Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$
Evaluate $$\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$$
First thing I saw is $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ so the integral is the same as:
\begin{gather*}
\int_0^1\ln^2{\left(x^2+x+1\right)} \, \mathrm{d}x
+ \int_0^1\ln^2{\left(x^2-x+1\right... | I would maybe try:
$$x^4+x^2+1=(x^2+\frac12)^2+\frac34$$
and so:
$$\int_0^1\ln^2(x^4+x^2+1)dx=\int_0^1\ln^2\left[(x^2+\frac12)^2+\frac{\sqrt{3}}{2}^2\right]dx$$
Now we know that:
$$\tan^2u+1=\sec^2u$$
so by letting:
$$\left[\frac{2}{\sqrt{3}}\left(x^2+\frac12\right)\right]=\tan(u)$$
$$\frac{4}{\sqrt{3}}xdx=\sec^2udu$$
... | {
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"url": "https://math.stackexchange.com/questions/3777828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way. EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$
I use the fact... | $$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$
$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$
$$\color{red}{\sin(-x)}+\sin x=0$$
and we're done by the double implications all thr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{\pi/2} \frac{\arctan{\left(\frac{2\sin{x}}{2\cos{x}-1}\right)}\sin{\left(\frac{x}{2}\right)}}{\sqrt{\cos{x}}} \, \mathrm{d}x$ Evaluate: $$\int_0^{\frac{\pi}{2}} \frac{\arctan{\left(\frac{2\sin{x}}{2\cos{x}-1}\right)}\sin{\left(\frac{x}{2}\right)}}{\sqrt{\cos{x}}} \, \mathrm{d}x$$
I believe there is a ... | $$\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx \overset{\tan\frac{x}{2}\to x}= 2\int_0^1 \frac{x\arctan\left(\frac{4x}{1-3x^2}\right)}{\sqrt{1-x^2}(1+x^2)}dx$$
$$=2\underbrace{\int_0^1 \frac{x(\arctan x + \arctan(3x))}{\sqrt{1-x^2}(1+x^2)}dx}_{\m... | {
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"url": "https://math.stackexchange.com/questions/3778663",
"timestamp": "2023-03-29T00:00:00",
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + ... | The remainder of $(x - 1)^{100}$ divided by $(x - 1)(x - 2)$ will be $(x - 1) (2 - 1)^{99} = x - 1$. The remainder of $(x - 2)^{200}$ divided by $(x - 1)(x - 2)$ will be $(x - 2)(1 - 2)^{199} = 2 - x$ Therefore, the total remainder will be 1.
| {
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"timestamp": "2023-03-29T00:00:00",
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Distance from plane to origin via Lagrange multipliers Let's say I have to find the least distance between origin and the plane $$x-2y-2z = 3$$
I know distance from origin to any $x-y-z$ plane is $\sqrt{x^2 + y^2 + z^2}$
so the constraint will be $$g(x, y, z) = x - 2y - 2z -3$$
however, what will my $f(x,y,z)$ be? Why ... | By C-S $$\sqrt{x^2+y^2+z^2}=\frac{1}{3}\sqrt{(1^2+(-2)^2+(-2)^2)(x^2+y^2+z^2)}\geq\frac{1}{3}(x-2y-2z)=1.$$
The equality occurs for $(x,y,z)||(1,-2,-2)$ and $x-2y-2z=3,$ id est, occurs,
which says that we got a minimal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $y= \frac1{1+x} + \frac2{1+x^2} + \frac4{1+x^4} + \dots + \frac{2^n}{1+x^{2^n}} $, then find $\frac{\mathrm dy}{\mathrm dx}$.
If $y= \dfrac1{1+x} + \dfrac2{1+x^2} + \dfrac4{1+x^4} + \dots + \dfrac{2^n}{1+x^{2^n}} $, then find $\dfrac{\mathrm dy}{\mathrm dx}$.
I am stuck up in this question. I tried taking log on b... | Hint:
$$\dfrac1{1+y}+\dfrac1{1-y}=\dfrac2{1-y^2}$$
Set $y=x,x^2,x^4,\cdots,x^{2^n}$ to find
$$y+\dfrac1{1+x}=\dfrac{2^n}{1-x^{2^{n+1}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
| $$\dfrac{a\cdot\dfrac{x^2}a+b\cdot\dfrac{4xy}b+c\cdot\dfrac{4y^2}c+d\cdot\dfrac{2z^2}d}{a+b+c+d}$$
$$\ge\sqrt[a+b+c+d]{\left(\dfrac{x^2}a\right)^a\left(\dfrac{4xy}b\right)^b\left(\dfrac{4y^2}c\right)^c\left(\dfrac{2z^2}d\right)^d}$$
$\left(\dfrac{x^2}a\right)^a\left(\dfrac{4xy}b\right)^b\left(\dfrac{4y^2}c\right)^c\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782030",
"timestamp": "2023-03-29T00:00:00",
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$(1−x^n) = (1−x)(1 + x + x^2 +\cdots + x^{n−1})$ Is this expression generally true?
$$(1−x^n) = (1−x)(1 + x + \cdots + x^{n−1})$$
The closest Identity I could find in "Mathematical Handbook of Formulas and Tables", Schaum's Outline is:
$$x^{2n+1} - y^{2n+1} = (x-y)(x^{2n} +x^{2n-1}y + x^{2n+2}y^2 + \cdots +y^{2n})$$
Wh... | Here are the formulæ I learnt in med-school:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+xy^{n-2}+y^{n-1}),$$
in other words, it is $x-y$ times the sum of all homogeneous monomials in $x$ and $y$, with total degree $n-1$.
The sum of two powers requires an odd exponent:
$$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x... | {
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"url": "https://math.stackexchange.com/questions/3782382",
"timestamp": "2023-03-29T00:00:00",
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The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is The coefficient of $x^{n}$ in the expansion of $\frac{(2-3 x)}{(1-3 x+\left.2 x^{2}\right)}$ is
$(a) \quad(-3)^{n}-(2)^{n / 2-1}$
(b) $2^{n}+1$
$(c) 3(2)^{n / 2-1}-2(3)^{n}$
(d) None of the foregoing numbers.
Now, $1-3 x+2 x^{2}... | Your approach is kind of brute-force, but you can already get the answer using $2(2^{n+1}-1)-3(2^{(n+1)-1}-1) = (4-3)2^n-2+3=(b).$
Note that there’s a $-1$ in the exponent of the term multiplied with $-3$ because to get an $x^n$ term, when the first term is $x^1$, you need the $x^{n-1}$ term.
| {
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Does $\lim_{n\to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ exist? Let $x \in \mathbb{R}.$ Is is true that the following limit exists : $$\lim_{n \to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$$ What is the value of this limit?
I tried the Integral test for convergence, but nothing ca... | We can expand $\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ as $$-\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n})^m}{m}$$ which converges for $\frac{x^2\sin^2k}{2n} \le 1 \to x^2 \le 2n \to |x| \le \sqrt{2n}$.
Then the summation in your question would become $$-\sum_{k=1}^n\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n}... | {
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If $f(x)$ is differentiable for all real numbers, then what is the value of $\frac{a+b+c}{2}$?
If $f(x)=\begin {cases} a^2 + e^x & -\infty <x<0 \\ x+2 & 0\le x \le 3 \\ c -\frac{b^2}{x} & 3<x<\infty \end{cases}$, where $a,b,c$ are positive quantities. If $f(x)$ is differentiable for all real numbers, then value of $\f... | In order to be differentiable everywhere, $f$ must be continuous. As $f$ is a piecewise continuous function:
To solve for $a$, in the last step using the fact that $a$ is positive
$$\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)$$
$$\lim_{x\rightarrow0^-}(a^2+e^x)=\lim_{x\rightarrow0^+}(2+x)$$
$$a^2+e^0=2+0$$
$... | {
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"url": "https://math.stackexchange.com/questions/3788665",
"timestamp": "2023-03-29T00:00:00",
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In $\triangle ABC$, if $AB=x+2$, $BC=x$, $AC=x-2$, and $C = 120^\circ$, then $x=5$ I need help with this Law of cosines problem:
Prove that when $AB = x + 2,\ BC = x,\ AC = x - 2\;$ and $\;C = 120^\circ$ while $ABC$ is a triangle, then $x = 5$.
