Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $1 < \frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}} < 2$ for $a, x > 0$
Let $a>0$, show that for $x>0$, $1<f(x)<2$, where $$f(x)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}.$$
Source (added 20221017): It was the last question in math exam of Jiangxi Province, 200... | Here is my proof for $f(x) < 2$ without calculus:
It suffices to prove that, for all $a, b, c > 0$ with $abc = 8$,
$$\frac{1}{\sqrt{1 + a}} + \frac{1}{\sqrt{1 + b}} + \frac{1}{\sqrt{1 + c}} < 2.$$
WLOG, assume that $a \le b \le c$.
We split into two cases:
*
*$a + b \ge 6$:
We have $b \ge 3$. Thus,
$$\mathrm{LHS} < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my... | You should be trying to get from $\frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^{\bf{n+1}}]$ because that is the formula applied to $n+1$
$$\frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Simplifying Expression Factorial Expression I'm confused as how I'm meant to simplify this:$$\frac{(n-2)!}{(n-2-r)!}$$
I have other factorial questions where the variable isn't present in the top factorial like the question above and I'm trying to figure out how I simplify.
Thanks
| \begin{align*}
\frac{(n-2)!}{(n-2-r)!} &=\frac{(n-2)!}{(n-(2+r))!} \\
&= \frac{(n-2)(n-(2+1))\dotsm(n-(2+(r-1)))(n-(2+r))!}{(n-(2+r))!}\\
&=(n-2)(n-3)\dotsm(n-1-r)\\
&=\prod_{k = 0}^{r-1}(n-(2+k))\\
&=\prod_{k=0}^{r-1}(n-2-k)
\end{align*}
For example, let $r = 3$. Then
\begin{align*}
\frac{(n-2)!}{(n-2-3)!}&=\frac{(n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve without involving hyperbolic function. How to solve this integral without involving hyperbolic functions?
$$\int \frac{1}{4-5\sin^2 x}dx$$
The answer is $\frac{1}{4}(\ln (\sin x+2 \cos x)-\ln(2\cos x-\sin x))+c$
| \begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{dx}{4 \cos^2 x + 4 \sin^2 x - 5 \sin^2 x}\\
&= \int \frac{dx}{4 \cos^2 x - \sin^2 x}\\
&= \int \frac{\sec^2 x}{4 - \tan^2 x} \, dx
\end{align*}
Let $u = \tan x, du = \sec^2 x \, dx$. So
\begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{du}{4 - u^2}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to g... | Opposed to all of these answers, I will simply go ahead and solve the cubic:
$$0=x^3+x^2-x-2$$
Applying the cubic formula, I get:
$$x=\frac16\left(-2+\sqrt[3]{4}(\sqrt[3]{42+3\sqrt{177}})+\sqrt[3]{42-3\sqrt{177}})\right)$$
According to Wolfram|Alpha.
Then I guess you would either do the floor function after getting it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all values of $(1+i)^{1+i}$ $$z=(1+i)^{1+i}$$
I'm having trouble with this one. I got as far as the below, but then I got stuck. Could someone give me a hint??
$$\ln(z)=(1+i)\ln(1+i)$$
After reading the hints + suggested answer:
$$z=(1+i)^{1+i}$$
$$z=e^{(1+i)ln(1+i)}$$
$$z=e^{(1+i)ln|1+i|+(arg(1+i)+2\pi k)i}$$
$$... | $\forall n\in \mathbb{Z}$,
\begin{align*}
e^{2n\pi i} &= 1 \\
1+i &=
\exp \left[ \frac{\ln 2}{2}+i\pi \left( 2n+\frac{1}{4} \right) \right] \\
(1+i)^{1+i} &=
\exp \left[
\frac{1+i}{2} \ln 2+i(1+i) \left( 2n+\frac{1}{4} \right) \pi
\right] \\
&=
\exp \left[
\frac{\ln 2}{2}-\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1655886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generating function with G'(1) and G''(1) reducing to 0/0 = undefined My question is about an analysis of an algorithm in D. E. Knuth's book The Art of Computer Programming, Vol. 1. More specifically, it is about section 1.2.10, equations 20 to 22.
First we have a generating function
$$G(z)=\frac{1}{n}z+\frac{1}{n}z^2... | The essential tool is the binomial formula
$$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$
Expanding $(1+z)^{n+1}$ in the second last formula gives
$$(1+z)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} z^k = 1 + (n+1) z + \frac{(n+1) \cdot n}{2} z^2 + \frac{(n+1) \cdot n \cdot (n-1)}{6} z^3 + \ldots$$
Hence,
$$(1+z)^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Problem evaluating $ \int_{0}^{\pi/2}\frac{x}{\tan x}dx $ I am trying to evaluate $\int_0^{\pi/2}\frac{x}{\tan x} \, dx$. This is how I started:
$$ \int_0^{\pi/2} x\cot x \, dx$$
Integrating by parts we get:
$$x\ln\left|\sin x\right| - \int_0^{\pi/2}\ln\left|\sin x\right| \, dx $$
Since $x$ is between $0$ and $\pi/2$... | Let $\displaystyle{I = \int^{\frac{\pi}{2}}_0 x \cot x \, dx}$. Now
\begin{align*}I &= \int^{\frac{\pi}{4}}_0 x \cot x \, dx + \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} x \cot x \, dx = \int^{\frac{\pi}{4}}_0 x \cot x \, dx + I_1\end{align*}
For the integral $I_1$, set $x = \frac{\pi}{2} - u, dx = - du$ while for the limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find two square roots a 2 by 2 matrix filled with twos Given matrix $A = \begin{pmatrix} 2 && 2 \\ 2 && 2\end{pmatrix}$, I want to find two square roots of A.
I have to go about this with only very introductory-type tools, those covered in an introductory matrix operations chapter.
*
*My Approach
*
*Since I kno... | Note that line 1 and 2 give $a^2 = 2-bc = d^2$ and line 3 and 4 give $b = 2/(a+d) = c$. Now since $b(a+d) =2\neq0$ we can't have $a=-d$, thus $a=d$. Therefore our four equations simplify to
*
*$a^2+b^2=2$
*$2ab=2$
Rewriting the second we get $b = 1/a$ and substituting in the first gives $a^2 +a^{-2} = 2$ or $a^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | Let $a,b,c$ be roots of $$x^3+mx-n=0$$
From Vieta formula we get:
$$ 0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$
so $$\boxed{a^2+b^2+c^2 =-2m}$$
Further $$x^3 = n-mx\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n $$
so $$\boxed{a^3+b^3+c^3 =3n}$$
Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$
so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 6
} |
integration of $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}*\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{1+x}+2\sqrt{({1+x}... | Since $x= \sin t$, you have that $t=\arcsin x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 ... | There is no difference between $4 \ln(2+ \cos x) - 2\cos x + C$ and $4 \ln(2+ \cos x) - 2\cos x -4 + C$. Either way it means $4 \ln(2+ \cos x) - 2\cos x$ plus some quantity not depending on $x$, which could be any number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Relationships between points lines and planes Develop the Cartesian equation of a plane with $x$-intercept $a$, $y$-intercept $b$ and $z$-intercept $c$.
