n = \tan^{-1}(\frac{b_n}{a_n})$
$\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How do I solve for a matrix in this linear algebra expression? In the problem I am working on I have an expression that looks like
$$B^T\operatorname{diag}(Bx) y$$
where $B$ is a known $m \times n$ matrix, $x$ is an unknown $n \times 1$ vector of variables, and $y$ is a known $m \times 1$ vector, all of real numbers.
T... | Note that $x \mapsto B^T \operatorname{diag}(Bx)y$ is a linear transformation on $x$. To find the matrix of this transformation (that is, the matrix $P$ for which $Px = B^T \operatorname{diag}(Bx)y$), we can see what happens to the standard basis vectors in order to get the columns of $P$.
Long story short: let $B_k$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recursively counting divisors of a number I want to make a recursive function f that counts all (not only prime) different divisors of a given natural number:
$f(n): = |{a ∈ ℕ | ∃ b ∈ ℕ : a . b = n }| $ ; with $ f(0)=0 $
for example $ f(3) = 2$, $f(6) = 4$, $f(16) = 5 $ etc.
Theoretically, how could I do that?
Than... | You need to use the Möbius function $\mu(n)$:
Recurrence:
Let $z=1$
$$T(n, k) = \text{ if } k = 1 \text{ then } 1 - \sum_{i=2}^{i=n} \mu(i) \frac{T(n, i)}{i^{(s - 1)}} \text{ else }
\text{ if } \text{mod(n, k)} = 0 \text{ then } T(n/k, 1) \text{ else } 0 \text{ else } 0.$$
Then:
$$\lim\limits_{s \rightarrow z} T(n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can thi... | $$\sqrt { 7+4\sqrt { 3 } } =\sqrt { 7+2\cdot 2\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } \right) }^{ 2 }+{ 2 }^{ 2 }+4\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } +2 \right) }^{ 2 } } =\sqrt { 3 } +2\\ \\ $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 0
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Find all positive integer roots of $\,5xy=19x+96y$
Find all positive integer roots of : $5xy=19x+96y$
I tried using decomposition technique but no success...,it seems suitable factorization of this equation is IMPOSSIBLE!!
Handy calculations show that it has as many as 6 answers.
| We have $19x=(5x-96)y$, so $y=0\bmod19$ or $x=4\bmod19$.
If $y=19n$, then we have $\frac{96}{x}=5-\frac{1}{n}$. So $x\le24$, and $x(5n-1)=96n\ (*)$. Checking $n=1,2,3,4,5$ we find that $n=1$ works giving solution $(24,19)$. $n=2,3,4$ do not work. $n=5$ gives the solution $(20,95)$. If $n>5$ then since $5n-1,n$ are copr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Evaluate $\int \frac{1}{3+4\tan x} \, dx$ I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$
I tried to use the universal trig substitution, that is
$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$
and after manipulation I got the integral to be
$$\int \frac{2(1 - t^... | For sure, as shown in comments and answers, the tangent half-angle substitution is not the easiest way for this problem. However, computing $$I=\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt$$ is doable noticing that the roots of the denominator are $-\frac 13$, $i$,$-i$ and $3$. Using it, by partial fraction decom... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Upper bound for $\gcd(a,b)$ if $\frac{a+1}{b}+\frac{b+1}{a}\in\Bbb{N}$
Suppose that $a,b$ are two positive integers so that $\frac{a+1}{b}+\frac{b+1}{a}$ is also a positive integer.Find the best upper bound for $\gcd(a,b)$.
My work:
$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a(a+1)+b(b+1)}{ab} \in \Bbb{N} \implies ab|a^2+b... | For the equation. $$\frac{m+1}{n}+\frac{n+1}{m}=a$$
If are solutions of the equation Pell. $p^2-(a^2-4)s^2=1$
Then the formula is:
$$n=2(p-(a+2)s)s$$
$$m=-2(p+(a+2)s)s$$
And another solution:
$$n=\frac{2p(p+(a-2)s)}{a-2}$$
$$m=\frac{2p(p-(a-2)s)}{a-2}$$
If are solutions of the equation Pell. $p^2-(a^2-4)s^2=4$
Then t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve this System of Equations? How do I begin to solve this system? $$x^2=y+a$$$$y^2=z+a$$$$z^2=x+a$$
Do I take the square roots of $x,y$ and $z$? If so, we get $$x=\pm\sqrt{a+y}$$$$z=\pm\sqrt{a\pm\sqrt{a+y}}$$$$y=\pm\sqrt{a\pm\sqrt{a\pm\sqrt{a+y}}}$$
What do I do now? I'm confused as to what to do.
