Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
In an arithmetic sequence, the third term is 10 and the fifth term is 16. I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to... | You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
\begin{align*}
a_3 & = a_1 + (3 - 1)d = 10\\
a_5 & = a_1 + (5 - 1)d = 16
\end{align*}
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\lim_{x \to 0} \frac{\sin(x+a) -\sin(a)}{\sin2x}$. I am having trouble proving out the below:
$$\lim_{x \to 0} \frac{\sin(x+a) - \sin(a)}{\sin2x}$$
I have got as far as below using $\sin(a) + \sin(b)$ in numerator and $\sin(2a)$ in the denominator but am not sure how to expand further, or if these are ... | Continue with this:
$$\frac{\sin x \cos a}{2\sin x\cos x} + \frac{\sin a[\cos x - 1]}{2\sin x\cos x}$$
$$\frac{\cos a}{2\cos x} - \sin a\frac{2\sin^2\frac{x}{2}}{4\sin\frac{x}{2}\cos\frac{x}{2}\cos x}$$
$$\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}$$
then
$$\lim_{x\to0}\frac{\cos a}{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
showing that $\frac{a+b}{2}\cdot\frac{c+d}{2} =\big( \frac{a+b+c+d}{4}\big)^2$ I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right? and $a,b,c,d >0$
$$ = \frac{a+b}2 \cd... | Let
*
*$x=\frac{a+b}{2}$
*$y= \frac{c+d}{2}$
then
$$\frac{a+b}{2}\cdot\frac{c+d}{2} =\big( \frac{a+b+c+d}{4}\big)^2\iff xy=\left(\frac{x+y}{2}\right)^2$$
which is false in general indeed by AM-GM
$$\frac{x+y}{2}\ge \sqrt{xy} \iff xy\le \left(\frac{x+y}{2}\right)^2$$
and equality holds if and only if $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| With $M = \left(
\begin{array}{ccc}
0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 0 \\
\end{array}
\right)$ and $v = (a,b,c)$ we have
$$
v^{\top}I_3 v + 2\rho v^{\top}M v \ge 0\Rightarrow v^{\top}\left(I_3+2\rho M\right)v \ge0
$$
so choosing $\rho$ such that $I_3+2\rho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How to prove that $\frac{1}{2\pi}\int^{2\pi}_0 e^{\cos \theta}\,d\theta = \sum\limits^\infty_{n=0}\frac{1}{(n!2^n)^2}$
How to prove that $$\frac{1}{2\pi}\int^{2\pi}_0 e^{\cos \theta}\,d\theta = \sum\limits^\infty_{n=0}\frac{1}{(n!2^n)^2}.$$
It seems Laurent expansion doesn't work well here. I visited similar question... | Taylor series of $$e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$$
For $$x = \cos \theta$$
$$e^{\cos \theta} = \sum\limits_{n=0}^{\infty}\frac{\cos^n\theta}{n!}$$
Let's integrate
$$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{n=0}^{\infty} \frac{1}{2\pi n!}\int\limits_{0}^{2\pi}\cos^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Positive Integers to make a square How many integers $n$ make the expression $7^n + 7^3 + 2\cdot7^2$ a perfect square?
Factoring $7^2$ we have that $7^n + 7^3 + 2\cdot7^2 = 7^2\cdot(7^{n-2} + 9)$.
How do we prove that the 2nd factor only has $1$ solution when $n = 3$?
| Given $7^n+7^3+2\cdot7^2$
Note that $7^3+2\cdot7^2=441=21^2$
So, our equation becomes,
$$7^n+441=p^2\mbox{ for some integer p} $$$$7^n=p^2-441$$$$7^n=(p+21)(p-21)$$
Since $7$ is a prime number, $(p+21)$ and $(p-21)$ should be degrees of $7$
So, the only possible value of $n$ is $3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding n-th term of a matrix I have to find the n-th power of the following matrix $$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}
$$ I computed a few powers $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$... | No, that is not true, we can also use other way :
$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}$
$A=B+C$
$C=\begin{pmatrix} 0& 0&1\\ 0 & 1& 0 \\ 1 & 0&0 \end{pmatrix}$
$B=\begin{pmatrix} 2 & 0&0\\ 0 & -4& 0 \\ 0 & 0&2 \end{pmatrix}$
$A^n = (B+C)^n = \sum_{k=0}^{n} C_n^k B^k C^{n-k} $
$= \frac{1}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$ $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$
I tried using L'Hospital rule, which yielded :
$$\lim_{x \rightarrow \infty}\frac{5^{x+1} \ln 5+7^{x+1} \ln 7}{5^x \ln 5 }$$
But I'm at the dead end... If I divide numerator and denominator by... | $\dfrac{5^{x+1}+7^{x+1}}{5^{x}+7^{x}}=\dfrac{5(\frac{5}{7})^x+7}{(\frac{5}{7})^x+1}\to \dfrac{5\cdot0+7}{0+1}=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Tight upper tail bound for Normal distribution The following is a well-known chain of inequalities for the tail of the normal distribution when $a = 1:$
$$
\Big(\frac{1}{x} - \frac{a}{x^3}\Big) \phi(x) \leq
\Big(\frac{x}{a + x^2}\Big) \phi(x) \leq
\Phi(-x) \leq \frac{1}{x}\phi(x), \qquad x > 0.
$$
where $\phi(x) = \fr... | Let
$$ G(x) = \left(\frac {1} {x} - \frac {a} {x^3}\right)\phi(x) - \Phi(-x)$$
for $a \in (0, 1)$. Then
$$ \begin{align} G'(x)
&= \left(-\frac {1} {x^2} + \frac {3a} {x^4}\right)\phi(x)
+ \left(\frac {1} {x} - \frac {a} {x^3}\right)(-x)\phi(x) + \phi(-x) \\
&= \left(-\frac {1} {x^2} + \frac {3a} {x^4} - 1 + \frac {a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this integral $\int_0^1\frac{(\arctan{x})^2\ln({x+1/x+2})}{(1+x)^2}dx$ $$I=\int_0^1\frac{(\arctan{x})^2\ln({x+1/x+2})}{(1+x)^2}dx=-\dfrac{\pi^3}{96}+\dfrac{5\pi}{16}\ln^22-\dfrac{\pi}{4}G-G+\dfrac{\pi}{2}\ln2+\dfrac{7}{16}\zeta(3)$$
Where G is the Catalan's Constant.
Using integration by parts we have:
$$v... | Break up the remaining integral as follows
\begin{align}
&\int_0^1\frac{\tan^{-1}{x}\ln({x+\frac1x+2})}{(1+x)(1+x^2)}dx\\
=& \ \frac12 \int_0^1\frac{(1-x)\tan^{-1}{x}\ln({x+\frac1x+2})}{(1+x)(1+x^2)}dx+\frac12 \int_0^1\frac{\tan^{-1}{x}\ln({x+\frac1x+2})}{1+x^2}dx\\
=&\ \frac12\left( -\frac{\pi^3}{96}+\frac{5\pi}{16}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Two six-sided dice are rolled (associated with the random variables $X$ and $Y$) . Find the probability distribution of $X | \max\{X,Y\} = z$. By definition,
$$ P( X = x \mid \max\{X,Y\} = z ) = \frac{P(X = x, \max\{X,Y\} = z )}{P(\max\{ X,Y \} = z)}.$$ (am I right?)
