Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which ar... | Let $s=\sin 20° \quad → \sin (3×20°) = \frac{\sqrt{3}}{2} = 3s-4s^3$
Let $c=\cos 20° \quad → \cos (3×20°) = \frac{1}{2} = -3c+4c^3$
$LHS = 4s^2 - 2\sqrt{3}\,s + 1 = 4s^2 - 4s\,(3s-4s^3) + 1 = (1-4s^2)^2$
$\displaystyle RHS = \left(\frac{1}{2c}\right)^2 = \left(-3+4c^2\right)^2 = (1-4s^2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that the rational function $f(x)/g(x)$ has a partial fraction decomposition in the case when $g(x)$ factors into distinct linear factors. To be honest i don't even know where to start. I thought about using diagonalizable matrices and characteristic polynomials, but the class hasn't gotten there yet so there shou... | We proceed by strong induction. Given that $n = 1,$ we have that $f(x) = a$ for some real number $a$ so that $$\frac{f(x)}{g(x)} = \frac a {x - c_1},$$ as desired. We will assume inductively that for any polynomial $f(x)$ of degree $\leq n - 1,$ we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 ... | Note that $y=bx-b^2$ is a solution, where
$$
b=\frac{-x_0\pm\sqrt{x_0^2-4y_0}}{2}
$$
This gives us two distinct solutions whenever $x_0^2\neq 4y_0$. If instead $x_0^2=4y_0$, we can still find a second solution $y=x^2/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
how should Cayley's resolvent be used in order to see if a given quintic equation is solvable or not?
Hi. I chose a simple equation to work with: $x^5+x^4+x^3+x^2+x+1=0$ (which is solvable and $x=-1$).
the amounts of $p$, $q$, $r$ and $s$ where $\frac{3}{5}$, $\frac{14}{25}$, $\frac{87}{125}$ and $\frac{2604}{3125}$ r... | Hint:
We have the following factorization:
$$
z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=
(z^2 + 22z + 169)(z^2 - 26z + 361)(z - 1)(z - 25).
$$
Of course, the splitting field of
$$
x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}
$$
is $\Bbb Q(\zeta_6)$, so that the Galois group is solvable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n... | Let $z=e^{ \frac{i\pi}{n}},$ for example. If we can show it for this point, it is true in general since this is and its conjugate are the closest to the point $z=1$.
$$|z-1|^2 = (\cos \frac{\pi}{n} -1)^2 + \sin^2 \frac{\pi}{n} = 2 \left( 1-\cos \frac{\pi}{n} \right).$$
$$|z-1|^2=2\left[ 1-\left(1-\frac{1}{2!}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
... | Perhaps it's easier if we use the sum notation. Notice that your first equation can be used to show
$$(n+1)^3 - 1 = \sum_{k=1}^n(k+1)^3 - k^3 = \sum_{k=1}^n (3k^2 + 3k + 1) = 3\sum_{k=1}^n k^2 + \frac 32n(n+1) + n,$$
where the first equality comes from the telescopic property.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| x>0;
$f(x):=\sqrt{x^2+6}= x\sqrt{1+6/x^2}=$
$x(1+3/x^2+$
$(1/2)(-1/2)(1/2!)(6/x^2)^2+ O(1/x^6))=$
$x(1+3/x^2-(6^2/8)/x^4+O(1/x^6));$
$x^2(x^2-xf(x)+3)=$
$x^2(- 3+3+(9/2)/x^2+O(1/x^4))$;
Take the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ ... | Using first approximation for $\sin x\approx x$ for $x$ near $0$, the limit can be rewritten without change of variables as $$\begin{aligned} &\lim_{x\to \infty}x^3\left(\frac{1}{x+2}-\frac{2}{x+1}+\frac{1}{x}\right)\\ = &\lim_{x\to \infty}x^3\left[\left(\frac{1}{x}-\frac{1}{x+1}\right)-\left(\frac{1}{x+1}-\frac{1}{x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove that $\int \sec x \, dx = \ln \left\vert\frac{1+\sin x}{1-\sin x}\right\vert^\frac{1}{2}+C = \ln \vert \sec x+\tan x\vert + C$ When our teachers give the rule
$$\int \sec x \, dx = \ln \vert \sec x+\tan x\vert + C$$
We ignore the other solution when we say
$$\sec x = \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} =... | You can also show they are the same using trigonometry:
$$\begin{aligned}\ln|\sec x+\tan x|
&=\ln\left\lvert\frac{1+\sin x}{\cos x}\right\rvert\\
&=\ln\left\lvert\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right\rvert\\
&=\ln\left\lvert\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right\rvert\\
&=\ln\left\lvert\frac{1+\sin x}{1-\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Big Oh and Little oh notation I have to prove or disprove the following:
$$2n^3 + 5n^2 -3 = o(n^3)$$
$$\frac{n+2}{\sqrt{n^2+4}} = O(1)$$
My attempt at the first question
$$2n^3 + 5n^2 - 3 \leq c |g(x)|$$
$$2n^3+5n^2-3 \leq c |2n^3+5n^3-3n^3|$$
$$2n^3+5n^2-3 \leq c |4n^3|$$
$$2n^3+5n^2-3 \leq 4 |n^3|$$
evaluating when n... | *
*$2n^3 + 5n^2 -3 = o(n^3)$ means $ \frac{2n^3+5n^2-3}{n^3} \to 0$ as $n \to \infty.$
But this is fals, since $ \frac{2n^3+5n^2-3}{n^3} \to 2$ as $n \to \infty.$
*$\frac{n+2}{\sqrt{n^2+4}} = O(1)$ means that the sequence $(\frac{n+2}{\sqrt{n^2+4}})$ is bounded.
This is true, since $0 \le \frac{n+2}{\sqrt{n^2+4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it.
A hint or a detailed answer would be ... | Step 1: Using extended Euclidean algorithm, find $g(x)$ and $h(x)$ such that $$g(x)(x^2+2x-4)+h(x)(x^2+x+1)=1$$.
\begin{align}
x^2+2x-4=x^2+x+1+(x-5)\implies& \begin{cases} q = 1\\
r=x-5\\
s=1\\
t=-1\end{cases}\\
x^2+x+1=(x+6)(x-5)+31\implies &\begin{cases} q = x+6\\
r=31\\
s=-(x+6)\\
t=1-(x+6)(-1)=x+7\end{cases}\\
x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to factor $x^6-4x^4+2x^3+1$ by hand? I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which ot... | Here's a possible way to do it:
$x^6-4x^4+2x^3+1 = (x^6+2x^3+1)-4x^4 = (x^3+1)^2 - 4x^4$
$(x^3+1)^2-4x^4 = [x^3+1-2x^2][x^3+1+2x^2]$
$x^6-4x^4+2x^3+1= [(x^3-x^2)+(1-x^2)][x^3+2x^2+1]$
Then, we have:
$x^6-4x^4+2x^3+1 =[x^3+2x^2+1][x^2(x-1)+(1-x)(1+x)]$
$x^6-4x^4+2x^3+1 = (x-1)(x^2-x-1)[x^3+2x^2+1]$
So that gives you a d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Maximum of $\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$ If $a>b>c>0$ such that $7a+8b=15c+24\sqrt[3]{abc}$, find the maximum value of
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$$
I tried to use $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3$ so that
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=(\frac{a}{c}+\frac{b}{a}+\frac... | Notice that:
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b} = \frac{(a-b)(b-c)(a-c)}{abc}$$
Now, let $x=a-b > 0$ and $y=b-c>0$. Then $a-c=x+y$ and from the condition
$$7a+8b=15c+24\sqrt[3]{abc}\Leftrightarrow \sqrt[3]{abc}=\frac{7x+15y}{24}$$
Therefore, using AM-GM:
$$
\begin{aligned}
\frac{(a-b)(b-c)(a-c)}{abc} &= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}$ for $a+b+c=1$. Source: RMO 2019, question 3
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}.$$
I tried using Holder's ... | Hint: Use AM-GM to prove the following ineq: for all $a,b,c>0$,
$$\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq \frac{3a+b+c}{25abc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi... | This is equivalent to the equation
of the sum of cosines of the tree angles of triangle,
\begin{align}
\cos \alpha+\cos\beta+\cos\gamma
&=\frac{3}{2}
,
\end{align}
it is well-known that this sum can be expressed
in terms of $r$ and $R$,
the radii or inscribed and circumscribed circle, respectively as
\begin{align} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Any shortcuts for integrating $\frac{x^6}{(x-2)^2(1-x)^5}$ by partial fractions? Is there a faster way to get the partial fraction decomposition of this $\frac{x^6}{(x-2)^2(1-x)^5}$?
