Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to approach this divisibility problem on polynomial Prove that the polynomial $x^{9999} + x^{8888} + x^{7777} + ... + x^{1111} + 1$ is divisible by $x^9 + x^8 + x^7 + ... + x + 1$.
I have no idea how to approach this problem, some help would be appreciated.
| $$ ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} \right) = x^{9999} - x^{9989}$$
So,
$$ x^{9999} - x^{9979} = x^{9999} - x^{9989} + x^{9989} - x^{9979} = ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} + x^{9980} - x^{9979} \right) $$
Continuing in this way, we see that:
$$ x^9 + x^8 + \ldots + x + ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function
Convert trigonometric function into algebraic function.
$$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$
My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma... | If $\cos^{-1}y=u,\cos u=y,0\le y\le\pi$
If $\sin(2\cos^{-1}y)=\sin2u=2\sin u\cos u=\begin{cases}2u\sqrt{1-u^2} &\mbox{if }0\le y\le\dfrac\pi2 \\
-2u\sqrt{1-u^2} & \mbox{if } y>\dfrac\pi2 \end{cases} $
Here $u=\dfrac{1-x^2}{2x}$
For real $\cos^{-1}y,-1\le\cot(2\tan^{-1}x)\le1$
$\iff\dfrac\pi4\le2\tan^{-1}x\le\dfrac{3\p... | {
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"url": "https://math.stackexchange.com/questions/3213326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$? I need to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$. I have changed the form of the function, but for some reason I'm still not getting the right answer.
My Steps:
$$\frac{x^3+64}{x+4}$$
apply $(a+b)^3=(a+b)(a^2+2ab+b^2)$ formula
$$=\frac{(x+4)(x^2+8x+16)}{x+4}$$
$$=x... | There is no value $f(-4)$ since $-4$ is not in the domain of the function by mere fact of the division by zero that results. Even if the expression simplifies, it doesn't change the removable discontinuity being present.
In any event, if you want to find an almost-equivalent expression for $f$ that gets rid of the disc... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Differentiating $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
Find the derivative with respect to $x$ of
$$\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$$
The partial solution to this problem is given as follows:
$y=\left(\frac{x}{\sqrt{1-x^2}}\right)$
Then:
$\frac{dy}{dx} = (1-x^2)^{-3/2}$
Can you show the step... | Continuing to work through the solution provided I have arrived at an answer. It is as follows:
$\frac{d}{dx} \frac{x}{\sqrt(1-x^2)}$
$\frac{d}{dx} x(1-x^2)^{-1/2}$
Using the Product Rule (u' * v) + (v' * u) where u = $x$ and v = $(1-x^2)^{-1/2}$ and combining with the Chain Rule for the derivation of v' yields the fo... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}... | Note that
$$\begin{align}
\int \frac {\sin (2x)}{(\sin(x)+\cos(x))^2} dx
&=\int \frac {2\sin(x)\cos(x)}{\cos^2(x)(1+\tan(x))^2} dx\\
&=\int\frac {2\tan(x)}{(1+\tan(x))^2} dx\\
&=\int\left(1-\frac {1+\tan^2(x)}{(1+\tan(x))^2}\right) dx\\
&=x+\frac{1}{1+\tan(x)}+c.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Solve $\frac {dy}{dx} + y\tan x=y^3 \sec x$ Solve
$$\dfrac {dy}{dx}+y\tan x= y^3 \sec x$$
My Attempt:
$$\dfrac {dy}{dx} + y\tan x= y^3 \sec x$$
Dividing both sides $y^3$,
$$y^{-3} \dfrac {dy}{dx} + y^{-2} \tan x= \sec x$$
Put $y^{-2} = z$
$$(-2)y^{-2} \dfrac {dy}{dx} = \dfrac {dz}{dx}$$
$$y^{-3} \dfrac {dy}{dx} = \df... | $$\frac{y'}{y^3}+\frac{\tan x}{y^2}=\frac{1}{\cos x}$$
$$ z'-2z\tan x=-\frac{2}{\cos x}$$
The integrating factor would be now:
$$e^{-2\int \tan x}dx=e^{\ln(\cos^2 x)+C}$$
So multiply both sides by $\cos^2 x$
$$(z\cos^2 x)'=-2\cos x\Rightarrow z\cos^2 x=-2\sin x+C$$
$$y=\sqrt{\frac{\cos^2 x}{-2\sin x+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ab+ac+bc=9$ and $a,b,c \geq 1,$ what is the maximum value of $a^2+b^2+c^2?$ If $ab+ac+bc=9$ and $a,b,c \geq 1,$ could anyone advise me how to find the maximum value of $a^2+b^2+c^2?$
I have shown that $(a+b+c)^2=a^2+b^2+c^2+18,$ so does it suffice to maximise $a+b+c ?$
Thank you.
| $$\sum_{cyc}(a-1)(b-1)\geq0$$ gives
$$a+b+c\leq6$$ and by your work we are done!
The equality occurs for $(a,b,c)=(1,1,4).$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find a limit for $\frac{({1+x})^\frac{1}{x} - e}{x}$ as x tends to 0 I have done it thus far:
$$\lim_{x \to 0}\frac{{(x+1)}^\frac{1}{x}-e}{x} = \bigg[\frac{0}{0}\bigg] = \frac{((x+1)^\frac{1}{x}-e)'}{x'}=({x+1})^\frac{1}{x} \cdot \left(\frac{\ln(x+1)}{x}\right)' = \\({x+1})^\frac{1}{x} \cdot \frac{\left(\frac{x}{x+1}-(... | $$\begin{align}
\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)
&=\lim_{x\to0}\left(\frac{x-(x+1)(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))}{x^2(x+1)}\right)\\
&=\lim_{x\to0}\left(\frac{-x^2+\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{3}+O(x^4)}{x^3+x^2}\right)\\
&=\lim_{x\to0}\left(\frac{-\frac{x^2}{2}+\frac{x^3}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of six digit numbers divisible by $3$ but none of the digits is $3$ Find number of six digit numbers divisible by $3$ but none of the digits is $3$
My try:
Let the six digits are $a,b,c,d,e,f$ such that
$$a+b+c+d+e+f=3p$$
where $1 \le p \le 18$
Now since $a \ge 1$ we have by Stars and Bars Technique number of so... | The digits we can use to form these numbers are $$\{0, 1,2,4,5,6,7,8,9\}\equiv \{0, 1, 2,1, 2,0,1,2,0\}\pmod3$$
Let $S$ be the sum of the last five digits,
$$
\begin{array}{c|lcr}
S\bmod3 & \color{blue}0 & \color{red}1 & \color{green}2 \\
\text{Probability} & 1/3 & 1/3 & 1/3 \\
\end{array}
$$
Observe that the first dig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $CD\parallel AB$ with square $FEBD$ As the figure shows, $FEBD$ is a squre. $AE=GE$, $FA=FB$ and $CD=BA$. Show that $CD\parallel AB$.
I have found that $\triangle BGD$ must be an equilateral triangle, but I have no proof yet. Please help. It's better to offer a synthetical solution without much computation. ... | Let $FEBD$ be the square with vertices $F=(-1,0)$, $E=(0,1)$, $B=(1,0)$, $D=(0,-1)$ and $A=(a,b)$ a point such that $AF = BF$.
