Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$ I have read in a few places that$$\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$$where $F_i$ denotes the $i$-th Fibonacci number. The series is a generalization of the more known $$\sum_{k=0}^{\infty} F_{k}z^... | Try using Binet’s formula for the Fibonacci numbers:
$$F_n=\frac{1}{\sqrt{5}}\bigg(\frac{1+\sqrt{5}}{2}\bigg)^n-\frac{1}{\sqrt{5}}\bigg(\frac{1-\sqrt{5}}{2}\bigg)^n$$
Combining this with the formula for the sum of a geometric series, we have that your sum is equal to
$$\begin{align}
\sum_{k=0}^\infty F_{mk}z^k &= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like... | Use the Chebyshev formulas: $$\int_0^{\frac{\pi}{4}} \frac{( \color{red}{\sin{(5x)} \color{black}{)^2}}}{\sin^2{x}} -\frac{( \color{blue}{\cos{(5x)} \color{black}{)^2}}}{\cos^2{x}} \,{dx}$$
$$=\int_0^{\frac{\pi}{4}}\frac{\left( \color{red}{ 16
\sin^5 (x)- 20 \sin^3( x) + 5\sin (x) } \right)^2}{\sin^2{x}} -\frac{ \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Implicit differantiation Let z and w be differantiable functions of x and y and they satisfy the following equations.
$$ xw^3+yz^2+z^3=-1$$ $$ zw^3-xz^3+y^2w=1$$
Find $\frac{\partial z}{\partial x}$ and its value at $(x,y,z,w)=(1,-1,-1,1)$.
I don't know what to do with two implicit functions. So I thought maybe I can w... | Since you have
\begin{align}
&x w^3 + y z^2 + z^3 = -1 \\
&z w^3 - x z^3 + y^2 w = 1,
\end{align}
take the derivative with respect to $x$ for both equation, you have
\begin{align}
&x (3 w^2 \frac{\partial w}{\partial x}) + w^3 + y (2 z \frac{\partial z}{\partial x})
+ 3 z^2 \frac{\partial z}{\partial x} = 0 \\
&z (3 w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick... | The equation is quadratic in $c$.
If $b, c$ are positive integers, we have
$$c = \frac{18b + \sqrt{3[(2b^2+1)^3 - 54]}}{6b^2 + 3}.$$
So, $(2b^2+1)^3 - 54 = 3m^2$ for some positive integer $m$.
Since $3 | 54$ and $3 | 3m^2$, we know that $3 | (2b^2 + 1)$. Let $x = \frac{2b^2+1}{3}$ and $y = \frac{m}{3}$.
We have $x^3 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Find $\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$ (I need a review of my resolution please :) ) Find the limit:
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
This is what I have, i'm not sure about my answer (I'm ... | Too long for comments.
Your work is fine but I think that there is a faster solution.
When I see a problem such as $\lim_{x \to a} f(x)$, my first reaction is to let $x=y+a$ and consider $\lim_{y \to 0} g(y)$.
If we do it for your problem, we have
$$\lim_{x \to 1} \cos(\pi x) \, \sqrt{\frac{(x-1)^2}{(x^2-1)}} \, \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series wh... | May be interesting to consider the most general case of
$$S_p=\sum _{n=1}^{p } \left(\frac{1}{a n+b}-\frac{1}{c n+d}\right)$$ Using the digamma function,
$$S_p=\frac{c \psi \left(\frac{b}{a}+p+1\right)-a \psi \left(\frac{d}{c}+p+1\right)-c \psi
\left(\frac{b}{a}+1\right)+a
\psi\left(\frac{d}{c}+1\right)}{a c}$$ E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Find $n$ such that $1-a c^{n-1} \ge \exp(-\frac{1}{n})$ I am trying to find the integer $n$ such that
\begin{align}
1-a c^{n-1} \ge \exp(-\frac{1}{n})
\end{align}
where $a>0$ and $c \in (0,1)$.
I know that finding it exactly is difficult. However, can one find good upper and lower bounds it.
It tried using lower bound ... | You could chain you're inequalities in the following way:
$$1 - ac^{n - 1} \geq exp\left (\frac{-1}{n} \right) \geq 1 - \frac{1}{n}$$
Then
$$-ac^{n - 1} \geq -\frac{1}{n}$$
$$\iff ac^{n - 1} \leq \frac{1}{n} $$
Now, we can move all of the constants on one side and all of the variables to the other:
$$\frac{a}{c} \leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2... | Yes, the solution is correct. The only minor stylistic change I would personally consider is not going into as much detail when factoring out the 2 in the second line, and mention that you're multiplying your fraction by $\frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove that $TK=TO$
Given $\triangle ABC$ such that $\angle A=90^\circ$ inscribed in circle with center $O$. Let $D$ be the feet perpendicular from $A$ to $BC$ and $M$ be the mid-point of $BD$. Draw the line $AM$ and let it intersect the circumcircle at $X$. Let $K$ be the point on $AX$ such that $OK//XC$. Lastly, deno... | Our solution relies on removing all the "annoying" points; essentially, $T$ and $K$ do not have many properties that we can use, so we attempt to rid them from our equations.
As you noted, we only need to have $\triangle ABX \sim \triangle TOC$, and then we are done. Since $\angle TCO = \angle BAX$, we only need to pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Primes Powers and Mods The Question is below:
For which primes $p$ is $(p − 1)^p + 1$ a power of $p$?
I think the answer is $2$ and $3$, none of the others work.
Here is what I have:
let $p^k-1=(p-1)^p$. Then we have $$p^{k-1}+...+p+1 \equiv0 \pmod{p-1}.$$
Please also include a proof of your answer
| If $(p-1)^p+1=p^k$ then
\begin{align*}
k&=\log_p \left(1+(p-1)^p\right)\\
&=\log_p \left(p^p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)\right)\\
&=p+\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)
\end{align*}
clearly
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)<0$$
since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help w... | I wrote Benedict W. J. Irwin's simplified recurrence in the form
$$
na_n = 1 + 2a_{n - 2} + (n - 1)a_{n - 1} ,
$$
which gave me for the generating function $G(x) = \sum\nolimits_{n = 1}^\infty {a_n x^n }$ the ODE
$$
\frac{1}{{1 - x}} + 2xG(x) + (x - 1)G'(x) = 0.
$$
The particular solution we are looking for is
$$
G(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Showing that $\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$
Show that $$\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$$
I understand the result of an arithmetic sequence $(\sin1^\circ)(\sin3^\circ)(\sin5^\circ)…(\sin89^\circ)$, how about the geo... | We first prove that
$$ (\sin x)^4(\sin 2x)^2 \leq \left(\frac{3}{4}\right)^3. $$
Indeed, applying the double angle formula $\sin 2x = 2\sin x\cos x$ and substituting $t = \sin^2 x$, we have
$$ (\sin x)^4(\sin 2x)^2 = 4t^3(1-t) $$
and the right-hand side is maximized at $t = \frac{3}{4}$ with the value $(3/4)^3$ as desi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Is the inequality true for all $n\geq 2$? Let $x,y,z>0$. I am wondering if the following inequality is true?
$$\sum_{cyc}\frac{x^n}{y^2+yz+z^2}\geq\frac{x^{2n-2}+y^{2n-2}+z^{2n-2}}{x^n+y^n+z^n},\qquad n\geq 2$$
If not, is it known for which $n$ it is true?
