Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^... | This graph helps to show what's happening.
It's not an answer but a companion to Claude's, in that it helps illustrate the turning points he identified and shows the fact that there's one solution. Note the scales are different, I had to stretch the Y axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Factoring question involving 4 integers. For all non-negative integer values of $a$, $b$, $c$, $d$ given that $ac+bd+bc+ad=42$ and $c^2-d^2=12$, then determine all possible values of $a+b+c+d$.
This was a question on one of my previous tests and I am not sure how to do it.
| $c^2 - d^2 = 12 \iff (c+d)(c-d)=12$. The only integers $c,d$ satysfying this equality are $c=4, d=2$. Putting these values into $ac + bd + bc + ad = 42$ we get that $7a+7b = 42 \iff a+b=6$. That means that there are 5 quadruples (a,b,c,d), namely (1,5,4,2), (2,4,4,2), (3,3,4,2), (4,2,4,2), (5,1,4,2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Ordered triples of n-powerful integers
Let’s say that an ordered triple of positive integers (a, b, c) is n-powerful
if:
*
*$a \le b \le c$,
*$gcd(a, b, c) = 1$ and
*$a^n + b^n + c^n$ is divisible by $a + b + c$.
For example, $(1, 2, 2)$ is 5-powerful.
a) Determine all ordered triples (if ... | Let $P_i = a^i + b^i + c^i$
Hint for part 2: Apply Newton's Identities.
Show that $P_{2007} = A P_{2004} + B P _{2005} + C (a+b+c) $ where $A,B,C$ are constants to be determined.
Not sure about part 1 as yet, but there are solution sets of $(1,1,1)$ and $(1,1,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Finding $a$ yielding minimum value for quadratic root expression $(x_1+2x_2)(x_2+2x_1)$ The problem is:
We have the expression $(x_1+2x_2)(x_2+2x_1)$, where $x_1$ and $x_2$ are the roots of $$f(x)=x^2+ax+a+\frac{1}{5}$$
Find the value(s) of $a$ yielding the least possible value for this expression.
My solution is: ... | Given $x_1$ and $x_2$ are the roots of $\displaystyle f(x)=x^2+ax+a+\frac{1}{5} = 0$
So $\displaystyle x_{1}+x_{2} = -a$ and $\displaystyle x_{1}\cdot x_{2} = a+\frac{1}{5}$
Now $$\displaystyle (x_1+2x_2)(x_2+2x_1) = x_{1}\cdot x_{2}+2\left[x_{1}^2+x_{2}^2\right]+4x_{1}\cdot x_{2} = 5x_{1}x_{2}+2\left[\left(x_{1}+x_{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2... | Notice, we have $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2x^4-2x^2+1}}dx$$
$$=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2\left(x^2-\frac{1}{2}\right)^2-\frac{1}{2}+1}}dx$$
$$=\frac{1}{2}\int_{1}^{2}\frac{2x(x^2-1)}{x^4\sqrt {2\left(x^2-\frac{1}{2}\right)^2+\frac{1}{2}}}dx$$ Now, $$x^2-\frac{1}{2}=t\implies 2xdx=dt$$
$$=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Where does this sequence $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.... converge? The given sequence is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and so on.
the sequence is increasing so to converge must be bounded above.Now looks like they would not excee... | Trick: Let $X = \sqrt{ 7 + \sqrt{ 7 + ... } } $. We have $X = \sqrt{ 7 + X } $ and so $X^2 = 7 + X $. Now you solve the quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$
| HINT: Integrate by parts
$$\int\dfrac{dx}{(x^4-1)^2}=\dfrac1{x^3}\int\dfrac{x^3}{(x^4-1)^2}-\int\left(\dfrac{d(1/x^3)}{dx}\cdot\int\dfrac{x^3}{(x^4-1)^2}\right)dx$$
$$=-\dfrac1{x^3}\cdot\dfrac1{4(x^4-1)}-\dfrac34\int\dfrac{dx}{x^4(x^4-1)}$$
Now $\dfrac1{x^4(x^4-1)}=\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}=\dfrac1{x^4-1}-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
$y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms. Prove that if $y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms.Then $\frac{dy}{dx}=\frac{1}{1+(x+n)^2}-\... | We can write $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+x+1}\right) = \tan^{-1}\left(\frac{1}{1+(x+1)\cdot x}\right) = \tan^{-1}\left(x+1\right)-\tan^{-1}(x)$$.
Similarly $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) = \tan^{-1}\left(\frac{1}{1+(x+2)\cdot (x+1)}\right) = \tan^{-1}\left(x+2\right)-\tan^{-1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| The identity is true.
\begin{align*}
\cos x + \sin x \tan \frac{x}{2} & =(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})+(2\sin \frac{x}{2}\cos \frac{x}{2})(\frac{\sin \frac{x}{2}}{\cos\frac{x}{2}})\\
& = \cos^2\frac{x}{2}+\sin^2 \frac{x}{2}\\
& = 1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Does the elliptic function $\operatorname{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\big|\frac{1}{2}\right)$ have a closed form? Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,
can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$... | To explain how an algebraic equation for $Z:=\operatorname{cn} \left(\frac{2K}{3}|m\right)$ can be obtained, notice that
*
*The usual parity properties imply that
$$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\operatorname{cn} \left(2K-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(-\frac{2K}{3}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... |
If $a,b,c,n\in\Bbb Z_{\ge 1}$, $a+b+c\mid a^2+b^2+c^2$, then $$a+b+c\mid a^n+b^n+c^n$$
is true when $n\nmid 3$, but not necessarily when $n\mid 3$.
$$x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2$$
$$\implies x+y+z\mid 2(xy+yz+zx)$$
$$\implies x+y+z\mid (x^k+y^k+z^k)(xy+yz+zx)$$
for all $k\ge 1$ (to see why, check cases when $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
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How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| A solution by Sumit Ray
$ac=-bd$
$\frac{a}{b} = -\frac{d}{c} = k$
$a=bk \text{ and }d=-ck$
$a^2+b^2=1\implies b^2 = \frac{1}{k^2+1}\implies c^2 = \frac{1}{k^2+1}$
Thus $b^2 - c^2 =0$
Now \begin{align*}ab+cd &= b^2\cdot k-c^2\cdot k\\&= k(b^2-c^2)
= 0\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to app... | Based on Vieta's formula and Brian and GAVD answers:
From Vieta formula, we got:
$a+b+c+d=7$
$ab+ac+ad+bc+bd+cd=3$
$abc+abd+acd+bcd=21$
$abcd=1$
Then:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$= (a+b+c+d-d)(a+b+c+d-c)(a+b+c+d-b)(a+b+c+d-a)$
$= (7-d)(7-c)(7-b)(7-a)$
Simplify and we got:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$=(7−a)(7−b)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so ... | Note that$$\frac{x^3}{x^2 + y^2} = x\cdot\frac{x^2}{x^2 + y^2}$$ In absolute value, this is $\le |x|.$ As $(x,y)\to (0,0),|x| \to 0.$ Same thing for $y^3/(x^2 + y^2).$ So $f(x,y)$ is the difference of two functions that both $\to 0$ as $(x,y) \to (0,0);$ hence $\lim_{(x,y)\to (0,0)}f(x,y) = 0.$ This shows $f$ is conti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution.
Let $x,y,z$ be positive real numbers such that $xyz =1$. P... | Let me try to cover the motivation for writing something like
$$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}8 + \frac{1+z}8$$
for doing AM-GM, seemingly out of the blue.