$AF=BF$
By the formula of distance between two points
$\begin{array}{} (a+1)^{2}+b^2=4 & ⇒ & b^{2}=4-(a+1)^{2} & (I) \end{array}$
$\begin{array}{} CD∥AB & ⇒ & m_{CD}=m_{AB} \end{array}$
$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x... | Note that from the expansions $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ and $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have
$$\begin{align}
\cos(\sin(x))&=\cos\left(x-\frac{x^3}{6}+O(x^5)\right)\\\\
&=1-\frac12\left(x-\frac{x^3}{6}+O(x^5)\right)^2+\frac{1}{24}\left(x-\frac{x^3}{6}+O(x^5)\right)^4+O(x^6)\\\\
&=1-\frac12x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x... | $$\int\frac{x^3}{1+x^2}dx = \int\left(\frac{x^2}{2(1+x^2)}\right)2xdx$$
Using $u = x^2 \implies du = 2xdx$
$$ = \frac{1}{2}\int\left(1 - \frac{1}{1+u}\right)du = \frac{1}{2}(u - \ln(1+u)) = \boxed{\frac{1}{2}(x^2 - \ln(1+x^2))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Continuity and differentiability of $f(x) = \frac{x}{1+x} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)}+...$ Here's the given function:
$$f(x) = \dfrac{x}{1+x} + \dfrac{x}{(x+1)(2x+1)} + \dfrac{x}{(2x+1)(3x+1)}+...$$
My question regarding this function: is this function considered a continuous and a differenciable... | Note that all the terms are $0$ when $x=0$!. Your calculation is correct if $x \neq 0$ and $x \neq -\frac 1 n$ for any $n$. Continuity and differentabilty are subject to intepretation. One interpretation is that since $f$ is not defined in any open interval containing $0$ it is not continuous, hence not differntiable, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225464",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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To find the sum of the given infinite series This is a question from the book 'Differential Calculus' by Joseph Edwards.
Prove that if $x$ be less than unity
$$\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} \ldots = \frac{1+2x}{1+x+x^2}$$
As the numerators are negative derivative of the ... | Start with the equality
$$
\frac{1+x}{1+x^3}\cdot \frac{1+x^2}{1+x^6}\cdot \frac{1+x^4}{1+x^{12} }\cdots=\frac{(1-x)^{-1}}{(1-x^3)^{-1}}\tag 1
$$
To understand the numerator, note that you can prove by induction that
$$
\prod_{i=0}^{n-1}(1+x^{2^i})=\sum_{i=0}^{2^n-1}x^i
$$
Letting $n\to\infty$, you get
$$
\prod_{i=0}^\... | {
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Primes number $n,n+2,n+6,n+8,n+12,n+14$
Find all natural number $n$ such that all the following numbers are primes :
$$n,\;\; n+2,\;\;n+6,\;\;n+8,\;\;n+12,\;\;n+14$$
are all prime numbers
| As $n$ is a prime, $n\ge 2$.
When $n=2$, $n+2=4$ is not prime.
When $n=3$, $n+6=9$ is not prime.
When $n=5$, all of them are prime.
$(n+6)-n\equiv 1$ (mod $5$)
$(n+2)-n\equiv 2$ (mod $5$)
$(n+8)-n\equiv 3$ (mod $5$)
$(n+14)-n\equiv 4$ (mod $5$)
At least one of these integers is a multiple of $5$.
When $n>5$, at least ... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof of $(px+1)^{p^{k}}=(p^{k+1}x+1)$ mod $p^{k+2}$ for any odd prime $p$ I need to prove this identity using only simple arithmetics and combinatorics (or to find a counter-example):
$(px+1)^{p^{k}}=(p^{k+1}x+1)$ mod $p^{k+2}$, which I verified using several values of $p$ and $k$ using GAP.
I tried the following (us... | Lemma 1: For $k>0,$ if $x\equiv y\pmod{p^k}$ then $x^p\equiv y^p\pmod{p^{k+1}}.$
Proof: If $x\equiv y\pmod {p^k}$ then $x^{p-1}+x^{p-2}y+\cdots+xy^{p-2} + y^{p-1}\equiv px^{p-1}\pmod{p^k}$ so $$x^p-y^p=(x-y)\left(x^{p-1}+x^{p-2}y+\cdots+xy^{p-2} + y^{p-1}\right)$$ is divisible by $p^{k+1}$ and hence $x^p\equiv y^{p}\pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do.
I would really appreciate any help
| Since
$$
\left( {1 + x + x^2 + x^3 + x^4 } \right)^{\,17} = \left( {{{1 - x^5 } \over {1 - x}}} \right)^{\,17}
$$
you might be interested to know that the general
case of what are somewhere called r-nomial coefficients corresponds to the O.G.F:
$$
F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Eliminate $\alpha,\beta,\gamma$ from the system of equations Eliminate $\alpha,\beta,\gamma$ from the following system of equations.
$$a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma)=0$$
$$a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma)=0$$
$$a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma)=0$$
My try:
Squaring and adding first two equation... | We have the following system of equations
\begin{align}
a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma) &= 0, \tag{1}\\
a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma) &= 0, \tag{2}\\
a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma) &= 0. \tag{3}
\end{align}
From (1) and (2), we have
$$a^2 + b^2 + 2ab \cos \alpha\cos \beta + 2ab \sin \alpha... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$
$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$
I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as
$x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?
| $$\lim_{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}=\lim_{x\to-\infty} \sqrt[3]{\frac{x^3+x}{x^3}} =\lim_{x\to-\infty} \sqrt[3]{1+\frac{1}{x^2}} = \sqrt[3]{1+\lim_{x\to-\infty}\frac{1}{x^2}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve ODE $y'' = y^3 -y y'$ I want to solve $y'' = y^3 -y y'$ with boundary conditions $y(1) = 1/2$ and $y(2) = 1/3$ but not sure how to start. If someone could give me a complete step by step explanation, it would be greatly appreciated because I want to fully understand it.
Edit:
By the hint $(y'+\frac{y^2}{2})'=y^3... | Hint:
$$(y'+\frac{y^2}{2})'=y^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x =... | Hint This is a classic problem and probably a duplicate here, but I was unable to find another instance of this question as written. Edit Found one.
First, note that $\sin 18^\circ = \cos 72^\circ$; we find the latter.
Recall that (for $n \geq 2$) the sum of the $n$th roots of unity sum to zero, and taking real parts l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is this valid when deriving quadratic equation? When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres th... | You can look at it differently:
$$ax^2+bx+c=0 \iff \\
x^2+\frac bax+\frac ca=0\iff \\
\left(x+\frac b{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}=0 \iff \\
\left(x+\frac b{2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \iff \\
\left[\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}\right]\cdot \left[\lef... | {
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"timestamp": "2023-03-29T00:00:00",
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Does there exist a function $f_{\Box,\Box}(\Box)$ making the formula $a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$ true? Let $a$ and $b$ denote the resistances of two resistors. If they're put in series, the total resistance is $a+b$. If they're put in parallel, the total resistance is $$a \oplus b := \frac... | Let me assume that $b\ge0,c\ge0, b+c>0$.
I am looking for $f$ satisfying $f_{b,c}(a) = f_{c,b}(a)$.
The equation in the OP is then equivalent to
$$
(b+c) f_{bc}(a)^2 + ((b+c)^2-2ab-2bc-2ca)f_{bc}(a)- (b+c)(ab+bc+ca)=0.
$$
Setting $d:=ab+bc+ca$ and dividing by $b+c$ this is equivalent to
$$
f_{bc}(a)^2 + (b+c -\frac{2d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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Geometry: Find the radius of a circle given parallel secants Suppose a circle has two parallel chords of lengths $a$ and $b$, and the chords are separated by a distance of $c$. Using only the usual high school geometry theorems (i.e. no trig or calculus), can we derive a formula for the radius?
I've tried drawing rad... | I just answered this question in chat, so I thought it would be worth posting the answer here.
