Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove I... | An alternative approach: $h(x)=x+\frac{1}{x}$ is positive and log-convex over $(0,1)$, since
$$\frac{d^2}{dx^2}\log h(x)=\frac{(1+3x^2)+(x^2-x^4)}{(x+x^3)^2}>0.$$
In particular $h(x)^3$ is convex and the claim follows from Jensen inequality, as already remarked by Micheal Rozenberg.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Envelope of a family of curves
Show that the envelope of the family of curves $$\frac{x}{a}+\frac{y}{4-a}=1$$ is the parabola $$\sqrt{y}+\sqrt{x}=2$$
I know how we can get the envelope, but I could not get the required relation. I differentiated the family of curves w.r.t $a$. I got
$$a=\frac{1}{2}(x-y+4)$$
I then s... | Calculate $y$ from last equation:
$$
y = \frac{2x+8 \pm \sqrt{(2x+8)^2 - 4(16+x^2-8x)}}{2}
=
\frac{2x+8 \pm 8\sqrt{x}}{2} = x \pm 4\sqrt x + 4.
$$
Now if you want to get form $\sqrt x + \sqrt y = 2$, you need to assume that $0 \leq x,y \leq 4$, hence $y = x - 4\sqrt x + 4 = (2 - \sqrt x)^2$. Now taking square root o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\lim_{n\to\infty}\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$ I am struggling to find the limit of $(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n$ as $n$ goes to $\infty$.
I know that the limit of $(1+\frac{t}{n})^n$ as $n$ goes to $\infty$ equals $e^t$ and that my limit should approach ... | In the same spirit as other answers, you could even go beyond the limit.
Starting with $$A=\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$$
$$\log(A)=n \log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )$$ Now, using long division (or better, Taylor series)
$$\frac{n}{(n+1)^2(n+2)}=\frac 1 {n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given the ratios of distances from three corners of a rectangle, find the coordinates of a point with said ratios.
I think the picture can explain it better than words, but I'm wondering how to figure this out. Given three ratios of distances from corners, not lengths (in the picture I set the base, $base=1$, to be th... | Segment $1$ and segment $\alpha$ create a circle of possibilities called the Circle of Apollonius, with endpoints of its diameter at $\frac{W}{1+\alpha}$ and $\frac{W}{1-\alpha}$. $\beta$ and $H$ work similarly. You can then intersect these circles to find two solutions.
To find points where we can be via $\alpha$ and ... | {
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"timestamp": "2023-03-29T00:00:00",
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max and min of $f(x,y)=\frac{xy+y^2}{x^2+y^2}$ The function is symmetric because $f(x,y)=f(-x,-y)$. It's defined in $\mathbb{R}^2-\{{(0,0)}\}$.If I calculate the gradient , it's null in the origin. Have I to prove $f$ is limited in $\mathbb{R}^2$?
| By the arithmetic mean-geometric mean inequality
$$
|xy| \leq \tfrac{1}{2}(x^2 + y^2)
$$
so
$$
\left|\frac{xy + y^2}{x^2 + y^2}\right| \leq \frac{|xy|}{x^2 + y^2} + \frac{|y|^2}{x^2 + y^2} \leq \frac{1}{2} + 1 = \frac{3}{2}
$$
Since $|f(x,y)| \leq \frac{3}{2}$, we know
$$
-\frac{3}{2} \leq f(x,y) \leq \frac... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving a Linear First Order Differential Equation
$u_x+x^2y^4u_y=0 , u(1,y)=cos(2y) \\
\frac{dy}{dx}= x^2y^4 \implies \int y^{-4} dy = \int x^2 dx \implies -\frac{y^{-3}}{3}= \frac{x^3}{3} +C \\
C= -1/3(x^3+y^{-3}) \\
u(x,y)=f(C)=f(-1/3(x^3+y^{-3})) \\
\text{Given auxiliary condition: } u(1,y)= f(\frac{-1}{3}(1+y^... | $$u_x+x^2y^4u_y=0 \tag 1$$
With the method of characteristics :
$$\frac{dx}{1}=\frac{dy}{x^2y^4}=\frac{du}{0}$$
A first family of characteristic curves comes from necesserarily $du=0 \quad\implies\quad u=c_1$
A second family of characteristic curves comes from $\frac{dx}{1}=\frac{dy}{x^2y^4}$ which is a separable ODE e... | {
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Prove That $ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $ for x, y, z $\gt 0$, prove that: $$ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $$
| Also, we can use Minkowski and AM-GM
$$\sum_{cyc}\sqrt{\frac{x}{y}+\frac{y}{z}}\geq\sqrt{\left(\sum_{cyc}\sqrt{\frac{x}{y}}\right)^2+\left(\sum_{cyc}\sqrt{\frac{y}{z}}\right)^2}\geq\sqrt{3^2+3^2}=3\sqrt2\geq3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite series for $e$... How do you prove that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$? Here I am assuming $e:=\lim_{n\to\infty}(1+\frac{1}{n})^n$. Do you have any good PDF file or booklet available online on this? I do not like how my analysis text handles this...
| First prove by the ratio test that the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!}$ converges and denote the sum by $S$.
Then note that
$$\begin{align*}
\left( 1 + \frac{1}{n} \right)^n & = \sum_{k=0}^n \binom{n}{k} \cdot 1^{n-k} \cdot \left( \frac{1}{n} \right)^k = \sum_{k=0}^n \frac{n (n-1) \ldots (n-k+1)}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\the... | We have
\begin{eqnarray*}
\cos( 2 \theta) =2 \cos^2(\theta)-1 \\
\cos( 4 \theta) =4 \cos^3(\theta)-3 \cos( \theta)
\end{eqnarray*}
Let $z=\cos( \theta) $ then $\cos(3 \theta) = \cos(2 \theta)$ becomes $4z^3-2z^2-3z+1=0$. This will factorise $(z-1)(4z^2+2z-1)=0$, so $\color{red}{\cos(\frac{2 \pi}{5} )=\frac{-1+\sqrt{5... | {
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Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.
For any real number $a\ge -1$ and every natural number n, the statem... | This is not an answer to the question about your proof. I am offering an alternative proof,using Taylor's Theorem. Note that, for integers $n$ and $k$ with $n> k\geq 0$,
$$(1+a)^n-\sum_{r=0}^k\,\binom{n}{r}\,a^r=(k+1)\binom{n}{k+1}\,\int_{0}^a\,(1+x)^{n-k-1}\,(a-x)^k\,\text{d}x\,.$$
Your question is a particular case... | {
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"source": "stackexchange",
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Can the difference of 2 undefined limits be defined? Is this limit defined or undefined?
$$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$
When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it'... | You write $x\to 0+$ but it's often written as $x\to 0^+$. You asked if your limit is defined or undefined. I'll answer that with another two ways.
$$\lim_{x\to 0^+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)=$$
$$=\lim_{x\to 0^+}\left(\frac{\sqrt{1+2x}-1}{x}\cdot \sqrt{x}\right)=$$
Use the definition of a der... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 4
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Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\f... | With Bernoulli, but rewriting to make it simpler:
\begin{align}
&\Bigl(1-\frac1{n+1}\Bigr)^{n+1}>\Bigl(1-\frac1{n}\Bigr)^{n}\iff\Bigl(\frac n{n+1}\Bigr)^{n+1}>\Bigl(\frac{n-1}{n}\Bigr)^{n}\\
\iff&\Bigl(\frac{n^2}{n^2-1}\Bigr)^{n}>\frac{n+1}n=1+\frac1n\iff\Bigl(1+\frac{1}{n^2-1}\Bigr)^{n}>1+\frac1n.
