Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
McLaurin expansion for $f(x)=e^{\sin{x}}$ of the 4:th order. By the 4:th order, they mean using the 4:th derivative. But the differentiation gets a bit ugly quite fast, so instead of computing all the derivatives , I should be able to use the standard expansions:
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+B(t) \\ \sin{x}=... | You already have
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+O(t^5)$$
and
$$\sin{x}=x-\frac{x^3}{3!}+O(x^5),$$
So just substitute:
$$e^{\sin x} = 1 + \left(x-\frac{x^3}{3!}+O(x^5)\right) + \frac{1}{2!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^2 \\+ \frac{1}{3!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^3 +\frac{1}{4!}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.
| Infinite Sum
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+n+1}
&=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}-\frac1{n+\frac12+i\frac{\sqrt3}2}\right]\tag1\\
&=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}+\frac1{-n-\frac12-i\frac{\sqrt3}2}\right]\tag2\\
&=\frac1{i\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute the Integral via Residue Theorem and Laurent Expansion Let $f(z)=\dfrac{z^{15}}{(z^2+1)^2(z^4+2)^3}$ . My goal is to compute $\displaystyle I=\int_{|z|=4}f(z) \,dz$.
Since all the singular points (except $∞$) lie in the circle $|z|<4$, we obtain $I=-2\pi i\operatorname{Res}(f,∞)$. Thus, we just need to find the... | You want to find $Res_{w = 0} \left[\frac 1{w^2}f\left(\frac 1w \right)\right] = Res_{w = 0} \left[\dfrac 1{w(1+w^2)^2(1+2w^4)^3}\right]$ . Now we have:
$$a_{-1} = \frac{1}{2\pi i} \int_{C_0} \frac {dw}{w(1+w^2)^2(1+2w^4)^3} = \frac{1}{(1+0^2)^2(1+2\cdot 0^4)^4} = 1$$
by Cauchy Integral Formula, as $\frac{1}{(1+w^2)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solut... | Let $\Lambda =$ the limit we need to find. Then, $ \ \Box\ \Lambda = 1$.
Proof: We will begin our proof using the following Lemma. $$\forall a, b\in\mathbb{R}, \ \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}.\tag1$$
Substitute $a = x$ and $b = x + \sqrt{x}$ into the Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:
$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$
Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$
| We have: $a_1a_2 = 1 $ , $a_2a_3 = 2$ ,..., $a_na_{n+1}$ = $n$.
Thus , $a_n = \dfrac{a_na_{n-1}}{a_{n-1}}= \dfrac{n-1}{a_{n-1}}= a_{n-2}\dfrac{n-1}{a_{n-1}a_{n-2}}=a_{n-2}\dfrac{n-1}{n-2}= \dfrac{(n-3)(n-1)}{(n-2)a_{n-3}}= a_{n-4}\dfrac{(n-1)(n-3)}{(n-2)(n-4)}= ....=\dfrac{(n-1)(n-3)\cdots 3\cdot 1}{(n-2)(n-4)\cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Equation of the sphere that passes through 4 points Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing on four points. But I did not... | Using the equation for points on spheres:
$\qquad(x-a)^2+(y-b)^2+(z-c)^2=r^2$
Using coordinates of the four points provided, we have four simultaneous equations to solve for $a, b, c, d$.
\begin{cases}
(-5-a)^2+(4-b)^2+(1-c)^2=r^2 \\
(3-a)^2+(4-b)^2+(-5-c)^2=r^2 \\
(0-a)^2+(0-b)^2+(4-c)^2=r^2 \\
(0-a)^2+(0-b)^2+(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Third degree polynomial with zeros $6$ and $-5+2i$ that passes through $(2,-63)$?
Find a third degree polynomial with real coefficients which has $6$ and $-5+2i$ as zeros and $f(2)=-63$
I tried to substitute in the equation
$$ f(x)=ax^3 +bx^2 + cx +d = 0$$
by the two given zeros and use $f(2)= -63$. But I think we ne... | $$\begin{align}f(x)&=a(x-\alpha)(x-\beta)(x-\gamma)\end{align}$$
Using complex conjugate theorem
$$x=-5+2i\text{ is a root}\implies x=-5-2i\text{ is a root}\\(x+5)^2=-4\\x^2+10x+29=0$$
The other factor is $(x-6)$
So far, we have
$$\begin{align}a(x-6)(x^2+10x+29)&=0\end{align}$$
Use the fact that $f(2)=-63$
$$\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \fra... | Let $m-\frac{1}{m}=1$, $t-\frac{1}{t}=m$.
Thus, $x-\frac{1}{x}=t,$ which gives
$$m=\frac{1\pm\sqrt5}{2},$$
$$t^2-\frac{1\pm\sqrt5}{2}t-1=0$$ and
$$x-\frac{1}{x}=\frac{1\pm\sqrt5\pm\sqrt{22\pm2\sqrt5}}{4}.$$
From the last inequality we can get all $8$ distinct real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum ... | I did the following:
Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}$, $y=\sqrt{a}+\sqrt{c}-\sqrt{b}$ and $z=\sqrt{a}+\sqrt{b}-\sqrt{c}$. Then $\frac{x+y}{2}=\sqrt{c}, \frac{x+z}{2}=\sqrt{b}$ and $\frac{y+z}{2}=\sqrt{a}$. Rewriting the original expression, we get $$\frac{1}{2}\left(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| By parts: $u=2x+3$ and $\mathrm dv=\sqrt{3x+1}$, then
$$
\int(2x+3)\sqrt{3x+1}\;\mathrm dx=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{4}{9}\int(3x+1)^{3/2}\;\mathrm dx\\
=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{8}{135}(3x+1)^{5/2}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Lucas polynomials and primality testing Can you provide a proof or a counterexample for the claim given below ?
Inspired by Agrawal's conjecture in this paper I have formulated the following claim :
Let $n$ be a natural number greater than one . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \... | This is a partial answer.
This answer proves that if $n$ is a prime number, then $L_n(x) \equiv x^n \pmod {x^r-1,n}$.