$\frac{x^6}{(x-2)^2(1-x)^5} = \frac{A_1}{x-2} + \frac{A_2}{(x-2)^2} + \frac{B_1}{1-x} + \frac{B_2}{(1-x)^2} + \frac{B_3}{(1-x)^3} + \frac... | set $$x=1-\frac{1}{u+1}$$ to get
$$\int{\frac{{{x}^{6}}}{{{(x-2)}^{2}}{{(1-x)}^{5}}}dx}=\int{\frac{{{u}^{6}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du}$$
long division for the integrand gives:
$${{u}^{3}}-5{{u}^{2}}+17u-49+\frac{129{{u}^{2}}+324u+196}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$. Question: Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$.
My approach: After trying some examples I can conjecture that, the minimum value of $x^2+y^2$ i... | Why not to write
$$y=k-x \implies x^2+y^2=2x^2-2kx+k^2=2 \left(x-\frac{k}{2}\right)^2+\frac{k^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\lim\limits_{n\to\infty}\frac{1}{\sqrt{4n^2-1^2}}+\dots+\frac{1}{\sqrt{4n^2-n^2}}$. I want to compute $$\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4n^2-k^2}}.$$
I found it on this question and the exercise appears to be from previous years of a Latvian competition.
I tried writing $\frac1{\sqrt{4n^2-k^2... | You can rewrite it to become $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{n}{\sqrt{4n^2-k^2}} = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{1}{\sqrt{4-\left(\frac{k}{n}\right)^2}}$$
This is the Riemann sum for $\frac{1}{\sqrt{4-x^2}}$ from $0$ to $1$. Therefore, the limit of the sum is $$\int_0^1 \frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Showing moderate decrease property in the real line for $f(z)=\frac{a}{a^2+z^2}$. Let $f(z) = \frac{a}{a^2 + z^2}$ for $a>0$. Then $f$ is holomorphic in the horizontal strip
$|\Im(z)| < a$. I would like to show that in if $|\Im (z)| < a/2$, we have some constant $A>0$ such that
$$|f(x+iy)| \le \frac{A}{1+x^2}$$ for al... | Note that the follow must hold,
$$\frac{A}{1+x^2}- \frac{a}{\frac{3a^2}{4} + x^2} \ge 0$$
for all $x$, which is equivalent to
$$(A-a)x^2+ a(\frac34 aA-1) \ge 0$$
Thus,
$$A = \max( a, \frac4{3a})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\frac{1}{\frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)}(5x^2)$ to $\frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)x^2$ I was following the steps from the book Ordinary Differential Equations (Lesson 25B page 274) to find a particular solution of
$4y'' - 3y' + 9y = 5x^2, \quad (4D^2 -3D +9)y = 5x^2$... | Use $$\dfrac 1 {1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+\dots$$
$$4y'' - 3y' + 9y = 5x^2, \quad (4D^2 -3D +9)y = 5x^2\implies y_p=\dfrac 5 {4D^2 -3D +9} (x^2).$$
Hence
$$\begin{align}
y_p &= \frac{5}{{9}\bigr(1 - \frac{D}{3} + \frac{4D^2}{9} \bigr)}(x^2) =\frac 5 9 \frac{1}{\bigr(1 -( \frac{D}{3} - \frac{4D^2}{9}) \bigr)}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$ Let be $a,b,c,x,y,z>0$ such that $ax\ge \sqrt{(b^2+c^2)(y^2+z^2)}$. Prove that
$$(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$$
I tried to expand
$$a^2(y^2+z^2)+x^2(b^2+c^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
Here my idea was to use the condition after the means inequality:
$$a^2(y^2... | Let $\frac{b}{a}=p$, $\frac{c}{a}=q$, $\frac{y}{x}=u$ and $\frac{z}{x}=v$.
Thus, the condition it's $$(p^2+q^2)(u^2+v^2)\leq1$$ and we need to prove that:
$$(u-p)^2+(v-q)^2\geq(pv-qu)^2,$$ which is true by C-S twice:
$$(u-p)^2+(v-q)^2=\sqrt{\left((u-p)^2+(v-q)^2\right)^2}\geq$$
$$\geq\sqrt{\left((v-q)^2+(p-u)^2\right)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^6-6x^4+12x^2-11$ is irreducible over $\mathbb{Q}$ Extracted from Pinter's Abstract Algebra, Chapter 27, Exercise B1:
Let $p(x) = x^6-6x^4+12x^2-11$, which we can transform into a polynomial in $\Bbb{Z}_3[x]$:
\begin{align*}
x^6+1
\end{align*}
Since none of the three elements $0,1,2$ in $\Bbb{Z}_3$ is a ro... | Let $p(x) = x^6 - 6x^4 + 12x^2 - 11$
Substitute $x^2 = y$
$h(y) = y^3 - 6y^2 + 12y - 11$
Letting $g(y) = h(y+2)$
$$g(y) = (y+2)^3 -6(y+2)^2 + 12(y+2) - 11$$
$$g(y) = y^3 - 3$$
$g(y)$ is obviously irreducible, thus so is $h(y)$ and $p(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Second-order difference equation solution Say we have a second-order difference equation:
$$x_n = x_{n-1} + x_{n-2} $$
Many of the notes that I have found online regarding how to solve this type of equation will have a step such as "guess" $x_n=Ar^n$.
What is the intuition behind this procedure? How would one know to "... | We can transform the recurrence relation into a linear algebraic problem.
Let $\vec{a}_n=
\begin{pmatrix}
x_n\\
x_{n+1}\\
\end{pmatrix}$,
$\vec{a}_0=
\begin{pmatrix}
0\\
1
\end{pmatrix}$.
Then $\vec{a}_{n+1}=
\begin{pmatrix}
x_{n+1}\\
x_{n}+x_{n+1}
\end{pmatrix}=A\vec{a}_n
$
where $A
=\begin{pmatrix}
0 & 1\\
1 & 1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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please solve quadratic simultaneous equations. $x,y,z$ are real variables. $l,m,n$ are real positive constants.
Solve for real $x,y,z$:
$$\begin{align}
x^2+xy+y^2 &= l\\
y^2+yz+z^2 &= m\\
z^2+zx+x^2 &= n
\end{align}
$$
This problem has a relevance in Electrical Engineering.
| $$x^2+xy+y^2=l~~(1),~ y^2+yz+z^2=m ~~~(2),~ z^2+zx+x^2=n~~~(3)$$
Subtract (1) from (2) and (1) from (3) to get
$$(x-y)(x+y+z)=m-l~~~(4),~ (x-z)(x+y+z)=n-l~~~(5)$$
Let $(x+y+z)=w$, then (4) and (5) give
$$(x-y)=\frac{m-l}{w}~~~(6), ~(x-z)=\frac{n-l}{w}~~~(7)$$
From these two we get $$[3x-(x+y+z)]w=m+n-2l \implies x=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$ How to prove that
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$
$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$
where $\text{Li}_r$ is the polylogarithm function a... | I proved here
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$
Divide both sides by $1+a$ the integrate from $a=0$ to $a=1$ we get
$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$ show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$
Here is the algebric proof:
We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\... | You want to prove that for all $\varepsilon > 0$ there is $N > 0$ such that
$\sqrt{n+1}-\sqrt{n}<\varepsilon$ when $n>N$. Your algebraic manipulation shows that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
so you would like to solve the inequality $\frac{1}{\sqrt{n+1}+\sqrt{n}} < \varepsilon$.
