Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin... | Apparently your second question hasn't been answered, about what to do next. Here's what to do:
$$\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\frac{\frac{1}{\cos \alpha}}{\frac{1}{\cos \alpha}}\times\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\dots$$
I hope that's helpful.
If you need any mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}... | This is an elegant problem. You can use induction on the difference and skip the brute force computation.
Proof by induction:
Let S(n) be the claim that $f(n)-g(n)=4-4n$.
Base case: trivial.
Inductive case: Suppose S(n-1) is true. That is, suppose $f(n-1)-g(n-1)=4-4(n-1)$. Then $f(n)-g(n)=[3f(n-1)+8(n-1)]-[3g(n-1)+12]=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Change of variable with g(u,v) = (u-v+1,u+v) I want compute the integral of f(x,y) = 1 over $R = \{ (x,y) : 0\leq x \leq 1 \land 0\leq y \leq x^2 \} $ with the transformation $g(u,v) = (u-v+1,u+v)$ and check that is the same that compute without change of variable. i.e. $ \int_{0}^{1} \int_{0}^{x^2} dy dx = \frac{1}{3}... | $0 \leq x \leq 1, 0 \leq y \leq x^2$
Transformation $x = u - v + 1, y = u + v$
$J = \begin{vmatrix} 1 & -1 \\ 1 & 1\end {vmatrix} = \, 2$
For the upper bound of $v$,
$y = x^2 \implies u + v = (u-v+1)^2$ (which is equation of another parabola)
$v^2 - (2u+3)v + (u^2+u+1) = 0$ which is a quadratic in $v$.
$(v - \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3977126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\iint_D (x^4-y^4) \,dx\,dy$ How should I proceed? the question is :
$$\iint_D (x^4-y^4) \,dx\,dy$$
$$D= \left\{(x,y):1<x^2-y^2<4, \quad \sqrt{17}<x^2+y^2<5,\quad x<0,\ \ y>0\right\}$$
I've tried to solve it with change of variable:
$$u=x^2-y^2, \quad v=x^2+y^2$$ $$|J|=\frac{1}{8xy}, \quad \frac{1}{8}\iint uv\... | Oddly enough polar coordinates will do the trick and does it well as an alternative. The region is one integral if done angular first (which is easy to see because the region is between two circular arcs)
$$1<x^2-y^2<4 \implies \pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)<\theta<\pi-\frac{1}{2}\cos^{-1}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to minimise the area of the folded part of a piece of paper when it touches the other side I am studying maths purely out of interest and have come across this question in my text book:
A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on ... | If $\angle AQP = \theta$, $\angle APQ = 90^0 - \theta$.
Draw a perp from $A_1$ to line segment $AP$ and say it is point $A_2$ on line segement $AP$.
Then $A_1A_2 = b$ and $\angle A_2PA_1 = 2 \ \angle APQ = 180^0 - 2\theta \ $.
So $AP = L = \frac{A_1 A_2}{\sin(180^0 - 2 \theta)} = \frac{b}{\sin (180^0 - 2 \theta)} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $e^{2z} - 2e^z + 2= 0$ So I've started by looking at
\begin{align} e^{z} = x:\\
x^2 - 2x + 2 = 0
\end{align}
Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}
Then I did the following:
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2... | Yes, the solutions of $x^2-2x+2=0$ are $1+i$ and $1-i$. So, the solutions of $e^{2z}-2e^z+2=0$ are all those numbers $z$ such that $e^z=1+i$ or that $e^z=1-i$. But$$1+i=\sqrt2\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)=\sqrt2e^{\pi i/4}. $$Therefore$$e^z=1+i\iff z=\log\sqrt2+\frac{\pi i}4+2\pi in,$$for some $n\in\Bbb Z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0
Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$
This is from a recently closed question.
Notice that the fraction on the LHS is well defined. The d... | No calculus is required.
Using your substitution in your linked effort:
Let $x=\frac{1-p}{p}$ then $p=\frac{1}{1+x}, x>0$. The inequality becomes
$$\frac{1}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{x^n-1} = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac1 2-\frac{1}{2n}$$
$$\iff g(x) = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac 12 - \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$
If the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$ can be expressed as $2^m \cdot 3^n$, prove that $m=n$ and find $m$.
This may be considered a sub-problem from another problem. Although I have greatly reduced calcu... | If $a$ is a factor of $6^{12}$ and $a \equiv 1 \pmod 7$, then $\frac{6^{12}}{a} \equiv \frac{(-1)^{12}}{1} \equiv 1 \pmod 7$, and these pairs multiply to $6^{12}$. In the special case where $a = 6^6$ and $a = \frac{6^{12}}{a}$, verify that $a \equiv 1 \pmod 7$. Therefore $m = 12k + 6, k \in \mathbb N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty wei... | WLOG
$$\sqrt{x^2+12y}=33\cos^2t\text{ and }\sqrt{y^2+12x}=33\sin^2t$$
$$(33\cos^2t)^2-(33\sin^2t)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
$$x+y=23\implies x-y=99(\cos^2t-\sin^2t)=99(2\cos^2t-1)$$
$$\implies x=99\cos^2t-38\implies y=61-99\cos^2t$$
$$\implies(33\cos^2t)^2=x^2+12y=(99\cos^2t-38)^2+12(61-99\cos^2t)$$
Let $33\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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When will matrix multiplication become just "concatenation"? Saw below entertaining matrix multiplication examples.
Obviously they are just coincidences. But I am curious when does below hold?
\begin{equation}
\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pma... | Let $A$ denote the first matrix in the product, and let $B$ denote the second.
I will consider only the case in which the entries of $B$ have one digit.
In this case, the "gluing" property of this matrix multiplication can be written as
$$
AB = 10A + B.
$$
Note that this equation can be rearranged into
$$
AB - 10 A - B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 1
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Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degre... | The Quotient ring ${Z_2[x]/(x^4+x^2+x)}$ contains below elements.$${\{1,x,x^2,x^3,x^2+x,x^3+x^2,x^3+x^2+x,x+x^3\}}$$ if we square each of them, then we can see that for ${(x^2+x)^2=x}$, so ${\sqrt{x}=(x^2+x)=p }$(as defined above by Teresa)
*
*now ${(x^3+x+1)^2= x^6+x^2+1= x^3+x^2+x+x^2+1=x^3+x+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a Viable/New Proof for Pythagoras Theorem? A triangle with side lengths $a$, $b$, $c$ and a height ($h$) that intercepts the hypotenuse ($c$) such that $c$ is split into two side lengths, $c = m + n$, we can find Pythagorean theorem using simple trigonometry.
We start by finding the sum of tan ratios that are ... | This amounts to a well-known similar-triangles argument, splitting the area, that $(a/c)^2+(b/c)^2=1$, which is a bit more direct.
As @DavidK notes, your opening use of trigonometry is invalid, and its conclusion (in fact, something stronger) should instead be deduced as per @WillOrrick's suggestion, giving $\tfrac{a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.
Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.
We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?
| Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$
$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$
Wolfram says that this is indeed a decreasing function in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Integrating an Infinite exponent tower $$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx = ~~?$$
What I've tried so far:
$$ x^{2^{x^{2 \ldots}}} \overbrace{=}^{\text{def}} y \\ x^{2^{y}... | Consider
$$I=\int_0^1 y^{2^{-y}}\,dy=\int_0^\frac 12 y^{2^{-y}}\,dy-\int_1^\frac 12 y^{2^{-y}}\,dy=I_1-I_2$$
Each integrand will be developed as a series expansion around the lower bound to order $(n+1)$ and termwise integrated. This does not lead to very complicated expressions for the integrands and the result of int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$
The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \... | Horner's method for division:
$$(x^2+x+3)(x^3+7)=x^5+x^4+3x^3+7x^2+7x+21$$
\begin{array}{*{7}{r}}
& 1 & 1 & 3 & 7 & 7 & 21 \\
+ & \downarrow & -1 & 0 & -3 & -4 & -3 \\
\hline
\times -1 & \color{red}1 & \color{red}0 & \color{red}3 & \color{red}4 & \color{red}3 & \color{cyan}{18}
\end{array}
so the quotient is $\;x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different.... | Both are correct, but the answer you obtain depends on the definition $\cot^{-1}(x)$, in particular its range, since inverse trigonometric functions are multi-valued.