I need help to get to that answer. With law of cosines I got this
$$(x + ... | It was a silly mistake. Already fixed it.
$(x + 2)^2 = x^2 + (x-2)^2 -2x(x-2)\cos 120°$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I show the quotient of the $k$th partial sums of $\sum\limits_{n=1}^{k} n$ and $\sum\limits_{n=1}^{k} n^2$ is $\frac{3}{2k+1}$? I've found using pen and paper that any trivial case of the sum of a sequence of integers from $1$ to $k$ divided by the sum of the squares of these integers is equal to $\frac{3}{2k+... | There are explicit formulas for the partial sum (that can e.g. be shown by induction):
$$
\sum_{n=1}^k n = \frac{k(k+1)}{2}
$$
and
$$
\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}.
$$
Take the ratio, and you get your formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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Method of summation for third order difference series: $2+12+36+80+150\dots$ I am unable to understand how one derived the formula for the $n$-th term $= an^3 +bn^2 + cn + d$, where the degree of the polynomial depends on the step at which we get a constant A.P. Here its at $2$nd step so degree $=2+1=3$.
But how do we ... | In this table, where $A_n$ is the $n^{th}$ term of your series,
$$
\begin{array}{c|c|c|c|c}
\style{font-family:inherit}{{n}} & \style{font-family:inherit}{A_n}
& \style{font-family:inherit}{B_n} & \style{font-family:inherit}{C_n} & \style{font-family:inherit}{D_n}\\\hline
0 & 2 ... | {
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"timestamp": "2023-03-29T00:00:00",
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On the golden ratio and even perfect numbers (Note: This post is an offshoot of this earlier MSE question.)
Here is my question in this post:
Is $I(2^{p-1}) - 1 > 1/I(2^{p-1})$ true when $I(2^{p-1}) = \sigma(2^{p-1})/2^{p-1}$ is the abundancy index of $2^{p-1}$ and $6 \neq 2^{p-1}(2^p - 1)$ is an even perfect number ... | You have written a proof for the follwing claim :
Claim For $p \geq 3$, $\frac{7}{4} \leq I(2^{p-1}).$
$$2^{p-1} \geq 4 \implies \frac{1}{2^{p-1}} \leq \frac{1}{4} \implies 2 - \bigg(\frac{1}{2^{p-1}}\bigg) \geq 2 - \frac{1}{4} = \frac{7}{4}.$$
I've found no errors here.
(One can also say "Since $I(2^{p-1})$ is incre... | {
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"url": "https://math.stackexchange.com/questions/3792694",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
... | $$ \int_0^1 \frac{\arctan x \ln(1+x^2)}{1+x} dx=\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}+{G\ln2}$$
$$\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x^2})}{1-x}dx=\frac{\pi^3}{192}-\dfrac{G\ln 2}{2}$$
$$\int_0^1\frac{\arctan x\ln(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 2
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Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those fo... | If $A= \begin{pmatrix}a & b \\
c & d\end{pmatrix}$
then we get from $A^TA= I$ that $a^2+c^2=1= b^2+d^2$ and also $ab+cd = 0$. Then $ab = -cd$ so squaring it gives $a^2b^2 = c^2d^2$. I'll let you finish the rest.
| {
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For $\triangle ABC$, show that $ac\cos B+ab\cos C-bc\cos A-a^2 \le \frac{c^2}{8\cos^2(90^\circ-C)}$
Triangle $\triangle ABC$ has sides $a$, $b$, and $c$, and circumradius $R$. Prove that
$$ac \cos B + ab \cos C - bc \cos A - a^2 \le \frac{c^2}{8\cos^2(90^\circ - C)}$$
When does equality occur?
I came across this ques... | Answer to the second question (equality).
Triangle $ABC$ has sides $a$, $b$, and $c$, corresponding angles
$\alpha,\beta,\gamma$, semiperimeter $\rho$, inradius $r$ and
circumradius $R$. Prove that
\begin{align} R^2-a^2+b^2+c^2\ge0\tag{1}\label{1}. \end{align}
When does equality occur?
By dividing \eqref{1} by $R^2... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int e^{2x}(7+e^x)^{1/2}\,dx$ $\int e^{2x}(7+e^x)^{1/2}\,dx$
Let $u=7+e^x \rightarrow du=e^xdx$
So the integral becomes:
$\int u^{\frac{3}{2}}-7u^{\frac{1}{2}}du$ and so our answer is
$\frac{2}{5}(7+e^x)^{\frac{5}{2}}-\frac{14}{3}(7+e^x)^{\frac{3}{2}}+C$
But this is not what wolfram says. Did I make a mistake... | You could have made life simpler getting rid of the radical
$$u=\sqrt{7+e^x}\implies x=\log \left(u^2-7\right)\implies dx=\frac{2 u}{u^2-7}\,du$$
$$e^{2x}\sqrt{7+e^x}\,dx=2u^2(u^2-7)\,du$$
$$\int e^{2x}\sqrt{7+e^x}\,dx=2\int (u^4-7u^2)\,du=\frac25u^5-\frac 73 u^3+ C$$
Back to $x$
$$\int e^{2x}\sqrt{7+e^x}\,dx=\frac{2}{... | {
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"timestamp": "2023-03-29T00:00:00",
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How many ways are there to distribute 6 passengers into three different hotels? 6 individuals want to go to 3 different hotels such that each hotel can select zero through 6 people all states are possible. From the passenger's angle, we know that there are $3^6=729$ different ways to do this task. But from the perspect... | The number of ways to choose $a$ people for the first hotel, $b$ for the second hotel, and $c$ for the third hotel, with $a+b+c=6$, is the multinomial coefficient
$$\binom{6}{a,b,c}= \frac{6!}{a! b! c!}$$
so the total number of possible arrangements is
$$\sum_{a+b+c = 6} \binom{6}{a,b,c}$$
where the summation is over a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit using Taylor expansion : which term do we expand? I want to check the limit $\displaystyle{\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )}$ using the Taylor expansion.
I have done the following:
$$\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\l... | We have that
$$\frac{n^2-2n+1}{n^2+1}=1-\frac{2n}{n^2+1}$$
then we can start by by first order Taylor's expansion for $\log(1+x)$ to obtain
$$\ln\left (\frac{n^2-2n+1}{n^2+1}\right )=\ln\left (1-\frac{2n}{n^2+1}\right )=-\frac{2n}{n^2+1}+O\left(\frac1{n^2}\right)$$
then
$$\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2... | {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)=\big\lfloor x\lfloor x\rfloor\big\rfloor$ for all $x\geq 0$, then for an integer $n$, solve for $x\geq 0$ such that $f(x)=n$.
Let $f(x) = \big\lfloor x \lfloor x \rfloor \big\rfloor$ for $x \ge 0.$
(a) Find all $x \ge 0$ such that $f(x) = 1.$
(b) Find all $x \ge 0$ such that $f(x) = 3.$
(c) Find all $x \ge 0$... | If $x$ were allowed to be negative this would be a real pain but if $x \ge 0$ then $[x] \ge 0$
If $[x] = n$ then $n \le x < n+1$ and $n^2 \le nx \le n^2 + n$ with the second equality holding only if $n= 0$....
So if $0 \le x < 1$ then $[x[x]] = 0$.
If $x \ge 1$ then $[x] = n \ge 1$ and $n \le x < n+1$ so $1\le n^2 \le ... | {
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Find $\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}$.
Calculate $$\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}\,.$$
I broke the sum into partial fractions and after writing 3-4 terms of the sequence I could see that it cancels but I wasn't able to arrive at the exact expression. I understand that it's trivial but for whatever... | $\mathcal{Hint}$
After writing $r=1$ separately, use the fact that the numerator can be written as $r(r-1)+2r+1$.