We obviously start with the first term. Now clearly, to get rid of the denominator, it would be great to consider the terms $(1+y)$ and $(1+z)$. As there is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| Given $$\displaystyle \lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$
Now Let $$\tan^{-1}(x) = y\Rightarrow x=\tan y\;,$$ Then when $x\rightarrow 0,$ Then $y=\tan^{-1}(0)\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{y\cdot \tan y}-\frac{1}{\tan^2 y} = \lim_{y\rightarrow 0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know ... | If $n>1$ is an integer we have
\begin{align*}
n-1&<n\\
\implies \frac{1}{n-1}&>\frac{1}{n}\\
\iff 1-\frac{1}{n-1}&<1-\frac{1}{n}\\
\iff \frac{n-2}{n-1}&<\frac{n-1}{n}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
transformation of uniformly distributed random variable f(x)=1/2pi into Y=cosx Let $X$ be a uniformly distributed function over $[-\pi;\pi]$. That is
$
f(x)=\left\{\begin{matrix}
\frac{1}{2 \pi} & -\pi\leq x\leq \pi \\
0 & otherwise
\end{matrix}\right.\\
$
Find the probability density function of $Y = \cos X$.
I f... | We deal with your question about why we multiply by $2$. We are interested in finding the cumulative distribution function $F(y)$ of $Y$. The only interesting part is when $-1\lt y\lt 1$, because $F(y)=1$ if $y\ge 1$ and $F(y)=0$ when $y\le -1$.
It is geometrically perhaps a little easier to find $G(y)=\Pr(Y\gt y)$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,coul... | As $A+B+C=\pi,\cos(A+B)=\cdots=-\cos C$
and using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$2\cos A\cos B\cos C=\cos C[\cos(A-B)+\cos(A+B)]$$
$$=-\cos(A+B)\cos(A-B)-\cos^2C$$
$$=-(\cos^2A-\sin^2B)-\cos^2C$$
$$=1-(\cos^2A+\cos^2B+\cos^2C)$$
So, the equation reduces to $$\sum(\cos A-\cos B)^2=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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simultaneous equation help Can someone help me with to solve this system of equations ?
$$
\left\{
\begin{array}{c}
y=x+1 \\
y^2+2x^2=2
\end{array}
\right.
$$
| $$\begin{cases} y=x+1 \\ { y }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow \begin{cases} y=x+1 \\ { \left( x+1 \right) }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow 3{ x }^{ 2 }+2x-1=0\quad \Rightarrow x=\frac { -1\pm 2 }{ 3 } ,y=1+\frac { -1\pm 2 }{ 3 } \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve equation $m^2=n^5-4$ in $\mathbb{N}$ I have an exercise to solve an equation like in the title.
My goal until now is that both $m$ and $n$ are odd, but then I can not continue. Can you help me?
Thanks a lot!
| Consider $m^2+4 (\bmod 11):$
$m=0 (\bmod 11) \Rightarrow m^2+4 = 4 (\bmod 11),$
$m=1,10 (\bmod 11) \Rightarrow m^2+4 = 5 (\bmod 11),$
$m=2, 9 (\bmod 11) \Rightarrow m^2 +4= 8 (\bmod 11),$
$m=3, 8 (\bmod 11) \Rightarrow m^2+4 = 2 (\bmod 11),$
$m=4, 7 (\bmod 11) \Rightarrow m^2+4 = 9 (\bmod 11),$
$m=5, 6 (\bmod 11) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Get number of additions that lead to a specific sum with given summands Suppose you have $n$ variables. Each of these variables (e.g. $a$) have their own interval between $0$ and $a_i$ (whole numbers). The only rule is that all these variables have to add up to a given value $s$.
An example:
$n = 3$, so the three varia... | Let me work out for this particular case. You should be able to put it into a formula.
If you can't, I'll see later.
We can use stars and bars with inclusion-exclusion.
Without any restrictions, there are ${6+3-1\choose 3-1} = {8\choose 2}= 28$ solutions.
To subtract cases that violate, (say) the $a_i$ restriction, pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
| Stiring Numbers of the Second Kind is also good for building sum of powers formula.
$x^n = \sum _{k=0} ^n S(n,k) (x)_k$
Example, for sum of x^2 formula, "integrate" the falling factorial form.
$$x^2 = (1)(x)_1 + (1)(x)_2$$
$$\sum_{k=0}^{n-1} x^2
= \frac{(n)_2}{2}+\frac{(n)_3}{3}
= \frac{n^3}{3} - \frac{n^2}{2} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Computing matrices to a power of $6$ Compute
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$
How would I solve this question. I found out that it's square would be $\begin{pmatrix} 2 & -2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix}.$ What next? Is there any easier way?
| Set
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
we seek $A^6$, where
$A = \begin{bmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{bmatrix} = \sqrt{3} I + J; \tag{2}$
we note that we may write $A$ as
$A = 2(\dfrac{\sqrt{3}}{2} I + \dfrac{1}{2}J); \tag{3}$
noting further that
$\dfrac{\sqrt{3}}{2} = \cos \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-... | When the denominator contains factors like $(ax^2 + bx + c)^p$ with $b^2 - 4ac < 0$ (such as $x^2 + 2x + 4$ in your case), then the fractions that appear in the expansions are
$$\frac {A_1 x + B_1} {ax^2 + bx + c} + \frac {A_2 x + B_2} {(ax^2 + bx + c)^2} + \dots + \frac {A_p x + B_p} {(ax^2 + bx + c)^p} .$$
In your ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int\frac{\left(1+x\right)\sin x}{\left(x^{2}+2x\right)\cos^{2}x-\left(1+x\right)\sin2x}dx$ (Q) $\displaystyle \int\frac{\left(1+x\right)\sin x}{\left(x^{2}+2x\right)\cos^{2}x-\left(1+x\right)\sin2x}dx$
Tried a lot to expand denominator and reduce it to bring its derivative on top , but all in vain .
| Let $$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x-(1+x)\sin 2x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x+1)\cos^2 x-(1+x)\sin 2x-\cos^2 x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+2)\sin x\cdot \cos x-(1-\sin^2 x)}dx$$
So $$\displaystyle I = \int\frac{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$ Without L'Hopital or Calculus? What is: $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$?
Thanks in advance
Much appreciated!
| $$
\begin{align}
\lim_{x\to1}\frac{\left(2x^2-1\right)^{\frac13}-x^{\frac12}}{x-1}
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-(x+1)^{\frac12}}{x}\tag{1}\\
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-1}{x}-\lim_{x\to0}\frac{(x+1)^{\frac12}-1}{x}\tag{2}\\
&=\lim_{x\to0}\frac{2x^2+4x}{x\left(\left(2x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | Hint:
$$\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right) = \frac{1}{2}\lim _{x\to 1}\left(\left(\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\right)\left(\frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}\right)\left(\frac{\sqrt{x} + 1}{\sqrt{x} + 1}\right)\right)$$
Now use the following equation to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| If $x^3 + \frac{1}{x^3} = 18$ then squaring gives $x^6 + 2 + x^{-6} = 18^2$ so $x^6 + x^{-6} = 322$. Squaring again gives $x^{12} + x^{-12} = 322^2 - 2 = 103682$
Then $(x + x^{-1})(x^{11} + x^{-11}) = (x^{12} + x^{-12}) + (x^{10} + x^{-10})$.
Now to me it seems clear that since we have
$$
(x+ x^{-1})^3 = (x^3 + x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Show that $g(x)=\sqrt{x^{2}+4}$ is continuous at $x=1$ Let $g(x)=\sqrt{x^{2}+4}$ from $\mathbb{R}\rightarrow \mathbb{R}$. I want to show that $g$ is continuous at $x=1$.