$\displaystyle(1)\qquad\sum_{cyc}\dfrac{x^2}{y^2+yz+z^2}\overse... | A proof for $n=4$.
We need to prove that:
$$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\frac{x^6+y^6+z^6}{x^4+y^4+z^4}.$$
Now, by AM-GM $$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\sum_{cyc}\frac{x^4}{y^2+\frac{y^2+z^2}{2}+z^2}=\frac{2}{3}\sum_{cyc}\frac{x^4}{y^2+z^2}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{x^2}{y+z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating limit of the function at $\frac{\pi}{2}$ I'm trying to solve this
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$
I have tried using the L'Hôpital's rule
But I'm stuck at
$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$
Since the above equation is not in the $\frac{0}{0... | Since we have an indeterminate form of type $(0/0)$, we can apply the l'Hopital's rule:
$$\color{blue}{\lim_{x \to \frac{\pi}{2}} \frac{\cos{\left(x \right)}}{\left(x - \frac{\pi}{2}\right)^{3}}} = \color{magenta}{\lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}\left(\cos{\left(x \right)}\right)}{\frac{d}{dx}\left(\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}... | Hint : Substitute $xy=c^2\to x=\frac{c^2}{y}$ then the integral becomes $$ I= \int_0^{\infty}\frac{\ln(c^2)-\ln y}{y^2+by+c^2}dy\\2I= \int_0^{\infty}\frac{\ln(c^2)}{y^2+by+c^2}dy$$ and since $$ y^2+by+c^2= \left(y+\frac{b}{2}\right)^2+\left(c^2-\frac{b^2}{4}\right)$$ and using elementary integral formula of $$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| $$ 8-3\sqrt{7} = a^2 + b^2 - 2ab $$
Let $3\sqrt{7}= 2ab$
$$ab = 1.5\sqrt{7}$$
$$b = \frac{1.5\sqrt{7}}{a}$$
$$a^2 + b^2 = 8$$
$$a^2 +\frac{15.75}{a^2} = 8$$
$y = a^2$
$$y + \frac{15.75}{y} = 8$$
$$y^2 + 15.75 = 8y$$
$$y^2 + 15.75-8y = 0$$
Solve and get
$$y = \frac{7}{2}$$
$$y = \frac{9}{2}$$
Case 1
$$y = \frac{7}{2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {x... | Let $x=y\rightarrow0^+$ and $z\rightarrow a$.
Thus, the left side is closed to $0$, but the right side is greater than $0$,
which says that our inequality is not true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$. I'm trying to solve this:
Which of the following is the closest to the value of this integral?
$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge.
I've found a lower bound by manually calculating $\int_{0}... | We know that,
${\displaystyle\int}\sqrt{\dfrac{1}{3x}+1}\,\mathrm{d}x$=$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{3}}}}{\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$
Substitute $u=\sqrt{\dfrac{1}{x}+3}$ and $\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\frac{1}{x}+3}x^2}$,i.e $\mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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On the series $\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )}$ For the values of $a$ for which the following series makes sense, prove that
$$\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} = \frac{\left ( 2 \cos ... | Indeed, following Jack's comment we have
\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} &= \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\left ( a+n \right )! \left ( a-n \right )!} \\
&=\frac{1}{\left (2a \right )!}\sum_{n=-\infty}^{\infty} \binom{2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What method can I use to compute the limit of this series? Let
$$
\begin{cases}
a_1 &=1 \\a_{n+1}&=\frac{1}{a_1+a_2+\cdots +a_n}-\sqrt{2}
\end{cases}
$$
Then $\sum_{i=1}^{\infty}{a_i}=?
$
I assume that the limit exists, and then I get the limit is equal to $\frac{\sqrt{2}}{2}$
$\sum_{k=1}^{n}{a_k=A_n,}A_{n+1}-A_n=\frac... | We recall that $\ A_n= \displaystyle \sum_{i=1}^n a_i$.
We have: $\ A_1 = 1 \ \text{ and } \ \forall n \in \mathbb N \ , \ A_{n+1}=A_n+\dfrac{1}{A_n}-\sqrt{2}$
Let $\ f(x)=x+\dfrac{1}{x}-\sqrt{2}$.
We have:
$f\circ f(x)-\dfrac{\sqrt{2}}{2} = \left(x-\dfrac{\sqrt{2}}{2}\right) \dfrac{ (\sqrt{2}-x)(1+\sqrt{2}-x)(x+1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sufficient to show the cases when $x = 0$, $y=0$, and $(x,y) \ne (0,0)$?
For a fixed $k \in \mathbb{N}$, define $f_k: \mathbb{R}^2 \rightarrow \mathbb{R}$ by:
$$
f_k(x,y)=
\begin{cases}
\dfrac{x^2(x+y^2)}{x^2+y^{2k}} &, (x,y)\neq (0,0)\\
0 &, (x,y)=(0,0)\\
\end{cases}
$$
Show that $f_1$ is not differentiable at $(0,0)... | So far you have shown that
$$\lim\limits_{(x,y)\to(0,0)}|\frac{f_k(x,y)-f_k(0,0)-\frac{\partial f_k}{\partial x}(0,0)x-\frac{\partial f_k}{\partial y}(0,0)y}{\sqrt{x^2+y^2}}|=\lim\limits_{(x,y)\to(0,0)}|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|$$
The function is differentiable at $(0,0)$ if and only if the abo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e... | \begin{align}
& \left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) \\[10pt]
= & \Big(\bullet\Big) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left( \bullet \cdot \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left(\left( \sum_{j=0}^\infty \frac{a^j}{j!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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How to evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx$? I have an exercise to evalute the following integral for all $n\geq 1 $
$$I(n)=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n+1} x+\cos^{2n+1} x}dx$$
I attempted to find the closed form for the integral above in the following manner, where I ... | Utilize the decomposition
$$\frac{\sin x +\cos x}{\sin^{2n+1}x +\cos^{2n+1}x}
=\frac{2^{n+1}}{2n+1}\sum_{k=1}^n\frac{(-1)^{k+1}\sin\frac{a_k}2 \cos^{n-1}a_k}{\csc a_k+\cot a_k \sin 2x}
$$
with $a_k=\frac{2\pi k}{2n+1}$ and integrate piecewise to arrive at
\begin{align}
\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)... | $$\int \frac{1}{\cos2x+3}dx=\int \frac{1}{\frac{1-\tan^2x}{1+\tan^2x}+3}dx$$
$$=\int \frac{1+\tan^2x}{2\tan^2x+4}dx$$
$$=\frac12\int \frac{\sec^2x\ dx}{\tan^2x+2}$$
$$=\frac12\int \frac{d(\tan x)}{(\tan x)^2+(\sqrt2)^2}$$
$$=\frac12\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+C$$
$$=\bbox[15px,#ffd,borde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) ... | Just to show you something a little bit different, we can do this with the binomial theorem.
$(a+b)^k = a^k + k a^{k-1}b + \frac {k(k-1)}{2} a^{k-2}b^2 + \cdots$
You learned this in algebra / pre-calculus with integers. It actually works for all real numbers.