I've found that
$$P(\max\{X,Y \} = z) = \frac{2z - 1... | Assuming X and Y are independent and have a uniform distribution $\mathbf X, \mathbf Y \sim \mathcal{U\{1,6\}}$ (which makes sense for two six-sided dices) we should start with the most immediate conclusions we can make:
$$\eqalign{
P(\mathbf X = x, \mathbf Y = y) &= P(\mathbf X = x) P(\mathbf Y = y) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos... | Direct way:
Factor out $\cos^2\theta$ from numerator and denominator of the l.h.s. fraction and do some trigonometry:
$$\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\dfrac1{\cos^2\theta}+2\tan\theta}{1-\tan^2\theta}=\frac{1+\tan^2\theta+2\tan\theta}{1-\tan^2\theta}=\frac{(1+\tan\theta)^2}{1-\tan^2\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\lim _{x \to 0} [2x^{-3}(\sin^{-1}x - \tan^{-1}x )]^{2x^{-2}}$=? $$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]^{\frac{2}{x^2}}$$
How to find this limit?
My Try: I tried to evaluate this $$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]$$ to understand the nature o... | Recall that $$\lim_{x \to \infty} \left( 1 + \frac{k}{x}\right)^x = e^k.$$
We can add $o(x^{-1})$ terms to the parentheses without changing this. If we want to evaluate, for example, $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}}\right)$$ where $\eta > 0$, then note that for any $\epsilon$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove using only the epsilon , delta - definition Prove using only the epsilon , delta - definition
That :
$$ \lim_{x\to 3} \dfrac{\left( \dfrac{1}{x} - \dfrac{1}{3} \right)}{x-3}=L \in \mathbb{R} $$
My Try :
Given $\epsilon > 0 $, there exists a delta such that
$$ 0<|x-3|< \delta$$ implies $$\left|\dfrac{\left( \d... | If $x\ne 3$ then
$$\dfrac{\dfrac{1}{x} - \dfrac{1}{3}}{x-3}=\dfrac{\dfrac{3-x}{3x} }{x-3}=\dfrac{3-x}{3x(x-3)}=\dfrac{-1}{3x}.$$ So $L=-\dfrac19.$ Now, for $x>0$ it is
$$\left|\dfrac 19-\dfrac {1}{3x}\right|=\dfrac{|x-3|}{9x}.$$
Consider $t <1.$ Then
$$|x-3|<t \implies \dfrac{|x-3|}{9x}<\dfrac{t}{18}$$ (note that $|x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a period if $f(x+k)=1+(2-5f(x)+10f(x)^2-10f(x)^3+5f(x)^4-f(x)^5)^{\frac{1}{5}}$
Let $f$ be a real valued function with domain $\mathbb{R}$ satisfying,
$$f(x+k)=1+(2-5f(x)+10f(x)^2-10f(x)^3+5f(x)^4-f(x)^5)^{\frac{1}{5}}$$
for all real $x$ and some positive constant $k$, then find the period
of $f(x)$ if it i... | Write $g(x) = (1-f(x))^5+1/2$ and you have $$g(x+k)=-g(x)$$
so $$g(x+2k) =-g(x+k)=g(x)$$
so $g$ has period $2k$.
So $$(1-f(x+2k))^5 = (1-f(x))^5\implies 1-f(x+2k) = 1-f(x)$$ and so $f$ has also period $2k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $k > 1$, then $\frac1{(k - 1)^2}-\frac1{(k + 1)^2}=\frac{4k}{(k^2 - 1)^2}$, hence simplify $\sum\limits_{k = 2}^n\frac k{(k^2 - 1)^2}$ I have the following problem:
Prove that if $k > 1$, then
$$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$
Hence simplify
$$\sum\limits_{k = 2}^n \dfrac{k}{... | Here are some guidelines:
*
*Expand $(k+1)^2-(k-1)^2$, or use the rule $a^2-b^2=(a-b)(a+b)$.
*Notice that $(k-1)(k+1)=k^2-1$, thus $(k-1)^2(k+1)^2=?$
*Once you show that $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2},$$ you can use it so that the sum becomes\begin{align*}
\sum_{k=2}^n\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arrangement of 15 balls including 3 each of 5 different colors in a triangle Fifteen balls including 3 each of 5 different colors are arranged in a triangle as shown. How many ways can this be done if rotations are allowed?
I was thinking the answer should be
$15!/(3*(3!)^5)$ as we can arrange 15 balls in 15 position... | We use the Polya Enumeration Theorem. The cycle index here is
$$Z(G) = \frac{1}{3} a_1^{15} + \frac{2}{3} a_3^5.$$
We then obtain for five colors, three each
$$[A^3 B^3 C^3 D^3 E^3] Z(G; A+B+C+D+E)
\\ = [A^3 B^3 C^3 D^3 E^3]
\frac{1}{3} (A+B+C+D+E)^{15}
\\ + [A^3 B^3 C^3 D^3 E^3]
\frac{2}{3} (A^3+B^3+C^3+D^3+E^3)^5
\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
My (attempted) proof:
Suppose that $a^2 \mid b$ and $b^3 \mid c$, where $a$, $b$, and $c$ are integers.
Therefo... | For some integers $k,l$, we have $a^2k =b, b^3l = c$.
Hence $c^3 = b^9l^3 = a^{18}k^9l^3 = (a^{10}k^5)a^8k^4l^3 = b^5a^4(a^4k^4l^3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I ... | $$\sum\limits_{n=0}^{+\infty} \frac{n+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \frac{(2n+1)+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \bigg(\frac{1}{(2n)!}+\frac{1}{(2n+1)!}\bigg) = \frac{1}{2} \sum\limits_{k=0}^{+\infty} \frac{1}{k!} = \frac{e}{2}$$
Maybe the details are not explai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
find angle between two lines between $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6$ how to find to angle between these two lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$
i tried so far like this
$y-\sqrt{3}x-5=0$
$y=\sqrt{3}x+5$
in the form of $y=mx+b$
got the value for $m_1=\sqrt{3}$
and for
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6}... | Recall that to find the angle we can refer to the lines parallel to the given and passing through the origin, that is
*
*$y=\sqrt{3}x$
*$\sqrt{3}y=x$
then, using parametric form, the direction vectors are
*
*for $y=\sqrt{3}x$ $$v_1=(1,\sqrt 3)$$
*for $\sqrt{3}y=x$ $$v_2=(\sqrt 3,1)$$
finally we can compute ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\, dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\, dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 ... | Both of the answers are correct. You have forgotten to add the integration constants in the solutions.
The first solution $\dfrac 12 \ln \left(\dfrac 43 x^2 -\dfrac 43 x +\dfrac 43\right) +\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C$ can be written as $\dfrac 12 \ln \left( x^2 - x +1\right)+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\int_1^2\frac{3x-5}{x^3}~dx$. I want to find the value of
$$\int_1^2\frac{3x-5}{x^3}~dx$$
I'm not sure how to proceed, due to the difference in degree between $3x-5$ and $x^3$. I've tried the substitution $u = 3x-5$, giving the equality
$$\int_1^2\frac{3x-5}{x^3}~dx = 9\int_{-2}^1 \frac{u}{(u-5)^3}~... | Notice that
\begin{equation}
\frac{3x - 5}{x^3}
=
\frac{3x}{x^3}
-
\frac{5}{x^3}
=
\frac{3}{x^2}
-
\frac{5}{x^3}
=
3 x^{-2} - 5x^{-3}
\end{equation}
So the integration becomes
\begin{equation}
3 \int x^{-2} - 5 \int x^{-3}
=
3 \frac{x^{-3}}{-3}
-
5
\frac{x^{-4}}{-4}
\end{equation}
Can you take it from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqr... | You could have noticed that the $x^2$ simplify in the first term,
$$\frac{x^2+\cdots}{x^2\sqrt x}=\frac1{\sqrt x}+\cdots=x^{-1/2}+\cdots$$
So with a little more care,
$$\int(x^{-1/2}-4x^{-3/2}+10x^{-5/2})\,dx=\frac21x^{1/2}-\frac21(-4)x^{-1/2}-\frac2310x^{-3/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}$ $$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}$$
i am a 12th class student and this question was given by our mathematics teacher a year ago.