If you write $\cot^{-1}(x) = \tan^{-1}(1/x)$ [as a calculator seldom has $\cot^{-1}$] and use the range $-\pi/2 < \tan^{-1}x < \pi/2$, we see that $\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$ I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$
$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$
$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(... | You may use:
$$\prod_{k=0}^{n} \frac{1}{x+k}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k-1} \frac{k{n \choose k}}{x(x+k)}=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k} [\frac{1}{x}-\frac{1}{x+k}]$$
So $$\int \prod_{k=0}^{n} \frac{1}{x+k} dx=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k}[\ln x- \ln (x+k)]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Constructing factor-/quotient groups
Given:$$\Bbb Z_8 =\{0,1,2,3,4,5,6,7\}$$ and $$H=\{0,2,4,6\} = \left<[2]\right>$$, construct the factor group $\frac{\Bbb Z_8}{H}.$
Definition:
For a normal subgroup $H$ of $G$, the group $\frac{G}{H}$ is called the factor-/ or quotient group of $G$ modulo $H$. The set $\frac{G}{H... | Those two cosets do indeed form $\Bbb Z_8/H$ (under the operation $(a+H)+(b+H):=(a+b)+H$). Well done! You could improve your answer by noting that $0+H=H$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$ Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2... | By substracting the two equations you get
$$ x^2(1-x^2) + y^2(y^2-1) = 0$$
$$ -(x^2-\frac12)^2 + (y^2-\frac12)^2 = 0 $$
so
$$ x^2 - \frac12 = \pm (y^2-\frac12) $$
that is
$$ x^2 = y^2 $$
or $$ x^2 = 1- y^2 $$
Analysing theses two cases it's easy to solve the equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
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Evaluate the integral $\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx$ I try to calculate the integral
\begin{align}
f(y,z):=\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx
\end{align}
on the set $y>0$ and $0<z<1$. I could ... | \begin{align}
f(y,z)=&\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-y(\sinh^2\frac{u}{2}+z\cdot \sinh\frac{u}{2}\cosh\frac{u}{2} ) \right)\mathrm du\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u -1+z\cdot \sinh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integral involving Airy function I met the following integral when I was reading a paper:
$$\int_0^\infty Ai(y)dy=\frac{1}{3},$$
where
$$Ai(y)=\frac{1}{\pi}\int_0^\infty \cos(\alpha y+\frac{\alpha^3}{3})d\alpha.$$
The paper adopted one asymptotic result of the Airy function,
$$\int_0^x Ai(y)dy \sim \frac{1}{3}-\frac{1}... | You can obtain it using the integral representation
$$
\operatorname{Ai}(y) = \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3} - \frac{{y^3 }}{{3t^3 }}} \right)dt} ,\quad |\arg y|<\tfrac{\pi}{6}
$$
(cf. http://dlmf.nist.gov/9.5.E6). Indeed, by the Fubini theorem and the Gauss multiplicati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$
How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$
| $t=\arctan x\iff \tan t=x$.
Recall that $1+x^2=1+\tan^2 t=\frac{1}{\cos^2 t}$, so $\cos^2 t=\frac{1}{1+x^2}$.
Now, $$\cos(2\arctan x)=\cos 2t=\cos^2t-\sin^2t=2\cos^2t-1=2\frac{1}{1+x^2}-1=\frac{1-x^2}{1+x^2}$$ and it is a tautology.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$
Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$
I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maxi... | For the maximum, you can easily prove that $4$ of the $5$ product terms cannot be negative. Then, the maximum is achieved either all $5$ are positive or only $3$ are positive. In the former case, simple $AM-GM$ gives the maximum of $32.$ For the latter case, assume $a+b\leq 0$ and $b+c\leq 0$ while the other three are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Given Orthocenter of a triangle and length of three segments, need to find area. Let, $O$ be an orthocenter of the triangle $\triangle ABC$, where $AO = 7, AB = 9, CO = 4$. Find the combined area of the triangles $\triangle AOC$ and $\triangle AOB$. Options given are $7\sqrt{3}, 14\sqrt{3}, 21\sqrt{3}, 28\sqrt{3}$.
I s... | $$\triangle=[AOB]+[AOC]=\frac{1}{2}\cdot BC\cdot AO$$
So we need to find $BC=a$
We use the fact that perpendicular bisector from circumcenter to any side is half the length of segment joining the opposite vertex to the orthocenter.
In the diagram below, not to scale, we see that
$$\left( \frac{c}{2} \right)^2 + \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem usefu... | less beautiful
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
- \frac{ 1 }{ 2 } & 1 & 0 & 0 \\
- \frac{ 1 }{ 2 } & - \frac{ 1 }{ 3 } & 1 & 0 \\
- \frac{ 1 }{ 2 } & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
355/113 and small odd cubes An important approximation to $\pi$ is given by the convergent $\frac{355}{113}$.
The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.
$$\frac{355}{113} = \frac{7^3+12}{5^3-12}$$
Is this a consequence of some general formula, such as a... | For a coprime pair of integers $1 \le p,q$, consider the following diophantine equation:
$$\frac{x^3+z}{y^3-z}= \frac{p}{q}$$
with $x,y,z$ integer unknowns. This is equivalent to
$$z= \frac{y^3p-x^3q}{p+q}$$
Thus a solution exists if and only if there exist $x,y$ such that
$$y^3p-x^3q \equiv 0 \pmod{(p+q)}$$
Now, note ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Tangent integration $\int \tan (2x)\tan (3x)\mathrm{d}x$ I know how to calculate $\displaystyle\int \tan (x) \tan (2x)\tan (3x)\mathrm{d}x$ with the tangent sum formula
$\tan 3x\tan 2x\tan x =\tan3x -\tan 2x -\tan x$. But I don't know how I can solve
$$\displaystyle\int \tan (2x)\tan (3x)\mathrm{d}x$$
I tried u sub, tr... | Proceed as follows
\begin{align}
\int \tan 2x\tan 3x {d}x
&=\int \frac{\sin 2x\sin 3x }{\cos 2x\cos 3x }dx\\
&=\int \frac{(2\sin x\cos x)[\sin x(4\cos^2x-1)] }{(2\cos^2x-1)[\cos x(4\cos^2x-3) ]}dx\\
&=2\int \frac{(1-\cos^2x)(4\cos^2x-1)}{(2\cos^2x-1)(4\cos^2x-3)}dx\\
&=\int \left( -1 -\frac1{2\cos^2x-1} + \frac2{4\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | Modulo $7,$ $4^{2^n}+2^{2^n}+1 \equiv 2+4+1\equiv 0$ when $n$ is odd, and
to $4+2+1\equiv 0$ when $n$ is even.