$(27 + 3)^\frac 13 = 27^\frac 13 + \frac 13 (27^{-\frac 23... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integ... | $$x^3dx+(y+2)^2dy=0$$
$$x^3dx=-(y+2)^2dy$$
$$x^3=-(y+2)^2y'$$
Integratation gives:
$$\int x^3dx=-\int (y+2)^2 y'dx$$
$$\int x^3dx=-\int (y+2)^2 dy$$
Substitute $u=y+2$
$$\int x^3dx=-\int u^2du$$
$$\dfrac {x^4}4+\dfrac {u^3}{3}=C$$
$$\dfrac {x^4}4+\dfrac {(y+2)^3}{3}=C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)$.
Let $a$, $b$, and $c$ be real numbers. Prove that $$(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\,.$$
In solution of this author take Let $a=\tan x, b=\tan y, c=\tan z$ with $-\f... | Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(a^2+1)(b^2+1)(c^2+1)=((a+b)^2+(ab-1)^2)(c^2+1)=$$
$$=((a+b)c+ab-1)^2+(a+b-(ab-1)c)^2\geq(ab+ac+bc-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't see... | This question has already been posted on this website. See my solution via Pell's equation here, where I wrote that the answer is $337.$ The problem appeared on the 1994 British Mathematical Olympiad Round 2
Edit: as suggested by Batominovski, I am copying my old solution here:
A while ago, I found this problem as the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Fundamental group of Klein Bottle It is well know that the fundamental group of the Klein Bottle $G$ is defined by
$$G=BS(1,-1)=\langle a,b: bab^{-1}=a^{-1}\rangle.$$
I know, for example that $BS(1,2)$ can be defined as the group
$$BS(1,2)=\langle A,B\rangle $$
where
$$A=\left(
\begin{array}{cc}
1 & 1 \\
0 & 1 \\
\... | Finally , $G$ can be described as the group of $2\times 2$ matrices generated by two matrix $A,B$ such that $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$ for all $k$. Let $A=\left(
\begin{array}{cc}
a & b \\
c & d\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
\lambda & 0 \\
0 & \mu \\
\end{array}
\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
What is the pdf of $\frac{|x-y|}{(x+y)(2-x-y)}$ when $x,y$ are i.i.d uniform on $[0,1]$? If $x,y$ are i.i.d uniform random variables on $[0,1]$.
I know that the PDF of $|x-y|$ is:
$$f(z) = \begin{cases}
2(1-z) & \text{for $0 < z < 1$} \\
0 & \text{otherwise.}
\end{cases}$$
I know that the PDF of $x+y$ is
$$f(t)=f(z) = ... | Let $$Z = \frac{|X-Y|}{(X+Y)(2-X-Y)}.$$ We first determine the support of $Z$ for $X, Y \sim \operatorname{Uniform}(0,1)$. A quick glance shows that the minimum value is $0$; the maximum is attained at $(X,Y) \in \{ (1,0), (0,1) \}$; therefore, $Z \in [0,1]$. Then we want $$\Pr[Z \le z] = \int_{x=0}^1 \int_{y=0}^1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $ $$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $$ if $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ find $\frac{A^2}{B^2}$
My Attempt
$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty \big(e^{\pi n \b... | Here is an approach using the residue theorem.
The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are:
$$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that the elements of the sequence are divisible by $2^n$ I am trying to prove the following:
Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.
The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal e... | Using induction, it's easy to prove that $A_n=(3+\sqrt{11})^n+(3-\sqrt{11})^n$ (roots of the characteristic polynomial of the corresponding recurrence $t^2-6t-2$ are $3\pm\sqrt{11}$).
Define
$$
B_n:=A_{2n+1}~\text{for}~n\geq 0.
$$
Then,
$$
B_n=(3+\sqrt{11})^{2n+1}+(3-\sqrt{11})^{2n+1}.
$$
It's easy to check that $B_0=6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove an inequality for positive real numbers Prove that :-
$$\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x} \geq x+y+z$$
Where $x,y,z$ are positive real numbers
My attempt :-
L.H.S = $\frac{x^3 z +x y^3 + y z^3 }{xyz} $
We need to show that
$x^3 z + x y^3 + y z^3 \geq xyz (x+y+z) $
I tried the AM-GM
$3(x^3 z + x y^3 + y z... | We have by Cauchy-Schwarz-Bunyakovsky inequality:
$$ (a^2+b^2+c^2)(x^2+y^2+z^2)\geq (ax+by+cz)^2$$
or writen like this:
$$(a+b+c)(x+y+z)\geq (\sqrt{ax}+\sqrt{by}+\sqrt{cz})^2$$
we have a folloving:
$$(y+z+x)(\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x}) \geq (x+y+z)^2$$
and thus a conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^... | In your approach, there's a shorter way to arrive at T, but anyway taking off from there :
$$ T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}} \\
= \frac{({\sin }{2^\circ}.{\sin }{6^\circ}.{\sin }{{10}^\circ}......{\sin }{{86}^\circ})^2}{{{2^{45}}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? Three dice are rolled. One is fair and the other two have 6 on two faces.
Find the probability of rolling exactly 2 sixes.
My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+... | Note that the official answer is correct if you make the (unnatural) assumption that the two non-standard dice have seven sides (two of which show $6$).
In that case the answer is $$\frac 27\times \frac 27\times \frac 56+2\times \frac 27\times \frac 57\times \frac 16=\frac {20}{147}$$
To be sure, this was arrived at by... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Definite integration $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx$ How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$
The actual integral that I encountered is:
$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)... | Assume $a>0$ and $b \in \mathbb{R}$.
Let's first make the substitution $u = ax+b$ to get $$\int_{-\infty}^{\infty} \frac{\arctan (ax+b)}{\cosh(\pi x)} \, \mathrm dx = \int_{-\infty}^{\infty} \frac{\arctan u}{a\cosh \left(\pi \left(\frac{u-b}{a} \right) \right)} \, \mathrm du.$$
Following the general approach that Iaros... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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The values of the parameters for which $\frac{ab}{c}+\frac{a(a+1)b(b+1)}{2!c(c+1)}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{3!c(c+1)(c+2)}+\dots$ converges. Determine the values of the parameters for which $\frac{ab}{c}+\frac{a(a+1)b(b+1)}{2!c(c+1)}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{3!c(c+1)(c+2)}+\dots$ converges.
Then $\frac{a_{n+1... | $$
\frac{(a+n)(b+n)}{(c+n)(1+n)} = 1 - \frac{1+c-a-b}{n} + O(1/n^2)
$$
as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of the Inscribed Angle Theorem using vectors I am trying to prove the inscribed angle theorem using vectors. I fixed the dots $A=(\cos\theta,\sin\theta)$, and $B=(\cos\varphi,\sin\varphi)$, and I took another point $C=(\cos\psi,\sin\psi)$ in the biggest arc $AB$.
My idea was to calculate $\dfrac{\langle A-C,B-C\r... | You expanded your terms all the way. Take a few steps back:
\begin{align}
\langle A-C, B-C\rangle &= \langle (\cos\theta-\cos\psi,\sin\theta-\sin\psi),(\cos\phi-\cos\psi,\sin\phi-\sin\psi)\rangle\\
&= (\cos\theta-\cos\psi)(\cos\phi-\cos\psi)+(\sin\theta-\sin\psi)(\sin\phi-\sin\psi)\\
&= -2\sin\frac{\theta-\psi}2\sin\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + ... | Since $(x - 1)^{99} \equiv (x - 1)\;\mod (x - 2)$, we get that
$(x - 1)^{100}\equiv (x-1)^2\;\mod (x-2)(x-1). \quad(*)$
Since $(x-2)^{199}\equiv -1\;\mod (x-1)$, we get that
$(x-2)^{200}\equiv -(x-2)\;\mod (x-1)(x-2). \quad(**)$
So, by adding $(*)$ and $(**)$, it follows that
$(x - 1)^{100} + (x - 2)^{200} \equiv (x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$ Given $\sinh x = 8/14$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$.