The horizontal position of a vertical chord of length $c$ in a circle of radius $r$ would be $\pm\sqrt{r^2-\frac{c^2}4}$.
Thus, the distance, $d$, between chords of length $c_1$ and $c_2$ (with $c_1\ge c_2$) would be
$$
d=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
integration of a gaussian with $x^2$ I need to integrate
$$\int_{-\infty}^{\infty} x^2 e^{-ax^2} \qquad \text{where } a\in R$$
The book does the following:
I don't understand what's happening. I tried solving the integral using integration by parts and this is what I got
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^... | Good job for identifying the components used for integration by parts.
If you are familiar with normal distribution, here's an intepretation for $a>0$,
Let $\frac{1}{2\sigma^2}=a$
\begin{align}\int_{-\infty}^\infty x^2 \exp(-ax^2) \, dx &=\sqrt{2\pi \sigma^2}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}}x^2 \exp\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Range of rational expression having algebraic terms
Find the range of $\displaystyle f(x) = \frac{x^2+x-1}{x^2-x+2}$ subjected to $-1 \leq x\leq 1$.
Plan
\begin{align} y = \frac{x^2+x-1}{x^2-x+2}&\implies yx^2-yx+2y=x^2+x-1\\&\implies(y-1)x^2-(y+1)x+(2y+1)=0.\end{align}
For $y=1$, we have $x=3/2$.
For $y\neq 1$,
$$(y... | I will propose an alternative solution using calculus (as this tag has been added by OP).
Rewrite $$f(x)=\frac{x^2+x-1}{x^2-x+2}=1+\frac{2x-3}{x^2-x+2}$$ and note that $f$ is defined in the interval $[-1,1]$.
Now we find the stationary points. Thus $$f'(x)=\frac{2(x^2-x+2)-(2x-3)(2x-1)}{(x^2-x+2)^2}=0\implies2x^2-6x-1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me pleas... | On completing the square, we have
\begin{align}
x= (x^2+6x+7)^2+6(x^2+6x+7)+7 &= ((x+3)^2-2)^2+6((x+3)^2-2)+7\\
&= (x+3)^4-4(x+3)^2+4+6(x+3)^2-5\\
&= (x+3)^4+2(x+3)^2-1= ((x+3)^2+1)^2-2 \geq 1-2=-1
\end{align}
Further, we alsoknow that $(x+3)^2$ is monotonically increasing for $x\geq -3$. Thus, we get that
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of series $\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$ How to evaluate series
$$\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$$
I tried to split the summation...but I failed. Please help
| $$\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$
$$\frac{1}{x}\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n+1)!}$$
let $x\rightarrow \sqrt{x}$
$$\frac{1}{\sqrt{x}}\sinh \sqrt{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{(2n+1)!}$$
let $x\rightarrow x^5$
$$\frac{1}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Rotation Matrix and Triple Angle Formulas? Define $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Observe that
$$
(R_\theta)^2 = \begin{pmatrix} \cos^2\theta-\sin^2\the... | If $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Then we can write
$$R_{\theta} = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix} + \begin{pmatrix} 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\f... | $\int \frac{x^4}{(2-x^2)^{3/2}}dx$
Comparing with $\int x^m (a+bx^n)^p dx$
$m=4,a=2,b=-1,n=2,p=-\frac{3}{2}$
Here $\frac{m+1}{n}+p=\frac{5}{2}-\frac{3}{2}=1$, integer.
So we first transform the integral $ x^m (a+bx^n)^p $ by factoring $x^n$ i.e., by $x^2$ in this way
$\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int {x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Question about the proof of simple algebra rule $\frac{1}{\frac{1}{a}} = a$ I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is:
\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}
And the proof:
\begin{eqnarray*}
1 = a \frac{1}{a} \L... | Let $x=\frac{1}{a}$. Then:
\begin{eqnarray*}
1 = x \frac{1}{x} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}
Better? You're right, replacing $a\to\frac{1}{a}$ is a minor abuse of notation, because ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one a... | Your problem is equivalent to the following:
In any acute $ABC$ we have $AB+BC+CA>AH+BH+CH$ , where $H$ is the orthocenter of the triangle. But this inequality is true for all points from the tringle's interior.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Closed form for sum of powers I wonder if it is possible to evaluate explicitly the sum
$$S(N):=\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}\left(1-\frac{2j}{N}\right)^{N+1},\quad N\in\mathbb{N}.$$
In the large $N$ limit the result is straightforward if one approximates the argument of the sum with the exponenti... | The finite sum can be expressed in terms of
Bernoulli polynomial $B_p(x)$. Bernoulli polynomials obey many interesting relations, the one we need
is
$$B_p(x+1) - B_p(x) = px^{n-1}$$
Let $J = \lfloor \frac{N-1}{2}\rfloor$ and using this relation, we find
$$\begin{align}S(N)
&=\sum_{j=1}^J\left(1-\frac{2j}{N}\right)^{N+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral test to find constant for which series is convergent Problem statement:
Use the integral test to find the value of $b$ for which the following series is
convergent
$$ \sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}.$$
I evaluated the corresponding integral, which gave the following ... | Hint $\sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}$ is convergent implies
$$\lim_{r \to \infty} \frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)} =0$$
This yields $b =\mbox{ something}$.
You can then use the integral test to decide if the series converges for this value of $b$.
Alternately
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $P=17x^2+17y^2+16xy$ Let $x,y$ be positive real numbers such that $4x^2+4y^2+17xy+5x+5y\ge 1$. Find minimum value of $$P=17x^2+17y^2+16xy$$
My idea:
I see that $x=y=\frac{\sqrt 2 -1}{5}\rightarrow P=2(3-2\sqrt 2)$
So I proved $P\ge 2(3-2 \sqrt 2)(4x^2+4y^2+17xy+5x+5y)\ge2(3-2\sqrt 2)$
Is my appro... | Yes, your answer is true.
The solution can be the following.
By AM-GM
$$1\leq4(x+y)^2+5(x+y)+9xy\leq4(x+y)^2+5(x+y)+\frac{9}{4}(x+y)^2,$$ which gives
$$x+y\geq\frac{2(\sqrt2-1)}{5}.$$
Thus, by C-S we obtain:
$$17(x^2+y^2)+16xy=8(x+y)^2+\frac{9}{2}(1^2+1^2)(x^2+y^2)\geq8(x+y)^2+\frac{9}{2}(x+y)^2=$$
$$=\frac{25}{2}(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Gravity of a Circular Ring at a co-planar external point. I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.
I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable... | Considering similar problems it is usually simpler to consider the potential rather than force. The latter can be found latter as the negative of the potential gradient. Assuming the masses of the test point-like body and the ring be $m$ and $M$, respectively, we have in spherical coordinates with the origin at the cen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$ Evaluate $$ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$$
I have tried Weierstrass substitution and tried to split into two integrations, but it gets really messy.
Is there a better way to approach this problem? I feel that comple... | After substitution $t=\tan{x}$ use
$$\frac{a^2+b^2t^2}{(a^4+b^4t^2)(1+t^2)}=\frac{1}{a^2+b^2}\left(\frac{1}{1+t^2}+\frac{a^2b^2}{a^4+b^4t^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfra... | Your answer is almost right - you just need to show there exists $a$ and $b$ where $\frac{5}{3}$ is attainable (this happens at $a = b = \frac{1}{2}).
