Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Easiest example of rearrangement of infinite leading to different sums I am reading the section on the rearrangement of infinite series in Ok, E. A. (2007). Real Analysis with Economic Applications. Princeton University Press.
As an example, the author shows that
is a rearrangement of the sequence
\begin{align} \frac{... | The following variant of the alternating harmonic is computationally easy. Consider the series
$$ 1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{8}-\frac{1}{8}+\cdots.$$
The sum is $0$. For the partial sums are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of a circle given two points and a line that passes through its center Find the equation of the circle that passes through the points $(0,2)$ and $(6,6)$. Its center is on the line $x-y =1$.
| First, find the center $(a, b)$. It is equidistant from the two points, so
$$
\sqrt{a^2 + (b - 2)^2} = \sqrt{(a - 6)^2 + (b - 6)^2}
$$
or
$$
a^2 + (b - 2)^2 = (a - 6)^2 + (b - 6)^2.
$$
Now, since the center is on the line $x - y = 1$, we know that $a - b = 1$, or
$$
a = b + 1.
$$
Substitute this into the quadratic equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| Let $$f(x)=(x-b)^3 + (b-c)^3 + (c-x)^3 -3(x-b)(b-c)(c-x) $$
The $f$ is a polynomial of degree at most $3$. Moreover, $x^3$ appears twice, with coefficients $\pm 1$, thus cancel. It follows that $f$ is at most quadratic and it is obvious that
$$f(c)=f(b)=f(0)=0$$
Since $f$ is at most quadratic and has three roots, $f \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 12,
"answer_id": 6
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Solve these equations simultaneously
Solve these equations simultaneously:
$$\eqalign{
& {8^y} = {4^{2x + 3}} \cr
& {\log _2}y = {\log _2}x + 4 \cr} $$
I simplified them first:
$\eqalign{
& {2^{3y}} = {2^{2\left( {2x + 3} \right)}} \cr
& {\log _2}y = {\log _2}x + {\log _2}{2^4} \cr} $
I then had:
$\eqal... | The equation wil be $\log_2 y= \log_2 x + \log_2 2^4$.
so $y$ will then be equal to $16x$.
So $4x$ will be $\dfrac y4$.
Substituting,so $3y=\dfrac y4+6$
$\dfrac {11y}4= 6$
So $y=\dfrac {24}{11}$. And $x=\dfrac {3}{22}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac... | The substitution $x^2=4k^2+1$ was wrong because $x^2=4k+1$, but nevertheless the next line is correct as if I've substituted the correct thing. The actual problem of the proof comes from when I said "$p$ has infinitely many solutions". I was thinking that $p$ is just an integer, and clearly $2(2^{n-3}-3)$ outputs a pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
The limit of $\lim\limits_{x \to \infty}\sqrt{x^2+3x-4}-x$ I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?
$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$
I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by... | Note that
\begin{align}
\sqrt{x^2+3x-4} - x & = \left(\sqrt{x^2+3x-4} - x \right) \times \dfrac{\sqrt{x^2+3x-4} + x}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{(\sqrt{x^2+3x-4} - x)(\sqrt{x^2+3x-4} + x)}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{x^2+3x-4-x^2}{\sqrt{x^2+3x-4} + x} = \dfrac{3x-4}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{3-4/x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = \Phi$? Consider:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}$$
How could one prove that $n$ is of such form that:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = {\sqrt{5\,}-1 \over 2} \implies n = \Phi$$
where $\Phi$ denotes the golden ratio?
Edit:
I notic... | HINT:
Just square both sides to get $$\frac1{(n+1)^2}=\frac n{n+1}\implies n^2+n-1=0$$ assuming $n+1\ne0$
Alternatively, $\sqrt{n+1}=(n+1)\sqrt n\implies 1=\sqrt{n(n+1)}$ assuming $n+1\ne0$
Now, square both sides to get $n^2+n=1$
So, $n=\frac{-1\pm\sqrt5}2$
As $\sqrt{\frac n{n+1}}$ is considered to be $\ge 0,\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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mathematical induction (sum of squares of first $2n$ numbers) I need to prove with mathematical induction the formula for the sum of the squares of the first $2n$ numbers.
The equation:
$$1^2 + 2^2 + 3^2 + \cdots + (2n)^2 = \dfrac{n(2n+1)(4n+1)}{3}.$$
The equation stands for $n \geqslant 1$.
Here is my solution:
$n=1$... | You are forgetting to add the term $(2k+1)^{2}$ in the LHS,i.e. for $n=k+1$ the LHS is $$1^{2}+2^{2}+\cdots +(2k)^{2}+(2k+1)^{2}+(2k+2)^{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $1 + \frac{1}{x} + \frac{1}{x^2}$ have a global minimum, where $x \in \mathbb{R}$? Does the function
$$f(x) = 1 + \frac{1}{x} + \frac{1}{x^2},$$
where $x \in {\mathbb{R} \setminus \{0\}}$, have a global minimum?
I tried asking WolframAlpha, but it appears to give an inconsistent result.
| $g(x)=1+\frac{1}{x}+\frac{1}{x^2},x\ne 0$
$g'(x)=-\frac{1}{x^2}-2\frac{1}{x^3}=0\Rightarrow \frac{-2}{x^3}=\frac{1}{x^2}\Rightarrow x=-2$
$g''(x)=\frac{2}{x^3}+\frac{6}{x^4}|_{x=-2}> 0$
$\Rightarrow g(-2)$ is the minimum.
Edit:This shows that $g$ has a local minimum at $x=-2$ but this is also the global minimum because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $ax - a^2 = bx - b^2$ for $x$ Method 1
Solve for x
$$ax - a^2 = bx - b^2$$
Collect all terms with x on one side of the equation
$$ax - bx = a^2 -b^2$$
Factor both sides of the equation
$$(a -b)x = (a+b)(a - b)$$
Divide both sides of the equation by the coefficient of $x$ (which is $a-b$)
$$x = a + b$$ (whe... | In the first method you come to the equation
$$(a-b)x=(a+b)(a-b)$$
In the second method, you come to the equation
$$(a-b)(x-(a+b))=0$$
which is just the first equation rearranged and factored. If we divide the first equation by $(a-b)$ without factoring, we still collect the solution $a=b$, because notice that it does... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ I have tried this as :
$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$
$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\l... | $$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
Generalising, $$\frac{1}{2^n-2^{n-1}+1}+\dots +\frac{1}{2^n}>\frac{1}{2^n}+\dots +\frac{1}{2^n}\text{ ($2^{n-1}$ terms)}$$
Hence, we can add an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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calculate riemann sum of sin to proof limit proposition $$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})$$
I'm having trouble expressing $\sin(x)$ differently here in order to calculate the riemann sum.
I want to show that this converges to $\frac{2}{\pi}$ so it equals to $\int_0^1 \sin(x\pi)$.