I have to show that for any $\epsilon>0$, there exists a $\delta>0$ such that $|x-1|< \delta \implies |g(x)-g(1)|<\epsilon$.
So I do some rough work f... | As suggested in a comment of imranfat, multiply by the conjugate, i.e.
\begin{align*}
\sqrt{x^2+4}-\sqrt{5}=\frac{(\sqrt{x^2+4}-\sqrt{5})(\sqrt{x^2+4}+\sqrt{5})}{\sqrt{x^2+4}+\sqrt{5}}=\frac{x^2-1}{\sqrt{x^2+4}+\sqrt{5}}=\frac{(x-1)(x+1)}{\sqrt{x^2+4}+\sqrt{5}}.
\end{align*}
Now, if $|x-1|<\delta$, then $|x+1|=|x-1+2|\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is row reduced echelon form? How to row reduce this matrix? I'm not being able to grasp the concept of row reduced echelon form. Please, explain how to row reduce one of the the following matrices.
$A =
\begin{bmatrix}
1&3&4&5\\3&9&12&9\\1&3&4&1
\end{bmatrix}$
$B=
\begin{bmatrix}
1&2&1&2\\0&1&0&1\\-1&2&0&3... | In order to obtain the reduced row echelon form (rref) of a matrix, we apply some row operations. According to this article, for the first case we have:
$\begin{array}{l} \begin{bmatrix} 1 & 3 & 4 & 5\\ 3 & 9 & 12 & 9\\ 1 & 3 & 4 & 1 \end{bmatrix}\overset{R_2:=3R_1 - R_2}{\to}\begin{bmatrix} 1 & 3 & 4&5\\0&0&0&6 \\ 1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any... | Remembering that $\sqrt{2}^2=2$ we get, from where you left off.
$$
7+5\sqrt{2} = x^3+3x^2y(\sqrt{2})+3xy^2(2)+y^3(2)(\sqrt{2})\\
= x^3+6xy^2+3x^2y(\sqrt{2}) +2y^3 (\sqrt{2})\\
= x^3+6xy^2+(3x^2y+2y^3)(\sqrt{2})
$$
So, we see that $7=x^3+6xy^2$ and $5=3x^2y+2y^3$. Which still looks like a 3 pipe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to solve this limit 2 variables $\lim_{(x,y) \to (4,1)} \frac{y \sqrt x - 2y - \sqrt x + 2}{4 - x + x \sqrt y - 4 \sqrt y}$ Please anybody can help me solve this?
$$\lim \frac{y \sqrt x - 2y - \sqrt x + 2}{4 - x + x \sqrt y - 4 \sqrt y}$$
with $(x,y) \to (4,1)$
Thank you!
| $$\frac{y\sqrt x-2y-\sqrt x+2}{4-x+x\sqrt y-4 \sqrt y}=\frac{\sqrt x (y-1)-2(y-1)}{\sqrt y(x-4)-(x-4)}=\frac{(y-1)(\sqrt x-2)}{(x-4)(\sqrt y-1)}=\frac{\sqrt y+1}{\sqrt x+2}$$
You can show that the function $f(x,y)=\frac{\sqrt y+1}{\sqrt x+2}$ is continuous at the point $(x,y)=(4,1)$ :
$$|f(x,y)-f(4,1)|=|\frac{\sqrt y+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What are the $\sin$ and $\cos$ of discrete calculus? I'm getting acquainted with Discrete Calculus, and I really like taking functions that arise in traditional calculus and finding what their counterparts in discrete-land are.
For example, if we define our difference operator
$$\Delta f(n) = f(n + 1) - f(n)$$
(the ana... | Note that if $f(n)$ is a discrete sinusoid, then
$$
f(n+1)=f(n)+\Delta f(n)
$$
and
$$
\Delta f(n+1) = \Delta f(n) + \Delta^2 f(n) = \Delta f(n) - f(n),
$$
or
$$
\left(\begin{matrix}f(n+1) \\ \Delta f(n+1)\end{matrix}\right)=\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}f(n)\\ \Delta f(n)\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
| Hint: Expand the fraction with $(x+1)$
$$y=\frac{(1+x+x^2)(x+1)}{(x+1)(1-x+x^2)}=\frac{(1+x+x^2)(x+1)}{1+x^3}$$
Now use the geometric series you wrote down for $$\frac{1}{1-(-x^3)}$$
$$y=(1+x+x^2)(x+1)(1+(-x^3)+O(x^6))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Proving a progression inequality Prove that $$1+ \frac{1}{2^3} + \cdot \cdot \cdot + \frac{1}{n^3} < \frac{5}{4} $$
I got no idea of how to approach this problem.
| Since:
$$\begin{eqnarray*} \frac{1}{n^3} &=& \frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{2n+1}{2n^3(n+1)^3}\\&\leq&\frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{1}{4}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to prove $(6666\ldots66)^2 + 8888\ldots88 = 4444\ldots44$ (with $n$ 6s and 8s, $2n$ 4s) How to prove that, if $n$ is a positive integer, then
$$
(\underbrace{666 \ldots 6}_{n \text{ copies of } 6})^2 +
\underbrace{888 \ldots 8}_{n \text{ copies of } 8} =
\underbrace{444 \ldots 4}_{2n \text{ copies of } 4}?
... | Given $$\underbrace{(666666666666\ldots)^2}_{n~\text{times}}+\underbrace{(888888888888\ldots)}_{n~\text{times}}$$
Now we can write $$\underbrace{666666666666\ldots}_{n~\text{times}} = (6+6\cdot 10+6\cdot 10^2+\cdots+6\cdot 10^{n-1})$$
$$\displaystyle =6\left[\frac{10^n-1}{10-1}\right] = \frac{2}{3}\left[10^n-1\right]$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
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Discrete mathematics - Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$. Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.
This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give... | The equation is equivalent to
$$
c^2+1=(a^2-1)(b^2-1)
$$
Since the left side is either $1$ or $2$ mod $4$, and the right side is either $0$ or $1$ mod $4$, they must both be $1$ mod $4$, which means that $a,b,c$ must be even.
Since any prime which is $3$ mod $4$ is a Gaussian prime, no prime which is $3$ mod $4$ can di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Given area of square $= 9+6\sqrt{2}$ Without calculator show its length in form of $(\sqrt{ c}+\sqrt{ d})$ $\sqrt{9+6\sqrt{2}}$ to find length
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$.