| Proceeding from Sonnhard's solution, besides $z = (1 \pm \sqrt{1+4a})/2$ from the factor $-z^2 + z + a$, we get the two cubics
$$z^3 + \frac{1\pm r}{2} z^2 + \frac{\pm r - 2a - 1}{2} z + \frac{a \mp ra -2}{2}$$
where $r = \sqrt{4a-7}$. If you're interested in real solutions, you need $
a \ge -1/4$ for the first pair ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distance from a set to a point There is this exercise I cannot understand well. It asks me for the distance between this set in $\mathbb{R}^3$
$$U = \{(x, y, z)\ |\ ax + y - 2z = 0, z = 0 \}$$
and the point $(0, b, 1)$.
Also it tells me to say "what is that".
Any hint or help?
| This is a least-squares problem. Note that $U$ contains the origin and is parametrized by
$$\left\{ \gamma \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} : \gamma \in \mathbb R \right\}$$
The projection matrix that projects onto $U$ is
$$P := \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} \left(\begin{bmatrix} 1\\ -a\\ 0\end{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 ... | I do wonder why it took so long... I added all pairs from the first ten squares on a piece of scrap paper in a few minutes. Maybe that's a long time.
\begin{array}{c|c|c}
&& +1 & +4 & +9 & +16 & +25 & +36 & +49 & +64 & +81 & +100 \\ \hline
1 && 2 \\ \hline
4 && 5 & 8\\ \hline
9 && 10 & 13 & 18 \\ \hline
16 && 17 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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How to find specific variables that cause vectors to be linearly independent / dependent The vectors
$v= \begin{bmatrix}
5\\
2\\
7\\
\end{bmatrix}, u = \begin{bmatrix}
4\\
4\\
13+k\\
\end{bmatrix}, \text{and } w = \begin{bmatrix}
-4\\
-2\\
-6\\
\end{bmatrix} $
are linearly independent if and only if $k \neq$ what?
How ... | As Fly by Night mentioned, these vectors will be linearly dependent if and only if $$\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} = 0.$$
\begin{align*}
\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} &= 5[(4)(-6) - (13+k)(-2)] - 4[(2)(-6) - (7)(-2)] + (-4)[(2)(13... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A common term for $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \text{if } n\ \text{ is odd. } \end{cases}$ When I was answering a question here, I found a sequence as a recursive one as given below.
$a_1=1$, and for $n>1$,
$$a_n=\begin{cases}
2a_{n-1} & \text{if } n\ \text{ is even, }\... | Note that in binary we have
$$\begin{align*}
a_1&=1\\
a_2&=10\\
a_3&=101\\
a_4&=1010\\
a_5&=10101\;,
\end{align*}$$
displaying a pattern easily shown by induction to be real. Now note that the binary expansion of $\frac23$ is $\frac23=0.\overline{10}_{\text{two}}$, so that
$$\begin{align*}
2\cdot\frac23&=1.\overline{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $\ln(1+x) \approx A+Bx+Cx^2$, differentiate twice both sides and show that $\ln(1+x) \approx x-\frac{1}{2}x^2$
Question: $\ln(1+x) \approx A+Bx+Cx^2$, for $-1<x\leq1$, where $A,B,C$ are constants.
Differentiate twice both sides of the approximation above and hence show that $$ \ln(1+x) \approx x-\frac{1}{2}x^2$$
U... | \begin{align}
\ln(1+x) & = A + Bx +Cx^2. & & \text{When } x = 0 \text{ then } \ln 1 = A. \\
& & & \text{So } A=\ln 1 = 0. \\[10pt]
\frac 1 {1+x} & = B + 2Cx. & & \text{When } x=0 \text{ then } \frac 1 {1+0} = B. \\
& & & \text{So } B = 1. \\[10pt]
\frac {-1}{(x+1)^2} & = 2C. & & \text{When } x=0 \text{ then } \frac{-1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all pairs such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$
Find all pairs $(x, y)$ of positive integers such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$. If there are too many to write, write a generic form.