Is there any... | Use that $$\sum_{k=1}^n \sin kx=\frac{\sin\dfrac{kx}2\sin (k+1)\dfrac x2}{\sin\dfrac{x}2}$$
ADD One can deduce the above in several ways. The first is to note it is the imaginary part of $$\sum_{k=0}^n e^{ikx}=\frac{e^{(n+1)ix}-1}{e^{ix}-1}$$
Another choice is to use $$\cos b-\cos a=2\sin\frac{b-a}2\sin\frac{b+a}2$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$\implies 0^2=1+2 (ab+bc+ca)\implies ab+bc+ca=-\frac12$$
Again, $ab+bc+ca=a(b+c)+bc=a(-a)+bc\implies bc-a^2$
$$\implies bc-a^2=-\frac12\implies bc=\frac{2a^2-1}2$$
$$\implies abc=a\cdot \frac{2a^2-1}2=\frac{2a^3-a}2$$
Now use Second derivative test for the extreme values of $f(a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to use double angle identities to find $\sin x$ and $\cos x$ from $\sin 2x $? If $\sin 2x =\frac{5}{13}$ and $0^\circ < x < 45^\circ$, find $\sin x$ and $\cos x$.
The answers should be $\frac{\sqrt{26}}{26}$ and $\frac{5\sqrt{26}}{26}$
Ideas
The idea is to use double angle identities. One such identity is $\sin 2x... | Suppose we had a right triangle with an angle $2x$, and $\sin2x=\frac{5}{13}$. Further suppose that the hypotenuse of the triangle was 13. We can deduce that the side oppoites $2x$ must be 5. Applying the Pythagorean theorem to find the other side we have
$$13^2=5^2+(\text{adjacent side})^2$$
$$169-25=(\text{adjacent s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
| suppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular
than $\cos A=\frac cb$ and $\sin A=\frac ab$
using pythgo. theo. $\;b^2=a^2+c^2$
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {\frac {c^2}{b^2}}{1-\frac {a}{b}}$$
$$\dfrac {\frac {c^2}{b^2}}{\frac {b-a}{b}}$$
$$\dfrac {c^2b}{b^2(b-a)}$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Question about modular arithmetic and divisibility If $$a^3+b^3+c^3=0\pmod 7$$
Calculate the residue after dividing $abc$ with $7$
My first ideas here was trying the 7 possible remainders and then elevate to the third power
$$a+b+c=x \pmod 7$$
$$a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc=x^3\pmod 7$$
$$3(a+b+c)(ab+bc+ac)-3abc... | HINT: What are the possible cubes in modulo 7? Hence what combinations of these cubes allow for $a^3+b^3+c^3=0$ (mod 7). From this $abc$ (mod 7) should be clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it? Studying for my Algebra exam, and this question popped out with no solution in a previous exam:
Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 ... | Hint: If a matrix $A$ has column vectors $a_1,..,a_n$, then
$$A\cdot e_i=a_i$$
where $e_1=\pmatrix{1\\0\\0\\ \vdots}$, $e_2=\pmatrix{0\\1\\0\\ \vdots}$, $e_3=\pmatrix{0\\0\\1\\ \vdots}$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| Hint: $(a+1)^2 + a = 0$ and so $\frac{1}{(a+1)^2} = -\frac{1}{a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
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Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqr... | Note the identity $$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3xyz$$
So that if $x+y+z=0$ we have $x^3+y^3+z^3=3xyz$
This is often useful with cubic expressions like this. Take $x=-S;\text{ } y=\sqrt[3] {5+2 \sqrt {13}};\text{ }z=\sqrt[3]{5-2 \sqrt {13}}$
Then we have$$-S^3+5+2\sqrt{13}+5-2\sqrt{13}=-3S\sqrt[3]{25-52... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac ... | HINT:
$$\text{As }\frac{d(\tan mx)}{dx}=m\sec^2mx,$$
$$\int\sec^2mx= \frac{\tan mx}m+C$$
here $m=\frac12$
or using Weierstrass substitution $\tan \frac x2=t,$
$\frac x2=\arctan t\implies dx=2\frac{dt}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$
$$\int \frac{dx}{1+\cos x}=\int dt=t+K=\tan\frac x2+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How find the value $\sum\limits_{k=1}^{\infty}(C_{3k}^k)^{-1}$ find the value
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{3k}^{k}}$$
and long ago,I have see this and is easy
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}$$
where
$$C_{n}^{k}=\dfrac{n!}{(n-k)!k!}$$
| For #2:
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}=\frac{1}{3}+\frac{2\pi}{9\sqrt{3}}$$
Because,
$$\sum_{n=0}^\infty\frac{x^n}{C^n_{2n}}=\frac{4(\sqrt{4-x}+\sqrt{x}\arcsin(\frac{\sqrt{x}}{2}))}{\sqrt{(4-x)^3}}$$
Which is because
$$\text{If we set, } A_n=\frac{1}{C^{n}_{2n}}$$
Then
$$(4n+2)A_{n+1}=(n+1)A_n$$
And so,
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to solve the given initial-value problem? Solve the given problem which the input function $g(x)$ is discontinuous?
$y''+4y = g(x)$, $y(0) = 1$, $y'(0) = 2$, where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,& x>\frac{\pi}{2} \end{cases}$$
And the given answer is,
$$y = \begin{cases} \cos 2x+\frac... | Note: It is best to use Laplace Transforms for these sorts of problems and this is where the LT shows it's usefulness, but that is not allowed, so we will use undetermined coefficients instead.
We are given:
$\tag 1 y''+4y = g(x), y(0) = 1, y'(0) = 2$
where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sq... | This is meant to follow up on Ethan's comment about using Herschfeld's theorem to prove that the expression converges.
Theorem (Herschfeld, 1935). The sequence
$$
u_n = \sqrt{a_1 + \sqrt{a_2 + \cdots + \sqrt{a_n}}}
$$
converges if and only if
$$
\limsup_{n\to\infty} a_n^{2^{-n}} < \infty.
$$
The American Mathem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 5,
"answer_id": 4
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Find coefficient of $x^{100}$ in the power series expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ I'm trying to find to coefficient of $x^{100}$ of
$$\sum_{n=0}^{∞}a_n x^{n}\ =\frac{1}{(1-x)(1-2x)(1-3x)}.$$
I used the sum:
$$\frac{1}{1-x}\ = 1+x+x^2+\ldots.$$
So : $$\frac{1}{(1-x)(1-2x)(1-3x)}= \left(1+x+x^2+\ldots\right)\... | Use partial fractions: $$\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{1}{2} \cdot \frac{1}{1-x} - 4 \cdot \frac{1}{1-2x} + \frac{9}{2} \cdot \frac{1}{1-3x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to prove that $\left(\sum\limits_{k=1}^{n}a_{k}\right)^2\ge\sum\limits_{k=1}^{n}a_k^3$? Let $$a_{n}\ge a_{n-1}\ge\cdots\ge a_{0}= 0,$$ and for any $i,j\in\{0,1,2\dots,n\},j>i$, there is
$$a_{j}-a_{i}\le j-i.$$
Prove that
$$\left(\sum_{k=1}^n a_k \right)^2\ge\sum_{k=1}^n a_k^3.$$
My idea is by mathematical induction... | We can solve this using induction, like you suggested. The case $n=1$ is quite easy: $(a_1)^2\geq(a_1)^3$ since $0\leq a_1\leq 1$.
Now assume that for some $n > 0$, the property is true for $n-1$. Let $S_n=\sum_{k=0}^n a_k=\sum_{k=1}^n a_k$.
\begin{align*}
(S_n)^2-\sum_{k=1}^n (a_k)^3 &= (a_n)^2 + 2a_nS_{n-1} + (S_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\rig... | \begin{align*}
(x+y)^4+x^4+y^4&=2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=2(x^4+2x^3y+2x^2y^2+x^2 y^2+2 x y^3+y^4)\\
&=2(x^4+2(xy+y^2)x^2+(xy+y^2)^2)\\
&=2(x^2+xy+y^2)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\cos 18^\circ$ without using the calculator I only know $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$ as standard angles but how can I prove that
$$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$
| Let $\theta = 18^\circ$.
