Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the value of $x^5 + \frac{1}{x^5}$ - question about correctness of method The task is: if $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.
I used the binomial formula and proved that $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the th... | $x+\frac{1}{x}=1$ gives $x^2-x+1=0,$ which gives $x^3+1=0$ or $x^3=-1$.
Thus, by your work $$x^5+\frac{1}{x^5}=-x^2-\frac{1}{x^2}=1.$$
Now, about your last question.
You used the following statement.
If $x+\frac{1}{x}=1$ so $\left(x+\frac{1}{x}\right)^2=1$.
It's true because if $a=b$ so $a-b=0$ and from here
$$a^2-b^2=... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $x, y \in \mathbb Z$. Prove that if $7 | (x^2 + y^2 ),$ then $7 | x$ and $7 | y$
Let $x, y \in \mathbb{Z}$. Prove that if $7 | (x^2 + y^2 ),$ then $7 | x$ and $7 | y$.
How do I prove this. Sorry for not providing how I approached the problem because I have no idea how to approach it.
| Well, you could if worst comes to worst just try every option:
$x \equiv i \pmod 7$ where $i=0,....6$ and $y\equiv j\pmod 7$ where $0..., 6$.
The $x^2\equiv i^2 \equiv 0, 1, 4, 2\pmod 7$. (Might be worthing noting that as $6,5,3 \equiv -1,-2,-3$ then $6^2,5^2, 3^2\equiv 1^2,2^2,3^2$). So $y^2\equiv j^2\equiv 0,1,4,2\... | {
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"timestamp": "2023-03-29T00:00:00",
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Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\... | first of all, notice that the substitution $x+1 = t, t \to \infty$ simplifies things quite a bit:
$$L = \lim_{t \to \infty} \left( (t(t^2-1))^{1/3} + ((t-1)^{t-1}\cdot t^t \cdot (t+1)^{t+1})^{1/3t} - 2t\right)$$
Factor out $t$ from the expression in the bracket and do some manipulations to get
$$\begin{align}
L &= \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove the trigonometry identity: prove the trigonometry identity:
$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$
of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.
|
$1. \ \cot^3A = \dfrac{1}{\tan^3A}\\$
$2. \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\$
$3. \ \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A$
So, the RHS becomes,
$\begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A -... | {
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"timestamp": "2023-03-29T00:00:00",
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An integer $n \geq 2$ is called square-positive- proof? An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.
So I have found the first four square-positive numbers, but I need to prove that why it $4$ is ... | Let $b-2, b-1, b, b+1$ be four consecutive numbers. The sum is $4b -2$. Let $m^2 = 4b-2$ Then $\frac {m^2}2 = 2b-1$. And $2|m^2$. So $2|m$. Let $m=2k$ then $4k^2 = 4b -1$ and $2k^2 = 2b-1$. That's impossible as the LHS is even and the RHS is odd. So $4$ is not square positive.
=====
In general:
........
If the f... | {
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"timestamp": "2023-03-29T00:00:00",
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Non-linear system of two equations (Method of Moments for Generalized Pareto) In a Method of Moments Estimation problem involving the generalized Pareto Distribution, the following system of 2 non-linear equations arise
\begin{align*}
\bar{X} &= \frac{\alpha \beta}{\alpha - 1}. \\
\frac{1}{n}\sum_{i=1}^{n} X^2_i &= \f... | $$
\begin{cases}
\frac{a b}{a-1}=x\\
\frac{a b^2}{a-2}=y\\
\end{cases}
$$
multiply the first equation by $b$
$$
\begin{cases}
a b^2=(a-1)bx\\
a b^2=(a-2)y\\
\end{cases}
$$
thus
$$(a-1)bx=(a-2)y\to b=\frac{(a-2) y}{(a-1) x} $$
Now plug this in the first, the very first equation
$$ab=x(a-1)\to a\cdot \frac{(a-2) y}{(a-1)... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the angle between matrices? I understand the formula:
$cos\left(θ\right)=\frac{tr\left(AB^T\right)}{\sqrt{tr\left(AA^T\right)\cdot tr\left(BB^T\right)}}$
Only one thing is not clear to me,the number I get is the angle itself?
Or the number should be obtained by the cos function?
I will give an example ... | Neither. With $\theta$ your angle, you have $\cos \theta = \frac{1}{\sqrt{3}}$, and so $\theta = 2 \pi n \pm \arccos\left(\frac{1}{\sqrt{3}}\right)$, where $\arccos\left(\frac{1}{\sqrt{3}}\right) \approx 0.955$ (radians).
| {
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"timestamp": "2023-03-29T00:00:00",
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$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$ Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$
I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it g... | There was a good answer posted by @Albusdumbledore that was deleted. I am just reproducing that answer and will delete mine if the other is undeleted.
The suggestion was to write $a = q, b = q r$ and $c = q r k$. Naturally, since $a,b,c \ne 0$, we also have $q,r,k \ne 0$. Plugging this into the relation, one concludes... | {
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Prove by induction $\left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$
What would be the best way to solv... | Hint: Prove that
$$
\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)
\left(1-\frac{1}{2^{n+1}}\right)
\ge
\frac{1}{4}+\frac{1}{2^{n+2}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Number of sequences of the form $x_1x_3x_5$
Find the number of sequences containing $5$ distinct terms $x_1,x_2,x_3,x_4,x_5$ from the set $A=\left\{1,2,3,4....20\right\}$ such that $x_1<x_2>x_3<x_4>x_5$.
My try:
The total number of sequences with $5$ distinct terms is $5!\times \binom{20}{5}$, among which there are $... | Note that for any set of five distinct numbers, the number of ways of arranging them so that they satisfy the given condition must be the same as for any other five distinct numbers. Thus we really only need to count the number of ways of ordering $\{1,2,3,4,5\}$ to satisfy the given condition.
Note that $5$ has to be ... | {
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The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I ma... | If $a,b$ are sides of a triangle and $x$ the mesure of angle between them, then the area of it is $${a\cdot b \cdot \sin x\over 2}$$
We use that formula here.
We have $AD = AE =DE =DC=a $ so$$\frac{[\Delta ADE]}{[\Delta DEC]} = {{a^2\sin 60 \over 2}\over {a^2 \sin 30 \over 2}} = \sqrt{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$ Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2... | Subtract $1/(x-1)$ from LHS and note that,
$$\frac 1{x+1}-\frac 1{x-1}=-\frac 2{x^2-1} \\ \frac 2{x^2+1}-\frac 2{x^2-1}=-\frac 4{x^4-1}\\ \frac 4{x^4+1}-\frac 4{x^4-1}=-\frac 8{x^8-1}$$
and so on. Do you see the telescoping pattern?
To finish, note that,
$$\frac {2^n}{x^{2^n}+1}-\frac {2^n}{x^{2^n}-1}=-\frac{2^{n+1}}{x... | {
"language": "en",
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Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square. Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
I found these pairs work by mod 3 and mode 4:
Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$
Case 2, $2^m \le 3^n$, $(0,0), ... | This is a partial solution concerning $m\ne 1$.