(Start with $4^{2^n}\equiv 2$
and $2^{2^n}\equiv 4$ when $n=1.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Taylor expansion of $\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$ Taylor Series Expansion of$$\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$$
My try-
$$\cos(\ln (n+1))= \cos\left (\ln \ n+ \ln\left(1+\frac{1}{n}\right)\right)$$$$=\cos (\ln\ n) \cos\left(\ln\left(1+\frac{1}{n}\right)\right)-\sin(\ln \ n)\sin\left(\ln \left(1+\frac{1... | Near $x=0$, we have
$$
\log (x + 1) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots ,
$$
$$
\cos (x) = 1 - \frac{1}{2}x^2 + \cdots
$$
and
$$
\frac{1}{{\sqrt {x + 1} }} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 -\frac{5}{16}x^3 + \cdots .
$$
Thus
\begin{align*}
& \frac{{\cos (\log (x + 1))}}{{\sqrt {x + 1} }} = \\ &= \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve the equation $x=1-5(1-5x^2)^2$ Solve the equation
$$x=1-5(1-5x^2)^2$$
###My work
Let $f(x)=1-5x^2$. Then we have tha equation $f(f(x))=x$. But in this case we don't use the equation $f(x)=x$ because $f(x)$ is not monotonic function
| You are on the right track; a modification of your approach will get you the rest of the way.
Let $y = f(x) = 1 - 5x^2$. Then $f(f(x)) = x$ implies $f(y) = x$, hence the solution to the simultaneous system $$\begin{align} y &= 1 - 5x^2, \\ x &= 1 - 5y^2, \end{align}$$ will yield the desired $x$-values. To this end, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3... | We have:
$$\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{y}{3} \tag{1}$$
$$\left( \frac{3}{6} \right)^y+\left( \frac{4}{6} \right)^y+\left( \frac{5}{6} \right)^y =\frac{z}{3}\tag{2}$$
$$\left( \frac{3}{6} \right)^z+\left( \frac{4}{6} \right)^z+\left( \frac{5}{6} \right)^z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to calculate the absolute extrema for this equation Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.
$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$
I have figured out the derivative, which is
$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\... | Hint: You can use a difference of squares with the derivative:
$$ \frac{\pi}{8} \sec^2 \left( \frac{\pi x}{8} \right) - \frac{4}{x^2} = 0 $$
$$ \Rightarrow \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) + \frac{2}{x}\right) \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) - \frac{2}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
how to determine a transformation matrix for transforming x and y with specific equation? I need to determine a matrix that can be used to:
a) transform $x$ and $y$ using equations
$$
\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right.
$$
Then I need to:
b) Transform a triangle $(0,0),(1,0),(1... | $\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right. \Rightarrow \left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right.$
$\left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right. \Rightarrow \begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Nested sums - reciprocal of a sum : Find exact value $$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
\sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right )^2}$$
(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)
At first I thought it has something to do with Rie... | (This is not a solution.)
The question of convergence is raised in the comments, so I show a proof here.
From $k > 0$ and
$$ k \leq j \leq k^2 $$
we have
$$ \frac{1}{\sqrt{k}} \geq \frac{1}{\sqrt{j}} \geq \frac{1}{k} \text{,} $$
using the monotonicity of the square root and standard results about reciprocals and ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Determine the following sum Let $S(n)$ be the sum of the digits of n in its binary representation. For example the binary notation for 19 is $10011$ and $S(19)=3$. Determine $\sum_{n=1}^{\infty} \frac{S(n)}{n(n+1)}$?
I've made several tables trying to pinpoint any useful pattern, but I've had minimal luck. The best I'd... | The key idea is to expand over digits and find a recurrence. Let $$\require{cancel}F(N)=\sum_{n=1}^{2^N}{\frac{S(n)}{n(n+1)}}=\sum_{n=1}^{2^N}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)}$$ and \begin{align*}
G(N)&=\sum_{n=0}^{2^{N-1}-1}{\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)}-\left(\frac{1}{2^N+1}-\frac{1}{2^N+2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$ $\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?
$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$
10/inf < 1 ---... | Using
$$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$$
we come to limit $\frac{a^n}{n!}$ for $a \gt 1$. Of course exists natural $k=\lfloor a\rfloor \leqslant a \leqslant \lfloor a\rfloor +1$. So we have
$$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
sequence limits - correctness of the adjustment I would like to ask why the following modification of the example is not allowed $$\displaystyle\lim_{n\rightarrow \infty} \frac{1+2+ \cdots + 2^n}{1+5+ \cdots + 5^n} = \lim_{n\to \infty}\left(\frac{2^n}{5^n}\right)\cdot \frac{\left(\frac{1}{2^n}+\frac{2}{2^n}+.....+1\rig... | To actually evaluate the limits I'd go as follows:
$$\frac{1+2+\ldots+2^n}{1+5+\ldots+5^n}=\frac{\frac{1-2^{n+1}}{1-2}}{\frac{1-5^{n+1}}{1-5}}=4\frac{1-2^{n+1}}{1-5^{n+1}}=4\left(\frac25\right)^{n+1}\;\frac{\frac1{2^{n+1}}-1}{\frac1{5^{n+1}}-1}\xrightarrow[n\to\infty]{}4\cdot0\cdot1=0$$
As for the second one:
$$\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove $a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$? We have $ad-bc=1$ . prove
$$a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$$
To solve it, I multiplied the inequality by $2$ and added $2(ad-bc)=2$ to it :
$$2a^2+2b^2+2c^2+2d^2+2ac+2bd+2ad-2bc\ge2+2\sqrt3$$
$$(a+c)^2+(b+d)^2+(a+d)^2+(b-c)^2\ge 2+2\sqrt3$$
From here I don't know how to pr... | by AM-GM $$a^2+b^2+c^2+d^2\ge 2\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{{(ac+bd)}^2+{(ad-bc)}^2}$$ $$=2\sqrt{{(ac+bd)}^2+1}$$ Let $ac+bd=t$ we have to prove $$2\sqrt{t^2+1}+t\ge \sqrt{3}$$ Can you end it now?
we have to prove $$2\sqrt{t^2+1}\ge \sqrt{3}-t$$ if $\sqrt{3}-t\le 0$ the inequality is obvious other wise squaring b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct re... | Here it is a slightly different approach from @amirali.
According to the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$, we conclude that
\begin{align*}
\lim_{\theta\to0}\frac{1-\cos(\theta)}{\theta\sin(\theta)} & = \lim_{\theta\to 0}\frac{2\sin^{2}(\theta/2)}{2\theta\sin(\theta/2)\cos(\theta/2)}\\\\
& = \lim_{\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$ Someone please help me to solve this integral
$$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$$
I have tried to use the formula
$$ab^2=\frac{(a+b)^3+(a-b)^3+2a^2}{6}$$
to reduce the original integral to three integrals
$$\frac{1}{6}\int_0^1\frac{\ln^3(x+x^2)}{1-x}\,dx+\frac{1}{6}\int_0... | $
\displaystyle\int_0^1\frac{\ln^2(x)\ln(1+x)}{1-x}dx=\displaystyle\int_0^1\displaystyle\int_0^1\frac{x\ln^2(x)}{(1+xy)(1-x)}dxdy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(\displaystyle\int_0^1\frac{\ln^2(x)}{1-x}-\frac{\ln^2(x)}{1+yx}dx\right)dy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(2\zeta(3)-\displaystyle\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{... | Here is a solution with elementary math.