So, $$(-1)^r\frac{r^2+r+1}{r!}=(-1)^r\left(\frac{1}{(r-2)!}+\frac{2}{(r-1)!}
+\frac{1}{r!}\right)$$
Can you finish now?
$\mathbf{Edit:-}$
Writing a few terms for $r=2$ onwards, we get
$$\left(\frac{1}{0!}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can $\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$ where $R$ and $x$ are positive constants, be solved using substitution? While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-... | $$\frac{1}{\sqrt{R^2+x^2-2 R x \cos (\theta )}}=\frac{1}{\sqrt{R^2+x^2}}\frac{1}{\sqrt{1-k\cos (\theta )}}$$ with $k=\frac{2Rx}{R^2+x^2}$.
$$\int \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 2 {\sqrt{{1-k}}} F\left(\frac{\theta }{2}|\frac{2 k}{k-1}\right)$$
$$\int_0^{2\pi} \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Quadratic form reduction on the $n$-dimensional complex space $\mathbb{C}^n$ How to reduce the quadratic form
$$\sum_{1 \leq k < l \leq n}(k + il)x_kx_l$$
to the canonical form on the $n$-dimensional complex space $\mathbb{C}^n$?
I started with its symmetric matrix
\begin{align*}
S = \frac{1}{2}\begin{pmatrix}
0 & 1 + ... | This might be helpful
If we define
$$
\Phi_1:=\sum_{1\leq k<l\leq n}(k+il)x_kx_l
$$
and
$$
\epsilon_{k,l}=\left\{\begin{array}{cc}
1\textrm{ , if }k<l\\
i\textrm{ , if }k>l\\
0\textrm{ , else }
\end{array}\right\}\textrm{, }i=\sqrt{-1}
$$
then
$$
\Phi_1=\sum^{n}_{k,l=1}\epsilon_{kl}kx_kx_l=\frac{1+i}{1-i}\sum_{k<l}(k+l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c}\geq 0$ for positive $a$, $b$, $c$ with $abc=1$. I recently saw the following inequality,
$$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0 \tag1$$
for all $abc=1$ and $a,b,c \in \mathbb{R}_+ \setminus \{0 \}$.
To prove this inequality, ... | Now, use AM-GM:
$$ab^2+a\geq2ab,$$
$$b^2+1\geq2b$$ and $$a^2b+bc+c\geq3\sqrt[3]{a^2b^2c^2}=3.$$
After summing we'll get your last inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is $3^{99} \pmod{100}$? I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we ... | We have $ 3^4 \equiv 81 \ (\text{mod} \ 100), \\ 3^5=243 \equiv 43 \ (\text{mod} \ 100),\\ 3^{10}=43^2 =1849 \equiv 49 \ (\text{mod} \ 100).$
Then $ \ 3^{20} =49^2=2401 \equiv 1 \ (\text{mod} \ 100) \Rightarrow 3^{80} \equiv 1 \ (\text{mod} \ 100).$
Also $3^9=3^4 \cdot 3^5 = 81 \cdot 43=3483 \equiv 83 \ (\text{mod} \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding intersection line question without the $z$ axis. What is the intersection line of these two planes:
$$ P_1 : 3x+4y+7z=5$$
$$P_2: 2x-5y=8$$
I will first multiply one by $4$ and the second by $5$ just to get rid of $y$:
$$15x+20y+35z -25 = 0$$
$$8x-20y-32=0$$
Adding those two, in order to get the intersection pla... | $P_1 : 3x+4y+7z=5$
$P_2: 2x-5y=8$
You can either find the cross-product of the normal vector of the two planes and a point on the intersection to find the equation of the line or you can do as below -
For $z = t$,
$3x+4y=5-7t$ and $2x-5y=8$
Equating the two, you get -
$x = \dfrac {57-35t}{23}, y = \dfrac{-14-14t}{23}, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\operatorname{Var}(Y)$ when $f_{X}(x)=3x^{2}I_{\{0Let $X$ be a random variable with the following pdf:
$$
f_{X}(x)=3x^{2}I_{\{0<x<1\}}
$$
let $Y$ be a random variable so $\left(Y\mid X=x\right)\sim \operatorname{Uni}[96x,100x]$. Calculate $\operatorname{Var}(Y)$.
What I did: If $(Y\mid X=xt)\sim \operatorname{Uni... | The entire calculation is unnecessary. Rather, apply the law of total variance.
Note for all positive integers $k$, $$\operatorname{E}[X^k] = \int_{x=0}^1 3x^{k+2} \, dx = \frac{3}{k+3}.$$ Then $$\begin{align}
\operatorname{Var}[Y] &= \operatorname{Var}[\operatorname{E}[Y \mid X]] + \operatorname{E}[\operatorname{Var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Maximum of $(\sin{x} + \sqrt3)(\cos{x}+1)$ The problem is
Maximum of $(\sin{x} + \sqrt3) (\cos{x}+1)$
I tried to set $\sin(t) = y$, $\cos(t) = x$, so that the problem turned out to be finding
Max $(y + \sqrt3)(x + 1)$
s.t. $x^2 + y^2 =1 $
I tried Lagrange multipliers, but it turns out to be a quartic equation.
I also... | We can proceed by tangent half angle identities to obtain
$$f(t)=\left(\frac{2t}{1+t^2} + \sqrt3\right) \left(\frac{1-t^2}{1+t^2} + 1\right) \implies f'(t)=-4\frac{\sqrt 3t^3+3t^2+\sqrt 3t-1}{(1+t^2)^3}$$
and since
$$g(t)=\sqrt 3t^3+3t^2+\sqrt 3t-1 \implies g'(t)=3\sqrt 3t^2+6t+\sqrt 3=3\sqrt 3\left(t+\frac{\sqrt 3}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$ I tried to compute $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$
\begin{equation}\begin{aligned}
&\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x&\\
&=\int_{0}^{\pi} \frac{2(a-1)+4 \sin ^{2} \frac{x... | This result $I=2\pi/a$ is true only if $a>1$, for a<1 $I=0$.
When $a<1$, $t=\frac{a+1}{a-1} \tan x \implies t=0 (x=0) ~\text{but}~ t=-\infty (x=\pi/2)$. Then
$$I=\left(1+\frac{|a-1|}{a-1}\right)\frac{\pi}{a}$$
When $a=1$ the orininal integral is $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this integral $\int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt$ If the question was $\int^{\infty}_{-\infty} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt$, I'd have gone with completing the square of $(t+ \frac{t^{2}}{2s^{2}} )$ and using the normal density trick, but I am stumped ... | Note
\begin{align}
I &= \int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt
= \int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-\frac1{2s^2}(t+ s^2 )^2 +\frac12s^2}dt\\
& \overset{x=t+s^2}=\frac{e^{\frac{s^2}{2}}}{2s^2} \int^{\infty}_{-\infty} (x-s^2)^2 e^{-\frac{x^2}{2s^2}} dx
= \frac{e^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the $26^{th}$ digit of a $50$ digit number divisible by $13$. $N$ is a $50$ digit number (in the decimal scale). All digits except the $26^{th}$ digit (from the left)
are $1$. If $N$ is divisible by $13$, find the $26^{th}$ digit.