First we consider the cases $m\ge 2$.
For case 1 ($2^m \ge 3^n$), write $2^m = 3^n + k^2$.
Taking modulo $3$, we have $(-1)^m \equiv k^2 \pmod 3$. This gives $m$ is even.
Write $m = 2s$, then $3^n = 2^m-k^2 = (2^s-k)(2^s+k)$, and both $2^s \pm k$ are powers of $3$.
Summin... | {
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"timestamp": "2023-03-29T00:00:00",
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if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then...
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$
Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remain... | So a is in the first quadrant
$b = \pi-a$
$c = a + 2\pi$
$d = 3\pi-a$
$\sin(\frac{b}{2}) = \sin(\frac{\pi}{2}-\frac{a}{2}) = \sin(\frac{\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{\pi}{2})\sin(-\frac{a}{2}) = \cos(\frac{a}{2})$
$\sin(\frac{c}{2}) = \sin(\frac{a}{2}+\pi) = \sin(\frac{a}{2})\cos(\pi)+\cos(\frac{a}{2})\sin(\pi)... | {
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"timestamp": "2023-03-29T00:00:00",
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If $f$ is continuous on $[0,1]$, prove that $\lim_{n\to\infty}\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi}{2}f(0)$. The solution given:
$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx+\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx$$
okay, my first question here: wh... | First for your highlighted question: note that
$$
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Recurring Sequence with Exponent Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have ... | I love to telescope.
If
$a_n = ua_{n-1} + vc^{n}
$,
then
$\dfrac{a_n}{u^n}
= \dfrac{a_{n-1}}{u^{n-1}} + v(c/u)^{n}
$.
Let
$b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+vd^n
$
where
$d = c/u$.
Then
$b_n-b_{n-1}
=vd^n
$.
Summing,
$\begin{array}\\
b_m-b_0
&=\sum_{n=1}^m (b_n-b_{n-1})\\
&=\sum_{n=1}^m vd^n\\
&=v\dfrac{d-d^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples? We have a group of $16$ people, $8$ men, $8$ women.
We divide them to $8$ couples.
Let $X$ Be the number of couples make of $2$ men.
Calculate: $P(X = 3)$
I am not sure how to approach this.
I thought tha... | Complicated problem.
Answer will be
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
For the denominator, not only do you have to consider how many ways there are of choosing the 1st couple, then the 2nd, ...
You then have to adjust for over-counting, re each grouping into 8 couples will be counted $8!$ ways.
There... | {
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Which matrix belongs to the stabiliser of a vector? We consider the action of $GL_2(\mathbb{R})$ on the plane $\mathbb{R}^2$ by linear transformation. Which of these matrices belong to the stabiliser of the vector $\begin{pmatrix} 1\\3\end{pmatrix}$? | $\begin{pmatrix} -1 & 1 \\ 4 & -1 \end{pmatrix}$
B: $\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix}$
C: $\begin{pmatrix} 1 & 0 \\ -3 & 2 \end{pmatrix}$
D: $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
What I have tried so far:
Given a group action of a group $G$ on a set $X$. For $x \in X$, the stabiliser of $x$, d... | {
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How to evaluate $\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2$ in a particular way. How to evaluate:
$$\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2,$$
without splitting the expression into more sums.
Here $H_n^{\left(m\right)}=\sum _{k=1}^n\frac{1}{k^m}$ is the ha... | To find the desired series we must first consider the following integral.
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx$$
To evaluate it one can make use of the following trilogarithm identity.
$$\operatorname{Li}_3\left(\frac{x}{x-1}\right)=-\operatorname{Li}_3\left(x\rig... | {
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Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........? Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?
I am not well versed wit... | Let $a_n = (1+\sqrt{3})^n + (1 - \sqrt{3})^n$ for $n \ge 0$.
Since $1 \pm \sqrt{3}$ are roots of $(\lambda - 1)^2 - 3 = \lambda^2 - 2\lambda - 2$, $a_n$ satisfies a recurrence relation:
$$a_{n+2} = 2(a_{n+1} + a_n)$$
Define two auxiliary sequences $b_n, c_n$ by
$\begin{cases}
b_n &= 2^{-(n+1)} a_{2n}\\
c_n &= 2^{-(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910795",
"timestamp": "2023-03-29T00:00:00",
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System of congruences with moduli not coprime and chinese remainder theorem application Find all solutions, if any, to the system of congruences
$$\Bigg\{
\begin{array}{c}
x \equiv 1 (\mod 6)\\
x \equiv 7 (\mod 9)\\
x \equiv 4 (\mod 15)\\
\end{array}$$
we can see that
$ x \equiv 1 \pmod 6 \implies
... | It is correct, but you can make the solution a bit shorter using the explicit inverse isomorphism in the C.R.T. for two coprime moduli $a$ and $b$, given a Bézout's relation $ua+vb=1$:
$$x\equiv \begin{cases}\alpha\mod a,\\\beta\mod b,\end{cases}\iff x\equiv \beta ua+\alpha vb\mod ab$$
As $4\cdot 3^2-7\cdot 5=1$, th... | {
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$\sum_{k=2}^\infty \frac{1}{1+(-1)^k \sqrt k}$ converges or diverges? $\frac{1}{1+\sqrt k (-1)^k}= \frac{1-(-1)^k \sqrt k}{1-k}$. Not quite sure how to move on from here.
| Rewrite
\begin{align}\sum_{k=2}^{2N+1}\frac{1}{1+(-1)^k \sqrt k}&= \sum_{k=1}^N\left(\frac{1}{1+(-1)^{2k} \sqrt {2k}}+\frac{1}{1+(-1)^{2k+1} \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1}{1+\sqrt {2k}}+\frac{1}{1- \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1- \sqrt {2k+1}+1+\sqrt {2k}}{(1+\sqrt {2k})(1- \sqrt ... | {
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Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{... | Let $\cos^{-1}\dfrac35=y,\cos y=?$
Using Principal values, as $1>\dfrac35>0; 0<y<\dfrac\pi2$
$\tan y=\dfrac{+\sqrt{1-\left(\dfrac35\right)^2}}{\dfrac35}=?, y=\tan^{-1}\dfrac43$
$$\pi-2\tan^{-1}2=2\left(\dfrac\pi2-\tan^{-1}2\right)=2\cot^{-1}2$$
Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
and
Inv... | {
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"answer_id": 3
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How to prove that $n!>2^n$ holds for all $n>3$? I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{p... | Since $n\ge 4$ we have $n! = 1\cdot 2\cdot 3\cdot 4 \cdot 5\cdots n = 24 \cdot 5 \cdots n > 16 \cdot 5 \cdots n\ge 2^4\cdot 2 \cdots 2 = 2^42^{n-4}=2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Number theory in sequence $x_{n+1}=x_n^3-2x_n^2+2$ $x_1=5, x_{n+1}=x_n^3-2x_n^2+2$
Prove that, there is no prime $p=4k+3(k>1)$ and $p\mid x_n^2-3x_n+3$
I think I can have $p\mid t^2+1$ and then I have QED.
