Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute integral $\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z$ How to Compute
$$\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z $$
The Mathematica 9.01 give a result is $$-\frac{\pi}{24}+\frac{1}{4}(\sqrt{3}+\log(7+4\sqrt{3})).$$ I want to know how to... | You are computing the average distance from a vertex in a unit hypercube. Assuming that $X,Y,Z$ are three independent random variables, uniformly distributed over $(0,1)$, and computing the probability density function of $W=X^2+Y^2+Z^2$ (that is a piecewise-simple function), it follows that
$$\mathbb{E}[\sqrt{W}] = \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve $x^{3}-3x=\sqrt{x+2}$ Solve for real $x$
$$x^{3}-3x=\sqrt{x+2}$$
By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe... | You shouldn't dismiss Karo's graph. Drawing a graph to get a feel for the equation (when you don't know how to proceed) is most helpful. And this isn't nonsense, in fact the graph is the key for the solution, as it makes it clear that we can consider only $x$ for which $-2 \leq x \leq 2$. This can be conveniently rewri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Get variables with Matrix I try to get the variables for this equation:
$$\begin{cases}
6x_1 + 4x_2 + 8x_3 + 17x_4 &= -20\\
3x_1 + 2x_2 + 5x_3 + 8x_4 &= -8\\
3x_1 + 2x_2 + 7x_3 + 7x_4 &= -4\\
0x_1 + 0x_2 + 2x_3 -1x_4 &= 4
\end{cases}$$
So i started with:
$$ \begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
... | As Git Gud points out, it should be 1, not -1. Now if you correct that and proceed further, the third row also will become zero, and you'll get
$\left(\begin{matrix}
6 & 4 & 8 & 17 & -20\\
0 & 0 & -2 & 1 & -4\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{matrix}\right)$
so that
$6x_1 + 4x_2 + 8x_3 + 17x_4 = -20\\
-2x_3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
| $$\frac1{a^2}+\frac1{b^2}=4$$
$$a^2b^2\left(\frac1{a^2}+\frac1{b^2} \right)=4a^2b^2$$
$$b^2+a^2=(2ab)^2$$
$$a^2-2ab+b^2=(2ab)^2-2ab$$
$$(a-b)^2=(2ab)^2-2ab=(2\cdot3)^2-2\cdot3=30$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
prove that $ a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right) $ good evening I want to show that
$(1)a;b\in\mathbb {R^*_+}:a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right)
$
$
\begin{array}{l}
\frac{a}{b} + \frac{b}{a} \ge 2 \\
\frac{... | Do you keep the first true in the case of
$a;b\in\mathbb {R^*_+} and : a^2 + b^2 \ge 2$
$ (2)x;y;z\in\mathbb {R^*_+}:$
$\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
Thank you for all helping and guidance. ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why does $u$-substitution not work here? $$ \int{\frac{1}{2y}dy} $$
Method 1:
$$\int{\frac{1}{2y}dy} = \frac{1}{2}\int{\frac{1}{y}dy} = \frac{1}{2}\ln|y|+C$$
Method 2 ($u$-substitution):
$$\int{\frac{1}{2y}dy} = \int{\frac{1}{u}dy} = \frac{1}{2}\int{\frac{1}{u}(2)dy}= \frac{1}{2}\int{\frac{1}{u}du}= \frac{1}{2}\ln|u|... | $$\ln |2y| = \ln 2 + \ln |y|$$
$$ \frac{1}{2}\ln|2y|+C = \frac 12 \ln |y| + \underbrace{\frac 12 \ln 2 + C}_{\large =\, C'}$$
So your results are equivalent up to a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to find integral $\underbrace{\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$ Find the integral
$$\int\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$$
where $n$ define the number of the square
I know this if
$0 \le x\le 2$, then let $$x=2\cos{t},0\le t\le\dfrac{\pi}{2}$$
so... | I think you have the wrong approach, take a look at this
$$f(x) = \sqrt{2+\sqrt{2+...+\sqrt{2+x}}} $$
Square both sides
$$(f(x))^2 = 2+\sqrt{2+...+\sqrt{2+x}} $$
$$(f(x))^2 = 2+f(x) $$
$$(f(x))^2 -f(x)-2= 0 $$
This has two solutions $f(x)=2$ and $f(x)=-1$
Note that $f(x)$ cannot be -1, since square root is always p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $ \int ^2 _1 \frac{\ln x}{x^2} \text{dx}$ $\displaystyle \int ^2 _1 \dfrac{\ln x}{x^2} \text{dx}$
My working so far:
$u=\ln x, v'=x^2$
$u'=x^{-1}, v=\dfrac{1}{3}x^3$
Integrating by parts:
$\displaystyle \dfrac{1}{3}x^3\ln x-\int\dfrac{1}{3}x^3\cdot x^{-1}$
$\displaystyle \left[\dfrac{1}{3}x^3\ln x-\int\dfrac{1... | You should have $v'=x^{-2}$, so $v=-x^{-1}$, now go from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is $f '(x)$ for $f(x)=(x-3)^3$?
What is $f '(x)$ for $f(x)=(x-3)^3$?
I'm thinking it is $3x^2 - 18x + 27$
but my textbook says it is $3x^2 - 18x - 27$
| Note that
$$
f(x)=U^n\\
f^{\prime}(x)=nU^{\prime}U^{n-1}
$$
Then
$$\begin{align}
f^{\prime}(x)&=3(x-3)^{2}\\
&=3(x-3)(x-3)\\
&=3(x(x-3)-3(x-3))\\
&=3(x^2-3x-3x+9)\\
&=3(x^2-6x+9)\\
&=3x^2-18x+27\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$
Let $n>2$ an integer and $x_k>0$ with $x_1\cdot x_2\cdots x_n=1$
Show that $$\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$$
I tried an induction without succeed, I do not really have idea to approach this inequality.... | it is true that $n\ge 6$ the inequality is always true because
$ \frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$ is always true when $x_k \ge 1$, proof:
$\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1 \iff x_k \ge \dfrac{1}{2\cos(\frac{2\pi}{n})} \iff 1 \ge \dfrac{1}{2\cos{(\frac{2\pi}{n})}} \iff \cos{(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
linear recursion $y_n=A \cdot y_{n-1}$ Let $a,b, \in \mathbb{R}$. Let $x_0=a, x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$ for $n \geq 2$
(i) Write the recursion in the form $y_n=A \cdot y_{n-1}$ where $A$ is a $2 \times 2$ matrix and $y_i=\begin {pmatrix} x_i \\ x_{i-1} \end{pmatrix}$
(ii) Find a diagonal matrix $D$ a... | Because $\lim_{n \rightarrow \infty}(-\frac{1}{2})^n=0$, we have
$$\lim_{n \rightarrow \infty}D^n= \lim_{n \rightarrow \infty} \begin {pmatrix} 1 & 0 \\ 0 & (-\frac{1}{2})^n \end{pmatrix}=\begin {pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$
To simplify the notation, we Define $X^\infty:=\lim_{n \rightarrow \infty}X^n$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}= $? How to calculate the following limit : $$\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}$$
Thanks in advance.
| Notice that by the Taylor series we have
$$a^x=\exp(x\ln a)=_0 1+x\ln a+\frac{\ln^2 a}2x^2+o(x^2)$$
and
$$x\tan x= x^2+o(x^2)$$
so we find easily that
$$\lim_{x\to0}\frac{10^x-2^x-5^x+1}{x\tan x}=\frac{\ln^2(10)-\ln^22-\ln^25}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Area of the intersection of two discs: Integral solution? Here is the problem :
We consider two circles that intersect in exactly two points. There $O_1$ the center of the first and $r_1$ its radius. There $O_2$ the center of the second and $r_2$ its radius. We note $d=O_1O_2$. Question: Express the area of the in... | I derived my own proof with simple geometry and trigonometry. I ended up with a slightly different formula:
$$
A=r_1^2 \text{arccos}\left(\frac{d^2-(r_2^2-r_1^2)}{2dr_1}\right) + r_2^2 \text{arccos}\left(\frac{d^2+(r_2^2-r_1^2)}{2dr_2}\right)-d\sqrt{r_1^2-\left(\frac{d^2-(r_2^2-r_1^2)}{2d}\right)^2}.
$$
The last term c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Equations of lines tangent to an ellipse Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find th... | You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point. The vertical line does that and you've already found it. Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).