Here is an alternative solution. Let $c = 1+a+b$. The expression becomes $\frac{a}{c-a} + \frac{b}{c-b} + \frac{1}{c-1}$. We claim that for fixed $c$, this is minimised ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of matrix to a power and diagonalized matrix to a power give different results I have the following problem. I wish to find
$$
\lim_{n \to \infty} \psi M^n 1
$$
where
$$
\psi = \begin{bmatrix} 1/4 & 1/2 \end{bmatrix}, \quad
1=\begin{bmatrix} 1 & 1 \end{bmatrix}^T
$$
and
$$
M=
\begin{bmatrix}
\lef... | I got $$\frac{1}4e^{-2L}\left(3\cosh(L\sqrt{2})+2\sqrt{2}\sinh(L\sqrt{2})\right)$$ in two different ways.
This is close to your complicated example. I get a different coefficient for $\sinh(L\sqrt{2}).$ (Your coefficient was $\sqrt{2}.$)
In general, if $M=\begin{pmatrix}a&2b\\b&a\end{pmatrix},$ it has eigenvectors $\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential ... | Start with
$$ (x^2-y^2-y)\,\mathrm{d}x - (x^2-y^2-x)\,\mathrm{d}y=0$$
We note that this is
$$
(x^2-y^2)\,\mathrm{d}(x-y)+(x\,\mathrm{d}y-y\,\mathrm{d}x)=0
$$
So $(u,v)=(x+y,x-y)$ is a reasonable choice to try. Simplifying the resulting equation gives
$$
(2u+1)v\,\mathrm{d}u=u\,\mathrm{d}v
$$
which is a separable equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must... | Hint
$$x^2=2x-2$$
$$x^3=2x^2-2x=2(2x-2)-2x=2x-4$$
$$x^4=(2x-2)^2=4-8x+4(2x-2)=-4$$
$$x^6=-4(2x-2)=?$$
$$x^9=(-4)^2x=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equi... | Solve $x^2+5x+6 \equiv (x+2)(x+3) \pmod {187}$.
Besides the two 'in your face' solutions, $x \equiv -2 \pmod{187}$ and $x \equiv -3 \pmod{187}$, we can buttress the argument given by José Carlos Santos to find all four solutions.
We want to find integers $x, k, j$ satifying
$\;\text{L1:}\quad x + 2 = 11k$
$\;\text{L2:}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For $a,b,c,d > 0$ and $abcd = 1$, show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$
Question:
For $a,b,c,d > 0$ and $abcd = 1$, show that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$$
My Attempts:
Trivial to see that... | Mixing Variables and $uvw$ help.
I'll post my solution, which I found eight years ago.
Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$.
Thus, $$f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$$
$$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Another way to calculate the integral $ \int_{0}^{1} (x-x^{2})^{3/2} dx $ $$\int_{0}^{1} (x-x^{2})^{3/2} dx=\dfrac{1}{8}\int_{0}^{1}[1-(2x-1)^{2}]^{3/2}dx=\dfrac{1}{8}\int_{-\pi/2}^{\pi/2}\dfrac{cos^{4}\theta}{2}d\theta=\dfrac{1}{8}\int_{0}^{\pi/2}cos^{4}d\theta=\dfrac{1}{8}(\dfrac{3\pi}{16})=\dfrac{3\pi}{128}$$
The ch... | There is also the following way.
Let $x=\cos^2t$, where $t\in\left[0,\frac{\pi}{2}\right]$.
Thus, $$\int\limits_0^1(x-x^2)^{\frac{3}{2}}dx=\int_{\frac{\pi}{2}}^0\sin^3t\cos^3t(-2\sin{t}\cos{t})dt=2\int\limits_0^{\frac{\pi}{2}}\sin^4t\cos^4tdt=$$
$$=\frac{1}{8}\int\limits_0^{\frac{\pi}{2}}\sin^42tdt=\frac{1}{32}\int\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2... | for all $m\in \mathbb{N}$ and $x\geq 1$ we have $x^{2m+1}\geq 1$ then $2x^{2m+1}\geq x^{2m+1}+1$
Notice that $\frac{x^{m+1}+1}{x^m+1} = x - \frac{x-1}{x^m+1} \geq x$ because $x\geq 1$
Then $$2\left(\frac{x^{m+1}+1}{x^m+1}\right)^{2m+1}\geq 2x^{2m+1}\geq x^{2m+1}+1$$
We obtain easily that $$\frac{x^{m+1}+1}{x^m+1} \geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Show that $76^{2019} \equiv 45 \pmod{101}$ Using Euler's theorem I see that $76^{2019} = 76^{19} \pmod{101}$, how do I proceed?
| Here is an approach that requires minimum computation. We have:
$$76^{19} \equiv (-25)^{19} \equiv -5^{38} \pmod{101}$$
How we have that by quadratic reciprocity:
$$\left( \frac{5}{101} \right)=\left( \frac{5}{101} \right) \cdot \left( \frac{1}{5} \right) = \left( \frac{5}{101} \right) \cdot \left( \frac{101}{5} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \g... | Use the fact that $ GM \leq AM $. For six numbers $ x_1, x_2, \dots, x_6 $, you have
$$
(x_1 x_2 \dots x_6)^{1/6} \leq \frac{x_1 + x_2 + \dots + x_6}{6}
$$
Then, you have
$$
x y z \leq \frac{x^2 y + x y^2 + y^2 z + y z^2 + z^2 x + z x^2}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this l... | Substitute $x= r \cos \theta$ and $y= r \sin \theta$, then you have
$$\lim_{r \to 0}\frac{r \cos \theta \ln(1+r \sin \theta)}{2 r^2 \cos ^2 \theta + r^2 \sin^2 \theta}=\lim_{r \to 0}\frac{ \cos \theta \ln(1+r \sin \theta)}{2 r \cos ^2 \theta + r \sin^2 \theta}\stackrel{l'Hospital}{=}\lim_{r \to 0}\frac{\frac{\cos \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Prove that $\cos^4(\theta)-\sin^4(\theta)=\cos(2\theta)$ The problem is to prove that $\cos^4(\theta)-\sin^4(\theta)=cos(2\theta)$.
So, here are my steps so far:
$(\cos^2\theta)^2-(\sin^2\theta)^2=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos\the... | There are a couple of nice ways we can do it. Firstly, note that:
$$\cos^4(\theta)-\sin^4(\theta)=\bigl(\cos^2\theta-\sin^2\theta\bigr)\bigl(\cos^2\theta+\sin^2\theta\bigr)$$
and since:
$$\cos^2\theta+\sin^2\theta=1\tag{1}$$
we can say that:
$$\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta=\cos2\theta$$
which is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Finding solutions to a system of equations using elementary symmetric polynomials Find the value of a, b, and c given:
$a^2 + b^2 + c^2 = 129$
$ab + ac + bc = -4$
$(a^2)(b^2) + (a^2)(c^2) + (b^2)(c^2)$ = 984
I attempted this problem using elementary symmetrical polynomials and found that the second equation can be writ... | \begin{align}
a^2 + b^2 + c^2 &= 129
\tag{1}\label{1}
,\\
ab + bc + ac &= -4
\tag{2}\label{2}
,\\
a^2b^2 + b^2c^2 + a^2c^2&=984
\tag{3}\label{3}
.
\end{align}
\begin{align}
\eqref{2}^2:\quad
a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)
&=16
,\\
abc(a+b+c)&=-484
,\\
\eqref{1}+2\cdot\eqref{2}:\quad
(a+b+c)^2&=121
.\\
a+b+c&=\pm 11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, ... | Geometric progression
\begin{align}
(a-b)\sum_{i=1}^{n}a^{n-i}b^{i-1}&=(a-b)a^{n-1}\left(\frac{1-\left(\frac{b}{a}\right)^n}{1-\frac{b}{a}}\right)\\
&=a^n-b^n.