This question was asked in RMO $1990$ and is very similar to this question and the s... | After your editions, your approach is correct. Here's an alternative one:
The number $N$ consists of $24$ ones followed by the two digits $1a$ (the $2$-digit number $10+a$) followed by another $24$ ones, so with the number $M$ consisting of $24$ ones, $M:=\sum_{k=0}^{23}10^k=\frac{10^{24}-1}{9}$, we have
$$N=M\cdot10^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
t... | You know that all power series that converge to the function $f$ agree. So you can just write for $u = x - 1$:
$\begin{align*}
\frac{3 x + 4}{x + 1}
&= \frac{3 u + 7}{u + 2} \\
&= 3 + \frac{2}{1 + u / 2} \\
&= 3 + 2 \sum_{n \ge 0} (-1)^n \left( \frac{u}{2} \right)^n \\
&= 3 + \sum_{n \ge 0} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Sums of powers of two.. with one restriction. For a positive integer $n,$ let $a_n$ denote the number of ways of representing $n$ as a sum of powers of 2, where each power of 2 appears at most three times, and the order of the terms does not matter. For example, $a_8 = 5$ because of the representations
\begin{align*}
8... | Let's prove @Oldnboy's conjecture with a hint by @JyrkiLahtonen to a technique the OP might not know well. Restate $n=\sum_jc_j2^j$, with $0\le c_j\le3$, as $x^n=\prod_jx^{c_j2^j}$, so $a_n$ is the $x^n$ coefficient in$$\begin{align}\prod_{j\ge0}(1+x^{2^j}+x^{2\cdot2^j}+x^{3\cdot2^j})&=\prod_j\frac{1-x^{4\cdot2^j}}{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$ This is what the answer says:
Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,
and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly ... | Observe that
$$\sqrt{x} - \sqrt{2-2x} = 1 \implies x = 1 + 2-2x + 2\sqrt{2-2x} \implies 3x - 3 = \sqrt{2-2x}$$ $$ \implies 9(x^2 - 2x + 1) = 2-2x \implies 9x^2 - 16x +7 = 0 \implies (x -1)(9x - 7) = 0 $$
Thus $ x = 1, \frac{7}{9} $. Now, $$ \sqrt{\frac{7}{9}}-\sqrt{2-\frac{14}{9}} \neq 1 $$
Thus $x=1 $ is the solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. ... | For $n=2$ we have
$$\sum_{i=1}^{2}(1-\frac{1}{(i+1)^{2}})=(1-\frac{1}{4})+(1-\frac{1}{9})=\frac{59}{36}$$ but $$\frac{n+2}{2n+2}\big|_{n=2}=\frac{2+2}{2(2)+2}=\frac{2}{3}\neq\frac{59}{36}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overri... | Setting up some vector identities:
$$
\overrightarrow{AB} = \overrightarrow{B} -\overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{A} = \overrightarrow{0}
$$
Proof:
$$
\begin{align}
&\overrightarrow{A}\times \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the e... | If $z^4=(2+3i)^4$ then $Z^4 = 1$ where $Z = \frac{z}{2+3i}$.
Hence the solutions set is
$$\{(2+3i), -(2+3i), i(2+3i), -i(2+3i)\}=\\ \{\sqrt{13} e^{i \phi},\sqrt{13} e^{i (\phi + \pi)},\sqrt{13} e^{i (\phi + \pi/2)},\sqrt{13}e^{i (\phi - \pi/2)}\}$$
where $\phi$ is such that $\cos \phi = \frac{2}{\sqrt{13}}, \sin \phi =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
} |
To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\fr... | By your work $abc=1$.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, $$xyz=1$$ and we need to prove that:
$$\sum_{cyc}\frac{1}{x^2}>x+y+z$$ or
$$\sum_{cyc}(x^2y^2-x^2yz)>0$$ or
$$\sum_{cyc}z^2(x-y)^2>0,$$
which is true because $x=y=z$ is impossible.
Indeed, let $x=y=z$.
Thus, $a=b=c=1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$ Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$?
The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills.
I asked a teacher/tutor and he proposed setting the expression equal to X and working backwards, squ... | This is what the OP means by the 'setting equal to X' method.
Let $x = (1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$, and thus:
$$x^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow x = \cancel{-\sqrt{2}}, \sqrt{2}$$
The reason why we have crossed out $-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
For any odd prime $p$, a quadratic is solvable mod $p^2$ if it is solvable mod $p$, and $p$ does not divide the discriminant and leading coefficient.
Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p$ does not divide $a$, and does not divide $D=b^2-4ac$, and such that
$$ax^2+bx+c\equiv0\pmod{p}... | Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p\nmid a$ and $p\nmid D$, where $D=b^2-4ac$. Let $x$ be an integer such that
$$ax^2+bx+c\equiv0\pmod{p},\tag{1}$$
so $ax^2+bx+c=mp$ for some integer $m$.
Exercise: Show that $p$ does not divide $2ax+b$, because $p\nmid a$ and $p\nmid D$ and $p$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{... | The sum of square form
$$2(x^{12}-x^9+x^4-x+1)$$
$$=x^6(x^3-1)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{4}\right)^2+(x-1)^2+\frac{5}{8}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I pro... | In the induction step, we assume that it holds for $n = k$.
That means that $\exists a \in \mathbb{Z} : k^3 + 5k = 3a$.
Then, for $(k+1)$, we get
$$
\begin{split}
(k+1)^3 + 5(k+1)
&= k^3 + 3k^2 + 8k + 6 \\
&= (k^3 + 5k) + 3k^2 + 3k + 6 \\
&= 3a + 3k^2 + 3k + 6 \\
&= 3(a + k^2 + k + 2).
\end{split}
$$
So, if $k^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?
In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both b... | NOT AN ANSWER. Commentary too long to fit in a Comment
*
*The equation: $2^a+2^b=10^c+10^d;\ 0\le a<b,\ 0\le c<d$. If $b>a,\ d>c$ were not already stated, it is provable that $a\ne b,\ c\ne d$ from which we could validly assume WLOG $b>a,\ d>c$.
*Although $a=0$ and $c=0$ are contemplated by the question as stated, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Transformation of roots of a polynomial Suppose there is a cubic polynomial in x with roots A,B,C and another cubic polynomial (in t) with roots
1/(A-1), 1/(B-1), 1/(C-1)
is to be found.
My text book mentions two ways of doing this. One is to use vieta's relations, which is time consuming.
Another way mentioned is ... | Let $A$, $B$ and $C$ be roots of the polynomial $x^3+ax^2+bx+c.$
Thus, $$A+B+C=-a,$$ $$AB+AC+BC=b,$$ $$ABC=-c$$ and we obtain:
$$\sum_{cyc}\frac{1}{A-1}=\frac{\sum\limits_{cyc}(A-1)(B-1)}{\prod\limits_{cyc}(A-1)}=\frac{b+2a+3}{-c-b-a-1},$$ which gives a coefficient before $x^2$ in new polynomial:
$$\frac{2a+b+3}{a+b+c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three ad... | \begin{align}
(y+z-x)(y-z)&=y^2-z^2-xy+xz=a^2-b^2\\
(x+y-z)(x-y)&=x^2-y^2-zx+zy=b^2-c^2\\
(z+x-y)(z-x)&=z^2-x^2-yz+yx=c^2-a^2
\end{align}
Let $u:=y-z$, $v:=y+z$, $\alpha^2:=a^2-b^2$, $\beta^2:=b^2-c^2$, $\gamma^2:=c^2-a^2$, then
\begin{align}
(v-x)u&=\alpha^2\\
(x+u)(u+v-2x)&=-2\beta^2\\
(x-u)(v-u-2x)&=2\gamma^2
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answ... | We have that
$$a=\left(2+\frac{1}{2}\right)\left(3+\frac{1}{3}\right)\left(4+\frac{1}{4}\right)\left(5+\frac{1}{5}\right)=$$
$$=\frac12\left(4+1\right)\frac13\left(9+1\right)\frac14\left(16+1\right)\frac15\left(25+1\right)=$$
$$=\frac{5\cdot 10\cdot17\cdot26}{120}$$
and similarly
$$b=\left(2+\frac{1}{5}\right)\left(3+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$
Let [$x$] denote the greatest integer less than or equal to $x$. Find the value of $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$$
I was able to find out the general pattern in the series i.e. $3(1)+ 5(2)+ 7(3)+\ldots+87(43)$ but was unable to proceed... | Hint 1: $[\sqrt{n^2}], [\sqrt{n^2 +1}], .........[\sqrt{(n+1)^2 - 1}]$ all equal the same the same thing: $n$.
Hint 2: $(n+1)^2 - 1= n^2 + 2n$ so from $[\sqrt{n^2}]$ to $[\sqrt{(n+1)^2 - 1}]=[\sqrt{n^2 + 2n}]$ there are $2n+1$ terms.