this is part of... | We have that
$$(1-x^k)=(1-x)(\overbrace{1+x+\ldots x^{k-1}}^{k\,terms})$$
therefore
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}=$$$$=\lim_{x\to 1} {\left(\frac{(1)(1+x)\cdots(1+x+\ldots+x^{2n-1})}{({(1)(1+x)\cdots(1+x+\ldots+x^{n-1})})^2}\right)}=\frac{(2n)!}{(n!)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality and limitation
Suppose $\{a_k\}$ ,$\{b_k\}$ and $\{\xi_k\}$ are non negative, and for all $k\ge 0$ , we have $$a_{k+1}^2\le(a_k+b_k)^2-\xi_k^2$$
Prove
1.$$\sum_{i=1}^{k}\xi_i^2\le(a_1+\sum_{i=0}^kb_i)^2$$
2. If $\{b_k\}$ also satisfy $\sum_{k=0}^{\infty}b_k^2\lt+\infty $, then $$\lim_{k\to\i... | Let us first prove by induction that
$$\tag{1} a_1 +\sum_{i=0}^k b_i \ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 \le i \le k$, then we have
$$a_1 + \sum_{i=0}^{k+1} b_i \ge a_{k+1} + b_{k+1} \ge a_{k+1} + \big(\sqrt{a_{k+2}^2 + \xi_{k+1}^2} - a_{k+1} \big) \ge a_{k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving the supremum I'm given these two sets
$A\subset (0,+\infty ),$ inf$A=0$ and $A$ is not upper bounded
$B=\left \{ \frac{x}{x+1}:x\in A \right \}$
and I have to find the supremum.
Here's the solution my book gives to prove that $supB=1$:
If $y\in B$ then there is a $x \in A$ s.t. $y=\frac{x}{x+1}<1$
We choose $ε>... | Note that
$$ \frac{x}{x+1} = \frac{x+1-1}{x+1} = 1- \frac{1}{x+1}, $$
so
$$ \frac{x}{x+1} < 1 - \epsilon \quad \Leftrightarrow \quad 1-\frac{1}{x+1} < 1 - \epsilon \quad \Leftrightarrow \quad \frac{1}{x+1} > \epsilon, $$
and inverting this gives (this is allowed since $x > 0,$ so $x+1 > 1 > 0)$
$$ x+1 < \frac{1}{\epsil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
zero Jacobian matrix determinant and local inverse Consider the mapping $f: \mathbb R^2 \backslash \{(0,0)\} \to \mathbb R^2$ given by
$$\begin{aligned}
f(x,y) = \begin{pmatrix}
(x^2-y^2)/(x^2+y^2) \\
xy/(x^2+y^2)
\end{pmatrix}
\end{aligned}$$
Does $f$ have a local inverse at every p... | Note that for any $z\in \mathbb R^2\setminus \{(0,0)\},$ $f(rz)=f(z)$ for all $r>0.$ This implies $f$ is constant on each ray from the origin. Such a function cannot be injective on any nonempty open set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Domain of $f(x)=\log_3\frac{x-1}{x+1}+\sqrt{2x+1} -\frac{1}{x}$ $$f(x)=\log_3\frac{x-1}{x+1}+\sqrt{2x+1} -\frac{1}{x}$$
First:
$\sqrt{2x+1}\ge0$,
$1.$ $x_1\ge-\frac{1}{2}$
$\frac{1}{x}$ ,
$2.$ $x_2\neq0$
We can't have log of negative number so:
$\log_3\frac{x-1}{x+1}\ge 0$
$3.$ $x_3\neq-1$
$x_1$ tells me that $x$ sho... | $$\log_3\frac{x-1}{x+1}\ge 0$$
This is wrong
You need $$\frac{x-1}{x+1}> 0 $$ and $$x\ne -1$$
Similarly, solve $2x+1\ge 0$ and $x\ne0$ take intersection of all the solutions you get
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find angle UFO in the picture attached I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.
*
*I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success.
*I als... | Consider a regular 36-gon $A_1A_2\ldots A_{36}$ inscribed in a circle of radius $R$. Inscribed angle over any side is $5^\circ$. We can see our configuration as it is shown on the picture.
It suffices to prove that $UF$ is parallel to the diagonal $A_{13}A_{34}=EA_{34}$; then we have $\angle NFU=\angle NEA_{34}=25^\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 0
} |
How do I simplify $\frac{\log_7 32}{\log_7 8\cdot\sqrt2}$? So far I have got $\log_7 2^5 - \log_7 2^3 + \log_7 2^{1/2}$ and am unable to proceed. Am I, on the right track so far? and how do I proceed? thank you!
| Your expansion is not correct. If $b, x, y > 0$, with $b \neq 1$, then
\begin{align*}
\log_b (xy) & = \log_b x + \log_b y\\
\log_b \left(\frac{x}{y}\right) & = \log_b x - \log_b y
\end{align*}
However, you have
$$\frac{\log_7 32}{\log_7 8\sqrt{2}}$$
which is a quotient of logarithms, not the logarithm of a quotient,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Express arccos in terms of arctan I'm trying to express $\arccos(x)$ in terms of $\arctan$ because the software I'm using only recognizes $\arctan$.
I know that $\arcsin(x)+\arccos(x)=\frac{\pi}{2}$
and $\arcsin(x)=2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$.
Does that mean that $\arccos(x)=\frac{\pi}{2}-2\arctan\le... | $y = \arccos x\\
\cos y = x\\
\sec^2 y = \frac {1}{x^2}\\
\tan^2 y + 1 = \frac {1}{x^2}\\
\tan y = \sqrt {\frac {1}{x^2} - 1} = \frac {\sqrt {1-x^2}}{x}\\
\arccos x = \arctan \frac {\sqrt{1-x^2}}{x}$
Except the range of $\arccos x$ will be $[0, \pi]$ and the range of $\arctan x$ is $(-\frac {\pi}{2}, \frac{\pi}{2})$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial ... | You're right that in the end $$\lim_{N\rightarrow\infty}\left(\frac{1}{N}-\frac{1}{N+1}\right)=0$$ But this is not the sum itself, for this, note that $$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots\\=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 1
} |
Solution to the differential equation $(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$ Question: Find the solution to the differential equation $(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$
The question prompts me to let $G=(2x^2+xy-2y^2)dx+(3x^2+2xy)dy$ and prove that $e^\frac{y}{x}\frac{G}{x}$ is an exact differential. But what is an exact dif... | Hint.
Note that
$$
\frac{\partial}{\partial y}\left(\frac{e^{\frac{y}{x}} \left(2 x^2+x y-2 y^2\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\right)}{x^2}\\
\frac{\partial}{\partial x}\left(\frac{e^{\frac{y}{x}} \left(3 x^2+2 x y\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find the points at which the following functions intersect? And how many times do they intersect? I have a function $f(x) = 4x^3 − 2x^2 − 4x − 2 $
And a line $y = 4x -1$
And I need to know where they intersect, I know that I have to do $4x^3 − 2x^2 − 4x − 2 = 4x -1$
But I can't find the x-values and I'm not al... | The
discriminant
of the cubic $ax^3+bx^2+cx+d$ is
\begin{align}
\Delta(a,b,c,d) &=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2
,
\end{align}
in particular for the cubic
\begin{align}
4x^3-2x^2-8x-1
\tag{1}\label{1}
\end{align}
the discriminant is
\begin{align}
\Delta(4,-2,-8,-1)&=6832>0
,
\end{align}
hence, \eqref{1} has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
| By using Euler's 3rd substitution $x = \frac{1-y^2}{3+y^2}$ ($y > 0$,
so $y = \sqrt{\frac{1-3x}{x+1}}$), we have
$(1-3x)(x+1) = \frac{16y^2}{(3+y^2)^2}$,
$\sqrt{(1-3x)(x+1)} = \frac{4y}{3+y^2}$,
$\mathrm{d} x = -\frac{8y}{(3+y^2)^2} \mathrm{d}y$, and hence
\begin{align}
I &= \int \Big(\frac{1}{8} - \frac{1}{8y^2}\Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
General Solution for a second order ODE So we have this problem in mathematical physics.