Lemma : For every positive integer $n$, there are integers $a_0,a_1,\cdots, a_{n-1}$ such that
$$L_n(x)=x^n+\sum_{k=0}^{n-1}a_kx^k$$
Proof for lemma :
The claim is true for $n=1$ and $n=2$ :
$$L_1(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality with $xyz=1$ Given $x,y,z>0$ such that $xyz=1$
Prove that: $\frac{x^7}{x^8+1}+\frac{y^7}{y^8+1}+\frac{z^7}{z^8+1}\leq \frac{3}{2}$
P/s: I tried to solve it by AM-HM but I failed
| since
\begin{align*}
&\frac34-\frac34\cdot\frac1{x^{12}+x^6+1}-\frac{x^7}{x^8+1}\\
={}&\frac{x^6(x-1)^2(3x^{12}+2x^{11}+x^{10}-x^8-2x^7-2x^5-x^4+x^2+2x+3)}{4(x^{12}+x^6+1)(x^8+1)}
\end{align*}
and
\begin{align*}
2x^{11}+2x&\geqslant2x^7+2x^5,\\
x^{10}+x^2&\geqslant x^8+x^4,
\end{align*}
we get
$$\frac{x^7}{x^8+1}\leqsl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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IAS math problem 2017-paper 1 4a Reduce the following equation into standard form and hence determine the nature of coincoid $x^2+y^2+z^2-zx-xy-yz-3x-6y-9z+21=0$
I have reduced the equation into $(x-3/2)^2 +(y-3)^2 + (z-9/2)^2 -zx -xy-yz = 21/2$ from here on I can sense to reduce it to any of the standard 3D form that ... | just to find the nature of the surface does not require an orthonormal coordinate change. Take
$$ u = x+y+z, \; \; \; v = -x + y, \; \; \; w = -x - y + 2z \; \; . $$
$$ 6u = \frac{3}{4} (v-1)^2 + \frac{1}{4} (w-3)^2 + 18 $$
which is a paraboloid. The $6u$ stuff does not show this, but it is a paraboloid of revolution.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of c... | Let's try to bound the sequence here after we call it $a_n$. Note that the following inequalities hold:
$$a_n={{(\sqrt{n^2+n}-\sqrt{n^2+1})^n}\over{n^2}}x^n$$$$={{(n-1)^n}\over{n^2}(\sqrt{n^2+n}+\sqrt{n^2+1})^n}x^n$$$$\le {{(n-1)^n}\over{n^2}(2n)^n}x^n$$$$\le{{n^n}\over{n^2}(2n)^n}={{({x\over 2})^n}\over{n^2}}=b_n$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How prove this inequality $a_{n}>\frac{2n}{n+2}$ Let sequence $\{a_{n}\}$ such $a_{1}=1$,and such
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{(n+1)^2}$$\show that
$$a_{n}>\dfrac{2n}{n+2}\tag{1}$$
I try use induction,then It's a pity that failed, becase
$$a_{n+1}>\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}$$
it need to show
$$... | Note that for any $n \in \mathbb{N}_+$,$$
\frac{2n + 2}{n + 3} > \frac{2n}{n + 2} \Longleftrightarrow (n + 1)(n + 2) > n(n + 3),
$$
thus it suffices to prove that$$
a_n \geqslant \frac{2n + 2}{n + 3}. \quad \forall n \in \mathbb{N}_+ \tag{1}
$$
For $n = 1$, $\displaystyle a_1 = 1 = \frac{2 + 2}{1 + 3}$, so (1) holds. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up ... | The error is right in the end after you obtain the system of equations. You have
$$
x_1 + \frac{2}{7}x_3 = 0 \\ x_1 + \frac{2}{17}x_2 = 0
$$
You want to write $x_2$ and $x_3$ in terms of the common element $x_1$ so
$$
x_2 = -\frac{17}{2}x_1 \\ x_3 = -\frac{7}{2}x_1
$$
This means that the vector you are looking for is
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^1 \frac {1- \tan(x)}{\sqrt{x}+\tan(x)}dx$ I am having an extremely hard time finding the anti-derivative of this function:
$$f(x)=\dfrac{1-\tan(x)}{\sqrt{x}+\tan(x)}$$
while the lower bound is 0 and the upper bound is 1.
I tried to expand by $\cos(x)$ so it would look like
$$f(x)=\dfrac{\cos(x)-\sin(x)... | This is not an answer but it is too long for a comment.
If you had "extremely hard time finding the anti-derivative of this function", now we are two !
I do not think that we could find the antiderivative
$$I=\int\frac {1- \tan(x)}{\sqrt{x}+\tan(x)}\,dx$$ even using special functions.
What I first did is to let $x=y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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How to find $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$? The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?
| \begin{align}
\int \sqrt{x^8 + 2 + x^{-8}} dx &= \int \sqrt{\left(x^4+\frac1{x^4}\right)^2} dx \\
&= \int \left|\underbrace{x^4+\frac1{x^4}}_{\ge0}\right| dx \\
&= \int \left(x^4+\frac1{x^4}\right) dx \\
&= \frac{x^5}5-\frac1{3x^3}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Want to use Herbert Wilf's snake oil method to show $\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$ I was trying to use the snake oil method to show that $$\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$$
tl;dr I wasn't able to and desperately need help
My approach was to try to establish a more ... | Following request by OP for snake oil method we evaluate
$$\sum_{0\le 2k\le q+1} {q+1\choose 2k} {m+k\choose q}$$
and introduce
$$F(z, u) = \sum_{q\ge 0} \sum_{m\ge 0}
z^q u^m \sum_{0\le k, 2k\le q+1}
{q+1\choose 2k} {m+k\choose q}
\\ = \sum_{q\ge 0} \sum_{m\ge 0}
{m\choose q} z^q u^m
+ \sum_{q\ge 0} \sum_{m\ge 0}
z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very e... | $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\sqrt{\frac{a^3b^3}{abc^2}}+\sqrt{\frac{b^3c^3}{bca^2}}+\sqrt{\frac{c^3a^3}{cab^2}}=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{\frac{ab^2c}{ac}}+\sqrt{\frac{bc^2a}{ba}}+\sqrt{\frac{ca^2b}{cb}}=a+b+c$$
Above, we have twice used the known inequality $x^2+y^2+z^2\ge xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$
I'm studying the convergence of the series
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$$
*
*$\frac {\sqrt{n^2+1}-n}{\sqrt{n}}>0, \forall n \ge 1$
*$\lim_{n \to +\infty}\frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\lim_{n \to +\infty}\fr... | Note that
$$ \frac {\sqrt{n^2+1}-n}{\sqrt{n}} \cdot \frac { \sqrt{n^2+1}+n }{\sqrt{n^2+1}+n}= \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}\sim \frac{1}{2\sqrt{n^3}}$$
thus
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\sum_{n=1}^\infty \frac { 1 }{\sqrt{n^3+n}+n\sqrt n}$$
converges by comparison test with $$\frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I... | Well you know that $(\tan(x))'= \frac 1 {\cos^2(x)}$
$$(\tan(x))'= \frac {\sin^2(x)+\cos^2(x)} {\cos^2(x)}$$
$$(\tan(x))'= {1+\tan^2(x)}$$
Integrate
$$\tan(x)= \int {1+\tan^2(x)}dx$$
$$\tan(x)-x=\int {\tan^2(x)}dx$$
$$\tan(x)-2x= \int -1+ {\tan^2(x)}dx$$
$$2x-\tan(x)= \int 1- {\tan^2(x)}dx=\int 1- {\frac {\sin^2(x)}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | a nice problem! squaring the first equation we obtain
$$xy+xz+yz=0$$
raise to the power $3$ the last equation we have
$$x^2(y+z)+y^2(x+z)+z^2(x+y)+xyz=0$$
and this can be written as
$$x^2(1-x)+y^2(1-y)+z^2(1-z)+xyz=0$$
and this is
$$x^2+y^2+z^2-x^3-y^3-z^3+xyz=0$$ therefore $xyz=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 0
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What is the number of ways to put 40 identical balls in 6 boxes while no box contains more than 14 balls. I believe that I have to use the Inclusion–exclusion principle but I am not sure how. I tried to use this formula: $\binom{n+k-1}{n-1}$ for every step of the inclusion–exclusion solution but it didn't work out.
Th... | Jack D'Aurizio gives the clue, it can be solved by using the generating functions. In this case, generating function is
$$G(x) = (1+x+x^2+\ldots+x^{14})^6$$
and we are seeking the coefficient of $x^{40}$. Now, if we manipulate the expression by using:
$$1+x+x^2+\ldots+x^{14} = \frac{1-x^{15}}{1-x}\ and\ \frac{1}{1-x} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Calculating the integral $\int \sqrt{1+\sin x}\, dx$. I want to calculate the integral $\int \sqrt{1+\sin x}\, dx$.