Show that the distance $d$ from the origin to this plane is given by $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
In the picture below I have included... | Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. The plane passes through the point $A$, so its equation is $(X-A)\cdot N =0$. Note that $A \cdot N = abc$, so this can also be written as:
$$
bcx + cay + abz = abc
$$
The line through the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| You can write
$$2^8+2^{11}+2^n=(2^4)^2+2.2^4.2^6+2^n$$
now, note that if $n=12$, follows that
$$(2^4)^2+2.2^4.2^6+2^{12}=(2^4)^2+2.2^4.2^6+(2^6)^2=(2^4+2^6)^2$$
Thus, $n=12$ to solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
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Proving that $1^n+2^n+3^n+4^n$ $(n\in \Bbb N)$ is divisible by 10 when $n$ is not divisible by 4 I was solving some math problems to prepare for math contests and came across this one:
Prove that $1^n+2^n+3^n+4^n$ $(n\in N)$ is divisible by 10 if and only if $n$ is not divisible by 4.
So, from what I understand, we hav... | $1^n+2^n+3^n+4^n$ is divisible by $10$ if and only if it's divisible by $2$ and $5$.
Clearly $1^n+2^n+3^n+4^n$ is even for all $n\in\mathbb Z^+$, so let's work with mod $5$.
If $n=4k$ for some $k\in\mathbb Z^+$, then:
$$1^k+16^k+81^k+256^k\equiv 1^k+1^k+1^k+1^k\equiv 4\not\equiv 0\pmod{5}$$
If $n$ is odd, then:
$$1^n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to prove the following binomial identity How to prove that
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
| Let be $$b_i=\frac{1}{2^{2i}}\binom{2i}{i}\quad \text{and}\quad a_i=\frac{i}{2^{2i-1}}\binom{2i}{i}.$$
Observing that $$a_i=2ib_i$$ and $$a_{i+1}=\frac{i+1}{2^{2i+1}}\binom{2i+2}{i+1}=\frac{i+1}{2}\frac{(2i+2)(2i+1)}{(i+1)(i+1)}\frac{1}{2^{2i}}\binom{2i}{i}=(2i+1)b_i$$
we have
$$
\begin{align*}
a_{i+1}-a_i&=(2i+1)b_i-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | Setting the AM $x+3=\dfrac1y$ to find
$$\lim_{y\to0}\dfrac{(1-5y^2+4y^4)^{1/5}-(1-3y)}y$$
$$=\lim_{y\to0}\dfrac{(1-5y^2+4y^4)-(1-3y)^5}y\cdot\dfrac1{\lim_{y\to0}\sum_{r=0}^4\{(1-5y^2+4y^4)^{1/5}\}^r(1-3y)^{4-r}}$$
Using Binomial expansion, this becomes
$$\lim_{y\to0}\dfrac{5\cdot3y+y^2\left(-5-\binom52\cdot3^2\right)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 4
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How to find a set of vectors spanning the solution space of $Ax=0$, where How to find a set of vectors spanning the solution space of $Ax=0$, where
Basically I have tried many times to solve it and my answer consistently comes in the following form:
$\pmatrix{1 \\ -1 \\-1\\0}$
While my book gives an answer of:
$\pmatr... | Reducing:
$$\begin{pmatrix}1&0&1&0\\1&2&3&1\\2&1&3&1\\1&1&2&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&2&2&1\\0&1&1&1\\0&1&1&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&1&1&1\\0&0&0&\!\!-1\\0&0&0&0\end{pmatrix}$$
The general solution to the homogeneous system $\;A\vec x=\vec0\;$ is
$$x_4=0\;,\;\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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integrate $\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$
$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$$
$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}\cdot\frac{\cos 2x-1}{\cos 2x-1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x... | Remember the bisection/duplication formulas and
note that $$\cos2x=2\cos^2x-1=1-2\sin^2x,$$ so your integral is
$$
\int_{0}^{\pi/4}-\frac{\sin^2x}{\cos^2x}\,dx
=
\int_{0}^{\pi/4}\frac{\cos^2x-1}{\cos^2x}\,dx
=
\int_{0}^{\pi/4}\left(1-\frac{1}{\cos^2x}\right)\,dx
$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Issue with the Euclidean Algorithm in $\mathbb{F}_2[x]$ So I have two polynomials in $\mathbb{F}_2[x]$ that I am trying to find the GCD of: $f(x)=x^6+x^5+x^4+x^3+x+1$ and $g(x)=x^5+x^3+x^2+x$.
So I start the algorithm:
$f(x)=g(x)(x+1)+x^3+1$
$g(x)=(x^3+1)(x^2+1)+x+1$
$x^3+1=(x+1)(?)+...?$
This is where I am having an i... | You want $(x+1)h(x)=x^3+1+r(x)$, where $\deg r(x)<1$. As you note, you see that $h(x)$ needs to be of the form $x^2+\dots$. But then you get an $x^2=1\cdot x^2$ term in $(x+1)h(x)=(x+1)(x^2+\dots)$ you need to cancel, so $h(x)$ needs to also have an $x$ term. You then have to cancel the $x=1\cdot x$ term in $(x+1)h(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Chain rule for differentiation I've been given this problem:
$y= \sqrt{7+6x^3}$
Using the chain rule am I right in suggesting that
$$u = 7+6x^3$$
$$y = \sqrt{u}$$
| Yes, there are two ways to see it: $$\frac{{\rm d}y}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}u} \frac{{\rm d}u}{{\rm d}x} = \frac{1}{2\sqrt{u}}18x^2 = \frac{9x^2}{\sqrt{7+6x^3}},$$or calling $f(x) = \sqrt{x}$ and $g(x) = 7+6x^3$, we have that $y(x) = f(g(x))$, so: $$y'(x) = f'(g(x))g'(x) = \frac{1}{2\sqrt{g(x)}}g'(x) = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-... | $$\text{z}=\sqrt{\sqrt{1-4i\sqrt{3}}}=\left(\sqrt{1-4i\sqrt{3}}\right)^{\frac{1}{2}}=\left(\left(1-4i\sqrt{3}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$$
$$\left(1-4i\sqrt{3}\right)^{\frac{1}{4}}=\left(7e^{-\arctan\left(4\sqrt{3}\right)i}\right)^{\frac{1}{4}}=\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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$1=\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$ $$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$
this look amazingly, square root of zero is $1$.
$$s=\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$$
$$s=\sqrt{n+s}$$
$$s=\frac{ 1\pm \sqrt{1+4n^2}}{2}$$
$$ \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\frac{1\pm \sqrt{1+4n^2}}{2... | First, you must prove if the limit exist.
If $s=1$, then $s=\sqrt{n+s}$ means $1=\sqrt{n+1}$
If $s=0$, then $s=\sqrt{n+s}$ means $0=\sqrt{n+0}$
When n=0, both are correct, but the limit is unique, so the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ I'm really at loss with this problem. I should prove that
$$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$
Only thing I managed to do by working the left side was
$$\sum_{n=1}^\infty \frac{1}{n^2} = 1+\sum_{... | So if I got it right it goes something like this
$$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$
$$=1+\sum_{n=1}^\infty (\frac{1}{n^2} - \frac{1}{n(n+1)})$$
$$=1+\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty (\frac{1}{n(n+1)})$$
$$=1+\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty (\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt... | The coefficient of $x^7$ will be ${n\choose 7}{2^{n-7}\over 3^7}$ and the coefficient of $x^8$ will be ${n\choose 8}{2^{n-8}\over 3^8}$ and not what you have got .you just missed out the factor of 3 that plays a role as a coefficient.So equating it you get
$${n\choose 7}{2^{n-7}\over 3^7}={n\choose 8}{2^{n-8}\over 3^8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
| This equation reduces to:
$$ \frac{(3^a)^4 + (3^a)^2 + 1}{(3^a)^2 + 3^a + 1} = \frac{7}{9} $$
Let $3^a = x$
$$ \frac{x^4 + x^2 + 1}{x^2 + x +1} = \frac{7}{9}$$
$$ 9x^4 + 2x^2 - 7x + 2 = 0$$
Can you factorise the above equation and then find the values of a by substituting back in to $3^a = x$? Looks like a tricky facto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A definite integral that surely needs contour integration: $\int_0^{\infty} \frac{1}{x^2 + a^2}\cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right)\, dx$ During my Master Thesis work I came up with an integral which I am going to consider as a hard challenge. I have been trying for days to crack it, but still nothing. The i... | Assume that all the parameters are real and nonnegative.