Nonvertical lines passing through ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Find the value of a + b + c + d Let $a$ and $b$ be the roots of the equation: $x^2 - 10cx - 11d = 0$ where $c$ and $d$ be the roots of $x^2 - 10ax - 11b = 0$. Find the value of $a+b+c+d$, assuming that they all are distinct.
I first tried making an equation with roots $(a+b)$ and $(c+d)$ to get the sum of the roots, ho... | We have
$a+b=10c$
and
$c+d=10a$, hence
$$b=10c-a$$
and
$$d=10a-c$$
We also have $ab=-11d$ and $cd=-11b$, hence
$$10ac-a^2=-11(10a-c)$$
and
$$10ac-c^2=-11(10c-a)$$
Subtracting these gives
$$a^2-c^2=-11(10c-a)+11(10a-c)=121(a-c)$$
Assuming $a\not=c$, we get
$$a+c=121$$
Going back to the formulas for $b$ and $d$, we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what ... | $\cos 2\theta = 2\cos^2 \theta - 1 \Rightarrow 1 + \cos 2\theta = 2\cos^2 \theta$. Therefore:
$$\dfrac{\sin 2x}{1 + \cos 2x} \times \dfrac{\cos x}{1 + \cos x}\\
= \dfrac{\not 2\sin x \not\cos x}{\not 2 \not\cos^2 x} \times \dfrac{\not\cos x}{2 \cos^2 (x/2)}\\
= \dfrac{\not 2\sin(x/2)\not\cos(x/2)}{\not 2\cos^{\not 2} (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Asymptotic value of Fibonacci numbers It is well known that $F_n\sim\frac{\phi^n}{\sqrt{5}}$, where $\phi=\frac{1+\sqrt{5}}{2}$. Does someone know a better estimate? With proof please.
I'm trying to calculate the following limit:
Let $u_1=1, u_2=C, u_3=C$ and
$$
u_n=\frac{C^{F_{n-1}}}{2^{F_{n-3}}3^{F_{n-4}}\cdots(n-2)... | If we let
$$
a_n = 2^{\large F_{n-3}} 3^{\large F_{n-4}} \cdots (n-2)^{\large F_{1}} = \prod_{k=2}^{n-2} k^{\large F_{n-1-k}}
$$
then
$$
\log a_n = \sum_{k=2}^{n-2} F_{n-1-k} \log k. \tag{1}
$$
Using Binet's formula
$$
F_m = \frac{\varphi^m - (-\varphi)^{-m}}{\sqrt{5}},
$$
$(1)$ becomes
$$
\begin{align}
\sqrt{5} \log a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$. I have problem with this equation:
$$x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$$
Any ideas on beatiful solving?
| $$
\begin{array}{l l l}
x-\sqrt{(x^2-36)}& =& {(x-6)^2\over 2x+12}\\
(2x+12)\left(x-\sqrt{(x^2-36)}\right)&=&(x-6)^2\\
12x+2x^2+(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2\\
(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2-12x-2x^2\\
(-12-2x)^2(x^2-36)&=&\left((x-6)^2-12x-2x^2\right)^2\\
4x^4+48x^3-1728x-5184&=&x^4+48x^3+504x^2-1728x+1296\\
3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Product in terms of $n$ of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$ What is the following product in terms of $n$?
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$
Thank you.
| As an alternative answer that has its uses, this product of fractions can be written as a definite integral:
$$\begin{align}
\frac12\frac34\frac56\frac78\dots\frac{2n-1}{2n}&=\frac{(2n-1)!!}{(2n)!!}\\
&=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2n}{x}\,\mathrm{d}x
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Square roots modulo powers of $2$ Experimentally, it seems like every $a\equiv1 \pmod 8$ has $4$ square roots mod $2^n$ for all $n \ge 3$ (i.e. solutions to $x^2\equiv a \pmod {2^n}$)
Is this true? If so, how can I prove it? If not, is it at least true that the maximum (over $a$) number of square roots of $a$ mod $2^n... | When $n\ge 3$, the number of solutions of $x^2\equiv 1\pmod{2^n}$ is $4$. The solutions are $x=\pm 1\pmod{2^n}$ and $x\equiv \pm 1+2^{n-1}\pmod{2^n}$.
Proof: We want $x^2-1\equiv 0\pmod{2^n}$, that is, $(x-1)(x+1)\equiv 0\pmod{2^n}$. Since $x$ must be odd, the gcd of $x-1$ and $x+1$ is $2$. Either all of the $2$'s come... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3 - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$ and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
| Someone will probably show you a smart way to do this, but ...I guess you could just use that $\alpha^2 = 1 - 2\alpha$ and then
$$\begin{align}
4(4\alpha^3 - 3\alpha)^2 + &2(4\alpha^3 - 3\alpha) - 1 \\ &= 4(4\alpha^6 + 9\alpha^2 - 24\alpha^4) + 8\alpha^3 - 6\alpha - 1 \\
&= 16\alpha^6 + 36\alpha^2 - 96\alpha^4 + 8\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/846203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that $\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$ The following question came up in chat
Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that
$\displaystyle\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$
Now the integral c... | Anticlimax ahead:
$$\sum_{k=0}^\infty \frac{1}{4^k(k!)^2} < \sum_{k=0}^\infty \frac{1}{4^k\,k!}$$
follows since $4^k(k!)^2 \geqslant 4^k\, k!$ for all $k$ and the inequality is strict for $k\geqslant 2$.
For the inequality
$$\sum_{k=0}^\infty \frac{1}{5^k\,k!} < \sum_{k=0}^\infty \frac{1}{4^k(k!)^2},$$
we note that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Matrices as linear transformations I am reading a proof which claims:
A matrix of $m\times n$ is a linear transformation from $m$
vector-space to $n$ vector-space, And therefore, by the dimension theorem: $m = \dim\ker A + \dim\Im A$
Isn't it the opposite?
For example: $A\in M_{4\times 2}(\mathbb{R})$ is a linea... | Let us take an example:
$$ \begin{pmatrix} \color{blue}1& \color{red}2 & \color{green}3 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} \color{blue}{10} \\ \color{red}{20} \\ \color{green}{30}\end{pmatrix} = \begin{pmatrix}\color{blue}1 \cdot \color{blue}{10} + \color{red}2\cdot \color{red}{20} +\color{green}3 \cdot \color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My m... | $$ \frac{dC}{dt}=\frac{dA}{dt}\frac{dC}{dA} $$
$$C=2\pi r= 2\frac{\pi r^2}{r}=2\frac{A}{r} $$ (since $A=\pi r^2 $) and
$$A=\pi r^2 \implies r=\sqrt{\frac{A}{\pi}}$$
Therefore using the previous 2 results $$C=2 \frac{A}{\sqrt{\frac{A}{\pi}}}=2\frac{A}{\sqrt{\frac{A}{\pi}}}\times \frac{\sqrt{\pi}}{\sqrt{\pi}}=2\frac{A\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all values that make the expression a perfect cube Find all the positive integers $n$ such that $n^3-n$ is a perfect cube.
| If $n > 1$, then $3n^2-4n+1 = (3n-1)(n-1) > 0$.
Hence, $(n-1)^3 = n^3-3n^2+3n-1 < n^3-n < n^3$.
Since $n^3-n$ is strictly between two consecutive perfect cubes, $n^3-n$ is not a perfect cube.