Let's denote
$$S_n = \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$$
You have
$$2S_n = 1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}$$
Then
\begin{align}
S_n = 2S_n - S_n &= 1+\left(\frac{3}{2}-\frac{1}{2} \right)+\ldots+\left(\frac{2n-1}{2^{n-1}} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$ I hit a wall on this question. Below are my steps
$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$
Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$
$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)... | You are right up to
$$\frac{7}{2a^{2}}(\int \frac{du}{2u+a^{2}}-\int \frac{du}{2u-a^{2}})
= \frac7{4a^2}\ln\frac{2u+a^2}{|2u-a^2|}= \frac7{4a^2}\ln\frac{x^2+a^2}{|x^2-a^2|}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequality $\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1}$ From this question How do you calculate this limit? $\lim_{n\to\infty}( \frac{1^2}{n^3+1^2} + \frac{2^2}{n^3+2^2} ... + \frac{n^2}{n^3+n^2})=?$
I have not understood the reason of the @xpaul's answer for the $... | Since the sum ranges from $k=1$ to $k=n$, we have $k\geq 1$ and $k\leq n$; thus $n^2\geq k^2\geq 1$. Translate by $n^3$ to get $n^3+n^2\geq n^3+k^2\geq n^3+1$, then invert (changing the direction of the inequalities) to get $\dfrac1{n^3+n^2}\leq \dfrac1{n^3+k^2}\leq \dfrac1{n^3+1}$. Multiply by $k^2$ to get $\dfrac{k^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Related rates +implicit differentiation cone problem I'm stuck at the following related rates problem:
As shown in the diagram the apex angle of a cone is $2\theta$ and the slope length is 10cm. The rate of increase of $\theta$ is 0.01 radians per second. The cone starts with $\theta=\frac \pi 6$.
Find the initial rate... | Your answer is correct. The book is incorrect by a factor of $10$.
To simplify your calculation, we have $$V(t) = \frac{1000\pi}{3} \sin^2 \theta(t) \cos \theta(t) = \frac{1000\pi}{3} (\cos \theta(t) - \cos^3 \theta(t)).$$ At time $t = 0$, $\theta(0) = \pi/6$ and $\theta'(0) = \frac{1}{100}$, so $$\begin{align}
V'(0) ... | {
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"url": "https://math.stackexchange.com/questions/4044762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the marginal PDF I am having a hard time trying to find the marginal of the following joint pdf.
$$f_{U,V}(u,v) =2 \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1}$$
where $ u<v<\frac{u+\theta+1}{2}$ and $\theta < u <\theta+1$.
This was m... | Your support is: $\theta\lt u< \theta +1$ and $u< v< (u+\theta+1)/2$.
That is equivalently: $\theta\leq v<\theta+1$ and $\max(\theta,2v-\theta-1)\leq u<v$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$ If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
The solution goes as follows:
$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+... | Well it's an attempt to delete the square root .
My only idea is to use Bernoulli's inequality and play with the coefficient $\sqrt{2}$.We have :
$$\sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}\leq \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}\frac{xyz}{(x+y)(y+z)(x+z)}\Big)\Big)$$
Remains to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Are there nontrivial rational solutions to $\sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{1-z^2}$? Obviously $(a,0,a)$,$(-a,0,a)$,$(0,a,a)$,$(0,-a,a)$ are solutions.
I tried finding solutions brute-forcing this problem, but I discovered there are no solutions with numerator and denominator smaller than 100'000.
I can prove that ... | Here is a way to generate solutions for $k=3$.
It is well known that every number whose prime factors are all of the form $6m+1$, is also of the form $n=a^2+3b^2$. Then
$$(2n)^2=4a^4+24a^2b^2+36b^4\\
=(2a^2-6b^2)^2+3(4ab)^2\\
=(a^2+6ab-3b^2)^2+3(a^2-2ab-3b^2)^2\\
=(a^2-6ab-3b^2)^2+3(a^2+2ab-3b^2)^2$$
Note two of the u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$ (Venezuela 1960)
If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $$\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$$
I tried solving it as follows:
The condition we want proved is: $\sqrt{3x^2+3-3yz}+\sqrt{3y^2... | The inequality ia actually very easy with just one hidden point kept in mind
Note that $$\sum \sqrt{x^2+1-yz}\ge \sum \sqrt{x^2+1-\frac{{(y+z)}^2}{4}}$$ Now use the fact $$x^2+1-\frac{{(y+z)}^2}{4}=x^2+\frac{{(x+y+z)}^2}{3}-\frac{{(y+z)}^2}{4}=\frac{1}{12} {(4x+y+z)}^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac... | I think the following step is wrong:$$\lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}.$$
I think you have used the false identity $$ \ln\cos2x \equiv 2\ln\cos x - 2\ln\sin x. $$
It is fal... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.
I solved it as follows:
$x+2\sqrt{x}-1\ge 0$
$(\sqrt{x}+1)^2-2\ge 0$
$(\sqrt{x}+1)^2\ge 2$
$... | The requirement $x\ge0$ is obvious. The expression is real iff $(\sqrt{x}\pm1)^2\ge2$ for both choices of $\pm$, i.e. neither $\sqrt{x}\pm1$ is in $(-\sqrt{2},\,\sqrt{2})$, i.e .$\sqrt{x}\notin[0,\,1+\sqrt{2})\cup[0,\,\sqrt{2}-1)=[0,\,1+\sqrt{2})$, since $\sqrt{x}\ge0$. The solution set is $[3+2\sqrt{2},\,\infty)$, as ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Two strange integrals I was asked to solve the two integrals
$$
\int\frac{1}{x^2e^{1/x}-x}dx
$$
$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$
I think the first one is not soluble with the known methods of calculus and the second one is involved with complex roots of cubic equation.
Any improvements appreciated.
Thanks.
| *
*For the integral
$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$
note that the cubic polynomial in the denominator has one real root $r= -0.5437$ and a pair of complex roots, which allows the factorization
$$x^3-x^2+x+1= (x-r)(x^2+(r-1)x-1/r)
$$
Then, decompose the integrand as
\begin{align}
\frac{x^2+1}{x^3-x^2+x+1}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Comparison of the largest eigenvalues of two symmetric matrices Let $$A=\left(\begin{array}{ccccc|cccc}0&\cdots&0&0&0&0&0&a\\\vdots&\cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\0&\cdots&0&0&0&0&0&a\\0&\cdots&0&0&b&0&0&0\\0&\cdots&0&b&0&b&0&0\\\hline 0&\cdots&0&0&b&0&\frac{1}{\sqrt{2}}&0\\0&\cdots&0&0&0&\frac{1}{\s... | Not an answer, just too long for a comment.
Well, I decided to try and brute-force the computation of the characteristic polynimial for $A$ and got the following:
let $D_n$ be the determinannt of $tI_n-A$ (asumme $n\geq6$ so $A$ will contain the required $a$'s), so:
$$D_n = tD_{n-1} -a^2t^{n-6}(t^4-6b^2t^2+b^4)$$
and:
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation with $ \sqrt[5]{}$ How to show that this equation $ \sqrt[5]{20-2x} + \sqrt[5]{7-x}+\sqrt[5]{3x+5}=2$ have 3 solutions 9, -6 and -25
Wolframsays the equation have no solutions !
| If $\sqrt[5]{20-2x}= a$ etc., $a+b+c=2$
$$a^5+b^5+c^5=32=(a+b+c)^5$$
Use How to factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$? to discover that $(a+b)(b+c)(c+a)=0$ should give three solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$
Attempt:
First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...