I have done the following:
\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin ... | You have $$\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin x}}\, dx=\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx\end{equation*}$$ which is true up to your last equality where you forgot your $\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{... | $$
\sqrt{n-2\sqrt{n}}-\sqrt{n}=\sqrt{n\left(1-\frac{2}{\sqrt{n}}\right)}-\sqrt{n}=\sqrt{n}\left(\sqrt{1-\frac{2}{\sqrt{n}}}-1\right)
$$
and
$$
\sqrt{1-\frac{2}{\sqrt{n}}}\underset{(+\infty)}{=}1-\frac{1}{\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)
$$
Finally
$$
\sqrt{n-2\sqrt{n}}-\sqrt{n}\underset{(+\infty)}{=}1+o\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Proving that a limit does not exist with absolute values I'm to prove that the following limit does not exist
$\lim_{x\to -2} \frac{\vert 2x +4\vert -\vert x^3 +8 \vert}{x+2}$
From here, I have taken the method of finding $\lim_{x\to -2^+}$ and $\lim_{x\to -2^-}$ to show that they're not equal
However my problem is tha... | Your simplification is only correct for one side. If you include more steps, we can pinpoint what went wrong
For $x > -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{(2x+4)-(x^3+8)}{x+2} = \frac{(x+2)(2 - (x^2+2x+4))}{x+2} \\ = -(x^2+2x+2) $$
For $x < -2$
$$ \frac{|2x+4|-|x^3+8|}{x+2} = \frac{-(2x+4)+(x^3+8)}{x+2} = \frac{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A and B are two matrices such that $(A+B)^3=A^3+3A^2B+3AB^2+B^3$ then $ AB=BA$ Let $A$ and $B$ be two invertible matrices in $M_2(\mathbb{R})$such that $(A+B)^3=A^3+3A^2B+3AB^2+B^3$ then prove or disprove that $ AB=BA$
My working:
$$(A+B)^3=A^3+3A^2B+3AB^2+B^3$$
$$\implies BA^2+B^2A+ABA+BAB =2A^2B+2AB^2$$
Now what sho... | Counter-example :
$$
A=
\left(\begin{array}{cc}
2 & 0 \\
& \\
0 & 3 \\
\end{array}\right),
B=
\left(\begin{array}{cc}
1 & 0 \\
& \\
1 & -10 \\
\end{array}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)... | Yes it is correct, indeed
$$(x + b/2)^2 - (b/2)^2=x^2+bx+\frac{b^2}4-\frac{b^2}4-=x^2+bx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?
How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$
I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
|
Since $$\ln(x+1)= x-\frac{x^2}{2}+O(x^3)~~~and ~~~e^x= 1+x+\frac{x^2}{2}+O(x^3)$$
we get
$$(1+x)^{\frac{1}{x}}= \exp\left(\frac{1}{x}\ln(1+x)\right) = \exp\left(\frac{1}{x}(x-\frac{x^2}{2} +O(x^3))\right) \\=\exp\left(1-\frac{x}{2} +O(x^2)\right) =e\exp\left(-\frac{x}{2}+O(x^2)\right) = e(1- \frac{x}{2}+O(x^2)) $$
He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
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Find $f(x)$ if $f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$ $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function satisfying $$f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$$
if $f'(0)=2$, find the function
My Try:
we have $$f\left(\frac{x+y}{3}\right)-\frac{f(y)}{3}=\frac{2+f(x)}{3}$$ $\implies$
$$\f... | Using $x=y=0$ in the functional equation we can see that $f(0)=2$. Further note that $$f(x+h)=\frac{2+f(3x)+f(3h)}{3}, f(x) =\frac{2+f(3x)+f(0)}{3}$$ so that $$\frac{f(x+h) - f(x)} {h} =\frac{f(3h)-f(0)}{3h}$$ Taking limits as $h\to 0$ we get $f'(x) =f'(0)=2$ and thus $f(x) =2x+c$ and from $f(0)=2$ we get $c=2$ so that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence i... | Your proof doesn't use any properties of the sequence: you saying that the sequence is Cauchy is no proof. Also, your first sentence is wrong, as the sequence is increasing.
What you want to do is
$$
\frac {3n+5}{2n+6}-\frac {3m+5}{2m+6}=\frac{8 (n-m)}{(2m+6)(2n+6)}
\leq \frac{8 (n-m)}{4mn}=\frac2m-\frac2n.
$$
Now you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2... | You could try the generating form
$$F(x)=\frac{x^4}{2.3.4}+\frac{x^6}{4.5.6}+...\\
\frac{d^3F}{dx^3}=x+x^3+x^5+...=\frac x{1-x^2}$$
Try to integrate the last expression three times, then take the limit as $x\to1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$
My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive nu... | Alternatively, denote: $y=ax,z=abx$, then:
$$\begin{cases}x+ax+abx=0 \\ \frac{1}{x}+\frac{1}{ax}+\frac{1}{abx}=0\end{cases} \Rightarrow \begin{cases}1+a+ab=0 \\ 1+\frac{1}{a}+\frac{1}{ab}=0\end{cases} \Rightarrow \begin{cases}a=-\frac{1}{b+1} \\ b^2+b+1=0\end{cases} \Rightarrow \emptyset.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 8
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What is the number of integral values of $k$ for which the equation $|x^2 - 5|x| + 6|=k$ has four solutions is? I have done the sum by first plotting the graph of the function in the Left Hand Side of the equation and then plotted the line $y=k$. For the equation to have $4$ solutions, both these two curves must inters... | Let $f(x)=|x^2-5|x|+6|$. Note that $f(x)=f(-x)$.
$f(0)=6$. So, when $k=6$, the number of solutions of $f(x)=k$ is odd.
When $k\ne6$, the number of solution is double the number of solution of $f(x)=k$ for $x>0$. It suffices to find $k$ such that $f(x)=k$ has two solutions in $(0,\infty)$.
For $x>0$, $f(x)=|x^2-5x+6|$.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the... | Equation of normal is
$$y-y_0=-\frac{1}{f'(x_0)}(x-x_0)$$
In your case as you said $x_0=\sqrt c$ and $y_0=2c$.
$$f'(x_0)=2x_0=2\sqrt c$$
When we put all this back to the equation of normal we get:
$$y-2c=-\frac{1}{2\sqrt c}(x-\sqrt c)$$
$$y=-\frac{1}{2\sqrt c}(x-\sqrt c)+2c$$
$$2y\sqrt c=-x+\sqrt c+4c\sqrt c$$
$$2y\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac... | Well, we can use that:
$$\cos\left(\arctan\left(x\right)\right)=\frac{1}{\sqrt{1+x^2}}\tag1$$
And:
$$\sin\left(\text{arccot}\left(x\right)\right)=\frac{1}{x\cdot\sqrt{1+\frac{1}{x^2}}}\tag2$$
So in your case, we need to solve:
$$\frac{1}{\sqrt{1+x^2}}=\frac{1}{\frac{3}{4}\cdot\sqrt{1+\frac{1}{\left(\frac{3}{4}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Dividing one polynomial by another How is this done? For example, how would one simplify the following?
$$\frac{x^3-12x^2+0x-42}{x^2-2x+1}$$
I can do it with long division, but it never makes intuitive sense to me. Either an explanation of the long division algorithm or a new way of solving this would be much appreci... | The long division algorithm is a way of taking the numerator, and split it into multiples of the denominator, separate those out into their own fractions, and simplify away the denominator. For instance, we have $x(x^2 - 2x+1) = x^3-2x^2+x$, so
$$
\frac{x^3 - 12x^2 + 0x-42}{x^2-2x+1} = \frac{(x^3-2x+x) - 10x^2 - 1x-42}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Alternate proof for Viète's infinite product of nested radicals I am looking for alternate proof for Viete's infinite product of nested radicals. (Reference - Wikipedia)
Basically we need to find $\lim_{n\to \infty}\prod_{k=1}^{n} T_k$ where $$T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}$$ and $T_1 = \sqrt{\frac{1}{... |
Answer: we have $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\sqrt{2}}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$
*
*First check that for all $k$ w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does the constant 1 drop out in this very simple algebraic simplification I was integrating $\int \frac{x^3}{x^2+1}\, dx$, and my answer was
\begin{align}
&\frac{x^2+1}{2}- \frac{\ln(x^2+1)}{2} + C\\&
\text{ Rewrite/Simplify:}\\&=\frac{x^2-\ln(x^2+1)}{2} + C
\end{align}
How exactly does the top simplify into th... | Let $D= C+\frac12$ which is a constant.
\begin{align}
\frac{x^2+1}{2}- \frac{\ln(x^2+1)}{2} + C=\frac{x^2-\ln(x^2+1)}{2} + D
\end{align}
Here, they just reuse the notation and write $D$ as $C$. It is just an arbitrary constant.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding the maximum value without using derivatives
Find the maximum value of $$f(x)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$$ without using derivatives.