\end{align}
Now, by ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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inequality with parameter - designate param that we get R for which param of A set of solutions is R?
$$-3 < {x^2 - ax - 2\over x^2 - x +1} < 2$$
hm
my trying:
$x^2-x+1$ is always + so I can multiply by $x^2-x+1$ instead of $(x^2-x+1)^2$
result should be set of $(-1 ;2 $ )
| We need
$$x^2-ax-2<2x^2-2x+2$$ or
$$x^2+(a-2)x+4>0,$$ for which we need
$$(a-2)^2-4\cdot1\cdot4<0.$$
Also, we need $$x^2-ax-2+3x^2-3x+3>0$$ or
$$4x^2-(a+3)x+1>0,$$ for which we need $$(a+3)^2-4\cdot4\cdot1<0.$$
Now, solve this system.
I got $$-2<a<1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A proof regarding floor I have been trying to prove that:
$$x \geqslant 1 \implies \text{floor}(x) \geqslant \frac{x}{2}$$
I know this claim is true because I have compared the graph of $x/2$ and $\text{floor}(x)$. However, I just can't seem to get the right intuition has to how to go about proving this. I'm not askin... | Either $\lfloor x \rfloor = 2k$ or $\lfloor x \rfloor = 2k+1$ for some integer $k \ge 0$.
CASE $1. \quad$ $x = 2k + \epsilon$ where $0 \le \epsilon < 1$
\begin{align}
\lfloor x \rfloor \ge \dfrac x2
&\iff 2k \ge k + \dfrac{\epsilon}{2} \\
&\iff k \ge \dfrac{\epsilon}{2} \\
&\iff \text{$(k=\epsilon = 0)$ or ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2... | a^n - b^n = (a - b)(a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + ab^(n-2) + b^(n-1)) - just prove it.
So, a^n - 1 = (a - 1)(a^(n-1) + a^(n-2) + ... + a + 1).
May be it can help you
| {
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Solving Geometric Progression. Question
$\text{While solving recurrence relation ,}$
$$T(n)=T(\sqrt{n})+n$$
I got stuck in following Geometric series.
$$T(n)=n^{\frac{1}{2^{0}}}+n^{\frac{1}{2^{1}}}+n^{\frac{1}{2^{2}}}+...n^{\frac{1}{2^{k-1}}}$$
where $k=\log \log n$
Please help me out.
| We must specify the numbers $n$ for which the recurrence holds (it cannot hold for $n=1$).
Suppose there is a real number $B$ and a function $T:[\sqrt{2},\infty)\rightarrow\mathbb{R}$ such that the following two properties hold:
1) $|T(x)|\leq B$ for all $x \in [\sqrt{2},2)$.
2) $T(x) = T(\sqrt{x}) + x$ for all $x... | {
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A special improper integral I need to evaluate the following integral:
$$\int_{0}^\infty dx \left(\frac{x^2}{2x^2+1}\right)^a\left(\frac{x^2}{2x^2+t_y^2}\right)^b\left(\frac{x^2}{2x^2+t_z^2}\right)^c\frac{1}{x^3}$$
where, $a$, $b$, $c$, $t_y$ and $t_z$ are real.
I have succeeded to perform this integral only for two s... | $$\int_0^\infty\left(\dfrac{x^2}{2x^2+1}\right)^a\left(\dfrac{x^2}{2x^2+t_y^2}\right)^b\left(\dfrac{x^2}{2x^2+t_z^2}\right)^c\dfrac{1}{x^3}~dx$$
$$=-2^{-a-b-c}\int_0^\infty\left(\dfrac{1}{1+\dfrac{1}{2x^2}}\right)^a\left(\dfrac{1}{1+\dfrac{t_y^2}{2x^2}}\right)^b\left(\dfrac{1}{1+\dfrac{t_z^2}{2x^2}}\right)^c~d\left(\df... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Ordinary Generating Function for the number of solution :$ x_1 + x_2 + \cdot\cdot\cdot + x_k = n$ Let $a_n$ be the number of solutions of the equation:
$$x_1 + x_2 + \cdot\cdot\cdot + x_k = n$$
where $x_i$ is the positive odd integer.
Hereto I would like to find the ordinary generating function for the sequence $a_n$ ... | The generating function is
$$
\begin{align}
&\left(\sum_{j=0}^\infty x^{2j+1}\right)^k\tag1\\
&=\left(\frac{x}{1-x^2}\right)^k\tag2\\
&=x^k\sum_{j=0}^\infty(-1)^j\binom{-k}{j}x^{2j}\tag3\\
&=\sum_{j=0}^\infty\binom{k+j-1}{j}x^{2j+k}\tag4\\
&=\sum_{n=k}^\infty\binom{\frac{n+k-2}2}{\frac{n-k}2}x^n\ [2|(n-k)]\tag5
\end{al... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
| Taking the long way:
\begin{align}
f_{n}(x) &= (x-1)(x-2)\cdots(x-n) \\
f_{1}(x) &= x-1 \\
f_{2}(x) &= x^2 - 3x + 2 \\
f_{3}(x) &= x^3 - 6x^2 + 11x - 6\\
f_{4}(x) &= x^4 - 10x^3 + 35x^2 = 50x + 24 \\
f_{5}(x) &= x^5 - 15x^4 + 85x^3 - 215x^2 + 274x - 120.
\end{align}
From here it is determined that $[x^n] \, f_{n}(x) = ... | {
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"timestamp": "2023-03-29T00:00:00",
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Values of $a,b,c$ such that $\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$ Find the values of $a,b,c$ such that $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$
Here's what I have got so far
Using L'Hospital's rule,
$$\lim_{x \to 0}\frac{a+b-\cos x+x\sin x-c\cos x}{5x^4}=1$$
So,$a+b-1=0$
Again,
$$\lim_{x \to ... | Since $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$ is given, we need
$$(x(a+b-\cos x)-c\sin x)'_{x=0}=0$$ or
$$(a+b-\cos{x}+x\sin{x}-c\cos{x})_{x=0}=0$$ or
$$a+b-c-1=0.$$
Thus, $$\lim_{x \to 0}\frac{a+b-(1+c)\cos{x}+x\sin{x}}{5x^4}=1.$$
Hence, we need $$(a+b-(1+c)\cos{x}+x\sin{x})'_{x=0}=0$$ or
$$((1+c)\sin{x... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there a matrix that is NOT I, but also has eigenvalue of 1? I am looking for a matrix that is not I, but also has eigenvalue of 1. Are there any? Can someone please show me an example and how it has eigenvalue 1.
| Here is a simple example: set
$D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag 1$
we have
$D \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \tag 2$
and
$D \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \b... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find, with proof, all integers $k$ satisfying the equation Find, with proof, all integers satisfying the equation
\begin{equation}
\frac{k-15}{2000} + \frac{k - 12}{2003} + \frac{k - 9}{2006} + \frac{k - 6}{2009} + \frac{k - 3}{2012} = \frac{k - 2000}{15} + \frac{k - 2003}{12} + \frac{k - 2006}{9} + \frac{k - 2009}{6} ... | They are just moving all the terms including $k$ to the right and all the ones that are constants to the left to make $b$. The term $\frac {k-2000}{15}$ becomes $\frac 1{15}k-\frac {2000}{15}$ and the second half goes to the right. The $\frac 1{15}$ in the expression for $a$ comes from the first term on the right, th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Common tangent of two ellipses Find the common tangent to: $4(x-4)^2 +25y^2 = 100$ and $4(x+1)^2 +y^2 = 4$.