Then $2\theta = 36^\circ$ and $3\theta = 54^\circ$.
Note that $90^\circ-3\theta = 2\theta$.
Thus, $\sin(90^\circ - 3\theta) = \sin (2\theta)$ or $\cos(3\theta) = \sin(2\theta)$
So, $4\cos^3\theta-3\cos\theta = 2\sin\theta\cos\theta$.
Since $\cos \theta$ can't be zero, we can divide by it to ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Possible to solve this coupled system of vector equations? Let $\gamma, \omega, c$ be positive constants, let $\mathbf{Q}_{a}$ and $\mathbf{Q}_{b}$ be three-dimensional vectors, and let $\mathbf{B}(\mathbf{r})=\mathbf{B}(x,y,z)$ be a vector field. Let $\mathbf{B}_{\circ}= \mathbf{B}(\mathbf{r}_{\circ})$ for some positi... | to long for a comment, this is not an answer (yet)
Begin with your equations:
$$
-\frac{\omega^2}{\gamma}\mathbf{a}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{a}+\omega\mathbf{b}\times\mathbf{B}_{\circ} \tag{1}
$$
$$
-\frac{\omega^2}{\gamma}\mathbf{b}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{b}-\omega\mathbf{a}\times\mathbf{B}_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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"Direct Proof" of the Steiner-Lehmus Theorem The Steiner-Lehmus Theorem is famous for its indirect proof. I wanted to come up with a 'direct' proof for it (of course, it can't be direct because some theorems used, will, of course, be indirect). I started with $\Delta ABC$, with angle bisectors $BX$ and $CY$, and set t... | Your method looks fine but I don't know what you mean by "easily disproved by a counterexample."
If $a,b,c>0$ then
$$
\frac{b}{c}=\frac{a+c}{a+b} \iff b^2+ab=c^2+ac \iff b=c
$$
so indeed it is sufficient to prove that. (There are algebraic counterexamples if you let $c=-a-b$, which seems to be why this type of proof is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $(1-\cos x)/\sin x = \tan x/2$ Using double angle and compound angles formulae prove,
$$
\frac{1-\cos x}{\sin x} = \tan\frac{x}{2}
$$
Can someone please help me figure this question, I have no idea how to approach it?
| $$
\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} = 1-2\sin^2\frac{x}{2}
$$
and
$$
\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}
$$
so
$$
\frac{1-\cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/440457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$ let $x,y,z>0$,and such that
$x^2+y^2+z^2=1$,prove that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
Does this have a nice solution? Thank you everyone.
| This answer is neither nice nor complete. At the end, I still need to plot a very complicated function in one variable to conclude the inequality is true.
In any event, here is my attempt.
Let $a,b,c$ be the three elementary symmetric polynomials associated with $x,y,z$, i.e:
$$
\begin{cases}
a &= x + y + z\\
b &= xy +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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Induction of inequality involving AP Prove by induction that
$$(a_{1}+a_{2}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)\geq n^{2}$$
where $n$ is a positive integer and $a_1, a_2,\dots, a_n$ are real positive numbers
Hence, show that
$$\csc^{2}\theta +\sec^{2}\theta +\cot^{2}\theta... | Define $x = a_1 + a_2 \cdots + a_{n-1}$ and $y = \frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}$.
By the induction hypothesis, we have $xy \geq (n-1)^2$.
We need to prove $(x + a_n)(y + \frac1{a_n}) \geq n^2$
But clearly by induction hypothesis,
$(x + a_n)(y + \frac1{a_n}) = xy + ya_{n} + \frac{x}{a_n} + 1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$ Recently I found a claim saying that
$$
\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2
$$
from what I can see this seems to be true. I already know that
$\int_... | First of all, we use compound angle formula and IBP.
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin (a x) \sin (b x)}{x^{2}} d x \\
=& \frac{1}{2} \int_{0}^{\infty}[\cos (a-b) x-(\sin (a+b) x] d\left(-\frac{1}{x}\right) \quad \text{(Via compound angle formula)}\\
\stackrel{IBP}{=} &-\left[\frac{1}{2 x}(\cos (a-b) x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 6,
"answer_id": 2
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Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\l... | $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}$ -- it is true only if these limits exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Use a known Maclaurin series to derive a Maclaurin series for the indicated function. $$f(x)=x\cos(x)$$
I'm not quite sure how to do this. I did two others, which I presume is the right way to do it, as follows:
\begin{align}
e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\\
e^{-x/4}&=1-\frac{x}{4}+\frac{x... | The well known Maclaurin series is $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots$
We can use this as follows, (look over @AWertheim's answer before viewing spoiler)
$$ f(x) = x \cos x = x \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right) = x - \dfrac{x^3}{2!} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding minimal polynomial of a matrix Let $A\in M\underset{nxn}{(\mathbb{R})}$ so that $\forall\lambda\in\mathbb{R}: A\neq \lambda I$. It also given that $A^4-5A^2+4I=A^3-7A-6I=0$.
Find the minimal polynomial.
I got the polynomial: $x^4-x^3-5x^2+7x+10=0$. After factorization I got:
$(x+2)(x+1)(x^2-4x+5)=0$.
I tried ... | Recall that if a polynomial $P$ annihilates a matrix $A$ then the minimal polynomial of $A$ divides $P$ and then the spectrum of $A$ is includ in the set of roots of $P$.
We have
$x^4-5x^2+4=(x-1)(x-2)(x+2)(x+1)$ annihilates $A$ and then $\mathrm{sp}(A)\in\{1,2,-2,-1\}$
and
$x^3-7x-6=(x-3)(x+2)(x+1)$ annihilates $A$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ $a,b,c$ belongs to natural prove that $\log_5 {abc}\geq2$ If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ with $a, b, c\in \mathbb N$, prove that $\log_5 {abc}\geq2$.
The equations I could form are:
1) $f(0)>0$ and ... | We need to show that $abc\ge25$. Since both roots are real and distinct, we have that $b^2-4ac>0$ and so $b^2>4ac$. Since both roots are in (0,1), their average $\frac{b}{2a}<1$ and therefore $b<2a$. Since the larger root $\frac{b+\sqrt{b^2-4ac}}{2a}<1$, we have that $b+\sqrt{b^2-4ac}<2a$ and therefore $\sqrt{b^2-4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$? EDIT:
My question was poorly worded.
I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$.