I was thinking of rewriting the divisibility in a simpler way such as $xy^2 + y + 7\mid x... | Hint:
$$\begin{array}{ccc}xy^2+y+7&|&x^2y+x+y\\
xy^2+y+7&|&x^2y^2+xy+y^2\\
xy^2+y+7&|&y^2-7x
\end{array}$$
Compare $y^2-7x$ with $xy^2+y+7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\f... | The gradient of $f(x)=\dfrac{x^2-1}{x+3}$ near $x=1$ is $$f'(x)=1-\dfrac{8}{(x+3)^2},$$which is positive in the range $0.9<x<1.1$; so, in this range, $|f(x)|$ ranges from $0$ (at $x=1$) to $\max\{|f(0.9)|,|f(1.1)|\}=0.0512...<1/13.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displa... | I think your method is efficient and pretty neat.
Continuing from where you left.
Observe that
$$\dfrac{b-c}{a}\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)=\dfrac{b-c}{a}\left(\dfrac{b(a-b)+c(c-a)}{(a-b)(c-a)}\right)=\dfrac{(b-c)^2 \cdot (a-b-c)}{a(a-b)(c-a)}$$
Now, using $a+b+c=0$ we get,
$$\dfrac{b-c}{a}\left(\dfrac{b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the least-squares solution unique? I am looking for a line closest to $(-5, -2)$, $(-2, 0)$, $(-1, 0)$, $(2, 3)$, $(5, 4)$ using the least square solution. So I set the line as $$ax+by+c=0$$ let $a=1$ (where $a$ is not $0$ obviously) and got
$$\begin{pmatrix}
-2 & 1 \\
0 & 1 \\
0 & 1 \\
3 & 1 \\
4 & 1\\
\end{pmatri... | By inspection, no the least squares is not unique. The nullspace $\mathcal{N}\left( \mathbf{A}^{*} \right)$ is not trivial. The column space has dimension $m=5$, and we only have 2 linearly independent vectors.
To continue the problem, a guess was made about the notation and true problem. Start with the $m=5$ data poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Rearrange the series $ \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ to converge to $1$. I have studied the Riemann's theorem about rearrangement of conditionally convergent series. Also I have seen other rearrangements of the given series on this site that converge to different sums $\ln2,\;\frac{3}{2}\ln2 $, etc. But Iam ... | The positive terms are $$ 1, \frac 1 3, \frac 1 5, \frac 1 7, \frac 1 9, \frac 1 {11}, \frac 1 {13}, \ldots $$ Their sum diverges to $+\infty$. The negative terms are $-1$ multiplied by $$ \frac 1 2, \frac 1 4, \frac 1 6, \frac 1 8, \frac 1 {10}, \frac 1 {12}, \ldots $$ Their sum diverges to $-\infty$.
$1 + \dfrac 1 3... | {
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How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative? No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative.
This is what I have tried:
$(a * b) * c = a * (b * c) [Associativity];
$(a^2 - ab + b^2) * c... | $$1 * (2 * 3)=1 * (2^2-2*3+3^2)=1 * 7=1^2-1*7+7^2=43$$
$$(1 * 2) * 3=(1^2-1*2+2^2) * 3=3 * 3=3^2-3*3+3^2=9$$
Therefore, $*$ is not associative. Always remember to guess and check values of $a,b,c$ to see if an operation is associative before trying to prove it!
Now, to show you why this is not associative, let's expan... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and ... | $$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$
Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$.
$4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say
$$\sqrt{4a^2} =
\begin{cases}
2a, & \text{if $a \ge 0$} \\
-2a, ... | {
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The double ${-6}$, from ${|3x + 7| = 11}$ Solve for $x$ such that $|3x + 7| = 11$.
Answer. "${x = \frac {4}{3}, -6, -6}$". First rewrite the absolute value equation as two separate linear equations. In the first equation, assume that the ${3 + 7}$ is positive and set it equal to 11. In the second equation, also equal t... | ${3x + 7 = 11}$
$-(3x + 7) = 11$
$3x + 7 = -11$
$3x = -18$
$x = -6$
| {
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"url": "https://math.stackexchange.com/questions/1845060",
"timestamp": "2023-03-29T00:00:00",
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Induction Sum of all odd Numbers Show that
$\sum_{k=1}^{n}(2k-1)=n^2$
Beginning: n=1
$\sum_{k=1}^{1}(2k-1)=(2*1-1)=1=1^2$
Let $\sum_{k=1}^{n}(2k-1)=n^2$ be true, then for n=p+1
$\sum_{k=1}^{p+1}(2k-1)=(p+1)^2$ has to be true too.