I have been getting two answers which has made me confused. I keep getting $\sqrt{65/7}$ or $\sqrt{4/7}.$
That's what I got: $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^... | You are right! Just use $$\cosh^2 x = 1+\sinh^2 x.$$
Now it only amounts to keeping track of what $\sinh x$ is and carefully doing the calculation. Is it $\sinh^2 x = \left(\frac{8}{13}\right)^2$ or is it $\sinh^2 x = \left(\frac{8}{14}\right)^2$?
UPDATE: With $\sinh x = \frac{4}{7}$:
$$\cosh^2 x = 1+ \left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elegant way of finding the least perimeter of triangle A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?
We have $\angle A = 2\angle B$
$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $
$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\s... | Found an another solution:
we have $a^2=b(c+b)$
A triangle of smallest perimeter means $gcd(a,b,c)=1$
In fact $gcd(b,c)=1$ since any common factor of $b,c$ would be a factor of $a$ as well.
A perfect square $a^2$ is being expressed as the product of two relatively prime integers $b$ and $c$.
it must be the case that bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Determine $\sqrt{1+50\cdot51\cdot52\cdot53}$ without a calculator? I've tried a lot of things but failed to do it, I've calculated the result inside the square root which is $7027801$ using substitution and factoring but $\sqrt{7027801}$ isn't possible to simplify.
| Including $50\cdot51\cdot 52\cdot 53$ inside the square root suggests that you should choose a value for $x$. Suppose you choose the highest value and set $x=53$. Then
\begin{align}1+x(x-1)(x-2)(x-3)&=1+x(x-3)\cdot(x-1)(x-2)\\&
=1+(x^2-3x)(x^2-3x+2)
\end{align}
Now let $y=x^2-3x$. The above equation becomes
$$1+y(y+2)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
| Your application of AM-GM is wrong. The statement for AM-GM states that for positive integers $a_1, a_2, \dots, a_n$:
$$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1a_2\dots a_n}$$
with equality at $a_1 = a_2 = \dots = a_n$. Observe that
$$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6} \geq \sqrt[6]{16x^4y^4z^4} = 16 \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is The coefficient of $x^{n}$ in the expansion of $\frac{(2-3 x)}{(1-3 x+\left.2 x^{2}\right)}$ is
$(a) \quad(-3)^{n}-(2)^{n / 2-1}$
(b) $2^{n}+1$
$(c) 3(2)^{n / 2-1}-2(3)^{n}$
(d) None of the foregoing numbers.
Now, $1-3 x+2 x^{2}... | You're on the right track. Consider splitting the fraction up using partial fractions.
\begin{align*}
\frac{2-3x}{1-3x+2x^2} &= \frac{2-3x}{(1-2x)(1-x)}\\ &=\frac{1}{1-2x} + \frac{1}{1-x}\\&=\sum_{n=0}^\infty 2^nx^n + \sum_{n=0}^\infty x^n\\&=\sum_{n=0}^\infty \left (2^n+1\right )x^n
\end{align*}
Hence, the coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Method of summation for third order difference series: $2+12+36+80+150\dots$ I am unable to understand how one derived the formula for the $n$-th term $= an^3 +bn^2 + cn + d$, where the degree of the polynomial depends on the step at which we get a constant A.P. Here its at $2$nd step so degree $=2+1=3$.
But how do we ... | $2+12+36+80+150\dots = (1+1) + (4 + 8) + (9 + 27)+(16+64)+(25+125)+...=\sum (k^2+k^3)$
$$\sum _{k=1}^n \left(k^3+k^2\right)=\sum _{k=1}^n k^3+\sum _{k=1}^n k^2=\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1)=$$
$$=\frac{1}{12} n (n+1) (n+2) (3 n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Use generating functions to solve the non-homogenous recurrence relation Let $a_0=0, a_1=2,$ and $a_2=5$. Use generating functions to solve the recurrence equation: $$a_{n+3} = 5a_{n+2} - 7a_{n+1}+3a_n + 2^n$$ for $n\geq0$.
This is a book problem from Applied Combinatorics. I am really confused about tackling $2^n$ par... | You made a mistake somewhere in the generating function derivation (hard to tell where since you did not include this part), I've got
\begin{align}
A(x)&=2x+5x^2+\sum_{n \geq 3}a_{n}x^n\\
&=2x+5x^2+5\sum_{n \geq 3}a_{n-1}x^n-7\sum_{n \geq 3}a_{n-2}x^n+3\sum_{n \geq 3}a_{n-3}x^n+\sum_{n \geq 3}2^{n-3}x^n\\
&=2x+5x^2+5x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those fo... | Let a 2x2 matrix $A$ be orthogonal i.e $AA^t = A^tA = E$. Notice that
$$ E= \left (\begin{array}{cc}a & b \\ c & d\end{array} \right )\left (\begin{array}{cc}a & c \\ b & d\end{array} \right ) \Rightarrow \|(a,b)\| = \|(c,d)\| = 1,\ (a,b)\perp (c,d) $$
Conversely, suppose a 2x2 matrix $A$ has orthonormal columns $(a,b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit using Taylor expansion : which term do we expand? I want to check the limit $\displaystyle{\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )}$ using the Taylor expansion.
I have done the following:
$$\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\l... | Expand the logarithm
$$
\ln\left(\frac{n^2-2n+1}{n^2+1}\right)=\ln\left (1-\frac{2n}{n^2+1}\right)=-\frac{2n}{n^2+1}+\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $a+\sqrt{a}=b+\sqrt{b}$ is $a=b$? If $a+\sqrt{a}=b+\sqrt{b}$, does this automatically mean that $a=b$?
I first tried to square both sides but that seemed to get me nowhere.
$$a-b=\sqrt{b}-\sqrt{a}$$
Can we just conclude that $a$ has to be equal to $b$ to make this expression to be true?
| $a-b=\sqrt{a}-\sqrt{b} \implies (\sqrt{a}-\sqrt{b})(-1+\sqrt{a}+\sqrt{b})=0\implies \sqrt{a}=\sqrt{b}$ or $1=\sqrt{a}+\sqrt{b}$. So it might not be the case that $a = b$.
I realize my answer above is for a different problem. So back to this one. We have $a - b = \sqrt{b} - \sqrt{a} \implies a - b +\sqrt{a}-\sqrt{b} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Partial sums over rational functions I recently came across the result that
$$\sum_{n=2}^\infty \frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{2}$$
I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.
If I plug the sum into Wolfram Alpha it gives
$$\frac{3k^4-k-2}{6k(k+1)(k^2+k... | In a word: the terms telescope in a nice way.