What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for sma... | To show that the inequality holds for $x \geq 0$ note that
$$ \log(1+x) = \int_1^{1+x} \frac{1}{t}dt. $$
Approximate the function $t \mapsto \frac{1}{t}$ by its tangent at $t = 1 + \frac{x}{2}$ to get
$$ \frac{1}{t} \geq -\frac{1}{(1+\frac{x}{2})^2}(t - 1 - \frac{x}{2}) + \frac{1}{1 + \frac{x}{2}} = -\frac{4 t}{(x + 2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The indu... | To see that $\binom{2n}{n} < 2^{2n}$, apply the binomial theorem $$2^{2n} = (1+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n}.$$
To see that $2^n < \binom{2n}{n}$, write it as a product $$\binom{2n}{n} = \frac{2n}{n} \cdot \frac{2n-1}{n-1} \cdot ... \cdot \frac{n+1}{1},$$ where each factor is $\ge 2$, and for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 3
} |
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}... | $$\sin\alpha + \cos\alpha = 0.2$$
$$\sin\alpha + \sqrt {1-\sin^2\alpha}= 0.2$$
$$\sqrt{1-\sin^2\alpha}= 0.2-\sin\alpha$$
$$1-\sin^2\alpha=0.04-0.4\sin\alpha+\sin^2\alpha$$
$$2\sin^2\alpha-0.4\sin\alpha-0.96=0$$
$$\sin^2\alpha-0.2\sin\alpha-0.48=0$$
$$\sin\alpha=\frac{0.2\pm1.392...}{2}$$
$$2\sin\alpha=1,592...$$
$$2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Simple looking log problem How would I solve this for $x$?
The original problem is
$$x+x^{\log_{2}3}=x^{\log_{2}5}$$
I have tried to reduce it down to this,
$$x^{\log_{10}3}+x^{\log_{10}2}=x^{\log_{10}5}$$
I have been stuck on this for a while now, can't seem to figure out the trick to this.
Sorry if I tagged this wron... | Oops!
My answer was wrong,
because I wrote the first term of the equation as
$1$
instead of $x$.
I will correct this now,
and hope I do not make any other errors.
The equation is
$x+x^{\log_{2}3}=x^{\log_{2}5}$.
Eliminating the obvious root of
$x=0$,
this becomes
$1+x^{\log_{2}3/2}=x^{\log_{2}5/2}$
or
$1=x^{\log_{2}5/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
n is+ve integer, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$, $y$ being positive integers and $(x \neq y)$ I wanted to know, how can i solve this.
For a given positive integer n, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$ and $y$ b... | If $(x,y)$ is a solution of $\frac{1}{x}+\frac{1}{y}$ = $\frac{1}{n}$ then $x>n$ and $y>n$.
We can write the equation as
$n(x+y)=xy$ = $(x-n)(y-n) = n^2$
and for any $k$ positive $k|n^2$, therefore the solutions are given by $x-n=k$ and $y-n=\frac{n^2}{k}$ or $x=n+k$ and $y=n+\frac{n^2}{k}$.
The only case for $x=y$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Differentiability of function involving absolute values I need to check the differentiability at $x=-10$ of the following function.
$f(x)=\cos|x-5|+\sin|x-3|+|x+10|^3-(|x|+4)^2$
Now,
\begin{align*}
\text{LHD} &= \lim_{x\to -10^-}\frac{f(x)-f(-10)}{x+10}\\
&= \lim_{x\to -10^-}\frac{\cos(x-5)-\sin(x-3)-(x+10)^3-(4-x)^2-... | Let's concentrate on the $|x+10|^3$ component since near $x=-10$ we have $$|x-5| =5-x $$$$|x-3|=3-x$$ and $$|x|+4=-x+4$$ and the components involving these quantities are readily seen to be differentiable.
Now $|x+10|$ is continuous at $x=-10$ but not differentiable there. We have
$$|x+10|^3= -(10+x)^3 \text{ for }x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| Related technique. You can use the Laplace transform technique. Recalling the Laplace transform
$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$
Taking $ f(x) = \frac{\sin(x)^3}{x^3} $ gives
$$ F(s)= \frac{\pi \,{s}^{2}}{8}+\frac{3\,\pi}{8}- \frac{3( {s}^{2}-1) }{8}\,\arctan \left( s \right) +\frac{( {s}^{2}-9)}{8}\,\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 2
} |
Providing a closed formula for a linear recursive sequence I am studying for an exam in linear algebra and I have some trouble solving the following:
Let $(a_n)$ be a linear recursive sequence in $\mathbb{Z}_5$ with
\begin{align}
a_0 = 2, a_1 = 1, a_2 = 0 \text{ and } a_{n+3} = 2a_{n+2} + a_{n+1} + 3a_n \text{ for } ... | Hint: Use elementary row operations on your matrix $A$. You shall get identity matrix as $D$. Represent $T$ by elementary matrixes
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Euler's phi function $\phi(n)$ is even for all $n \geq 3$; when is it not divisible by $4$? Problem 1: Show that $\phi(n)$ is even for all $n \geq 3$.
Proof: Assume that n is a power of 2, let us say that $n=2^k$, with $k \geq 2$. By the Phi Function Formula, we have $\phi(n) = \phi(2^k)=2^k - 2^{k-1}=2^k(1/2)=2^{k-1}... | HINT: If $p$ is prime, $\varphi(p^k)=p^k-p^{k-1}=p^{k-1}(p-1)$. If $m$ and $n$ are relatively prime, then $\varphi(mn)=\varphi(m)\varphi(n)$. And every $n\ge 1$ is a product of powers of distinct primes. These facts make the first question very easy to answer and are sufficient to answer the second question as well.
Y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Geometric Identities involving $π^2$ Are there any known geometric identities that have $π^2$ in the formula?
| The sphere of radius $r$ in $\mathbb R^4$ has volume (4-dimensional volume) $$ \frac{\pi^2 \; r^4}{2} $$ Here, we refer to the set $$ x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq r^2. $$
The sphere of radius $r$ in $\mathbb R^5$ has volume (5-dimensional volume) $$ \frac{8 \pi^2 \; r^5}{15} $$ Here, we refer to the set $$ x_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdo... | Let $S_n(p) = \displaystyle \sum_{k=1}^n \dfrac1{k^p}$ and $S(p) = \lim_{n \to \infty} S_n(p)$. We then have
$$S_{2n+1}(p) = \sum_{k=1}^{2n+1} \dfrac1{k^p} = 1 + \sum_{k=1}^n \left(\dfrac1{(2k)^p} + \dfrac1{(2k+1)^p}\right) < 1 + \sum_{k=1}^n \dfrac2{(2k)^p} = 1 + \dfrac{S_n(p)}{2^{p-1}}$$
Letting $n \to \infty$, we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be... | Firstly, if either $a$ or $b$ is a perfect cube, then we have $$ (\sqrt[3]{a}-N)^2 = 49 + 20 \sqrt[3]{6} $$
Convince yourself that the LHS must involve 2 different surd terms, hence this is not possible. Thus, neither $a$ nor $b$ are a perfect cube.
Like you did, expand the terms and consider what they are. Clearly $a,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | Clearly, $\sqrt{(4-\frac{9}{2})^2}$ is what? Well, $4-\frac{9}{2}$ is $-\frac{1}{2}$. But we all know $(-\frac{1}{2})^2$ is $\frac{1}{4}$ since $(-x)^2=x^2$. That means that $\sqrt{(4-\frac{9}{2})^2}$ is $\frac{1}{2}$. So the assumption that $\sqrt{(4-\frac{9}{2})^2} = 4 - \frac{9}{2}$ leads to $-\frac{1}{2}=\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 8
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What are the coefficients of the polynomial inductively defined as $f_1=(x-2)^2\,\,\,;\,\,\,f_{n+1}=(f_n-2)^2$?