\end{align}
Can you complete the answer now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(x^3 + \frac{a}{x^2})^5 = -270$, find $a$ I am working on my scholarship exam practice and not sure how to begin. Please assume math knowledge at high school or pre-university level.
Let $a$ be a real constant. If the constant term of $(x^3 + \frac{a}{x^2})^5$ is equal to $-270$, then $a=$......
Could you please giv... | From the binomial theorem, we know that the sum of the powers of $x^3$ and $a/x^2$ must add to $5$. So if our term in the expansion is $k (x^3)^p (\frac{a}{x^2})^q$, $p+q = 5$. Moreover, we want our term to be constant (no nonzero powers of $x$), so we have $3p - 2q = 0$. Solving these two expressions, we have $p = 2, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that inequality $27+7\left(ab+bc+ca\right)\le 8\left(\sum \sqrt{a+3b}\right)$ For the positive reals $a,b$ and $c$ so that $a+b+c=3$. Show that $$27+7\left(ab+bc+ca\right)\le 8\left(\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\right)$$
That inequality has been created by Imad Zak, i think it is interesting problem.
We ca... | Let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, by Holder
$$\sum_{cyc}\sqrt{a+3b}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+3b)}\right)^2\sum\limits_{cyc}(a+3b)^2}{\sum\limits_{cyc}(a+3b)^2}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+3b)\right)^3}{\sum\limits_{cyc}(10a^2+6ab)}}=\sqrt{\frac{32(a+b+c)^6}{27\sum\limits_{cyc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}... | A sketch of the correct method:
Since $x-x^2$ has order $1$, you have to expand $\sin u$ up to degree $6$, and substitute $x-x^2$ in $u, u^2,\dots u^6$, truncating the expansions of the different $(x-x^2)^k$ at degree $6$, which leads to a recursive computation:
\begin{align}
(x-x^2)^2&=x^2-2x^3+x^4& \quad(x-x^2)^3&=x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$
*
*Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we ... | Note that the domain of $\sqrt {1+x}+\sqrt {1-x}$ is $[-1,1]$
On $(0,1]$ we have $\sqrt{1+x}>1$
On $[-1,0)$ we have $\sqrt {1-x}>1$
At $x=0$ we have $\sqrt {1+x}+\sqrt{1-x}=2>1$
Thus $\sqrt {1+x}+\sqrt {1-x}>1$ on its domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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The sum of an infinite series containing a finite series in each denominator
Evaluate
$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\cdots+\frac{1}{\left[\frac{k(k+1)}{2}\right]^2}+\cdots$$
to $\infty$, where $k$ is the $k$th term of the series.
Using Microsoft Excel, I found that the sum of the fir... | $$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$
$$\frac {1}{k^2(k+1)^2}= \frac {1}{k^2}+\frac {1}{(k+1)^2} - \frac {2}{k(k+1)}$$
$$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$
$$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Factoring out the expression I'm sorry, it must pretty basic, but...
How to factor out this expression $x^2 + 5x - 5 = 0$ ?
(The method I normally use to factor expressions doesn't work there because when we factor expression above so that format will be $(ax + b)(qx + c)$ $b$ and $c$ definitely won't be integers, an... | Since $x^2+5x-5$ is a monic polynomial whose roots are $\frac{1}{2} \left(-5+3\sqrt{5}\right)$ and $\frac{1}{2} \left(-5-3\sqrt{5}\right)$,$$x^2+5x-5=\left(x-\frac{1}{2} \left(-5+3\sqrt{5}\right)\right)\left(x-\frac{1}{2} \left(-5-3\sqrt{5}\right)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the series is conditionally convergent $$ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $$
I have done the$ \sum_{n=1}^\infty \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ part , and showed it divergent using Gauss test .
But i am not able to ... | $(\frac{1}{2})^2 < \frac{1}{2}$,
$(\frac{4}{5})^2 < \frac{3}{4}$,
$(\frac{7}{8})^2 < \frac{5}{6}$, $\dots \dots$
$(\frac{3n-2}{3n-1})^2 < \frac{2n-1}{2n}$
Multiplying all , we get
$\prod_{n=1}^n(\frac{3n-2}{3n-1})^2 <\prod_{n=1}^n \frac{2n-1}{2n}$$ ......(1)$
Now,
$(\frac{1}{2}) < \frac{2}{3}$,$(\frac{3}{4}) < \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$
$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is
$$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$
My Try
I tried this using the conventional ... | We have:
$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$
as you mentioned. Now note that
$$f(a)=Aa^2+Ba+C\\
f(b)=Ab^2+Bb+C\\
f(c)=Ac^2+Bc+C$$
Now it is easy to solve this system of equations for $A$, $B$ and $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x)=\frac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$ let $f(x)=\dfrac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$
my try : $x:=\tan t$ then $f(\tan t)=\dfrac{\tan t}{1+\tan^2 t}=\sin t \ \ \ \text{or} \ \ -\sin t$
Now what ?
| $f(x)=\dfrac{x}{\sqrt{x^2+1}}$
Let $y=f(x)$
Let's start by making x a function of y, which will give you inverse.
$y=\dfrac{x}{\sqrt{x^2+1}}$
This simplifies to,
$y^2(x^2+1)=x^2$
From here,
$x=\dfrac{y}{\sqrt{1-y^2}}$
Therefore,
$f^{-1}(x) = \dfrac{x}{\sqrt{1-x^2}}$
Hence,
$f^{-1}(\sin x) = \dfrac{\sin x}{\sqrt{1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+... | Setting $x=\sqrt 2i$ (so $x^2=-2$) yields
$$-4D-E+\sqrt2(-2D+E)i=2+\sqrt 2 i,\quad\text{whence }\begin{cases}\;2D+E=-2,\\-D+E=1,
\end{cases}\iff D=-1,\;E=0.$$
Next, setting $x=0$, you get
$$2=4A-2C-E, \quad\text{ so }\; C=-\tfrac13.$$
Last, multiply both sides of the decomposition by $x$ and let $x\to+\infty$, obtainin... | {
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"url": "https://math.stackexchange.com/questions/3276113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $x^2-5y=3$ has no integer solutions. Here is my attempt:
$x^2-5y=3$ is the same as $2x^2-10y=6$, which is equivalent to $2x^2=10y+6$.
This means that 2 divides $x^2$. Since 2 is prime, then 2 divides x, or $x=2k$ for some integer $k$.
Then we have that $(2k)^2-5y=3$, or $4k^2=5y+3$. This implies that 2 divi... | Hint: $x^2 - 5y = 3 \iff x^2 = 5y + 3$. Now observe that the units digit of a perfect square must be one of the following digits: $0, 1, 4, 5, 6, 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| Hint: Consider $f(x)=\sqrt[3]{x}$. So $$f'(2)=\lim_{x \to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2}=\frac{1}{3}x^{-\frac{2}{3}}\Bigg|_{x=2}=\frac{1}{3}\frac{1}{2^{\frac{2}{3}}}$$ So $$\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2} \frac{1}{x^2+2x+4} \longrightarrow f'(2).\frac{1}{12}$$ as $ x \t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi... | You just divided twice by $7$.
$$\sqrt[7]{(-\sqrt3-i)^5}=\sqrt[7]{32\left(\cos(5\cdot210^{\circ})+i\sin(5\cdot210^{\circ})\right)}=$$
$$=2^{\frac{5}{7}}\left(\cos\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)+i\sin\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)\right),$$ where $k\in\{0,1,2,3,4,5,6\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $a,b,c \in \mathbb{R^+}$ such that $a^2+b^2+c^2+abc=4$ and $a+b+c=3$
Find $a,b,c \in \mathbb{R^+}$ such that $$a^2+b^2+c^2+abc=4\;\;\;{\rm and}\;\;\;a+b+c=3$$
Seeing $a^2+b^2+c^2+abc=4$,I substituted $a=2\cos{A}$,$b=2\cos{B},c=\cos{C}$ with $A+B+C=180°$.