I have found the derivatives of the above two equations:
$\dfrac{dy}{dx}=\dfrac{16-4x}{25y}$ and $\dfrac{dy}{dx}=\dfrac{-(4x+4)}{y}$
What do I do next?
| Attempt:
1)$\dfrac{(x-4)^2}{5^2} + \dfrac{y^2}{2^2}=1;$
2) $(x+1)^2 +\dfrac{y^2}{2^2} =1;$
Draw them.
1) Major axis $5$, minor axis $2.$
Centred at $(4,0).$
2) Major axis $2$, minor axis $1.$
Centred at $(-1,0).$
The only common tangents :
1) At $(4,2)$ for ellipse $1.$
2) At $(-1,2)$ for ellipse $2.$
$y=2$;
Check your... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| Let $f(x)=\sqrt{x+2}-\sqrt[3]{x+20}$ and $g(x)=\sqrt[4]{x+9}-2$, then
\begin{align*}
\lim_{x\rightarrow 7}\dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}&=\lim_{x\rightarrow 7}\dfrac{f(x)-f(7)}{x-7}\lim_{x\rightarrow 7}\dfrac{x-7}{g(x)-g(7)}\\
&=f'(7)\times\dfrac{1}{g'(7)}.
\end{align*}
Alternatively, if we let
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
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Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so... | Here is one way to evaluate your integral (there are no doubt other ways).
Let
$$I = \int^\infty_0 \frac{\ln^2 x}{a^2 + x^2} \, dx, \quad a > 0.$$
Setting $x = \dfrac{a}{u}$ then $dx = -\dfrac{a}{u^2} \, du$ and we have
\begin{align*}
I &= a \int^\infty_0 \frac{\ln^2 \left (\frac{a}{u} \right )}{\frac{a^2}{u^2} + a^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Expectation of the Minimum of 2 random variables Let Z = min{|X|,|Y|}, with $X , Y \sim Normal(0,1)$.
Then, show that $E(Z) = \frac{2(\sqrt{2} - 1)}{\sqrt{\pi}}$
So far I got
$F_{Z}(z) = 1-[1 - F_{|X|}(z)][1-F_{|Y|}(z)]$
$F_{Z}(z) = 1-4F_{X}(-z)F_{Y}(-z)$
$F_{Z}(z) = 1-\frac{4}{2\pi}\int_{-\infty}^{-z}\int_{-\infty}^... | We know that $F_{|X|}(x)=Pr(-x\leq{X}\leq{x})=F_{X}(x)-F_{X}(-x)$
Hence, $f_{|X|}(x)=f_{X}(x)+f_{X}(-x)=2f_{X}(x)$
Therefore,
$f_{|X|}(x)=\sqrt{\frac{2}{\pi}}e^{\frac{-x^{2}}{2}}$ and similarly,
$f_{|Y|}(y)=\sqrt{\frac{2}{\pi}}e^{\frac{-y^{2}}{2}}$
Then, by symmetry of Normal curve, we can simplify to:
$E[min(|X|,|Y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$ The question:
Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.
Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.
\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^... | If the sum of a finite number of non-negative expressions is $0$, each of them has to be zero.
In other words, when $a,b,c\geq0$
$a+b+c=0 \implies a=b=c=0$
You have done the hard part by showing that it can be written as the sum of three squares.
This means that $a=b$, $a=1$ and $b=1$. What does that tell you about $x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
If $a=2$,what is the solution set of the equation?
Options
*
*{$3,6$}
*{$2$}
*{$3$}
*{$6$} Correct Answer
I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equatio... | You factored it correctly, but your mistake is considering both roots of $x$. When we say $\sqrt{x}$, we refer to the principal root of $x$; that is, the $positive$ square root of $x$ (issue in red, corrections in blue):
For 3:
$\sqrt {3-2} = 3-4$
$\sqrt {1} = \color{red}{±}1$
$\color{blue}{\sqrt {1} = -1}$
$\color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Finding the modulo class of $n \mod 7$ from that of $n^2 + n^3 \equiv 0\pmod 7$ The remainders of $n\mod 7$ will have the set of equivalence classes with residue:
$\{0,1,2,3,4,5,6\}$.
The remainders of $n^2\mod 7$ will have the set of equivalence classes with residue: $\{0,1,4,2\}$.
The remainders of $n^3\mod 7$ will ... | Since $7$ is prime and $n^2+n^3\equiv0\pmod{7}$ you have that either $n^2\equiv0\pmod{7}$ or $n+1\equiv0\pmod{7}$. Thus
$$
n\equiv0\pmod{7}
\qquad\text{or}\qquad
n\equiv6\pmod{7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be ... | $$\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$$
Implies that $\alpha$ and $\beta$ are the roots of $x^2 + 3x + 1$, thus:
$$x^2 + 3x + 1 = (x - \alpha)(x - \beta)$$
Expand $(x - \alpha)(x - \beta)$ to find $-\alpha - \beta = 3$ and $\alpha\beta = 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root modulo of $p$ How to efficiently find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root of $p$ in the range of $[0,p-1]$?
$p$ is a prime.
Is is true that it only has three roots?
So if one of the primitive roots is given, how to efficient found al... | COMMENT (posted because is long for write as a comment).- There are three roots if and only if $p=3m+1$ and only a root if and only if $p=3m+2$.
Since $x^{p-1}-1=0$, if $x^3-1=0$ has three roots then
$\dfrac{x^{p-1}-1}{x^3-1}\in\mathbb F_p$ and if only one root (obviously equal to $1$) then $\dfrac{x^{p-1}-1}{x^3-1}\no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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probability of rolling at least N of C when rolling X dice I have been looking this up but am struggling to find a straightforward answer.
I am still confused about the process.
So basically what is the probability of rolling at least three of a particular number say X when rolling five dice.
I believe the way to go is... | First choose one to be shown $3$ times. There are $6\choose{1}$ ways to choose this die. Then use the standard binomial to get the probability of that selected die showing up exactly $3$ times.
$${6\choose{1}}\cdot{5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 \approx .193$$
And for the same number being shown exact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^{50}$ in the polynomial. What is the coefficient of $x^{50}$ in the polynomial
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}]$
The solution given in one of the books was ,
We have ,
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}... | To get one $x^{50}$ term, you take the $x$ term (coefficient: one) from fifty sets of parentheses and a single constant term from the single remaining set of parentheses. Adding all of these up gives you the total coefficient of $x^{50}$.
As a simple analogy, consider the $x^2$ term in $(x+1)(x+3)(x+5)$.