To find $N$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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What are the number of ordered m-tuples of integers,such that sum of square of elements is a given integer? Given a non-zero integer $n$,the problem is to find a m-tuple of integers,$(x_1,x_2,x_3,...,x_m)$such that the following equation is satisfied---$$\sum_{i=1}^mx_i^2=n$$
I have no idea how to approach the problem,... | Given $$\sum_{i=1}^m x_i^2=n$$
We can use Euclid's formula for generating Pythagorean triples
$$A=x^2-y^2\qquad B=2xy\qquad C=x^2+y^2$$
We can find a case for $m=2$ easily if we solve the $C$-function for $(y)$ and test a defined range of $m$-values to see which yield integers. For example
$$C=x^2+y^2\implies y=\sqrt{C... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Easy way to find the partial fraction I always have trouble trying to find the partial fraction, especially for complicated ones.
For example, this is what I will do to find the partial fraction of $\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$
*
*$\displaystyle \frac{A}{x}+\frac{B}{2x+3}+\frac{C}{(2x+3)^2}+\f... | $A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $
With $ x = 0, 27 A = 27, A = 1 $.
This gives us $Bx(2x+3)^2+Cx(2x+3)+Dx = -x^2 - 12x $.
Dividing by $x$, this gives us $B(2x+3)^2+C(2x+3)+D = -x - 12 $.
With $x = - \frac{3}{2}$, $D = - 10 \frac{1}{2}$.
This gives us $ B(2x+3)^2 + C (2x+3) = -x - \frac{3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factorize this determinant?
The question is to factorize $$\det\begin{pmatrix}(x^2+1)^2 & (xy+1)^2 & (xz+1)^2 \\ (xy+1)^2 & (y^2+1)^2 & (yz+1)^2 \\ (xz+1)^2 & (yz+1)^2 & (z^2+1)^2 \end{pmatrix}.$$
I have a hint which is considering the factorization of $\det\begin{pmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c... | By calculation (I got this slightly mixed up in my comment),
$$\begin{pmatrix} 1 & 2x & x^2 \\ 1 & 2y & y^2 \\ 1 & 2z & z^2 \end{pmatrix} \begin{pmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{pmatrix}^\top = \begin{pmatrix} 1 + 2x^2 + x^4 & 1 + 2xy + (xy)^2 & 1 + 2xz + (xz)^2 \\ 1 + 2xy + (xy)^2 & 1 + 2y^2 + y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$
Question: Given that $X$ and $Y$ are two random variables satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$
Calculate
$$\mathbb{E}(X-Y \mid 2X+Y).$$
I tried to use the fact t... | We use two property:
First: $E(2X+Y|2X+Y)=2X+Y$
Second: $(X-dY,2X+Y)$ is bi-variate normal(for $d\neq - \frac{1}{2}$), if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent(by set $\rho=0$ in bivarite distribution of joint $(X-dY,2X+Y)$
Correlations_and_independence). so $E(X-dY|2X+Y)=E(X-dY)=0$.
$$E(2X+Y|2X+Y)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Evaluating $\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln(1+x)-x^2}$ using L'Hôpital Considering this limit, assigned to my high school students,
$$\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln \left(1+x\right)-x^2}=\left(\frac00\right)=\lim_{x\to 0}\frac{\frac{d}{dx}\left(x\sin \left(x\right)-2+2\cos \left(x\right)\right)... | The derivative of the denominator is
$$
\ln(1+x)+\frac{x}{1+x}-2x=\frac{(1+x)\ln(1+x)-x-2x^2}{1+x}
$$
Then after the first step you find
$$
\frac{x\cos x-\sin x}{(1+x)\ln(1+x)-x-2x^2}(x+1)
$$
The factor $x+1$ can be disregarded, because its limit is $1$.
At the next application, the numerator will become
$$
\cos x-x\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Given a positive integer $k$, describe all positive integers $n$ such that $\langle n\rangle=k$.
Question: For any positive integer $n$, let $\langle n \rangle$ denote the integer nearest to $\sqrt{n}$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n\rangle=k$.
(b) Show that $... | The condition states that $ k- \frac{1}{2} \leq \sqrt{n} \leq k + \frac{1}{2}$
Squaring, we get
$ k^2 -k + \frac{1}{4} \leq n \leq k^2 + k + \frac{1}{4}$
Now, use the fact that $n $ is integer to conclude
$ k^2 - k + 1 \leq n \leq k^2 +k$
Your part b) is correct.
A slightly nicer solution to b) is to look at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)... | $x^{40}(1+x^5+x^7)^{20}=x^{40}(1+x^5(1+x^2))^{20}=$
$x^{40}(1+20x^5(1+x^2)+$
$(20)(19)/2![(x^5)(1+x^2)]^2+$
$(20)(19)(18)/3![x^5(1+x^2)]^3+..).$
Need the coefficient of $x^{17}$ in the above expansion.
Which term is relevant?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx$ $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx$$
which is W3, Jozsef Wildt International Mathematical Competition 2019.
| A glance at the following equality
$$\frac{\cos x+1}{x\sin x+1}\mathrm{d}x=\frac{\mathrm{d}(x+\sin x)}{x\sin x+1}$$
inspires us to calculate that
$$\frac{\mathrm{d}}{\mathrm{d}x}\frac{x+\sin x}{x\sin x+1}=\frac{\cos x(\cos x+1-x^2)}{\left(x\sin x+1\right)^2}.$$
Let $y:=\frac{x+\sin x}{x\sin x+1}$, and compared with the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function woul... | We'll prove that $\max(x^2+y^2)=4.$
Indeed, let $xy\geq0$.
Thus, by C-S $$2=x^4+y^4+3xy\geq x^4+y^4=\frac{(1+1)(x^4+y^4)}{2}\geq\frac{(x^2+y^2)^2}{2},$$
which gives $$x^2+y^2\leq2<4.$$
Now, let $xy\leq0.$
Thus, by C-S again and AM-GM we obtain:
$$2=x^4+y^4+3xy\geq\frac{(x^2+y^2)^2}{2}-\frac{3(x^2+y^2)}{2},$$
which give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluating Series Integral How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$
I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.
| \begin{align*}
J&=\int_0^\infty \frac{\ln^2 x}{1+y^2}\,dx\\
A&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{\ln^2(x)+\ln^2(y)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{2\ln^2(x)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=2J\Big[\arctan x\Big]_0^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$ Find the limit:
$$\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$$
I applied the identity:
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$
by multiplying the numerator and denominator by the complementary part.
$$\lim\limits_... | With binomial series, it's $$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac6{n^2}+\dfrac1{n^6}\right)^{1/3}-n^2\right]$$
$$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac13\dfrac6{n^2}+O\left(\dfrac1{n^4}\right)\right) -n^2\right]$$
$$=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the values of $x$ in terms of $a$ in $x^2+\frac{(ax)^2} {(x+a)^2} =3a^2$ The question is as follows.