Therefore,$cosA+cosB+cosC=\frac 32$ have thousands of so... | If we put $c=3-a-b$ we get quadratic equation on $a$ with parameter $b$, if $b \ne 2$:
$$(2-b)a^2-a(b^2-5b+6)+(2b^2-6b+5)=0$$
which has a solution iff it discriminat is nonnegative: $$(b-2)(b+2)(b-1)^2\geq 0$$
so $b>2$ or $b=1$...
*
*Clearly if $b=1$ we get $a^2-2a+1=0$ so $a=1$ and then $c=1$.
*If $b>2$ then with ... | {
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How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
| Substitute $x = \sqrt{2}y$ and divide by $\sqrt{2}^5$. The resulting equation is
$$y^5 + (y+1)^5 = 0.$$
Now substitute $y = -1/(z+1)$ and multiply by $(z+1)^5$ and you get
$$z^5 - 1 = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $z^4*\overline{z}=-1+\sqrt{3}i$ need some help solving this problem:
$z^4*\overline{z}=-1+\sqrt{3}i$$\DeclareMathOperator{\cis}{cis}$
The answer is: $z_k=(2*{2^{1/5}})^{1/5}\cis(\frac{2\pi+6\pi*k}{15})$,
$k=0,1,2,3,4$.
I tried using De Moivre to solve it but I got the wrong answer :
I substituted $z=r*\cis(\al... | Since $r^5\operatorname{cis}3\alpha=2\operatorname{cis}\frac{2\pi(1+3n)}{3}$ with $n\in\Bbb Z$, $r=2^{1/5}$ while $\alpha=\frac{2\pi(1+3n)}{9}$ with $0\le n\le4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Speed of convergence of a special sequence. Let $a\in(0,1)$.
Consider a following sequence:
$$b_{n}=(\sum_{k=1}^{n}\frac{(-1)^k\cos(\ln(k))}{k^{a}})^{2}+(\sum_{k=1}^{n}\frac{(-1)^k\sin(\ln(k))}{k^{a}})^{2}$$
How fast does this series converge to its limit $b$?How to estimate $|b_{n}-b|$?
I know that the limit exists ... | Hint:
Consider the complement of $b_n$ to $b$, i.e.
$$
\eqalign{
& \bar b_{\,n} = \left( {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}
+ \left( {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)^{\,k} \sin \left( {\ln k} \righ... | {
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If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$. Question : If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$.
My Attempt
Corresponding Equation of Inequality, $f(x) =5x^2 - 2kx + 1 =... | We have $f(x) =5x^2 - 2kx + 1=5(x^2-2\frac{k}{5}x +\frac{k^2}{25})-\frac{k^2}{5}+1=5(x-\frac{k}{5})^2 - \frac{k^2}{5}+1 \geq 1-\frac{k^2}{5}$
So it takes it's minimum at $\frac{k}{5}$ which is $1-\frac{k^2}{5}$
We want $1-\frac{k^2}{5}<0$ in order for the inequality to have solution, hence $k<-\sqrt{5}$ and $k>\sqrt{5... | {
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Evaluating $\sum_{r=1}^n \frac{\tan(x/2^r)}{2^{r-1}\cos(x/2^{r-1})}$
I was asked to find the sum of
$$\sum_{r=1}^n \dfrac{\tan \dfrac{x}{2^r}}{2^{r-1}\cos\dfrac{x}{2^{r-1}}}$$
I proceeded by breaking $\tan x$ into $\sin x$ and $\cos x$ and writing numerator as follows
$$\sin\left(\frac{x}{2^{r-1}}-\frac{x}{2^r} \r... | Start with $$\cos(x/2) \cos(x/2^2) \cos(x/2^3)....\cos(x/2^n)= \frac{\sin x}{2^n\sin(x/2^{n})}$$
Take $\ln$ of both sides, then
$$\sum_{r=1}^{n} \ln \cos (x/2^{r}) = -\ln \sin (x/2^{n})+\ln \sin x-n\ln 2$$ D.w.r.t x and get
$$\sum_{r=1}^{n} \frac{1}{2^{r}} \tan (x/2^{r})=\frac{1}{2^{n}} \cot (x/ 2^{n})- \cot x=g(x) ~~... | {
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ambiguity in solving algebraic equations. Given $x+\frac{1}{x}=2 \tag{1}$
$x^3+\frac{1}{x^4}=1 \tag{2}$
Find the value of $$x^4+\frac{1}{x^3}$$
If equation $\bf{2}$ is multiplied by $\bf x$ we get $x^4+\frac{1}{x^3}=x$ now, we just need to find the value of $x$ which from $\bf{1}$ is equal to $1$.
But the answer to t... | The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.
Thus the system does not have a solution.
| {
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The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$ This attractive identity that
$$\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}~~~(*)$$
emerged while doing numerics at Mathematica with harmonic numbers, binomial coefficients
and sums involving them.
H... | Let $$S_n=\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}] {n \choose k}.$$
Notice that $$H_k^2+H_k^{(2)} =\left(\sum_{j=1}^{k} \frac{1}{j}\right)^2+\sum_{j=1}^{k} \frac{1}{j^2}=2 \sum \sum_{1 \le i \le j \le k} \frac{1}{ij}$$
So $S_n$ can be re-written as
$$S_n=2\sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} \su... | {
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\... | $2n+1$ is odd and $10\cong -1\pmod {11}$. Thus $10^{2n+1}\cong (-1)^{2n+1}\cong -1\pmod{11}$.
| {
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shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | Let $c$ be a real number such that $3x+4y=c$ has common points $(x,y)$ with the circle and parallel to $3x+4y=12$.
Thus, by C-S we obtain:
$$|c|=|3x+4y|\leq\sqrt{(3^2+4^2)(x^2+y^2)}=\sqrt{25\cdot4}=10,$$ which says that
$3x+4y=10$ is a tangent to the circle,
which is parallel to the line $3x+4y=12$ and nearest to thi... | {
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Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left... | Let
$$x=y-\frac1y$$
so that
$$y=\frac{x\pm\sqrt{x^2+4}}2.$$
Then
$$f(x)=f\left(y-\dfrac1y\right)=y^2-\frac1{y^2}=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac2{x\pm\sqrt{x^2+4}}\right)^2\\
=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac{x\mp\sqrt{x^2+4}}2\right)^2=\pm x\sqrt{x^2+4}
.$$
Note that, as raised b... | {
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Proving a reversed AM-HM inequality
Problem:
Given $n \ge 3$, $a_i \ge 1$ for $i \in \{1, 2, \dots, n\}$. Prove the following inequality:
$$(a_1+a_2+\dots +a_n) (\frac{1}{a_1}+\frac{1}{a_2} + \dots +\frac{1}{a_n}) \le n^2+\sum_{1\le i<j\le n}|a_i-a_j|.$$
This inequality seems like a reversed form of AM-HM inequality ... | I will prove this via induction. Without loss of generality, assume $a_1 \le \cdots \le a_n$. It turns out this inequality remains true for $n = 1, 2$ as well, reducing to $1 \le 1$ in the first case. We actually need to prove the $n = 2$ case in order to elicit intuition for the inductive step.
Base Case $n = 2$
If $a... | {
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Find all irreducible monic polynomials of degrees $2$ in $\Bbb{Z}_{4}[x]$. Premise: not a native speaker.