N... | You can combine the answers from S.H.W and José Carlos Santos to figure out a solution if you don't know the sum $\sum^\infty_{k=0} 2^{-k}$.
$$
\begin{align}
\sum^\infty_{k=2} 2^{-k} &= \sum^\infty_{k=0}2^{-k} - 1 - \frac{1}{2} \tag{S.H.W} \\
&=2^{-2}\sum^\infty_{k=0}2^{-k} \tag{José Carlos Santos}
\end{align}
$$
If we... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true? Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)... | It can be shown that
\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{1}{{n^s }}} & = \prod\limits_p {\frac{1}{{1 - \frac{1}{{p^s }}}}} = \exp \left( { - \sum\limits_p {\log \left( {1 - \frac{1}{{p^s }}} \right)} } \right) \\ &= \exp \left( {\sum\limits_p {\sum\limits_{n = 1}^\infty {\frac{1}{{np^{ns} }}} } } \right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting th... | With $n:=2m+1$, the polynomial is
$$(2m)^2(2m+2)^2(4m^2+4m+2)^2(16m^4+32m^3+24m^2+8m+2)
\\=2^7m^2(m+1)^2(2m^2+2m+1)(8m^4+16m^3+12m^2+4m+1).$$
The last two factors are always odd and the remaining ones form at least a multiple of $2^9$.
More precisely, if the multiplicity of the factor $2$ in the prime factorization of... | {
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If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$? If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
I tried to solve it as follows:
I state that $p=ab$
$p^2=(2b+15)(2... | $a^2=2b+15$ and $b^2=2a+15$
Subtracting, $a^2-b^2 = -2(a-b)$. As $a \ne b$,
$a+b = - 2$
Also adding both equations, $a^2+b^2 = 2(a+b)+30 = 26$
$(a+b)^2 = a^2+b^2+2ab \implies 4 = 26 + 2ab$
$ab = -11$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $2^k +k \equiv 0 \pmod {323}$ Find all $k$ such that
$$2^k + k \equiv 0 \pmod{323}.$$
I noticed that $323 = 17\cdot 19$ so I thought about using the Chinese Remainder theorem by considering $2^k+k$ modulo 17 and 19. I got $k \equiv 16 \pmod{17}$ and $k\equiv 18 \pmod{19}$, which gives $k\equiv 322 \pmod{323}$.... | Okay to solve $2^k \equiv \pmod {17}$ we know by FLT than $2^{16}\equiv 1 \pmod {17}$ and as $2^4 =16 \equiv -1$ that $2^8\equiv 1 \pmod {17}$.
So if $k \equiv 0, 1,2,3,4....,7\pmod {16}$ we have $2^k\equiv 1,2,4,8,-1,-2,-4, -8\pmod {17}$ and by CRT will be a unique solution for each pair. $k\equiv a \pmod {16}$ and $k... | {
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Where does the Integral went wrong .... $ \displaystyle \int _{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x $ This is how I solved :
Let $$ \displaystyle y = \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x
$$
Then I used expansion of $\displaystyle \ln (1+x) = \... | Your mistake is thinking $\int_0^{\pi/2}\cos^nx\mathrm{d}x$ is a Beta function; it's only half of one, viz.$$\operatorname{B}(a,\,b)=2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\mathrm{d}x.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$. Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$
Or notice that $(x+y)^3 = x^3 + 3... | If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}
| {
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Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$ Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+... | We’ll let $a = x + 1$
Transform the inequality by substitute $x$ with $a - 1$.
After some calculations we the get
$$ -a^2 + 6a + 3 \leq 8\sqrt{a} $$
Square both sides and we get
$$ a^4 - 12a^3 + 30a^2 - 28a + 9 \leq 0$$
Which is just $(a-1)^3(a - 9) \leq 0$.
Since $a > 1$ , this statement is true for all $a \in [1,9]$.... | {
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"timestamp": "2023-03-29T00:00:00",
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From the given information, how do we use the Remainder Theorem to reach the answer? Why does the sequence for the polynomial alternate when it equals $-30$, and not alternate when it equals $10$? Why is the denominator when the sum of the geometric series is equal to $30$, $1+r$, and not $1-r$? I am unable to correc... | The polynomial is a cubic whose coefficients are a geometric sequence (in the order of descending powers, i.e. starting with $x^3$ and ending with the constant term), meaning that our polynomial must take the form
$$p(x) = ax^3 + arx^2 + ar^2x + ar^3.$$
Remainder theorem states that the remainder of $p(x)$ when divided... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$ With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?
$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$
Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count ... | Either you don't select $2$ or you select $2$. In the first case you have 3 choices: $(2, 2, 1); (2, 1, 2); (1, 2, 2)$. For each such choice you have $\frac{5!}{2!2!}$ allocations. If you select $2$, you have two paths:$(2, 2, 0), (2, 0, 2), (0,2,2)$, in which case you have $\frac{5!}{2!2!}$ allocations and $(1, 1, 2),... | {
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Congruency application I want to prove whether the following is true using congruencies: If $n$ is odd and $3\not | n$, then $n^2\equiv 1\pmod{24}$.
I tried a direct proof.
Let $n=2k+1$ for an integer $k$ such that $3\not| n$, or $3c=n+r$ such that $r=1$ or $r=2$ for an integer $c$. Then in the first case, $3c=n+1$ imp... | It's the same as proving that $n^2-1 \equiv 0 \pmod 24$.
Since $n \equiv 1 \pmod 2 \implies (n-1)(n+1) \equiv 0 \pmod 8$ (as one of $n-1, n+1$ will be $2 \pmod 4 $ and the other will be $0 \pmod 4$)
and since $n \neq 0 \pmod 3$, one of $n-1,n+1$ must be $0\pmod 3 $ $\implies $(n-1)(n+1) \equiv 0 \pmod 3 $
$\implies n^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the maximal area between hyperbola and chord $AB$?
Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?
Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sq... | Disclaimer: I am not sure since I am still new to calculus.
Find the minimum value of $f(t) = \displaystyle\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx$.
The fundamental theorem of calculus can be applied here:
\begin{align*}\dfrac{d}{dt}f(t) &= \dfrac{d}{dt}\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx \\ f'(t) &=\left(\sqrt{(t + h... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the area of a sector within a triangle.
For part (a) of the question I am getting an answer of $33.6^{\circ}$ of $0.586 radians$ which I am pretty sure is right.
Part (b) is where I am having difficulties because my answer is not matching the mark scheme.
For b (i) I have an answer of $4.516 cm$
for (ii) I hav... | $|AB|=5, |AC|=|BC|=3=|AX|=|YB|$ and since $|ABC|$ is isosceles, sum of angles in $|ABC|$ is $180°$, then $2\theta_1+\theta_2= 180°$, from cosine rule
$$ \cos(\theta_2) = \frac{ 3^2+3^2-5^2}{2×3^2}$$
$$\theta_2 = \cos^{-1}(\frac{-7}{18})$$
$$\theta_1 = 90° - \frac{1}{2}\theta_2$$
$$\theta_1 = 90° -\frac{1}{2} \cos^{-1}(... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$ Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$
This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$... | No matter what, by simply calculating one has $P(\sqrt 3, 0, 0) = \sqrt 3$. On the other hand,
We know a minimum must exist by Weierstrass's theorem. This is because $Z := \{(a, b, c) \in \mathbb R ^3\mid a, b, c \geq 0, a^2 + b^2 + c^2 = 3\}$ is a closed bounded set and $P$ is continuous on $Z$.
Write $$P(a,b,c) = S... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $x^x+y^y\ge x^2+y^2$ for $x,y>0$ and $x+y\le 2$. We may prove the inequality for $x,y\in (0,1]$.
Note that, for $0<x\le 1$, it holds that
\begin{align*}
x^x&=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\cdots\\
&\ge1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\\
&\ge x^2.
\end{align*}
Similarily, for $y \in (0,1]$, it holds that $$... | Remarks: @Q. Zhang gave a nice proof. I give an alternative proof here.
Fact 1: $f(v) = v^v - v^2$ is strictly decreasing on $(0, 1)$.
The proof is given at the end.
Fact 2: $u^u \ge \frac{u^2 - u + 2}{3 - u}$ for all $u$ in $(0, 2)$.