Hint 3: So we have $\color{blue}{[\sqrt {1}] + [\sqrt 2]+ [\sqrt 3]} + \color{purpl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Power series $\ \frac{x}{(x-2)(x-3)}$, radius of convergence I got a little trouble with this function while expanding this to power series:
$$\ f(x) = \frac{x}{(x-2)(x-3)} $$
After partial fraction decomposition:
$$\ f(x) = \frac{3}{x-3} - \frac{2}{x-2} $$
I can see here geometric series and rewrite as the sums:
$$\ \... | Hint:
We can write $$\dfrac3{x-3}=-\dfrac1{1-\dfrac x3}$$ right?
So, we need $|x|<$min$(2,3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving inequality using Taylor polynomial of arctan I have to prove that $ x - \frac{x^3}{3} < \arctan(x) < x - \frac{x^3}{6} $ is true if $ 0 < x \leq 1 $
I tried to convert the second inequality into the one I'm trying to prove. Since $ \arctan(x) $ is a monotonically increasing function, applying it to each part of... | We have that
$$f(x)=\arctan x- x+\frac16 x^3 \implies f'(x)=\frac{x^2(x^2-1)}{2(x^2+1)}\le 0$$
with $f(0)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| To check, you can see if the LHS equals the RHS by using
$$\cos^2 x = \dfrac {\cos 2x + 1}{2} \\ \sin^2 x = \dfrac {1- \cos 2x}{2}$$
Squaring both sides
$$\cos^4 x = \dfrac 14(\cos^2 2x + 2 \cos 2x +1) \\ \sin^4 x = \dfrac 14(1-2 \cos 2x + \cos^2 2x)$$
Adding we get
$$\cos^4 x + \sin^4 x = \left(\dfrac 14 {\cos^2 2x} +... | {
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"url": "https://math.stackexchange.com/questions/3854825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $x^2 \sqrt{1 + x^2}$ How may one evaluate $\int x^2 \sqrt{x^2 + 1}\ dx$? I tried parts by integrating $\sqrt{x^2 + 1}$ but that seems to lead me down a rabbit hole of endless computations.
| Substitute $x=\tan(u)$ then we have $$\int x^2 \sqrt{x^2 + 1}\ dx$$
$$=\int \tan^2(u)\sec^3(u)du=\int(\sec^2(u)-1)\sec^{3}(u)du$$
$$=\int \sec^5(u)du-\int \sec^3(u)du.$$
Now applying the reduction formula $$I_{n}=\int \sec^{n}(x)dx=\frac{1}{n-1}\tan(x)\sec^{n-2}(x)+\frac{n-2}{n-1}I_{n-2}$$ gives $$=\int \sec^5(u)du-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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reduce a differential equation $y'+\dfrac{x}{y}=0$ I want to reduce a differential equation.
$y'+\dfrac{x}{y}=0$
I reduced this. But my answer don't much with Wolfram alpha. Please tell me what is wrong.
$y'=-\dfrac{x}{y}=-\dfrac{1}{\left( \dfrac{y}{x}\right) }$
When I put, $u=\dfrac{y}{x}$,
$\dfrac{dy}{dx}=-\dfrac{1}{... | Not sure exactly what our OP langhtorn means by "reduction", or how the calculations presented in the text of the question itself work, but here is my solution:
$y' + \dfrac{x}{y} = 0; \tag 1$
multiply through by $y$:
$yy' + x = 0; \tag 2$
$\dfrac{1}{2}(y^2)' + x = 0; \tag 3$
$(y^2)' = -2x; \tag 4$
integrate:
$y^2 = -x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how ... | If you use $\tan(x)=t$, you end with
$$I = \int\dfrac{\cos^3(x)}{\sin (x) + \cos (x)}dx=\int \frac{dt}{(t+1) \left(t^2+1\right)^2}$$ Using partial fraction decomposition
$$\frac{1}{(t+1) \left(t^2+1\right)^2}=\frac{1-t}{4 \left(t^2+1\right)}+\frac{1-t}{2 \left(t^2+1\right)^2}+\frac{1}{4
(t+1)}$$ does not seem too ba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Mathematical Induction Proof for $(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$ Mathematical Induction Proof $$(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$$
Base case: $n = 2$, $$(a^2_1+a^2_2)(b^2_1+b^2_2) \geq (a_1b_1
+ a_2b_2)^2$$
$$(a^2_1+a^2_2)(b^2_1+b^2_2) ... | By the assumption of the induction twice and by AM-GM we obtain:
$$\sum_{i=1}^{n+1}a_i^2\sum_{i=1}^{n+1}b_i^2=\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2+a_{n+1}^2\sum_{i=1}^nb_i^2+b_{n+1}^2\sum_{i=1}^na_i^2+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2\sqrt{a_{n+1}^2b_{n+1}^2\sum_{i=1}^na_i^2\sum_{i=1}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - ... | $$\frac{2}{2-\sqrt[4]2}=\frac{\sqrt[4]{8}}{\sqrt[4]{8}-1}$$ and use $$(x-1)(x+1)(x^2+1)=x^4-1$$ for $x=\sqrt[4]{8}.$
Now, about the WA's bang.
$$7\sqrt{\frac{64}{49}+\frac{72\sqrt2}{49}}=\sqrt{64+72\sqrt2}=\sqrt{8\sqrt2(4\sqrt2+9)}=$$
$$=2\sqrt[4]8\sqrt{(2\sqrt2)^2+4\sqrt2+1}=2\sqrt[4]8(2\sqrt2+1)$$ and we made denesti... | {
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"source": "stackexchange",
"question_score": "2",
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Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find... | Thanks to Peter Franek I would try to calculate the limit.
The sum can be rewritten as:
$$\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3} = \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) $$
$$\ a_n=\frac{n}{k^2+n^2} = \frac{n}{n^2}\cdot\frac{1}{\frac{k^2}... | {
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"source": "stackexchange",
"question_score": "2",
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In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm.
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.
What I... | By your work $$(2x)^2+(2y)^2=4^2,$$ which gives $$x^2+y^2=4.$$
Also, $$AD=\sqrt{DO^2+AO^2}=\sqrt{25y^2+4x^2}$$ and
$$BC=\sqrt{25x^2+4y^2}.$$
Now, let $PABC$ be parallelogram.
Thus, $P\in DC$, $AP=BC$, $$DP=DC-PC=10-4=6$$ and $$\measuredangle DAP=\measuredangle Q=45^{\circ}.$$ Thus, by the law of cosines for $\Delta DAP... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of right isosceles triangles that can be formed with points lying on the curve $8x^3+y^3+6xy=1$ Number of right isosceles triangles that can be formed with points lying on the curve $$8x^3+y^3+6xy=1$$
MY ATTEMPT :
We have,
$$8x^3+y^3+6xy=1$$
adding both the sides $$6xy^2+12x^2$$ and simplifying we get ,
$$y^2+y... | Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we see that our equation it's
$$(2x+y-1)(4x^2+y^2+1-2xy+2x+y)=0,$$ which gives $$2x+y-1=0$$ or
$$(2x-y)^2+(2x+1)^2+(y+1)^2=0.$$
The second gives a point $A\left(-\frac{1}{2},-1\right)$ and two other points are placed on the line $2x+y-1=0$.