| Let the side of the square be $(\sqrt c+\sqrt d)$ then its area is given as $$(\sqrt c+\sqrt d)^2=9+6\sqrt 2\tag 1$$
$$c+d+2\sqrt{cd}=9+6\sqrt 2$$ by comparing the corresponding rational & irrational parts, we get $$c+d=9\tag 2$$
$$2\sqrt{cd}=6\sqrt 2\tag 3$$
Now, we know $$(\sqrt c-\sqrt d)^2=(\sqrt c+\sqrt d)^2-2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating th... | Consider $$B = \sqrt c - \sqrt{c-1},\quad D = \sqrt {c+1} - \sqrt{c}$$
By mean value theorem, there exist $$b\in (c-1, c),\quad d\in (c, c+1)$$ that satisfy
$$\frac1{2\sqrt b} = B,\quad \frac1{2\sqrt d} = D$$
and since $b\le d$, $$\begin{align*}
\dfrac1{2\sqrt b} &\ge \dfrac1{2\sqrt d}\\
\sqrt c - \sqrt{c-1} &\ge \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How to show this limit exists I know the limit is 0, but how would I prove the limit? $$\lim_{(x,y)\to(0,0)} {\frac{x^3y^2}{x^4+y^6}}$$
| As
\begin{align*}
x^4+y^6 = \frac{1}{2} x^4 + \frac{1}{2} x^4 + y^6 \geq \frac{3}{\sqrt[3]{4}}x^{8/3}y^2,
\end{align*}
then
\begin{align*}
\Big|\frac{x^3y^2}{x^4+y^6}\Big| \leq \frac{\sqrt[3]{4}}{3}|x|^{1/3} \rightarrow 0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why are the derivatives of these two equations different? I am trying to implicitly differentiate the following: $\frac {x}{x-y}\ =\ y^2-1 $
I originally multiplied the whole equation by $(x - y)$ and the result of implicitly differentiating the resulting equation ($\ x=xy^2-x-y^3+y $ ) was $\frac {dy}{dx}\ = \frac {2-... | Contrary to what was said in the comments, you lose no information by your manipulation. In the original equation, you knew $x\ne y$. You don't lose anything by multiplying by $x-y$. You have correctly differentiated both expressions, so we have:
$$
\frac{dy}{dx}=\frac{2-y^2}{2xy-3y^2+1}, \text{ and} \\ \frac{dy}{dx}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$. Let $2$ positive real numbers $x,y$ satisfy $x\leq 1,y\leq 1$ are chosen at random.Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$.
Since the $x+y\leq ... | For the conditional probability you need to divide by the "given" area.
in this case $ Area(x^2+y^2 \ge \frac 14)=1-\frac{\pi}{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to Solve for $x$ in the equation $\sin(2x) = -\frac{1}{2}$ within the interval of $[0, 2\pi]$? So far, I only have the solutions $11\pi/12$, and $7\pi/12$ (I think there are a total of $4$). I think this has something to do with trig identities, and should be able to be solved without using a calculator. However, I... | If $$\sin(2x) = -\frac{1}{2}$$ then
\begin{align*}
2x & = \arcsin\left(-\frac{1}{2}\right) + 2n\pi & 2x & = \pi - \arcsin\left(-\frac{1}{2}\right) + 2m\pi\\
2x & = -\frac{\pi}{6} + 2n\pi & 2x & = \pi - \left(-\frac{\pi}{6}\right) + 2m\pi\\
x & = -\frac{\pi}{12} + n\pi & 2x & = \frac{7\pi}{6} + 2m\pi\\
& & x & = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How do I transform, with long division, this polynomial into the answer given? I need to transform:
$ \frac {x^5}{(x-2)(x+2)(x^2+4)} $
into
$ \frac{-2x}{x^2+4} + x + \frac{1}{x-2} + \frac{1}{x+2} $
How can I solve it?
Thanks.
| Let me try. We have $$\frac{x^5}{(x-2)(x+2)(x^2+4)} = \frac{x^5}{x^4-16}= x + \frac{16x}{x^4-16} = x + 2x\left(\frac{1}{x^2-4} - \frac{1}{x^2+4}\right) = x -\frac{2x}{x^2+4} +\frac{2x}{(x+2)(x-2)} = x -\frac{2x}{x^2+4} + \frac{1}{x-2} + \frac{1}{x+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$$
That's
$$\frac{x^2+x\cdot \sin x}{-1+\left(1-2\sin^2\frac{x}{2}\right)} = \frac{x^2+x\cdot \sin x}{-2\sin^2\frac{x}{2}}$$
Let's split this
$$\frac{x\cdot x}{-2\cdot\sin\frac{x}{2}\cdot ... | Notice, $$\lim_{x\to 0 }\frac{x^2+x\sin x}{-1+\cos x}$$
$$=\lim_{x\to 0 }\frac{x^2+x\sin x}{-1+1-2\sin ^2\frac{x}{2}}$$
$$=-\frac{1}{2}\lim_{x\to 0}\frac{x\sin x+x^2}{\sin ^2\frac{x}{2}}$$
$$=-\frac{1}{2}\lim_{x\to 0}\frac{x^2\left(\frac{\sin x}{x}+1\right)}{\sin ^2\frac{x}{2}}$$
$$=-2\lim_{x\to 0}\frac{\frac{\sin x}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About Integration $\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx$ What i want to prove is following integral
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx=\frac{1}{2} \sqrt{\left(1+\sqrt{2}\right) \pi }
\end{align}
can you give some explicit method to obtain this result?
| Let your integral be labelled
$$I_1=\int_{-\infty}^{\infty} e^{-x^2}\cdot \cos(x^2)\space dx,$$
and a second integral
$$I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot\sin(x^2)\space dx.$$
It follows that
$$I_1-i\cdot I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot(\cos(x^2)-i\cdot \sin(x^2))\space dx.$$
Apply Euler's formula in c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How do I rationalize higher index roots? Often I've found exercises like
$$\frac{2-\sqrt[6]{3x+64}}{5x}$$
Where I need the rationalize. But I am not sure how to do it with indexes greater than $2$. How do I rationalize higher index roots? (the above one is just an example).
That is, perform
$$\frac{2-\sqrt[6]{3x+64}}{5... | (See my comments on original for links that explain this in detail... effectively, if we have a sum or difference involving an $n^{th}$ root we take either the sum of difference of $n^{th}$ roots and evaluate, using the given terms as the first factor and then plug in values for the remaining factor... it's messy, but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Are these boundaries true? I want to find the double integral
$$\iint xy \, dx\,dy $$
over the region
bounded by the positive $y$- axis, the line $y=\sqrt3 \ x$ and the circle
$x^2+y^2=4$.
My solution is that $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.
Is that correct?
|
My solution is that $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.
If you express it in that kind of language, it would be best to be explicit that you're putting the integral with respect to $y$ on the outside.
This would be
$$
\int_0^2 \left( \int_0^{y/\sqrt 3} xy \, dx \right) \, dy
$$
and that is not correct. I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Here is an algebraic approach.
$x^2 - 3x + 1 = 0$ implies $\dfrac1x = 3-x$ and so $x^4 + \dfrac{1}{x^4}=x^ 4+(3-x)^4$.
The extended Euclidean algorithm for $\gcd(x^ 4+(3-x)^4,x^2 - 3x + 1)$ gives
$$
47 = 1\cdot(x^ 4+(3-x)^4)+ (-2 x^2 + 6 x - 34)\cdot(x^2 - 3x + 1)
$$
and so $x^4 + \dfrac{1}{x^4}=47$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$
Evaluate the indefinite integral $$\int \frac{x-1}{9x^2-18x+17} \, dx .$$
This is an exercise from a book I'm studying. It gives the answer as:
$$\ln(9x^2 -18x+17)^\frac{1}{18} +C .$$
This is an easy integral. You just notice that the numerator... | $$\int \frac{x-1}{9x^2-18x+17}dx=\dfrac{1}{18}\log(9x^{2}-18x+17)+Cst$$
indeed,
*
*Substitute $u=9x^2-18x+17$ and $du=(18x-18)dx$
\begin{align}
\int \frac{x-1}{9x^2-18x+17}dx&=\dfrac{1}{18}\int \dfrac{1}{u}du\\
&=\dfrac{1}{18}\log(u)+Cst
\end{align}
*
*Substitute back for $u=9x^2-18x+17$:
\begin{align}
\int \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is... | $$\displaystyle \lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$$
$$=-2\cdot\dfrac1{\lim_{x\to1}\{1+\cos(x-1)\}}\cdot\left(\lim_{x\to1}\dfrac{\sin(x-1)}{x-1}\right)^2\cdot\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
Set $\sqrt x-1=y\implies x=(y+1)^2$ in
$$\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
to ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Solve $y\frac{dy}{dx}+x=\sqrt{x^2+y^2}$ using line integrals I'm aware of the fact that the ODE
\begin{align}y\frac{dy}{dx}+x=\sqrt{x^2+y^2},\end{align}
can be solved using substitution methods. I've worked out the solution quite easily by setting $u=\sqrt{x^2+y^2}$ and everything works out. But what if I rewrite the e... | Another simplest way :
From your given equation we have ,
$$2(y\,dy+x\,dx)=2\sqrt{x^2+y^2}\,dx$$
$$\implies \frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}=2\,dx$$
$$\implies 2\sqrt{x^2+y^2}=2x+2C \implies \sqrt{x^2+y^2}=x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What... | First, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.