But $p\mid x_n^2-3x_n+3$ means that $p\mid (2x_n-3)^2+3=t^2+3$
What should I do next?
| I found this to be a quite interesting (and challenging to solve) question. It's asking about the primes $p$ where
$$p \mid x_n^2 - 3x_n + 3 \tag{1}\label{eq1A}$$
As you showed, multiplying by $4$ gives
$$p \mid 4x_n^2 - 12x_n + 12 = (2x_n - 3)^2 + 3 \tag{2}\label{eq2A}$$
Since $p \neq 3$ (note $x_n \equiv 5 \pmod{72}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Typesetting Multiline Inequalities If we write
$$\begin{align}
a&<b\\
&=c\end{align}$$
does this mean
$$a<b=c$$
or
$$a<b; a=c$$
?
How do we write the other in multiline format?
(Edited to Add)
What about
$$\begin{align}
a&<b\\
&=c\\
&=d\\
&=f\end{align}$$
?
| Generally,
\begin{align*}
a &< b \\
&= c
\end{align*}
is to be interpreted as $a < b = c$. This can be convenient for a series of derivations or inequalities, when putting them on the same line could be ugly or hard to follow. To pull an example from a homework of mine:
\begin{align*}
f(z)
&= \frac{1}{(z-ia)^2} \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits I must use the two standard limits $\lim\limits_{n \rightarrow \infty} \frac{e^{\alpha_n}-1}{\alpha_n} = 1$ and $\lim\limits_{n \rightarrow \infty} \frac{\ln(1+\beta_n)}{\beta_n} = 1$ if $\alpha_n, \beta_n \rightarrow 0$ so m... | $$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$
$$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$
$$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifyi... | Case 1: Let $ab+bc+ca\ge a+b+c$, then the AM-HM of $a,b,c$ with weights as $1/b,1/c,1/a$ gives
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{1/a+1/b+1/c)^2}{1/(ab)+1/(bc)+1/ca)}=\frac{(ab+bc+ca)(1/a+1/b+1/c)}{a+b+c} \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Case 2: Let $(ab+bc+ca) \le a+b+c$, then AM-HM of $1/b,1/c,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so tha... | Its wrong ! the step $$A(x+4)+B(x-1) = x$$ is not true instead you should have $$Ax(x+4)+B(x-1)=x$$ As you can see we would then have to take $A=0$ and the purpose of partial fractions is gone
I prefer $$\frac{x}{x^2-3x+4}=\frac{1}{5}(\frac{1}{x-1})+\frac{4}{5}(\frac{1}{x+4})$$
another way is to write $x=l(2x-3)+m$ (be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $ \sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$ Problem proposed for JBMO practice in symmetrical inequalities (Chebyshev, rearrangement):
For every positive real numbers $a, b, c$, for which $a + b + c = 3$ we have:
$$\sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + ... | By AM-GM, we have $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \le a + b + c = 3$. Thus,
$\mathrm{RHS} \ge \frac{1}{3}$.
By Chebyshev's sum inequality, we have
\begin{align}
a^3 + b^3 + c^3 - (a+b+c) &= (a-1)(a^2 + a) + (b-1)(b^2+b) + (c-1)(c^2+c)\\
&\ge \frac{1}{3}(a-1 + b-1 + c-1)(a^2+a + b^2+b + c^2 + c)\\
& = 0.
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Solving $(\sin x-\cos x)^2+\tan x=2\sin^2x$ From this post (Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$), which is my post, I have tried solving the last equation. I have gotten a solution(s). Would like to see if I did correctly.
$$\bigl(\sin\left(x\right)-\cos\left(x\right)\bigr)^2+\tan\left(x\right)=2... | Alternatively
\begin{align}
& (\sin x-\cos x)^2+\tan x-2\sin^2x\\
=& (1-\sin2x )+\tan x -(1-\cos 2x) \\
=&- \frac{2\tan x}{1+\tan^2 x}+ \tan x+\frac{1-\tan^2x}{1+\tan^2 x}\\
= &\frac1{1+\tan^2x}(\tan x-1)^2(\tan x+1)=0
\end{align}
Thus, $\tan x= \pm1$, leading to the solutions
$$x =\frac\pi4 +\frac{\pi n}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac... | Multiply the two series $\ln(1+x) = x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \cdots$ and $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$.
For instance, the coefficient of $x^2$ in the product is $(-\frac{1}{2} x^2) \cdot 1 + x \cdot (-x) = -(1 - \frac{1}{2}) x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent? I used the definition.
$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$
$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$
I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$
$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\... | Once you know that $$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2},$$ you already know that $$(1/2, 1/2, -1) \cdot (1, \cos x, \cos^2 \tfrac{x}{2}) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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matrix and linear transformations With regards to this (Matrix associated to a linear transformation with respect to a given basis)
I saw Ivo Terek solution which goes:
\begin{align}L(1,1,1) &= (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
2\cdot (1,0,0) \\ L(1,1,0) &= (0,8,2) = 2\cdot (1,1,1) + 6\cdot (1,1,0) -
8\cdot... | I suppose you understand the first equality in
$$L(1,1,1) = (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
2\cdot (1,0,0)$$
well, as $L$ is defined by $L \left( \begin{bmatrix}x_{1}\\x_{2}\\x_{3} \end{bmatrix} \right) = \begin{bmatrix}4x_{3}\\3x_{1}+5x_{2}-2x_{3}\\x_{1}+x_{2}+4x_{3} \end{bmatrix}$ in the linked question.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Surface integral over ellipsoid $ax^2+by^2+cz^2=1$ I'm not sure how to compute the integral
$$\int_{s} (a^2x^2+b^2y^2+c^2z^2)^{-1/2} d\vec{S}\cdot \vec{n}$$
over the surface of the ellipsoid
$$ax^2+by^2+cz^2=1$$, $$z>0$$
Where $\vec{n}$ is the unitary normal vector to the surface.
Any help is appreciated. Thank you.
| We propose to attack the problem with a blunt instrument: on the surface
$$\vec{r}=\langle x,y,z\rangle=\left\langle x,y,\pm\sqrt{\frac{1-ax^2-by^2}{c}}\right\rangle$$
so
$$\begin{align}d\vec{r}&=\left\langle1,0,\mp\frac{\frac{ax}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dx+\left\langle0,1,\mp\frac{\frac{by}c}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ ... | My third solution:
Actually $a^2 + b^2 + c^2 + 6 - 3(a+b+c) + (abc - 1) \ge 0$ for all $a, b, c \ge 0$.