$\sum_{k=1}^{p+1}(2k-1)=1+3+...+(2(p+1)-3)+(2(p+1)-1)$
$=1+3+...+(2p-1)+(2p+1)$
$=1+3+...+... | Hint:
$1$ $ +3+5+\cdots+2k-1=x$
$2k-1+2k-3++\cdots+3+1=x$
$2k+2k+\ldots+2k=2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. Here it goes:
We want to evaluate $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$.
Here's my work:
Let $z=e^{i\theta}$ so that $d\theta=dz/iz$ and ... | Edit
$$I(a)=\int_{0}^{2\pi}\frac{1}{a+\cos x}dx=\int_{-\pi}^{\pi}\frac{1}{a-\cos x}dx=2\int_{0}^{\pi}\frac{1}{a-\cos x}dx$$
We know $\cos x=\large \frac{1-\tan^2\frac x 2}{{1-\tan^2\frac x 2}}$,thus
$$I(a)=2\int_{0}^{\pi}\frac{{1+\tan^2\frac x 2}}{(a-1)+(a+1)\tan^2\frac{x}{2}}dx=\frac{2}{(a+1)}\int_{0}^{\pi}\frac{1+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
On the proof that $\sum\limits_{k=0}^{n-1}\frac {a^k}{(1+a^k x) (1+ a^{k+1}x)}=\frac 1 {1-a} \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$ Question:-
Find the sum to $n$ terms of the following series
$$\frac{1}{(1+x) (1+ax)} + \frac{a}{(1+ax) (1+a^2 x)} + \frac{a^2}{(1+a^2 x) (1+a^3 x)} + \cdots$$
My solution:-... | Since the general term is $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)} \ ,\qquad j=0,1,2,\dots,$$
we can decompose it into partial fractions as follows (so as to using finite telescoping series). Suppose $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)}=\frac{A}{1+a^jx}+\frac{B}{1+a^{n+1}x},$$
then $$A=\frac{a^j}{1-a^{j+1}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| We have
$$\frac{1}{\cos (\theta)} - \cos (\theta) = n \qquad \qquad \qquad \qquad \sin (\theta) + \frac{1}{\sin(\theta)} = m$$
Let $x := \cos (\theta)$ and $y := \sin(\theta)$. Hence, $x^2 + y^2 = 1$. The two equations above can be rewritten as
$$\frac{1}{x} - x = n \qquad \qquad \qquad \qquad y + \frac{1}{y} = m$$
or,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Show that $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$
Show that the improper integral $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$, for ... | The improper integral can be evaluated as the limit of an integral over $[1/n,1]$ as $n \to \infty.$
Making the change of variables $u = 1/x,$ we get
$$\alpha\int_{1/n}^1 \left\lfloor\frac{1}{x}\right\rfloor dx = \alpha\int_{1}^{n } \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1}\int_{k}^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Is difference of two consecutive sums of consecutive integers (of the same length) always square? I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks.
If $x$ is th... | If you start at 1, then yes.
The sum of the first $n$ integers is $\frac{n(n+1)}{2}$.
Note that the next $n$ integers are simply $n^2 + \frac{n(n+1)}{2}$. That is, you can remove $n$ from each of the subsequent n integers, and just wind up with the sum of the first $n$ integers again.
For example:
$y = 5$
$x = 1 + 2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 9,
"answer_id": 4
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Area of a square whose one part is in circle A square has two of its vertices on a circle and the other two on a tangent to the circle. If the diameter of the circle is $10$ cm, then what is the area of the square is?
My solution:
I figured out this diagram
What to do after this ?
| Observe the diagram below:
Let $x =$ the length of one side of the square.
Since the diameter of the circle is $10 \ \text{cm}$, we know that $OA=OC=OD=5 \ \text{cm}$, since they are all radii of circle $O$. Thus, $OB = x-5$. We also know that $AB = \frac{x}{2}$. Using the Pythagorean theorem:
$$
\begin{align*}
(AB)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
| As per Parth Kohli's comment we can write $$f\left(x - \frac{1}{x}\right) = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right).$$
That is, $f(x) = x^3 + 3x$ so $f(-x) = (-x)^3 + 3(-x) = -(x^3 + 3x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Fractions in Questions and Answers
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