And for all $y\neq \pi/4+k\pi/2$,
$$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$
Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:
$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$
And since $2\arctan z \in ]-\pi/2,\pi/2[$, you c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 5
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Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2... | To find this decomposition, we have to factorize $x^4 + 1$. As $x^4 + 1$ does not have real roots, it does not have real linear factors. This gives the ansatz
$$ x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d) $$
We have
$$
(x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (b + d + ac)x^2 + (bc + ad)x + bd $$
Comparing the coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels l... | I can obtain something a bit weaker. Since the square root is a concave fucntion, we have that
$$
1/3(\sqrt{A}+\sqrt{B}+\sqrt{C})\leq \sqrt{1/3 (A+B+C)}.
$$
This implies
$$
\frac{\sqrt{3}}{3}\left(\sqrt{A}+\sqrt{B}+\sqrt{C}\right)\leq \sqrt{A+B+C}.
$$
Now, take $A=a/(b+c)$, $B=b/(a+c)$ and $C=c/(b+a)$. Since $a,b,c>0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Limit Infinity Minus Infinity form without using infinite series Expansions Evaluate $$L=\lim_{x \to 0} \frac{1}{\sin^2x}-\frac{1}{\sinh^2x}$$ If we take L.C.M and use LHopital's Rule it becomes quite Tedious. I tried in this way.
$$
\sinh x=\frac{e^x-e^{-x}}{2}
$$
Now
$$\lim_{x \to 0}\frac{e^x-e^{-x}}{x}=2 \implies ... | For every $x$ with $0<|x|<\pi$ we have
$$
\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{x-\sin x}{x^2\sin x}\cdot\frac{x+\sin x}{\sin x}
$$
Using l'Hopital's rule we have:
\begin{eqnarray}
\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}&=&\lim_{x\to0}\frac{(x-\sin x)'}{(x^2\sin x)'}=\lim_{x\to0}\frac{1-\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.
I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity laws to solve this problem. But i get a fraction:
$8x^2... | Multiply both sides by $8$:
$$(8x-1)^2\equiv 25\pmod{75}$$
So $5\mid 8x-1$. I.e. $3x\equiv 1\equiv 6\pmod{5}\stackrel{:3}\iff x\equiv 2\pmod{5}$. Assuming this,
$$\iff \left(\frac{8x-1}{5}\right)^2\equiv 1\pmod{3}$$
$$\iff \frac{8x-1}{5}\equiv \pm 1\pmod{3}$$
$$\iff 8x-1\equiv \pm 5\equiv \pm 2\pmod{3}$$
$$\iff \left(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^... | $$x+y=35$$
$$y=35-x$$
Then plug this into the function you want to maximize to create a single variable function. So we want to maximize $$f(x)=x^2(35-x)^5.$$
To find all local extremes, we need to take the derivative, set it equal to $0$, and solve for $x$. Using the product rule, we get
$$f'(x)=2x\cdot (35-x)^5+x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Proof verification - Sum formulas in trig We have $a: \mathbb{R}^2 \to \mathbb{R}^2$ represent rotation about $(0,0)$ over an angle $\alpha$. We know that $a$ is a linear map and we know that $a$ corresponds to the matrix $A = \begin{pmatrix}
\cos(\alpha) & -\sin(\alpha) \\
\sin(\alpha) & \cos(\alpha) \\
\end{pma... | Rotating by $\alpha + \beta$ degrees can be thought of as first rotating by $\beta$ degrees, then rotating by $\alpha$ degrees. In other words, if $R_\theta:\Bbb R^2 \to \Bbb R^2$ is the rotation by angle $\theta$, then
$$
R_\beta (R_\alpha(x)) = R_{\alpha + \beta}(x)
$$
That is, the combined rotation is the compositi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show (with epsilón - delta) that $\lim_{x -> \infty}(\sqrt{n^2+n}-n)=\frac{1}{2}$ I need show that $\forall \epsilon > 0$ exist $M \in \mathbb{N}$ such that $|\sqrt{n^2+n}-n-\frac{1}{2}| < \epsilon$ for $n \geq M$.
I tried considering $|\frac{(\sqrt{n^2+n}-n)\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n)}-\frac{1}{2}|=|\frac{n-\sqrt... | Consider $\sqrt{n^2+n}-n-\frac{1}{2}$. Multiply top and missing bottom by
$\sqrt{n^2+n}+n+\frac{1}{2}$. We get
$\frac{-1/4}{\sqrt{n^2+n}+n+\frac{1}{2}}$.
The absolute value of this is less than $\frac{1}{8n}$. Now finding an appropriate $M$ is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Multiplication in GF(256) I would like to compute the following:
10100011 * 01100011 in GF(256) using the AES irreducible polynomial.
So first we get the polynomials:
10100011 = x^8 + x^6 + x + 1
01100011 = x^7 + x^6 + x^2 + 1
Multiplying these out, I got the result:
x^56 + x^48 + x^42 + x^36 + x^16 + x^1... | The exponentiation law is:
x^a * x^b = x^(a+b)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Equation - what first? I have this equation:
$$
(x+y)(x^x + y^y) = 2009.
$$
I must designate all pairs of integers satisfying the equation.
What first? I tried multiply brackets , but to no avail
| hint: $2009=7^2\times 41$ so keeping in mind that $x+y\lt x^x+y^y$ we're looking for integers $x$ and $y$ such that
$$(x+y)(x^x+y^y)=7^2\times 41$$
This means finding integers such that $x+y=41$ and $x^x+y^y=49$ or $x+y=7$ and $x^x+y^y=7\times 41$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Answer to some trigonometry series I have two series as bellow
$$\frac{2}{N}\sum_{n=0}^{N-1}\cos^2(\frac{2\pi n(l-q)}{N}) \quad l\neq q \quad(1)$$
$$ \sum_{q=0}^{N-1}cos(\frac{2\pi nq}{N})cos(\frac{2\pi rq}{N}) \quad n\neq r \quad (2)$$
Where $l$, $q$, $r$, $n$ and $N$ are integers. I have computed the two series... | For Reference, I am putting two formulas I've used in the solution of these questions :
(Sum of sines and cosines with angle are in A.P.)
$\displaystyle \sum_{k=0}^{n} \sin(k\beta) = \dfrac{\sin \dfrac{n\beta}{2}}{\sin \dfrac{\beta}{2}} \cdot \sin \left [ (n+1)\dfrac{\beta}{2} \right]$
$\displaystyle \sum_{k=0}^{n} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x... | As you've already noted, we can set $z^2 +z +1 = x^2 +x +1 = y^2 +x +1= \lambda^2$. We can do this because $x^2 +x+1>0 $ , $\forall x \in \mathbb{R}$.