Then the equation $$\int_{0}^{\infty} \frac{1}{x^2 + a^2} \, \cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right) \, dx = \frac{\pi}{2a} \, \exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right)$$ holds iff $a >0$ and $b \ge c$.
To see why this is the case, consider the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Cauchy-Shwarz inequality in vector analysis Vectors $x$ and $y$ are related as follows $$\mathbf{x}+\mathbf{y(x \cdot y)}=\mathbf{a}.$$
Show $$\mathbf{(x \cdot y)}^2=\mathbf{\frac{|a|^2-|x|^2}{2+|y|^2}}$$
I think we need to proceed using Cauchy-Shwarz inequality.
$\mathbf{y(x \cdot y)}=\mathbf{a}-\mathbf{x}$
$\mathbf{y... | We have $a = x + \def\<#1>{\left<#1\right>}\<x,y>y$, hence
\begin{align*}
|a|^2 &= \<a,a>\\
&= \<x + {\<x,y>y, x+ \<x,y>y}>\\
&= \<x,x> + 2\<x,y>^2 + \<x,y>^2\<y,y>\\
&= |x|^2 + (2 + |y|^2)\<x,y>^2\\
\iff |a|^2 - |x|^2 &=(2 + |y|^2)\<x,y>^2\\
\iff \<x,y>^2 &= \frac{ |a|^2 - |x|^2 }{2 + |y|^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding vertical asymptotes of $\frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$ I'm trying to find the vertical asymptotes for $$f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$$
If I understand correctly, the vertical asymptote exists at $x=a$ when a value $a$ is found such that $f(a)$ increases to $∞$.
So, we mus... | Hint:
By factoring the numerator we have
\begin{align}
3x^4+3x^3-36x^2&=3x^2(x^2+x-12)\\
&=3x^2(x+4)(x-3)
\end{align}
Then
$$\frac{3x^4+3x^3-36x^2}{x^4 - 25x^2 + 144}=\frac{3x^2(x+4)(x-3)}{(x+3)(x-3)(x+4)(x-4)}=\frac{3x^2}{(x+3)(x-4)}\qquad x\neq -4, x\neq 3$$
So, the limit of the function when $x\to -4$ and $x\to 3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I simplify this rational expression? I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$
I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$
$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
then I am stuck.
| $$\begin{align}
& \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(5^{k+1} - 1) + 3(4 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(1\cdot5^{k+1} + 4 \;\cdot\;5^{k+1} - 1)}{4}\\
= & \frac{3(5\cdot5^{k+1} - 1)}{4}\\
= & \frac{3(5^{k+2} - 1)}{4}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
Question:
calculate:
$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$
Attempt at a solution:
This can be written as:
$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \fra... | $$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$
$$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$
$$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, other... | Let $u=\sqrt{2-x}$ then we simply want
$-2\int \sqrt{2-u^2}du$ which is simple after $u=\sqrt{2}\sin{v}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How can I find this sum? I'm doing some examples related to convolution (digital signal processing). I post my problem here because it is actually mathematics problem.
I have to calculate this sum:
$$\sum_{k\ = \ n-5}^{n+5} e^{-|k|}$$
Any suggestion?
|
We show the following is valid:
\begin{align*}
\sum_{k=n-5}^{n+5}e^{-|k|}=
\begin{cases}
{\displaystyle \frac{e^{-5}-e^{6}}{1-e}e^{-n}}&\qquad\qquad |n|>5\\
\\
{\displaystyle \frac{e^{-n-5}-e^{n-5}-1-e}{1-e}}&\qquad\qquad |n|\leq 5
\end{cases}
\end{align*}
Since
\begin{align*}
e^{-|k|}=
\begin{cases}
e^{-k}&\qquad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the asymptotes of the graph $y=\frac{x^2-5x-4}{x^2-5x+4}$? I just came across this question asking me to find the asymptotes of
$$y=\frac{x^2-5x-4}{x^2-5x+4}$$
I typed this into my graphics calculator but I've never seen this type of graph before so this is something new. Currently studying functions.
Wou... | Hint.
$$\begin{aligned}
f(x) &=\frac{x^2-5x-4}{x^2-5x+4}\\
&= \frac{x^2-5x+4 - 8}{x^2-5x+4}=1-\frac{8}{x^2-5x+4}\\
&=1-\frac{8}{(x-1)(x-4)}\\
&=1+\frac{8}{3} \left( \frac{1}{x-1} -\frac{1}{x-4} \right)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?
It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
| \begin{align}
\sin^3(x)+\cos^3(x) & = (\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
& \{\sin(x)+\cos(x) = \sqrt{2}\sin(x+\pi/4)\} \\
& = \sqrt{2}\sin(x+\pi/4)(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
& \{\sin(2x) = 2\sin(x)\cos(x)\} \\
& =\sqrt{2}\sin(x+\pi /4)(\sin^2(x) + \cos^2(x) -\sin(2x)/2)) \\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Demidovich problems 556 and 557 Exist two problems:
$$
№556: \lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}};
$$
$$
№557: \lim_{x\rightarrow0}(\frac{a^{x+1}+b^{x+1}+c^{x+1}}{a+b+c})^{\frac{1}{x}}
$$
And $(a>0,b>0,c>0)$. I sure that have one solution for both.
| $$\lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}}$$
On substituting $x=0$, the limit is of the form $1^{\infty}$.
Hence,
$$\lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^{\frac{1}{x}}=\lim_{x\rightarrow0}e^{\left(\frac{a^x+b^x+c^x}{3}-1\right)\frac1x}=\lim_{x\rightarrow0}e^{\left(\frac{a^x-1}{x}+\frac{b^x-1}{x}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cute Diophantine equation (simplify the expression) Find the largest integer $n$ less than $1000$ of the form
$n=(x+\sqrt{x^2-1})^{\frac{4}{3}}+(x+\sqrt{x^2-1})^{\frac{-4}{3}}$
for some positive integer $x$.
| Write $$u:=x+\sqrt{x^2-1}=e^{3t}\quad(t\geq0)\ .$$
Then
$$x={1\over2}\left(u+{1\over u}\right)=\cosh(3t)$$
and
$${n\over2}={1\over2}(u^{4/3}+u^{-4/3})=\cosh(4t)\ .$$
It follows that
$$8x^4-8x^2+1=\cosh(12t)=4\left({n\over2}\right)^3-3\left({n\over2}\right)\ $$
(see the comment by san below), so that $x$ and $n$ are r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is it true that for any $x>2$ and $k,j \in N$ with $k>j$, $p_k(x)$ is larger than $p_j(x)$? Let $p_k(x)=(x+\sqrt{x^2-4})^k-(x-\sqrt{x^2-4})^k$.