The only remaining positive integer is $n = 1$, which yields $n^3-n = 0 = 0^3$.
Therefore, the only positive integer $n$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1... | You're almost there.
$$\frac{1}{m^2 + (N-m)^2} = \frac{1}{2m^2 +N^2 -2mN} = \frac{1}{2m(m-N) +N^2}\ge \frac{1}{N^2}$$
Now $$\sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{m^2 + (N-m)^2} \ge \sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{N^2} = \sum_{N=1}^{\infty}\frac{N-1}{N^2}$$
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Determinant involving recurrence Evaluate
$$\left| A \right| = \left| {\matrix{
{x + y} & {xy} & 0 & \cdots & \cdots & 0 \cr
1 & {x + y} & {xy} & \cdots & \cdots & 0 \cr
0 & 1 & {x + y} & \cdots & \cdots & 0 \cr
\cdots & \cdots & \cdots & \cdots & \cdots & \vdots \cr
0 & ... | The problem is that the recurrence formula should be
$$D_n=(x+y)D_{n-1}-xyD_{n-2}$$ then
$$D_n=\frac{x^n-y^n}{x-y}(x+y)-xy\frac{x^{n-1}-y^{n-1}}{x-y}=
\frac{(x^n-y^n)(x+y)-xy(x^{n-1}-y^{n-1})}{x-y}=
\frac{x^{n+1}-y^{n+1}+x^ny -xy^n-x^{n}y+xy^{n}}{x-y}=
\frac{x^{n+1}-y^{n+1}}{x-y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that a specific $w$ cannot be the root of an quadratic with integer coefficients. Let $w$ be the only real root of $x^3-x-1=0$. Show that
$w$ cannot satisfy the quadratic $ax^2 + bx + c$ ,where $a,b,c\in \Bbb Z$.
I have written
$$w^3=w+1$$
but I can't go any further than this. Thank you.
| Note that, by Euclidien division
$$x^3-x-1=Q(x)(ax^2+bx+c)+x \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1$$
So, if $w$ is a root of both $x^3-x-1=0$ and $ax^2+bx+c=0~$, then we would have
$$
w \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1=0
$$
So if $a^2+a c-b^2\ne0~$ then
$$
w=\frac{b c-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$13\mid4^{2n+1}+3^{n+2}$ How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?
| You can use congruence:
$$
4^{2n+1}+3^{n+2}\equiv 4(16^n)+9(3^n) \equiv 4(16^n)+(13-4)(3^n) \equiv 4(16^n-3^n) \equiv 4(3^n-3^n) \equiv 0\ (\text{mod }13).
$$
Also you can prove it directly:
$$
4^{2n+1}+3^{n+2} = 4(16^n-3^n)+(13)(3^n) = 13[4(16^{n-1}+16^{n-2}3+16^{n-3}3^2+\ldots+3^{n-1})+3^n]
$$
which proves that $4^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculate cosh(x) given sinh(x) Given the value of sinh(x)
for example sinh(x) = 3/2
How can I calculate the value of cosh(x) ?
| Use the identity $\cosh^2x-\sinh^2x \equiv 1$. If $\sinh x = \frac{3}{2}$ then
$$\cosh^2x - \left(\frac{3}{2}\right)^{\! 2} = 1$$
$$\cosh^2x - \frac{9}{4} = 1$$
$$\cosh^2x = \frac{13}{4}$$
It follows that $\cosh x = \pm\frac{1}{2}\sqrt{13}$. Since $\cosh x \ge 1$ for all $x \in \mathbb{R}$ we have $\cosh x = \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Finding domain of $\sqrt{4-2\sqrt{x^2 - 1}}$ Find the (maximum) domain of: $\sqrt{4-2\sqrt{x^2 - 1}}$
Well, I guess it is $2 - \sqrt2 \sqrt{x^2 - 1}$ = $2- \sqrt{2(x^2 - 1)}$
from here, $x^2 - 1 \ge 0$ and then $ x \ge 1$ and $ x \ge -1$
What do you guys think?
| You are right on the first restriction, we need $|x|\ge 1$. We also need $4-2\sqrt{x^2-1}\ge 0$. This inequality is equivalent to $2\sqrt{x^2-1}\le 4$, or equivalently $x^2-1\le 4$, that is, $|x|\le \sqrt{5}$.
Thus a way of describing the (maximum) domain is that it is $\{x: 1\le |x|\le \sqrt{5}\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $c$ if $a,b, \; c$ satisfy $c = (a+bi)^3 - 107i$
Find $c$ if $a,b, \; c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$
I can try expanding the cube, but that seems too direct. What other ways are there to go about this?
| Since $c \in \mathbb Z$, the right-hand side must have an imaginary component of $0$. Expanding, we get
$$c = (a+bi)^3 - 107i = (a^3-3ab^2) + (3a^2b-b^3-107)i$$
Therefore $3a^2b-b^3-107 = 0$, which implies that $(3a^2-b^2)b=107$. Since $107$ is prime, we conclude that $b$ is either $1$ or $107$.
Case 1: $b=1$. Then $3a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a... | You have shown that $a^3 \equiv b^3 \pmod p$. Remark that $(3, p-1)=1$ because $p=3k+2$. Thus we can write $3m + (p-1)n = 1$ for some integers $m, n$. Use this to show that $a\equiv b \pmod p$, so that $a^2+ab+b^2 \equiv 3a^2 \equiv 3b^2 \equiv 0 \pmod p$, and conclude from there.
(P.S. I have to commend you for posti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find a sequence Find the function for the sequence $a_0 = 0, a_1 = 1$ and $a_{n}=a_{n+10}+a_n$ for all $n>0$.
| Let's build such function:
$$
f(x) = \sum_{n=1}^\infty F_nx^n,
$$
where $F_n$ are Fibonacci numbers.
Then
$$
f(x) = x+\sum_{n=2}^\infty (F_{n-1}+F_{n-2})x^n = x + \sum_{k=1}^\infty F_kx^{k+1}+\sum_{m=0}^\infty F_mx^{m+2}
\\= x+xf(x)+x^2f(x),
$$
or much wide:
$$
f(x) = x+x^2+2x^3+3x^4+5x^5+8x^6+13x^7+\cdots \\
= x \;\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
N! ends with exactly 30 zeros? How many values of N exist, such that N! ends with exactly 30 zeros?
| The number of zeros at the and of $n!$ is just the number of factors of $5$ in $n!$ (since the number of factors of $2$ in $n!$ is always larger than that). So, we need all $n$ such that the numbers $1$ up to $n$ have a total of $30$ factors of $5$. A first guess would be $30\cdot 5=150$, but then we forget that the mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving that one of $a(1-b), b(1-c), c(1-a) \le \frac{1}{4}$ how can a prove that at least one of those is less than or equal to 1/4.
$$\forall a,b,c\in \mathbb R^+, \ a(1-b)\leq 1/4 \lor b(1-c) \leq 1/4 \lor c(1-a) \leq 1/4.$$
help please!
| We can assume $1-a, 1-b, 1-c \geq 0$, since otherwise we are done.
By the AM-GM inequality (see http://en.wikipedia.org/wiki/AM-GM_inequality), we have $abc(1-a)(1-b)(1-c) \leq (\frac{a+b+c+(1-a)+(1-b)+(1-c)}{6})^6= (\frac{1}{2})^6 = \frac{1}{64}$.