Find the values of $x$ in terms of $a$ in
$x^2+\dfrac{(ax)^2}{(x+a)^2} =3a^2 $
My solution:
Multiply both sides by $(x+a)^2$ and expand. On rearranging we get
$x^4+2ax^3-a^2x^2-6a^3x-3a^4=0$
Now dividing by $a^4$... | It's $$x^2+\frac{a^2x^2}{(x+a)^2}-\frac{2ax^2}{a+x}+\frac{2ax^2}{a+x}=3a^2$$ or
$$\left(x-\frac{ax}{a+x}\right)^2+\frac{2ax^2}{a+x}=3a^2$$ or
$$\left(\frac{x^2}{a+x}\right)^2+\frac{2ax^2}{a+x}-3a^2=0$$ or
$$\left(\frac{x^2}{a+x}\right)^2+\frac{2ax^2}{a+x}+a^2-a^2-3a^2=0$$
or
$$\left(\frac{x^2}{a+x}+a\right)^2-(2a)^2=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes:
If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?
So what I tried was I know that since th... | Another way.
Let $y=mx+n$ be an equation of the tangent.
Thus, $$mx-y+n=0,$$ which since equations of our circles they are
$$(x+1)^2+(y-2)^2=4$$ and $$(x-1)^2+y^2=\frac{8}{5},$$ we obtain that the point $(-1,2)$ is placed on the line $y=mx+n$
and the distance from $(1,0)$ to the line is equal to $\sqrt{\frac{8}{5}}.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find without L'Hospital's rule: $\lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}}$ $A = \sqrt{17 - 2x^{2}} \sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} \\
\Rightarrow \lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}} \\
= \lim\l... | Yes, that works.
You can also use the fact that, if $f(x)=\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}$, then $f(2)=9$, $f'(2)=0$, and $f''(2)=-\frac{22}3$. So, near $0$,$$f(x)=9-\frac{11}3(x-2)^2+O\bigl((x-2)^3\bigr)$$ and therefore,\begin{align}\lim_{x\to2}\frac{\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}}{(x-2)^2}&=\lim_{x\to2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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approaching $\int \sqrt{x^5+2}\; dx$ I recently came across this integral $\int\sqrt{x^5+2}\; dx$. From Wolframalpha i can see that it has a closed form. how does one get to that closed form? what techniques should i approach?
| As said in comments, using Taylor or the binomial theorem, we have
$$\sqrt{x^5+2}=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{
2^n\, n!} x^{5 n}$$
$$\int \sqrt{x^5+2}\,dx=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{
2^n\,(5n+1)\, n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Given $\cos C(\sin A +\sin B)=\sin C \cos(A-B)$, prove that triangle $ABC$ is equilateral.
Given $\cos C(\sin A +\sin B)=\sin C\cos(A-B)$, prove that triangle $ABC$ is equilateral.
My attempt:
By applying some trigonometric identities, I have:
$$\cos C(\sin A +\sin B)=\sin C \cos(A-B)\\
\iff \cos C \cdot2\sin \fr... | I got the same result!
Now take $A=30^{\circ}$, for example, and we see that our triangle is not equilateral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Estimate expected payoff of rolling a dice, with choice of rolling up to $50$ times. This is an extended question of the classical rolling dice and give face value question.
You roll a dice, and you'll be paid by face value. If you're not satisfied, you can roll again. You are allowed $k$ rolls.
In the old question... | Let $E_k$ be the expected payoff, if you're allowed to roll $k$ times, with the rules as you've described them. We can compute $E_k$ recursively.
With just $1$ roll, you must take what you get, since there are no more rolls. The expected value is therefore $$E_1 = \frac{1+2+3+4+5+6}{6} = 3.5$$
With $2$ rolls, if your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Calculate $\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$ I want to calculate $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$$
where $|z|=2$ is run counterclockwise.
What I did: $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}=\int_{|z|=2}\frac{1}{(z+1)(z^2+1)} $$
Using Cauchy's integral formula
$$f(-1)=\frac{1}{2\pi i}\int_{|z|=2} \frac{f(z)}{z+1}dz$$
... | The poles $-1,\,\pm i$ all have modulus $1<2$, so we get contributions from all of them. The poles are also all first-order. The $z=-1$ residue is $\frac{1}{(-1)^2+1}=\frac12$. The $z=\pm i$ residue is $\frac{1}{(1\pm i)(\pm 2i)}=\frac{-1\mp i}{4}$. So the integral is $2\pi i(\frac12+\frac{-1}{2})=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let the reduced row echelon form of A = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$. Determine A if ... Let the reduced row echelon form of $A$ be $R$ = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$.
Determine A if the first, third, and si... | Let $C_{i}$ be the $i$th column. By observing $R$ we notice that $C_{4} = 3C_{3} + 4C_{1}$ and $C_{5} = -C_{6} + 2C_{3} + 5C_{1}$.
The idea is to find a linear combination of the other columns which each
have only one non-zero entry ,to then express the values in the column that we're evaluating. You did right for c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
S... | The squared distance from $(2,1)$ to an arbitrary point $(x,y)=(x,4x/3+20/3)$ of the straight line is
$$(x-2)^2+\left(\frac43x+\frac{20}{3}-1\right)^2=\frac{25}{9}x^2+\frac{100}{9}x+\frac{35}{9}.$$
As the vertex of that parabola occurs at $x=-2$, the square of the minimal distance is $25$, hence the distance is $5$. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Find x that satisfies the equality (matrix determinant): This is the exercise.
Find the value of $x\in\mathbb{R}$ that satisfies
$$
\begin{vmatrix}
x & -1\\
3 & 1-x
\end{vmatrix} =
\begin{vmatrix}
1 & 0 & -3 \\
2 & x & -6 \\
1 & 3 & x-5
\end{vmatrix}
$$
This is what I've done.
\begin{align*}
\det (A) &= \de... | There's nothing wrong – just that $x=\frac{3\pm\sqrt{33}}4$, so simplifying the determinants after substituting $x$ in might take some work. The determinants on both sides should evaluate to $\frac{9\mp\sqrt{33}}8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Checking the argument for calculation of the limit $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
In computing this limit, i have used following steps:
$\lim_{x\to \frac{\pi}{6}} \f... | We can not replace $\cos\dfrac\pi6$ with $\cos x$ just because $x\to\dfrac\pi6$
Better use Prosthaphaeresis Formula $$\sin x-\sin\dfrac\pi6=2\sin\dfrac{x-\dfrac\pi6}2\cos\dfrac{x+\dfrac\pi6}2$$
Then apply $\lim_{h\to0}\dfrac{\sin h}h=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding a polynomial whose roots are connected to the roots of a different polynomial Suppose we have a polynomial function $$f(x) =x^5-4x^4+3x^3-2x^2+5x+1$$ Function $f$ will have 5 roots which can be denoted by $a, b, c, d, e$. I was interested in trying to find a degree 10 polynomial whose roots are given by $abc, a... | $f(x) = x^5-4x^4+3x^3-2x^2+5x+1$
$f$ has $5$ roots donated by $a$, $b$, $c$, $d$ and $e$
The elementary symmetric functions of the roots are
$a+b+c+d+e = 4$
$de+ce+be+ae+cd+bd+ad+bc+ac+ab = 3$
$cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$bcde+acde+abde+abce+abcd = 5$
$abcde = -1$
Let $z = abc$, Computing the elementa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
the range of the vector dot product $\overrightarrow{P M} \cdot \overrightarrow{P N}$
$ABCDEF$ is a equilateral regular hexagon with $AB=2 \sqrt2$. Let $P$ be an arbitrarily point of the hexagon and $MN$ be a movable chord of the circumcircle of the hexagon with $MN=4$. What would the range of the vector dot pro... | Continue with
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=
f(x, t)=x^2+4x\sin t+4\sqrt6\cos t+6,$$
Note that $f_x’=f_t’=0$ yields $(x,t)=(0,0),(0,\pi),(0,2\pi)$, which leads to extreme values $6\pm 4\sqrt6$.