Is it true that for any $x>2$ and $k,j \in N$ $k>j$, $p_k(x)$ is larger than $p_j(x)$?
| Yes, it is true.
We can have
$$\begin{align}\frac{d}{dk}&p_k(x)\gt 0\\&\iff \left(x+\sqrt{x^2-4}\right)^k\ln \left(x+\sqrt{x^2-4}\right)-\left(x-\sqrt{x^2-4}\right)^k\ln\left(x-\sqrt{x^2-4}\right)\gt 0\\&\iff \left(\frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\right)^k\gt\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The case $x < - 3$ in the absolute value equation $|x + 3| + |x - 2| = 5$ In the absolute value equation $|x + 3| + |x - 2| = 5$, why do we replace $|x + 3|$ by $-x - 3$ rather than $3 - x$ when $-\infty < x < -3$?
$$|x+3|+|x-2|=5$$
What is the result set?
$$\begin{array}{c|c|c|c}
& \hphantom{xxx}-3 & \hphantom... | Reminders about absolute value:
$$|x|:=\begin{cases} x&\text{when}~x\geq 0\\ -x&\text{when}~x<0\end{cases}$$
For your specific question, finding the values of $x$ such that $|x+3|+|x-2|=5$, it behooves us to break into cases so that we may remove the absolute value signs.
$|x+3|+|x-2| = \begin{cases} x+3+x-2&\text{when... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
the system of diophantine equations: $x+y=a^3$; $xy=\dfrac{a^6-b^3}{3}$ has only trivial solutions. Without using Fermat's Last Theorem, how can one prove that the following system of diophantine equations has only trivial solutions:
$$x+y=a^3$$
$$xy=\dfrac{a^6-b^3}{3}$$
We suppose of course that $\gcd(x,y)=\gcd(a,x)=\... | If $a$, $b$, $x$ and $y$ are integers with $\gcd(x,y)=\gcd(a,b)=1$ such that
$$x+y=a^3\qquad\text{ and }\qquad xy=\frac{a^6-b^3}{3},$$
then $b^3=a^6-3xy$ and $a^6=(x+y)^2=x^2+xy+y^2$, so
$$b^3=(x^2+xy+y^2)-3xy=x^2-2xy+y^2=(x-y)^2.$$
It follows that for some integer $c$ we have $b=c^2$ and $x-y=c^3$. Then
$$x=\frac{a^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}... | We have $$\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}=\dfrac{\sin x \cos x+\sin x +\cos x +1}{\sin x \cos x}=\dfrac{1/2 \sin 2x+\sqrt{2}\sin(\frac{\pi}{4}+x)+1}{1/2\sin 2x}$$
$=1+\dfrac{2\sqrt{2}\sin(\frac{\pi}{4}+x)+2}{\sin 2x}$ is at its minimum when $x=\pi/4$ since we want the denominator closest to 1.
At... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Writing complex numbers in form $a+bi$ Can $\sqrt{i+\sqrt{2}}$ be expressed as $a+bi$ with $a,b \in \mathbb{R}$? In general, what kinds of expressions can be rewritten in that form?
| An efficient approach relies on use of Cartesian-to-polar coordinate transformation. To that end, let $z=x+iy=\sqrt{x^2+y^2} e^{i\arctan2(x,y)+i2\ell \pi}$. Then, the square root of $z$ is given by
$$\begin{align}\sqrt{z}&=\sqrt{x+iy}\\\\&=\sqrt{\sqrt{x^2+y^2}\,e^{i\arctan2(x,y)+i2\ell \pi}}\\\\
&=(-1)^{\ell}(x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | HINT :
Dividing the both sides by $\cos^2\theta\ (\not=0)$ gives
$$6\tan^2(x)+\tan(x)-1=\frac{5}{\cos^2(x)}=5(1+\tan^2(x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
Integrating $\sqrt{r^2 - x^2 - z^2}$ I need to integrate the following piece:
$\int \sqrt{R^2 - x^2 - z^2} \, dx$.
This is what I tried:
*
*$\int \sqrt{R^2 - x^2 - z^2} \, dx = \int (R^2 - x^2 - z^2)^\frac{1}{2} \, dx$
*Substitute $u = (R^2 - x^2 - z^2)^\frac{1}{2}, du = \frac{du}{dx}dx = \frac{1}{2}(R^2 - x^2 - z... | Hint:
Let $R^2-z^2=a^2$. Then, substitute $x=a\sin\theta$.
$$I=\int\sqrt{a^2-x^2}dx=\int a^2\cos^2\theta d\theta$$
Use $\cos^2\theta=\frac{1+\cos2\theta}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
=... | $$ \left(1-\frac{n}{n^2 - n + 1} \right)^n \leq \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) \leq \left(1-\frac{n}{n^2} \right)^n
$$
$$\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right)^n = \frac{1}{e}$$
$$\begin{align*} \lim_{n \to \infty} \left(1-\frac{n}{n^2 - n + 1} \right)^n
&= \lim_{n \to \infty} \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\s... | Multiply the LHS by product of the reciprocal of RHS, and the RHS.
$$\begin{array}{lll}
\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}&=&\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}\cdot\frac{\sec x + \csc x}{\tan x + \cot x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{\sec^2 x - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
How to simplify this surd: $\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$ $$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$
We have to find the value of $x$.
Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.
I think the answer l... | try converting expr inside surd into a square.
$$x = \sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$$
$$ = \sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}} $$
$$ = \sqrt{\frac{1+3+2\sqrt{3}}{2^2}}+\sqrt{\frac{1+3-2\sqrt{3}}{2^2 }} $$
$$ = \sqrt{\left(\frac{1+\sqrt{3}}{2}\right)^2}+\sqrt{\left(\frac{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the value of a trigonometric function given the value of $\tan(\alpha)\tan(\beta)$ Given:
$$\tan{\alpha} \tan{\beta} = -\frac{b^2}{a^2}$$
where $a$ and $b$ are constants, find:
$$\cos^2(\frac{\alpha - \beta}{2})$$
in terms of $a$ and $b$.
Here is my attempt:
$$\frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\be... |
actually this is related to ellipse, the question was "for the ellipse x^2/a^2+y^2/b^2, find the locus of the centroid of the triangle formed by the center and the points of intersection of chord of the ellipse which subtend right angle at the origin". while solving this problem i encountered this trigonometry problem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| $$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
$$\sin^6 x+\cos^6 x=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)=1-3\sin^2x\cos^2x$$
Thus your expression simplifies to
$$2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$$
Which is $0$, for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24... | A calculus approach. Let $$f(x) = \frac{x}{2}+\frac{1}{x}.$$ So $$f(x)-x = -\frac{x}{2}+\frac{1}{x}$$ is the sum of two decreasing functions on $]0,+\infty]$. Hence $f(p_{n})-p_n<0$, the sequence is decreasing.