Using partial fractions, we have
$$
\frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{3}\left(\frac{n - 2}{n^2 - n + 1}- \frac{n-1}{n^2 + n + 1} + \frac{1}{n - 1} - \frac{1}{n + 1}\right)
$$We need to be a bit careful because the harmonic series diverges, so we should only group terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c}\geq 0$ for positive $a$, $b$, $c$ with $abc=1$. I recently saw the following inequality,
$$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0 \tag1$$
for all $abc=1$ and $a,b,c \in \mathbb{R}_+ \setminus \{0 \}$.
To prove this inequality, ... | Hint: A common trick in an inequality like this is to substitute:
$$a = \frac xy, b = \frac yz, c = \frac zx$$
which transforms your final inequality to:
$$(x^3z+y^3x+z^3y) + (x^2y^2+z^2x^2+y^2z^2)\geq 2xyz(x+y+z).$$
But the above is easy: you can suitably use AM-GM and prove that two expressions in parentheses is at l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= ... | Yes, it is necessary to show.
As $\tan$ has a periodicity $\pi$, it's enough to check the signs for $\dfrac{\alpha}{2}$ in the ranges, $\left[0,\dfrac{\pi}{4}\right], \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right] $ and $\left[\dfrac{3\pi}{4},\pi\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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What is $3^{99} \pmod{100}$? I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we ... | You can do this using the Chinese remainder theorem and by calculating some powers.
$3^{99} = (3^3)^{33} \equiv 2^{33} \pmod{25} = (2^7)^4 \times 2^5 \equiv 3^4 \times 2^5 = 3^3 \times 3 \times 2^5 \equiv 3 \times 2^6 \equiv 17$.
$3^{99} \equiv (-1)^{99} \equiv 3 \pmod{4}$.
So we need the unique number satisfying those... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Calculating $\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$ without using Beta function and Euler sum. Is it possible to show that
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$
without using the Beta function
$$\text{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and the generalized E... | \begin{align}J&=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1-x}dx\\
&\overset{\text{IBP}}=\underbrace{\left[\left(\int_0^x \frac{\ln^2 t}{1-t}dt-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1}_{=0}+\\&\int_0^1 \frac{1}{1-x}\left(\underbrace{\int_0^x \frac{\ln^2 t}{1-t}dt}_{u(t)=\frac{t}{x}}-\int_0^1 \frac{\ln^2 t}{1-t}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Factoristaion of $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$ My question revolves around the polynomial
$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$$
I know that it can be factorised into
$$(x^2+x+1)(x^6+x^3+1)$$
but what method can we employ to obtain this result in the first place? Any method that works will be of great help to me.
Than... | Method 1: Use what you know about polynomials of the form $x^n-1$. For instance, your polynomial is $\frac{x^9-1}{x-1}$. So if you can factor the numerator, many of those factors would be inherited to your polynomial. And indeed, we have
$$
x^9-1=(x^3)^3-1=((x^3)^2+x^3+1)(x^3-1)\\
=(x^6+x^3+1)(x^2+x+1)(x-1)
$$
Put this... | {
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"url": "https://math.stackexchange.com/questions/3802087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of series $\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$ How to deduce if series is convergent (depending on parameter $a$):
$$
\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}?
$$
If $a=e,$ we have that $\lim_{n\to\infty}a_n=0,$ but it is not sufficient to ... | If $a \neq e$, then $$\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$$
does not tend to $0$ so the series cannot converge.
If $a=e$, you have
$$\left[\left(1+\frac{1}{n}\right)^{n+1}-e\right]\sin{n} = \left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)}-e\right]\sin{n}=e\left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)... | {
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"url": "https://math.stackexchange.com/questions/3805075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$ prove $$\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$$ where $a,b,c>0$ and $a\ge b\ge c$
My try it seemed qite simple but i couldnt apply the rearrangement inequality directly. so i tried manipulating the inequality.
the inequality can be written as $$a^2(b-c)+... | Your first step is wrong:
We need to prove that:
$$\sum_{cyc}(a^3b^2-a^3bc)\geq0$$ and it indeed gives a proof:
$$\sum_{cyc}(2a^3b^2-2a^3bc)\geq0$$ or
$$\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)\geq\sum_{cyc}(a^3c^2-a^3b^2)$$ or
$$\sum_{cyc}a^3(b-c)^2\geq(ab+ac+bc)(a-b)(b-c)(c-a),$$ which is obvious.
We can get $$\sum_{cyc}(a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$ If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$
My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get :
$$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$
I don't know what to do nex... | Suppose $t = \frac{a+b}{2}$ and $c = \max \{a,b,c\}.$ Let
$$f(a,b,c) = \frac{1}{abc}+36 - \frac{21}{ab+bc+ca}.$$
We have
$$f(a,b,c) -f(t,t,c) = \frac{1}{abc}-\frac{4}{c(a+b)^2}+\frac{84}{(a+b)(a+b+4c)}-\frac{21}{ab+bc+ca}$$
$$ = \frac{(a-b)^2}{a+b}\left(\frac{1}{abc(a+b)}-\frac{21}{(ab+bc+ca)(a+b+4c)}\right) \geqslant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using rnfequation to show that $\left(\frac{\frac{-1+\sqrt{5}}{2}+\sqrt{\frac{-\sqrt{5}-5}{2}}}{2}\right)^5=1$, in PARI/GP I want to understand how can the code below shows that $\left(\frac{\frac{-1+\sqrt{5}}{2}+\sqrt{\frac{-\sqrt{5}-5}{2}}}{2}\right)^5=1$
{
V=rnfequation(T^2-5,x^2-Mod((-T-5)/2,T^2-5),1);
a=V[2];
b=x... | By using the property that each side of a regular decagon inscribed in a circumference is the golden part of the radius, we get that
$\cfrac{\text{side}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{2}\;$ and $\;\sin 18^\circ=\cfrac{\text{side / 2}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{4}\;.$
Moreover,
$\cos 18^\circ=\sqrt{1-\sin^... | {
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"timestamp": "2023-03-29T00:00:00",
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$|z-1|+|z-2|+|z-3|=6$ in the Argand Plane In the argand plane $$C:|z-1|+|z-2|+|z-3|=6 ~~~(1)$$ represents a bounded curve which is a rounded blob mimicking an "ellipse".
Using the the inequality $$|Z_1+Z_2+Z_3| \le |Z_1|+|Z_2|+|Z_3|, ~~~~(2)$$ we can see that $$6=|z-1|+|z-2|+|z-3|\ge |3z-6| \implies |z-2| \le 2. ~~~(... | Our claim is that the circle
$$C_2:|z-2|=\sqrt\frac{44}{3}-2$$
lies entirely inside $C$. It is not hard to show that this value is the unique real $r$ for which $2\pm ir$ are on $C$.
We can calculate that
$$|z-2|^2=\frac{|z-1|^2+|z-3|^2}{2}-1.$$
We have that
$$\frac{|z-1|^2+|z-3|^2}{2}\geq\left(\frac{|z-1|+|z-3|}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810271",
"timestamp": "2023-03-29T00:00:00",
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Find the $26^{th}$ digit of a $50$ digit number divisible by $13$. $N$ is a $50$ digit number (in the decimal scale). All digits except the $26^{th}$ digit (from the left)
are $1$. If $N$ is divisible by $13$, find the $26^{th}$ digit.