(pqr method)
Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
By Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$
which is written as $p^3 - 4pq + 9r \ge 0$, we have $\frac{p^3 + 9r}{4p}\ge q$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$ Which change of variable can be used to show
$$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$
(direct integration isn't allowed, must use change of variable)
| $$I=\int_0^\infty\frac{1}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}dx+\int_1^\infty\frac{1}{1+x^2}dx$$
now:
$$I_1=\int_1^\infty\frac{1}{1+x^2}dx$$
$$u=\frac1x,du=-1/x^2dx\Rightarrow dx=-\frac{1}{u^2}du$$
$$I_1=-\int_1^0\frac{1}{1+(1/u)^2}\frac{du}{u^2}=\int_0^1\frac{1}{1+u^2}$$
$$\therefore I=2\int_0^1\frac{1}{1+x^2}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian
Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$
$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
| Hint:
Let $7x-24y=z\iff x=?$
Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$
As $y$ is real, the discriminant must be $\ge0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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partial fractions when the fraction cannot be decomposed I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can... | $\frac{1}{(x^2+1)^2}$ is already in "partial fractions form", so...
"Is it just manipulation and trial and error?"
Yes.
It's not super trivial, but the result is true.
I did:
\begin{align} \frac{1}{(x^2+1)^2}
&= \frac{1 + x^2 - x^2}{(x^2+1)^2} \\\\
&= \frac{1}{(x^2+1)}\ - \frac{x^2}{(x^2+1)^2} \\\\
&= \frac12\left[\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when serie... | First of all $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\frac{\ln{x}}{\left(x-1\right)\sqrt{x+1}}\cdot\frac{1}{\sqrt{x-1}}\underset{x\to 1}{\sim}\frac{1}{\sqrt{2}\sqrt{x-1}} $, and we know that $ \int_{1}^{a}{\frac{\mathrm{d}x}{\sqrt{x-1}}} $ converges for any $ a>1 $.
We also have that $ \frac{\ln{x}}{\left(x-1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3... | Edit
First see the comment of NinadMunshi immediately following this answer. I have edited the answer accordingly.
Continuing Infinity_hunter's answer, and letting
$$x_n~ \text{denote}~
\sqrt{12 + \sqrt{12 + \cdots \text{n times} }}
$$
To show convergence, all that is necessary is to show that
*
*the sequence is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Evaluate the integral $\int_0^{\pi/2}\frac{\tan x}{\tan(\frac{x}{2})} \, dx$ Evaluate the integral.
$$\int_0^{\frac{\pi}{2}}\frac{\tan{x}}{\tan(\frac{x}{2})}\,dx$$
I tried to solve it with $u = \tan{x/2}$, but i got divergent part of the solution. How can I integrate it, such that, when boundaries are plugged in, the r... | Write $$\frac{\tan x}{\tan \frac{x}{2}} = \frac{\sin x}{\cos x} \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\sin\frac{x}{2} \cos\frac{x}{2}}{\cos x}\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\cos^2\frac{x}{2}}{\cos x} = \frac{1 + \cos x}{\cos x}$$Then your integral reduces to $$\int_0^{\frac{\pi}{2}} (\sec x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Can each root equation be rearranged to a corresponding polynomial equation? Let $n\in\mathbb{N}_+$.
Can each root equation of $n$ unknowns be rearranged to a polynomial equation of $n$ unknowns whose solution set contains the solution set of the root equation?
If not, is this true at least for root equations of one or... | By irrational equation, I assume you meant one with one or more variable under a radical as assigned here. The technique of raising both sides to the same power may be invoked repeatedly if needed until there are no radicals. Take, for example.
$$A=\sqrt{x}+\sqrt{y}\implies A^2=\big(\sqrt{x}+\sqrt{y}\big)^2 = 2 \sqrt(x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Determinant of $n\times n$ Matrix Linear Algebra So, I have a matrix
$$
A = \begin{pmatrix}
0 & 1 & 1 & ... & 1 \\
1 & 0 & x & ... & x \\
1 & x & 0 & ... & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x & x & ... & 0
\end{pmatrix}
$$
I need to evaluate it's determinant. At first I calculated $\... | First factor out $1+x(n-2)$ from all columns except the first to get
$$\det A(x)=\begin{vmatrix}
n-1 & 1 & 1 & \cdots & 1 \\
1-x & -x & 0 & \cdots & 0 \\
1-x & 0 & -x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1-x & 0 & 0 & \cdots & -x
\end{vmatrix}$$
and note that this cancels the multp... | {
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"url": "https://math.stackexchange.com/questions/3960714",
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"source": "stackexchange",
"question_score": "3",
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Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqr... | Here is a solution that does not involve derivatives.
By AM-GM, we have that $a+b \geq 2\sqrt{ab}$. Hence it suffices for us to prove that $2\sqrt{ab}+\dfrac{1}{1+\sqrt{ab}} \geq \dfrac{5}{2}.$
From the given condition, $\dfrac{1}{a} + \dfrac{1}{b} =2 \Rightarrow a+b=2ab \Rightarrow 2ab \geq 2\sqrt{ab} \Rightarrow \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how ... | Denote $t:=\sqrt{45x^2-30x+1}$. Then we observe that
$$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$
As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$
Since $t\geq 0$, it follows that $t=4$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$
Hence $(3x+1)(x-1)=0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation of Cosine Series Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $
My approach is as follow
$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k ... | Move the last term of the sum out of the sum. The remaining sum will match your reference.
$$\sum_{k=0}^n f(k) = \left( \sum_{k=0}^{n-1} f(k) \right) + f(n) \text{.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \in... | From the binomial theorem, we know $(1+y)^n\approx1+ny$ for $y\approx0$. Divide both numerator and denominator by $x^{2/3}$. The limit is$$\lim_{x \to \infty}{\frac{\left(1+\frac1x\right)^{\frac{2}{3}}-\left(1-\frac1x\right)^{\frac{2}{3}}}{\left(1+\frac2x\right)^{\frac{2}{3}}-\left(1-\frac2x\right)^{\frac{2}{3}}}}=\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Solve $ \int\frac{1}{\sin(x)-\cos(2x)}dx $ ... Weierstrass substitution :
$$\tan(\frac{x}{2})=t$$ $$\sin(x)=(\frac{2t}{1+t^2})$$ $$\cos(x)=(\frac{1-t^2}{1+t^2})$$ $$dx=(\frac{2\,dt}{1+t^2})$$
Than : $$\cos(2x)=\cos^2(x)-\sin^2(x)$$
P.s I tried using these substitutions, but I couldn't get so far .