Recall the law of cosines and observe that $z^2 +z +1=\lambda^2$ can be rewritten as $\lambda^2=z^2+ 1^2 -2 \cdot z \cdot 1 \cdot \cos 120^\circ $, which is the above l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically... | Using Prosthaphaeresis Formulas,
$$\sin2x+\sin4x-\sin6x=2\sin2x\cos2x-(2\sin2x\cos4x)$$
$$=2\sin2x(\cos2x-\cos4x)=4\sin2x\sin3x\sin x$$
Now from this, $\sin(2n+1)x=(2n+1)\sin x+\cdots+2^{2n}(-1)^n\sin^{2n+1}x$
If $\sin(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer
$x=\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l Problem :
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of $n$ such that $l$ is non zero finite real number. Find value of $l$.
My approach :
$$\lim_{x \to 0} \frac{x^n}{\cos(\s... | In order to simplify matters we are going to look at the reciprocal of the given expression. The denominator can be developed into a series as follows: From
$$\cos x=1-{1\over2}x^2+{1\over24}x^4+?x^6$$
and
$$\eqalign{\cos(\sin x)
&=1-{1\over2}x^2\left(1-{1\over 6}x^2+?x^4\right)^2+{1\over24}x^4(1+?x^2)^4+?x^6\cr
&=1-{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Factoring a polynomial (multivariable)
Factor $ (a - b)^3 + (b - c)^3 + (c-a)^3$ by SYMMETRY.
Okay, this is the problem. Let $f(a) = (a - b)^3 + (b-c)^3 + (c-a)^3$ obviously, if you let $a = b$ then, $f(b) = 0$, thus $(a - b)$ is a factor of $f(a)$. Then someone said :
If $(a - b)$ is a factor then $(b - c)$ and $(a... | The symmetry to use is not to change $f(a,b,c)$ to $f(b,a,c)$.
Instead, change $f(a,b,c)$ to $f(b,c,a)$ or $f(c,a,b)$.
This is the same as replacing $(a - b)^3 + (b - c)^3 + (c - a)^3$ by either
$(b - c)^3 + (c - a)^3 + (a - b)^3$ or $(c - a)^3 + (a - b)^3 + (b - c)^3$,
which clearly makes no change in the value.
So t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Problem related to integration I tried using partial fractions but I am not sure whether it is the right approach or not. I need help with this problem
Evaluation of Integral $$\int\frac{3x^4-1}{(x^4+x+1)^2}dx $$
| Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\left(x^3+1+x^{-1}\right)^2}dx = \int\frac{3x^2-x^{-2}}{(x^3+1+x^{-1})^2}dx$$
Now Put $x^3+1+x^{-1} = t\;,$ Then $(3x^2-x^{-2})dx = dt$
So Integral $$\displaystyle I = \int\frac{1}{t^2} dt = -\frac{1}{t}+\mathcal{C} = -\frac{1}{x^3+1+x^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Fourier sine series for $x^3$ It is asked to find the Fourier Sine Series for $x^3$ given that
$$\frac{x^2}{2} = \frac{l^2}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \cos\left(\frac{n \pi x}{l} \right)$$
integrating term by term. (This result was found in another exercise). As suggested, I integrat... | 1) You can simply conclude that $C=0$ as the equation must hold when $x=0$.
2) The answer of wolfram is different from you in appearance. The reason is that in the answer by wolfram, $x$ is also written in sin Fourier series and then combined with other terms in your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^... | The first two are equations of circles. Two circles may have no point in common, or only one tangency point, or two points of intersection or all points if they're actually the same circle.
It's easy to check this is not the first nor the last case. Now, subtract the first equation from the second one:
$
\begin{cases}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Given that $\pi=2\int_{-1}^1\sqrt{1-x^2}dx$ use the properties of the integral to compute $\int_{-2}^{2}(x-3)\sqrt{4-x^2}dx$ in terms of $\pi$. This is part c of the exercise set 2.4 in Apostol's Calculus 2nd Ed. We are only supposed to use the properties of the integral such as: linearity, additivity with respect to t... | By linearity
$$\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx = \int_{-2}^{2} x \sqrt{4-x^2}dx - 3 \int_{-2}^{2} \sqrt{4-x^2}dx$$
But $\int_{-b}^b f(x) dx = 0$ for odd $f(x)$ (Optional Exercise 25), and $f(x) = x \sqrt{4-x^2}$ is odd.
\begin{align}
\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx &= - 3 \int_{-2}^{2} \sqrt{4-x^2}dx\\
&= -6 \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Deriving Frobenius number for two denominations? I would like to know if the following question has an intelligent solution:
Determine the largest bet that cannot be made using chips of $7$ and $9$ dollars.
After not being able to solve it I found a solution online which writes out all combinations of $7$ and $9$ up to... | A very nice explanation is given at Cut the knot.
You want the number for $p, q$ with $\gcd(p, q) = 1$ (otherwise it makes no sense). Consider the $q$ sequences:
$\begin{align}
&0 + 0, 0 + q, 0 + 2 q, \dotsc \\
&p + 0, p + q, p + 2 q, \dotsc \\
&\vdots \\
&(q - 1) p + 0, (q - 1) p + q, \dotsc
\end{align}$
T... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$ $$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$
I have simplified above to:
$$\frac{3-\sqrt[3]{3}}{(\sqrt[3]{3}+1)(\sqrt[3]{3}-1)}$$
What is equal to:
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}$$
WolframAlpha says this can be shown as... | $$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}=\frac{(3^{1/3})^3-3^{1/3}}{3^{2/3}-1}=\frac{3^{1/3}(3^{2/3}-1)}{3^{2/3}-1}=3^{1/3}(=\sqrt[3]{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplify $(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$? In calculating the length of a deltoid one gets the following string of trigonometric functions:
$$(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$$
$$=4\big(\sin^2(2t)+2\sin(t)\sin(2t)+\cos^2(2t)-2\cos(t)\cos(2t)+\sin^2(t)+\cos^2(t)\big)$$
$$(\sin^2x+\cos^2x=1)... | Notice,
$$4(1+2\sin t\sin 2t-2\cos t\cos 2t+1)$$
$$=8(1+\sin t\sin 2t-\cos t\cos 2t)$$
$$=8(1-(\cos t\cos 2t-\sin t\sin 2t)$$
using $\color{red}{\cos A\cos B-\sin A\sin B=\cos (A+B)}$,
$$=8(1-\cos(t+2t))$$ $$=8(1-\cos 3t)$$
using $\color{red}{\cos A=1-2\sin^2\frac{A}{2}}$,
$$=8\left(1-\left(1-2\sin^2 \frac{3t}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
Th... | Clearly $(x-y)^2 \geq 0$
So $x^2-2xy+y^2 \geq 0 \Rightarrow x^2+y^2 \geq 2xy $
Since $x $ and $y$ are positive integers , $x \neq 0 $ and $y \neq 0$.
Thus we can divide by $xy$.
So we have $\frac{x^2+y^2}{xy} \geq \frac {2xy}{xy} \Rightarrow \frac {x}{y}+\frac{y}{x} \geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 10,
"answer_id": 0
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What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| Clearly
$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$
$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$
Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,-\frac{3}{4} ,1,3$.
Then check the validity for some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Given integer roots of $x^2+mx-n=0$ and $x^2-mx+n=0$. Show $6 \mid n$. Suppose that $m$ and $n$ are integers such that both the quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
I figured out the roots of the equations through quadratic formula-for the first one i... | Claim 1:
Suppose both $m$ and $n$ were odd, then both $x^2+mx-n=0$ and $x^2-mx+n=0$ cannot have any integer solutions.