The proof is given at the end.
(Note: $\frac{u^2 - u + 2}{3 - u}$ is the Pade $(2,1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to proceed solving $(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$? I have$$(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$$
By noticing that $7+4\sqrt3=\frac1{7-4\sqrt3}$ one way to solve the equation is using substitution $(7+4\sqrt3)^m=t$ and solve for $t$ in $t+\frac1t=14$.
But I'm trying to use a little different approach:
We have $7+4\sqr... | $m = 1$ is a solution, which is easily seen. Suppose that $m \neq 1$. Clearly, $m = 0$ is not a solution Because $u + \dfrac{1}{u} \neq 2$.
First, suppose that $m > 0$. Then, we would have
$$u^m - u = \dfrac{1}{u} - \dfrac{1}{u^m}.$$
Here, we can use $a^k - b^k = \left( a - b \right) \left( a^{k - 1} + a^{k - 2}b + \cd... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Dirichlet problem on the unit disk using Poisson’s formula I’ve been trying to do the following exercise, of course without success, because I’m struggling with the integral. First things first, here's my exercise:
Writing the unit disk as $D \subset \mathbb{R}^2$, we define $g \in C(\partial D)$ by $g(x,y)=$ $\begin{c... | You can use the Poisson formula for the unit disk on the form
\begin{equation*}
u(r\cos\theta,r\sin\theta) = \frac{1-r^2}{2\pi} \int_0^{2 \pi} \frac{h(\phi)}{1 - 2r \cos(\theta - \phi) + r^2} \, d\phi
\end{equation*}
where $h(\phi)$ is the function giving the boundary data, in this case (since $y=\sin\phi$ on the un... | {
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"url": "https://math.stackexchange.com/questions/4090465",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove the identity $\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}$ Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})... | Since the roots on the LHS get "annihilated", it makes very much sense to look for natural $m,n$ such that
\begin{eqnarray*}
17+2\sqrt{30} & = & \left(\sqrt n + \sqrt m\right)^2 & = & n + m + 2\sqrt{nm}\\
17-2\sqrt{30} & = & \left(\sqrt n - \sqrt m\right)^2 & = & n + m - 2\sqrt{nm}
\end{eqnarray*}
From this approach yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
| Set $x = 1 + y^2 \Rightarrow y = ±\sqrt{x - 1}$ so that $1 + \tan^2 A, 1 + \tan^2 B, 1 + \tan^2 C$ give the roots of $x^3 - ax^2 + b = 0$.
This gives $±(x-1)^{3/2} - a(x-1) +b=0$. By Vieta, the product of the roots is the constant term, but we do not have a polynomial yet. Squaring both sides of $±(x - 1)^{3/2} = a(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2... | Everything is OK.
First: Accept
$$a^2≥b^2$$
Then accept
$$a^2≤b^2$$
You will see that the roots have not changed.
We have
$$abx^2-(a^2+b^2)x+ab=(a x - b) (b x - a)$$
Small Supplement:
Note that,
$$\begin{cases} ±|a^2-b^2|=±(a^2-b^2), a^2≥b^2 \\ ±|a^2-b^2|=±(b^2-a^2), a^2≤b^2 \end{cases}$$
$$\implies \pm(a^2-b^2)=\mp(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Solving an SDE with Ito's Lemma I want to solve the initial value problem
$$ dX_t = \left(\frac{b^2}{4} - X_t\right)dt + b\sqrt{X_t}dw$$
I have the initial condition $X_0 = x > 0$. Note this process stops when $X_t = 0$.
I'm pretty sure I need to apply Ito's formula here but I'm not sure how. I tried making the substit... | Yes, we need to apply the Ito's formula. We denote $\tau = \inf\{t\geq 0 : X_t = 0\}$ and denote $W$ the Brownian motion.
For $t < \tau$, we can apply Ito's formula to the function $\phi(x) = \sqrt{x}$ using the Ito process $X$. Thus,
\begin{align}
dZ_t = d(\sqrt{X_t}) &= \frac{1}{2\sqrt{X_t}}dX_t - \frac{1}{8(X_t)^{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092987",
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"source": "stackexchange",
"question_score": "1",
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What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
| HINT Divide by $x^3y^3$ on both sides. You will get a new equation in terms of $1/x$ and $1/y.$ Can you finish from here?
| {
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"source": "stackexchange",
"question_score": "2",
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How to show that $5/(1+2^{2/3}) = 1+2\sqrt[3]{2}-2^{2/3}$? I was trying to simplify the solution, $\frac{1}{5}(1+2\sqrt[3]{2}-2^{2/3})$, from a problem I done and saw that mathematica suggested $\frac{1}{1+2^{2/3}}$. I was wondering if anyone could show how to approach simplification of such expressions manually (it ap... | This is the cubic formula from basic algebra:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
$$\frac{a^3+b^3}{a+b}=a^2-ab+b^2$$
Observe,
$$\begin{align}(1+2^{2/3})×(1+2\sqrt[3]{2}-2^{2/3})&=1^3+2^2\\
&=5.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| Yet an other answer following the same natural lines.
The cases when $x=0$ and/or $y=0$ are clear from the OP. We search for other, "new" solutions.
Let $a=(x,y)$ be the lcm of $x,y$, taken now to be $\ge 1$,
so we can and do write $x=aX$, $y=aY$ with relatively prime integers $X,Y$.
Then the given equation is equivale... | {
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"url": "https://math.stackexchange.com/questions/4100375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$ Integrate $$\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$$
I tried $2\cdot\int\frac{3x^4+2x^2+4x+2}{(1-2x)^3}dx\:\:$ now after partial fractions i have:
$$\begin{align}
&\>\>\>\>\>\frac{3x^4+2x^2+4x+2}{(1-2x)^3}\\&=\frac{A}{(1-2x)}+\frac{B}{(1-2x)^2}+\frac{C}{(1-2x)^3}\\[1ex]... | Substitute $t=1-2x$. Then $x=\frac{1-t}2$
\begin{align}
& \int \frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx\\
=& \int \frac{6(\frac{1-t}2)^4+ 4(\frac{1-t}2)^2+ 8(\frac{1-t}2)+4}{t^3}(-\frac{dt}2)\\
=& -\frac1{16} \int \frac{3t^4 -12t^3+26t^2-60t+75}{t^3}dt
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\sum_{cyc}\frac{ab}{c^5}$, where $a$, $b$, $c$ are the roots of $x^3-px+1=0$. How to avoid a large expansion?
Given the polynomial $x^{3}-p{x}+1=0$, evaluate $\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}$ in terms of $p$ if $a,b$ and $c$ are the roots of the polynomial.
My attempt involved:
$$\frac{... | As Martin R noted we have to find the value of $$t_6=p^6+q^6+r^6$$ where $p.q.r$ are the roots of $x^3-px^2+1$. Now let $t_k=p^k+q^k+r^k$ for $k\ge 3$ indeed from $x^3=px^2-1$ we get the recurrence $$t_k=pt_{k-1}-3$$ Now we have to find $t_6$ when we know the value of $t_2=p^2$....
The rest should be easy
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the integral: $\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_r^2(\theta) \, d\theta$ Let, $\displaystyle Q_r(\theta) = \frac{r \sin \theta}{1 - 2r\cos \theta + r^2}$ Find the integral: $\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_r^2(\theta) \, d\theta$.