We are asked to find the general solution for the ODE
$$
(x^2)y''+(x)y'-(n^2)y=0
$$
for integer $n$.
But this is what I tried to do. I let
$$
y=x^p\\
y'=px^{p-1}\\
y''=p(p-1)x^{p-2}
$$
Then substituted it back to the original equa... | Here's another possible approach:
Let us consider your differential equation
$$x^2\cdot\frac{d^2y}{dx^2}+x\cdot\frac{dy}{dx}-n^2\cdot y=0$$
Let $t=\ln(x)\rightarrow x=e^t$.
Then,
$$\begin{align}
\ \frac{dy}{dx} & =\frac{dt}{dx}\cdot\frac{dy}{dt} \\
& = \frac{d}{dx}\left(\ln(x)\right)\cdot\frac{dy}{dt} \\
& = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating the square root of 2 Since $\sqrt{2}$ is irrational, is there a way to compute the first 20 digits of it?
What I have done so far
I started the first digit decimal of the $\sqrt{2}$ by calculating iteratively so that it would not go to 3 so fast. It looks like this:
\begin{align}
\sqrt 2 & = 1.4^{2} \equi... | On a similar note to the answer by R. Romero: in the special case of taking the square root of an integer $N$, it is fairly straightforward to calculate the continued fraction representation of $\sqrt{N}$.
In the particular case $N=2$, we have:
$$ \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \ddots}}}. $$
(Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 14,
"answer_id": 0
} |
Number of sequences formed from $1, 1, 1, 2,3,4,5,6$ in which all three $1$s appear before the $6$ I want to find number of sequences which contains $2,3,4,5,6$ exactly once and $1$ exactly three times. Also all three $1$s should be placed before $6$.
Example $1)1,1,1,2,3,4,5,6\\2)1,2,3,4,1,1,6,5$
I think that $6$ can ... | Here are some simpler approaches.
Method 1: Place the $2$, $3$, $4$, and $5$ in $$\binom{8}{4}4! = \frac{8!}{4!}$$ ways. Once you have done so, there is only one way to place the three $1$s and $6$ in the remaining positions so that all the $1$s appear before the $6$.
Method 2: Choose four positions for the three $1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Absolute Value Inequalities (Quadratics) I am currently struggling with the solutions of absolute value inequalities that involve quadratics. This is the example problem:
$$x|x + 5| \geq -6$$
I am able to find the solutions, but I struggle in interval notation. I considered graphing the two quadratic functions and find... | $$|x + 5| = \begin{cases}
x + 5 & \text{if $x \geq -5$}\\
-x - 5 & \text{if $x < -5$}
\end{cases}
$$
Case 1: $x \geq 5$
\begin{align*}
x|x + 5| & \geq -6\\
x(x + 5) & \geq -6\\
x^2 + 5x & \geq -6\\
x^2 + 5x + 6 & \geq 0\\
(x + 2)(x + 3) & \geq 0
\end{align*}
The expression on the left-hand side equals zero when $x = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the sum of squares of the real roots
Let $r_1,r_2,r_3,\cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+\cdots+r_n^2$ is $$(A)\,3\quad(B)\,14\quad(C)\,8\quad(D)\,16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't kno... | Hint: the real roots are the roots of the first factor:
$$
\begin{align}
x^8\color{red}{+2x^4-2x^4}-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \\
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Undetermined coefficient in recurrence relation I am given $3x^2(x+2)y''+7xy'-2y=0, x \geq 0$. I am asked to solve this differential equation with a series solution around $x=0$. Note, however that $x=0$ is a regular singular point since:
$$
P(x) = 3x^2(x+2) \implies 3(0)^2(0+2)= 0,
$$
$$
\lim_{x \to 0}x\frac{Q(x)}{P(x... | The collected coefficients for the power $n+r$ are
$$
0=6(n+r)(n+r-1)a_n+3(n+r-1)(n+r-2)a_{n-1}+7(n+r)a_n-2a_n\\
=[6(n+r)^2+(n+r)-2]a_n+3(n+r-1)(n+r-2)a_{n-1}
$$
with the convention that $a_{-1}=0$. Then the indicial equation for $n=0$ is
$$
0=6r^2+r-2=(2r-1)(3r+2)
$$
with recursion
$$
a_n=-\frac{3(n+r-1)(n+r-2)}{(2n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the integral $ \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3)\, dy dx .$
Calculate the integral $$ \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3) dy dx .$$
My attempt: Notice that $ x^2y^3 $ is an odd function with respect to $ y $, so \begin{align*} \int_{0}^{1}\int_{-\sqrt{1-x^... | It's fine.
I would perhaps compute it using polar coordinates:$$\int_{-\frac\pi2}^\frac\pi2\int_0^1r^3\cos^2(\theta)+r^6\cos^2(\theta)\sin^3(\theta)\,\mathrm dr\,\mathrm d\theta.$$Using the same argument as yours, this is just$$\int_{-\frac\pi2}^\frac\pi2\int_0^1r^3\cos^2(\theta)\,\mathrm dr\,\mathrm d\theta,$$which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The length of a rectangle is 6m longer than the width. If the area of a rectangle is $84^2$m, find the dimensions of the rectangle. I'm in grade 11 math.
The length of a rectangle is $6m$ longer than the width. If the area of a rectangle is $84~\text{m}^2$, find the dimensions of the rectangle. I don't know where to st... | From the given information, you have the following system of equations to solve:
\begin{align*}
\begin{cases}
L = 6 + W\\
LW = 84\\
\end{cases}
\end{align*}
Where $W$ denotes the width and $L$ denotes the length.
If we substitute the first relation into the second, it results into the following equation:
\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ ... | I think the step when you arrive at $a^3-a^2>b^2-b^3$ is incorrect as you multiply inequalities whose expressions can be negative.
Alternatively, note that $a^3+a\ge 2a^2$ (this is equivalent to $a(a-1)^2 \ge 0$) and analogously $b^3+b\ge 2b^2$. Therefore
$$a^3+b^3+a+b\ge 2(a^2+b^2) > a^2+b^2+a+b,$$
where the second in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Understanding a particular group/semigroup operation Let $\odot$ be the binary operation defined by
$$ x\odot y := (x+y)+(x\cdot y)$$
where $+$ and $\cdot$ are the usual operations of addition and multiplication from whatever ring you're working with. It's straightforward to show that $\odot$ is both commutative and as... | Suppose we are working on $\mathbb{Z}_n$. Your conjectures are:
*
*If $n$ is prime, then $\{x\in\mathbb{Z}_n\mid x\not\equiv -1\pmod{n}\}$ is a group under $\odot$.
*If $n$ is not prime, then $\{x\in\mathbb{Z}_n\mid x\not\equiv -1\pmod{n}\}$ is not a group under $\odot$.
(Note that $-1$ is a zero in the correspondi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Two different ways of substitution leads to two different results for $\int\frac{x\ln x^2+1}{x^2+1}dx$, how's that? I was working on a solution for
$$\int\frac{x\ln (x^2+1)}{x^2+1}dx$$ and my approach was the following:
first substituting $u = x^2+1$, which means $du = 2x dx$, and therefore:
$$\int\frac{\ln u}{2u}du$$
... | $$I=\int\frac{x\ln(x^2+1)}{x^2+1}dx$$
$$u=x^2+1$$
$$dx=\frac{du}{2x}$$
$$I=\frac{1}{2}\int\frac{\ln(u)}{u}du$$
$$v=\ln(u)$$
$$du=udv$$
$$I=\frac{1}{2}\int vdv=\frac{v^2}{4}+C=\frac{\ln^2(u)}{4}+C=\frac{\ln^2(x^2+1)}{4}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)}$ I was thinking about the following sum -
$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4... | We can prove the following involving harmonic numbers:
$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} {n\choose r}
= H_{n+1} - 1.$$
Start by writing for the LHS
$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} \frac{r+1}{n+1} {n+1\choose r+1}
= \frac{1}{n+1}
\sum_{r=1}^n \frac{(-1)^{r+1}}{r} {n+1\choose r+1}
\\ = \frac{1}{n+1}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
7 people are stopped randomly and asked about their birthdays 7 persons are stopped on the road at random and asked about their birthdays.If the probability that 3 of them are born on Wednesday, 2 on Thursday and the remaining 2 on Sunday is $\frac {K}{7^6 }$ , then K is equal to
I tried solving using Bernoulli triads... | This is like rolling a seven sided die $7$ times and getting $3$ twos, $2$ fives and $2$ sixes. There are $7^7$ possible outcomes and $\frac{7!}{2!2!3!}$ ways to get this particular outcome.