This question was asked in RMO $1990$ and is very similar to this question and the s... | Another way is to use the trick from Wikipedia (that doesn't solve your solution)
Taking $N$ from the right, and applying the sequence $(1, −3, −4, −1, 3, 4)$ as instructed on the page (multiply the digits from the right by the given numbers in sequence), we get
$0$ for the 6 first digits from the right ($1-3-4-1+3+4=0... | {
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"timestamp": "2023-03-29T00:00:00",
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Sequence and polynomial power relationship I recently encountered this relationship between polynomial powers and a certain associated sequence and I am seeking any help or idea that might answer why the relationship is true.
Let $P(x)$ be a polynomial, say for instance $P(x)=1+3x+2x^2$. Consider the consecutive powers... | In the product $$(1 + 6 x + 13 x^2 + 12 x^3 + 4 x^4)(2x^2+3x+1)$$ if you try to compute the coefficient of $x^3$, note that $x^3$ can be formed in the following ways
$12x^3$ from the first bracket, $1$ from the second bracket
$13x^2$ from the first bracket, $3x$ from the second bracket
$6x$ from the first bracket, $... | {
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"url": "https://math.stackexchange.com/questions/3811282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $I=\int\frac{\sin(x)+\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$ I'm interested in the following problem:
Evaluate the indefinite integral$$I=\int\frac{\sin(x)+\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$$
Here's what I did:
We note that
$$I = \int\frac{\sin(x)}{\sin^4(x)+\cos^2(x)}~dx+\int\frac{\cos(x)}{\sin^4(x)+\cos^2(x)}~d... | You can factorize
$$
u^4-u^2+1=(u^2+\sqrt{3} u+1)(u^2-\sqrt{3} u+1),
$$
so that the partial fraction decomposition is
$$
\frac{1}{u^4-u^2+1}=\frac{1}{4 \sqrt{3}}\left[\frac{(2 u+\sqrt{3})+\sqrt{3}}{u^2+\sqrt{3} u+1}-\frac{(2 u-\sqrt{3})-\sqrt{3}}{u^2-\sqrt{3} u+1}\right],
$$
and we have
$$
\int\frac{1}{u^4-u^2+1}du=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
t... | Hint: $f(x)=\frac{3x+4}{x+1}=3+\frac{1}{x+1}$ and from here we obtain $f^{(n)}(x)=\frac{(-1)^n n!}{(x+1)^{n+1}}, n \geqslant 1.$
Now, knowing derivative, Taylor series will be .. can you finish?
Addition:
It should be
$$\frac{3x+4}{x+1}=\frac{7}{2}-\frac{1}{4}(x-1)+ \cdots$$
and
$$\frac{1}{x+1}=\frac{1}{2}-\frac{1}{4}(... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there a way to solve the equation $\sin x = x\ln x$ analytically?
Is there a way to solve the equation $\sin x = x\ln x$ numerically or analytically?
The only way I have been able to solve this is using a graphic calculator like Desmos, but is there another way to solve this?
| This is a transcendental equation; this means no hope for a closed form solution and numerical methods are reaquired. So, considering that you looks for the zero's of function $$f=\sin (x)- x\log(x)$$ for which $x=0$ is a trivial solution. By inspection, the solution is between $1$ and $2$. Being lazy (myself), make a ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$ This is what the answer says:
Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,
and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly ... | Notice that $\sqrt{x}-\sqrt{2-2x} \leq \sqrt{x}$. Then if $x<1$ we have $\sqrt{x}<\sqrt{1}=1$. It follows that $\sqrt{x}-\sqrt{2-2x} <1$ for $x<1$. So if $x<1$ it is not a solution. If $x=1$ we have $\sqrt{x}-\sqrt{2-2x}=\sqrt{1}-\sqrt{2-2}=1 $. So $x=1$ is a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving infinite nested square roots of 2 converging to finite nested radical Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)
Converging to finite nested radical of ... | Somehow I got the answer from my subsequent post (after a long homework for cyclic infinite nested square roots of 2)
Sivakumar Krishnamoorthi (https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi), Solving cyclic infinite nested square roots of 2 as cosine functions, URL (version: 2020-09-26): https://... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$ I have to evaluate the following limit
$$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$$
My solution is:
$$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }=\lim_{x \to 0}{\frac{\pi \cos x}{x \cdot x} }=+\infty$$
But the correct result is $\frac{\pi}{... | By some elementary inequalities, near $x=0$ we have
$$
\cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have
$$
\frac{\sin(\pi\cos(x))}{x\sin(x)} = \frac{\sin(\pi(1-x^2/2)+ O(x^4))}{x\sin(x)}
$$Use the sine-reflection and angle-addition:
$$
\frac{\sin(\pi(1-x^2/2)+O(x^4))}{x\sin(x)} = \frac{\sin(\pi/2 \cdot x^2+O(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. ... | So the claim is
$$\sum_{i=1}^{n} (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$$
Base case is true, now assuming that it is true for all naturals less than $n+1$ we will show that it is true for $n+1$.
$$\sum_{i=1}^{n+1} (1-\frac{1}{(i+1)^2}) = \frac{n+3}{2n+4}$$
is to be proved
Instead of working with summation,we will est... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rate of convergence for a sequence (Preferably without Taylor series) I am trying to solve the following problem:
Knowing that the sequence $(a_{n})$ with:
$$a_{n+1}=\frac{1}{2}(a_{n}+\frac{3}{a_{n}})$$
converges to $\sqrt{3}$, find it's rate of convergence.
After doing some searching, I found this formula from wikiped... | @ClaudeLeibovici notes this iteration is by the Newton-Raphson method; so, under mild conditions (which apply here), the convergence is quadratic (i.e. the order of convergence is $2$) so $\mu=0$. @user's work makes this easy to check. With $x:=\tfrac{a-\sqrt{3}}{a+\sqrt{3}}$ we have$$a_n-\sqrt{3}=2\sqrt{3}x^{2^n}\unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overri... | Since $A$, $B$ and $C$ are collinear, you can write $C$ as $A+\lambda(B-A)\require{cancel}$. So\begin{align}A\times B+B\times C+C\times A&=A\times B+B\times(A+\lambda(B-A))+(A+\lambda(B-A))\times A\\&=\cancel{A\times B+B\times A}+\lambda B\times(B-A)+\cancel{A\times A}+\lambda(B-A)\times A\\&=-\lambda B\times A+\lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In triangle ABC a point X is taken on AC and a point Y is taken on BC if AY and BX meet at O This question is from pre collage mathematics.
The question goes on like this:
In $\triangle ABC$ a point $X$ is taken on $\overline{AC}$ and a point $Y$ is taken on $\overline{BC}$. If $\overline{AY}$ and $\overline{BX}$ meet... | $\frac{A(ABY)}{A(ACY)}=\frac{A(BXY)}{A(CXY)}=\frac{BY}{CY}$
$\frac{y+z}{x+A(OXY)+A(CXY)}=\frac{z+A(OXY)}{A(CXY)}$
$A(CXY)\times(y+z)=(x+A(OXY)+A(CXY))\times(z+A(OXY))$
$y\times A(CXY)=xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2+A(CXY)\times A(OXY)$
$A(CXY)=\frac{xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2}{y- A(OXY)}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$.
If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.
My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.
First I proved ... | $$x^5-x^3+x-2=x^5+x^2-(x^2-x+1)-(x^3+1)=$$
$$=(x^2-x+1)(x^3+x^2-x-2).$$
Now, we see that our equation has an unique real root $1<\alpha<2$ and from here $$[\alpha]=1.$$
Now, about your new problem.