Moreover, check the boundary values at $x=\pm\sqrt2$,
$$f(\pm\sqrt2,t)= 2\pm4\sqrt2 \sin t+4\sqrt6 \cos t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix, and Differential Equation Solution Verification I given the following system:\begin{equation}\mathbf{X'}=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\mathbf{X}\end{equation}
Every variable in the system is a matrix. I am then given that $\mathbf{X}$ is a column matrix.\begin{align}\frac{d}{dt}\begin{pmatri... | Your solution looks correct to me.
$$\begin{align}\frac{d}{dt}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}&=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}\end{align}$$
You can also write it this way:
$$\begin{align}\begin{pmatrix}-1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{... | Starting from J.C.Santos result
$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$
I'd multiply num and den by $\left(2x+\sqrt{x^2+3}\right)$ to get
$$\lim_{x\to1}\frac{\left(4x^2-x^2-3\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$
that is
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Doubt in solution of APMO 1998 Inequality problem Question -
Let $a, b, c$ be positive real numbers. Prove that
$$
\begin{array}{c}
\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt[3]{x y z}} \\
(\text { APMO } 1998)
\end{array}
$$
My doubt -
in pham kim hung s... | $$\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)
=2+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).$$
And each term in parentheses satisfies the inequality you have proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Real solutions for $\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|$ Do there exist real numbers $a,b,c,d$ such that
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|?$$
| Consider the points $M=(a,b)$ and $N=(c,d)$. By the triangle inequality
$$OM-ON\leq MN \leq MP+NP $$
which gives us
$$\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\leq MN\leq |a-c|+|b-d| $$
Since $$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<\sqrt{a^2+b^2}-\sqrt{c^2+d^2} $$ we have the reverse inequality:
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<|a-c|+|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x... | Consider a linear function $ f(x)=ax+b$
$$ f(x^2+x) = ax^2+ax+b$$
$$ f(x^2-3x+2)= ax^2-3ax +2a+b$$
$$ f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$
$$a=3, b=-4$$
$$ f(x) = 3x-4$$
$$f(2016)=6044$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
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Show that $\frac{1}{x+2}$ is continuous for $(-2,0]$. I must use an $ε-δ$ proof to prove the function $\frac{1}{x+2}$ is continuous for $(-2,0]$.
I set $x \in (-2,0]$, and $y\in (-2,0]$ s.t $|x-y|<δ$.
I am having a hard time determining what precisely I should set $y$ to, as
$|\frac{1}{x+2}-\frac{1}{y+2}|=|\frac{(y+2)... | Hint
For $\vert y-x \vert \le \frac{x +2}{2}$ you have
$$-\frac{x +2}{2} \le y-x \le \frac{x +2}{2}, \text{ hence } 2 + y \ge \frac{x}{2} + 1 = \frac{x+2}{2}$$
and therefore
$$0 \le \frac{1}{2+y} \le \frac{2}{2+x}$$
Finally
$$\left\vert\frac{1}{x+2}-\frac{1}{y+2}\right\vert=\left\vert\frac{(y+2)-(x+2)}{(x+2)(y+2)}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing values of x fall into a range or domain, when y is also part of the equation I am told to consider the relation:
$$\frac{x^2}{\:20}+\frac{y^2}{5} = 1$$
(i) Show that the values of x for which the relationship is defined are given by:$-\sqrt{20}<=\:x\:<=\:\sqrt{20}$
(ii) Similarly, find the values of y for whic... | Your equation is
$$\frac{x^2}{20} + \frac{y^2}{5} = 1 \implies \frac{x^2}{20} = 1 - \frac{y^2}{5} \tag{1}\label{eq1A}$$
You have $\frac{y^2}{5} \ge 0 \implies -\frac{y^2}{5} \le 0 \implies 1 - \frac{y^2}{5} \le 1$. Thus, $\frac{x^2}{20} \le 1 \implies x^2 \le 20 \implies -\sqrt{20} \le x \le \sqrt{20}$. Similarly, sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Fixed points of $2$-dimensional linear system
Consider $$\begin{aligned} \dot{x} &= ax + by\\ \dot{y} &= bx + ay\end{aligned}$$ where $a>0$ and $b<0$. Find the fixed points and classify them.
I haven't been able to find any fixed points other than the trivial $(x,y)=(0,0)$ solution. Any help would be appreciated!
| Fixed points are going to be whatever coordinates in the plane that make the derivative zero. If you set this up as a matrix differential equation you get
$$
\frac{d}{dt} \left [ \begin{array}{c}
x\\ y\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
b & a \\
\end{array} \right ]\left [ \begin{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$... | $$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy=2\operatorname{Li}_4\left(\frac12\right)+\frac{133}{64}\ln2\zeta(3)-\frac{37}{768} \pi ^2 \log ^2(2)+\frac{77}{384}\log^42-\frac{3197\pi^4}{92160}-\frac{C^2}{2}-\frac{1}{8} \pi C \log (2)+\frac{3}{2}\text{Li}_4(2)+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y... | Also, uvw helps.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition gives $$3u^2-2v^2=1,$$ which does not depend on $w^3$.
In another hand, we need to prove that $f(w^3)\geq0$, where $f$ is a concave function:
$f(w^3)\geq0$ is a quadratic inequality of $w^3$ with a coefficient before $w^6$ is eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $ Question -
Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$
\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \... | Yes, your second inequality is also true.
Indeed, after expanding we need to prove that:
$$a+b+c+d+ac+bd+\sum_{cyc}abc\geq\sum_{cyc}a^2bd+abcd\sum_{cyc}ab+a^2b^2c^2d^2+abcd.$$
Now, by AM-GM $$4=a+b+c+d\geq4\sqrt[4]{abcd},$$ which gives $$abcd\leq1.$$
Thus, by AM-GM again we obtain:
$$ac+bd\geq2\sqrt{abcd}\geq abcd+a^2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Linear independence of complex basis of vectors. I understand that for a $\mathbb{C}^n$ as a real vector space, we choose $$\left\{\pmatrix{1\\0\\\vdots\\0},\pmatrix{\mathrm i\\0\\\vdots\\0},\pmatrix{0\\1\\\vdots\\0},\pmatrix{0\\\mathrm i\\\vdots\\0},\dots,\pmatrix{0\\0\\\vdots\\1},\pmatrix{0\\0\\\vdots\\\mathrm i}\rig... | See a little case for this, for example, to show the independence of
$$S = \left\{
\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} i \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ i \end{pmatrix}
\right\}$$
suppose that $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ can be written as
$$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Precalculus limits as x approaches negative infinity Example 1 $\displaystyle \lim _{x\rightarrow -\infty }\frac{\sqrt{9x^{2} +2}}{2x-9}$
Example 2
$\displaystyle \lim _{x\rightarrow -\infty }\frac{6x^{2} -x}{\sqrt{9x^{4} +7x^{3}}}$
I'm having difficulty understanding why in Example 1, we would divide the numerator and... | It would be far more convenient if, in the limit, almost every term were zero except for the "most important" term. For polynomials, the most important term is the leading term (highest power of $x$), so let's factor that one out of the two polynomials we can see. Recall that $\sqrt{a^2} = |a|$ (because the square ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Why is $p$ necessarily greater than $r$ in this number theory problem? From 1998 St. Petersburg City Mathematical Olympiad, presented in Andreescu & Andrica NT: SEP:
Let $n$ be a positive integer. Show that any number greater than $n^4/16$ can be written in at most one way as the product of two of its divisors having ... | We have $a>c\ge d>b$, and $p=a+b$, $r=c+d$, $q=a-b$, $s=c-d$. These are all positive (as the question deals with only one sign for the divisors for uniqueness of solution, so we take the $+$ve route), although with the one exception of $s$ which could equal zero if $c=d$.