You take the $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$
My Attempt:
Given
\begin{align}
a(a+7b+49c)+c(a-b+c) &< 0 \\
49a \left( \dfrac {a}{49} + ... | $$a(a+7b+49c)+c(a-b+c)\lt 0$$
is equivalent to
$$ac\lt \frac{-a^2-7ab+bc-c^2}{50}$$
So, we get
$$\begin{align}b^2-4ac&\gt b^2-\frac{4}{50}(-a^2-7ab+bc-c^2)\\\\&=\frac{1}{50}(50b^2+4a^2+28ab-4bc+4c^2)\\\\&=\frac{1}{50}\left(4\left(a+\frac{7b}{2}\right)^2+(b-2c)^2\right)\ge 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find a limit of a function W/OUT l'Hopital's rule. I've got an expression: $\lim_{x\to 0}$ $\frac {log(6-\frac 5{cosx})}{\sin^2 x}$
The question is: how to find limit without l'Hopital's rule?
| Another way using Taylor series
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$$
$$\frac 1 {\cos(x)}=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^5\right)$$
$$6-\frac 5 {\cos(x)}=1-\frac{5 x^2}{2}-\frac{25 x^4}{24}+O\left(x^5\right)$$
$$\log\left(6-\frac 5 {\cos(x)} \right)=-\frac{5 x^2}{2}-\frac{25 x^4}{6}+O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$,
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the soluti... | Using the formulas of half-angle substitution, also called Weierstrass substitution formulas,
$$\cos(u)=\dfrac{1-t^2}{1+t^2} , \ \sin(u)=\dfrac{2t}{1+t^2} \ \ \ \text{where} \ \ \ t:=\tan(\tfrac{u}{2}),$$
one has to prove, setting $a=\tan(\alpha/2)$ and $b=\tan(\beta/2)$, that :
$$2\dfrac{2a}{1+a^2}+2\dfrac{2b}{1+b^2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How many solutions does this system of equations have? I have $2$ equations :
$$-\frac{a}{2}x-\frac{b}{2}+cx+d=x+2$$
$$-2ax-2b+cx+d=2x+1$$
with $a, b, c, x \neq 0$
We have to find all possible solutions of $a, b, c, d$ that make the equations true for all $x$.
I found one solution
$a=-\frac{2}{3}$,
$b=\frac{2}... | Write the equations as:
$$
\begin{cases}
\begin{align}
(a-2c+2)x+b-2d+4 = 0 \\
(2a-c+2)x+2b-d+1 = 0
\end{align}
\end{cases}
$$
A polynomial is identical $0$ (i.e. for all $x$) iff all its coefficients are $0\,$, which gives the system to solve:
$$
\begin{cases}
\begin{align}
a - 2c + 2 = 0 \\
2a - c + 2 = 0 \\[7px]
b -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Ratio of parts of an intersected segment in a rectangle
In rectangle $ABCD$, points $E$ and $F$ lie on sides $BC$ and $CD$ respectively. Point $F$ is the midpoint of $CD$ and $BE=\frac13BC$. Segments $AC$ and $FE$ intersect at point $P$. What is the ratio of $AP$ to $PC$? Express your answer as a common fraction.
I'... | $A = (0,0)$
$B = (x,0)$
$C = (x,-y)$
$D = (0,-y)$
$E = (x,-y/3)$
$F = (x/2,-y)$
$P = (tx,-ty)$
$P = \lambda E + (1-\lambda)F$
$tx = \lambda x + (1-\lambda)x/2 = \lambda x + x/2 - \lambda x/2 = (\lambda + 1/2 - \lambda/2) x $
$t = \frac{(\lambda + 1)}{2}$
$-\frac{(\lambda + 1)}{2}y = \lambda (-y/3) + (1-\lambda)(-y) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And bec... | your general term is equivalent to $1/n^2$, so it converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the general term of the sequence, starting with n=1 Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.
$$\frac{3}{2^2 - 1^2}, \frac{4}{3^2 - 2^2} , \frac{5}{4^2 - 3^2}, \cdots$$
Can you help me with this,I know how to solve the problem with ... | Observe the series first.
$$ \frac{1\color{green}{+2}}{(1\color{green}{+1})^2-(1)^2}, \frac{2\color{green}{+2}}{(2\color{green}{+1})^2-(2)^2}, \frac{3\color{green}{+2}}{(3\color{green}{+1})^2-(3)^2}, ....
\text{upto} \frac{n\color{green}{+2}}{(n\color{green}{+1})^2-(n)^2} $$
So now,
General Term or $ n^{th} $ term c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$25! \pmod {78125}$ $25! \pmod{78125}$ is a problem I'm working on.
Since $78125$ looked very divisible by $5$, I checked, and found that $78125 = 5^7$.
Then I thought, if there are seven factors 5 in $25!$, then $25! \equiv 0 \mod 78125$, but I only found $25$ to be divisible by 5 six times, so I don't think that got ... | If $\frac{25!}{5^6}\equiv k \pmod{5}$, then $5\mid\frac{25!}{5^6}-k$, i.e. $5^7\mid 25!-k\cdot 5^6$ so $25!\equiv k\cdot 5^6 \pmod{5^7}$. What remains is to find $k$.
For that, note $\frac{25!}{5^6}=(1\cdot 2 \cdot 3 \cdot 4) \cdot (6
\cdot 7 \cdot 8 \cdot 9) \cdot (11 \cdot 12 \cdot 13 \cdot 14) \cdot (16 \cdot 17 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Yet another way: let $a=\frac{1+i\sqrt{7}}{2}$ and $b=\frac{1-i\sqrt{7}}{2}$ then $a+b=1, ab=2$ so $a,b$ are the roots of the quadratic $x^2-x+2=0\,$.
Therefore $a^2=a-2\,$, then successively:
*
*$a^3 = a\cdot a^2 = a \cdot(a-2) = a^2-2a = (a-2)-2a=-a-2$
*$a^4= a \cdot a^3 = a(-a-2) = -a^2-2a=-(a-2)-2a=-3a+2$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
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$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
| Using series expansion:
$$\cos x = 1 - \frac{x^2}{2}+\frac{x^4}{24}+ O(x^6)$$
$$\frac{1}{1-\cos x}=\frac{1}{x^2}\frac{2}{1 -\frac{1}{12}x^2 + O(x^4)}$$
$$\frac{1}{1-\cos x}-\frac{2}{x^2}=\frac{2}{x^2}\left(\frac{ \frac{1}{12}x^2 + O(x^4) }{1 -\frac{1}{12}x^2 + O(x^4)}\right) \to \frac16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?
| Write $x^2 - 194xy + y^2$ in matrix form:
$$
\pmatrix{x & y}
\pmatrix{ 1 & -97 \\ -97 & 1}
\pmatrix{x \\ y}
$$
This matrix has eigenvectors $(1,\pm 1)$, which leads us to consider $u=x+y$ and $v=x-y$.
Then $x^2 - 194xy + y^2 + 9408=0$ becomes $49 v^2 - 48 u^2+ 9408=0$.