$f(x)$ is strictly increasing on $]\sqrt{2},2]$ (easy with the derivative), thus if $x \in ]\sqrt{2},2] $, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Resolve $ \frac{120}{x+y} + \frac{60}{x-y} = 6;\,\frac{80}{x+y} + \frac{100}{x-y} = 7$ I want to resolve this system of equations:
$$\begin{cases} \frac{120}{x+y} + \frac{60}{x-y} = 6 \\\frac{80}{x+y} + \frac{100}{x-y} = 7\end{cases}$$
I came to equations like
$$x - \frac{10x}{x-y} + y - \frac{10y}{x-y} = 20$$
and
$$-... | You could also solve by eliminating one of the variables:
Multiply the first equation by 100 and the second by 60:
$$ \frac{12000}{x+y} + \frac{6000}{x-y}= 600$$
$$ \frac{4800}{x+y} + \frac{6000}{x-y}= 420$$
Subtract them from each other and get rid of $\frac{6000}{x-y}$:
$$ x+y = \frac{720}{180} = 40$$
Substitute $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\fra... | Solve to get
$$t = \frac{5x}{2y-2x} = \frac{3(x+y)}{y}$$
This gives the quadratic equation:
$$ 6x^2 - 5xy - 6y^2 = 0 $$
Factorize to get:
$$(3x + 2y) (2x - 3y) = 0$$
yielding
$$\frac{y}{x} = \frac{-3}{2} \,\,\,\text{or} \,\,\,\frac{2}{3}$$
Thus unless some sort of restriction is placed on $x$ or $y$
(for e.g both... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$? For a prime number $p \neq 2$, $3$, does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$?
A case where this is true as to suggest plausibility: $13 = 2^2 + 9 \times 1^2$.
| Finish:
For $p=12k+1$, we have $p=x^2+y^2$, and, if $gcd(xy, 3)=1$, then $p\equiv x^2+y^2\equiv 2 \pmod 3 $, a contradiction.
Conversely, $x^2+9y^2\equiv 0$ or $1 \pmod 3$, and $x^2+9y^2\equiv 0, 1, 2\pmod 4$. If this number is prime, then we have only one possibility:
$$x^2+9y^2\equiv 1 \pmod {12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Linear Algebra Identity Proof Prove that the following identity holds for all $x,y$ are in $\mathbb{R}^n$:
$x\cdot{}y = \frac{1}{4}(\left\|x+y\right\|^2 - \left\|x-y\right\|^2)$
| We have
$$
\begin{align}
\frac14 \left(\left\|x + y\right\| ^2 - \left\|x - y\right\|^2\right) &= \frac14 \left((x + y)\cdot(x+y) - (x - y)\cdot(x-y)\right)\\
& = \frac14(x \cdot x + 2x \cdot y + y \cdot y - (x \cdot x - 2 x \cdot y + y \cdot y))\\
&= \frac14(4x \cdot y)\\
&= x \cdot y
\end{align}$$
and so we are done.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse function of $x^2-4$ The function $h$ is defined by $$h(x)=x^2-4$$. for $$x\leq0$$
Find an expression for $h^{-1}(x)$
My attempt,
Let $h^{-1}(x)=a$
$x=h(a)$
$x=a^2-4$
$a=\sqrt{x+4}$
$h^{-1}(x)=\sqrt{x+4}$
Am I wrong? Is the answer $-\sqrt{x+4}$?
| There is a misstep in your procedure.
$y = x^2 - 4 \implies x^2 = y + 4\implies \sqrt{x^2} = \sqrt{y+4}$
Note that $\sqrt x^2 = x$ but $\sqrt{x^2} \neq x$ in general. Actually, $\sqrt{x^2} = |x|$, and combined with $|x| = -x$ for $x\leq 0$ we have
$-x = \sqrt{y+4}\implies x=-\sqrt{y+4}\implies h^{-1}(x) = -\sqrt{x+4}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.
| Have a look at the scene to come up with an idea:
(Large Version)
The green surfaces belong to the cone, the red and blue surfaces are spheres around $P$.
E.g you could try to find the intersection between cone and sphere and fiddle with the radius to shrink it to a point.
We can describe the intersection by the syste... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$? How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$ ? Explain your answer.
I started by factoring $2016$. I found the factors to be $36$, but I couldn't go further.
| We know that $2016 = 2^5\cdot 3^2\cdot 7^1$ and so $2016^{2016} = 2^{2016\cdot 5}\cdot 3^{2016\cdot 2}\cdot 7^{2016}$
We want $x^y$ to be equal to $2016^{2016}$. It is clear then by the prime factorization of $2016$ and the fundamental theorem of arithmetic that $x$ must be of them form $x=2^{a}\cdot 3^b\cdot 7^c$, im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Differential equation with nasty coefficients $ x^2(1-x)^2 y'' + (Ax + b)y = 0 $ I have encountered a differential equation on the form
$$ x^2(1-x)^2 y'' + (Ax + b)y = 0 $$
My math background is too limited to even know where to begin, so any help of solving the equation (if a solution exist?) would be greatly apprecia... | In fact it just belongs to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=262.
Let $y=x^p(1-x)^qu$ ,
Then $y'=x^p(1-x)^qu'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Roots of a quadratic equation If $ a $ , $b$ are the roots of $ x^2+x+1=0 $ , then the equation whose roots are $a^k $ and $b ^k$ where k is a positive integer not divisible by 3 is
$a)$ $x^2 - x + 1 = 0$
$b)$ $x^2 + x+1 = 0$
$c)$ $ x^2 -x -1 =0$
$d)$ None of the above
My attempt : I was able to solve the question , ... | Since $x^3-1=(x-1)(x^2+x+1)$, you have $a^3=1$ and $b^3=1$.
If $k=3h+1$, then $a^k=a(a^3)^h=a$ and $b^k=b(b^3)h=b$.
On the other hand, $a^2=b$ and $b^2=a$, so if $k=3h+2$, you have $a^k=a^2(a^3)^h=b$ and $b^k=b^2(b^3)^h=a$.
Why is $a^2=b$? It is clear that $(a^2)^3=(a^3)^2=1$, so $a^2$ is a root of $x^3-1$. However, $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
finding recursive formula and show it converges to a limit Suppose we are playing cards and we start with $1000$ dollars. Every hour we lose $\frac{1}{2}$ of our money and then we buy another $100$ dollars. I am trying to find $x_n$ for the amount of money the player has after $n$ hours.
I think we can just take $x_n =... | $x_n = \frac{x_{n-1}}{2} + 100 = (\frac{\frac{x_{n-2}}{2} + 100}{2}) + 100 =\frac{x_{n-2}}{4} + 100.(\frac{1}{2} + 1)= \frac{\frac{x_{n-3}}{2} + 100}{4} + 100.(\frac{1}{2} + 1)=...$
$\implies x_n = \frac{x_0}{2^n} + 100.(\frac{1}{2^{n-1}} + \frac{1}{2^{n-2}} + ... + \frac{1}{2} + 1 ) $
where $x_0 = 1000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | Using Henry W.'s answer $$A_n=\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = 2^n\frac{(n!)^2 }{(2n)!}$$ $$\log(A_n)= 2\log(n!) + n \log( 2) - \log \big((2n)!\big)$$ Now, using Stirling approximation
$$\log(p!)\approx p\log(p)-p+\frac 12\log(2\pi p)$$ $$\log(A_n)\approx \frac{1}{2} \log (\pi n)-n \log (2)$$ $n$ var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Finding area bounded by two graphs of functions I know that someone posted this same question before here, because i found it, but i couldn't find the answer about this example that i was looking for, so i will post everything i did so far:
I have to find area bounded by graphics of functions
$y_1=x\sqrt{4x-x^2}$
$y_2... |
$P=\int_0^1 \sqrt{4x-x^2}dx - \int_0^1 x\sqrt{4x-x^2}dx $
You should have
$$P=\int_{0}^{1}\sqrt{4x-x^2}\ dx-\int_{0}^{1}x\sqrt{4x-x^2}\ dx\color{red}{-\left(\int_{1}^{4}\sqrt{4x-x^2}\ dx-\int_{1}^{4}x\sqrt{4x-x^2}\ dx\right)}$$
Also, $\arcsin(-1/2)=-\pi/6$ because the range of usual principal value of arcsine is $[-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of the series $\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$ I need to test the convergence and find the sum of the following series:
$\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$
But i am not really sure what kind of series is this?