We cannot have $p<r$, since then $p^2-r^2<0$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the all positive integer solutions $(a,b)$ to $\frac{a^3+b^3}{ab+4}=2020$.
Find the all positive integer solutions of given equation
$$\frac{a^3+b^3}{ab+4}=2020.$$
I find two possible solutions, namely $(1011,1009)$ and $(1009,1011)$, but the way I solve the equation was messy and I don't know if there are an... | Note that if $p\mid a^2-ab+b^2$, where $p$ is a prime natural number s.t. $p\equiv 2\pmod{3}$, then $p\mid a$ and $p\mid b$. For $p=2$, the claim is easily seen by inspection. Let now $p>2$. We prove by contradiction. Suppose that $p\nmid a$ or $p\nmid b$. It follows immediately that $p\nmid a$ and $p\nmid b$. Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Is it possible to find an expression for $\frac{d^n}{dx^n}e^{-x^2}$? I am trying to find a general form to derivatives of the function $e^{-x^2}$. I tried to do it by finding a pattern for the first derivatives but with no success.
Any tip is welcome.
| Set $u(x)=e^{x}$, $v(x)=-x^2$ and use the famous Faà di Bruno's Formula.
$$
\frac{d^{n}}{d x^{n}} u(v(x))=
\sum_{m_1+m_2+\ldots+m_n=n}
\frac{n !}{m_{1} ! m_{2} ! \cdots m_{n} !}
\cdot
u^{\left(m_{1}+\cdots+m_{n}\right)}(v(x))
\cdot
\prod_{j=1}^{n}\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
$$
For $n=2$ we have
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ $(a>0)$
I just can analyse $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}=a\left(\frac{1}{1}-\frac{1}{1+a}+\frac{1}{2}-\frac{1}{2+a}+\frac{1}{3}-\frac{1}{3+a}...+\frac{1}{n... | Let's call the original sum $\lim_{n \to \infty} V_n$.
Asymptotic solution for $V_n$ with $a>0$: the first sum is Harmonic, so it is $\log n + O(1)$. The second sum is
$$
\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k+a} = \lim_{n \to \infty} S_n
$$
Each value in the (argument, value) tuple in this sum, $(1, \frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n } = 1$? Question:
Find the limit
\begin{equation}
A = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }
\end{equation}
The series originated from the asymptotic analysis in this question. I can show that ... | I think if you set $b_{k+1} = a_{k+1} - a_k$, where $a_k = k \log k$ you could rewrite the denominator as $b_{k+1} = \log (1+\frac{1}{k})^k + \log (k+1) < 1+ \log (k+1) < 2 \log (k+1) < 2(k+1)$ hence the sum you have is $S_n = \sum_{k=1}^{n}\frac{1}{b_k} > \frac{1}{2}\sum_{k=1}^{n} \frac{1}{k+1}$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation ... | Partial Fractions and Telescoping Sums
$$
\begin{align}
\sum_{n=4}^\infty\sum_{k=2}^{n-2}\frac1{\binom{n}{k}}
&=\sum_{k=2}^\infty\sum_{n=k+2}^\infty\frac1{\binom{n}{k}}\tag1\\
&=\sum_{k=2}^\infty\frac{k}{k-1}\sum_{n=k+2}^\infty\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}\right)\tag2\\
&=\sum_{k=2}^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$
Let $x$ and $y$ be positive real numbers such that
$$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$
I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I ... | Cauchy-Schwarz implies $$((x+2)+2(y+2))\left(\frac 1{x+2}+\frac 1{y+2}\right)\geq (1+\sqrt{2})^2 $$ $$\Rightarrow x+2y+6\geq 3(1+\sqrt{2})^2,$$where equality is achieved when $$x+2=3(1+\sqrt{2}),y+2=\frac 3{\sqrt{2}}(1+\sqrt{2}).$$ This shows that the minimum of $x+2y$ is $3+6\sqrt{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An interesting property of a particular set of triples - multiplying two and adding the other always gives 1
Find all triples of real numbers such that multiplying any two in a
triple and adding the third always gives $1$.
When will this be the case? How can we find all such triples?
So far, I've let the numbers be... | You have three equations and three variables to solve for. The three equations are $$ab + c=1$$ $$ac +b=1$$ $$bc + a=1$$
Using the first equation and solving for $c$, we get $$c = 1-ab$$
This now reduces to two equations $$a(1-ab)+b=a+b-a^2b=1$$ $$b(1-ab)+a=a+b-ab^2=1$$
Subtracting the second equation from the first, I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A tricky Inequality problem $ \dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1;\ a_1 , a_2 , \ldots , a_n > 0
$ show that $ \sqrt{a_1} + \sqrt{a_2} + \cdots + \sqrt{a_n} \ge (n-1) \left(\dfrac{1}{\sqrt{a_1}}+ \cdots + \dfrac{1}{\sqrt{a_n}}\right)$
I have tried AM - GM and this problem is from Ine... | Here's a method that only involves algebraic manipulations and does not require the AM-GM inequality or any other more advanced inequalities. Not the most elegant of solutions however, but I believe it works.
$\dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1 \Rightarrow \left(1-\frac{a_1}{1+a_1}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$. I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:
The minimum distance $r$ (which will be the radius o... | We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $y = e^{-x^2}$ as well as a single point of intersection in the first quadrant. That is to say, $$b \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$ For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limit... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\frac{\sum\limits_{cyc}(a^3+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b)}+\frac{3}{4}\geq\frac{(ab+ac+bc)\sum\limits_{cyc}(a+b)^2(a+c)^2}{\prod\limits_{cyc}(a+b)^2}$$ or
$$\frac{27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3}{9uv^2-w^3}+\frac{3}{4}\geq\frac{3v^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac... | $$y'+\frac{xy}{1+x^2} =x$$
$$2x\dfrac {dy}{dx^2}+\frac{xy}{1+x^2} =x$$
Substitute $u=x^2$
$$2\dfrac {dy}{du}+\frac{y}{1+u} =1$$
$$2\sqrt {1+u}{y'}+\frac{y}{\sqrt {1+u}}=\sqrt {1+u}$$
$$(2\sqrt {1+u}{y})'=\sqrt {1+u}$$
Integrate:
$$2\sqrt {1+u}{y}=\int \sqrt {1+u}du$$
$$\sqrt {1+u}{y}=\frac 13({1+u})^{3/2}+K$$
$${y}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| $12(8^x+27^x)=14(12^x+18^x)$
$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$
$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$
We put :$t=(\frac{3}{2}) ^x$
$\Rightarrow $ $\frac{1}{t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$ Let $m$ and $n$ be natural numbers. Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$
I interpreted this statement as $$\frac{m}{n} < \sqrt{2} \rightarrow... | Here is a somewhat different approach: assume that $\dfrac mn < \sqrt 2$. Then
\begin{align*}
m^2 &< 2n^2\\
2m^2 &< 4n^2 \\
2m^2 + 3mn &< 3mn + 4n^2 \\
m(2m + 3n) &< n (3m + 4n) \\
\frac mn &< \frac{3m + 4n}{2m + 3n}
\end{align*}
and
\begin{align*}
m^2 &< 2n^2\\
9 m^2 - 8m^2 &< 18 n^2 - 16n^2 \\
9m^2 + 16n^2 &< 8m^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
To find $A$ given that $2I + A +A^2 = B$ where $B$ is given. How to find a matrix $A$ such that the following holds:
$$2I + A +A^2 = B,$$ where the matrix $B$ is given. I tried with char poly of $B$ but not getting any idea.