Let $\{f_n(x)\}_{n\in \Bbb N}$ be a sequence of polynomials defined inductively as
$$\begin{matrix} f_1(x) & = & (x-2)^2 &
\\ f_{n+1}(x)& = & (f_n(x)-2)^2, &\text{ for all }n\ge 1.\end{matrix}$$
Let $a_n$ ... | HINT:
$f_1(x)=(x-2)^2=4-4x+x^2$
$f_2(x)=(f_1(x)-2)^2$
$=(x^2-4x+4-2)^2=(2-4x+x^2)^2=2^2+2\cdot2(-4x)+\cdots=4-4^2x+\cdots$
$f_3(x)=(f_2(x)-2)^2$
$=(4-2^4x+\cdots-2)^2=(2-2^4x+\cdots)^2=2^2-2\cdot2\cdot4^2x+\cdots=2^2-4^3x+\cdots$
$f_4(x)=(f_3(x)-2)^2$
$=(2^2-4^3x+\cdots-2)^2=(2-4^3x+\cdots)^2=2^2-2\cdot2\cdot2^6x+\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many zeros does $z^{4}+z^{3}+4z^{2}+2z+3$ have in the first quadrant? Let $f(z) = z^{4}+z^{3}+4z^{2}+2z+3$. I know that $f$ has no real roots and no purely imaginary roots.
The number of zeros of $f(z)$ in the first quadrant is $\frac{1}{2\pi i}\int_{C}\frac{f'(z)}{f(z)}\, dz$ by the Argument Principle where $C = ... | Here's an alternative idea, which can be applied to any quartic.
Let $z=x+iy$, expand the powers of $z$, and separate the real and imaginary parts to get
$$x^4+x^3+4x^2+2x+3-(6x^2+3x+4)y^2-y^4=0$$
and
$$(4x^3+3x^2+8x+2)y-(4x+1)y^3=0$$
Since we looking for solutions with $x,y\gt0$, we can rewrite the second equation as
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true? Can someone explain?
| A rotation matrix is clearly a linear transformation, so by rotating the elementary basis vectors, a two-dimensional rotation matrix for angle $x$ is
\begin{align}
\begin{bmatrix}
\cos(x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{bmatrix}.
\end{align}
Squaring this matrix is equivalent to applying the rotation twice, so
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 8,
"answer_id": 7
} |
Calculate the integer part I have to calculate the integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
| Calvin Lin provided an excellent proof. I am just adding in some details.
Obviously, $\sqrt{ 5} - \sqrt{2} > 0$
Also, $\sqrt{2} > 1$
$2 \sqrt{2} > 2$
$1 + 2 \sqrt{2} + 2 > 5$
$1^2 + 2 \sqrt{2} + (\sqrt{2})^2 > (\sqrt{5})^2$
$(1 + \sqrt{2})^2 > (\sqrt{5})^2$
Taking the positive square root, we have $1 + \sqrt{2} > \sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
|
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$
AB , BC , CD are lengths of sides of triangle and AD , BE , CF are lengths of medians of triangle
HINT :
start with appollonius theorem and add all
Apollonius Theorem
so you will get relation as
3 (sum of square of sides) = 4 ( sum of squares of medians )
Proo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\... | Hint: Telescope your rationalized sum by adding terms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?
I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?
| Let positive integer $d$ divides both $a+b,a^2+b^2$
$\implies d$ divides $(a^2+b^2)+(a+b)(a-b)=2a^2$
Similarly, i.e., $d$ divides $(a^2+b^2)-(a+b)(a-b)=2b^2$
$\implies d$ divides $2a^2,2b^2\implies d$ divides $(2a^2,2b^2)=2(a^2,b^2)=2(a,b)^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
How to find the sum $\sum\limits_{k=1}^n (k^2+k+1)k!$? How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$
What I tried as follows :
$$\sum^n_{k=1} (k^2+k+1)k!$$
=$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$
Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$... | HINT:
$$(k^2+k+1)k!= (k+1)(k+2) k!+A (k+1)k!+B k!=(k+2)!+A (k+1)!+B k!$$
$$\implies k^2+k+1=(k+1)(k+2)+A(k+1)+B=k^2+k(3+A)+2+A+B $$
Equating the coefficients of $x,A+3=1\implies A=-2$
Equating the constant terms, $A+B+2=1\implies B=-1-A=-1-(-2)=1$
So, $$(k^2+k+1)k!= (k+2)!-2(k+1)!+ k!=(k+2)!-(k+1)!-\{(k+1)!-k!\}$$
Can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descarte... | I'm assuming that you are looking for a factorization of the polynomial
$$
f = x^4 + 12x^3 + 44x^2 + 72x + 36
$$
in $\mathbb Q[x]$.
By the Lemma of Gauss and the fact that $f$ is monic, this is the same as looking for a factorization in $\mathbb Z[x]$.
Since there are no rational roots, the only remaining possibility i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 1
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Why can we use inspection for solving equation with multiple unknowns? In our algebra class, our teacher often does the following:
$a + b\sqrt{2} = 5 + 3\sqrt{2} \implies \;\text{(by inspection)}\; a=5, b = 3
$
I asked her why we can make this statement. She was unable to provide a satisfactory answer. So I tried prov... | HINT:
Assuming $a,b,x,y$ are rationals $\sqrt2(b-y)=x-a$ rational which is only possible
if $b-y=0$
As for $b-y\ne0,\sqrt 2=\frac{x-a}{y-b}$ which is rational
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463577",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3... | Let
$$D = \frac{(x+h)^{4/3}-x^{4/3}}{h}$$
Then
$$\begin{align}D^3 &= \frac{(x+h)^4-3 (x+h)^{8/3} x^{4/3} + 3 (x+h)^{4/3} x^{8/3} - x^4}{h^3}\\ &= \frac{4 x^3 h + 6 x^2 h^2 + 4 x h^3 + h^4 -3 (x+h)^{4/3} x^{4/3} \left [ (x+h)^{4/3}-x^{4/3}\right ]}{h^3}\\ &= \frac{4 x^3 + 6 x^2 h + 4 x h^2 +h^3}{h^2} - \frac{3 (x+h)^{4/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
| Let $\mathbf v(x) = (\sin x \cdot \sin 2x, \cos x \cdot \cos 2x) \in \mathbb R^2$. Compute:
$$\begin{align} |v|^2 &= \sin^2 x \cdot \sin^2 2x + \cos^2 x \cdot \cos^2 2x \\
&= (\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \\
&= 1 - \sin^2 x \cdot \cos^2 2x - \sin^2 2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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How to calculate this expression? evaluate the expression [1]:
$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)} $$
where $x$ is a real number, $0\le x\le1$, and $x$ is rounded to 3 digits.
For example, when $x=0.500$, the expression is [2]:
$$\left(\frac11 -\frac1{1.5}\right)+\left(\frac... | For $x=0$, the sum evaluates to $0$, and for $x=1$, it is $1$.
For $0<x<1$, you can derive a power series expression for the given sum as below:
\begin{align}
\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{x+n}\right)=&\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n}\left(1+\frac{x}{n}\right)^{-1}\right)\\
\ =& \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
| I am assuming that you mean
$\frac{\sqrt{x^2+4x^4}}{8+x^2}$.
Since $x^2$ and $x^4$ are positive,
the limit is the same for
$x \to \infty$ and
$x \to -\infty$.