I know one thing that $Q_r(\theta)$ is the imaginary part of the series $\displaysty... | Integrate both sides of the equation
$$\left( \frac{2r\sin \theta}{1 - 2r\cos \theta + r^2}\right)’
=-\frac{1+r^2}{1 - 2r\cos \theta + r^2}+\frac{(1-r^2)^2}{(1 - 2r\cos \theta + r^2)^2}
$$
to establish
$$
\int_0^\pi \frac{(1-r^2)^2}{(1 - 2r\cos \theta + r^2)^2}d\theta
= \int_0^\pi \frac{1+r^2}{1 - 2r\cos \theta + r^2}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1 - \lim_{x \to \infty} \left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2}$ I'm currently working on my master thesis and I need to solve this limit.
I forgot almost everything about limits since last time a I saw them was basically in high school.
Anyway, I solved this lim... | We shall use the expansion $\log(1+x) = x + O(x^2)$ as $x\rightarrow 0$.
Write $$
\begin{align*}\left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2} &= \exp\left\{kx^2\log \left(1-\frac{2p}{x^2}+\frac{2p}{x^2e^{x^2/2}}\right)\right\} \\
&= \exp\left(kx^2\left(-\frac{2p}{x^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try. Examine the convergence of:
$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$
Pointwise convergence (for every $ x \in \mathbb{R}$ ):
$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$
Uniform convergen... | If the series converged uniformly on $\mathbb{R}$, then for any $\epsilon > 0$ there exists $2N+1 \in \mathbb{N}$ such that for all $x \in \mathbb{R}$,
$$\left|\sum_{n= 2N+1}^\infty x \frac{\sin n^2 x}{n^2} \right| < \epsilon$$
In particular, for $x_m = \frac{\pi}{2} + 2m\pi$ we have $\sin n^2x_m = \sin \left( n^2\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Prove that: $x^x\cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $ Prove that:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $$
for every $x, y > 0$
Where I got stuck:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} \iff \ln(x^x \cdot y^y) \geq \ln \left[\left(\frac{x+y}{2} \right)^{x+y} \right] $$
$$x\... | First, rewrite your inequality as $$\frac{x\ln(x)+y\ln(y)}2\geq \left(\frac{x+y}2\right)\ln\left(\frac{x+y}2\right)$$
Now if $f(z)=z\ln z$, then $f'(z)=1+\ln z$ and $f’'(z)=\frac1z$ which is larger than $0$ for $z\in(0,\infty)$. Hence, $z\ln z$ is convex in the domain $(0,\infty)$.
As $x,y\in(0,\infty)$, we can conclud... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ I am trying to find the area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ using Double integrl in Polar coordinates.
Now as shown in th... | Try setting your integral as
$A=2\int_0^\frac{\pi}{4}\int_{\sin\theta}^{\cos\theta}rdrd\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$x \mathcal{o}\left(\frac{1}{x^3}\right) =\mathcal{o}\left(\frac{1}{x^2}\right) $? Is this correct:
$$x \mathcal{o}\left(\frac{1}{x^3}\right) = \mathcal{o}\left(\frac{1}{x^2}\right)$$
or the expression $x \mathcal{o}\left(\frac{1}{x^3}\right)$ cannot be simplified?
Here $\mathcal{o}(\cdot)$ stands for little $\mathcal{... | $f(x) = o\left(\dfrac{1}{x^2} \right)$ if $\lim_{x \rightarrow \infty} \ x^2f(x) = 0.$
Define $g(x) = \dfrac{1}{x}f(x)$, so $g(x) = \dfrac{1}{x}o\left(\dfrac{1}{x^2}\right)$, but
$$\lim_{x \rightarrow \infty} \ x^3g(x) =\lim_{x \rightarrow \infty} \ x^2f(x) = 0$$ so $g(x) = o\left(\dfrac{1}{x^3}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show this $x_{n+1}-x_{n}\ge 2\pi$ let $f(x)=e^x\cos{x}-\sin{x}-1$,and $n$ be postive integer,such $x_{n}$ be a root of $f(x)=0$ ,and
$\dfrac{\pi}{3}+2n\pi<x_{n}<\dfrac{\pi}{2}+2n\pi$,show that
$$x_{n+1}-x_{n}\ge 2\pi,\forall n\in ^{+}\tag{1}$$
My try: since
$$\dfrac{7\pi}{3}+2n\pi<x_{n+1}<\dfrac{5\pi}{2}+2n\pi$$
and
$... | We have $f'(x) = \mathrm{e}^x(\cos x - \sin x) - \cos x < 0$ on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Thus, $f(x)$ is strictly decreasing on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Also, $f(\pi/3 + 2n\pi) > 0$ and $f(\pi/2 + 2n\pi) < 0$. So, $f(x)=0$ has exactly one real solution (namely, $x_n$)
on $(\pi/3 + 2n\pi, \pi/2 + 2n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Series convergence by Gauss Test I want to try to prove that the $$\sum_{n=1}^\infty \frac{n!e^n}{n^{n+p}}$$ series convergent for $p>1.5$ with Gauss Test but failed? Gauss Test said if
$\frac{a_n}{a_{n+1}}$ can be represnted as
$\frac{a_n}{a_{n+1}}=\lambda+\frac{\mu}{n}+b_n$
where $\sum_{n=1}^\infty b_n$ absolutely c... | Proof with Stirling formula :
You have
$$\frac{n!e^n}{n^{n+p}} \sim \sqrt{2\pi n} \frac{n^ne^n}{e^nn^{n+p}} \sim \frac{\sqrt{2\pi}}{n^{p-\frac{1}{2}}}$$
which is the general term of a convergent series, since $p>1,5$. So the given series converges.
Proof with Gauss test :
Let $$a_n = \frac{n!e^n}{n^{n+p}}$$
You have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How does one treat big O when Taylor expanding $\sin(\sin x)$? I am working on finding the Taylor (Maclaurin expansion) of $\sin(\sin(x))$ to the third order.
We have $$\sin{t} = t - \frac{t^3}{6} + \mathcal{O}(t^5)$$
If I then set $t = \sin(x)$, and expand once more, I end up with a big O which is
$$\mathcal{O}((x - \... | The first line can be written as $\sin t = t - \frac{t^3}{6} + h(t) t^5$ for some function $h$ such that $|h(t)| \le C$ for $t$ near $0$.
Then,
\begin{align}
\sin \sin x
&= \sin x - \frac{(\sin x)^3}{6} + h(\sin x) (\sin x)^5
\\
&= \left(x - \frac{x^3}{6} + h(x) x^5\right)
- \frac{1}{6} \left(x - \frac{x^3}{6} + h(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of possible paths on a lattice grid given a restriction
On a $6 \cdot 6$ lattice grid, an ant is at point $(0,0).$ There is a teleportation pad at $(2,2)$ and $(3,3).$ When an ant reaches either teleportation pad, it teleports the ant from the pad it is on to the other pad, and then the pad disappears. (Meaning... | *
*How many paths are there that avoid both teleportation pads?
*How many paths are there from $(0,0)$ to the first pad at $(2,2)$?
*How many paths are there starting from the second pad at $(3,3)$ to $(5,5)$? (Hint: compare with #2 above.)
*How many paths are there from $(0,0)$ to the second pad at $(3,3)$, avoi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $b$ and $d$ such that $2x^4+ax^3+bx^2+cx+d $ has roots $x=\sqrt2-3$ and $x=3i+2$
I've got the following function:
$$f(x)=2x^4+ax^3+bx^2+cx+d$$
Given the two roots $x=(\sqrt2-3)$ and $x=(3i+2)$, I need to compute values $b$ and $d$.
Can anyone point me in the right direction of how to start this?