So $$P = \frac{\frac{7!}{2!2!3!}}{7^7}$$
But if $$P = \frac{K}{7^6} = \frac{\frac{7!}{2!2!3!}}{7^7}$$
Then $$K = \frac{\frac{7!}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Formal Epsilon Argument Proof for Sequence
Using a formal epsilon argument, show that $$\lim_{n \to \infty} \sqrt{n^2 + 2} - \sqrt{n^2 -2} =0 .$$
This is what I have, but i'm unsure if it's safe to assume that $\sqrt{n^2 - 2} >0$ or if I made any other errors.
Proof:
$$\left |\sqrt{n^2 + 2} - \sqrt{n^2 - 2} - 0\right... | It looks fine.
To make sure $\sqrt{n^2-2}>0$, you can impose conditions such that $N \ge 2$ and $N \ge \frac2{\epsilon}$.
For example, Let $N = 2 + \lceil \frac2{\epsilon}\rceil .$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't find $\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$ I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$$
I don't know whi... | Recall the formulas:
$$
\begin{align}
a^2 -b^2 &= (a - b)(a + b)\\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
\end{align}
$$
Using them we can get the following:
$$
\begin{align}
\sqrt[3]{1+2x} + 1 &=
\frac{1 + 2x + 1}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} =
\frac{2(x + 1)}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} \\
\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to ... | The second equation can be written $ab=2\sqrt{3}$ which gives $b = \frac{2\sqrt{3}}{a}$. If we substitute back into the first equation we get $a^2 - \frac{12}{a^2} = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to... | Just a kind of summary.
We have
\begin{align*}
S_n&=\sum_{j=1}^n\frac{x^{2^{j-1}}}{1-x^{2^{j}}}\\
&=\sum_{j=1}^n\left(\frac{x^{2^{j-1}}}{1-x^{2^{j-1}}}-\frac{x^{2^j}}{1-x^{2^j}}\right)\tag{1}\\
&\,\,\color{blue}{=\frac{x}{1-x}-\frac{x^{2^{n}}}{1-x^{2^{n}}}}\tag{2}
\end{align*}
Comment:
*
*In (1) we use the identit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Complex number equations ( process) Among the exercises I was solving these are the ones I don't understand and I don't know if the process or solutions are ok, so please correct whenever I went wrong. Thanks everyone.
(I'm supposed to solve by converting to trigonometric form when it's convenient and then finding root... | I would write $$z^6-1=iz^3$$ substituting $$t=z^3$$ we get
$$t^2-it-1=0$$
now you have a quadratic to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove central binomial coefficient upper bound I am trying to prove that $\binom{2n}{n} < \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) < 4\sqrt{n(n+1)}$ which is false)
| Straight induction would be
$$
{{2n+2}\choose{n+1}}
=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}
<\frac{2(2n+1)}{n+1} \frac{4^n}{\sqrt{2n}}
$$
So, we only have to prove that
$$
\frac{2(2n+1)}{n+1} \frac{1}{\sqrt{2n}} < \frac{4}{\sqrt{2n+2}}
$$
which unfortunately is never true.
Fortunately, straight induction works... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\df... | $$f(z)=\left|z-\frac{1}{z}\right|=\frac{\left|z^2-1\right|}{2}$$
since we know $|z|=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Can we interchange columns in a determinant? Can we interchange columns in a determinant like this and preserve the value of it? For example:
$$
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 \\
2 & 1 & 1 & 1 & 0 \\
3 & 1 & 1 & 0 & 0 \\
4 & 1 & 0 & 0 & 0 \\
5 & 0 & 0 & 0 & 0
\end{pmatrix}
$$
to
$$
\begin{pmatrix}
1 & 1 & 1 & 1 & ... | Yes, we are allowed to interchange the columns of a matrix, but we must introduce a minus sign upon each swapping of a pair of columns. So, your calculation above is correct. The rules of row operations are the same as the rules of column operations because, as @Paul says in the comments above, $\det(A) = \det(A^T)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\alpha$ and $\beta$ are roots of $(x^2)-(4x)-1=0$, find $\sqrt[3]{\alpha}+ \sqrt[3]{\beta}$ My question in handwriting
https://i.stack.imgur.com/4vPBs.jpg
If $\alpha$ and $\beta$ are roots of this equation
$$(x^2)-(4x)-1=0$$
Then find $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$
Please do not use the $\Delta = b^2-4ac$ met... | Let $\sqrt[3]{\alpha}+\sqrt[3]{\beta}=x$.
Thus, since $\alpha+\beta=4$ and $\alpha\beta=-1,$ we obtain
$$x^3=\alpha+\beta+3\sqrt[3]{\alpha\beta}(\sqrt[3]{\alpha}+\sqrt[3]{\beta})=4+3\cdot(-1)x$$ or
$$x^3+3x-4=0$$ or
$$x^3-x^2+x^2-x+4x-4=0$$ or
$$(x-1)(x^2+x+4)=0,$$ which gives $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find polynomial $p(n)$ such that $\displaystyle \sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$ converges
Find a polynomial $p(n)$ such that series:
$$\sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$$
converges.
Attempt. If $p(n)=q^3(n)$ for some polynomial $q(n)$ then
$$\sqrt[4]{n^4+... | We have:
\begin{align}
\sqrt[4]{n^4+2n^2}
&=n\sqrt[4]{1+\frac 2{n^2}}\\
&=n\left(1+\frac 1{2n^2}+O\left(\frac 1{n^4}\right)\right)\\
&=n+\frac 1{2n}+O\left(\frac 1{n^3}\right)
\end{align}
Clearly, $p(n)\sim n^3$, hence $p(n)=n^3+an^2+bn+c$ and
\begin{align}
\sqrt[3]{p(n)}
&=n\sqrt[3]{1+\frac an+\frac b{n^2}+\frac c{n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Domain and range of $g(x) = \frac{3x^2+4}{x-2}$
$$g(x) = \frac{3x^2+4}{x-2}$$
for the domain, I find, $\{x|x\ne 2\}$'
and for the range
$y = \dfrac{3x^2+4}{x-2}$ and when $x=2-\epsilon$, $y \to \dfrac{3(2-\epsilon)^2 + 4}{(2-\epsilon)-2} = \dfrac{K}{-\epsilon} \to -\infty$
When $x = 2+\epsilon$, $y \to \dfrac{3(2+\ep... | Fix $y \in \mathbb{R}$. Then you look for $x$ such that
$$\frac{3x^2+4}{x-2}=y.$$
Solve this equation: you get
$$3x^2+4=xy-2y \Rightarrow 3x^2-yx+2y+4=0,$$
which gives
$$x=\frac{y\pm\sqrt{y^2-24y-48}}{6}. $$
So your equation admits real solutions only if $y^2-24y-48 \geq 0$. But this is $y^2-24y+144-192 = (y-12)^2-192... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
There are 3 sections in a question paper with 5 questions each. There are 3 sections in a question paper each containing 5 questions. A candidate has to solve only 5 questions, choosing at least one question from each section. In how many ways can he make his choice?