Since $9408 =2^6×3×7^2$, this implies that $u=7u_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\fra... | \begin{align*}
\sum_{k=2}^{N}\dfrac{1}{2k(2k-1)}\leq\sum_{k=2}^{N}\dfrac{1}{2k(2k-(k/2))}=\dfrac{1}{3}\sum_{k=2}^{N}\dfrac{1}{k^{2}}<\dfrac{1}{3}\sum_{k=2}^{\infty}\dfrac{1}{k^{2}}<\infty,
\end{align*}
so $\{S_{2N}\}$ is convergent, so is $\{S_{2N+1}\}$ because $\lim_{N}(S_{2N+1}-S_{2N})=0$ (so they have the same limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this fo... | For your convenience, you could first write
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies\left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}$$
Thus, with the aid of the chain rule:
$$\left(\arctan\sqrt\frac{1+x}{1-x}\right)'=\frac1{1+\frac{1+x}{1-x}}\cdot\frac1{2\sqrt\frac{1+x}{1-x}}\cdot\frac2{(1-x)^2}=$$
$$=\frac{1-x}2\cdot\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(xy - 1) + f(x)f(y) = 2xy - 1$ Using induction, I proved that $f(x) = x$ and $f(x) = -x^2$ work, but only for rational numbers.
How can I prove them for all real numbers?
| So we assume $$f(xy - 1) + f(x)f(y) = 2xy - 1\tag1$$ for all real $x,y$. The RHS is not constant, so constants aren't solutions. Substituting $y=0$ in (1) gives
$$f(-1)+f(x)f(0)=-1,$$ and as $f(x)$ is not a constant, we get
$$f(0)=0,\quad f(-1)=-1\tag2.$$
Letting $y=1$, we obtain $$f(x-1)+f(x)f(1)=2x-1\tag3.$$ Replacin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5... | For $x=y=z=\frac{1}{3}$ we get a value $\frac{3}{2}.$
We'll prove that it's a minimal value.
Indeed, let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}.$
Hence, $a+b+c=3$ and
$$\sum_{cyc}\frac{2-x}{3+x}-\frac{3}{2}=\sum_{cyc}\left(\frac{2-\frac{a}{3}}{3+\frac{a}{3}}-\frac{1}{2}\right)=\frac{3}{2}\sum_{cyc}\frac{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation us... | How does $\cos^2y=\dfrac12$ imply $$\dfrac{n\pi}2\pm\dfrac\pi6?$$
In fact $\cos^2y=\dfrac12\iff\cos2y=0$
$\implies2y=m\pi+\dfrac\pi2,4y=\pi(2m+1)$ where $m$ is any integer $\ \ \ \ (1)$
Again, $\cos^2y=\cos^2\dfrac\pi4\iff\sin^2y=\sin^2\dfrac\pi4$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$y=n\pi\pm\dfrac\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is there a criterion for showing that a two variable continuous function $f:\Bbb S^1\times\Bbb S^1\to \Bbb R^2$ pass from the origin? Is there a criterion for showing that a symmetric ($f(x,y)=f(y,x)$) two variable continuous function $f:\Bbb S^1\times\Bbb S^1\to \Bbb R^2$ pass from the origin? i.e.
$$\exists\,(x,y)\... | Let $\alpha(\theta)=\left\lgroup
\begin{array}{c}
\cos \theta
\\
\sin \theta
\end{array}
\right\rgroup \in \mathbb{S}^1$ and $\beta(\varphi)=\left\lgroup
\begin{array}{c}
\cos \varphi
\\
\sin \varphi
\end{array}
\right\rgroup\in\mathbb{S}^1$ parametrizations of $\mathbb{S}^1$ whit $0\leq \theta\leq 2\cdot \pi$ and $0\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral of x² from 0 to b - using archimedes sum of squares - Apostol's I'm reading Apostol's Calculus book and in the first chapter is presented the way archimedes found the sum of the square and how it can be used to calculate the integral of $x²$. But I'm not able to follow some steps of the proof.
from the book:
W... | You may want to read this. It really is not an easy or intuitive inequality to find.
Alternatively, you can consider this:
Eventually, you would want $n$ to approach $+\infty$.
So let's consider
\begin{align}
S_{\text{big}} &= \frac{b^3}{n^3} \cdot \left(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}\right)\\
&=b^3\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to solve this equation using logarithms for the equation :
$ 3^{x^2} + 3^x = 90 $
my solution was :
$3^{x^2} + 3 ^x = 3^{2^2} + 3^2$
so $x=2$
but i want to know if there is any solution by using logarithms ?
when using wolframAlpha th solution was $ x= -2.02356 $
or $x = 2 $
but How ?
| If we divide both sides by $3$ first, we get $$3^{x^2-1} + 3^{x-1} = 30$$
Then, we can write it as $$(3^{x-1})^{x+1}+3^{x-1} = 3^{x-1}(3^{x-1})^{x}+3^{x-1}= 30$$
Now, if we bracket $3^{x-1}$, we get $$3^{x-1}(3^{x^2-x}+1)=30$$
From here, if there exists an integer solution, $30$ must be factorized as $3 \cdot 10$ so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I solve the following problem with floor functions?
Ideally I'd like to find an analytical solution, though I doubt one exists. If an analytical doesn't exist, any insight into tackling this problem numerically would be greatly appreciated. I can't think of anything other than naive brute-force.
Given $c$, fi... | Define $\{x\} = x - [x]$ for $x \in \mathbb{R}$, then$$
d = \{\sqrt{c}\}, \quad h = \{2d\}.
$$
For $c \in \mathbb{N}_+$, suppose $m \in \mathbb{N}_+$ satisfies $m^2 \leqslant c < (m + 1)^2$. Note that $\displaystyle c \neq \left(m + \frac{1}{2}\right)^2$ since $c \in \mathbb{N}_+$.
Case 1: $\displaystyle m^2 \leqslant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor limits with sine I'm having troubles calculating these two limits (I prefer to write a sigle question including both of them, instead of two different ones).
This one
$$\\ \lim_{x\rightarrow 0} \frac{ x-\sin^2(\sqrt x)-\sin^2(x)} {x^2} $$ I tried expanding with Taylors at different orders but the... | If you're uncertain about $\sqrt{x}$, substitute $\sqrt{x}=t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{t^2-\sin^2(t)-\sin^2(t^2)}{t^4}
$$
Of course we just need Taylor up to degree $4$:
\begin{align}
\sin^2(t)&=\left(t-\frac{t^3}{6}+o(t^3)\right)^2=t^2-\frac{t^4}{3}+o(t^4)\\[4px]
\sin^2(t^2)&=(t^2+o(t^2))^2=t^4+o(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding The Basis Of An Intersection Of Two Subspaces
Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$
I have forgot the algorithm for finding inters... |
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
Call $u=\begin{bmatrix} 1 & 2 & -1\\
28 & 28& 28\\
2 & 2 & 2\\
39& 39&39\end{bmatrix}$ and $\xi=\begin{bmatrix}x\\y\\z\\w\end{bmatrix}$.
Then $(x,y,z,w)\in U$ if and only if $\operat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$... | What if I add extra information regarding the vector $V$ as the following,
Question: Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and ${V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}}$ , find $A^{-1}V$.