Reason:
From the first equation we have $n=x(x+m)$. Thus if $x$ is even, then $n$ must be even and if $x$ is odd then $x+m$ is even, either way $n$ would be even which contradicts the assumption that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving linear system equation How would I solve this linear system equation?
$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$
First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.
| Write your system into normal form:
\begin{cases}
x-y+2w=4\\
x-3z=-6\\
x+2y-9z-4w=-7
\end{cases}
Now, depending on the tools you have available, there are several possibilities. The most efficient is Gaussian elimination: the matrix of the system is
\begin{align}
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Olympiad problem about finding minimum value with $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$ Let $x,y,z$ be positive real numbers such that $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$.
Find the minimum value of
$$\frac{x^2y^2} {z^3(x^2+y^2)}+\frac {y^2z^2} {x^3(y^2+z^2)}+\frac {z^2x^2} {y^3(z^2+x^2)}$$
I'm pretty sure that the answer... | If $x=y=z=\sqrt3$ then we get a value $\frac{\sqrt3}{2}$.
We'll prove that it's a minimal value, for which it's enough to prove that
$$\sum_{cyc}\frac{y^2z^2}{x^3(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{\frac{x^2y^2+x^2z^2+y^2z^2}{x^2y^2z^2}}$$ or
$$\sum_{cyc}\frac{y^3z^3}{x^2(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove this inequality? $\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2$? How can I prove this inequality?
$$\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2.$$
| Summation by parts is enough. We have:
$$ \sum_{k=1}^{n-1} k\log k = \frac{n^2-n}{2}\cdot\log n-\sum_{k=1}^{n-2}\frac{k^2-k}{2}\cdot\log\left(1+\frac{1}{k}\right)$$
but $k\mapsto k\cdot \log\left(1+\frac{1}{k}\right)$ is an increasing function, hence:
$$ \sum_{k=2}^{n-1} k\log k \leq \frac{n^2-n}{2}\cdot \log n- \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even? How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even?
Soln: The way I tried approaching this problem was, I made a grid of $7 \times 7$ squares. So $49$ ordered pairs in ... | Your table is
$\begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{sum } & 1 &2 &3 &4 &5 &6 & 7\\ \hline \hline \hline 1 &\color{red}2 &3 &\color{red}4 &5 &\color{red}6 &7&\color{red}8 \\ \hline 2 & 3 &\color{red}4 &5 &\color{red}6 &7&\color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $p$ and $q = 2p + 1$ be odd primes. Show that $(−1) ^{\frac{p−1}{ 2}} 2$ is a primitive root modulo q. Let $p$ and $q = 2p + 1$ be odd primes. Show that $(−1) ^{\frac{p−1}{ 2}} 2$ is a primitive root modulo q.
We see that our number must have order $q-1$ to be a primitive root. If we check the divisors which are $1... | I.e. we want to prove $\left((-1)^{\frac{p-1}{2}}2\right)^{d}\not\equiv 1\pmod{q}$ for all $d\in\{1,2,p\}$.
If $(-1)^{\frac{p-1}{2}}2\equiv 1\pmod{q}$, then $\pm 2\equiv 1\pmod{q}$, so $q=3$, which is not of the form $2p+1$.
If $\left((-1)^{\frac{p-1}{2}}2\right)^2\equiv 1\pmod{q}$, then $4\equiv 1\pmod{q}$, i.e. $q=... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$. Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$
for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.
I tried many different ways to expand left side and estimate it but always got stu... | I will show that
$|\sqrt{n}-\frac{x}{y}|
>\frac1{(\sqrt{n}+\sqrt{n+1})y^2}
$.
For $n=2$,
$\sqrt{2}+\sqrt{3}
< 3.15
$,
so
$|\sqrt{2}-\frac{x}{y}|
>\frac1{3.15 y^2}
$.
I also show that if
$|\sqrt{n}-\frac{x}{y}|
<\frac1{(2\sqrt{n}+\epsilon)y^2}
$.
then
$y
<\sqrt{ \frac1{2\epsilon\sqrt{n}}}
$.
In general,
if $n$ is not a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$
Hi i was doing this question but i'm not sure i was right. Does this make sens... | No, this is not correct. It looks like you assumed that the first function described in the problem was the objective function (that which is to be optimized) and the second function is the constraint function.
But if you read the problem, you see that both $x^2 + y^2 = 1$ and $x+y+z=1$ are constraints. So you need a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Equation of a circle in matrix form I have an equation $ \left( x-3 \right)^{2}+\left( y-3 \right)^{2}=9 $, and am trying to apply a matrix rotation of 180 degrees to it, however, I am having difficulty transferring the equation of the circle into matrix form so to complete the transformation.
Thanks
| The equation $(x-3)^2+(y-3)^2=9$ is can be described using matrix and vector $[x \ y \ z]^{\text{T}}$ as follows. Then $z=1$.
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
The rotation matrix is
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculate area by using double integral Area $D$ confined by lines:
$x=\sqrt{4-{y}^{2}},\: y=\sqrt{3x}, \: x\geq 0$
Need to calculate $I = \iint\limits_D \, dx\,dy$
My steps:
*
*Draw $D$:
*Set boundaries:
$I = {I}_{1} + {I}_{2}$
${I}_{2} = \frac{1}{4}\left(\pi {r}^{2} \right)=\pi$
${I}_{1} = \int_{0}^{\sqrt{... | You selected the wrong area. You want the bit bound by all 3 lines mentioned.
You can get that by doing $2\pi$ take away your answer. Or recalculate the easier integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why two Inequalities are true $A$ is a subset of real numbers.
Consider the set $A$ of all real
numbers $x$ such that $x^2 \le 2$.
This set is nonempty and bounded from above, for example by 2. Call the $\sup(A),\; y$. Then $y^2 = 2$, because the other possibilities
$y^2 \lt 2$ and $y^2 \gt 2$ both lead to a c... | In (1), the equality part is easy and the second inequality is just an assumption. So, you only need to prove that $y^2-2zy+z^2 > y^2-4z$ which is equivalent to $z^2-2zy+4z > 0$ and, since $z>0$, this is equivalent to $z-2y+4 > 0$, or, in other words, $z > 2y-4$. So, you only need to prove $z > 2y-4$. Now, since $y$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Forgot my basic math... So a younger (college student) asked me for my help to solve a basic math question, and to my surprise, I've forgot some basic math rules, rendering me unable to answer the problem.
According to online math-generators the answer should'nt be what I'm getting.. And I am unable to follow the auto-... | Small mistake from step 1 to step 2. Notice that:
$$ \left(\frac{1}{2}\right) \times \left(\frac{2}{2a}\right) = \left( \frac{1\times 2}{2\times 2a} \right) = \left(\frac{2}{4a} \right) = \left( \frac{1}{2a} \right).$$
You will then end up with
$$ \left( \frac{1}{2a} \right) + \frac{\left(\frac{2a}{6}\right)}{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Applying Euler's substitution to evaluate the integral? $$\int \frac{dx}{\sqrt{x^2+a^2}} $$
let $$\sqrt{x^2+a^2}=x+t $$
$$a^2=2xt+t^2 $$
$$x=\frac{1}{2}\Big( \frac{a^2}{t}-t\Big) $$
$$dx=-\frac{1}{2}\Big(\frac{a^2}{t^2}+1\Big)dt$$
hence $$x+t=\frac{a^2+t^2}{2t} $$
this gives $$\int \frac{dx}{\sqrt{x^2+a^2}}=-\ln({\sqrt... | Hint
You are not wrong and you did a good job but, may be, you just forgot the integration constant at the end of the calculation.