Note that it is also given that $B$ is invertible.
P.S. $B = \begin{pmatrix}-2&-7&-4\\ \:12&22&... | According to the other posts, we may assume that $B=\begin{pmatrix}4&1&0\\0&4&0\\0&0&2\end{pmatrix}$. Note that $B$ is cyclic and that $AB=BA$; thus $A$ is a polynomial in $B$: $A=aI_3+bB+cB^2$.
Then $A$ is in the form -considered by copper.hat-
$A=\begin{pmatrix}u&p&0\\0&u&0\\0&0&v\end{pmatrix}$, where $2+u+u^2=4,2+v+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Asymptotic behavior of recursive sequence Suppose, the real sequence
$x_{n+1}=\frac{1}{2}(x_n+\sqrt{x_n^2+c})$ with c>0 is given.
Find the asymptotic behavior of this sequence.
I have shown, that this sequence goes to infinity as $n\to\infty$ per contradiction. I guess, that it holds
$x_n \approx \frac{1}{2}\sqrt{cn}$
... | You've already derived that
\begin{align}
x_{n+1}^2 - x_n^2 &= \frac 12 \left(\sqrt{1 + \frac{c}{x_n^2}} -1 \right) + \frac c4 \\
&= \frac{c}{2x_n^2} \frac{1}{\sqrt{1 + \frac{c}{x_n^2}} +1} + \frac c4.
\end{align}
Taking $n \to \infty$ we obtain
\begin{equation}
\lim\limits_{n\to\infty}\left[x_{n+1}^2 - x_n^2 \right] ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed... | A combinatorial proof is in order!
The given expression is the number of ways to choose an even-sized subset of $n$ people, and promote one of them to be the leader. Notice that $n\cdot 2^{n-2}$ is the number of ways to choose the leader first, then choose an odd-sized subset from the remaining $n-1$ people (as the num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find t... | $$f(x)=\frac{x+2}{(x+1)^2}$$
$$f'(x)=\frac{(x+1)-2(x+2)}{(x+1)^3}$$
$$=\frac{-x-3}{(x+1)^3}$$
so
$$f((-\infty,-1))=[f(-3),+\infty)$$
and
$$f((-1,+\infty))=(0,+\infty)$$
thus, the range is $$[-\frac 14,+\infty).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital Evaluating
$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to ... | Yet another approach: as it's an even function, assume $x>0$. Cut a sector in a radius-$\sqrt{2}$ angle subtending $\pi x$ radians at the centre so $\pi x-\sin\pi x$ is the area in the sector outside the triangle with the same vertices. We'll approximate the arc as a parabola, in Cartesian coordinates with the line seg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \si... | Here is another way to reach the goal,
The equation reduces to:
$$\begin{array}{l}
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+\cos ^{2} x+\sin x \cos x=0 \\
(\sin x+\cos x)^{2}+\cos x(\sin x+\cos x)=0
\end{array}$$
\begin{array}{l}
(\sin x+\cos x)(\sin x+2 \cos x)=0 \\
\Longrightarrow \tan x=-1 \text { and } \tan x=-2
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $a+b+c=0, ab+bc+ca=1$ and $abc=1,$ then find the value of $\frac ab+\frac bc+\frac ca.$ Clearly $a, b, c$ are the roots of the cubic equation: $x^3+x-1=0\tag{1}.$ We have to find:
\begin{align}
\frac ab+\frac bc+\frac ca&=\frac{a^2c+b^2a+c^2b}{abc}\\\\
&=a^2c+b^2a+c^2b\\\\
&=p,\text{ say}.
\end{align} ($p$ is not a ... | Both, $p$ and $q$, are solutions. For instance if you change $a,b,c$ for $a,c,b$, the problem is the same, but $p$ goes to $q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Question regarding the Quotient Rule, How does the textbook reach this intermediate step? $$\begin{align*}
\left(\frac{f(x)}{g(x)}\right)’ &= \frac{(x-3)^{1/3}\frac{1}{2}(x+2)^{-1/2}}{(x-3)^{2/3}} - \frac{(x+2)^{1/2}\frac{1}{3}(x-3)^{-2/3}}{(x-3)^{2/3}}\\
&= \frac{(x-3)^{-2/3}(x+2)^{-1/2}}{(x-3)^{2/3}}\cdot\left[\frac{... | $$\dfrac{(x-3)^{1/3}\frac12(x+2)^{-1/2}}{(x-3)^{2/3}}-\dfrac{(x+2)^{1/2}\frac13(x-3)^{-2/3}}{(x-3)^{2/3}}$$
$$=\dfrac{(x-3)\color{blue}{(x-3)^{-2/3}}{\frac12\color{blue}{(x+2)^{-1/2}}}-(x+2)\color{blue}{(x+2)^{-1/2}}\frac13\color{blue}{(x-3)^{-2/3}}}{(x-3)^{2/3}}$$
$$=\dfrac{\color{blue}{(x-3)^{-2/3}(x+2)^{-1/2}}}{(x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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integral calculation mistake I try to solve this:
$
\int_{0}^{\pi /2}\sin x \cos x\sqrt{1+\cos^{2}x } dx
$
This is what I do:
$ \cos x = t; -\sin x dx = dt; -\sqrt{1-u^{2}} dt $
$ \frac{-\sqrt{1-t^{2}}*t*\sqrt{1+t^{2}}}{-\sqrt{1-t^{2}}} dt $
$-\int_{0}^{1} t * \sqrt{1+t^{2}} dt$
$1+t^2 = a; 2tdt = da; tdt = da/2$
$-\i... | After your first substitution you should have $-\int_{1}^{0}t\sqrt{1+t^{2}}dt$= $\int_{0}^{1}t\sqrt{1+t^{2}}dt$.
The limits for your second substitution should be $a=1$ at ($t=0$) and $a=2$ (at $t=1$) so we have $\frac{1}{2}\int_{1}^{2} \sqrt{a}da=\frac{1}{3}a^{\frac{3}{2}}|^{a=2}_{a=1}=\frac{2\sqrt{2}}{3}-\frac{1}{3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
-8 \\
1
\end{bmatrix} = \begin{bmatrix}
3 \\
8
\end{bmatrix}$$
and
$$M\begin{bmatrix}
1 \\
5
\end{bmatrix} = \begin{bmatrix}
-8 \\
7
\end{bmatrix}... | Note that the given input vectors form a basis of $\Bbb{R}^2$. So we can express any vector in $\Bbb{R}^2$ in terms of the given vectors such as:
$$\begin{bmatrix}a\\b\end{bmatrix}={\small\left(\frac{b-5a}{41}\right)}\begin{bmatrix}-8\\1\end{bmatrix}+{\small \left(\frac{a+8b}{41}\right)}\begin{bmatrix}1\\5\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help with proof of Euler's criterion Problem
This problem is about finding square roots modulo a prime $p$.
(a) Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.
An integer $x$ is called a square root of $n$ mod $p$ when $x^2 \equiv n \pmod p$. An integer with a square r... | Let $g$ be a primitive root in modulo $p$. Then for an integer $x$ relatively prime to $n$ we have, $x\equiv g^i$(mod $p$), for some $i=1,2,\dots,p-1$. Similarly since, $gcd(n,p)=1$, so there are some $j=1,2,\dots,p-1$ such that $n\equiv g^j$(mod $p$).
Now, there is an integer $x$ relatively prime to $p$ satisfying $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find real numbers $s$ and $t$ such that $\Gamma$ is the graph of $r = 2s \cos (\theta + t)$ The question: Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of
$r = 2s \cos (\theta + t).$
I did this so far:
We can use the general circl... | From
$$r=2(h \cos \theta + k \sin \theta)$$
$$r=2\sqrt{h^2+k^2}\left( \frac{h}{\sqrt{h^2+k^2}} \cos \theta + \frac{k}{\sqrt{h^2+k^2}}\sin \theta \right)$$
We can pick $s=\sqrt{h^2+k^2}$
and we can pick $t$ to satisfy $\cos t = \frac{h}{\sqrt{h^2+k^2}}$ and $\sin t = -\frac{k}{\sqrt{h^2+k^2}}$.