Thus, the vertex of t... | {
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"url": "https://math.stackexchange.com/questions/3868541",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving for $\lambda$ and $\mu$ in $5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$
Solve for $\lambda$ and $\mu$ in the following equation:
$$5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$$
I have tried:
$$
5\sin(x) = 2\lambda \sin(x) -3\lambda+2\mu \cos(x)
$$
Comparing the coefficient of $\sin x$:
$$
2\lambda = 5
... | It's $$(2\lambda-5)\sin{x}+2\mu\cos{x}=3\lambda$$ or since the case $2\lambda-5=\mu=0$ is impossible,
$$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}\sin{x}+\frac{2\mu}{\sqrt{(2\lambda-5)^2+4\mu^2}}\cos{x}=\frac{3\lambda}{\sqrt{(2\lambda-5)^2+4\mu^2}}.$$
Now, take $$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}=\co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving $\tan^{-1}(x+1)+\tan^{-1}x+\tan^{-1}(x-1)=\tan^{-1}3$ solve $$\tan^{-1}(x+1)+\tan^{-1}x+\tan^{-1}(x-1)=\tan^{-1}3$$.
we have the formula;
let $$\tan^{-1}(\frac{x+y}{1-xy})=a$$ :when $xy<1$ $$\tan^{-1}x+\tan^{-1}y= a$$
when $x>0,y>0,xy>1$ $$\tan^{-1}x+\tan^{-1}y= \pi+a$$
when $x<0,y<0,xy>1$ $$\tan^{-1}x+\tan^{... | $$\tan \left(\arctan x+\arctan(x+1)\right)=\frac{-2 x-1}{x^2+x-1}$$
$$\tan \left(\arctan 3-\arctan(x-1)\right)=\frac{4-x}{3 x-2}$$
provided that $\left(x (x+1)<1\right)\land \left(3 (x-1)<1\right)$ that is
$\frac{1}{2} \left(-\sqrt{5}-1\right)<x<\frac{1}{2} \left(\sqrt{5}-1\right)\land x\ne -1$
The equation becomes
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Choose two sequences from a set such that the average of one sequence is larger than the other by 4th decimal point
Choose two sequences $a_n, b_m$ from $\{ \frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}, \frac{5}{3}, 2 \}$. Calculate their average $\bar{a_n}, \bar{b_m}$. Turned out that $\bar{a_n}$ is the same with $\bar{b... | Fill in the gaps as needed.
*
*The terms of the sequences are $ \{ \frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3},\frac{6}{3} \}$. So, we can write $ \bar{a} = \frac{ A}{ 3n}$, where $ n \leq A \leq 6n$. Likewise, $ \bar{b} = \frac{ B}{ 3m}$, where $ m \leq A \leq 6m$.
*$ \frac{A}{3n} \neq \frac{B}{3m} ... | {
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"source": "stackexchange",
"question_score": "3",
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Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3... | Hint: You are done, because $k^2+9k+18$ is always even, so that you have divisibility by $2\cdot 3=6$.
| {
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"url": "https://math.stackexchange.com/questions/3871936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Using the number e to find the limit of a sequence I have been trying to solve this problem for the past day and I just can't get my head around it, I know I need to use Euler's formula but how do I proceed after simplifying it.
$$
B_n = \left( \frac{n^2 -1}{n^2 +2} \right)^{2n^2-3}
= \left( \frac{n^2 +2 -3 }{n^2 +2}... | Set $y=n^2+2$, to get:
$$
\lim_{n\to\infty} \left(1 + \frac{-3 }{n^2 +2} \right)^{2n^2-3}= \lim_{y\to\infty} \left(1 + \frac{-3 }{y} \right)^{2y-7}\\
=\lim_{y\to\infty} \bigg(\left(1 + \frac{-3 }{y} \right)^y\bigg)^2\left(1 + \frac{-3 }{y} \right)^{-7}\\=(e^{-3})^2\cdot 1=e^{-6}
$$
| {
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"url": "https://math.stackexchange.com/questions/3873285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| Let $b+c=k\iff b=k-c$
$$3=b^2+bc+c^2=(k-c)^2+(k-c)c+c^2$$
$$\iff c^2-kc+k^2-3=0$$ which is a quadratic equation in $c$
As $c$ is real, the discriminant must be $\ge0$
i.e., $$(-k)^2\ge4(k^2-3)\iff k^2\le4\iff-2\le k\le2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x+1$ is always invertible in $\Bbb Z_{x^3}$.
Claim: For any $x > 1$ and $x \in \Bbb N$, $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Proof: We know that $x^3 +1 = (x+1)(x^2-x+1)$. Since $x > 1$ and $x \in \Bbb N$, we have $x^2-x+1 >0$ and $(x+1)(x^2-x+1) = x^3 +1 \equiv 1 \mod x^3$. Hence $x+1$ is always invertible... | Actually you have proved that, for any positive integer $x$, $x+1$ is invertible modulo $x^3$.
Indeed, $(x+1)(x^2-x+1)=x^3+1\equiv1\pmod{x^3}$.
No need to check that $x^2-x+1>0$ (which it is, and it may be just remarked). If you find $u\in\mathbb{Z}$ such that $xu\equiv 1\pmod{x^3}$, then you also find $v>0$ such that ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equatio... | The recurrent equation is
\begin{align}
a_n-\dfrac{1}{4}a_{n-1}=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots.
\end{align}
Solve the homogeneous equation,
$$a_n-\dfrac{1}{4}a_{n-1}=0.$$
The characteristic equation is
$$r-\dfrac{1}{4}=0$$
which gives
$$r=\dfrac{1}{4}.$$
The solution of homogeneous equation i... | {
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"question_score": "2",
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Proving that every positive integer is of the form $x^2+y^2-5z^2$
Prove that every positive integer can be written as
$$x^2+y^2-5z^2$$
with $x$, $y$ and $z$ are non-zero integers.
I made the following observations
*
*if a number is congruent to 0,1,2 mod 4 than it can easily be expressed in this by taking z to be z... | $$2=1^2+9^2-5\cdot4^2$$
$$4=20^2+3^2-5\cdot9^2$$
If $n \ge 3$ then $$2n=(n-2)^2+(2n-1)^2-5(n-1)^2$$
$$1=10^2+9^2-5\cdot6^2$$
If $n \ge 2$ then $$2n-1=n^2+(2n-2)^2-5(n-1)^2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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Find the equation of the circle which touches the curve $x^2+xy-y^2=4$ at $(2,2)$ and the line $3x-y+6=0$ For the first curve, the slope of the tangent is $3$
Let the circle be $x^2+y^2+2gx+2fy+c=0$
Then
$$\frac{2+g}{2+f} =-3$$
$$g+3f+8=0$$
Also the equation of tangent to a circle is
$$y=mx \pm r\sqrt{1+m^2}$$
From $y=... | You have used the equation for tangent to a circle centred at origin but you don't know the centre is at the origin.
The tangent to $x^2+xy-y^2=4$ at $A(2,2)$ is parallel to the line $y=3x+6$. Find the point $B$ on $y=3x+6$ so that $AB$ has slope $-1/3$ i.e. $AB$ is perpendicular to the tangent and the line. The centre... | {
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"question_score": "1",
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ wh... | By C-S $$\sum_{cyc}\frac{7+2b}{1+a}=\sum_{cyc}\left(\frac{7+2b}{1+a}-\frac{7}{2}\right)+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{4b+7(1-a)}{1+a}+\frac{21}{2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{11b+7c}{1+a}+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{(11b+7c)^2}{(11b+7c)(1+a)}+\frac{21}{2}\geq$$
$$\geq\frac{1}{2}\frac{324(a+b+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Difference between anti-derivative and indefinite integral My teacher gave me the following integral to evaluate:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx$$
After half an hour of uselessly fumbling around with trig identities I gave up and plugged it into an integral calculator: https://www.integral-calculator.com/. H... | For example, because $$\begin{align*}&(x^2-1)\sin^2x+2x\sin{x}\cos{x}+1\\&=(x^2-1)\sin^2x+2x\sin{x}\cos{x}+\sin^2x+\cos^2x\\&=x^2\sin^2x+2x\sin{x}\cos{x}+\cos^2x\\&=(x\sin{x}+\cos{x})^2.\end{align*}$$
Now, what does happen in the numerator?
We have the following:
$$\begin{align*}&(x^2-1)\cos{x}\sin{x}+2x\cos^2x-x\\&=(x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find the equation of line $AP$ in triangle $ABC$ where $P$ will make $PA=PB=PC$ I was recently doing COMC (Canadian Opening Mathematic Challenge) past year exams. Here is another question that I want to ask!