Then, if $a(1-b)> 1/4, b(1-c) > 1/4$ and $ c(1-a) > 1/4$, multiplying ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| Like Timbuc,
$$n^4-1=(n-1)(n+1)(n^2+1)=(n-1)(n+1)(n^2-4+5)$$
$$=\underbrace{(n-2)(n-1)(n+1)(n+2)}+5(n-1)(n+1)$$
As $n$ must divide exactly one of any five consecutive integers and $5\nmid n,(5,n)=1,$
$n$ must divide one of the multiplicand underbrace
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 15,
"answer_id": 7
} |
Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$. How to do this indefinite integral (anti-derivative)?
$$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$
I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.
| Let $x=\sqrt{7}\tan{u}\implies dx=\sqrt{7}\sec^2{u}du$
\begin{align}
I
=\int \frac{1}{(2x+1)\sqrt{x^2+7}}dx=\int \frac{\sec{u}}{(1+2\sqrt{7}\tan{u})}du\\
\end{align}
Let $t=\tan{\frac{u}{2}}\implies du=\frac{2}{1+t^2}dt$
\begin{align}
I
=\int\frac{\frac{1+t^2}{1-t^2}}{1+2\sqrt{7}\frac{2t}{1-t^2}}\frac{2}{1+t^2}dt=-2\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the triangular matrix and determinant. I have a 4x4 matrix and I want to find the triangular matrix (lower half entries are zero).
$$A=
\begin{bmatrix}
2 & -8 & 6 & 8\\
3 & -9 & 5 & 10\\
-3 & 0 & 1 & -2\\
1 & -4 & 0 & 6
\end{bmatrix}
$$
Here are the elementary row operations I performed to get it into triangular f... |
"I then did -21*row 4 + 6*row 3 to replace row 4 and got"
This is a determinant altering operation and not an elementary operation.
Don't write that $A$ equals something which isn't $A$.
Picking up where you errored and using the same idea you had one gets:
$$\begin{align} \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
| Assume that both $A, B > 0$ and $p \geq 0$ to begin with, then we can rewrite the inequality as:
$\left(\dfrac{A}{A+B}\right)^p + \left(\dfrac{B}{A+B}\right)^p \geq 2^{-p}$. So there are $3$ cases to consider:
*
*$p = 0$, then $LHS = 2 > 1 = RHS$, and the inequality holds.
*$p = 1$, then $LHS = 1 > \dfrac{1}{2} = R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Trigonometric Identities help How do you solve this? I can't figure out what I should do.
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero?
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{... | You just have to know that:
$\sin^4(a)=\sin^2(a)\sin^2(a)$
$\cos^4(a)=\cos^2(a)\cos^2(a)$
$\sin^2(a)+\cos^2(a)=1$
$$\begin{align}
\sin^4(a)+\cos^2(a)&=\cos^4(a)+\sin^2(a) \\
\sin^2(a)\sin^2(a)+\cos^2(a)&=\cos^2(a)\cos^2(a)+\sin^2(a) \\
\sin^2(a)\left(1-\cos^2(a)\right)+\cos^2(a)&=\cos^2(a)\left(1-\sin^2(a)\right)+\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+... | Here is an approach to compute $a_n$ using the observation that it is the number of ways to make change for $n$ using coins of denomination $2, 3, 4$.
Case: $n$ even
It may be noted that $3$ denomination coin can be used, but only as a multiple of $6$, so you have $n=2k=2p+4q+6r \implies k = p+2q+3r$. Thus this is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Area of a Curved Surface Find the area of the part o the surface $z=xy$ that lies within the cylinder $x^2+y^2=1$.
I'm not sure how to set up the surface integral to compute this.
| The formula of the are of the surface given as a graph of the function $z=f(x,y)$ over the region $(x,y) \in D$ is
$$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA$$
In this case $D$ is the disk of radius $1$ with center at $(0,0)$:
$D=\{(x,y): x^2+y^2 \leq 1\}$
$$z=f(x,y)=xy$$
$$f_x=y, f_y=x$$
So, we have the following:
$$A(S)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric functions of the acute angle Find the other five trigonometric functions of the acute angle A, given that:
\begin{align}
&\text{a)}\ \ \sec A = 2 \\[15pt]
&\text{b)}\ \ \cos A = \frac{m^2 - n^2}{m^2 + n^2}
\end{align}
Help me. I don't know how to solve this one. Thanks.
| Honestly, I suspect that equation b is a red herring.
\begin{align}
\cos{A}
&=\frac{1}{\sec{A}}\\
&=\frac{1}{2}
\end{align}
\begin{align}
\sin{A}
&=\sqrt{1-\cos^2{A}}\\
&=\frac{\sqrt{3}}{2}
\end{align}
\begin{align}
\tan{A}
&=\frac{\sin{A}}{\cos{A}}\\
&=\sqrt{3}
\end{align}
\begin{align}
\csc{A}
&=\frac{1}{\sin{A}}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}... | $$x=\frac{a}{b}$$
$$x+\frac{2}{3}=\frac{a+4}{b} \Rightarrow \frac{a}{b}+\frac{2}{3}=\frac{a+4}{b} \Rightarrow a+\frac{2}{3}b=a+4 \Rightarrow b=\frac{12}{2}=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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Can a Mersenne number ever be a Carmichael number? Can a Mersenne number ever be a Carmichael number?
More specifically, can a composite number $m$ of the form $2^n-1$ ever pass the test: $a^{m-1} \equiv 1 \mod m$ for all intergers $a >1$ (Fermat's Test)?
Cases potentially proved so far: (That are never Carmichael numb... | Partial Proof
I have found a proof for numbers with an even number of factors, and a prime exponent. ($m = 2^p-1$ and $p > 2$)
First note that $2^p-1 \equiv 3 \mod 4$ for $p > 1$. Next note the following table for $a*b \mod 4$:
$$
\begin{array}{c|lcr}
b & a = 0 & a = 1 & a = 2 & a = 3 \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 4,
"answer_id": 3
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Why these two series are convergent or divergent? I do not understand why $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac1k$$ is divergent but the other series $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}k$$ is convergent. For both cases the $\displaystyle\lim_{n \to +\infty} z_{n} = 0$. Coul... | We do not answer the specific question, but instead look at two related series,
$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots\tag{1}$$
and
$$1-\frac{1}{2}+\frac{1}{2}-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where... | I'm adding a second answer because the method is very different.
This proof uses the double angle formulas for sine and cosine. From
$$\sin 2x=2\sin x\cos x\qquad\cos2x=2\cos^2x-1$$
we get
$$\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\sin x\cos x}{2\cos^2x-1}>\frac{2x(1-x^2/3)(1-2x^2/3)}{2(1-x^2/3)^2-1},$$
using the bounds $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Count the number of ways n different-sided dice can add up to a given number I am trying to find a way to count the number of ways n different-sided dice can add up to a given number.
For example, 2 dice, 4- and 6-sided, can add up to 8 in 3 different ways: ($(2,6),(3,5),(4,4)$).
I've found a similar question but I can... | I can't provide a nice formula as in your referenced question, but I can show you at least half of the way.
Let's start with your
Example: There are two dice with 4 and 6 sides. How many ways are there, so that rolling both dice once result in 8 pips?
We encode the dies with polynomials and use the exponents to label... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $ $$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$
My approach is to calc
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$
However, I must do something wrong. The correct answer should be $2\ln(2)$.
$$\int_... | Another approach :
Setting $x\mapsto\frac1x$, we will obtain
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=\int_{0}^{1} \ln\left(\frac{2-x}{x}\right) dx=\int_{0}^{1} \bigg[\ln(2-x)-\ln x\bigg]\ dx.
$$
Note that
$$
\int\ln y\ dy=y\ \ln y-y+C,
$$
hence
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=2\ln 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequaliti... | By AM–GM, we have
$$14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}\geq 19\frac{a^2}{b}\quad (1)$$
$$2\frac{a^3}{b^2}+14\frac{b^3}{c^2}+3\frac{c^3}{a^2}\geq 19\frac{b^2}{c}\quad (2)$$
$$3\frac{a^3}{b^2}+2\frac{b^3}{c^2}+14\frac{c^3}{a^2}\geq 19\frac{c^2}{a}\quad (3)$$
Add all three equations together and conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\sum_{k=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4k+1)\pi}{2^{12}}\sin\frac{(4k+3)\pi}{2^{12}}}$ I need to evaluate
$$\sum_{n=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4n+1)\pi}{2^{12}}\sin\frac{(4n+3)\pi}{2^{12}}}$$
Please give me some hint!