Easy to see that for our root $\alpha$ we have $$1.205<\alpha<1.206,$$
which gives $$3<\alpha^6<4$$ and from here: $$[\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive integers $(a, b, c)$ are a primitive Pythagorean triple
Show that if $a = m^2 - n^2$ , $b = 2mn$, $c = m^2 + n^2$ , where $m$, $n$ are relatively prime, not both odd, and $m>n$, then $(a, b, c)$ is a primitive Pythagorean triple.
This is part one of a proof I am required to do.
I know that if $m$ and $n$ are... | Line up your ducks. And then shoot them.
Does
$(m^2 - n^2)^2 + (2mn)^2 {? \over=} (m^2+n^2)^2$
$m^4 - 2m^2n^2 + n^4 + 4m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$
$m^4 + 2m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$?
The answer is... yes, it does.
So $m^2-n^2, 2mn, m^2 + n^2$ are a pythogorean triple.
====
But are the a primativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\fr... | The following relation:
$$(ab+bc+ca)(\frac{1}{b}+\frac{1}{c}+\frac{1}{a})\geq(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ is easily obtained by C-S inequality.
Since you arleady know $abc=1$ , you can easily notice that $$ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$
and your question becomes much easier at this point.
What you s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Characteristic polynomials for a class of simple matrices Let $M_{k}$ be the kxk-matrix with entries $m(i,j,k)=0$ for $i+j<k-1$ and $m(i,j,k)=1$ else.
For example $M_{3}=\begin{bmatrix}0 & 0 & 1\\0&1&1\\ 1&1&1\end{bmatrix}.$
Let $F_{n}(x)$ denote the Fibonacci polynomials defined by $F_{n}(x)= F_{n-1}(x)+x F_{n-2}(x... | The squares $N_k = M_k^2$ are very nice matrices: we have (indexing from $1$ to $k$)
$$N_k(i, j) = \text{min}(i, j)$$
hence, for example,
$$N_4 = \left[ \begin{array}{cc} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{array} \right].$$
Define the polynomials $P_k(x) = \det (N_k - xI)$, and (it tu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Angle between the curves without differentiation $y^2=4(x+1)$ and $x^2=4(y+1)$
Are the two curves, the answer online uses calculus and we have not been taught that.
can this be solved any other way?
Thanks in advance.
| Write the product of the equations of the parabolas around the intersection points $(2\pm 2\sqrt 2, 2\pm 2\sqrt2):$ $-16[(x-(2\pm 2\sqrt{2}))^2\pm \sqrt{2}(x-(2\pm 2\sqrt{2}))(y-(2\pm 2\sqrt{2}))+(y-(2\pm 2\sqrt{2}))^2]+$ higher order terms, and use the $\tan \theta = {2 \sqrt{h^2-ab} \over a+b},ax^2 + 2hxy + by^2 + 2g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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Integrate $\int_0^3 \int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}(1) dzdydx $ How to integrate using cylindrical coordinates the following function?
$$\int_0^3 \int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}(1) dzdydx $$
My problem is not with the function itself but with arranging the respective boundaries with the ... | The boundary is a cylinder with a cap, that lies on the first octant. The projection on the $x$-$y$ plane is the positive part of $\frac14$ of a circle with radius $3$ (lying on the first quadrant). How do you find such area using polar coordinates?
$$A=\int_0^{3}\int_0^{\sqrt{9-x^{2}}}dxdy=\int_{0}^{\pi/2}\int_0^3rdrd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Wel... | Written as $\left(\frac{z+w}{|z||w|+zw}\right)^2+\left(\frac{z-w}{|z||w|-zw}\right)^2\ge\frac1{|z||w|}$, the inequality scales nicely, so we can assume wlog that $R=|z|=|w|=1$.
Set $z=e^{i\alpha}$ and $w=e^{i\beta}$. If $(zw)^2\ne1$, then $\cos^2(\alpha+\beta)\ne1$. Furthermore,
$$
\begin{align}
&\left(\frac{z+w}{1+zw}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{... | Hint: Break it into cases:
If $x \ge 1$, then $x^{12} \ge x^9$ and $x^4 \ge x$.
If $x \le 0$, then $x^{12} \ge 0$ and $x^9 \le 0$ and $x^4 \ge 0$ and $x \le 0$.
If $0 < x < 1$, then $x^{12} > 0$ and $x^4 > x^9$ and $1 > x$.
Can you finish each of these cases?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule?
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}... | We have that
$$\frac{\partial g}{\partial \theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{1}{\left(x+y^2\right)^2}(-32 r \sin \theta)-\frac{2y}{\left(x+y^2\right)^2}(3r\cos \theta)=$$
with $x=64$ and $y=6$ then
$$=-\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answ... | The sum of all four factors is the same in both cases. To maximize the product, we want the factors to be as close together as possible.
The factors $(2 + \frac12)(5 + \frac15)$ are closer to their average than $(2 + \frac15)(5 + \frac12)$, so $(2+\frac12)(5+\frac15) > (2+\frac15)(5+\frac12)$.
Similarly, $(3 + \frac13)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Sum of hitting probabilities not equal to 1? I have a Markov Chain with transition matrix
\begin{equation}
P=
\begin{pmatrix}
0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\
\frac{1}{5} & 0 & 0 & 0 & \frac{4}{5} \\
0 & 0 & 1 & 0 & 0\\
\frac{5}{6} & 0 & \frac{1}{6} & 0 & 0\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
\... | Nvm, I found the answer. I was not taking into account the ordering of the loops. eg looping between states $0$ and $1$ followed by $0$ and $3$ results in a different chain as compared to looping between $0$ and $3$ first then $0$ and $1$. $f$ should be changed as following
\begin{equation}
f=\sum_{i=0}^{\infty}\sum_{j... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using $u = \tan^{-1}(x)$ The problem asks to solve $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using the u-substitution $u = \tan^{-1}(x)$.
I was able to solve the integral using the $u$-substitution $v = \frac{x-1}{x+1}$ (see below); however, I was not able to make any progress when... | I think I have figured out a messy way to get the solution. Let $u = \tan^{-1}(x)$.
$\begin{align*}
\int \frac{1}{(x^2-1)^{2/3}(x+1)^{2/3}} \, dx &= \int \frac{1+\tan^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u)+1)^{2/3}} \, du \\
&= \int \frac{\sec^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u+\frac{\pi}{4})(1-\tan(u)))^{2/3}} \, du\\
&= \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y ... | The domain of $f$ is the interval $(-\infty,0)$.
Thus, $g:=f^{-1}$ should be such that its range is $(-\infty,0)$.
This way, one can see that
$$
g(s) = \frac{s-\sqrt{s^2+8}}{2}
$$
is the proper solution for the inverse of $f$,
(because we always have $g(s)<0$), but
$g(s) = \frac{s+\sqrt{s^2+8}}{2}$ would not be, becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.
If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.
I started with:
Suppose $a|(b+c)$ and $\gcd(b,c)=1$.
This means that $ak=b+c$, for some integer $k$.
And $1|b$ and $1|c$.