Thanks in adv... | This is a basic application of Vieta's formulas. Here is one way we can approach the problem:
Assuming rational coefficients, there are two conjugate pairs we are dealing with, namely:
*
*the surd conjugate pair: $\left(\sqrt 2-3\right)$ and $\left(-\sqrt 2 - 3\right)$, whose sum is $-6$ and product is $7$
*the comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral as a limit of sum: $\lim\limits_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)$
The value of$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)=\,?$$
Can the limit be partially applied to the denominator after converting the numerator into an integral?
I wrote this as $$\lim_{n\to\infty}\fr... | One the one hand
$$
\mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 + n + 2k}}} \le \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 }}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} = \int_0^1 {x\, dx} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Computing limit based on upper and lower Riemann sums I have a question where it is asking me to compute the $$\lim_{n \to \infty} \frac{(\sqrt{1}+\sqrt{2}+...+\sqrt{n})}{(n\sqrt{n})}$$ based on the integral of sqrt(n) from the interval of $[0,n]$. Based on my previous answer, I was able to use the upper Riemann sum si... | $\displaystyle\mathbf{Method}\mathbf{1}:$
$\displaystyle\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}}{n\sqrt{n}}=\lim_{n\rightarrow \infty} \frac{1}{n}\left( \sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots +\sqrt{\frac{n}{n}} \right) =\int_0^1{\sqrt{x}\text{d}x}=\frac{2}{3}x^{\frac{3}{2}}\mid_{0}^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\ell=6m-1$ is prime then $\ell\ne3\frac{j^2+3k^2}{j+3k}$ Let $\ell=6m-1$ for some integer $m\ge1$ be a prime and for any integer $1\le j\le (\ell-1)/2$, $$\dfrac{(j-m)(3j-3m+1)}{2}\neq\dfrac{\ell^2-1}{24}.$$ Then, I need a proof for the inequality that for integers $1\le j,k\le (\ell-1)/2$, $$\dfrac{(j-m)(3j-3m+1)}... | For $3|j+3k$, we must have $3|j$. Then, let $j=3t$ for some integer $t$. Then
$$\ell=\frac{3(j^2+3k^2)}{j+3k}=\frac{3(9t^2+3k^2)}{3(t+k)}=\frac{3(k^2+3t^2)}{t+k}.$$
So, $3|t+k$. Let $k=3u-t$; then
$$\ell=\frac{(3u-t)^2+3t^2}u=\frac{4t^2-6tu+9u^2}u.$$
So, $u|4t^2$. Consider any prime $p|u$. Also, since
$$u=\frac j9+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If bisector of angle $C$ of $\triangle ABC$ meet $AB$ in $D$ and circumcircle in $E$ prove that $\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$. If bisector of angle $C$ of $\triangle ABC$ meets $AB$ in point $D$ and the circumcircle in point $E$ then prove that $$\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$$.
My Attempt
Using the fact that... | Complete the cyclic quadrilateral $ACBE$ and note,
$$\angle ABE=\angle ACE=\angle BCE=\angle BAE=\frac{\angle C}{2}\implies AE=BE.$$
Using the Angle Bisector Theorem,
$$\frac{a}{b}=\frac{BD}{AD}\implies \frac{a+b}{b}=\frac{c}{AD}\implies AD=\frac{bc}{a+b}$$
Since $\triangle ADC\sim\triangle EDB,$
$$\frac{b}{AD}=\frac{B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| $$ \text{Let }S_{n}=1+x+x^2...+x^{n-1}$$
$$\implies xS_{n}=x+x^2....+x^{n}$$
Subtracting both equations,
$$S_{n}(1-x)=1-x^{n}$$
$$\implies S_{n}=\frac{1-x^n}{1-x}$$
Since it is an infinite series, $n\to\infty$ and it converges only when $|x|<1$. When $n\to\infty,x^n\to0$
$$\implies s_{\infty}=\frac{1-0}{1-x}=\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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How do I evaluate $\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$? I am trying to deduce if the following integral converges or diverges. If it converges I would also like to check for absolut convergence:
$$\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$... | It converges, but not absolutely.
$0<x\le1$:
For $0<x\ll1$ the integrand is approximately $\frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)}\approx x^{-1/2}\sin(x^{-3/2})$, which can be integrated and converges (but in a very oscillating manner):
$$
\int_0^1 x^{-1/2}\sin(x^{-3/2})dx
= E_{\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving $(x-2)^{x^2-6x+8} >1$ Solving this equation:
$(x-2)^{x^2-6x+8} >1$, by taking log on base $(x-2)$ both the sides, I get the solution as $x>4$. My work: Let $(x-2)>0$
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1 =0\implies x<2, or, x>4$$
But this doesn't appear to be the complete solution for instance $x=5/... | Be careful with $f(x)=\log_a x$, apart from $x>0, a>0$, there are two cases when $0 <a<1$ and $a>1$. In the former $f(x)$ decreases and increases in the latter.
So take
Case 1: $0<(x-2)<1 \implies 2<x<3$, then
$$(x-2)(x-4)\log_{(x-2)} (x-2) < \log_{(x-2)} 1=0 \implies (x-2)(x-4) <0 \implies 2<x<4.$$ Eventually $2<x<3.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Let $a$ and $b$ be two complex numbers such that $|a+b|=20$ and $|a^2+b^2|=16$ then find maximum and minimum value of $|a^3+b^3|$ Let $a$ and $b$ be two complex numbers such that
$|a+b|=20$ and
$|a^2+b^2|=16$
then find maximum and minimum value of $|a^3+b^3|$
My Attempt:
$|a^3+b^3|=|(a+b)((a+b)^2-3ab|)\leq |a+b|^3+3|ab... | We have $|a+b|=20\iff a+b=20e^{it}\iff ae^{-it}+be^{-it}=20$
So WLOG we can replace $a,b$ by phase shifted ones and assume $a+b=20$.
We also have $|a^2+b^2|=16\iff a^2+b^2=16e^{i\theta}$
$\begin{align}\Big|a^3+b^3\Big|
&=\Big|(a+b)^3-\tfrac 32(\underbrace{(a+b)^2-(a^2+b^2)}_{2ab})(a+b)\Big|\\
&=\Big|20^3-\frac 32((20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n$
Show that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n.$
I'm not very certain how to proceed on this problem. I know $x^2+y^2=(x+y)^2-2xy,$ so we essentially have $x+y=3n, (x+... | We are given
$$x + y = 3n, \; x^2 + y^2 - xy = 7n \tag{1}\label{eq1A}$$
As you showed,
$$\begin{equation}\begin{aligned}
(x + y)^2 - 3xy & = 7n \\
9n^2 - 3xy & = 7n \\
3xy & = 9n^2 - 7n \\
xy & = 3n^2 - \frac{7n}{3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
From this, plus the first part of \eqref{eq1A}, then us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Symmetric rational functions $x_1,x_2,x_3$ are the roots of the polynomial $x^3+px+q \in \mathbb{R}[x]$ with real coefficients.
Express through $p,q$ the following symmetric rational function
$$
\sum\left(x_1,x_2,x_3\right) = \frac{1}{x_1^2 + x_2^2} + \frac{1}{x_2^2 + x_3^2} + \frac{1}{x_3^2 + x_1^2}
$$
of $x_1,x_2,x_3... | $0=x_1+x_2+x_3$, $p=x_1x_3+x_1x_2+x_2x_3$ and $q=-x_1x_2x_3$
Now $x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=-2p$.
So your rational expression is
$$\frac{1}{-2p-x_1^2}+\frac{1}{-2p-x_2^2}+\frac{1}{-2p-x_3^2}$$
$$=-\frac{(2p+x_1^2)(2p+x_2^2)+(2p+x_1^2)(2p+x_3^2)+(2p+x_2^2)(2p+x_3^2)}{(2p+x_1^2)(2p+x_2^2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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