The final answer is $2^{10}$.
By CuriousGuest's answer, ... | Hint: note that $$\frac{\pi}{2^{11}}=\frac{(4n+3)\pi}{2^{12}}-\frac{(4n+1)\pi}{2^{12}}.$$
Then use formula for $\sin(\alpha-\beta)$ and you'll get a telescopic sum.
Edit: It turns out to be not exactly telescopic, but something like
$$\cot\frac{\pi}{2^{12}}-\cot\frac{3\pi}{2^{12}}+\cot\frac{5\pi}{2^{12}}-\cot\frac{7\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear... | In this particular case
$$\frac{x^2-3}{x^2+1}=\frac{x^2+1-1-3}{x^2+1}=\frac{x^2+1}{x^2+1}-\frac{4}{x^2+1}=1-\frac{4}{x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
How to find the limit $x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$ without using L'Hopital's rule? Let
$$x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$$
It is very simple to solve it using L'Hopital's rule, but problem is to solve this limit without L'Hopital's rule. Is there any way to do this?
| Well, this is essentally the long form of L'Hopital, but:
$$\begin{align}\frac{a^b-b^a}{a^a-b^b} &= \frac{a^b-b^b+b^b-b^a}{a-b} \cdot\frac{a-b}{a^a-b^b}\\
&=\frac{\dfrac{a^b-b^b}{a-b} - \dfrac{b^a-b^b}{a-b}}{\dfrac{a^a-b^b}{a-b}}
\end{align}$$
The three terms here converges to the derivative of the functions $x^b$, $b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Evaluating a limit using L'Hôpital's rule I know that it can be also evaluate using Taylor expansion, but I am intentionally want to solve it using L'Hôpital's rule:
$$ \lim\limits_{x\to 0} \frac{\sin x}{x}^{\frac{1}{1-\cos x}} =
\lim\limits_{x\to 0}\exp\left( \frac{\ln(\frac{\sin x}{x})}{1-\cos x} \right)$$
Now, fr... | Alternative approach using Taylor series:
Using $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ we get $\ln(\frac{\sin x}{x}) \sim_0 \ln(1-\frac{x^2}{6})$. This expression has the Taylor expansion $-\frac{x^2}{6}+o(x^2)$.
We can approximate the denominator by $1-\cos x=\frac{x^2}{2} + o(x^2)$.
Then we have:
$\frac{\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerat... | I want to share with you my method using rationalization. $$
\begin{aligned}
& \int \frac{\sqrt{x^{2}+1}}{x} d x \\
=& \int \frac{x^{2}+1}{x \sqrt{x^{2}+1}} d x \\
=& \int \frac{x^{2}+1}{x^{2}} d\left(\sqrt{x^{2}+1}\right) \\
=& \int\left(1+\frac{1}{x^{2}}\right) d\left(\sqrt{x^{2}+1}\right) \\
=& \int 1 d\left(\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers. I stumbled upon it on some olympiad papers. Tried to AM>GM but didn't get any idea to move forward.
| Macavity's way is certainly the most elegant. Here is a way to do it just using Cauchy-Schwarz and AM-GM. Use Cauchy-Schwarz inequality to get:
$$(x+y+z)\left(\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\right)\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
So to prove your inequality, it suffices to show that
$$ (a^{3/2}+b^{3/2}+c^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Explaining why $\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$ For $x>0$. I understand the technical operation of extracting $x^2$ out of the root, but is there a way proving it?
$$\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$$
| Assuming that $x>0$ we have that
\begin{align}
\sqrt{x^2+a} &=\\
\sqrt{x^2\left(1+\frac{a}{x^2}\right)} &=\\
\sqrt{x^2}\sqrt{1+\frac{a}{x^2}} &=\\
|x| \sqrt{1+\frac{a}{x^2}}&=\\
x\sqrt{1+\frac{a}{x^2}}.
\end{align}
In the first equality we must require that $x\neq 0$, since otherwise the division by $x^2$ is not define... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| First let's start with your denominator, $$\frac{3}{x^2}+\frac{2x+1}{x}$$
We can turn this into one term by multiplying the,
$$\frac{2x+1}{x} by \frac{x}{x}$$
and then combine terms to get,
$$\frac{2x^2+x+3}{x^2}$$
Now if we put this back into the first expression we now have,
$$\frac{2x+1}{\frac{2x^2+x+3}{x^2}}$$
Fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$
$a=?, b=?, c=?$
The solution should be general.
| HINT:
$\displaystyle\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+(c+a)+(a+b)}$
Similarly,
$\displaystyle F=\frac{a}{b+c}=\frac{b}{c+a}=\frac{a-b}{b+c-(c+a)}$
Either $a=b$ or $\displaystyle F=-1$
Hope these should help
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$? How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$?
I tried this way:
Let
$$f(x)=\sqrt{4+\sqrt{4-x}}$$
So, $x=f^2(x)=f^{2n}(x)$ where $n\in\mathbb{N}$. Then, I tried to prove that $f^k(x)=f^{k+1}(x)$ for any $k\in\m... | $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$, the replace $x$ at RHS by the same equaility, we get
$$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}}}}}$$.
Continue in this way then we know that $x$ is equal to the limit(provided it exists) of $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Mathematical way to solve integer numbers $217 = (20x+3)r+x$ Is there any mathematical way to find the integer numbers that solve the following equation:
$$217 = (20x+3)r+x$$
| $$217 = (20x+3)r+x\\
217=20xr+3r+x$$
Now
$$(5x+\frac{3}{4})(4r+\frac{1}{5})=20xr+3r+x+\frac{3}{20}$$
Therefore
$$217+\frac{3}{20}=(5x+\frac{3}{4})(4r+\frac{1}{5})$$
Multiplying both sides by $20$ we get
$$4343=(20x+3)(20r+1)$$
There are only few ways of writing $4343$ as a product of two integers, in each case solve. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate $\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$ I am trying to solve $$\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$$
My answer is slightly different to the memo:
$x=3\sin\theta\quad\iff\quad\theta=\arcsin\left(\frac x 3\right)\\
\text dx=3\cos\theta\ \text d\theta\\$
$\begin{align}I&=\int\frac{3\sqrt{1-\sin^2\theta}}{3\si... | $$\int\frac{\sqrt{9-x^2}}{x^2}dx$$
$x=3\sin t,dx=3\cos tdt$
$$\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2t}3\cos tdt=\int\frac{\cos^2t}{\sin^2t} dt\neq3\int\frac{\cos^2t}{\sin^2t} dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \... | For any polynomial $P(x)$ of degree at most $n$, we have $\frac{d^k}{dx^k}P(x) = 0$ for any $k > n$. As a result, for any constant $y$, we have
$$\sum_{k=0}^{n} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x)
= \sum_{k=0}^{\infty} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) = P(y)$$
The last equality is true because the express... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$
Let $x_1, x_2, x_3, x_4$ be nonnegative integers.
(a) Find the number of solutions to the following equation:
$$ x_1 + x_2 + x_3 + x_4 = 14 $$
I got $17 \choose 3$ for this. Is that correct?
(b) Find the number of solutions if... | The answer you got for the first question is right.
For the second, call a distribution bad if one or more of the $x_i$ is $\ge 7$. Our strategy is to count the number of bads, and subtract from the answer of a).
One can have $2$ of the $x_i$ equal to $7$. This can be done in $\binom{4}{2}$ ways.