Need a bit help to so... | Since $\cos2x=2\cos^2x-1=\frac{1-6t^2+t^4}{(1+t^2)^2}$, $\sin x-\cos 2x=\frac{-1+2t+6t^2+2t^3-t^4}{(1+t^2)^2}$, so your integral is$$\int\frac{2(1+t^2)}{-1+2t+6t^2+2t^3-t^4}dt.$$Sincce $-1+2t+6t^2+2t^3-t^4=-(t^2-4t+1)(t+1)^2$ (the repeated root is easily guessed with the rational root theorem), we seek a partial fracti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $ How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$
My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$ (\sin x + \sin x \cos x)^2 = (\frac{1... | You may continue with
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
knowing that $ \cos x = \frac1{\sqrt2} $ is a solution from inspection, as commented. So, factorize the equation as
$$ (\cos x - \frac1{\sqrt2})(\cos^3x +(2+\frac1{\sqrt2})\cos^2x+ (\frac12+\sqrt2)\cos x-1+\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integer solutions for the following equation: Find all positive integer solutions for the following equation:
$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$
I've tried simplifying the equation and then refactoring but I can't find any solutions.
| We write the given equation equivalently:
$$
\begin{aligned}
0 &= x^2y^2 - 2x^2y + 2xy^2 + x^2 - 4xy + y^2 + 2x - 2y - 3\ ,\\
0 &= x^2(y-1)^2 + 2x(y-1)^2 + (y-1)^2-4\ ,\\
4 &= (x+1)^2(y-1)^2\ .
\end{aligned}
$$
Now consider all possible ways to write $4$ as a product of two perfect squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Series representation of the reciprocal of prime In the article on unique primes in Wikipedia. It is stated that the representation of the reciprocal of a prime $p$ in the numeral base $b$ is periodic of period $n$ if
$$\displaystyle {\frac {1}{p}}=\sum _{i=1}^{\infty }{\frac {q}{(b^{n})^{i}}}$$ where $q$ is a positive... |
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds
Well, $(0.\overline{a_1a_2a_3\dots a_n})_b=0.a_1a_2a_3\dots a_na_1a_2a_3\dots a_na_1a_2a_3\dots a_n\dots$
$=\dfrac{a_1}{b}+\dfrac{a_2}{b^2}+\cdots\dfrac{a_n}{b^n}+\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
What I Tried: I thought... | You can first get rid of the denominator in each fraction by multiplying above and below by the conjugate expression of the denominator. Then you will have 6th powers above, and you can get rid of the cubic roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art o... | You're ever so slightly off. Notice the median of $(-5, -4, ..., 5, 6)$ is $\frac{-5+6}{2}=\frac 12$ which hints at trying $$f(\frac12+x)+f(\frac 12-x)\overbrace{=}^{y=x+\frac 12}f(y)+f(1-y)$$
We see that: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ a... | I would have calculated it like this:
Let $u=1-\sqrt{2+x} \Leftrightarrow x=(1-u)^2-2$ where $x\ge-2$. We then have
$$ \mathrm dx=2(1-u)(-1)\,\mathrm du$$
so
\begin{align*}
\int_2^7\!\frac{x}{1-\sqrt{2+x}}\,\mathrm dx
&=\int_{-1}^{-2}\!\frac{(1-u)^2-2}{u}\cdot2(1-u)(-1)\,\mathrm du
\\&=2\int_{-2}^{-1}\!\frac{((1-u)^2-2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$? $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$
$Domain: x\in (-\infty;-2]\cup(1;+\infty)$
$x = 2\longrightarrow$ Done
How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?
Any better way su... | Hint:
*
*Note that $$(x-1)\sqrt{\frac{x+2}{x-1}} = \sqrt{(x-1)(x+2)}$$ when $x>1$ and
$$\begin{align}(x-1)\sqrt{\frac{x+2}{x-1}} &= -(1-x)\sqrt{\frac{x+2}{x-1}} \\[1mm]&= -\sqrt{(1-x)^2\frac{x+2}{x-1}}\\[1mm]&=-\sqrt{(x-1)(x+2)}\end{align}$$ when $x \le -2$.
*Substitute $$t = \sqrt{(x-1)(x+2)}$$ Then, how will the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic i... | Update:
I found that this solution provides a neat way in general to solve this kind of problems. I will apply the method here:
$$(2n+1)2^{4n+5} \equiv 3 \pmod 7 \\ \iff g(n)=(2n+1)2^n\equiv (2n+1) 2^{4n+6} \equiv 3\cdot 2 \equiv -1 \pmod 7$$
Suppose $n=3k+i, i=0,1,2$, then
$$-1 \equiv g(3k+i) = (6k+2i+1)2^{3k+i} \equi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please... | Let $f(a,b)=a^3+b^3-6ab+11$. At the extrema, the two curves $f(a,b)=0$ and $k=a+b$ are tangential to each other, i.e.
$$\frac{f_a’}{f_b‘}=\frac{3a^2- 6b}{3b^2-6a}=1\implies (a-b)(a+b+2)=0
$$
which leads to the tangential points $a=b$ and $a+b=-2$. It is straightforward to verify the maximum $a+b<-2$ and the minimum is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$ Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$, except when $x$ is midway between two integers $n$ and $n+1$, when it is the smaller of these two integers.
We want to prove 2 cases he... |
The problem is why the ceiling function was changed to floor function?
Because $n$ is an integer, $\lfloor n\rfloor= \lceil n \rceil = n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there an other way for it? Consider two equations:
$x+y=2$$y+z=4$
Find the value of $(x+z)$.($x,y,z$ all are positive real numbers)
My Approach:
$\because x+y=2 \Longrightarrow y=0$ or $y=1$.
Case 1:
taking $y=0$$ \Longrightarrow$ $x=2$ ; $z=4$.$\therefore x+z = 2+4 = 6.$Case 2:
taking $y=1 \Longrightarrow$
$x=1$ ... | Given
$\qquad x+y=2\space\land\space y+z=4\qquad$
we can start by eliminating $y$.
$$
y+z=4\implies z=4-y\\
(x+y=2)\implies x=2-y\\
z-x=(4-y)-(2-y)=2\qquad\longrightarrow (z-x=2)\\
(x+z=4)-(z-x=2)\implies 2x=2\implies x=1\\
(x+z=4)+(z-x=2)\implies 2z=6\implies z=3\\
\therefore\quad x+z=1+3=4
$$
At this point we can als... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving state-space function with using of Runge-Kutta method I need to implement my own integration routine that will take state space function $f$, free variable $t$, and initial state $x(0)$ as input and produce the solution $x(t)$ as output. I thought that using Runge-Kutta method will be great. But I cannot unders... | You have a system $\dot{x} = f(x)$ with
$$
f(x) = A x = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix} = \begin{pmatrix}
x_2 \\
x_3 \\
-10 x_1 - 5 x_2 - 2 x_3
\end{pmatrix}
$$
The formula for Runge Kutta with step size $h$ is:
$$
\begin{align}
k_1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that the function $f(x) = \frac{x}{1+x^2}$ is continuous using the $\epsilon \delta$ definition I am attempting to prove that $f(x) = \frac{x}{1+x^2}$ is continuous. Here's my attempt so far.