You can take your last step
further and write
$\begin{align}
\frac{\sqrt{x^2+4x^4}}{8+x^2}
&= \frac{x\sqrt{1+4x^2}}{8+x^2}\\
&= \frac{x^2\sqrt{1/x^2+4}}{x^2(1+8/... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation for x, y and z: $\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$ I am having some trouble with this problem,
Solve for $x,y,$ and $z$.
$$\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$$
Here is my work so far,
$$x - y +z = x+y+z-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy}$$
$$2y-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy} =... | Even easier, your last equation can be written as:
$$
-\sqrt{y}\left(\sqrt{x} - \sqrt{y}\right) + \sqrt{z}\left(\sqrt{x} - \sqrt{y}\right) = 0
$$
or
$$
\left(\sqrt{z} - \sqrt{y}\right)\left(\sqrt{x} - \sqrt{y}\right) = 0
$$
Which in turn gives you the same conclusion as Calvin pointed out
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $x$ in the equation $ax^3+bx^2+cx=d$ $\begin{equation} \tag{A} ax^3+bx^2+cx=d \end{equation}$
We can define Delta for quadratic equation to check whether the equation has answer or not....for $f(x)$ which contains powers higher of $2$ for Is there any method to see how many acceptable roots the polynomial contain... | It happens that equations of the form $ax^3+bx^2+cx+d=0$ and cubic polynomials in general, do have their own so-called discriminant $\Delta$. It is defined as: $$
\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd.\,
$$
We have $3$ distinct cases:
*
*If $\Delta > 0$, then the equation has three distinct real roots.
*If... | {
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"url": "https://math.stackexchange.com/questions/468157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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question about applying a comparison test to a sequence In lecture, we were asked:
Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$ converge?
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$
In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosin... | We do an explicit comparison, since that's what the question seems to ask for, but there are simpler ways.
For $n\gt 1$, the top is positive and $\lt 2n^3+3n^3=5n^3$.
The bottom is trickier. Rewrite it as
$$\frac{1}{2}n^5+ \frac{1}{2}n^5-5n^3-n^2+2.$$
Note that if $n\gt 10$, then $\frac{1}{2}n^5-5n^3-n^2+2\gt 0$. This... | {
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"url": "https://math.stackexchange.com/questions/469345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Expected value sum of dots We throw $n$-times the die. Let $E_n$ be expected value sum of dots (got in all throws).
Compute
*
*$E_1$
*$E_2$
*$E_3$
*$E_4$
So i know how can I do it, for example in 1. I have:
*
*$E_1 = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \... | If you take $X=X_{1}+,...,+X_{n}$ where $X_{i}, i \in \{1,...,n\}$ is the number of dots on the throw $i$ then you need $E[X]$. So, because expectation of a sum is the sum of the expectations
$$E[X]=E\left[\sum\limits_{i=1}^{n}X_{i}\right]=\sum\limits_{i=1}^{n}E{\left[X_i\right]}$$
but $E[X_i]=E[X_j]$ for all $i,j$ bec... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | Let $7x=\pi\implies 4x=\pi-3x$
$$\frac1{\sin2x}+\frac1{\sin4x}=\frac{\sin4x+\sin2x}{\sin4x\sin2x}$$
$$=\frac{2\sin3x\cos x}{\sin(\pi-3x)\sin2x}(\text{ using } \sin2A+\sin2B=2\sin(A+B)\cos(A-B))$$
$$=\frac{2\sin3x\cos x}{\sin(3x)2\sin x\cos x}(\text{ using } \sin2C=2\sin C\cos C \text{ and }\sin(\pi-y)=\sin y)$$
$$=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 +... | Start from $\displaystyle\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{2\tan{15^\circ}}{1-\tan^2{15^\circ}}$.
If $x=\tan{15^\circ}$, then $\displaystyle\tan 15^\circ = x = \frac{2\tan{(\frac{15}{2})^\circ}}{1-\tan^2{(\frac{15}{2})^\circ}}$.
If $y=\tan{(\frac{15}{2})^\circ}$, then $x=\frac{2y}{1-y^2}$. Hence
$\displaystyle... | {
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"url": "https://math.stackexchange.com/questions/472594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $\sum_{r=1}^nr(r+1)$ Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $$\sum_{r=1}^nr(r+1)$$
I can obtain $$\sum_{r=1}^n3r^2+3r+1=(n+1)^3-1$$
and I think the next step is
$$3\sum_{r=1}^nr(r^2+1)+\frac13=\left((n+1)^3-1\right)$$
But how do I deal with the const... | We have
$$r(r+1)=\frac{1}{3}((r+1)^3-r^3-1)$$
so
$$\sum_{r=1}^n r(r+1)=\frac{1}{3}\sum_{r=1}^n((r+1)^3-r^3)-\frac{1}{3}\sum_{r=1}^n1=\frac{1}{3}((n+1)^3-1)-\frac{n}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
| If $2n$ has 28 divisors, then $2n$ is of the form $a^6.b$, where b has 4 divisors. Note that if chose $2^13$ or $2^26$ here, then the third multiple would have more than $30$ divisors, since it would be $3*13* or $2*27$ divisors required.
Since $3n$ does not have a divisor of the form $a^6$, we see that $a=2$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way? I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p ... | The polynomials $P_n(x)=P_n(x,1)$ are called the Cauchy-Mirimanoff polynomials.
They are defined for all $n$ natural $n\geq 2$ by:
$$ (X^n+1)-X^n-1 = X(X+1)^E(X^2+X+1)^eP_n(X) $$
Where $E=0$ if n is even, and 1 otherwise, and, as @KierenMacMillan notice, $e= 0,1,2 $ according to $n= 0,2,1\mod 3$.
We know that $P_{2p}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
| Let's observe the relation with $\;\sin(2\pi j/3)$ and rewrite this in an elementary way :
\begin{align}
\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}&= \sum_{k=1}^{\infty}\frac 0{3k+0}+\frac 1{3k+1}-\frac 1{3k+2}\\
&= \frac 2{\sqrt{3}}\Im\left(\sum_{j=1}^\infty\frac {e^{2\pi\,i\,j/3}}{j}\right)-\frac 1{1\cdot 2}\\
&= -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Performing and simplifying equations
The expression is $$\frac {5}{x-2} - \frac 3{x+7} + 2$$
I need to simplify it too.
| $$\frac {5}{x-2} - \frac 3{x+7} + 2$$
Find the common denominator!
$$\begin{align}\frac {5}{x-2} - \frac 3{x+7} + 2 & = \dfrac{5(x+7)}{(x - 2)(x+ 7)} - \dfrac{3(x - 2)}{(x-2)(x+7)} + \frac{2(x-2)(x+7)}{(x-2)(x+7)} \\ \\& =\dfrac{5(x+7) - 3(x-2) + 2(x-2)(x+7)}{(x - 2)(x+7)} \\ \\ \end{align}$$
Now, expand the numerator,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$
Could anyone tell me how to proceed?