Now we count the numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.
| Let $n_3 = \lfloor \frac{708}{3}\rfloor, \; n_5 = \lfloor \frac{708}{5}\rfloor, \; n_15 = \lfloor \frac{708}{15}\rfloor.\;$ Then using the hints in the comments your sum $S$ is
$$S=3\sum_{k=1}^{n_3}k + 5\sum_{k=1}^{n_5}k-15\sum_{k=1}^{n_{15}}k
=3\frac{n_3(n_3+1)}{2}+5\frac{n_5(n_5+1)}{2}-15\frac{n_{15}(n_{15}+1)}{2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How many ways are there to divide $100$ different balls into $5$ different boxes so the last $2$ boxes contains even number of balls? How many ways are there to divide $100$ different balls into $5$ different boxes so the last $2$ boxes contains even number of balls?
I tried to think about tylor function but got stuck.... | Let $g_{e}(x)=\displaystyle\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)^3\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)^2=(e^x)^3\left(\frac{e^x+e^{-x}}{2}\right)^2$
$\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle e^{3x}\left(\frac{e^{2x}+2+e^{-2x}}{4}\right)=\frac{1}{4}(e^{5x}+2e^{3x}+e^x)$.
The coefficient of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the values of $f(0)$, $f(4)$, $f(6)$ and $f(18)$
A function $f:\mathbb{R} \to \mathbb{R}$ is such that $f(2)=2$ and
$$f(x+1)+f(x-1)=\sqrt{3}f(x) \tag{1}.$$
Find the values of $f(0)$, $f(4)$, $f(6)$ and $f(18)$.
My approach: replace $x$ with $x+1$ in $(1)$ we get
$$f(x+2)+f(x)=\sqrt{3}f(x+1) \tag{2}.$$
Repla... | Without any additional conditions on $f$, it is possible to pick $x_0, f(x_0), f(x_0+1)$ arbitrary and then use $(1)$ (i.e. $f(x) = \sqrt 3f(x-1)-f(x-2)$ to go forward $f(x) = \sqrt 3f(x+1)-f(x+2)$ to go backwards) to extend this uniquely to $x_0+\mathbb Z$. Likewise, we can prescribe arbitrary values for $f|_{(0,2]}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Analogue of $\zeta(2) = \frac{\pi^2}{6}$ for Dirichlet L-series of $\mathbb{Z}/3\mathbb{Z}$? Consider the two Dirichlet characters of $\mathbb{Z}/3\mathbb{Z}$.
$$
\begin{array}{c|ccr}
& 0 & 1 & 2 \\ \hline
\chi_1 & 0 & 1 & 1 \\
\chi_2 & 0 & 1 & -1
\end{array}
$$
I read the L-functions for these series have special va... | We see that
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{(3z+1)^2}{\rm d}z
&=\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{(3z+1)^2}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=-\frac{4\pi^2}{27}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=0
\end{align}
and
\begin{align}
\frac{1}{2\pi i}\oint_... | {
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"url": "https://math.stackexchange.com/questions/906424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer Prove by induction that this number is an integer:
$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$
Progress
I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt... | Let $v_n = (3+\sqrt{5})^n - (3-\sqrt{5})^n$, then
$$
u_1 = 6, v_1 = 2\sqrt{5}
$$
And
$$
u_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n + (3-\sqrt{5})(3-\sqrt{5})^n = 6u_n + \sqrt{5}v_n
$$
$$
v_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n - (3-\sqrt{5})(3-\sqrt{5})^n = 6v_n +\sqrt{5}u_n
$$
If $v_n$ is an integer multiple of $\sqrt{5}$ and ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Sum the series $\sum_{n = 1}^{\infty}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3}$ This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as
\begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi... | Yet another approach using contour integration is to integrate the function $$f(z) = \frac{\pi \cot (\pi z) \coth (\pi x z)}{z^{3}} $$ around a circle centered at the origin that avoids the poles on the real and imaginary axes.
If we let the radius of the circle go to infinity discretely, the integral will vanish.
So s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 3,
"answer_id": 2
} |
Show the origin is the only critical point of $f$ $f(x,y,z)=\frac1 2(x^2+y^2+z^2)+a(xy+yz+zx)$
For $a\neq-\frac1 2$ and $a\neq1$, show that the origin is the only critical point of f.
I have found $\nabla f =\begin{pmatrix} x+a(y+z) \\ y+a(x+z) \\ z+a(x+y) \end{pmatrix} $
Equating this to zero to find critical points,... | Suppose $a\neq-\frac{1}{2}$ and $a\neq 1$. Sum the components of $\nabla f$ gives
$$
(x+y+z)+2a(x+y+z)=0
$$
If $x+y+z\neq 0$, then we would have $a=-\frac{1}{2}$. So $x+y+z=0$. Use this in $x=-a(y+z)$:
$$
x=-a(y+z)=-a(-x)=ax.
$$
If $x\neq 0$, then $a$ would be $1$. So $x=0$. Similarly, you can apply $x+y+z=0$ to $y=-a(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\sin( 2 \theta) = \cos( 3)$ and $\theta \leq 90°$, find $\theta$ Find $\theta\leq90°$ if
$$\sin( 2 \theta) = \cos( 3)$$
I know that $\sin 2\theta = 2\sin\theta\cos \theta$, or alternatively, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$.
Can somebody help me?
| $$\sin2\theta=\cos3^\circ=\sin87^\circ$$
$$\implies2\theta=180^\circ m+(-1)^m87^\circ\text{ where } m \text{ is any integer}$$
$$\implies\theta=90^\circ m+(-1)^m43.5^\circ$$
Check for even $m=2r$(say) and for odd $=2r+1$(say)
Find $m$ such that $\displaystyle\theta\le90^\circ$
Alternatively,
$$\cos3^\circ=\sin2\theta=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
$\ln (2 x-5)>\ln (7-2 x)$ Solve
$$\ln (2 x-5)>\ln (7-2 x)$$
The answer is given as $$3<x<7/2$$
This is what I have done $$\ln (2 x-5)-\ln (7-2 x)>0$$
$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$
However I am not able to understand how to get to the answer provided.
| Exponentiation is a one to one, increasing function so:
$\ln (2x-5) > \ln (7-2x)\\
e^{\ln (2x-5)} > e^{\ln (7-2x)}\\
2x-5>7x-2\\
...\\
x>3$
However there are the domains to consider as well:
$2x-5>0$ gives $x>\frac{5}{2}$ and $7-2x>0$ gives $x<7/2$. We have to put all of the inequalities $x>3$ and $x>\frac{5}{2}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How prove this there many infinite numbers postive ineteger $n$ such $n^3+1|n!$ Question:
show that: there exsit infinite postive integer $n$ such
$$n^3+1|n!$$
Buy the way,I know prove this this simple problem:
prove there are many infinte postive integer numbers $n$,such
$n^2+1|n!$
This problem we only consider... | I will prove there are infinitely many solutions of the form $n=3a^2$. For such $n$ we have $3 \nmid n+1$, hence $\gcd(n+1,n^2-n+1)=1$ and it is sufficient to show that both $n+1 \mid n!$ and $n^2-n+1 \mid n!$. We have $n+1 \mid n!$ whenever $n+1 = 3a^2+1$ is not prime, whereas $n^2-n+1 = 9a^4-3a^2+1$ factors as $(3a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\sec \theta + \tan \theta$. If $\tan \theta=x-\frac{1}{x}$, find $\sec \theta + \tan \theta$.
This was the question ask in my unit test.
My Efforts:
$\tan^2 \theta=(x-\frac{1}{x})^2$
$\tan^2 \theta= (\frac {x^2-1}{x})^2$
Now we can use identity $\sec^2 \theta= 1 + \tan^2 \theta$.