I have thought of a solution but I am over counting ... | He can choose the following combinations of questions from each section:
$(2,2,1);(2,1,2);(1,2,2);(3,1,1);(1,3,1);(1,1,3)$
Each of the first three combinations has $\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}$ ways. And each of the next three combinations has $\binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}$ ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$ by induction Here is my attempted proof:
$\forall n \in \mathbb{N}$, let $S_n$ be the statement: $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
Base case: $S_1$: $\frac{2^{4(1)}-(-1)^1}{17} = \frac{16+1}{17} = 1 \in \mathbb{N}$
Inductive step: $\forall n \geq 1$, $S_n$ holds
$... | Alternatively, assuming $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$:
$$\begin{align}\frac{2^{4(n+1)}-(-1)^{n+1}}{17} &= \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} =\\
&=\frac{2^{4} \cdot (2^{4n}-(-1)^n) + 2^{4}(-1)^n +(-1)^{n}}{17}=\\
&=2^4\cdot \underbrace{\frac{2^{4n}-(-1)^n}{17}}_{\ge 1}+(-1)^{n}\in \mathbb N.
\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many different passwords of length $10$ have $3$ A's and $7$ digits such that no two A's are adjacent? I have this question. Apparently my solution is wrong and I want to know why it's wrong. The question is:
How many different passwords of length $10$ have $3$ A's and $7$ digits such that no two A's are adjacent?... | As @JMoravitz stated in the comments, your answer is correct. Let's confirm it.
Method 1: We use the Inclusion-Exclusion Principle.
We choose three of the ten positions for the $A$s, then fill each of the remaining positions with one of the ten digits, which can be done in
$$\binom{10}{3}10^7$$
ways. From these, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\prod\limits_{k=1}^\infty \left(1+\frac1{2^k}\right) \lt e ?$ How would you prove that $$\displaystyle \prod_{k=1}^\infty \left(1+\dfrac{1}{2^k}\right) \lt e ?$$
Wolfram|Alpha shows that the product evaluates to $2.384231 \dots$ but is there a nice way to write this number?
A hint about solving the problem... | It can be easily shown that for any $x\in(0,1)$ we have
$$ \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9765}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9765}} <1+x < \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9450}}{1+x+\frac{x^2}{3}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n... | Write, as usual,
$$H_n=\sum_{k=1}^n\frac1 k.$$
Then
$$\frac11+\frac13+\frac15+\cdots+\frac1{2n-1}=H_{2n}-\frac{H_n}2.$$
In your notation,
$$S_{3n}=H_{4n}-\frac{H_{2n}}{2}-\frac{H_n}{2}.$$
But
$$H_n=\ln n+\gamma+O(1/n)$$
where $\gamma$ is Euler's constant, so
$$S_{3n}=\ln 4n-\frac{\ln 2n}{2}-\frac{\ln n}{2}+O(1/n)=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\sin81^\circ$ given $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$
If $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$, then $\sin81^\circ$ is equal to
\begin{align}
&a)\quad\frac{\sqrt{5}+1}{4}\\
&b)\quad\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}\\
&c)\quad\frac{\sqrt{10+2\sqrt{5}}}{4}
\end{align}
$$
\cos18^\circ=\sqrt{1-\frac... | $\sin81^\circ = \sin(45 + 36)\\
\sin45\cos 36 + \cos 45\sin 36$
$\cos 36 = 1 - 2\sin^2 18 = \frac {1+\sqrt 5}{4}$
$\sin 36 =$$ \sqrt {\frac {1-\cos 72}{2}}\\
\sqrt {\frac {1-\sin 18}{2}}\\
\sqrt {\frac {5-\sqrt 5}{8}}\\
\cos 45\sin 36 = \frac {\sqrt{5-\sqrt5}}{4}$
$\sin 81 = \frac {\sqrt 2 + \sqrt 10}{8} + \frac {\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof Verification: If $a\in \mathbb{Z}$, then $a^3≡a \pmod{3}$. So, I understand that congruence equation can be written as $a^3 -a = 3k$ ,$k\in \mathbb{Z}$. I also understand that this can be written as $a(a^2 - 1) = 3k$, which can be simplified further to show the multiplication of $3$ consecutive integers.
$(a-1)a... | You simply need to realize that if you have $k$ consecutive integers one of them has to be divisible by $k$.
The proof is almost too trivial to bother with.
If your consecutive numbers are $m, m+1, m+2, m+3,...... , m+(k-1)$ and if $m \equiv i \pmod k$ then $m+1 \equiv i+1 \mod p$ and $m+2 \equiv i+2 \mod p$ and so o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | Claim:
For all positive integers $n$, we have
$$\prod_{k=1}^n \left(1+\frac{1}{k^3}\right) < e$$
or equivalently
$$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1$$
Proof:
It suffices to show that
$$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1-\frac{1}{(n+1)^2}\tag{*}$$
holds for all positive integers $n$.
To p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
I would really appreciate some help solving this induction problem!!
$3.$ for all $n\ge1$, $\displaystyle\sum_{i=1}^n(2i)^2=\frac{2n(2n+1)(2n+2)}{6}$
I have
$$\frac{2n(2n+1)(2n+2)(12(n+1)^2)}6= \frac{2n(2n+1)2(n+1)12(n+1)(n+1)}6.$$
I think I need to do something with the $(n+1)$.
However, I'm not sure where to go fro... | So to prove this we must use induction and since the LHS is a sum, we add $(2(n+1))^2$.
We do this because if $n$ and $n+1$ every natural number works as $n$ is arbitrary. Therefore, we only need to prove it works for $n+1$.
$$\sum_{i=1}^{n+1}(2i)^2=\frac{2n(2n+1)(2n+2)}{6}+(2(n+1))^2$$
This gives:
$$\sum_{i=1}^{n+1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 ... | $$342342^{1001} \equiv2^{1001} \space \bmod5$$
$$2^{1001}=2 \times (2^4)^{250}\equiv2\times1^{250}\bmod 5=2$$
Case $342343^2 \bmod 3$ is even simpler. You can easily prove that $a^2\equiv 0 \bmod 3$ iff $a\equiv0\bmod3$. In all other cases $a^2\equiv 1 \bmod3$.
Number 342343 is not divisible by 3 so the result must be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$? Wolfram Alpha shows that $$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$
I tried to use the Fourier series
$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$
I am not sure how to continue from this point. I need some he... | HINT
By using the basic trigonometric identity
$$1-\cos(x)=2\sin^2\left(\frac x2\right)$$
yor given integral becomes
$$\begin{align}
\int_{0}^{2\pi}x^2\ln (1-\cos x)~dx &= \int_{0}^{2\pi}x^2 \ln\left(2\sin^2\left(\frac x2\right)\right)~dx\\
&=\int_{0}^{2\pi}x^2\ln(2)~dx+2\int_{0}^{2\pi}x^2 \ln\left(\sin\left(\frac x2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x_1, x_2$ are roots of $P(x)=x^2-6x+1$, prove that $5 \nmid x_1^n+x_2^n$
Given $x_1$ and $x_2$ are roots of $P(x)=x^2-6x+1$ prove using Binomial expansion that $x_1^n+x_2^n$ is not divisible by 5 if $n\in \Bbb N\setminus\{0\}$.
Source: list of problems for the preparation for math contests.
Notice: this problem ... | Here is a solution inspired by Pell's equations, which tangentially is related to Binomial expansion. $(3,2)$ is a solution for
$$x^2-2y^2=1 \tag{1}$$
but then any $(x,y)$ of the form
$$x=\frac{\left(3+2\sqrt{2}\right)^n+\left(3-2\sqrt{2}\right)^n}{2} \space\space\space\text{ and }\space\space\space
y=\frac{\left(3+2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\c... | Multiply by the conjugate: $(\cos(x) - \sqrt{3} \sin(x))(\cos(x) + \sqrt{3} \sin(x)) = 0$. Then we have $\cos^2(x)-3\sin^2(x)=0$. This is the same thing as $1-4\sin^2(x)=0$ or $\sin(x)=\pm \frac{1}{2}$.