So: I really don't know how to find all the polynomials except to calculate them all:
\begin{equation}x^{2}+ax+b \end{equation} \begin{equation} a,b ∈ {(0, 1, 2, 3)} \end{equation}
\begin{equation}
\begin{split}
1)... | It is easier to count all reducible monic polynomials of degree $2$ because they are just the product of two monic linear polynomials, which are $x,x-1,x-2,x-3$ in your cases. So there are ${4 \choose 2} = 6$ products of different linear polynomials. As Robert pointed out, there are also $2$ squares of them. There are ... | {
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Area boundes by two circles
Find the area bounded by $r \le 3\sin(t)$ and $r \le -3 \sqrt3 \cos(t)$.
Well, I've found the intersection point, which is $t=\frac{2 \pi}{3}$ . Then, I set up two integrals
$$
A=\frac{1}{2} \int_{\pi/2}^{2\pi/3} (3\sin(t))^2 dt
$$
and
$$
A=\frac{1}{2} \int_{2\pi/3}^{\pi}(-3 \sqrt3 \cos(t... | The correct integral expressions are
$$
A_s=\frac{1}{2} \int_{2\pi/3}^{\pi} [3\sin(t)]^2 dt=\frac{3\pi}{4}-\frac{9\sqrt{3}}{16}
$$
$$
A_l=\frac{1}{2} \int_{\pi/2}^{2\pi/3}[-3\sqrt3 \cos(t)]^2 dt =\frac{9\pi}{8}-\frac{27\sqrt{3}}{16}
$$
$$Area=A_s+A_l =\frac{15}{8}\pi-\frac{9}{4}\sqrt{3}.$$
The integration ranges are sp... | {
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Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$. Today I had a competition in Xiamen, China. I know how to do the questions except this strange equation.
Given $x, y$ that $xy-\dfrac{x}{y^2}-\dfrac{y}{x^2}=3$, work out $xy-x-y$.
Such a strange question, right? I have found the integral solutio... | It is, of course, possible to solve the cubic $$xy-\frac x{y^2}-\frac y{x^2}=3\implies y^3\color{red}{-\frac{3x^2}{x^3-1}}y^2\color{blue}{+0}y\color{green}{-\frac{x^3}{x^3-1}}=0$$ using Cardano's method, which however may not be suitable as a contest problem.
We have $$Q=\frac{3\cdot\color{blue}{0}-\left(\color{red}{-\... | {
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Limit of an oscillating sequence Let $a$ be any real number. Then, how do I compute the
$$
\limsup_n \left|\cos\big(a\sqrt{n^2+1}\big) \right|^{1/n}\:?
$$
Also, how do I compute the value when $a$ is any complex value? The sequence is harshly oscillating... so I cannot find a way to approach
| I'll find the limit in the most general case when $a$ is complex.
Let $a=x+iy$ and the $n^{th}$ term of the sequence be denoted by $s_n$
\begin{align*}
s_n &= \left| \cos \left( a \sqrt{n^2+1} \right) \right|^{1/n} \\
&= \left| \frac{ e^{ia\sqrt{n^2+1}} + e^{-ia\sqrt{n^2+1}} }{2} \right|^{1/n} \\
&=\left|e^{ia\sqrt{n^... | {
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$n^4 + 4^n$ is a not a prime Prove that $n^4 + 4^n$ is not a prime for all $n > 1$ and $n \in \mathbb{N}$.
This question appeared in the undergrad entrance exam of the Indian Statistical institute.
When $n$ is even the proof is simple.
For $n = 2m+1$ I am utterly stuck.
| Firstly, I’ll show that we can factorise $x^4+4y^4$.
$$x^4+4y^4=x^4-4x^2y^2+4y^4-4x^2y^2= \left(x^2+2y^2\right)^2-\left(2xy\right)^2= \left(x^2+2xy+2y^2\right)\left(x^2-2xy+2y^2\right)$$
Back to the question, when $n=2m+1$ that $m$ is a positive integer, we can substitute $x=n$ and $y=2^m$ into the polynomial above, we... | {
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Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| The Maclaurin series of the given function is the product of the Taylor series of $\ln(1+x)$ and $1+3x$, treating the former as if it were a regular polynomial:
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots$$
$$3x\ln(1+x)=3x^2-\frac{3x^3}2+\frac{3x^4}3-\frac{3x^5}4+\dots$$
$$(1+3x)\ln(1+x)=x+\left(3-\frac12\ri... | {
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Every sufficiently large positive integer is the average of $n$ distinct primes for certain $n \geq 2$? I want to generalize a stronger Goldbach's conjecture a little bit because that might help solve it.
I was thinking:
For all $n \geq 2$, every sufficiently large positive integer $x \geq b_n$ is the average of $n$ ... | For $n$ large enough Vinogradov's theorem gives $\sim C \frac{n^2}{\ln^3 n}$ solutions to $2n+1 = p_1+p_2+p_3$ which gives $\sim B \frac{n^3}{\ln^4 n}$ solutions to $2n = p_1+p_2+p_3+p_4$.
And the same method as Vinogradov gives $\sim A \frac{n^2}{\ln^3 n}$ solutions to $2n = 2q_1+q_2+q_3$.
Thus we have $\sim B \fra... | {
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Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3... | Its a hint
Split the series to two. One is sum of cubic of odd. And other one is 3*(sum of square of even )
| {
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Show that the vectors $v_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ are linearly independent... Show that the vectors $v_1 = \begin{pmatrix}
1 \\
2
\end{pmatrix}$ and $v_2 = \begin{pmatrix}
2 \\
3
\end{pmatrix}$ are linearly independent, and find the unique coefficients $... | Another way is just to note that two vectors are linearly independent as long as they're not multiples of each other.
The hint by @Integrate This leads to $\begin {pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}-3&2\\2&-1\end{pmatrix}\begin {pmatrix}a\\b\end{pmatrix}$, after inverting the $2×2$ matrix.
| {
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Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}... | By the binomial of Newton it's just $$\frac{1}{2}((1+4)^n+(1-4)^n).$$
| {
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Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\fra... | Yes, it's the same, in fact the first fraction can be rewritten as $\frac{-(b-2)}{2b}\cdot \frac{1}{b-2}=-\frac{1}{2b}$. The second, instead: $\frac{-(b-2)}{2b}\cdot \frac{1}{2b}=-\frac{1}{2b}$.
| {
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Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I... | Another way:
Set $x-1=y$
$x^r=(1+y)^r\equiv1+\binom r1y+\binom r2y^2\pmod{y^3}$
$x^m-x^n\equiv y(m-n)+y^2\left(\binom m2-\binom n2\right)\pmod{y^3}$
Replace $y$ with $x-1$
| {
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In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.
Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at po... | In the standard notation by the angle bisector theorem we obtain:
$$\frac{FE}{AF}=\frac{CF}{AC}=\frac{\frac{ab}{b+c}}{b}=\frac{a}{b+c}.$$
Thus, $$\frac{FE}{AE-FE}=\frac{a}{b+c},$$ which gives
$$FE=\frac{a}{a+b+c}\cdot AE=\frac{a}{a+b+c}\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c... | {
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Find $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$
Find two positive integers $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$
I have completely no idea where to enter this question.
| Note that $\text{lcm}(a,b)=1764=(2\cdot 3\cdot 7)^2$ and $85113=3^2\cdot 7^2\cdot 193$. We may assume that $7^2$ divides $b$ then $a^2=85113-b^2$ is divisible by $7^2$ which means that $7$ divides $a$. Let $A:=a/7$, $B:=b/7$ and solve the reduced diophantine equation
$$A^2+B^2=85113/7^2=1737$$
with $\text{lcm}(A,B)=2^2... | {
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how to prove this inequality $\sum\frac{x}{(n-1)y+1}+\frac{1}{\sum x}\ge\frac{n+1}{n}$ if $x_{i}>0$,show that
$$\sum_{i=1}^{n}\dfrac{x_{i}}{(n-1)x_{i+1}+1}+\dfrac{1}{\sum_{i=1}^{n}x_{i}}\ge\dfrac{n+1}{n}$$
I have proven $n=2$and $n=3$
$n=3$ case
let $x,y,z>0$.prove
$$\dfrac{x}{2y+1}+\dfrac{y}{2z+1}+\dfrac{z}{2x+1}+\d... | A proof for $n=4$.