Since
$2\sqrt{n+1}>\sqrt{n}+\sqrt{n+2} \forall n \in \m... | Since
\begin{align*}
\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}&=\sqrt{n+2}-\sqrt{n+1}-(\sqrt{n+1}-\sqrt{n})\\
&=\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}
\end{align*}
Then, the sum is telescopic:
\begin{align*}
\sum_{n=1}^N(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})&=\frac{1}{\sqrt{N+2}+\sqrt{N+1}}-\frac{1}{\sqrt{2}+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $2^{15}-2^3$ divides $a^{15}-a^3$ for all $a$ Show that for all $a$, $2^{15}-2^3$ divides $a^{15}-a^3$.
I was able to prove that this is true for all $a$, such that $\gcd(a,2^{15}-2^3)=1$, by using Euler's theorem, where I concluded that $a^{12}\equiv1 \pmod{2^{15}-2^3}$, since $a^{12}\equiv1$, mod all of $2^... | Similar to Ian's answer san the brute force: since
$$2^{15}-2^3=2^3\times 3^2 \times 5 \times 7 \times 13$$
we only need to show $a^{15}-a^3=a^3(a^{12}-1)$ is divisible by each of the above factors.
By Fermat's theorem
*
*$a(a^{12}-1)$ is divisible by $13$.
*$a(a^6-1)$ is divisible by $7$.
*$a(a^4-1)$ is divisib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $ \int_{a}^{b} \frac{\mathrm{d}{x}}{x^{2}} $ using Riemann sums only. How does one evaluate the following Riemann integral by using Riemann sums only?
$$
\int_{a}^{b} \frac{\mathrm{d}{x}}{x^{2}}.
$$
| Compute the integral as the limit of a Riemann sum:
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $ x+iy = \sqrt{\frac{a+ib}{c+id} } ,$Show that$ (x^2+y^2)^2 = \frac {a^2+b^2}{c^2+d^2} $ $ x+iy = \sqrt{\frac{a+ib}{c+id} } , $Show that $ ({x^2+y^2})^2 = \frac {a^2+b^2}{c^2+d^2} $
How do i do this ?I tried squaring both sides but x+iy expansion becomes difficult when squaring the next time .I also tried conjugatin... | $$x^2+y^2=|x+iy|^2=\left|\sqrt{\frac{a+ib}{c+id}}\right|^2=\left|\frac{a+ib}{c+id}\right|$$
Therefore,
$$(x^2+y^2)^2=\frac{|a+ib|^2}{|c+id|^2}=\frac{a^2+b^2}{c^2+d^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the volume of ice cream cone using cylindrical/spherical coordinates I'm stuck on what the boundaries are for the volume bounded by the cone $z=-\sqrt{(x^2+y^2)}$ and the surface $z=-\sqrt{(9-x^2-y^2)}$ $\,\,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordina... | By cylindrical coordinates the set up of the integral should be
$$\int_0^{2\pi}\int_{-3}^{-\frac3{\sqrt2}}\int_{0}^{\sqrt{9-z^2}} r\,\, dzdrd\theta+\int_0^{2\pi}\int_{-\frac3{\sqrt2}}^0\int_{0}^{-z} r\,\, dzdrd\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the general solution to $\sin(4x)-\cos(x)=0$
Question: Find the general solution to $\sin(4x)-\cos(x)=0$
My attempt:
$$ \sin(4x)-\cos(x)=0$$
$$ \Leftrightarrow \sin(4x) = \cos(x) $$
$$ \Leftrightarrow \sin(4x) = \sin( \frac{\pi}{2} -x)$$
From another post I learnt that you can equate $\sin(x) = \sin(y)$ on 2 ... | $\sin(4x)−\cos(x)=0$
$2\sin(2x)\cos(2x)-\cos(x)=0$
$4\sin(x)\cos(x)(1-2\sin^2(x))-\cos(x)=0$
One possible solution is $\cos(x)=0$
$4\sin(x)(1-2\sin^2(x))=1$
$8\sin^3(x)-4\sin(x)+1=0$
Now, let $\sin(x)=m$ and solve the resulting cubic...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of subsets from an ordered set where adjacent elements may or may not be tied together Assume we have an ordered set $S$ with a finite number of elements $S=\{1,2,3,\ldots,N\}$. I need to know the number of subsets where adjacent elements from the original set may either be tied together as one "unit" shown with... | Let $a_n$ be the count of these listings for given $n$.
Then $a_n$ is obtained as sum of those listings not ending in $n$ (there are $a_{n-1}$ of these), those ending in "$,n$" (there are $a_{n-1}$ of these), and those anding in "${-}n$" (there are $a_{n-1}-a_{n-2}$ of these).
Hence we have the recursion
$$a_n=3a_{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{... | $\sin^2A+\sin^2B+\sin^2C =1-\cos^2A+ \dfrac{1-\cos(2B)}{2} + \dfrac{1-\cos(2C)}{2}=1-\cos^2A+ 1 - \cos(B+C)\cos(B-C)=2-\cos^2A+\cos A\cos(B-C)=2$. From this you can see $\cos A = 0$,or $A = B-C \Rightarrow B = \dfrac{\pi}{2}$. Thus $B)$ is the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is In a triangle $ABC,$if $\angle A=30^\circ,b=10$ and $a=x$,then the values of $x$ for which there are $2$ possible triangles is given by
$(A)5<x<10(B)x<\frac{5}{2}(C)\frac{5}{3}<x<10(D)\frac{5}{... |
$$x^2=c^2+100-20c\frac {\sqrt 3}2$$
$$c^2-10\sqrt3 c+100-x^2=0$$
Two triangles exist -- equation has two roots:
$$\frac D4=75-(100-x^2)>0$$
$$x^2> 25$$
$$x > 5$$
$$x<b+c=10+c_1 \Rightarrow x<10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$... | Suppose one of the variables is zero. Then you have shown the maximum is $\frac12$.
So let us consider the case when none of the variables are zero. Then as both the objective and constraint are symmetric & homogeneous, we may set WLOG $z=1$ and $1\ge x\ge y>0$. Thus in this case we have the constraint $x^2+y^2+1 = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A conditional inequality which itself implies a sharper version of it Problem: Given that $m, n$ are positive integers such that $\sqrt{7} -\frac{m}{n} > 0$. Then show that $\sqrt{7}-\frac{m}{n} > \frac{1}{mn}$.
I have failed to do this fascinating problem.
My efforts: I tried to approach by contradiction.
Assume that ... | A more direct approach.
We write:
$$1\leq 7n^2-m^2 = (n\sqrt{7}+m)n \left(\sqrt{7}-\frac{m}{n}\right)$$
If $\sqrt{7}-\frac{m}{n}\lt \frac{1}{mn}$ (equality is not possible) then $n\sqrt{7}< m+\frac{1}{m}$ so $m+n\sqrt{7}\lt 2m+\frac{1}{m}$ and $7n^2-m^2\lt (2m+\frac{1}{m})n\frac{1}{mn}=2+\frac{1}{m^2}$.