*
*NOTE OF CAUTION: This gives you the answers to both the question and its conjugate. You'd have to plug in and c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Maximum length perimeter of a box whose diagonal is 10 unit long. Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?
Here is my solution:
Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of th... | You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=\sqrt{\frac{100}{3}}$.
To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Apostol Calculus, Method of Exhaustion In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 \ to\ b$. Using the fact that,
\begin{align}
&1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2\label{1} \\
&\Rightarrow \frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < \frac{b^... | First, there is a typo in your question. The inequality from the book is
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$
Picking it up from here. Note that the Apostol's ultimate goal in this proof is to somehow show that $A = \frac{b^3}{3}$. He basically just applied the law of trichotomy which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Global extrema for $\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$? Does the function
$$\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$$
have either a global minimum or a global maximum for $q \geq 5$ and $k \geq 1$?
WolframAlpha is unable to find any.
WolframAlpha computation for the global minimum
Wolf... |
can we do better than this?
No, we cannot.
Let $f(k)$ be your function.
Then, we have
$$f'(k)=\frac{ (q-4)q^{2 k + 1} +2 q^{k + 1} + 2 q^{2 k} - 1}{q^k(q - 1) (q^{k + 1} - 1)^2}\ln q$$
which is positive for $k\ge 1$ and $q\ge 5$.
So, we see that $f(k)$ is increasing for $k\ge 1$.
Since
$$f(1)=\frac{q-1}{{q}(q+1)},\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x$ intercept problem How would I find the $x$ intercept of $x^5-x^3+2=0$? I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.
| $$f(x)=x^5-x^3+2\implies f'(x)=5x^4-3x^2\implies f''(x)=20x^3-6x$$
The first derivative cancels twice at $x=0$ and at $x=\pm \sqrt{\frac{3}{5}}$.
$$f(0)=+2 $$
$$f\left(+\sqrt{\frac{3}{5}} \right)=2-\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 1.8141 \qquad f''\left(+\sqrt{\frac{3}{5}} \right)=6 \sqrt{\frac{3}{5}}$$
$$f\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the indefinite integral $\int {(1-x^2)^{-3/2}}dx$ Evaluate the indefinite integral
$$\int \frac {dx}{(1-x^2)^{3/2}} .$$
My answer: I have taken $u = 1-x^2$ but I arrived at the integral of $\frac{-1}{(u^3 - u^4)^{1/2}}$. What should I do next?
| Before edit: $\int \frac {dx}{(1-x^2)^3/2}$.
You can decompose the fraction as follows, then easily integrate:
$$\frac {2}{(1-x^2)^3}=\frac14\cdot \left[\frac {1}{1-x}+\frac{1}{1+x}\right]^3=\\
\frac{1}{4(1-x)^3} +\frac{3}{4(1-x)(1+x)}\left[\frac1{1-x}+\frac1{1+x}\right] + \frac{1}{4(1+x)^3}=\\
\frac{1}{4(1-x)^3} +\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why is $S_{n,3} = \frac{1}{6}(3^n - 3\cdot2^n+3)$? (Stirling)
Why is $S_{n,3} = \frac{1}{6}(3^n - 3\cdot2^n+3)$?
I know that $S(n,3)=3S(n-1,3)+S(n-1,2)$
Where we know $S(n,2)=2S(n-1,2)+1$
We can also see the latter recurrence leads to $S(n,2)=2^{n-1}-1$
So we get $S(n,2)=2s(n-1,2)+2^{n-1}-1$
I am new to Stirling, so ... | That is a standard linear recurrence for
$S(n,3)$ of the form
$a(n+1)=ca(n)+b(n)$.
Dividing by $c^{n+1}$, this becomes
$a(n+1)/c^{n+1}=a(n)/c^{n}+b(n)/c^{n+1}$.
Letting
$d(n)=a(n)/c^{n}$, this become
$d(n+1)=d(n)+b(n)/c^{n+1}$
which can readily be solved.
(added after a request)
$d(n+1)=d(n)+b(n)/c^{n+1}$
becomes,
usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving trigonometric identities I’ve had a bit of difficulty of this question:
(1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA
I tried to do:
(SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA
But then I’m kind of lost. Any help will be appreciated! Additionally, I am not allowed to move one side to anoth... | \begin{align*} &\frac{1 + \sin A + \cos A}{1-\sin A + \cos A} = \frac{1 + \sin A}{\cos A} \\ &\iff \cos A + \sin A \cos A + \cos^2 A = 1 - \sin^2 A + \sin A \cos A + \cos A \\ &\iff \cos A + \sin A \cos A + \cos^2 A = \cos A + \sin A \cos A + \cos^2 A, \end{align*} where we used $\sin^2 A + \cos^2 A = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
evaluating limits with 2 variables Evaluate the limit :
$\lim_{(x,y)\to (2,2)}$ $\frac{x^2 + y^2 - 8}{\sqrt{x^2 +y^2} - \sqrt{8}}$
*
*$−1$
*$\infty$
*$0$
*$1$
*none of the other choices
*does not exist
How is the answer number 5, I thought it should be "does not exist"?
| HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = r\cos\theta$, $y = r\sin\theta$, and hence $r^2 = x^2 + y^2$. So, $$\require{cancel}{x^2 + y^2 - 8 \over \sqrt{x^2 + y^2} - \sqrt{8}} = {r^2 - 8\over r - \sqrt8} = {\cancel{(r-\sqrt{8})}(r + \sqrt{8})\over \cancel{r-\sqrt8}} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Seeking methods to solve $ \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx $ As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:
$$ \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx $$
I'm seeking methods using the Feyn... | My approach
Let
\begin{align}
\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\
&= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\
&= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\s... | This is just another way of saying what the others told you.
$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}
\ne \lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}$$
The theorem is
IF $\displaystyle \lim_{x\to 0}f(x) = L$
and $\displaystyle \lim_{x\to 0}g(x)=M$, where $M, N \in \mathbb R$,
THEN $\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$
For convenience, let
$$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$
$$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$
$$y'=A'+B'$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2}... | In first $A'$ instead of $(-2x)$ you have $(-2-2x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Factoring cubic polynomial over R $z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4\cos^2(\frac{\pi}{14})$, $4\cos^2(\frac{3\p... | The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \;2 \cos \frac{8 \pi}{7} \; . \; \; $$
This is pretty easy if we take $\omega $ a 7th root of unity, then take $x = \omega + \frac{1}{\omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $\omega^6 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?
| Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $\sin x$:
$3\sin x+\cos x=2$ implies $1-\sin^2x=\cos^2x=(2-3\sin x)^2=4-12\sin x+9\sin^2x$, or
$$10\sin^2x-12\sin x+3=0$$
which solves to $\sin x=(6\pm\sqrt{36-30})/10=(6\pm\sqrt6)/10$. Both are valid solutions, since $|\sin x|=|6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domai... | $$\int_0^{\pi/4}\frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{\pi}{4}$$
can be shown by using partial fraction in OP's substituted integral.