Thank you.
| Try to prove by induction that:
$$
A^n=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}^n = \begin{pmatrix}1&0&0\\ \left\lceil\frac{n}{2}\right\rceil&n+1\bmod{2}&n\bmod{2}\\\left\lfloor\frac{n}{2}\right\rfloor&n\bmod{2}&n+1\bmod2\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 0
} |
Why $ \lim\limits_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1} $? Can anyone explain the following equation?
$$\lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$
| $$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} = \frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}}$$
Taylor expand in the denominator in the RHS:
$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} \approx \frac{1}{n+1} \frac{1}{1-(1-(k+1)/n)}=\frac{n}{n+1} \frac{1}{k+1}$$
The result follows from taking the limit as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4... | Subtract the formula for $a_{n-1}$ from that for $a_n$
$$
a_n-a_{n-1}=\frac14-\frac12a_{n-1}
$$
Multiply by $2^n$ and bring $a_{n-1}$ from the left to the right
$$
2^na_n=2^{n-2}+2^{n-1}a_{n-1}
$$
Using the formula for the sum of a geometric series, we get
$$
2^na_n=2^{n-1}+C
$$
Plug in $n=1$ to find that $C=-\frac12$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| Elementary hint:
Multiply the function by $$1=\frac{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}$$ and use the basic well known fact that $a^3-b^3=(a-b)(a^2+ab+b^2)$ to get $1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Find maximum and minimum value of function $f(x,y,z)=y+z$ on the circle Find maximum and minimum value of function $$f(x,y,z)=y+z$$ on the circle $$x^2+y^2+z^2 = 1,3x+y=3$$
We have that $$y=3-3x$$ So we would like find minimum and maximum value of function $$g(x,z)=3-3x+z$$ on $$x^2+ (3(1-x))^2 + z^2 = 1$$ And then I w... | Another way. Given that you have $g = 3- 3x + z$ to find extrema, with the constraint
$1 = x^2 + 9(1-x)^2 + z^2 = (10x^2 -18x + 9) + z^2 = 10(x-\frac{9}{10})^2 + z^2 + \frac{9}{10}$
Thus the constraint is $100(x-\frac{9}{10})^2 + 10z^2 = 1$
Now we can let $x - \frac{9}{10} = \frac{\sin(t)}{10}$ and $z = \frac{\cos(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
N... | As $\log_aa=1$ and $\log_a(a^m)=m\log_aa=m$
taking logarithm wrt $3,$
$$(x+2)\log_32+\frac{3x}{x-1}=2$$
$$\implies(x+2)\log_32=2-\frac{3x}{x-1}=-\frac{x+2}{x-1}$$
$$\implies (x+2)\left(\log_32+\frac1{x+1}\right)=0$$
We know if $a\cdot b\cdot c\cdots=0;$ at least one of $a,b,c,\cdots$ is zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Math contest proof equation problem Could someone help me with this?
If $m$ and $n$ are positive integers, then show that $$\frac{m}{
\sqrt n}+ \frac{m}{\sqrt[4]{n}} \neq 1$$.
| $$\frac{m}{\sqrt n}+ \frac{m}{\sqrt[4]{n}} = 1 \implies \frac{m}{\sqrt n}\left(1+\sqrt[4]{n}\right) = 1 $$
Or $m(1+\sqrt[4]{n}) = \sqrt{n}$. Squaring, $m^2(1+\sqrt[4]{n})^2 = n$.
Suppose $n$ is not a perfect fourth power, so $r = \sqrt[4] n$ is irrational. Then we have $(1+r)^2 = N$, for some factor of $n$. So $r =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
| Note that $3x - 5 = 0$ when $x = \frac{5}{3}$. When $x < \frac{5}{3}$, $3x - 5 < 0$, and when $x > \frac{5}{3}$, $3x - 5 > 0$.
Also, $x - 1 = 0$ when $x = 1$. When $x < 1$, $x - 1 < 0$, and when $x > 1$, $x - 1 > 0$.
Now you have three cases:
*
*$x < 1$,
*$1 < x < \frac{5}{3}$, and
*$x > \frac{5}{3}$.
In each ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Need help calculating integral $\int\frac{dx}{(x^2+4)^2}$ I need some help calculating the integral
$\int\frac{dx}{(x^2+4)^2}$
I tried integration by parts, but I could not arrive to an answer.
| As an alternative, note that
$$\int \frac{dx}{(x^2+a^2)^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \int \frac{dx}{x^2+a^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \frac{\arctan{(x/a)}}{a}$$
Carry out the differentiation and substitute $a=2$; I get
$$\frac18 \frac{x}{x^2+4} + \frac{1}{16} \arctan{\frac{x}{2}}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a,b,c > 0$ and $a+b+c = 1$, then $(\tfrac{1}{a} −1)(\tfrac{1}{b} −1)(\tfrac{1}{c} −1) \geq 8$
If $a, b, c$ are positive real numbers and $a+b+c = 1$, prove that
$$\left(\frac{1}{a} −1\right)\left(\frac{1}{b} −1\right)\left(\frac{1}{c} −1\right) \geq 8.$$
Thank you.
| Hint: $\dfrac1a-1=\dfrac{b+c}{a}\ge 2\dfrac{\sqrt{bc}}{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Formulas for the polygonal or figurate numbers? Here is an interesting formula for the reciprocal of the heptagonal numbers. Are there any other analogous formulas for the polygonal or figurate numbers?
$$
\sum_{n=1}^\infty \frac{2}{n(5n-3)} =\\
\frac{1}{15}{\pi}{\sqrt{25-10\sqrt{5}}}+\frac{2}{3}\ln(5)+\frac{{1}+\sqrt{... | figurate numbers
$$ \sum_{n=1}^{\infty} \frac{1}{\dbinom{n+r-1}{r}} = \sum_{n=1}^{\infty} \frac{r!(n-1)!}{(n+r-1)!} $$
$$ = r\sum_{n=1}^{\infty} \frac{\Gamma(r) \Gamma(n)}{\Gamma(n +r )} = r\sum_{n=1}^{\infty} \beta(r,n) = r\sum_{n=1}^{\infty} \int_0^1 x^{n-1}(1 - x)^{r-1} \ dx = r\int_0^1 (1-x)^{r-1} \sum_{n=1}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How prove this inequality $\pi<\frac{\sin{(\pi x)}}{x(1-x)}\le 4$ let $x\in (0,1)$ show that
$$\pi<\dfrac{\sin{(\pi x)}}{x(1-x)}\le 4$$
I idea we know
$$\sin{x}<x$$
and
$$x(1-x)\le\dfrac{1}{4}$$
But not usefull for this problem
| Let
$$y=\frac{\sin{(\pi x)}}{x(1-x)}$$
then
$$\dfrac{dy}{dx}=\dfrac{{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}{x^2(x-x^2)}$$
For the maximum/minimum values we need to know x that make
$${{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}=0$$
If$\quad$ $x=\frac{1}{2}$, we can easily find $\dfrac{dy}{dx}=0$
If$\quad$ $x\not=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Proving $\gcd( m,n)$=1 If $a$ and $b$ are co prime and $n$ is a prime, show that:
$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a+b$ is a multiple of $n$
Also enlighten me why $n$ has to be prime so that $\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a$+$b$ is a multiple of $n$?
| For odd $n$, we have
$$\frac{a^n+b^n}{a+b} = a^{n-1} - a^{n-2}b +-\dotsb -ab^{n-2} + b^{n-1} = \sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k.$$
The latter sum can be written as
$$\left(\sum_{k=0}^{n-2} (-1)^k(k+1)a^{n-2-k}b^k\right)(a+b) + (-1)^{n-1}nb^{n-1}$$
(verification by distributing $(a+b)$ over the sum is elementary). T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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A necessary and sufficient on the co-efficients of a quadratic to give an integer Le $f(n) := an^2 + bn + c$ for all integers $n$, where $a$, $b$, $c$ are rational. What are the necessary and sufficient conditions on $a$, $b$, and $c$ such that $f(n)$ be an integer for all $n$?
| We are told that $an^2+bn+c$ is an integer forevery integer value of $n$. Set $n=0$. This tells us that $c$ must be an integer.