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Explicit form of the function $F\left(p\right)=\int_{0}^{\, \pi/2}\ln\left(1+p\sin^2\left(t\right)\right)\text{d}t$ Let $x \in \mathbb{R}^{+}$, I wonder how to prove that
$$
F\left(x\right)=\int_{0}^{\, \pi/2}\ln\left(1+x\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\frac{1}{2}\left(1+\sqrt{1+x}\right)\right)
$$
It... | Consider
$$ F'(x) =\int_0^{\pi/2} \frac{\sin^2 t}{1+x\sin^2t}dt = \int_0^{\pi/2} \frac{1}{\csc^2t + x}dt $$
Let $u = \cot t$
$$ \begin{align}
F'(x) &= \int_0^\infty \frac{1}{(u^2+1)(u^2+1+x)}du \\
&= \frac{1}{x}\int_0^\infty \left(\frac{1}{u^2+1} - \frac{1}{u^2+1+x} \right) du \\
&= \frac{1}{x}\left(\frac{\pi}{2} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 ... | You're assuming that the limit of the sum of the first $n$ Fibonacci numbers exists as $n \to \infty$, which it doesn't. Which is to say that in order to apply your method, the series must be convergent but the series diverges, so your method is wrong. It is, however, related to the power series of the Fibonacci number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$\exp(z^2)=i$ solution check Find all complex solutions to the equation $\exp(z^2)=i$.
Attempt
We have $z^2 = x^2-y^2+2ixy$ where $z=x+iy$. Then
$$exp(x^2-y^2)=1 \text{ and } 2xy=\frac{\pi}{2}$$
So $x^2=y^2$ and $xy = \frac{\pi}{4}$ Thus $x = y$ and so $x=y= \pm \frac{\sqrt{\pi}}{2}$
I just don't feel very confident ab... | If $\exp(z^2) = i$ then $z^2 = \log i = \ln|i| + i\arg i = 0 + i(\frac{\pi}{2} + 2\pi n) = \frac{4n+1}{2}\pi i$ where $n$ is an integer. These are the numbers $\ldots, -\frac{7\pi i}{2}, -\frac{3 \pi i}{2}, \frac{\pi i}{2}, \frac{5 \pi i}{2}, \frac{9\pi i}{2}, \ldots$. Taking square roots (using that $\sqrt{i} = \pm\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$f$ is differentiable at x=1, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n} $ $f$ is differentiable at $x=1$, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n}$.
Here's what I got so far... | Just a suggestion (if I may) to make life easier and to get more than the limit.
Consider $$A=\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^{n^a}\implies \log(A)=n^a \log\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)$$ Assuming that $f(.)$ is continuoulsy differentiable at $x=1$, Taylor series gives ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac... | $$
\frac{a+b}{1+a+b} =\frac{a}{1+a+b} + \frac{b}{1+a+b}
$$
then prove
$$
\frac{a}{1+a+b} \leq \frac{a}{1+a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Find $ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2} $ I want to find the following limit:
$$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$
I guess it converges to 0 and I tried to prove it using the squeeze theorem b... | Note that
$$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2}
=\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$
indeed for Harmonic series and Euler–Maclaurin formula
$$\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Confusion about solving quadratics $x^2-8x=5$ I got into a little confusion, hopefully someone can clear it up for me.
So I'm solving $x^2-8x=5$ right, and here are my steps:
$$1: x^2-8x=5$$
$$2: x^2-8x+16=21$$
$$3: (x-4)^2=21$$
$$4: x-4=\pm\sqrt{21}$$
$$x=\pm\sqrt{21} +4$$
So I have a question, why is it plus or minus... | We have the equation
$$x^2 - 8x = 5$$
Add $16$ to both sides so we can factor the LHS in a nice way, so
\begin{align}
x^2 - 8x + 16 &= 5 + 16 \\
(x-4)^2 &= 21
\end{align}
Now take the square root on both sides (we can do this because both sides of the equation are positive terms)
$$ \sqrt{(x-4)^2} = \sqrt{21}$$
Noti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Infinite series summation
$$\sum^{\infty}_{n=0}\frac{1}{(4n+5)(4n+6)\cdots \cdots (4n+11)}$$
Try: $$\sum^{\infty}_{n=0}\frac{(4n+4)!}{(4n+11)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\frac{(4n+4)!\cdot 6!}{(4n+4+6+1)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\int^{1}_{0}x^{4n+4}(1-x)^6dx$$
$$ \frac{1}{6!}\int^{1}_{0}\left(\sum^{\... | The telescoping series, as answered by lab bhattacharjee, is the good way to go.
With regard to the integral you properly wrote, using long division and partial fraction decomposition, $$\frac{x^4(1-x)^6}{1-x^4}=-x^6+6 x^5-15 x^4+20 x^3-16 x^2+12x-16+\frac{4 x}{x^2+1}+\frac{16}{x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite integral of a rational fraction Can I find the value of $$\int_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?
I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?
| write $$\frac{x-1}{(x^2-2x-3)^2}$$ as $$-1/8\, \left( x+1 \right) ^{-2}+1/8\, \left( x-3 \right) ^{-2}$$
use that $$\frac{(x-1)}{(x^2-2x-3)^2}=\frac{x-1}{(x+1)^2(x-3)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-3}+\frac{D}{(x-3)^2}$$
it is
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{A(x+1)(x-3)^2+B(x-3)^2+C(x-3)(x+1)^2+D(x+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Using only the digits 2,3,9, how many six-digit numbers can be formed which are divisible by 6?
Using only the digits $2,3,9$, how many six-digit numbers can be formed which are divisible by $6$?
The options are:
(A) $41$
(B) $80$
(C) $81$
(D) $161$
The last digit must be $2$. But I faced problem when calculating the... | Let's find a divisability test for 6.
\begin{equation}
1 = 1 \mod 6\\
10 = 4 \mod 6\\
100 = 4*10 = 4 \mod 6,\\
\text{and so on for higher powers of 10}
\end{equation}
Thus, we find: a number X is divisable by 6 iff, cutting of the last digit, taking the sum of the other digits times 4 and adding the last digit the resu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my p... | From $a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$ one gets $\frac {a_n^2}{a_1^2 \ldots a_{n-1}^2}- \frac 4 {a_1^2 \ldots a_{n-1}^2}= (a_1^2-4)$ therefore $\frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {\frac 4 {a_1^2 \ldots a_{n-1}^2} + (a_1^2-4)} \tag 1$
Now, since $\lim a_n= +\infty$ it follows $\lim a_1^2 \ldots a_{n-1}^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Proving that $ (A + I)^n=(2^n-1)A + I $ Assume that $ A^2 = A$ , where $A$ is a square matrix.
Then prove $(A + I)^n=(2^n-1)A + I $
Could someone help ?
| By indution if $(A+I)^n=(2^n-1)A+I$ and $A^2=A$, we have
\begin{eqnarray}
(A+I)^{n+1}&=&(A+I)^n(A+I)=((2^n-1)A+I)(A+I)\\
&=&(2^n-1)A^2+A+(2^n-1)A+I\\
&=&2(2^n-1)A+A+I\\
&=&(2^{n+1}-2)A+A+I\\
&=&(2^{n+1}-1)A+I
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate pro... | There is the following formula.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\prod_{cyc}(a-b)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}-1\right)=-31.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\be... | The limit $$\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$$
does not exist.
Notice that as $x\to 0^+$, $1/x\to \infty$ .
Therefore sin(1/x) jumps up and down between $-1$ and $1$ infinitely many times.