Here is the question:
Triangle $ABC$ has its sides determinded in the following way: side $AB$ by line $3x-2y+3 ... | In other words, the point $P$
is the center of the circumscribed circle,
and
the coordinates of the circumcenter
is known to be found as
\begin{align}
P&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\tag{1}\label{1}
,
\end{align}
wh... | {
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Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots
Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.
Obviously as soon as you factor this equation to $$(4x^... | Observing and categorizing the coefficients can be effective here.
Rewrite the equation into
$$(4x^4+8x^3+a^2x^2)+(4ax^3+8ax^2+a^3x)+(4x^2+8x+a^2)=0\\\Longrightarrow x^2(4x^2+8x+a^2)+ax(4x^2+8x+a^2)+(4x^2+8x+a^2)=0\\\Longrightarrow (x^2+ax+1)(4x^2+8x+a^2)=0$$
Or in this way:
$$(4x^4+4ax^3+4x^2)+(8x^3+8ax^2+8x)+(a^2x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886264",
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"source": "stackexchange",
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Solving a functional equation with derivatives? Am I doing this correctly?
Given
\begin{align}
f(x) = \frac{x^n}{1+x^n} - kf(x-1)
\end{align}
, then evaluating its the derivative
\begin{align}
\frac{df}{dx}=\frac{nx^{n-1}}{(1+x^n)^2}-k\frac{df}{dx}(x-1).
\end{align}
Let $t=x-1$ so $x=t+1$ and $dx=dt$. Thus,
\begin{alig... | Formally a solution is given by
$$
f_0(x) = \sum_{j = 0}^\infty (-k)^j\frac{(x-j)^n}{1+(x-j)^n}
$$
One can expect convergence if $|k| < 1$. Mathematica can evaluate this for certain choices of $k$ and $n$. For example, if $k = 1/2, \, n = 2$, then
$$
f_0(x) = \frac{1}{6} \left(4+3 i \left(-\frac{1}{2}\right)^{x-i}
\... | {
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On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$ The topic of odd perfect numbers likely needs no introduction.
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and le... | Let me try to work backwards from
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
This can be rewritten as
$$\frac{q-1}{q(q+1)} \leq 3 - \bigg(I(q^k) + I(n^2)\bigg) = \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-2}{q(q-1)}.$$
We also have
$$\frac{2n^2}{q+1} \leq D(n^2) < \fr... | {
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How to find the dot product using the law of cosines I'm working with the following problem:
We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$
Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.
$$AC... | $$\cos\measuredangle BAC=\frac{3^2+(\sqrt7)^2-2^2}{2\cdot3\cdot\sqrt7}=\frac{2}{\sqrt7},$$
which gives $$\vec{AC}\cdot\vec{AB}=3\cdot\sqrt7\cdot\frac{2}{\sqrt7}$$
In your solution this statement is wrong: $-2=-6\sqrt7\cos{x}$.
It should be $$-12=-6\sqrt7\cos{x}.$$
About you last adding.
It should be
$$\vec{AB}\cdotp\ve... | {
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"question_score": "2",
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)... | Triangle inequality tells us $|z+i| + |z-i| \ge |(z+i)-(z-i)| = |2i| = 2$ with equality holding if and only if geometrically of $0$ is colinear and between $z+i$ and $z-i$.
The only values colinear between $z + i$ and $z-i$ must all have the same real value so the real part as $z$ has and if $0$ is one such number, the... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
I got this question on my exam today ... | By De Moivre's theorem we have
$$\cos(6\theta)+i\sin(6\theta)=(\cos(\theta)+i\sin(\theta))^6$$
$$=-\sin^6(\theta) + \cos^6(\theta) + 6 i \sin(\theta) \cos^5(\theta) - 15 \sin^2(\theta) \cos^4(\theta) - 20 i \sin^3(\theta) \cos^3(\theta) + 15 \sin^4(\theta) \cos^2(\theta) + 6 i \sin^5(\theta) \cos(\theta)$$ by the Binom... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b = 0$, $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \... | Let $\vec a=t \vec i$, $\vec b= s \vec j$ then $\vec c=(t\vec i-\mu s \vec j)/4$.
$$2|\vec b+ \vec c|=|\vec a- \vec b| \implies |s \vec j+(t/4) \vec i-(s\mu/4)\vec j|=|t \vec i- s\vec j|$$
$$\implies (4-\mu)^2s^2+t^2=t^2+s^2\implies (\mu^2-8\mu+12)s^2=3t^2~~~(1)$$
Next $$(\vec b-\vec c).(\vec a- \vec b)=0 \implies t^2... | {
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Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?
EDIT: The sum should have been infinite.
| Sum should be infinite, because for finite sum you should get:
$$\sum_{k=0}^{n}(\frac{1}{3})^k=1+\frac{1}{3}+\dots+(\frac{1}{3})^n=\frac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}},$$ where we used formula $1+q+\dots+q^n=\frac{1-q^{n+1}}{1-q}$ for $q=\frac{1}{3}$.
Therefore,
$$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=... | {
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Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1... | Hint: $3^4 \cdot 3^{4k+2} + 1= 3^4(3^{4k+2}+1)+1-3^4=3^4(3^{4k+2}+1)-80.$
| {
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$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4... | The last expression
$$A=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
leads to the result.
Since $n$ is large
$$\sqrt{n+1}\sim \sqrt{n} \qquad \sqrt{n-1}\sim \sqrt{n}\qquad \sqrt{n^2-1}\sim \sqrt{n^2}=n$$
$$A \sim -\frac{2\sqrt{n^3}}{\big(\sqrt{n}+\sqrt{n}+2\sqrt{n}\big)\big(... | {
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Distance between the curves from a particular point The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to
(A) 1
(B) 2
(C) 3
(4) 4
My appr... | Distance $D_1$ of P(u,v) from the circle is $D_1=\sqrt{(u+2)^2+v^2}-2=\sqrt{F}-2$
and $D_2=\sqrt{(u-2)+v^2}=\sqrt{G}$, we require
$$D_1=D_2 \implies \sqrt{(u+2)^2+v^2}-2=\sqrt{(u-2)+v^2} \implies \sqrt{F}-\sqrt{G}=2~~~(1)$$ We also have $$F-G=8u \implies \sqrt{F}+\sqrt{G}=4u~~~~(2)$$
From (1) and (2), we have $$\sqrt{F... | {
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Is the absolute value on matrices well-defined with respect to PSD cone? Consider for two matrices $A, B$ we say $A \geq B $ if $ A - B$ is positive semidefinite.
Also, there exists a definition of absolute value for a matrix $A$ that is $|A| = \sqrt{A^\dagger A}$.
I was studying Hermitian matrices(matrices like $A$ wi... | The statement is false. Consider
$$A=\begin{bmatrix} 6 & 0 \\ 0 & 2\end{bmatrix}, \quad B=\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}.$$
We have $$A+B = \begin{bmatrix} 7 & 2 \\ 2 & 3\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 5 & -2 \\ -2 & 1\end{bmatrix} \ge 0$$
so $A \ge B \ge -A$. On the other hand, we have
$$... | {
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | $ 1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ multiplying this by $ 2^2 $ gives the sum $ 2^2+4^2+\dots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3} $ and subtracting this from the given identity results int the required equality.
| {
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"source": "stackexchange",
"question_score": "1",
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Tips finding the gcd of $(n^2-3n-1, n-4)$
I need tips on how to find the $gcd(n^2-3n-1, n-4)$ for any $n \in \mathbb{N}$.
I tried the two following methods (but got stuck):
Since I need to find the $gcd(n^2-3n-1, n-4)$, then there is a $q \in \mathbb{Z}$ which is the divisor of both $(n^3-3n-1)$ and $(n-4)$, giving e... | Bear in mind that $\gcd(a,b) = \gcd(a\pm kb, b)$ for all integers $k$.