In order to prove that $f(x)$ is continuous, it is necessary to show that for every $\epsilon>0$ there exists a $\delta>... | When $c \neq 0$, use the bound
$$\tag{*}\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| = \frac{|x-c+c^2x-cx^2|}{(1+x^2)(1+c^2)} =\frac{|x-c+c^2x-cx^2+c^3 - c^3|}{(1+x^2)(1+c^2)} \\ \leqslant\frac{|x-c|+c^2|x-c|+|c||x^2-c^2|}{(1+x^2)(1+c^2)}= \frac{1+c^2+|c||x+c|}{(1+x^2)(1+c^2)}|x-c|$$
If $|x-c| < \frac{|c|}{2}$, then, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all positive integers $x$ such that $x(x+2021)$ is a perfect square I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you
I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$
| Start from $$x (x+2021)=y^2$$
multiply both sides by $4$ and add $2021^2$ to both sides
$$4 x^2+8084 x+2021^2=4 y^2+2021^2\to (2x+2021)^2=4y^2+2021^2$$
set $2x+2021=z$ and remember that $2021^2=43^2\times 47^2$. The equation becomes
$$z^2-4y^2=43^2\times 47^2\to (z+2y)(z-2y)=43^2\times 47^2$$
we get the following possi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thank... | A cubic has only one real root iff its discriminant is negative. (See Wikipedia.)
The discriminant of $1 + a^2 + ax - x^3$ is $-27 a^4 + 4 a^3 - 54 a^2 - 27$.
Now, $-27 a^4 + 4 a^3 - 54 a^2 = a^2 (-27 a^2 + 4 a - 54)$ and $-27 a^2 + 4 a - 54=-27 \left(a - \frac{2}{27}\right)^2 - \frac{1454}{27} < 0$ for all $a$.
Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Complex Integral : How can I evaluate $\int_{C_R} f(z) dz$? I have to calculate $I$ ,using complex integral.
\begin{equation}
I:=\displaystyle\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}.(a>0)
\end{equation}
Let $f(z)=\dfrac{\text{log}(z+ia)}{1+z^2}$.
$
C_R : z=Re^{i\theta}, \theta : 0 \to \pi.$
$C... | $$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=
\int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$
$$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$
$$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$
$$\frac{dI}{da}=\frac{2}{a^2-1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996356",
"timestamp": "2023-03-29T00:00:00",
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A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that $x_{2017} + x_{2023} = 990$ ----- (1) and $x_{n+1} = 3x_n + \... | After you get $x_{n+1} - 6x_n + x_{n-1}=0$ as in the other two answers, a conceptually easy way to establish the relationship between $x_n$ and $x_{n-3}+x_{n+3}$ is as follows:
Rewrite the recurrence equation as
$$(\mathbb E^2-6\mathbb E+1)x_n=0$$
where $\mathbb E$ is the forward shift operator $\mathbb E^{i} x_n = x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series Find limit without using L'Hospital or Taylor's series:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$
$$\displaystyle \lim_{x \to 0} \frac{... | Just a slight variant on Bernard's answer: rewrite the fraction as
$${\ln(1+\sin(2x^5))\over\sin(2x^5)}\cdot{\sin(2x^5)\over2x^5}\cdot\left(3x\over\sin(3x)\right)^3\cdot{x^2\ln5\over e^{x^2\ln5}-1}\cdot{2\cos^3(3x)\over3^3\ln5}$$
then use the familiar limits
$$\lim_{u\to0}{\ln(1+u)\over u}=\lim_{u\to0}{\sin u\over u}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem on inverse trigonometric function How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=... | The series transformations might seem at first glance rather arbitrary. But there are some nice aspects behind the curtain which can be sometimes conveniently used.
Let's assume we want to calculate
\begin{align*}
\color{blue}{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{2n^2}\right)}\tag{1}
\end{align*}
We know that tel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Nested equilateral triangles Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.
Firstly, I would be very grateful if someone can explain to me how I am su... | As others have done, I'll assume $AB=8,DF=5$ without loss of generality. Then equilateral triangles $\triangle ABC$ and $\triangle DEF$ have area $(\sqrt{3}/4)AF^2 = 16\sqrt{3}$ and $(\sqrt{3}/4)DF^2=(25/4)\sqrt{3}$ respectively. Since the triangles $\triangle FAE,\triangle DBF, \triangle ECD$ are all congruent, they ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Legendre 3 Square problem I got a little help on the way here yesterday but it seems like my question is dead.
Problem:
If $ n \in \mathbb{N} $ can be represented as $ n = n_1^2 + n_2^2 + n_3^2 ,\quad n_1, n_2, n_3 \in \mathbb{Z}, $ show that then $ n \neq 4^a(8m+7)$ for $a,m \in \mathbb{Z}.$
Solution:
$ 4^a(8m+7) \equ... | What you have done so far is to prove that:
Let $a \geq 1$. If $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}4 = 4^{a-1}(8m+7)$.
Applying this observation $a$ times, one notices that if $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}{4^a} = 8m+7$. Thus,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.
Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$
My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0... | This integral does have a closed form despite being very messy. We have \begin{align}I&=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}\,dx\\&=\frac14\int_{-\pi}^\pi\frac{\cos^3(u/2)}{\sin^2(u/2)+\cos^3(u/2)}\,du\\&=\frac14\oint_{|z|=1}\frac{\frac18(z^{1/2}+z^{-1/2})^3}{-\frac14(z^{1/2}-z^{-1/2})^2+\frac18(z^{1/2}+z^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denomin... | $$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$
plug $x=y+3$ so we get
$$\underset{y\to 0}{\text{lim}}\frac{\sqrt{16-y}-2 \sqrt[4]{y+16}}{\sqrt[3]{8-y}-y-2}$$
we have the following expansions as $y\to 0$
$$
\begin{array}{rll}
\sqrt{16-y} &\sim &4-\frac{y}{8} \\
2 \sqrt[4]{y+16} &\sim& \frac{y}{16}+4 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we hav... | From the well known
$$x-1<\lfloor x\rfloor \leq x \tag{1}$$
Thus
$$\frac{1+na^2}{a}-1<
\left\lfloor \frac{1+na^2}{a}\right\rfloor\leq
\frac{1+na^2}{a}$$
and
$$n<n+\frac{1}{a^2}<
\frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a}\leq
n+\frac{1}{a^2}+\frac{1}{a}\tag{2}$$
Finally, for $\color{red}{a>\phi}$
$$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:
*
*Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.
*Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $... | Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.
From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.
To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.