But i am not able to get the answ... | As you have written $\sec^2 \theta = 1 + \tan^2 \theta$. From this you could get
$$ \sec \theta + \tan \theta = \pm \sqrt{1+\tan^2 \theta} + \tan \theta = \pm \sqrt{1 + \left( x - \frac{1}{x} \right)^2} + x - \frac{1}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Laplace Transform assistance Find the inverse laplace transform of: $\frac{25}{(s-1)^2(s^2+4)}$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{s^2 + 4}$
$$25=A(s^2+4)(s-1)+B(s^2+4)+C(s-1)^2$$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{0}{s-1}+\frac{5}{(s-1)^2}+\frac{-15}{s^2 + 4}$
Maybe I don't know how to t... | $$25=A(s^2+4)(s-1)+B(s^2+4)+(Cs+D)(s-1)^2$$ is correct. Therefore we have $$25=s^3(A+C)+s^2(-A+B-2C+D)+s(4A+C-2D)+(-4A+4B+D)$$ Now, $$A=0, -A+B+C=0,4A-2C=0, -4A+4B+C=25$$ which gives $$A=-2, C=2, D=-3, B=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equat... | To motivate this to students who are just beginning to learn about remainders, it often helps to use real-world examples, e.g. if changing $\,n\,$ pennies (1 cent coins) into dimes (10 cent coins) leaves 8 cents, then changing these $\,n\,$ pennies into nickels (5 cent coins) leaves 3 cents, because we can first change... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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How to solve the ODE $y(y'(x)+a)=bx$ I've been very frustrated attempting to separate this guy. Since there are three separate items when you multiply through by the y, it is very difficult to make all sides work out to forms $f(y)dy$ and $f(x)dx$. I've attempted the trick of getting it in the form $y'dx/y$ to get the ... | We have $$0 = y(y'+a) - bx = \frac{1}{2}\frac{d}{dx}(y^2 + ay - bx^2) - \frac{a}{2}x^2\frac{d}{dx}\left(\frac{y}{x}\right)$$
Let us therefore consider the function
$$h(x) = \frac{y(y'+a) - bx}{y^2 + axy - bx^2}$$
Then $h \equiv 0$ and by integrating the last expression we get
$$\int \frac{y(y'+a)-bx}{y^2 + axy - bx^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove $\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$ if $a+b+c=2$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove: $$\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$$
Additional info:I'm looking for solu... | By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}=\sum_{cyc}\frac{ab\sqrt{\frac{3}{8}}\cdot2\sqrt{\frac{8}{3}(2c+a+b)}}{2(2c+a+b)}\leq\sqrt{\frac{3}{32}}\sum_{cyc}\frac{ab\left(\frac{8}{3}+2c+a+b\right)}{2c+a+b}=$$
$$=\sqrt{\frac{1}{96}}\sum_{cyc}\frac{ab(4(a+b+c)+6c+3a+3b)}{2c+a+b}=\sqrt{\frac{1}{96}}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}$
Prove that
\begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}
\end{equation}
I tried to use by parts method and ended with
\begin{equation}
\int \ln^2(\cos x)\,dx=x\ln^2(\cos x)+2\int x\ln(\cos x)... | In my post, I had found a reduction formula for the integral
$$
J_n=\int_{0}^{\frac{\pi}{2}} \ln ^{n}(\cos x) dx
$$
that
$$\begin{aligned}
J_n&= -\ln 2 J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{(-1)^{n-k}}{k !} \left(1-\frac{1}{2^{n-k-1}} \right)\zeta(n-k) J_k.
\end{aligned}
$$
Using it yields
$$
\begin{aligned}
\boxed{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 4
} |
How can I accurately compute $\sqrt{x + 2} −\sqrt{x}$ when $x$ is large? How can the values of the function $f(x) = \sqrt{x + 2} −\sqrt{x}$ be computed accurately when $x$ is large?
I have tried using Matllab. I am not able to understand when $x$ will be large.
| In fact, an asymptotic expansion of $f(x) = \sqrt{x+a} - \sqrt{x}$ about $x = \infty$ is $$f(x) \approx \frac{a}{2} x^{-1/2} - \frac{a^2}{8} x^{-3/2} + \frac{a^3}{16} x^{-5/2} - \frac{5a^4}{128} x^{-7/2} + \frac{7a^5}{256} x^{-9/2} - \frac{21a^6}{1024} x^{-11/2} + \cdots.$$ We can calculate this via the generalized bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
A $(3k+3)$-node graph with degrees $k+1,\dots,k+3$ has $k+3$ degree-$(k+1)$ nodes or $k+1$ degree-$(k+2)$ nodes or $k+2$ degree-$(k+3)$ nodes?
Graph $G$ has order $n=3k+3$ for some positive integer $k$ . Every vertex of $G$ has degree $k+1$, $k+2$, or $k+3$. Prove that $G$ has at least $k+3$ vertices of degree $k+1$ o... | Let $a,b,c$ denote the number of vertices of degree $k+1,k+2,k+3$ respectively.
Then we have $a+b+c=3k+3$
If $a\geq k+3$ or $c\geq k+2$ then we are done.
Now if $a< k+3$ and $c< k+2$ then we have to show that $b\geq k+1$.
Thus we have $a\leq k+2$ and $c\leq k+1$. Now we consider two cases.
Case $1$: k is odd.
Putting t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $\sqrt x = x/2$. If $f(x) = \sqrt x$ and $g(x) = x/2$, find the area of this limited area between $f(x)$ and $g(x)$.
I am having trouble solving this equation $\sqrt x = x/2$ that should give me the x values.
I know that the next step after this one is to solve $\int_a^b f(x)-g(x)~dx$ where $a= x_1$ and $b=x_2$
H... | $$\sqrt{x}=\frac{x}{2}\\x\geq 0\\(\sqrt{x}=\frac{x}{2})^2\\x=\frac{x^2}{4}\\4x=x^2\\x(4-x)=0\\x=0\\x=4\\ \int_{0}^{4}(\sqrt{x}-\frac{x}{2})dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}=\\(\frac{4^{\frac{3}{2}}}{\frac{3}{2}}-\frac{4^2}{4})-(\frac{0^{\frac{3}{2}}}{\frac{3}{2}}-\frac{0^2}{4})=\frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting ver... | $\cos2x=1-2\sin^2 x$, let $t=\sin x$ so $dt=\cos xdx$ in numerator.
SO:
$$I=\int\frac{dt}{(1-2t^2)^{3/2}}$$
Then use the substitution $u^2=(1-2t^2)$ so $udu=-2tdt$, so $\displaystyle dt=-\frac{udu}{2t}=\frac{-udu}{\sqrt2\sqrt{1-u^2}}$:
$$I=-\int\frac{udu}{\sqrt2\sqrt{1-u^2}u^3}=\frac{-1}{\sqrt2}\int\frac{du}{u^2\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $ \frac{dy}{dx} = y^2 - 9$ $ \frac{dy}{dx} = y^2 - 9$
This is separable so I rewrite it as $ \frac{1}{(y^2 - 9)}dy = dx$ then I get
$$\int \frac{1}{(y^2 - 9)}dy = \int dx = \int 1dx = x + c,$$ for some $c \in \mathbb{R} $
The left hand side:
$$\int \frac{1}{(y^2 - 9)}dy = \int \frac{1}{(y+3)(y-3)}dy = \int \lef... | $\frac{y-3}{y+3}=\pm e^{6x}e^{6c}=Ce^{6x}$ so $y-3=(y+3)Ce^{6x}$ Group the $y$ terms together to get $y(1-Ce^{6x})=3+3Ce^{6x}$. Now divide by $1-Ce^{6x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding asymptotes for $f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$ $$f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$$
I know that the horizontal asymptote is $1/3$. To find the vertical asymptotes, I set the denominator equal to zero and used the quadratic formula, and I got $-5$ and $2/3$, and this is wrong.