Thus the limit $$\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$$ does n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Limits of power of fractions I'm having trouble solving these limits:
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}\\\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}\\\lim_{x\to \infty} \left (\frac{2-3x}{1-3x}\right )^{x+4}$$
Are they even defined?
| Hint: Knowing that $$e^a =\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x$$
show by yourself that:
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}~~~~~~~~DNE$$
$$\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}= \lim_{x\to \infty} \left (\frac{3}{2}-\frac{11}{4x+2}\right )^{2x-1}=\infty$$
$$\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integral involving multiple variables $$\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx$$
So this is what I did:
$u = 3ax + bx^3$
$\dfrac{\mathrm du}{\mathrm dx}= 3a + 3bx^2$
$\mathrm du = 3a + 3bx^2 dx$
= $\displaystyle\int 3\cdot \frac{1}{\sqrt{u}}\mathrm du$
= $3\cdot \frac{u^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c... | $$\begin{align}I&=\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx\\&=\dfrac13\int\dfrac{3a+3bx^2}{\sqrt{3ax+bx^3}}\,\mathrm dx\\&=\dfrac13\int\dfrac{\mathrm d(3ax+bx^3)}{\sqrt{3ax+bx^3}}\qquad\text{Use substitution now, let }u=3ax+bx^3\\&=\dfrac13\int\dfrac{\mathrm du}{\sqrt u}\\&=\dfrac13\cdot 2\sqrt u+C\\&=\dfrac23\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove $\lim \limits_{x \to 1^-} \sum\limits_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2} \ $? $\lim \limits_{x \to 1^-} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2}$
The power $n^2$ is problematic. Can we bring this back to the study of usual power series?
I do not really have any idea for the moment.
| EDITED. Here is yet another answer based on my recent answer. Indeed, if $P$ is a non-constant polynomial with coefficients in $\mathbb{R}$ such that $P(n) \to +\infty$ as $n \to +\infty$, one immediately deduces from the result in the link that
$$ \lim_{x \uparrow 1^-} \sum_{n=0}^{\infty} (-1)^n x^{P(n)}
= \lim_{s \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre The line with equation $2x+y-5=0$ is a tangent to the circle with equation $(x-3)^2 + (y-p)^2=5$
Find the two possible values of $p$.
| The equation
$$(x-3)^2+(5-2x-p)^2=5$$
or
$$x^2+9-6x+4x^2+q^2-4qx=5$$
where $q=5-p $
must have only one solution.
$$5x^2-2x (2q+3)+4=0$$
$$\delta'=(2q+3)^2-20=0$$
thus
$$p=5-q=5-\frac {-3\pm 2\sqrt{5}}{2} $$
$$=\frac {13}{2}\pm \sqrt {5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$
Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$
My approach:
I was thinking to use the binomial f... | A combinatorial proof based upon lattice paths.
The number of lattice paths of length $n+1$ from $(0,0)$ to $Q_k=(n-k+1,k)$ consisting of $(1,0)$-steps and $(0,1)$-steps is $$\binom{n+1}{n-k+1}=\binom{n+1}{k}$$ since we have to choose precisely $n-k+1$ $(1,0)$-steps from a total of $n+1$ steps.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving $ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$
Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$
Well I first observe... | Let $\sqrt {x-1}=t \implies x= 1+t^2 \quad x^2=(1+t^2)^2$
$$\frac{(1+t^2)^2}{t^2} + t +\frac{t}{(1+t^2)^2} = \frac{t^2}{(1+t^2)^2} + \frac{1}{t} + \frac{(1+t^2)^2}{t}$$
$$(1+t^2)^4 + t^3(1+t^2)^2 +t^3 = t^4 + t(1+t^2)^2 + t(1+t^2)^4$$
$$(t - 1)(t^2 - t + 1) (t^2 + t + 1) (t^4 + 2 t^2 - t + 1)=0 \implies t=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\tan^{-1}\Big[\frac{\cos x-\sin x}{\cos x+\sin x}\Big]$, where $x<\pi$ Write in the simplist form
$$
\tan^{-1}\bigg[\frac{\cos x-\sin x}{\cos x+\sin x}\bigg],\quad x<\pi
$$
My Attempt:
$$
x<\pi\implies -x>-\pi\implies \frac{\pi}{4}-x>\frac{-3\pi}{4}
$$
$$
\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-... | The given expression is defined for $x\neq \frac{3\pi}{4}+k\pi$ with the given condition $x<\pi$.
When you simplify the expression by $\arctan$ you need to restrict the domain to $\frac{-\pi}{4}<x<\frac{3\pi}{4}$ which is not in contrast with the initial one.
Thus I don't see any kind of problem with your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Divergence of a Series $\sum_{n=1}^\infty (\frac{1}{n!})(\frac{n}{e})^n$ I'm trying to show that $$\sum_{n=1}^\infty \left(\frac{1}{n!}\right)\left(\frac{n}{e}\right)^n$$ diverges by the Corollary of Raabe's Test.
Corollary of Raabe's Test: Let $X=\{x_n\}$ be a sequence of nonzero real numbers and let $a=\lim_{n \to \... | The limit is $a=1/2$.
Start by the approximation
$$\ln \left( \frac 1e (1+x)^{1/x} \right) = -1 + \frac 1x \ln (1+x) = -1 + 1-\frac 12 x +o(x) = -\frac 12 x +o(x) $$
Plugging $x= \frac 1n$ you get
$$1- \frac 1e \left( 1+ \frac 1n \right)^n = 1- e^{\ln \left( \frac 1e (1+\frac 1n)^{n} \right)} = 1- e^{-\frac{1}{2n} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equ... | So $(a-b)(a^2 + ab + b^2) = 3^3*37$
So $(a-b)|3^3*37$.
So $a-b = 3^j*37^k$ where $j = 0,1,2,3$ and $k=0,1$
And $(a^2 + ab +b^2) = \frac {3^3*27}{a-b} = 3^{3-j}*37^{1-k}$
That's 8 possible systems of equation.
But there are some obvious things to note.
If $37|a-b$ then $a - b \ge 37$ and ... that just seems wrong.
Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any other idea than to guess them.
| The peculiar form of the polynomial, with the same $\,\sqrt{2}\,$ present in the imaginary parts of the coefficients, only, suggests splitting it into two terms for closer inspection:
$$
\begin{align}
P(z) &= z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10 \\
&= (z^4-2z^3+3z^2+4z-10) + i \,\sqrt{2}\,(2z^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=... | $$\frac{6+\sqrt{41}-2}{5}=\frac{10+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-6}{5}$$
with $0<\sqrt{41}-6<5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\... | A great approach on your part! Let me try another.
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} \int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2} $$
So we really only need to solve the integral in the form:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
| Hint
$$\frac{\sqrt{x+2}-2}{x-2}\cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}= \frac{\sqrt{x+2}^2-2^2}{(x-2)\sqrt{x+2}+2}$$
$$ \frac{x-2}{(x-2)\sqrt{x+2}+2}= \frac{1}{\sqrt{x+2}+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$. Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.
I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have n... | By C-S $$\sum_{cyc}\frac{x^2}{(x+y)(x+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(x+z)}=\frac{\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+3xy)}\geq\frac{3}{4},$$
where the last inequality it's $$\sum_{cyc}(x-y)^2\geq0.$$
The equality occurs for $x=y=z$, which says that we got a minimal value.