Just rewriting your last expression
$$A=-\log\left({\sqrt{x^2+a^2}-x}\right)=\log\left(\frac 1 {\sqrt{x^2+a^2}-x}\right)$$ $$A=\log\left(\frac 1 {\sqrt{x^2+a^2}-x}\times\frac{\sqrt{x^2+a^2... | {
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"question_score": "1",
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Show that the sequence is null? Show that $\frac{14n + 26}{49n^3 +14}$ is a null sequence.
So we need to show that: $\forall \epsilon > 0, \exists X \in \mathbb{R}, \forall n > X,|\frac{14n + 26}{49n^3 +14}| < \epsilon $.
$|\frac{14n + 26}{49n^3 +14}|= \frac{14n + 26}{49n^3 +14} < \frac{14n+26}{n}$ (since $n> 49n^3 +1... | You have made the bound too coarse.
For all $n \geq 1$ we have
$$
\frac{14n + 26}{49n^{2}+14} < \frac{14n + 26}{n^{2}} = \frac{14}{n} + \frac{26}{n^{2}};
$$
given any $\varepsilon > 0$, we have $\frac{14}{n} < \frac{\varepsilon}{2}$ if $n > \frac{28}{\varepsilon}$ and we have $\frac{26}{n^{2}} < \frac{\varepsilon}{2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A good way to solve this trigonometric equation $$\sin x+\cos x=\frac{1}{2}$$
What is the value of $\tan x$? I tried using
$$\sin2 x=\frac{2\tan x}{1+\tan^2x}$$ and $$\cos2 x=\frac{1-\tan^2x}{1+\tan^2x}$$ but we get a quadratic for $\tan\left(\frac{x}{2}\right)$ . So any better approach would be much appreciated. Than... | Given $$\sin x+\cos x=\frac{1}{2}$$
squaring both the sides, $$(\sin x+\cos x)^2=\left(\frac{1}{2}\right)^2$$
$$\sin^2 x+\cos^2 x+2\sin x\cos x=\frac{1}{4}$$
$$1+\sin 2x=\frac{1}{4}$$$$\sin 2x=-\frac{3}{4}$$
$$\frac{2\tan x}{1+\tan ^2x }=-\frac{3}{4}$$
$$3\tan^2 x+8\tan x+3=0$$
let $\tan x=t$ $$3t^2+8t+3=0$$
I hope you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
| If you have calculus, you can conclude this is impossible as follows, since $lim_{x, y \to \infty}x^2+xy+y^2=\infty$, there is a point where the minima is achieved. This point has vanishing partial derivatives, thus $\partial_x(x^2+xy+y^2)=2x+y=0$ and likewise $2y+x=0$, so $x=-2y=4x$ so $x=0$ and likewise $y=0$, so $x=... | {
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"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating m... | Set $u=\cfrac{1}{4x}, du=\cfrac{-1}{4x^2}$. Substitute:
$$\int \sqrt{1+u}(-4x^2)du$$
Since $4x=\cfrac{1}{u}, x=\cfrac{1}{4u}$,
$$=\int \cfrac{-1}{4u^2}\sqrt{1+u}du$$
Integrate by parts: $s=\sqrt{1+u}, ds=\cfrac{1}{2\sqrt{1+u}}du, dt=\cfrac{1}{u^2}du, t=\cfrac{-1}{u}$,
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Upper Triangularising a Matrix Kay and Wilson Linear Algebra I have been reading Linear Algebra by Richard Kay and Robert Wilson and am specifically looking at pages 156-158.
I understand the book's proof on page 156 of the proposition 10.10:
"Let $V= \mathbb{C}^n $ be the $n$ dimensional vector space over $\mathbb{C}$... | Let's see why the procedure work. Let's denote the usual standard basis of $\mathbb{R}^3$ by $\{e_1,e_2,e_3\}$. Then we have
\begin{align}
(A-4I)e_1&= f_1\\ (A-4I)e_3 &= f_2
\end{align}
So, $A(f_1) \in Range(A-4I) = Span\{ f_1, f_2\} $ and similarly $Af_2\in Range(A-4I) = Span\{ f_1, f_2\} $.
Now extend $Span\{ f_1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Can someone explain the integration of $\sqrt{v²+\tfrac14}$ to me? I am currently trying to integrate this root:
$$\sqrt{v^2+\frac{1}{4}}$$
According to several integration calculators on the web it is:
$$\frac{\operatorname{arsinh}(2v)}{8} +\frac{v\sqrt{v^2+\tfrac{1}{4}}}{2}$$
However, I just can't get my head around ... | Since $\sqrt{v^2+\frac{1}{4}}$ is of the form $\sqrt{x^2+a^2}$, we can do a trigonometric substitution using $\tan\theta$.
Let $v=\frac{1}{2}\tan u$, $dv=\frac{1}{2}\sec^2(u) du$
$$\int \sqrt {v^2+\frac{1}{4}}dv=\int \sqrt {\frac{1}{4}\tan^2 u+\frac{1}{4}}\cdot\frac{1}{2}\sec ^2 (u)du$$
Using the trig identity $1+\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Geometry question on square $ABCD$ is a square and $AB$ = 1. Equilateral triangles $AYB$ and $CXD$ are drawn such that $X$ and $Y$ are inside the square.How can I find the length of $XY$ ?
|
You can use also this approach if you like,also if, as already said, you can really solve it by simmetry arguments.
Area of $\Delta DPR = \cfrac {DP ^2 \cdot \sin P \cdot \sin D}{2 \sin R}=\cfrac {1}{4} \cdot \cfrac{ \sin 90 \cdot \sin 30 }{2 \sin60} =\cfrac { \sqrt {3}}{24}$.
( ----> Area of any triangle $\Delta ABC... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\... | Hints:
*
*For $0 \lt k \lt 1$ and $0 \lt x \lt 1$ you have $x \lt x^k \lt 1$
*$ \displaystyle \lim_{n\to\infty} \bigg(\dfrac{n^2}{n^2 + 1} \bigg) = \lim_{n\to\infty} \bigg(1-\dfrac{1}{n^2 + 1} \bigg) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Frobenius Method Indicial Equation I need to verify that the indicial equation only has one root.
$xy''+(1-x)y'+\frac{1}{2}y=0$
Attempt:
$y=\sum\limits_{m=0}^\infty {a_mx}^{m+r}$
$y'=\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}$
$y'=\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}$
Substitute:
$x\sum\limits_{... | The indicial must be at $x=0$ in this case. Dividing by the coefficient of the highest order derivative gives
$$
y'' + \left(\frac{1}{x}+1\right)y'+\frac{1}{2x}y = 0.
$$
The indicial equation is found by keeping the $\frac{1}{x}y'$ terms and $\frac{1}{x^{2}}y$ terms:
$$
y''+\frac{1}{x}y' = 0.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
| So i found the solution using the fact that Sin is odd, here is my solution:
$\sin\left (x-\frac{\pi}{6}\right ) + \cos\left (x+\frac{\pi}{4}\right ) = 0\Leftrightarrow \cos\left (x+\frac{\pi}{4}\right ) = -\sin\left (x-\frac{\pi}{6}\right )$
Since sin is an odd function, $-\sin\left (x-\frac{\pi}{6}\right ) = \sin\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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