The domain of $f(x)$ is $x \in [1,\infty)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D(f)$ and the maximum value is $f(1)= 2 - \sq... | Note that
\begin{align}
f(x) & = (\sqrt{x}-\sqrt{x-1})-(\sqrt{x+1}-\sqrt{x})\\
& = \frac{1}{\sqrt{x}+\sqrt{x-1}} - \frac{1}{\sqrt{x+1}+\sqrt{x}}\\
& = \frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})} \\
& = \frac{2}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}+\sqrt{x-1})}
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by 11 How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by $11$
I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is $5! = ... | Let the base-$10$ number $abcde$ be divisible by $11,$ where $\{a,b,c,d,e\}=\{2,3,4,5,6\}.$ The base-$10$ number $abcde$ is divisible by $11$ iff $11$ divides $(a+c+e)-(b+d).$
We cannot have $a+c+e\geq 11 +b+d$ because $a+c+e\leq 4+5+6=15$ and $11+b+d\geq 11+2+3=16.$
We cannot have $b+d\geq 11+a+c+e$ because $b+d\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conic section(parabola and ellipse) An ellipse and a parabola have a common focus $S$ and intersect in two real points $P$ and $Q$, of which $P$ is the vertex of the parabola. If $e$ be the eccentricity of the ellipse and $x$, the angle which $SP$ makes with the major axis, prove that
$$\frac{SQ}{SP}=1+\frac{4e^2\sin^... | Let the parabola be $r=\dfrac{a}{1+\cos (\theta-x)}$, then
$$\theta=x \implies r=\frac{a}{2}=SP$$
Now the the ellipse is
$$r=\frac{a(1+e\cos x)}{2(1+e\cos \theta)}$$
Take $Q$ as $\theta=y$,
$$SQ=\frac{a(1+e\cos x)}{2(1+e\cos y)}=\frac{a}{1+\cos (y-x)}$$
Let $(X,Y)=(\cos x,\cos y)$, then
$$\frac{1+eX}{1+eY} = \frac{2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643505",
"timestamp": "2023-03-29T00:00:00",
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Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem,... | we have $$x=\frac{9}{6-y}$$ and $$z=\frac{9}{6-x}$$ putting things together we have
$$6-\frac{9}{y}=\frac{9}{6-\frac{9}{6-y}}$$
simplifying we get
$$9\,{\frac { \left( y-3 \right) ^{2}}{y \left( -9+2\,y \right) }}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$
| Ok, so building off of what lab bhattacharjee said:
$$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
$$=-\int\sqrt{1-(x-1)^2}dx-\int\dfrac{2-2x}{\sqrt{2x-x^2}}dx+2\int\dfrac1{\sqrt{1-(x-1)^2}}dx$$
Ok, so I use #8 on the 1st integral, u-substitution on the 2nd, and #1 on the 3rd.
$$=-(\frac{(x-1)\sqrt{1-(x-1)^2}}{2}+\frac{1}{2}\arcs... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Sequence that converges to a square root Let there be any two numbers $x$ and $y$, where $x,y \in \mathbb R$ and $y \neq -1$.
Consider a sequence where $$a_1 = \frac{y+x}{y+1}$$ and $$a_{n} = \frac{a_{n-1}+x}{a_{n-1} + 1}.$$
For example, when $x=5$ and $y=50$: $$a_1 = \frac{50+5}{50+1} = \frac{55}{51},$$$$a_{2} = \frac... | Write the recurrence relation as:
$$a_{n}=1+\frac{x-1}{1+a_{n-1}}=1+\cfrac{x-1}{2+\cfrac{x-1}{2+\cfrac{x-1}{1+a_{n-3}}}}$$
Writing the 'infinite' representation (formally so far) for the limit we have:
$$a_{ \infty} =1+\cfrac{x-1}{2+\cfrac{x-1}{2+\cfrac{x-1}{2+\dots}}}$$
For $x=2$ we have:
$$a_{ \infty} =1+\cfrac{1}{2+... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$ Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that:
$$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$
| By C-S and Schur we obtain:
$$\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}\frac{a^2}{ab^2+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2b+a^2c)}\geq\frac{3(a+b+c)}{\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{a+b+c}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{1}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}.$$
Now, let $a^2... | {
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What is the value of $\delta $ if $\epsilon=0.01$?
Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2}, & \text{if
$(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$
Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0)
\vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-
1.$\delta <0.01... | Note that we have
$$\frac12|2x+3y|\le |x|+\frac32|y|\le \frac52\sqrt{x^2+y^2}<0.01$$
whenever $\sqrt{x^2+y^2}<\delta=0.004$
| {
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maximum value of $a^2b$ condition is given the maximum value of $a^2b$ subjected to the condition $a+b+\sqrt{2a^2+2ab+3b^2}=10$ given $a,b\geq 0$
solution i try
$\displaystyle \frac{a}{2}+\frac{a}{2}+b\geq \left(\frac{a^2}{4}b\right)^{\frac{1}{3}}$
for $2a^2+2ab+3b^2=2(a+b)^2-ab+b^2$
$a+b+\sqrt{2a^2+2ab+3b^2}=(a+b)+... | The condition gives
$$2a^2+2ab+3b^2=(10-a-b)^2$$ or
$$a^2+2b^2+20(a+b)=100$$ or
$$(a+10)^2+2(b+5)^2=250$$ or
$$2\left(\frac{a}{2}+5\right)^2+(b+5)^2=125.$$
Now, by AM-GM and Holder we obtain:
$$125=2\left(\frac{a}{2}+5\right)^2+(b+5)^2\geq3\sqrt[3]{\left(\frac{a}{2}+5\right)^4(b+5)^2}=$$
$$=3\left(\sqrt[3]{\left(\frac{... | {
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Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.
Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.
Let the image be $(a,b,c)$.
Equation of the line joining $(1,3,4)$ and $(a,b,c)$ is $(a-1,b-3,c-4)$ and it will be parallel to the plane given.
Now we have $\dfrac{a-1}{2}=\dfrac{b-3}{-1... | Let $P=a(1,3,4)$ then
$$2a-3a+4a+3=0\implies a=-1$$
thus $$P=(-1,-3,-4)$$
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Double integral over region bounded by nonparallel lines Problem
I'm having trouble with the following integral:
$\int \int_D \frac{x}{y} dx dy $ where D is the area bounded by $1 \leq 2x+y \leq 5, \quad 4x \leq y \leq 8x$
My attempt at a solution
Substitution:
$ u = 2x + y, \quad v= y \ \implies x = \frac{1}{2}(u-v)... | Hint. Integrate w.r.t. $y$ immediately:
$$\int_{4x}^{8x} \frac{dy}{y}=\log 2$$
Now the only thing you have left is finding the limits for $x$ from the two inequalities and integrating:
$$\log 2\int_{x_1}^{x_2} xdx$$
Note that the first integration is equivalent to a change of variable:
$$y=x t$$
$$\int_{4x}^{8x} \frac... | {
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$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$ Find period and Fourier expansion Given the following function:
$$f(t)=\frac{\cos(\frac{3}{2}t)}{1+\cos^2(\frac{t}{4})}$$
Find period and Fourier expansion.
I think the period is $T=4 \pi$, observing the functions.
As for the Fourier expansion I have no clue on ... | I will still show the evaluation of the integrals, since it's not always possible to use tricks, sometimes we need the general methods as well. (For what it's worth, in this case the integrals are very simple).
Edited because the period of the function is $4 \pi$ the correct expression should be:
$$a_k=\frac{1}{\pi}\in... | {
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Algebra Smallest Possible Value What is the smallest possible value of
$x^2+y^2-x-y-xy$?