Since $c$ is an integer, $an^2+bn+c$ is an integer for all $n$ if and only if $an^2+bn$ is an integer for all $n$.
Suppose that $an^2+bn$ is an integer for all $n$. Then in particular it is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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what to do next recurrence relation when solving exponential function? find gernal solution of :$a_n = 5a_{n– 1} – 6a_{n –2} + 7^n$
Homogeneous solution:
$$a_n -5a_{n– 1} + 6a_{n –2} = 7^n$$
put $a_n=b^n$:
$$b^n -5b^{n– 1} + 6b^{n –2} =0
\\b^{n-2} (b^2-5b^{} + 6b) =0
\\b^2-5b^{} + 6b =0
\\(b-2)(b-3)=0\\
b=2,3$$
$$a^h_{... | Your computation of the general solution of the homogeneous equation $a_{n}=5a_{n-1}-6a_{n-2}$ is correct. Note that the homogeneous equation does not have $7^n$ on the right.
We want to find a particular solution of the full equation
$$a_{n}=5a_{n-1}-6a_{n-2}+7^n.\tag{1}$$
We look for a solution of the shape $a_n=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Difference of number of cycles of even and odd permutations Show that the difference of the number of cycles of even and odd permutations is $(-1)^n (n-2)!$, using a bijective mapping (combinatorial proof).
Suppose to convert a permutation from odd to even we always flip 1 and 2. Then the difference can be easily calcu... | Here is an answer using generating functions that may interest you. Recall that the sign $\sigma(\pi)$ of a permutation $\pi$ is given by
$$\sigma(\pi) = \prod_{c\in\pi} (-1)^{|c|-1}$$
where the product ranges over the cycles $c$ from the disjoint cycle composition of $\pi$.
It follows that the combinatorial species $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$ How can I evaluate
$$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$
for $a,b>0$?
My methods:
Let $a,b > 0$ and let
$$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$
Then
$$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,d... | $$\begin{align}
I
= & \int_0^{\infty} e^{-ax^2 - bx^{-2}} dx\\
\stackrel{\color{blue}{[1]}}{=} &
\left(\frac{b}{a}\right)^{1/4}\int_0^{\infty} e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\
= &
\left(\frac{b}{a}\right)^{1/4}\left[ \int_0^{1} + \int_1^{\infty} \right] e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\
\stackrel{\color{blue}{[2]}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 4
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How prove this $ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$ let $a,b,c\ge 0$, show that
$$ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$$
my idea use the SOS methods, But I don't work at last. Thank you
| Since this expression is symmetric in $b$ and $c$, we can make the substitution $x=b+c$, $y=bc$ to obtain the equivalent inequality
$$
y^2+axy+a(x^3-3xy)+a^3x \geq 2a^2(x^2-y) \, ;
$$
the condition $b,c \geq 0$ is equivalent to requiring $x,y \geq 0$.
This inequality is quadratic in $y$. By collecting terms, we see tha... | {
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"url": "https://math.stackexchange.com/questions/498804",
"timestamp": "2023-03-29T00:00:00",
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found t... | $$
\left(%
\begin{array}{cc}
1 & 1
\\
0 & 1
\end{array}\right)
=
1 + A
\quad\mbox{where}\quad
A
=
\left(%
\begin{array}{cc}
0 & 1
\\
0 & 0
\end{array}\right)
$$
Notice that $A^{2} = 0$. It means that a function ${\rm f}\left(A\right)$ is
${\it linear}$ in $A$. For instance,
$\left(1 + \mu A\right)^{n} = \alpha + \beta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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} |
Integrals of $\sqrt{x+\sqrt{\phantom|\dots+\sqrt{x+1}}}$ in elementary functions Let $f_n(x)$ be recursively defined as
$$f_0(x)=1,\ \ \ f_{n+1}(x)=\sqrt{x+f_n(x)},\tag1$$
i.e. $f_n(x)$ contains $n$ radicals and $n$ occurences of $x$:
$$f_1(x)=\sqrt{x+1},\ \ \ f_2(x)=\sqrt{x+\sqrt{x+1}},\ \ \ f_3(x)=\sqrt{x+\sqrt{x+\sq... | I think there is highly have no chance to have an integer $n>2$ such that $f_n(x)$ is integrable in elementary functions.
The reasons are mainly because of the processes of eliminating the radicals.
For $f_0(x)$ and $f_1(x)$ , they are trivially integrable in elementary functions.
For $\int f_2(x)~dx$ ,
Let $u=\sqrt{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "80",
"answer_count": 3,
"answer_id": 2
} |
How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.
| As you stated, we must have $x^2 + 3y^2 = 7 $ and $3 = xy + y^2$.
This is an ellipse intersecting a hyperbola, so there are at most 4 points of intersection.
Substtuting $ x = \frac{3-y^2}{y}$ into the first equation, we get a quartic equation
$$ (3-y^2) + 3y^4 = 7y^2. $$
This has solutions $y = -\frac{3}{2}, -1, 1, \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| Reorder $x+y = 6$ to
$y=6-x$
Substituting $y$ in $(x^2+y^2)$ yields
$x^2+(6-x)^2 = 2x^2-12x+36$
The minimum occurs where the derivative equals $0$
$4x -12 = 0$
Therefore at the minimum,
$x=3$
Hence the minimum is $2\cdot 3^2-12\cdot 3+36 = 18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 2
} |
Polynomial long division: different answers when reordering terms When I use polynomial long division to divide $\frac{1}{1-x}$, I get $\;1 + x + x^2 +x^3 + x^4 + \cdots$
But when I just change the order of terms in the divisor: $\frac{1}{-x+1}$, the long division algorithm gives me a very different answer: $-\... | Seeing as you tagged this sequences and series, I presume you know that you have to be careful when talking about adding up an infinite sequence of terms.
There is an infinite series hidden in your work when you write
$$
\frac{1}{1-x}=1+x + x^2 +x^3 + \cdots,
$$
but this series only converges when $|x|<1$. This sum is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
| You don't need any integrations by parts for this. As others have suggested, if you substitute $u^2 = 1 -x^2$ and $2u\,du = -2x\,dx$ the integral becomes
$$-\int u\sqrt{2 - u}\,du$$
Now do another substitution $v = 2 - u$ and the integral becomes
$$\int (2-v)\sqrt{v}\,dv$$
$$= \int 2v^{1 \over 2} - v^{3 \over 2}\,dv$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Prove the Trigonometric Identity: $\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$ I'm doing some math exercises but I got stuck on this problem.
In the book of Bogoslavov Vene it says to prove that:
$$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$
It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\co... | Note that we must have that $\cos x \neq 1$, so x cannot be any even multiple of $\pi$. Similarly, $\sin x \neq 0$, so x cannot be any multiple of $\pi$.
Why do I note the above?
Because we have an identity, $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$$ which is senseless if $\cos x = 1,$ and/or $\sin x = 0$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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