Let $x+y+z+t=4u$.
Thus, by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x}{3y+1}+\frac{1}{x+y+z+t}=\sum_{cyc}\frac{x^2}{3xy+x}+\frac{1}{x+y+z+t}\geq$$
$$\geq\frac{(x+y+z+t)^2}{3\sum\limits_{cyc}xy+x+y+z+t}+\frac{1}{x+y+z+t}=\frac{16u^2}{3(x+z)(y+t)+4u}+\frac{1}{4u}\geq$$
$$\geq \frac{16u^2}{3\left(\fra... | {
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Trigonometric inequality $\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}$ for all $x>1/2$ In Hobson's book "A treatise on plane trigonometry", the following exercise is to be found
If $x>1/2$ then
$$\frac{1}{1+x+x^2}<\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}.$$
The first inequality is easy as it follo... | The given inequality is equivalent to
$$ \frac{1}{1-\sqrt{z-z^2}}-1\geq\frac{\tan(z)}{z}-1\geq\frac{1}{1+\sqrt{z-z^2}}-1 $$
for $z\in\left(0,\frac{4}{5}\right]$. This is fairly simple to prove by concavity/convexity: the LHS lies above the line $y=\frac{z}{2}$, the central term lies below it. The central term lies abo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of roots of a Cubic Polynomial Let $f(x) = x^3 + px +q$ prove that if $(\frac{p}{2})^2+(\frac{q}{3})^3 > 0$ and $p <0 $ Then the equation has 1 real solution
Getting $f'(x)$ and setting it to $0$
$$x = \pm\sqrt{\frac{-p}{3}}$$
using the positive value and substituting it to $f(x)$, we get that its $y$ value is
... | Let $q\geq0$.
Thus, $$f\left(\sqrt{\frac{-p}{3}}\right)=\sqrt{\frac{-p^3}{27}}+p\sqrt{\frac{-p}{3}}+q=$$
$$=\sqrt{\frac{-p^3}{27}}-3\sqrt{\frac{-p^3}{27}}+q=2\left(\frac{q}{2}-\sqrt{\frac{-p^3}{27}}\right)=$$
$$=\frac{2\left(\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3\right)}{\frac{q}{2}+\sqrt{\frac{-p^3}{27}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $ax^3+bx^2+cx+d=0$ using a substitution different from Vieta's? We all know, a general cubic equation is of the form
$$ax^3+bx^2+cx+d=0$$ where $$a\neq0.$$
It can be easily solved with the following simple substitutions:
$$x\longmapsto x-\frac{b}{3a}$$
We get,
$$x^3+px+q=0$$ where, $p=\frac{3ac-b^2}{3a^2}... | Possible substitution $x\to X+\frac{1}{z}$, where $X$ is some parameter and $z$ is new unknown, but additional need use some properties of cubic polynomial, which historically we received from Lagrange.
Thus we have solution of equation $f=ax+bx^2+cx^3$:
$\begin{cases}
X=\dfrac{-(a b + 9 c f) + \sqrt{(a b + 9 c f)^2-4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 3
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Combinatorics floral arrangement question A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind? Please show working.
I have tried to use ... | At least three of each kind means that $10=4+3+3=3+4+3=3+3+4$ and therefore the number of different bunches of $10$ flowers out of $6$ Banksias, $5$ Wattles and $4$ Waratahs should be
$$\binom{6}{4}\binom{5}{3}\binom{4}{3}+\binom{6}{3}\binom{5}{4}\binom{4}{3}+\binom{6}{3}\binom{5}{3}\binom{4}{4}=600+400+200=1200.$$
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiating $\frac{dy}{dx}=3x+2y+xy$? What would $y''$ be of $\frac{dy}{dx}=3x+2y+xy$? In other words, what is the result ($y''$) if I differentiate $y'=3x+2y+xy$?
Can I do the following:
Set $$f(x,y) = 3x+2y+xy$$
then
$$f'(x,y) = \frac{df}{dx} + \frac{df}{dy}\frac{dy}{dx}$$
$$ f'(x,y) = (3 + y)+(2+x)(y')$$
$$ f'(... | $\frac {dy}{dx} = 3x + 2y + xy$
Differentiate both sides using the rules of implicit differentiation:
$\frac{d}{dx}\frac {dy}{dx} = \frac{d}{dx} (3x + 2y + xy)\\
\frac {d^2y}{dx^2} = \frac{d}{dx} 3x + \frac{d}{dx}2y + \frac{d}{dx}xy\\
\frac {d^2y}{dx^2} = 3 + 2\frac {dy}{dx} + y + x\frac{dy}{dx}\\
$
We can substitute $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Angle between two vector
I tried to solve this by directly multiplies a-b and 6a+b equal zero, and substitute some of it with the |a| and |b| to get a.b. but somehow I get different answer from the answer sheet that is 60 degree
It is still not complete but after that I get some bizzare answer so I didn't write it
| Given $(\overrightarrow{a}-\overrightarrow{b})\perp(6\overrightarrow{a}+\overrightarrow{b})$, we know that
\begin{align}
(\overrightarrow{a}-\overrightarrow{b})\cdot(6\overrightarrow{a}+\overrightarrow{b})&=0\\
6\overrightarrow{a}\cdot\overrightarrow{a}-5\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{b}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Powers of two are not sums of squares of different integers greater than zero It seems that non-trivial sums of squares of different non-zero integers aren't powers of two. Tested for sums of $2,3,4$ terms.
Is there something known about this topic? Are there counter-examples?
| Couldn't find a counter-example for up to $4$ terms as well (maybe someone can show why), but for more terms there are for example
$$
1^2+5^2+7^2+9^2+10^2=2^8\\
2^2+3^2+5^2+7^2+13^2=2^8\\
3^2+5^2+6^2+9^2+19^2=2^9\\
\vdots \\
1^2+5^2+6^2+7^2+8^2+9^2=2^8\\
\vdots \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Prove $\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$ How to prove without using Euler sums that
$$I=\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$
where $\zeta$ is the Riemann zeta function.
We can relate this integral to some Euler sum as follows:
\begin{align}
I&=-\sum_{... | Start off with the substitution $x\to \frac{1-x}{1+x}$ to get:
$$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$
$$X=\int_0^1 \frac{\ln(1-x)\ln ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
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$X^5 + 1$ into irreducible factors over $\mathbb{Q}$ and $\mathbb{Z}_5$ I need to split this polynomial into irreducible factors. I managed to do this for $\mathbb{C}$ and $\mathbb{R}$, but I can't figure out how to solve this problem for $\mathbb{Q}$ and $\mathbb{Z}_5$.
| This answer only addresses $\mathbb{Z}_5$.
For factoring over (small) finite fields, the easiest choice is to just try all the points in the field.
\begin{align*}
&x :& &x^5+1 \pmod{5} \\
&0 :& &1 \pmod{5} \\
&1 :& &2 \pmod{5} \\
&2 :& &3 \pmod{5} \\
&3 :& &4 \pmod{5} \\
&4 :& &0 \pmod{5}
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.