Is this even possible to solve? Please help.
| $$2(x^2+y^2-x-y-xy)=(x-y)^2+(x-1)^2+(y-1)^2-1-1\ge-2$$
Alternatively, let $$x^2+y^2-x-y-xy=k\iff x^2-x(1+y)+y^2-y-k=0$$
As $x$ is real, the discriminant must be $\ge0$
$$\implies(1+y)^2\ge4(y^2-y-k)$$
$$\iff4k\ge3y^2-6y-1=3(y-1)^2-3-1\ge-3-1$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Prove determinant is negative Prove that the determinant $\Delta$ is negative
$$
\Delta=\begin{vmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{vmatrix}<0
$$
where $a,b,c$ are positive and $a\neq b\neq c$.
My Attempt:
Applying Sarrus' rule,
$$
\begin{matrix}
a&b&c&a&b\\
b&c&a&b&c\\
c&a&b&c&a
\end{matrix}
$$
$$
\Delt... | You can factor $$a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca),$$ where the first term is certainly positive. For the second term, observe $$a^2 + b^2 + c^2 - ab-bc-ca = \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2),$$ which is also positive.
| {
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"question_score": "2",
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inverse of a $2\times2$ matrix, Gaussian elimination with unknown $x$ $C=AB$ I need to find A so $CB^{-1}=A$ working in modulo 10
and C=$ \begin{bmatrix}2 & 5 \\3 & 1 \end{bmatrix}$ mod 10
with B=$ \begin{bmatrix}3 & 4 \\2 & x \end{bmatrix}$ mod10
calculating $B^{-1}$
I wrote$ \begin{bmatrix}3 & 4 \\2 & x \en... | To find $B^{-1}$, you can start by analogy with inverses over $\mathbb{Q}$: $$B^{-1}=\frac{1}{3x-8} \begin{pmatrix} x&-4\\-2&3\end{pmatrix}$$
Note that if $x$ is even, then $det(B)$ is even, and therefore $B$ has no inverse modulo $10$. Further, if $x=1$, then $det(B)=5$, and therefore $B$ has no inverse modulo $10$. ... | {
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"source": "stackexchange",
"question_score": "1",
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$\det (A+B)=\det (A-B)$, prove that $B^{-1}$ exists iff $b_{11}\neq b_{21}$
Let $A=\begin{pmatrix}
0 & 1 & 2 \\
0 & 1 & 2 \\
0 & 2 & 3
\end{pmatrix}\in \mathbb{R}^{3\times 3}$ and let $B=\left [b_{ij}\right ]\in \mathbb{R}^{3\times 3}$ such that $\det \left (A+B\right )=\det \left (A-B\right )$. Prove that $B$ is inv... | $$det(A+B) = det(B) +
det
\begin{pmatrix}
b_{11} & 1 & b_{13}\\
b_{21} & 1 & b_{23}\\
b_{31} & 1 & b_{33}
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & b_{12} & 2\\
b_{21} & b_{22} & 2\\
b_{31} & b_{32} & 3
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} & 1 & 2\\
b_{31} & 2 & 3
\end{pmatrix}
$$
$$det(A-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_x^{px}\frac{\sin^2 t}{t^2}\,\mathrm dt\leqslant\fracπ2\left(1-\frac1p\right)$ for $x\geqslant0,\ p>1$
Given $x\geqslant 0$ and $p>1$, define $$I (p,x) = \int_x^{px}\frac{\sin^2 t}{t^2} \,\mathrm{d}t.$$ Prove that
$$ I (p,x) \leqslant \frac{\pi}2\left(1-\frac1p\right).$$
So far I only figured out that $... | $\def\d{\mathrm{d}}$It will be proved that$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) \tag{1}
$$
for any $0 < a < b$.
Case 1: $a \geqslant \dfrac{2}{π}$. Then$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^b \frac{1}{x^2} \,\d x = \frac{1}{a} - \frac{1}{b} = \fr... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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Third Moment of Geometric Series? $\sum_{x=1}^{\infty}x(1-p)^{x-1} = \frac{1}{p^2}$
$\sum_{x=1}^{\infty}x^2(1-p)^{x-1} = \frac{2-p}{p^3}$
$\sum_{x=1}^{\infty}x^3(1-p)^{x-1} =$ ... ?
The textbook gives the first 2 moments, but not the third one. I could not find the forumla on Google as well. I think it'll be useful in ... | First, note that
$$\sum_{k=0}^\infty k(1-p)^{k-1} = \sum_{k=1}^\infty k(1-p)^{k-1}$$
This allows us to perform a little trick to calculate the moments. For the first moment, we have:
\begin{align}
\sum_{k=1}^\infty k(1-p)^{k-1} &= \sum_{k=0}^\infty k(1-p)^{k-1}\\
&=\sum_{k=1}^\infty (k-1)(1-p)^{k-2} \\
&=\sum_{k=1}^\i... | {
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)... | Let
$$g(t)=\frac{t^2+4t+5}{2t^2+8t+8}=\frac{1}{2}+\frac{1}{2t^2+8t+8},$$
then
$$
g'(t)=-\frac{2(2t^2+8t+8)(4t+8)}{(2t^2+8t+8)^2}=-\frac{1}{(t+2)^3}.
$$
Thus $g$ strictly decreases in the interval $[-1,1]$, and $g(-1)$ and $g(1)$ are maximum and minimum respectively.
| {
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"source": "stackexchange",
"question_score": "3",
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Evaluate $ \int_{0}^{\pi/2} \frac{\sin(nx)}{\sin(x)}\,dx $ For every $odd$ $n \geq 1$ the answer should be $\pi/2$
For every $even$ $n \geq 2$ the possible answers are :
$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $
$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/... | first we note that $I_0=0$ and $I_1 = \frac{\pi}2$.
now, using the trig identity
$$
\sin A - \sin B = 2\sin \frac{A-B}2 \cos \frac{A+B}2
$$
we obtain
$$
I_{n+2} = I_n + \frac2{n+1} \sin \frac{(n+1)\pi}2
$$
if $N$ is odd, therefore, we have $I_N =I_{N-2}=\dots = I_1 = \frac{\pi}2$.
for even values of $N$, say $N=2M$, w... | {
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"source": "stackexchange",
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Infinite series involving binomial and geometric How to find the following sum:
$$\sum_{x=0}^{\infty}\left(-\frac{1}{16}\right)^x \binom{2x}{x}$$
| \begin{eqnarray*}
\binom{2n}{n} = \binom{-1/2}{n}(-4)^{n}.
\end{eqnarray*}
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \left( \frac{-1}{16} \right)^n = \sum_{n=0}^{\infty} \binom{-1/2}{n} \left( \frac{1}{4} \right)^n =\frac{1}{\sqrt{1+\frac{1}{4}}} =\color{red}{\frac{2}{\sqrt{5}}}.
\end{eqnarray*}
| {
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Why, my result has the opposite sign to the correct result? I found this exercise of factoring:
$$5a^2 + 3a - 2$$ is a trinomial of the form:
$$ax^2 + bx + c = \frac{(ax + p)(ax + q)}{a}$$
I made this:
$$3 = p + q$$
$$-10 = pq$$
then, I need to get the values of $p$ and $q$.
$$p = -\frac{10}q$$
replace:
$$3 = -\frac{1... |
the total factoring was: $\;(a + 1)(5a - 2)$
You could get that directly by finding the roots using the quadratic formula $\,\dfrac{-3 \pm \sqrt{49}}{10}=\begin{cases}-1 \\ 2/5\end{cases}\,$.
Then the quadratic factors as $5\big(a-(-1)\big)\left(a - \dfrac{2}{5}\right)=(a+1)(5a-2)\,$.
[ EDIT ] In the general case $\... | {
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Number of Non negative integral solutions of $x+2y+3z+4w=12$ Find Number of positive integral solutions of $x+2y+3z+4w=12$
I tried to split the problem in to cases :
we have $2z+2y+4w$ is always Even and $12$ is Even number.