Thus, the equation $(3)$ will be true when either $f(x)=0$, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Integral of $\sqrt{4-x^2}$ by Riemann sum I was trying to compute $\int_{0}^{2}\sqrt{4-x^2}\,dx$ by Riemann sum like this
$$\sum_{i=1}^{n}\sqrt{4-x_i^2}\,\Delta x_i$$
where $x_i = \frac{2i}{n}$ and $\Delta x_i = \frac{2}{n}$, but I couldn't get something useful for the answer. I think it should be with another kind of ... | The right Riemann sum using the partition with points $x_k = 2 \sin \frac{\pi k}{2n}\,\,(k = 0,1,2,\ldots, n)$ is
$$\tag{*}S_n = \sum_{k=1}^n \sqrt{4 - 4 \sin^2 \frac{\pi k}{2n}}\left(2 \sin \frac{\pi k}{2n}- 2\sin \frac{\pi (k-1)}{2n}\right)\\ = 4\sum_{k=1}^n \cos \frac{\pi k}{2n}\left( \sin \frac{\pi k}{2n}- \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the Fourier Coefficient to this discrete time signal? I'm having trouble finding the fourier coefficients to this discrete time signal:
My approach:
I start by first finding the discrete period, $N$, and discrete frequency, $\Omega_0$
From the problem, we can easily identify that $N = 10$ and so $\Omega_{0... | Your second sum should index from $n=5$. Then here is a small simplification:
\begin{align}
F[k] &= \frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=5}^9e^{-ink\pi/5} \Big) \\
&=\frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=0}^4 e^{-5ik \pi/5} e^{-ink\pi/5} \Big) \\
&= \frac{7(1 - (-1)^k)}{20} \sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N? Pls help me with A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N? I amen't solve it. Help needed
| A number $\overline{xyz}$ in base $b$ is equal to $x\cdot b^2 + y\cdot b + z$ and so on for larger numbers continuing the pattern increasing the power of $b$ as we continue to the left. For instance, in base 10 we have $35013$ is equal to $3\cdot 10^4 + 5\cdot 10^3 + 0\cdot 10^2 + 1\cdot 10 + 3$.
So, we are told that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computation of density function of $Y=\frac1X-X$ where $X\stackrel{\mathrm d}= U[0,1]$
Calculate the density function of $Y=\frac1X-X$, where $X\stackrel{\mathrm d}= U[0,1]$.
I am confused about the technique to deal with the problem like this when after the transformation, I can not express the density of $Y$ in ter... | $1/X-X$ is a decreasing function of $X$ so $Y\in(0,\infty)$ for $X\in(0,1)$.
$$P(Y\le y)=\begin{cases}0,&y\le0\\P(1-X^2\le yX),&y>0\end{cases}$$
Now $X^2+yX-1$ is an upward opening parabola with zeroes at $\frac{-y\pm\sqrt{y^2+4}}2$, and thus $$\begin{align*}P(X^2+yX-1\ge0)&=\underbrace{P\left(X\le\frac{-y-\sqrt{y^2+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that sin A/2 * sin B/2 * sin C/2 = r/4R The other day I came across an identity in the book "Problems from the Book" and it was presented as well known:
$$\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{r}{4R}$$
However I wasn't familiar with thee identity so I tried proving it.
Here is par... | $$LHS = \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} $$
$$= \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-c)(s-a)}{ca}} \cdot \sqrt{\frac{(s-a)(s-b)}{ab}}$$
$$= \frac{(s-a)(s-b)(s-c)}{abc}$$
Note that $$r = \frac{\Delta}{s}$$
$$R = \frac{abc}{4 \Delta}$$
$$\Delta^2 = s(s-a)(s-b)(s-c)$$
Now $$ RHS... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$ $$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$
I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?
Thanks!
| Note $(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=(\sqrt[3]{2})^3-1=1$ and
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}
=& \sqrt[3]{\frac1{\sqrt[3]{4}+\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{(\sqrt[3]{2}+1)^3}}=
\frac{\sqrt[3]{3}}{\sqrt[3]{2}+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Determine if $x^3+y^3+z^3+t^3 = 10^{2021}$ has a solution I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions
PowersRepresentations[10^2021, 4, 3]
return
PowersRepresentations::ovfl: Overflow occurred in computation.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x ... | Easy, notice that $10^{2021}=100\times 10^{3\times 673}$. Next use your code, but for the factor 100.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 100, 0<x<y<z<t}, {x,y,z,t}, Integers]
yielding a single result
(*{{x -> 1, y -> 2, z -> 3, t -> 4}}*)
Now verify the solution
(x^3 + y^3 + z^3 + t^3 /. {x -> 1 10^673,y -> 2 10^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-
Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)
Developments-
Firstly applying... | I will use a well known inequality for my solution -
$a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$
$\implies (ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$
Now using AM-GM,
$a^2+b^2+c^2+2 (ab + bc + ca) \geq 3 [(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$9 \geq 3[(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$3^3 \geq 9 abc(a^2+b^2+c^2)$
$3 \geq abc(a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that $f$ verifies $f(1-x)=-f(x)$ I'm trying to show an identity verified by a function. I have this function which is defined for all real numbers as
$ f(x) = \sum_{k=0}^{p-1} \frac{(x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}$
I would like to show that $f(1-x) = -f(x)$.
First thing that tried:
$
\begin{align*}
f(... | We can rewrite the sum as
$$
\eqalign{
& f\left( x \right) = \sum\limits_{k = 0}^{p - 1} {{{\left( {x + k} \right)^{\,2m - 1} }
\over {\left( { - 1} \right)^{\,p + k} \left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} = \cr
& = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k} {{\left( {x + k} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0)
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) but the restritions to straight line in $\mathbb{R}^2$ is cont.
My attempt:
**Proving continuity **
if $f(x,y)=\... | Equation $y^6+x^2=0$ has only one solution in #\mathbb{R}^2$, and that is $(0,0)$.
To prove that the function is unbounded in the neighbourhood of $(0,0)$ consider (for $y\neq 0$)
$$ f(y^3,y) = \frac{y^3 \cdot y^2}{(y^3)^2+ y^6} = \frac{1}{2y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An Elegant New Proof of Herons Formula? A triangle with side lengths $a, b, c$ an altitude($h$), where the height($h_a$) intercepts the hypotenuse($a$) such that it is the sum of two side lengths, $a = u +v$ and height($h_b$) intercepts hypotenuse($b$) such that it is also the sum of two side lengths $b = x + y$, we ca... | It is certainly a correct proof of Heron's formula that I have not seen before. I don't believe that it is any simpler than the other proofs that I have seen, but I am still entertained by it. However, I do not know if this is an appropriate posting since you do not really have a question other than possibly 'is my pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove thi... | First approach. By the beta integral
\begin{align*}
\frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the last two digit of $5^{121}*3^{312}$ The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$
and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of $5^{143}*3^{312}$ are $25$.
Don't know why $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies th... | The first sentence says that whatever power greater than $1$ you pick, $5$ to that power ends in $25$. Note that the powers of $5$ are $5,25,125,625,3125\ldots $ The end digits are always $25$ after the first one because when you multiply some number of hundreds by $5$ you still have some number of hundreds and when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$
I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. Bu... | (Here's a suggested approach. If you're stuck, explain where you're stuck and what you've tried.)
*
*Factor it as $ 3 \times 2^m = (n-2) ( n+2)$.
*Hence $ \{ n-2, n+2 \} = \{ 3 \times 2^a, 2^b \}$.
*Hence $ 3 \times 2^a = 2^b \pm 4 $.