So if $n^3 - 3n -1 = P(n)(n-4) + R(n)$ then $\gcd(n^2 -3n-1, n-4) = \gcd(n^2-3n-1 -P(n)(n-4), n-r) = \gcd(R(n), n-4)$.
And if we use synthetic division
$n^2 -3n -1 = n^2(n-4) +4n -3n -1=$
$n(n-4) +n-1=$
$n(n-4) +(n-4) +4-1=$
$n(n-4) + (n-4) + 3=$
... | {
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"timestamp": "2023-03-29T00:00:00",
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| we have $5^2 \equiv 1\ (mod\ 3) \Rightarrow 5^{2(3n+3)} \equiv 1\ (mod\ 3) \Rightarrow 5^{6n+6} \equiv 1\ (mod\ 3) $
and we have $2^2 \equiv 1\ (mod\ 3) \Rightarrow 2^{2(n+1)} \equiv 1\ (mod\ 3) \Rightarrow 2^{2n+2} \equiv 1\ (mod\ 3) $ Thus $$5^{6n+6}-2^{2n+2} \equiv 0\ (mod\ 3)$$
| {
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"source": "stackexchange",
"question_score": "1",
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Find MacLaurin Series and radius of convergence $$\sin{2z^2}$$
$$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$
Let's replace z with $2z^2$ in the MacLaurin series:
$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n
\frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$
$... | Isn’t it quite easy? Start with the standard Laurent series
$$
\sin(z)=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\,,
$$
make your substitution $z\to2z^2$, and get
$$
\sin(2z^2)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}z^{4n+2}}{(2n+1)!}\,,
$$
in which your $a_n$-term is $(-1)^n2^{2n+1}z^{4n+2}/(2n+1)!$.
Now form $\mid... | {
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Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\si... | $x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$
So it is a sphere with radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.
In spherical coordinates, $x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$ where $\theta$ is the azimuthal angle and $\phi$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\e... | Note
\begin{align}
\sqrt{\sqrt{2}+1}&=\sqrt{(\sqrt2-1)(\sqrt2+1)} \cdot\sqrt{\sqrt{2}+1}
= \sqrt{\sqrt{2}-1} (1+\sqrt2)\\
\end{align}
with $a=b = \sqrt{\sqrt{2}-1}$, thus no rational simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Find the solution for Cauchy problem using transformation I'm learning differential equations for the first semester and I'm not sure how to even begin with the following problem:
Find the solution for Cauchy problem using transformation:
$\left\{\begin{matrix}
(1-t^2)x''-tx'+n^2x=0 & n \in \mathbb{Z}\\
y(0)=1 & y'(0... | In fact this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0220.pdf
Let $s=\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$ ,
Then $\dfrac{dx}{dt}=\dfrac{dx}{ds}\dfrac{ds}{dt}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}$
$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}\right)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Nicomachus theorem proof - what did I do wrong? The exercise asked me to proof
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$...$
I formulate the equation as
(1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$
Proof base case:
$$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$
Proof (a+1) case:
(2) $$(a+1)^3=a^3+3a^2+3a+1$$
(3) $$(a+1)^3=\sum_{i=0}^{(a... | Your attempt seems a little messy, so I hope you don't mind that I present an alternative approach to an inductive proof.
You want to show $\displaystyle (1+2+...+n)^2 = 1^3 + 2^3 +... + n^3$
Assume, for some $k$ that $\displaystyle (1+2+...+k)^2 = 1^3 + 2^3 +... + k^3$ as the inductive hypothesis and call the expressi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
General proof that the minimum value of $x+y$ from $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$ is $4+11+3+8=26$ I found this problem and the solution on Twitter (translated).
$x$ and $y$ is a (positive) integer that satisfies $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$. If $y$ is the smallest number, what is $x+y$?
The... | If I were starting out from scratch here, I'd note that:
*
*$\frac{a}{b}<\frac{x}{y}$ means $bx-ay>0$. Since we are working with integers, $bx-ay\geq1$.
*$\frac{x}{y}<\frac{c}{d}$ means $-dx+cy>0$. Since we are working with integers, $-dx+cy\geq1$.
So we have the system
$$\left\{
\begin{aligned}
bx-ay&\geq1\\
-dx+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$ Induction proof, need a hint for more efective proof. I am looking for a smarter way of proving that inequality by induction.
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$$
For $n + 1$:
$$4^{n} {{3... | This isn't induction.
It uses Stirling
and it is more general.
I will show that
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
&\sim \dfrac{\sqrt{2\pi an}(an/e)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$ Prove that for natural numbers $n_1 < n_2 < ... < n_k$:
$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$
I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2... | We can go further to even state the biconditional equality condition. The following is a quote of Proposition 3 from Sum of Cubes is Square of Sum by Barbeau and Seraj, with permission from me, one of the authors:
We prove by induction that:
For integers $a_k$, if $1 \leq a_1 < a_2 < \cdots < a_n$ then
$$a_1^3 + a_2^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Value of integration $\int_0^1 \sin^{-1}(3x-4x^3)dx$ I have stumbled across an definite integration problem.
The question is to evaluate $$I=\int_0^1 \sin^{-1}(3x-4x^3)dx$$
I carried out the integration as this. Let $x = \sin A$ so that A = $\sin^{-1}(x)$ then
$$\begin {align} I &=\int \sin^{-1}(3\sin A-4\sin^3A)dx\\ &... | Note
\begin{align}
I&=\int_0^1 \sin^{-1}(3x-4x^3)dx \>\>\>\>\>\>\>(x=\sin A)\\
&= \int_0^{\frac\pi2} \sin^{-1}(\sin3A)\cos AdA \\
&= \int_0^{\frac\pi6} \sin^{-1}(\sin3A)\cos AdA
+ \int_{\frac\pi6}^{\frac\pi2} \sin^{-1}(\sin(\pi-3A))\cos AdA \\
&= \int_0^{\frac\pi6} 3A\cos AdA
+ \int_{\frac\pi6}^{\frac\pi2} (\pi-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$? I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.
Is there a way to find the equations of the three lines separately?
I tried factorizing the equation but wasn't able... | If $b+c=-2$, you can write $b=-k$ and $c=k-2$, so
$$\begin{align}x^3+bx^2y+cxy^2+y^3&=x^3-kx^2y+(k-2)xy^2+y^3\\&=x^3-kx^2y+kxy^2-2xy^2+y^3\\&=(x^3-xy^2)-(kx^2y-kxy^2)-(xy^2-y^3)\\&=(x-y)(x(x+y)-kxy-y^2)\\&=(x-y)(x^2-(k-1)xy-y^2)\end{align}$$
You can factor the latter part by completing the square but the computation is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $n$ such that $(3-3w+2w^2)^{4n+3}+(2+3w-3w^2)^{4n+3}+(-3+2w+3w^2)^{4n+3}=0$ for $w\neq1$ a cube root of unity
Let
$${\left( {3 - 3\omega + 2{\omega ^2}} \right)^{4n + 3}} + {\left( {2 + 3\omega - 3{\omega ^2}} \right)^{4n + 3}} + {\left( { - 3 + 2\omega + 3{\omega ^2}} \right)^{4n + 3}}=0$$
If $\omega\ne1$ ... | Your sum is $A^{4n+3}(1+\omega^{4n+3}+\omega^{2(4n+3)})$, and $A\ne 0$. So we must have $1+\omega^{4n+3}+\omega^{2(4n+3)}=0$. And $\omega^3=1$. So this is
$$1+\omega^n+\omega^{2n}=0$$
which is true if and only if $n\equiv 1$ or $2\bmod 3$.
As to which of the four answers to select, that is a different puzzle...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+... | Just another way to do it
$$\frac{1+x^{2}}{1+x^{4}}=\frac{1+x^{2}}{(x^2+i)(x^2-i)}=\frac 12\left( \frac{1+i}{x^2+i}+\frac{1-i}{x^2-i}\right)$$ and you could even continue with partial fractions.
Without any simplifications
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac{2 (i+1) \tan ^{-1}\left(\frac{x}{\sqrt{i}}\right)-(i-1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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