Hence $x+z$ should be even number which implies
Case $1.$ $x$ is Even and $z$ is even
Letting ... | Minimum of $x+2y+3z=6\Rightarrow 4w=4$ since $8+6\gt12$ then $w=1$ necessarily.
Minimum of $x+2y+4w=7\Rightarrow 3z=3$ since $7+6\gt12$ then $z=1$ necessarily.
It follows $$x+2y=12-3-4=5$$ The only solutions of this equation are $(x,y)=(1,2),(3,1)$.
Thus there are only two solutions
$$(x,y,z,w)=(1,2,1,1),(3,1,1,1)$$
| {
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Limit of Recursive Sequence $y_{n+1}=y_n+\frac{1}{ny_n}$ with $y_1=1$ I came up with this sequence as I was playing around with another question on this site. And so I decided to ask the users if they can find a solution.
Let $$y_1=1$$
$$y_{n+1}=y_n+\frac{1}{ny_n}.$$
Evaluate $$\lim\limits_{n\to \infty} y_n \leq \infty... | The sequence diverges.
Suppose the sequence was bounded by some constant, i.e. $y_n < C$ for all $n$. Then in particular $\frac{1}{y_n} > \frac{1}{C}$ for all $n$ so:
$y_{2}>y_1 + \frac{1}{C}$
$y_{3} > y_2 + \frac{1}{2C} > y_1 + \frac{1}{C} + \frac{1}{2C}$
$y_{4} > y_3 + \frac{1}{3C} > y_1 + \frac{1}{C} + \frac{1}{2C... | {
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"source": "stackexchange",
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How i get the $x$ of the equation I have this:
$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}$ = 1
My main problem is that I want to multiply by something on both sides.
I will move $(5-\frac{5}{x})$ to side of 1, then:
$\frac{32}{20-16} = 1(5-\frac{5}{x}$)
$8 = 5 -\frac{5}{x}$ , now multiply by $x$
$8x = 5x - 5 $
$3x = 5$
... | This step is uncorrect
$$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=1\implies \color{red}{\frac{32}{20-16} = 1(5-\frac{5}{x})}$$
following this way it should be indeed (for $x\neq 1$)
$$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=1\implies \frac{1}{(5-\frac{5}{x})}\cdot\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=\frac{1}{(5-\f... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefin... | Hint...it would be a lot easier if you wrote $$(x,y)=(3\cos\theta,2\sin\theta)$$ then obtain an expression for the area of the rectangle in terms of $\theta$ and differentiate...
| {
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"question_score": "1",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdo... | To count the number of factors of $2079000 =2^3\cdot3^3\cdot5^3\cdot7\cdot11$, we would compute
$$
(3+1)(3+1)(3+1)(1+1)(1+1)=256
$$
However, we only want to count the number of factors divisible by $30=2\cdot3\cdot5$. Therefore, we want to compute the number of factors of $\frac{2079000}{30}=2^2\cdot3^2\cdot5^2\cdot7\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fraction and simplification
solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$
What are the possible answers ?
(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0
The answer from where i've referred this is (B), but when i simplify it I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1... | You can simply multiply by the LCD, $x(x-1)$ on both sides of the equation.
$$1+(x-1)=x$$
$$x=x$$
Therefore, there are infinitely many solutions for $x$, where $x\ne0,1$.
If $x=0, 1$, then the denominator(s) of the original expression would equal $0$, so the expression would be undefined.
In interval notation, the solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Probability of picking a certain fruit from a tree A tree has 20 fruits. 15 of which have no seeds and the rest do have seeds. A bird eats 5 of these fruits picked at random.
a) If i pick one fruit from whats left on the tree whats the probability it has seeds
b) Given that this one fruit i pick has seeds whats the pro... | Since there bird eats only five fruits, it's relatively easy to consider all possible cases. Let's write
*
*$P_0 = P(\text{bird eats zero fruits with seeds}) = \frac{15}{20} \frac{14}{19} \frac{13}{18} \frac{12}{17} \frac{11}{16} = \frac{1001}{15504}$
*$P_1 = P(\text{bird eats exactly one fruit with seeds}) =$ $5\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $B=(x,y)$ so triangle is equilateral Let $O=(0,0)$, $A=(3,4)$ and $B=(x,y)$ be three points in $xOy$. Find real numbers $x$ and $y$, so that $OAB$ is an equilateral triangle.
I'm really struggling with this one, can someone help?
| We need to have $|\overline{OB}|=|\overline{AB}|=|\overline{OA}|=5$. Thus from
$$ |\overline{OB}|^2=x^2+y^2=25$$
and $$|\overline{AB}|^2=(x-3)^2+(y-4)^2=25$$we reach
$$x^2-6x+9+y^2-12y+16=x^2+y^2 $$
or $$6x+12y=25\text{ or }x=2y-25/2.$$
Replace this value to $x^2+y^2=25$:
$$4y^2+50y+625/4+y^2=25$$
or $$y^2+10y-\frac{15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of rolling a 2 before a 6 and a 1 in that order
I am rolling a fair die. What is the probability that the first time I see a $2$ is before the first time I see a $6$ , and the first time I see a $6$ is before the first time I see a $1$ ?
So I thought about doing it with an infinite sum:
\begin{align}
\fr... | Your roll will look like this:
$\qquad (\text{1})$: $k$ rolls, each roll being one of $3,4,5$
$\qquad (\text{2})$: a $2$
$\qquad (\text{3})$: $m$ rolls, each roll being one of $2,3,4,5$
$\qquad (\text{4})$: a $6$
$\qquad (\text{5})$: $n$ rolls, each roll being one of $2,3,4,5,6$
$\qquad (\text{6})$: a $1$
For each $k$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big... | First of all $$\sin x\neq \sin (x/2)+\sin (x/2)$$
In fact $$\sin (a+b)=\sin a\cos b+\cos a\sin b$$
Substitute $a=b=x/2$ to get $$\sin x= 2\sin (x/2)\cos (x/2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Formula for the large derivative Is there any formula for the large number derivative?
I need to find $y^{(100)}$ at $x=0$, if $y=(x+1)2^{x+1}$
I tried to find a pattern, but 2nd and 3rd derivatives are already too hairy. I see no pattern, how it evolves.
1st derivative $2^{x+1}+\ln \left(2\right)\cdot \:2^{x+1}\left(x... | Let $u=x+1$ and $v=2^{x+1}$.
$u^{(1)}=1$, $u^{(k)}=0$ for $k\ge 2$.
$v^{(k)}=2^{x+1}(\ln2)^k$ for $k\ge 1$.
By the general Leibniz rule, (https://en.wikipedia.org/wiki/General_Leibniz_rule)
$$y^{(100)}=uv^{(100)}+\binom{100}{1}u^{(1)}v^{(99)}=(x+1)2^{x+1}(\ln 2)^{100}+100\cdot2^{x+1}(\ln 2)^{99}$$
$$y^{(100)}(0)=2(\ln ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $P_n$ is $n$th Legendre polynomial and $|x|\leqslant1$ then $\frac{1-x^2}{n^2}(P'_n)^2+P^{2}_{n} \le 1$ I am trying to prove theorem 17 (h) of chapter 7 from Kaplan's Advanced Calculus book. Being $P_{n}$ a legendre polynomial and $P^{\prime}_{n}$ its derivative, I need to prove:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}... | Given the identity
$$\begin{eqnarray*} \frac{1-x^2}{n^2}P_n'^2+P_n^2 &=& \frac{1-x^2}{n^2}P_{n-1}'^2+P_{n-1}^2\\&=&\left(1-\frac{1}{n}\right)^2\left(\frac{1-x^2}{(n-1)^2}P_{n-1}'^2+P_{n-1}^2\right)+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\\\\&\stackrel{\text{Ind.Hyp.}}{\leq}&\left(1-\frac{1}{n}\right)^2+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$ I'm stuck here:
Prove that $w=\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$
By using the definition of $z=\sin(w)$, I found that
$$\arcsin(z)=\frac{1}{i}\log(iz+\sqrt{1-z^2})$$
But where that $\frac{\pi}{2}$ comes from? How can I rearrange my expression... | In order to get to the final expression you need to use the log identity:
$$
\ln(a+b)=\ln(a)+\ln(1+\frac{b}{a})
$$
So in your case:
$$
\frac{1}{i}\left[\ln(iz+\sqrt{1-z^2})\right] = \frac{1}{i}\left[\ln(iz) + \ln\left(1+\frac{\sqrt{1-z^2}}{iz}\right)\right]
$$
We know that when $z>0$,
$$
\ln(iz)=\frac{i\pi}{2}+\ln(z)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I ... | $$b^2-4c<0\le(2x+b)^2$$
$$\implies0<(2x+b)^2-b^2+4c=4x^2+4bx+4c$$
$$\implies x^2+bx+c>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
Compute the intersection of perpendicular lines Lines Ax + By = C and Bx – Ay = 0 are perpendicular.