*If $ b = 0, 1, 2$, what can we say? What are the solutions with these small ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels? A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the large... | Another way to add up your diminishing circle circumferences
is to use the formulas $2\pi R = \frac LN$ and $N\pi = \frac{L}{2R},$
both of which are derived from $N = \frac{L}{2\pi R}.$
\begin{align}
\left(L - \frac LN\right)2\pi &+ \left(L - 2\frac LN\right)2\pi
+ \left(L - 3\frac LN\right)2\pi + \cdots \\
&= \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
I attempted to solve this question as follows:
$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le... | It is just Schur’s $^{[1]}$ Inequality of degree 3. You can search the web and read about it. It is pretty well known and someone named Aritra12 posted a handout on it recently on Aops.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verif... | Partial answer :
Well we can use Jensen's inequality remarking that :
$$f(x)=\ln\Big(\tan\Big(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\Big)\Big)$$
Is concave on $[\frac{-17}{100},\frac{17}{100}]$
So we have :
$$f(x)+f(-x)\leq 2f(0)$$
A bit of algebra and we get the result !
We can improve the reasoning for that we use a subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes I have come to a problem in a multivariable calculus book that I'm having trouble with.
The problem statement is :
"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{... | Because of the symmetry of spheres with respect to rotation, the statement “for any point of intersection between two spheres $A$ and $B$, the corresponding tangent planes are perpendicular” is equivalent to “between two spheres $A$ and $B$, there is a point of intersection at which the corresponding tangent planes are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).
By and large these things require considerations o... | Let
$$s_1=\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}
$$
$$s_2=\cos\frac{6\pi}{17} + \cos\frac{10\pi}{17} + \cos\frac{12\pi}{17} + \cos\frac{14\pi}{17}
$$
and note
$$s_1s_2=2(s_1+s_2) =2\cdot (-\frac12)=-1
$$
Thus, $s_1$ and $s_2$ satisfy $s^2+\frac12 s-1=0$, which yields $s_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Easiest way to solve $\frac{2}{9}x=\sin(\frac{\pi}{9}x)$ We wanted to find the points of intersection between 2 curves, so we have to solve this equation $\frac{2}{9} x= \sin(\frac{\pi}{9}x)$.
I couldn't figure out an easy way to find the points of intersection, they are $-\frac{9}{2}$, $0$, and $\frac{9}{2}$.
Any idea... | uhm the following way isn't that easy... you have to consider several properties of functions...
first off: $\sin(\dfrac{\pi x}{9}) = \dfrac{2x}{9} \iff \sin^{-1}(\dfrac{2x}{9}) = \dfrac{\pi x}{9}$
Let: $f(x) =\sin^{-1}(\dfrac{2x}{9}) - \dfrac{\pi x}{9} $
Our goal is to find where $f(x) = 0$
Notice $f(x) = -f(-x) \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4062283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that $(X-1,Y^2-4Y-XY+Y+4)=(X-1,Y-2) $ We consider the ideal $I=(X-1,Y^2-4Y-XY+Y+4)$ of the polynomial ring $k[X,Y]$. I need to prove that $(X-1,Y^2-4Y-XY+Y+4)=(X-1,Y-2) $. Since $Y^2-4Y-XY+Y+4=(Y-2)^2-Y(X-1)$, it is easy to see that $I \subseteq (X-1,Y-2) $, but I can't prove that $(X-1,Y-2) \subseteq I$.
Any h... | The claim is false since
$$\begin{align}
(x-1,f(x)) &= (x-1,\,f(x)\bmod x-1)\\
&= (x-1,\,f(1))\\
&= (x-1,\,y^2-4y+4)\ \ {\rm in\ OP}\\
&= (x-1,\,(y-2)^2)
\end{align}\qquad$$
but $\,(x\!-\!1,(y\!-\!2)^2)\neq (x\!-\!1,y\!-\!2),\,$ else $\,y\!-\!2 = (x\!-\!1)f + (y\!-\!2)^2g,\,$ so $\,y\!-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of converg... | Given $\epsilon>0$ be any number.
Now
\begin{align}
&\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|\\
&=\left|\frac{-4n-3}{4(4n^2+1)}\right|\\
&=\frac{4n+3}{4(4n^2+1)}\\
&<\frac{n+1}{(4n^2+1)}\\
&<\frac{n+1}{(n^2+1)}\\
&<\frac{n+1}{n^2}<\epsilon\quad &\text{whenever}\quad n^2\epsilon-n-1>0\\
&&\text{i.e whenever}\quad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\tan(x)\tan(x+r) = \frac{\tan(x+r) - \tan(x)}{\tan r} -1$ $\tan(x)\cdot \tan(x+r) = \dfrac{\tan(x+r) - \tan(x)}{\tan r} -1$
I tried L.H.S = $\dfrac{\sin(x)\cdot \sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{... | Hint:
Use $$\tan(a-b)=\frac{\tan a -\tan b}{1+\tan a \tan b} $$ with $a=x+r, b=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?
Its answer is easy if it were $x.y=12^3$ , because there are $28$ po... | Let's define an order that $a < b$ if the power of $3$ that divides $a$ is less than the power of $3$ that divide $a$ or those powers are equal then if power of $2$ that divides $a$ is larger then the power than divides $b$
Then wolog $x \le y \le z$ and we need to consider the cases
$x=y=z$
$x=y < z$
$x < y = z$
$x < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
I do not understand this particular step in a proof using the Well Ordering Principle Below is a proof using the Well Ordering Principle. I get lost starting at $(13)$...
$$
\begin{aligned}
P(c): &:=c^{3} \leq 3^{c} \\
& \equiv c^{3} \leq 3(c-1)^{3} \\
& \equiv c \leq \sqrt[3]{3} \times(c-1)
\end{aligned}
$$
I don't un... | You want to show that
$c^3 \le 3^c$
by induction.
This means that you want to show that
$(c-1)^3 \le 3^{c-1}$
implies that
$c^3 \le 3^c$.
By the induction hypothesis,
$3^c = 3\cdot 3^{c-1}
\ge 3(c-1)^3
$.
Therefore,
if you can show that
$3(c-1)^3
\ge c^3
$
then
$3^c \ge c^3
$.
Now we can proceed as above.
Taking cube r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the least $n$ such that $f(n)/g(n)=4/7$
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 1+f(n+1) \hspace{14pt} \text{otherwise}} $$
$$g(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 2+g(n+2) \hspace{14pt} \text{otherwise}}$$
for positive ... | I can not believe the number of arithmetic errors I made. All inexcusable. but.
Let $n > 0$ so there is a $k$ so that $(k-1)^2 < k \le n^2$ and let $j = k^2 - n$.
Then $$f(n) = f(k^2-j)= 1 + f(k^2-j+1) = 2+f(k^2 - j + 2) =.... =j + f(k^2 -j + j) = j + f(k^2) = j + k$$. (That holds if $j=0$.)
Now similarly if $j$ is ev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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