How do you find the vertica... | Note that
$$f(x)=\frac{(x+5)(x-2)}{(x+5)(3x-2)}.$$
Therefore
$$\lim_{x\to-5} f(x) = \lim_{x\to-5}\frac{(x+5)(x-2)}{(x+5)(3x-2)} =\lim_{x\to-5}\frac{x-2}{3x-2} = \frac{7}{17}.$$
Since this limit is actually not infinite, there is no vertical asymptote when $x=-5$.
In general, vertical asymptotes of a rational function o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The minimum of $x^2+y^2$ under the constraints $x+y=a$ and $xy=a+3$ I solved the following problem: If $x,y,a \in \mathbb{R}$ such that $x+y=a$ and $xy=a+3$, find the minimum of $x^2+y^2$
Here is my solution. $x^2+y^2=(x+y)^2 -2xy= a^2-2a-6$. The minimum value is obtained when $a=1$ and it is $-7$. Where did I go wrong... | Since $x,y$ are the real roots $X$ of $$(X-x)(X-y)=0\iff X^2-(x+y)X+xy=0\iff X^2-aX+(a+3)=0,$$ you need to have $$D=(-a)^2-4(a+3)\ge 0\iff (a-6)(a+2)\ge 0\iff a\le -2\ \text{or}\ a\ge 6.$$
So, considering $x^2+y^2=a^2-2a-6=(a-1)^2-7$ for $a\le -2\ \text{or}\ a\ge 6$ gives you that the minimum is $2$ at $a=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The limit of $\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$ as $n\to\infty$ The task is to calculate $$\lim_{n\to\infty}\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$$ I tried various estimates I know to find the dominating integrable function and nothing worked. Does anyone have any idea? Is this even an application of... | DCT as Discrete Cosine Transform? I cannot understand what you are meaning.
Anyway, your integral converges towards $\frac{\pi}{2}$, since:
$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx \geq \int_{0}^{1}\frac{\sqrt{n}}{1+nx^2}\,dx = \arctan(\sqrt{n})=\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right),$$
while the diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
| since
$$LHS=\dfrac{a^2}{3bc}+\dfrac{b^2}{3ca}+\dfrac{c^2}{3ab}=\dfrac{a^3+b^3+c^3}{3abc}=\dfrac{3abc}{3abc}=1$$
I think this condition $a^3+b^3+c^3=3abc$ is redundant.because
$$a+b+c=0\Longrightarrow a^3+b^3+c^3=3abc$$
because
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.
| $3ab>0$
Hence, $2ab>-ab$
Hence, $a^2+2ab+b^2>a^2+b^2-ab$
Hence $a+b>\sqrt{a^2+b^2-ab}$ because $a^2+b^2-ab>a^2+b^2-2ab=(a-b)^2>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$
$\int\f... | As Hassan Muhammad commented, you have a mistake. Since you set $x=\sqrt{u+\frac{1}{2}}$, you then have $dx=\frac{1}{2 \sqrt{u+\frac{1}{2}}}$ and then $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int \frac{2 u+3}{\sqrt{u+\frac{1}{2}} \left(4 u^2+3\right)}du$$ which is not very nice.
If you use Hippalectryon's idea, you e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Closed form of $\int_0^{\frac{\pi}{3}} \log^2(\sin x) \mathrm{d}x$ Inspired by the popularity of these kind of integrals appearing on MSE lately, I actually learned new methods to attack weird integrals by studying the beautiful answers on the similar past questions, so I conjecture $$\int_0^{\frac{\pi}{3}} \log^2(\sin... | $\def\B{{\text{B}}}\def\F{{\text{$_2$F$_1$}}}$I find the closed form is really nasty since it's involving the hypergeometric functions. Here is my approach. Let $I$ be the given integral and let $\sin x=\sqrt{t}$, then
\begin{align}
I&=\frac{1}{8}\int_0^{3/4}\frac{\ln^2t}{\sqrt{t}\cdot\sqrt{1-t}}\,dt\\
&=\frac{1}{8}\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the... | Since we have$$y^3-x^3=2(4x^2-3x+4)$$
there exists an integer $k$ such that
$$y-x=2k\iff y=x+2k.$$
So, we have
$$(x+2k)^3-x^3=2(4x^2-3x+4)\iff (3 k-4) x^2+(6 k^2+3) x+4k^3-4=0\tag1$$
Now we have
$$D=(6k^2+3)^2-4(3k-4)(4k^3-4)\ge 0\iff -12 k^4+64 k^3+36 k^2+48 k-55\ge 0$$$$\iff 12 k^4-64 k^3-36 k^2-48 k+55\le 0$$
Here, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
Also find the center and radius of the circle.
Here's my solution, I... | Lets' solve for:
X^2 + y^2 + ax + by + c = 0
Step 1: Add -x^2 to both sides
ax + by + x^2 + y^2 + c + - x^2 = 0 + - x^2
ax + by + y^2 + c = - x^2
Step 2: Add - y^2 to both sides
ax + by + y^2 + c + - y^2 = - x^2 + y^2
ax + by + c = - x^2 - y^2
Step 3: Add - by to both sides
ax + by + c + by = - x^2 - y^2 + - by
ax + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/931215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Meaning of congruence notation for Bernoulli Numbers I am studying Theorem 4(von Staudt's Theorem) in Borevich-Shafarevich's Number Theory(1966)(page 384) which states:
Let $p$ be a prime and $m$ an even integer. If $(p-1)\nmid m$, then
$B_m$ is $p$-integral(that is, $p$ does not appear in the denominator
of $B_m$... | The fraction $\frac{1}{n}$ is, by definition, the number which you can multiply $n$ by to get the answer $1$. Now the same definition works in congruence arithmetic. For example modulo $p$, as long as $n$ is not divisible by $p$ there is some number $m$ such that $nm \equiv 1$ mod $p$. In that case $\frac{1}{n}\equiv m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/931707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Maximization of $x+y$ when each is greater than $1$ and $xy = 16$. The product of two numbers $x$ and $y$ is $16$. We know $x\ge 1$ and $y\ge 1$. What is the greatest possible sum of the two numbers?
| Hint:
Since $x \cdot y = 16$ and $x,y \geq 1 > 0$ we have $y = \frac{16}{x}$. Now can you find the maximum of the function
$$f(x) = x + \frac{16}{x}$$
for $x \geq 1$ and $\frac{16}{x} \geq 1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial ... | Doing the same as Aditya (Weierstrass substitution), you arrive to $$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx=\int \frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}dt$$ Now, using partial fraction decomposition $$\frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}=\frac{1}{18 t^2}-\frac{32}{9 \left(t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the value of $f(x)$ for $x = 2 + 2^{2/3} + 2^{1/3}$ If $x = 2 + 2^{2/3} + 2^{1/3}$, then find the value of $f(x)=x^3 - 6x^2 + 6x$.
I am unable to get to the answer - end up with more than one term. Please help me solve this!
| We have $$x-2=2^{\frac23}+2^{\frac13}$$
Cubing we get, $$x^3-3x^2(2)+3x(2^2)+2^3=(2^{\frac23})^3+(2^{\frac13})^3+3\cdot2^{\frac23}\cdot2^{\frac13}(2^{\frac23}+2^{\frac13})$$
$$\iff x^3-6x^2+12x+8=2+2^2+6(x-2)$$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $x+y+z.$ Let $x,y,z$ are nonegative such that $(x - y)(y - z)(z - x) \geq 1.$
Find the minimum value of $x+y+z.$
| Clearly, two among $x-y, y-z, z-x$ should be negative. WLOG let $a=x-y, b=z-y$ be positive numbers. Then $a(-b)(b-a) \ge 1 \iff a^2b \ge ab^2+1$ and we need the minimum of $a+b+3y$. Obviously it is a good idea to set $y =0$, then we are left with minimising $a+b$.
$$a^2b-b^2a-1 \ge 0 \iff a \ge \frac{b^2+\sqrt{b^4+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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