Remark: handle the case w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that the size of the Turan graph $T_r(n)$ is at least $(1 - \frac{1}{r}) \binom{n}{2}$. A Turan graph $T_r(n)$ is defined as the complete $r$-partite graph of order $n$ such that the number of vertices in each of the $r$ classes is either $\lfloor \frac{n}{r}\rfloor$ or $\lceil \frac{n}{r} \rceil$. For fixed $n$ a... | The size $S$ of the Turán graph is $\frac 12\left(n^2-\frac {(n^2-s^2)}{r}-s\right)$, see my answer. It follows $$S- \left(1 - \frac{1}{r}\right) \binom{n}{2}=\frac{(n-s)(r-1)+s(s-1)}{2r}\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the val... | Consider the four corner/vertex points $(\pm 6\cos\theta, \pm 4\sin\theta)$ of rectangle lying on given ellipse: $4x^2+9y^2=144$
Now, the sides of the rectangle are length: $(2\cdot 6\cos\theta)$ & width $(2\cdot 4\sin\theta)$ which are in ratio $3:2$ as given in question therefore we have $$\frac{12\cos\theta}{8\sin\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Analogue of $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ for vectors The Brahmagupta–Fibonacci identity $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ allows us to write a product of squares as a sum of squares. Is there an analogue of this identity when $a, b, c, d$ are vectors in $\mathbb{R}^n$ and the ... | I assume (perhaps unreasonably closed-mindedly) that you want an analogue of the form
\begin{align}
\left(\left<a,a\right>^2 + \left<b,b\right>^2\right) \left(\left<c,c\right>^2 + \left<d,d\right>^2\right) = \left<v,v\right>^2 + \left<w,w\right>^2
\end{align}
that holds for any vectors $a, b, c, d \in \mathbb{Q}^n$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Consider $v=v_1+v_2$, where $v_1 \in M$ and $v_2 \in M^{\perp}$ $M = span\{\begin{pmatrix}8\\0\\-6\end{pmatrix}, \begin{pmatrix}8\\6\\-6 \end{pmatrix}\}$
I am trying to calculate $v_1$ and $v_2$ when $v=\begin{pmatrix}2 \\ 4 \\ 6 \end{pmatrix}$.
I know that since $v_1 \,\in \, M$ and $v_2 \, \in \, M^{\perp}$, $v_1\cdo... | It is quite easy to see that $w=\begin{bmatrix}6\\0\\8\end{bmatrix}$ is orthogonal to the basis vectors given for $M$. Thus $w \in M^{\perp}$. Moreover $\text{dim}(M)=2$ and $M \subset \Bbb{R}^3$, so $\text{dim}(M^{\perp})=1$. This means we can say that $\{w\}$ is a basis for $M^{\perp}$.
So we want to solve for $a,b,c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my ... | Note
\begin{align}
&\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}} \\
=& \sqrt{\frac{(2a+b)+b +2\sqrt{(2a+b)b}}{2}}\\
= &\sqrt{\frac{(\sqrt{2a+b}+\sqrt b )^2}{2}}\\
= &\sqrt{\frac{2a+b}2}+\sqrt {\frac b{2}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\lim_{n\to \infty} a_n=\frac{\sqrt{5}-1}{2}$ if $a_{n+1}=\sqrt{1-a_n}$ and $0If $0<a_0<1$ and $a_{n+1}=\sqrt{1-a_n}$, prove that:
$$\lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$$
Here what I do is that when $n\to \infty$, $a_{n+1}=a_n$
$\therefore a_n=\sqrt{1-a_n}\Rightarrow a_{n}^2+a_n-1=0$
which implies t... | $$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{\frac{\sqrt5-1}{2}-a_n}{\sqrt{1-a_n}+\frac{\sqrt5-1}{2}}$$ and
$$a_{n+2}-a_n=\sqrt{1-\sqrt{1-a_n}}-a_n=\frac{a_n\sqrt{1-a_n}\left(\frac{\sqrt5-1}{2}-a_n\right)\left(\frac{\sqrt5+1}{2}+a_n\right)}{(\sqrt{1-\sqrt{1-a_n}}+a_n)((1+a_n)\sqrt{1-a_n}+1)}.$$
Now, let $a_1<\frac{\sqrt5-1}{2}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
System of equations and recurrence relation I am trying to find the general solution for $N$ of the following system of equations
$$
\begin{cases}
(x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\
{x_n}^2 + {y_n}^2 = 1
\end{cases}
$$
with the initial values $x_0 = 1$ and $y_0 = 0$ and t... | Because it is very difficult to solve this relation, we can use complex numbers. Let
$$(x_n, y_n) = x_n + iy_n \quad \text{and} \quad z_n = x_n + iy_n$$
then we have
$$z_1 = -\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}$$
and the term $z_1$ is always given by the same expression with respe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the probablity that the sum of two dice is 4 or 6?
What is the probablity that the sum of two dice is 4 or 6?
The explanation I found is as follows:
Total number of outcomes $6 \times6 = 36$
Number of outcomes where the event occurs: $1+3, 2+2, 3+1, 1+5, 2+4, 3+3, 4+2$ and $5+1$ (Total $ 8$)
The probability t... | In developing an understanding of the sample space, you should think of the two dice as being distinct. Yes, in principle, the two dice may be indistinguishable. However, from the point of view of probability theory, the behaviour of each die is described by a different random variable. Hence the two dice are not, r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Deriving $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$ I tried but I could not get $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$ from $\psi_1(x) = \cos^3 x + \sin^2 x + \cos ... | To obtain the required decomposition it suffices to sum the left-hand and right-hand sides of the following trigonometric identities:
$$1=\sqrt{\pi}\varphi_0(x),$$
$$\cos x=\frac{\sqrt{\pi}}{\sqrt{2}}\varphi_1(x),$$
$$\sin^2 x =\frac 12(1-\cos 2x)=\frac{\sqrt{\pi}}{2}\varphi_0(x)-\frac {\sqrt{\pi}}{2\sqrt{2}}\varphi_2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist... | I believe you messed up by looking at $$\lim_{x \to 1}\left(\frac{1}{x-1} \color{red}{+} \frac{3}{1-x^3}\right)$$ instead of $$\lim_{x \to 1} \left( \frac{1}{x-1}\color{red}{-}\frac{3}{1-x^3} \right)$$
The value of the first one is $1$, but the second one does not exist.
We can factor $1-x^3$ as $(x-1)(-x^{2}-x-1)$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How find this nice minmum of this value $\frac{\prod_{i=1}^{n-1}(a_{i}+a_{i+1})\sum a_{i}}{\prod_{i=1}^{n}a_{i}}$ Let $n$ be give postive integer number. For any $a_{i}>0 (i=1,2,\cdots,n)$, find the minimum of the value
$$F_{n}(a_{1},a_{2},\cdots,a_{n})=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})\cdots (a_{n-1}+a_{n})(a_{1}+a_{2... | Here's some partial progress. Maybe someone will have further ideas.
By homogeneity, we can set $a_1 + \cdots + a_n = 1$ throughout. First dispense with boundary behavior. Fix $1 \leq k \leq n$ and note that
$$\frac{\prod_{i=1}^k (a_i+a_{i+1})}{\prod_{i=1}^n a_i} = \prod_{i=1}^{k-1} \frac{a_i + a_{i-1}}{a_i} \cdot \pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.