I know I can solve this using the theorem th... | One thing I like to do is figure if $k$ is a common (positive) divisor of $a$ and $b$, and $a|b+c$ then $k|b+c$. But we also have $k|b$ so $k| (b+c)-b = c$. So $k$ is a common (positive) divisor of $b,c$. But $\gcd(b,c) = 1$ so the only common (positive) divisor of $b,c$ is one so $k=1$. So the only common divisor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854471",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times). Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times). I don't know where to start... Should I use mathematical induction? But what should be my hypothesis? Should I calculate for n = 1, n = 2?
| First note
$$f(x) = |x+1| - |x-1|\\
= \begin{cases}
-2, & x< -1\\
2x, & x\in [-1, 1] \\
2, & x>1\\
\end{cases}$$
Proposition: $$\color{red}{f^{n+1}(x) = \left|2^nx+1\right| - \left| 2^nx-1\right|}$$.
Note: this is exactly equivalent to
$$f^{n+1}(x) = \begin{cases}
-2, & x< -\frac1{2^n}\\
2^{n+1}x, & x\in [-\frac1{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855284",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how ... | $$I-J = \int \frac{ \cos^3 x - \sin^3 x}{ \sin x + \cos x} dx = \int \frac{ (\cos x - \sin x)( \cos^2 x + \sin^2 x + \sin x \cos x )}{ \sin x + \cos x} dx$$
Substitute $$ \sin x + \cos x = t$$
$$t^2 = 1 - 2 \sin x \cos x$$
Or,
$$ \sin x \cos x = \frac{1-t^2}{2}$$
$$ I-J= \int \frac{(1+ ( \frac{1-t^2}{2}))}{t} dt$$
Ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Stuck on Mathematical Induction Proof I have the following question: Prove with mathematical induction that $3^n+4^n\le 5^n$ for all $n\ge 2$.
$$\text{Assume true: }3^n+4^n \le 5^n \text{. Prove that $3^{n+1}+4^{n+1} \le 5^{n+1}$} \\
= 3\cdot3^n+4\cdot4^n \\ =3\cdot3^n+4^n(3+1)\\
=3\cdot3^n+3\cdot4^n+4^n \\
=3(3^n+4^n... | From where you left off, we have
$$
5^{n+1}-2(3^n+4^n)+4^n=5^{n+1}-2\cdot3^n - 4^n\leq 5^{n+1}
$$
and your proof is finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - ... | We can use rationalization by a double step
$$\frac2{2 - \sqrt[4]{2}} \cdot \frac{2 +\sqrt[4]{2}} {2 + \sqrt[4]{2}} \cdot \frac{4+\sqrt {2}} {4+ \sqrt{2}}=\frac{(2 +\sqrt[4]{2})(4+\sqrt {2})}7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion.... | Write $u = \cos x$, $v=\sin x$. From $u^2 + v^2=1$ we get $u$, $v$ $\le 1$ so $u^3\le u^2$, $v^3 \le v^2$, and adding up we get $u^3 + v^3\le 1$. We have equality if and only if we have equality in the previous equalities, therefore $(u,v)=(1,0)$ or $(0,1)$.
It is interesting to sketch the curves $u^2+v^2=1$ and $u^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$. There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$.
Our professor gave us lengthy and a bit of a complex proof. I tried to construct my own, more simple proof. Could you please verificate it ?
Suppose there are $r,s$ such that$\sqrt{3}... | The proof is wrong .
For example $2+\sqrt{3}$ and $-2+\sqrt{3}$ are irrtionlas and give an integer that is not zero on subtraction
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find... | Attempts
Using the integral test, we can show the convergence since
$$I=\int \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}\,dk=\int \left(\frac{n}{k^2+n^2}+\frac{1}{k+n}\right)\,dk=$$
$$I=\log (k+n)-\tan ^{-1}\left(\frac{n}{k}\right)$$
Integrating between $k=1$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Riemann zeta function in the critical strip Hello wonderful people
It seems that the Riemann zeta function in the critical strip may be given by:
\begin{gather}
\zeta(s)=\frac{1}{s-1}+1-s \int\limits_{1}^{+\infty} \frac{x-[x]}{x^{s+1}} \mathrm{d}x
\end{gather}
Does someone have a nice and pedagogical proof for a pub... | If a "public of engineers" wants a much more detailed proof, see below.
We assume $Re(s) > 1$ until indicated otherwise.
\begin{align*}
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} &= \sum_{n=1}^{1} \frac{n}{n^s}
+ \sum_{n=2}^{\infty} \frac{1}{n^s}
= \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{n - (n-1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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In this linear equation system, find the $a,b,c$ values such that...
Consider this linear equations system:
\begin{align*} \begin{pmatrix} a & 1 & 1\\ 1 & a & 1 \\ 1 & 1 & 1
\end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=
\begin{pmatrix} b\\ c\\ 2 \end{pmatrix} \end{align*}
*
*Determine the $a,b,c$ va... | How I would solve the first question: A quick check shows that $\det A=(a-1)^2$, so there is a unique solution unless $a=1$.
A hands-on approach without any theory works too: Note that $x+y+z=2$ and so
$$b-2=(ax+y+z)-(x+y+z)=(a-1)x,$$
$$c-2=(x+ay+z)-(x+y+z)=(a-1)y,$$
which shows that if $a\neq1$ then
$$x=\frac{b-2}{a-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Functional equation involving integral
Find $f(x)$ given that it satisfies the equation $f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+1) dt$
My attempt:
Let $\int_0^{\pi/2}f(t)dt=c$
So
$$f(x)=x^2+c\sin(x+1)$$
Integrate w.r.t. $x$ from $0$ to $\pi/2$
$$\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}x^2dx+c\int_0^{\pi/2}\sin(x... | $$f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
differentiate twice wrt $x$
$$f''(x) = 2-\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
but
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt=f(x)-x^2$$
thus we get
$$f''(x)=2-f(x)+x^2$$
$$f(x) = x^2+a \cos (x)+b \sin (x)$$
Compute now
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm.
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.
What I... | Let $OC = a$, $OD = b$. So $OA=\frac{2}{5}OC$, $OB = \frac{2}{5} OD$.
(Note you have swapped labels $C$ and $D$ in figure)
Also let $AD=3x$, $BC=3y$, so that $QA=2x$, $QB=2y$.
We have $a^2+b^2=100$
By Pythagoras,
$$ (OA^2+OD^2)+(OB^2+OC^2)=9(x^2+y^2) $$
$$ \Rightarrow x^2+y^2=116/9 $$
By cosine-rule in $\triangle QAB$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$ Here is the question I want to answer:
Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = ... | You are asking about definition for $\Bbb{Q}(\sqrt{2},\sqrt{3})$.
Let me add some more.
Suppose $F$ is a field of a subfield $K$ and $\alpha$ is an element of $K$. Then the collection of all subfields of $K$ containing both $F$ and $\alpha$ is nonempty(why!).
Since the intersection of subfields is again a subfield, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Question about asymptotes.
Let $f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2}.$ Give a polynomial
$g(x)$ so that $f(x) + g(x)$ has a horizontal asymptote of $0$ as $x$
approaches positive infinity.
I've tried using that if the degree of the denominator is bigger than the degree of the numerator, the horizontal asymptote... | Perform the Euclidean division of the numerator by the denominator:
$$x^4+x^3+x^2+1=(x^2+3)(x^2+x-2) -3x+7. $$
We deduce
$$f(x)=3(x^2+3)+3\frac{-3x+7}{x^2+x-2}.$$
The fraction tends to $0$ as $x\to\infty$, hence we can take $\:g(x)=-3(x^2+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.