So $7n^2-m^2=1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | Hint: except $-1^2$ make pairs of other terms: $$-1^2+(-2^2+3^2)+(4^2-5^2)+(-6^2+7^2)+....$$ and then proceed further to get $-1^2+2+3+4+5+6+...$ and than solve it to get the result
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Numbers expressible as sum of 2 squares in 2 distinct ways I was trying this question here which goes like:
Find numbers which are squares and can be expressed as $x^2y^2-x^2-y^2+2$ for non-consecutive positive integers only.
Let the number be $a$
\begin{align*}
a^2 &= x^2y^2-x^2-y^2+2 \\
a^2-1 &= (x^2-1)(y^2-1) \\
&... | The condition $(x^2-1)(y^2-1)+1=a^2$ implies that the set $\{1,x^2-1,y^2-1\}$ is a Diophantine triple see (e.g. here), i.e. the product of any two of its distinct elements increased by $1$ is a perfect square. For fixed $x$, finding $y$ leads to Pellian equation $a^2 - (x^2-1)y^2 =-x^2$. It has infinitely many solution... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\sqrt 7 \pmod {2579}$ Find $\sqrt 7 \pmod {2579}$.
I think I understand how I would solve a very basic equation like this:
$x^2 = 1 \pmod 5$
make a table of all the possible solutions like this
$x=0 \implies x^2=0 \\
x=1 \implies x^2=1 \\
x=2 \implies x^2=4 \\
x=3 \implies x^2=4 \\
x=4 \implies x^2=1 \\
x=5 \impl... | First, check that there are solutions for $x^2 \equiv 7 \bmod 2579 $ using Euler's criterion:
If $p$ is a prime that does not divide $a$,
then
$\qquad x^2 \equiv a \bmod p$ has a solution
$\iff a^{\tfrac{p-1}{2}} \equiv 1 \bmod p$.
We compute and see that $7^{\frac{2579-1}{2}} \equiv 1 \bmod 2579$.
Next, we fin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Volume of the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0$ I want to find volume the region in $\mathbb{R}^3$ defined by $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0.$
I set it up as $\displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{-\sqrt{\... | $x^2+y^2+z^2 \leq 1$ is clearly a ball. And for $z^2 \geq x^2 + y^2$, you can think in this way, let $r^2 = x^2+y^2$, then $x^2+y^2 = r^2 \leq z^2$, which is a cone. The overlap of a cone and a ball is still a cone. So try to use spherical coordinates.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The possible values of $\gcd(a^2,b^2)+\gcd(a,bc)+\gcd(b,ca)+\gcd(c,ab)$
Let $A=\gcd(a^2,b^2)+\gcd(a,bc)+\gcd(b,ca)+\gcd(c,ab)$, where $a$, $b$ and $c$ are positive integers. What are the values that $A$ can take, when $a$, $b$ and $c$ range over all positive integers?
I have no clue how to start. Any kind of help wil... | The values that $ A $ can take are exactly the composite positive integers. I first noticed this by writing a computer program checking the value of $ A $ when $ a $, $ b $ and $ c $ are less than $ 30 $. The result looked amazing to me, as the defining expression of $ A $ didn't indicate it in a clear way. So, before ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all the parameters and such that the line $y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles in at least one point . Problem:
Find all the parameters and such that the line
$y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles
in at least one point.
My work:
We tr... | Let $P=(x_P,y_P)$ the intersection point between the line and the hyperbola.
The hyperbola has equation $y=\frac{1}{x}$ so the derivative (without implicit differentiation) is $y'=\frac{-1}{x^2}$ and, at $P$ it is $y'(x_P)=\frac{-1}{x_P^2}$.
This is the slope of the tangent to the hyperbola at $P$, and if we want that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question... | Need to show that
$$x^3+3x^2+5x+3 =0 \mod 3$$
That is
$$x(x^2+5)=0 \mod 3$$
It's true when $x=0 \mod 3$.
For $$x=\pm 1 \mod 3$$,
$$(3k + 1)((3k + 1)^2+5)=(3k + 1)(9k^2+6k+6)=0 \mod 3$$,
$$(3k - 1)((3k - 1)^2+5)=(3k - 1)(9k^2-6k+6)=0 \mod 3$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Eigenvalue of 3x3 matrix with real unknown constants how do i find the eigenvectors of this matrix which has unknown real constants a and b ?
$$
\begin{bmatrix}
a & -\sqrt{2}b & 0 \\
-\sqrt{2}b & 0 & -\sqrt{2}b \\
0 & -\sqrt{2}b & a \\
\end{bmatrix}
$$
Eigenvalues of the matrix are $\lambda_{1}=a, \lambda_{2}=\frac{a-\... | You have that $\frac{4\sqrt2 b}{-a+\sqrt{a^2+16b^2}}=\frac{a+\sqrt{a^2+16b^2}}{2\sqrt2 b}$ and $\frac{4\sqrt2 b}{-a-\sqrt{a^2+16b^2}}=\frac{a-\sqrt{a^2+16b^2}}{2\sqrt2 b}$. So your answer and the output from Mathematica are the same.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expand the function $f(z)=\frac{1}{(z-a)(z-b)}$ where $0 < |a| < |b|$ in a Laurent series in different annuli I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)... | Hint:
For $\left|z\right|\lt\left|c\right|$,
$$
\frac1{z-c}=-\frac1c\left(1+\frac zc+\frac{z^2}{c^2}+\frac{z^3}{c^3}+\dots\right)
$$
For $\left|z\right|\gt\left|c\right|$,
$$
\frac1{z-c}=\frac1z\left(1+\frac cz+\frac{c^2}{z^2}+\frac{c^3}{z^3}+\dots\right)
$$
Apply these to
$$
\begin{align}
\frac1{(z-a)(z-b)}
&=\frac1{b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $x^3-6x-6=0$
Evaluate the roots of
$$x^3-6x-6=0$$
I solved it using Cardano's method, but I'm looking for other elementary approaches through substitutions and properties of polynomials.
Thanks.
| Hint:
$x^3-6x-6=x^3-3bcx+b^3+c^3$
$$bc=-2, b^3+c^3=-6$$
$$b=-\sqrt[3]{2}, c=-\sqrt[3]{4}$$
So $(b+c)^3=b^3+c^3+3bc(b+c)$, then number $b+c$ - solution of $x^3-3bcx+b^3+c^3=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ I am being asked to find the sum of the following convergent series :
$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$
Attempting to generalize from partial sums yields nothing of interest:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43... | $\frac{2^n}{6^n}=(\frac{1}{3})^n$ and $\frac{3^n}{6^n}=(\frac{1}{2})^n$
Now, use the well known result that for $|x|<1$,
$$\sum \limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753313",
"timestamp": "2023-03-29T00:00:00",
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f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
| Hint: we have: $\dfrac{1+a\cos x}{x^2} - \dfrac{b}{\sin^2 x} \to 1$, and rewrite $\dfrac{b}{\sin^2 x} = \dfrac{b}{x^2}\cdot \dfrac{x^2}{\sin^2 x}$ then the limit on the left equals $\dfrac{1+a\cos x - b}{x^2} = 1$, this means you can assume L'hopitale rule meaning: $1 + a - b = 0$, and differentiate both numerator and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$ Prove that
$$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$$
I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\f... | Hint:
$$\arctan\frac {1}{1+k+k^2}=\arctan\frac{(k+1)-k}{1+(k+1)k}=\arctan(k+1)-\arctan k$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this ... | You have a little mistake with the binomial coefficient:
$$\frac{n^4}{\binom{4n}4}=\frac{4!n^4(4n-4)!}{(4n)!}=\frac{24n^4}{4n(4n-1)(4n-2)(4n-3)}\xrightarrow[n\to\infty]{}\frac{24}{256}=\frac3{32}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^... | Well very obviously Wolfram's answer is incoherent, since the constant term in the expansion at infinity and the limit should be equal. Your reasoning, however, fails on the last line, since $x\sqrt{a}$ does not equal $\sqrt{x^{2}a}$ as $x$ goes to minus infinity, but $-\sqrt{x^{2}a}$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another... | We imagine taking out the candies one at a time.