Another way:
We n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the ... | Note that by Taylor's series
*
*$\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4)$
*$\sin x^2=x^2+o(x^2)$
thus
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-x^2+\frac{x^4}{2}+o(x^4)}{x^4+o(x^4)}=\frac{\frac12+o(1)}{1+o(1)}\to \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $... | The first formula fails to find the probability after $n+1$, because:
When $n=2$, the sample space is: $(1,1),(1,2),(2,1),(2,2)$, while $2\le k \le 4$ and
$$k=2: (1,1); k=3: (1,2),(2,1); k=4: (2,2)$$
Note that when $k=4$, we are ignoring $(1,3),(3,1)$, because they are not in the sample space. So the two formulas must... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\si... | We can use the standard formula $$\int_{0}^{2\pi}\frac{dx}{1-a\cos x} =\frac{2\pi}{\sqrt{1-a^2}},|a|<1\tag{1}$$ Note that $$\frac{d} {dx} \frac{\sin x} {1-a\cos x} =\frac{\cos x-a} {(1-a\cos x) ^2} =-\frac{1}{a}\frac{a^2-1+1-a\cos x} {(1-a\cos x) ^2}$$ and integrating the above with respect to $x$ in interval $[0,2\pi]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
| When $u = 2x$, $du = 2dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx
= \int^{3\sqrt{3}}_0\frac{u^3/8}{(u^2+9)^{3/2}}\frac{du}{2},
$$
this allows us to do further substitution $u = 3 \tan t$ to get rid of the root sign in the denominator.
If you use $u = x^3$, $du = 3x^2 dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | It is easier to expand $$(1 + x + x^2 + x^3 + x^4 + x^5)^2$$
to get $$ 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10}$$
All we have to do is check the squares and twice the products to see if the coefficients are correct.
How do we see that P(x) is a perfect square?
When evaluated at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
Solve the equation in positive integers $k^m+m^n=kmn$
Solve the equation in positive integers $$k^m+m^n=kmn$$
My work so far:
1) Let $k=m$. Then
$$k^k+k^n=k^2n$$
If $k=1$ then $n=2$; if $k=2$ then $n=2$ or $n=3$
Let $k\ge 2$. We see that $n\ge2$.
$$k^{k-2}+k^{n-2}=n$$
2) Let $k=n$. Then $$k^m+m^k=k^2m$$. I find $m... | Suppose first that $k$ is the largest among $k$, $m$, and $n$. Then, $$k^m<k^m+m^n=kmn\leq k^3$$ implies that $k>1$ and $m<3$. If $m=1$, then we have $k+1=kn$, which is not possible (recalling that $k>1$). If $m=2$, then $k^2+2^n=2kn$ implies that
$$0\leq (k-n)^2=n^2-2^n\,,$$
yielding $n\in \{2,3,4\}$. Hence, in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$
| You can calculate the partial sum using
$$
\frac{1}{(3k+1)(3k+2)(3k+3)}=\frac{1}{2}\frac{1}{3k+1}-\frac{1}{3k+2}+\frac{1}{2}\frac{1}{3k+3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}... | The limit is indeed $\frac{1}{2}$. Due to Taylor's formula with integral remainder,
$$ \sum_{k=0}^{n}\frac{1}{k!} = e-\int_{0}^{1}\frac{(1-t)^n}{n!}\,e^t\,dt=e\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)=\exp\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)\tag{1} $$
while
$$ \left(1+\frac{1}{n}\right)^{n+x}=\exp\left[(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Equation of a circle cutting another given circle orthogonally What is the equation of the circle which touches the line $x+y=5$ at $(-2,7)$ and cut the circle $$x^2+y^2+4x-6y+9=0$$ orthogonally?
I tried to denote the center of circle as $(h,k)$ and radius as $r$. Orthonogallity implies
\begin{align*}
(h+2)^2+(k-3)^2&=... | The equation of tangent to a circle is given by $T=0$
Let the equation of required circle be $x^2+y^2+2gx+2fy+c=0$ and the tangent to that circle is at point $(-2,7)$. Hence using above information the equation of tangent at that point is given by $$(g-2)x+(f+7)y-2g+7f+c=0$$
But the equation of tangent is given to be $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}... | Setting $U=\dfrac{XY}{\sqrt{X^2+Y^2}}, V=\dfrac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ and using moment generating functions:
\begin{align}
M_{(U,V)}(u,v) & =\mathbb{E}\left[\mathrm{e}^{\langle\,(u,v)\,;\,(U,V)\,\rangle}\right] \\[10pt]
& =\iint_{\mathbb{R}^2}\exp\left(u\frac{xy}{\sqrt{x^2+y^2}}+v\frac{x^2-y^2}{2\sqrt{x^2+y^2}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
| Let $\tan{a}=x$, $\tan{b}=y$ and $\tan{c}=z$.
Thus, $$0=\sum_{cyc}\frac{\sin(a-b)}{\sin(a+b)}=\sum_{cyc}\frac{x-y}{x+y}=\frac{\sum\limits_{cyc}(x-y)(x+z)(y+z)}{\prod\limits_{cyc}(x+y)}=$$
$$=\frac{\sum\limits_{cyc}(x-y)(z^2+xy+xz+yz)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x-y)z^2}{\prod\limits_{cyc}(x+y)}=$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3... | Let $7 = 1 + 6$, then $$\begin{align} x(2x+7) + 3 &= x(2x + 1 + 6) + 3 \\ &= x(2x + 1) + 6x + 3 \\ &= x(2x + 1) + 3(2x + 1) \\ &= (2x+1)(x+3)\end{align}$$ as should be desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem i... | This is not a full proof yet, just an idea I had.
From Leibniz series we know that for $k \geq 1$:
$$\frac{\pi}{4} \geq 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1}$$
Squaring we have:
$$\frac{\pi^2}{6} \geq \frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are th... | Using Complex Addition
$$
e^{ia}+e^{i(\pi-b)}=\sqrt{\frac12}+i\sqrt{\frac32}=\color{#C00}{\sqrt2}e^{i\color{#090}{\pi/3}}
$$
*
*$b+a=\frac\pi2$ since $\left|e^{ia}+e^{i(\pi-b)}\right|^2=\color{#C00}{2}\implies\overbrace{\color{#C00}{2}=2+2\cos((\pi-b)-a)}^{\text{Law of Cosines}}$
*$b-a=\frac\pi3$ since $\frac{a+(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Understanding the formula for stereographic projection of a point. I was wondering about the equation of line I can write which can help me finding the coordinates of Point $P'$ in relation with coordinates of points on the sphere that is $P$.
Let the $P'(X,Y)$, then how can I find these following coordinates -
I trie... | Let $S$ be the point of tangency of the sphere and the plane,
and let $Q$ lie on line $NS$ such that angle $\angle NQP$ is a right angle.
Observe that triangles $\triangle NQP$ and $\triangle NSP'$ are similar,
with $NQ = a - z$ and $NS = 2a.$
Therefore $$\frac{P'S}{PQ} = \frac{2a}{a-z}. \tag1$$
But we also have $P = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \co... | Hint:
Since
$$
\begin{align}
\sin(y)
&=3\cos(x+y)\sin(x)\\
&=3\cos(x)\cos(y)\sin(x)-3\sin(x)\sin(y)\sin(x)
\end{align}
$$
we get
$$
\begin{align}
\tan(y)
&=\frac{3\sin(x)\cos(x)}{1+3\sin^2(x)}\\
&=\frac{3\tan(x)}{\sec^2(x)+3\tan^2(x)}\\[3pt]
&=\frac{3\tan(x)}{1+4\tan^2(x)}
\end{align}
$$
Then note that $1+4\tan^2(x)=4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
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.