One can similarly show $$\int_{\pi/4}^{\pi/2}\frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{\pi}{4}$$
Observe that the integrand has period $\frac\pi2$, the problem can be easi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$$
My try:
The limit can be written as follows:
$$\lim_{n\to\infty}\... | Hint: $$\frac{(k+1)^k}{k^{k-1}} = k \left(1+\frac{1}{k}\right)^k$$
and
$$ \left(1+\frac{1}{k}\right)^k = \exp\left(k \ln\left(1+\frac{1}{k}\right)\right) = \exp\left(1 + O(1/k)\right) = e + O(1/k)$$
Now, what can you say about $$\sum_{k=1}^n k (e + O(1/k))$$
?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find Minimum value of $\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$ Find Minimum value of $$f(x)=\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$$
My try: the domain of the function is $x \in [-1 \,\,\,1]$
Differentiating and equating it to zero we get
$$f'(x)=\frac{-21}{\sqrt{58-42x}}+\frac{70}{\sqrt{1-x^2}\sqrt{149-140\sqrt{1-... | Let $(x,y)$ be a point on the unit circle $x^2+y^2=1$. We have to minimize the function:
$$
\begin{aligned}
f(x,y)
&=
\sqrt{(7x-3)^2+(7y-0)^2} \ +\ \sqrt{(7x-0)^2+(7y-10)^2}
\\
&=
\operatorname{Distance}(\ (7x,7y)\ ,\ (3,0)\ )\
+\
\operatorname{Distance}(\ (7x,7y)\ ,\ (0,10)\ )
\\
&\ge
\operatorname{Distance}(\ (3,0)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find close expression for the sum $S_{n,k}=\sum\limits_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$ Find close expression for the sum $$S_{n,k}=\sum_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$$
For small $k$ I have got following
\begin{align}
&S_{n,0}=1,\\
&S_{n,1}=2,\\
&S_{n,2}=-n+2,\\
&S_{n,3}=-2n+2,\\... | $$S_{n,k}=\sum_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}=\sum_{i=0}^{k} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}$$
Let $[x^j] f(x)$ denote the coefficient of $x^j$ in the expansion of $f(x)$.
$$\large\therefore S_{n,k}=[x^k]\ \sum_{k=0}^{\infty}\sum_{i=0}^{k} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}x^k$$$$\large=[x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq \cos \vartheta$, for $\vartheta \in [0, \frac{\pi}{2}]$ For my thesis I need the inequality $1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq 2 \cos \vartheta$ for $\vartheta \in [0, \frac{\pi}{2}]$ which can be proved by exploiting the fact that $\cos \vartheta$ ... | The inequality obviously holds for $θ = 0$. For $θ \in \left(0, \dfrac{π}{2}\right]$, note that\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} 1 - \frac{2}{π} θ \sin θ \leqslant \cos θ\\
&\Longleftrightarrow 2\sin^2 \frac{θ}{2} = 1 - \cos θ \leqslant \frac{2}{π} θ \sin θ = \frac{4}{π} θ \sin\frac{θ}{2} \cos\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Greater of the angles $\alpha=2\tan^{-1}(2\sqrt{2}-1)$, $\beta=3\sin^{-1}\frac{1}{3}+\sin^{-1}\frac{3}{5}$
Find the greater of the two angles $\alpha=2\tan^{-1}(2\sqrt{2}-1)$ and $\beta=3\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{3}{5}$
My Attempt
$$
\alpha=2\tan^{-1}(2\sqrt{2}-1)=2\tan^{-1}(1.82)>2\tan^{-1}\sqrt{3}=2.\fr... | Note that $\sqrt{2}\approx1.414$.
$(2\sqrt{2}-1)^2=8-4\sqrt{2}+1=9-4\sqrt{2}>3\Leftrightarrow6>4\sqrt{2}\Leftrightarrow $ after raising to the second power $\Leftrightarrow 36>32$ which is true and $2\sqrt{2}-1>\sqrt{3}$
We know that $\tan\dfrac{\pi}{3}=\sqrt{3}$, we get that $(2\sqrt{2}-1)>\tan\dfrac{\pi}{3}$, then $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum\frac1{2-A_k}$ for $(A_k)$ the $n$th roots of unity
If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that
$$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$
What I did: I tried to use some of the following formulas:
$$1+ A_1 ... | Note that
$$
z^n-1=\prod_{k=0}^{n-1}(z-a_k)
$$
so that
$$
nz^{n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}\prod_{k=0}^{n-1}(z-a_k)
$$
and therefore,
$$
\frac{nz^{n-1}}{z^n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}
$$
Thus,
$$
\sum_{k=0}^{n-1}\frac1{2-a_k}=\frac{n2^{n-1}}{2^n-1}
$$
Subtracting the $k=0$ ($a_0=1$) term gives
$$
\sum_{k=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Mysterious polynomial sequence Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.
I'd like to find $P(n)$, $n\in \mathbb{Z}^+$
\begin{align}
P(0)&= 1\\
P(1)&= a\\
P(2)&= a^2+b\\
P(3)&= a^3+2ab\\
P(4)&= a^4+3a^2b+b^2\\
P(5)&= a^5+4a^3b+3ab^2\... | Try:
$$-\frac{2^{-n} \left(\left(a-\sqrt{a^2+4 b}\right)^n-\left(\sqrt{a^2+4
b}+a\right)^n\right)}{\sqrt{a^2+4 b}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $a^3 + b^3 +c^3, $ given $a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$ $a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$.
Find $a^3 + b^3 +c^3 = ?$
My approach is to use AM-GM inequality. Is it correct?
| If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$\frac{a}{3}, \frac{a}{3}, \frac{a}{3}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Identity involving difference of binomial coefficients I am trying to prove the following identity but not sure how to prove it.
[The followings are equivalent forms of the original equality I asked.]
$$
\binom{m+n}{s+1} - \binom{n}{s+1} = \sum_{i=0}^s \frac{m}{s-i+1}\binom{m+1+2(s-i)}{s-i}\binom{n-2(s-i+1)}{i}.
$$
$$
... | Remark. What follows answers one of several queries that appeared
at this post, which each query replacing the previous one. We suggest
making a list so that all the different varieties may be examined.
Starting from the claim
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Methods to solve $\int_{0}^{\infty} \frac{\cos\left(kx^n\right)}{x^n + a}\:dx$ Spurred on by this question, I decided to investigate for different functions on the numerator. Here, I went from $\exp(..)$ to $\sin(..) / \cos(..)$. I initially thought I could modify the result from $\exp(..)$ but got stuck. So I decided ... | A better and easier way to solve this :
$$\displaystyle I = \int_{0}^{\infty} \frac{\cos (kx^n)}{x^n + a} \ \mathrm{d}x$$
Then resolve it as
$$\displaystyle I = \sum_{r=0}^{\infty} (-1)^r \frac{k^{2r}}{(2r)!} \int_{0}^{\infty}\int_{0}^{\infty} x^{2nr} e^{-(x^n + a)t} \ \mathrm{d}t \ \mathrm{d}x$$
Time to do some subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Diagonal entries are zero, others are $1$. Find the determinant. $\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=?$
Attempt:
First I tried to use linearity property of the determinants such that $$\det\binom{ v+ku }{ w
}=\det\binom{v }{ ... | Notice that
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix} = (n-1)\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the closed form of $\sum_{k=1}^{n}\cos^{2m+1}\left(\frac{(2k-1)\pi}{2n+1}\right)$ Add it: Let $m,n$ be postive integers. Find the closed form of
$$f=\sum_{k=1}^{n} \cos^{2m+1}{\left(\dfrac{(2k-1)\pi}{2n+1}\right)}$$
for $m, n \in \mathbb{N}^{+}$.
Maybe use Euler
\begin{align}
2\cos{x} &=e^{ix}+e^{-ix} \\
\dfrac... | The function $\frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^\text{st}$ root of unity. So we need to account for the residues at $0$ and $\infty$.
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\sum_{k=1}^n\cos^{2m+1}\left(\pi\,\frac{2k-1}{2n+1}\right)
&=\frac12-\frac12\sum_{k=0}^{2n}\cos^{2m+1}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to c... | $x^2+1/4=x;$
Note $x>0.$
AM-GM:
$x^2+1/4 \ge 2\sqrt{x^2(1/4)}=x$;
Equality for $x^2=1/4$, hence
$x=1/2$ (discard $-1/2$ (why?)).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 7
} |
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.