Your $\frac{16}{40}\cdot\frac{15}{39}\cdot \frac{24}{38}$ calculates the provability of getting Yellow, Yellow, Other in that order. However, two Yellow and one Other can happen in two additional orders, Yellow, Other, Yellow or Other, Yellow, Yellow. Each of these turns... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines
Suppose we have some segment $AB$ of constant length that slides in such a way that its endpoints are moving along orthogonal lines. Let $P$ be a point in the segment so that $|AP| = a$ and $|PB| = b$. How can we find the curve alo... | WLOG let the orthogonal lines be the $x$-axis and $y$-axis respectively.
Let $P=(x,y), A=(0,y+k), B=(x+h,0), AP=a, PB=b$.
By Pythagoras' theorem,
$$\begin{align}
(y+k)^2+(x+h)^2&=(a+b)^2\\
(y+\sqrt{a^2-x^2})^2+(x+\sqrt{b^2-y^2})^2&=(a+b)^2\\
y\sqrt{a^2-x^2}+x\sqrt{b^2-y^2}&=ab\\
y^2(a^2-x^2)&=a^2b^2-2abx\sqrt{b^2-y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that
$$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$
Could you give me some hints on this?
I thought that Jensen's inequality might be of use ... | Another way:
By AM-GM we obtain:
$$\sum_{cyc}a\sqrt{b+c}=\sqrt{\sum_{cyc}(a^2b+a^2c+2ab\sqrt{(b+c)(a+c)}}\leq$$
$$\leq\sqrt{\sum_{cyc}(a^2b+a^2c+ab(b+c+a+c)}=\sqrt{2(a+b+c)(ab+ac+bc)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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What is the sum of this series: $1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 +\cdots$? Say I have a series like the following;
$$1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 + \frac{1 \times 6 \times 11}{5 \times 10 \times 15}x^3 + \cdots.$$
How do I find the sum of this?
I'm trying to find an expres... | The term for $x^n$ seems to be :
$$\frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)} $$
| {
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"url": "https://math.stackexchange.com/questions/1764428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate the limit $\lim\limits_{n\rightarrow \infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n})$ $$\displaystyle\lim_{n\rightarrow \infty} \left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n}\right)$$
I am not able to find any technique to proceed. It migh... | To complement Ákos' answer:
$\begin{align}&\sum_{n=2}^\infty\frac 1 {n^3-n}=\sum_{n=2}^\infty \frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\\
&=\frac 1 2 \left[ \left(\frac 1 2 +\frac 1 6 + \frac 1 {12} + \cdots \right)-\left(\frac 1 6 +\frac 1 {12} + \cdots \right)\right]\\
&=\frac 1 2 \times \frac 1 2=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding minimum distance between a circle and curve what is the minimum distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
in Question there is a circle and a curve and we have to find the least distance between them
| Hint: These are the two ellipses:
Solution:
The red ellipse seems to be a rotated ellipse, centered around the origin. The minimal distance then is the length of the major semi-axis.
So we try to find the rotation:
We can write the second the second equation as
$u^t Q u = 1$ with $u = (x, y)^t$ and
$$
Q =
\begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in... | Here's a quick method, not unrelated to your approach or to the other answers here.
The squares mod $4$ are $0$ and $1$ (can be verified easily by checking all four). Odd numbers are congruent to $1$ or $3$ mod $4$ and these each have square congruent to $1$ mod $4$. Hence the sum of two odd squares is congruent to $2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to show $\frac{19}{7}How can I show $\dfrac{19}{7}<e$ without using a calculator and without knowing any digits of $e$?
Using a calculator, it is easy to see that $\frac{19}{7}=2.7142857...$ and $e=2.71828...$
However, how could this be shown in a testing environment where one does not have access to a calculator?... | Since you know what the result should be, try remove terms one by one by computing residues $r_i$. That will reduce the size of integers, as long as yoou factorize while you can, which is easy because of the factorials (highly composite numbers).
First, notice that the first $3$ terms for $e$ give $\frac{5}{2}$.
Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
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Student is to answer 7 out of 10 questions in an examination
Student is to answer 7 out of 10 questions in
an examination. How
many if she must answer at least 3 of the first 5
questions?
Credit: A FIRST COURSE IN PROBABILITY - Sheldon Ross
University of Southern California
The answer is:
$$\binom{5}{3}\binom{5... | You are counting those cases in which she answers four or five of the first five questions more than once. You are counting them both among the $\binom{5}{3}$ selections of three of the first five questions and among the $\binom{7}{4}$ selections of four additional questions.
Suppose she answers four of the first five... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Deriving formula for asymptotes of a hyperbola I'm trying to find a precalculus-level derivation of the formula for the asymptotes of a hyperbola. My book says:
Solving $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
for $y$, we obtain
$y = \pm \frac ba \sqrt{x^2 - a^2}$
$ = \pm \frac ba \sqrt{x^2(1 - \frac {a^2}{x^2})}$
$ = ... | $$y^2=-b^2\left(1-\frac{x^2}{a^2}\right)$$
$$y^2=\frac{b^2}{a^2}(x^2-a^2)\tag{factor $1/a^2$ out}$$
$$y=\pm\sqrt{\frac{b^2}{a^2}(x^2-a^2)}$$
$$y=\pm\frac{b}{a}\sqrt{x^2-a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean t... | Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then
\begin{align*}
\sqrt{2}+\frac{b}{x} &= x \\
x\sqrt{2}+b &= x^{2} \\
x^{2}-\sqrt{2} \, x-b &= 0 \\
x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\
&= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\
&= \sqrt{b+1+\sqrt{2b+1}} \\
\end{align*}
T... | {
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"url": "https://math.stackexchange.com/questions/1773880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relation between $I_{k}$ and $I_{k+2}$
If $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,n\in \mathbb{N}$ Then Relation between $I_{k}$ and $I_{k+2}$
$\bf{My\; Try::}$ Given $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,$ Then $\displaystyle I_{k+2}=\int_{0}^{\frac{\pi}{2}}... |
Note:
$$\int_0^\frac{\pi}{2} x(\sin x+\cos x)^n dx=\frac{\pi}{4}\int_0^\frac{\pi}{2}(\sin x+\cos x)^ndx=I_n$$
(Put $x=\frac{\pi}{2}-x$ to prove this)
We have $$I_{n+2}-I_n=\frac{\pi}{4}\int_0^\frac{\pi}{2} (\sin x+\cos x)^n \sin 2xdx$$
Applying by parts, differentiating $(\sin x+\cos x)^n$ and integrating $\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.