I am trying to understand how their intersection is computed below:
$$ x = \frac{AC}{A^2 + B^2}$$
$$ y= \frac{BC}{A^2 + B^2}$$
I tried solving for y first but I am not sure how to continue:
$$ Ax+By-C= Bx-Ay $$
$$By+Ay... | If $B \ne 0$, then
$Bx - Ay = 0 \iff x = \dfrac ABy$
Substituting into $Ax+By=C$, we get
\begin{align}
\dfrac{A^2}{B}y+By=C
&\implies A^2y + B^2y = BC \\
&\implies y = \dfrac{BC}{A^2+B^2} \\
&\implies x = \dfrac ABy = \dfrac{AC}{A^2+B^2}
\end{align}
If $B=0$ and $A \ne 0$, then $Ax+By=C \implies Ax=C \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$ $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$
I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2... | It's simpler if you consider the limit you get by (formally) substituting $n=1/x$:
$$
\lim_{x\to0^+}\frac{1}{x^{\alpha}}\frac{\sqrt[5]{1+x}-\sqrt[5]{1+2x+x^2}}{\sqrt[5]{x^2}}
$$
If this limit exists (finite or infinite), then it's the same as the limit of your sequence. Now use Taylor:
$$
\frac{1+\dfrac{1}{5}x-1-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\... | $$\lim_{x \rightarrow \infty}\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
Using the "divide top and bottom by the highest power" method, the expression simplifies:
Top first term:
$$x\sqrt{x}=x^{3/2};\frac{x^{3/2}}{x^{3/2}}=1$$
Top second term:
$$\sqrt[3]{x+1}=(x+1)^{1/3}$$
$$\frac{(x+1)^{1/3}}{x^{3/2}}=\frac{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Some details about 'Collatz Conjecture'? Yes, there is no one who doesn't know this problem.My question is only about curiosity.
$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
On this problem, I caught something like this.I'm sure, We all realized that... | If you search a single formula for any $k$, here it is:
$$n_k=\frac{2^{l_1+l_2+...+l_k}}{3^k}-\frac{2^{l_2+l_3+...+l_k}}{3^k}-\frac{2^{l_3+l_4+...+l_k}}{3^{k-1}}-\frac{2^{l_4+l_5+...+l_k}}{3^{k-2}}-...-\frac{2^{l_{k-1}+l_k}}{3^3}-\frac{2^{l_k}}{3^2}-\frac{2^0}{3^1}$$
e.g.
$$19=\frac{2^{4+3+2+1+3+1}}{3^6}-\frac{2^{3+2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How do we count solutions of $a^2 + b^2 + c^2\equiv n\pmod p$? Let's count number of solutions of $a^2 + b^2 + c^2 \equiv n\pmod p$ for each integer $n$ and each prime $p$. I'd start by just writing a function:
$$V_n(\mathbb{C}) = \left\{ (a,b,c) \in \mathbb{C}^3: a^2 + b^2 + c^2 = n \right\}$$
Maybe we could arrange... | Partial solution.
You can also rewrite it as: $$\sum_{u+v+w=n}\left(1+\left(\frac u p\right)\right)\left(1+\left(\frac v p\right)\right)\left(1+\left(\frac{w} p\right)\right)$$
This is because $1+\left(\frac mp\right)$ is the number of solutions to $x^2\equiv m\pmod{p}$.
This can then be rewritten as:
$$\sum_{u+v+w=n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? I'd like to find the four independent units in (the ring of integers of ) $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{R}$ We also have that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$, as a field extension.
I just want to find... | Just a check that Pell unit provided by @Rene Schipperus is indeed independent of the other Pell units. For assume that we have
$$(3+2\sqrt{2})^m(2+\sqrt{3})^n(5+2\sqrt{6})^p=1$$
Applying the Galois maps $\sqrt{2}\mapsto - \sqrt{2}$ ,$\sqrt{3}\mapsto - \sqrt{3}$ we get
$$(3-2\sqrt{2})^m(2-\sqrt{3})^n(5+2\sqrt{6})^p=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to interpret $\left(\begin{smallmatrix}p_n & p_{n-1}\\ q_n & q_{n-1}\end{smallmatrix}\right)$? We know that \begin{equation*}
a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots+\cfrac{1}{a_n}}}}}=[a_0,a_1, \cdots, a_n]
\end{equation*}
If $\frac{p_n}{q_n}=[a_0,a_1, \cdots, a_n]$.
How to prove that $$
... | Consider the canonical map from $GL_2(\mathbb{R})$, the group of invertible $2 \times 2$ matrices, to the group $PGL_1(\mathbb{R})$, the group of projective linear transformations $\mathbb{P}^1_{\mathbb{R}} \to \mathbb{P}^1_{\mathbb{R}}$:
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \left(x \mapsto \frac{ax+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is of the form $k^{4}$ with $k$ an integer. Find the number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is not of the form $k^{4}$ with $k$ an integer.
Notice that
$$ n^{4} - 20n^{2} + 100 = (n^{2} - 10)^{2} $$
We... | $$n^{4} - 20n^{2} + 100= (n^2-10)^2$$
Thus the condition reduces to finding all $n$ such that $|n^2-10|$ is of form $k^2$. $n=1,3$ satisfy this. Let $n>3$ so that $|n^2-10|=n^2-10$. Suppose there does exist a $k$ satisfying this, then
$$n^2-10=k^2$$
$$(n-k)(n+k)=10$$
Now the highest power of $2$ in $10$ is $2^1$. Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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$\int \frac {4}{7\sqrt {5x-6}}dx$ Why can't I use the substitution
$$u=\sqrt {5x-6}$$
$$u^2=5x-6$$
$$2udu=5dx$$
$$dx=\frac {2u}{5}du$$
$$\int\frac {4}{7u}\cdot \frac {2u}{5}du=\int \frac {8u}{35u}du=\int \frac {8}{35}du=\frac {8}{35}+C$$
The answer is obviously incorrect, but what have I done wrong?
| $$\int\frac {4}{7u}\cdot \frac {2u}{5}du = \int \frac{8}{35} du =\frac{8}{35} \color{red}{u} + C = \frac{8}{35} \sqrt{5x-6} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Explain how to solve this trigonometric limit without L'Hôpital's rule? In my previous class our professor let us the following limit:
$$
\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)}
$$
He solved it by applying L'Hôpital's rule as follow:
$
\lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^... | $$\dfrac{\tan x-\sin x}{x-\sin x}=\dfrac1{\cos x}\cdot\dfrac{\sin x}x\cdot\dfrac{1-\cos x}{x^2}\cdot\dfrac1{\dfrac{x-\sin x}{x^3}}$$
$\dfrac{1-\cos x}{x^2}=\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{1+\cos x}$
Finally for $$\dfrac{x-\sin x}{x^3}$$ use Are all limits solvable without L'Hôpital Rule or Series Expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Binomial sum involving power of $2$
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$
Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\bi... |
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}2^{k+2}\binom{n}{k}}
&=\frac{1}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{k+2}2^{k+2}\tag{1}\\
&=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}2^k\tag{2}\\
&=\frac{1}{(n+1)(n+2)}\left(3^{n+2}-1-2(n+2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Why do we have $ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $ Show that
$$ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $$
I know $ 1 - \sin^2x = (1 - \sin x) (1 + \sin x) $ but I'm not sure how to apply it.
Or maybe we need $ 2 \sin^2 (x) = 1 - \cos(2x)$?
| Remember that $$ \sin 3x = 3\sin x-4\sin^3 x \implies \frac{\sin 3x}{3\sin x} = 1- \frac 43 \sin^2 x$$
Now you can see the telescoping product.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a simi... | To elaborate on the other answer, which is correct, though presented with no concrete examples, note that your matrix $A$ is a sum $A = I + N$, where $N$ is the nilpotent matrix
$$N = \begin{bmatrix} 0 & 1 & & \\ & 0& 1 & \\ & & 0& 1 \\ & & &0 \end{bmatrix}$$
The matrix $N$ has powers of a simple form, for example
$$N^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Orthogonal diagonalize this matrix how would one orthogonally diagonalize the matrix $A$ = $\begin{pmatrix}a&0&b\\ 0&a&0\\ b&0&a\end{pmatrix}$
where $b≠0$?
| This is a classic problem. Read up on "Jacobi rotations"
Here is the short answer:
Focus on the matrix $ \left( \begin{array}{c c}
a & b\\
b & a
\end{array}
\right)
$.
A Jacobi rotation is given by
$$ \left( \begin{array}{c c}
c & s\\
-s & c
\end{array}
\right)
$$
where $ c = \cos( \theta ) $ and $ s = \sin( \theta ... | {
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"url": "https://math.stackexchange.com/questions/2721139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric... | Yes, it works.
$$\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix} \begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \end{bmatrix} $$
Thus you are correct in choosing the third eigenvector as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Finding complex roots of fourth degree polynomial $z^4 + 8z^3 + 16z^2 + 9$ I have the equation: $$z^4 + 8z^3 + 16z^2 + 9 = 0$$
I need to find all the complex solutions and I've got no clue how to approach it. I've tried factoring but nothing came out of it. I'm still very new to the world of complex numbers so I'll ap... | We can use factorization $x^2+y^2 = (x+iy)(x-iy)$:
\begin{align}
z^4+8z^3+16z^2+9 &= z^2(z^2+8z+16) + 9 \\
&= z^2(z+4)^2+9 \\
&= (z(z+4)+3i)(z(z+4)-3i) \\
&= ((z+2)^2 - (4-3i))((z+2)^2-(4+3i))\\
&= (z+2-z_0)(z+2+z_0)(z+2-\overline {z_0})(z+2+\overline {z_0})
\end{align}
where $z_0^2 = 4-3i$.
An explanation for the las... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Symmetric inequality with three variables including radicals Let $a, b, c$ be positive reals. Show that the following inequality holds:
$$\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{b^2 + c^2}} + \frac{c}{\sqrt{c^2 + a^2}} \le \frac{3}{\sqrt2}$$
I managed to do the following:
The ineq. is equivalent to this:
$$\sqrt{\... | By C-S
$$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+b^2}}\right)^2\leq\sum_{cyc}\frac{a^2}{(a^2+b^2)(a^2+c^2)}\sum_{cyc}(a^2+c^2)\leq\frac{9}{2},$$
where the last inequality it's
$$9\prod_{cyc}(a^2+b^2)\geq8\sum_{cyc}a^2\sum_{cyc}a^2b^2,$$ which is
$$\sum_{cyc}c^2(a^2-b^2)^2\geq0.$$
Also, we can use Jensen here:
$$\sum_{cyc}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the surface area of the unit sphere $x^2+y^2+z^2=1$ Compute the surface area of the unit sphere $x^2+y^2+z^2=1$
The following is a solution the book suggests:
The upper hemisphere is the graph of the function $\varphi(x,y)=\sqrt{1-x^2-y^2}.$
A little calculation yields $$\sqrt{1+(\partial_x\varphi)^2+(\par... | This is an alternative solution in case one's interested in:
By the spherical coordinates we get $x=\cos\theta\sin\varphi, \; \;y=\sin\theta\sin\varphi, \; \;z=\cos\varphi$
And, since $$d A= \sqrt{\left[ \frac {\partial y \partial z}{\partial \theta \partial \varphi}\right]^2 +\left[ \frac {\partial z \partial x}{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of $6$-digit numbers made up of the digits $1$, $2$, and $3$ with no digit occurring $3$ or more times consecutively? Find the number of 6-digit numbers made up of the digits $1$, $2$, and $3$ that have no digit occur three or more times consecutively. (For example, $123123$ would count, but $123111$ would not.)... | Discussion of what's missing in your answer
There are four places a block of three can start in the number.
$$\begin{matrix}
(1): && d & d & d & \cdot & \cdot & \cdot \\
(2): && \cdot & d & d & d & \cdot & \cdot \\
(3): && \cdot & \cdot & d & d & d & \cdot \\
(4): && \cdot & \cdot & \cdot & d & d & d \\
\end{matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the su... | Here is another proof.
Say $$a_n = \frac{p_n}{q_n}$$
then from the identity we have that
$$p_{n+1} = q_n, q_{n+1} = 2q_n + p_n = 2q_n + q_{n-1}$$
$q_n$ are of the form
$$ A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n $$
($\sqrt{2} + 1$ and $1 - \sqrt{2}$ are roots of $x^2 = 2x + 1$)
Thus $$a_n = \frac{q_{n-1}}{q_n} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\bin... | Let $E_1$ be the event that no red ball is chosen, $E_2$ the event that no green ball is chosen, and $E_3$ the event that no blue ball is chosen.
The probability that at least ball of each color is chosen is
$1 - P(E_1 \cup E_2 \cup E_3).$
By the inclusion-exclusion principle,
$$
P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far
| This can be not the only way to proceed, but it is convenient to avoid working with $e^{(i \pm 1) x}$ terms too much. Don't be afraid by the amount of lines, I simply did almost all the steps.
$$I = \int \sin x \sinh x \ \mathrm{d}x =\\
\int \frac{e^{ix} - e^{-ix}}{2i} \cdot \frac{e^{x} - e^{-x}}{2} \ \mathrm{d}x =\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
In regards to a game scenario I am designing In the game there is a 6 sided die which has 5 different faces (i.e. two of the faces are the same.) Considering that the die is fair and has 1/6 chance to roll any one side on each individual roll. What is the average number of rolls until I see the 5 different faces of the... | In general, if your probability of "success" for an event is $p$, then the expected number of trials until success is $1/p$. If there are $k$ faces of a die you haven't seen, then the probability of seeing a face you haven't seen yet is $k/6$ and you'll have to throw the dice an expected amount $6/k$ before you see one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find k with little oh notation? For which values of $k$ are the following true (as $x \to 0$)?
(a) $\sqrt{1+x^2} = 1 + o(x^k)$
(b) $\sqrt[3]{1+x^2} = 1 + o(x^k)$
(c) $1 - \cos(x^2) = o(x^k)$
(d) $1 - \cos^2 x = o(x^k)$
How would you find $k$ for problems like these?
| For (a), have a look at $\sqrt{1+x^2}-1$, which we can transform with the standard trick of multiplying and dividing by the conjugate:
$$ \sqrt{1+x^2}-1=\frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{\sqrt{1+x^2}+1}=\frac{x^2}{\sqrt{1+x^2}+1}\sim \frac{x^2}2$$
A similar trick helps with (b). For (c) and (d), have a look at th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.