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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 do this problem.
In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.
So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.
How do I find the first term and the sum of the first 20 terms?
|
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 is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 \implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
& = \sum_{k = 1}^{n} [a_1 + (k - 1)d]\\
& = \sum_{k = 1}^{n} a_1 + \sum_{k = 1}^{n} (k - 1)d\\
& = a_1\sum_{k = 1}^{n} 1 + d\sum_{k = 1}^{n} (k - 1)\\
& = na_1 + \frac{d(n - 1)n}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
& = a_1 + a_2 + a_3 + \cdots + a_{n - 2} + a_{n - 1} + a_n\\
& = a_1 + [a_1 + d] + [a_1 + 2d] + \cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
\end{align*}
Since we obtain the same sum if we write the terms in reverse order, we obtain
\begin{align*}
S_n & = a_n + a_{n - 1} + a_{n - 2} + \cdots + a_3 + a_2 + a_1\\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + \cdots + [a_1 + 2d] + [a_1 + d] + a_1
\end{align*}
Adding the two expressions for $S_n$ yields
\begin{alignat*}{10}
S_n & = & a_1 & + & [a_1 + d] & + & \cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & \cdots & + & [a_1 + d] & + & a_1\\ \hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & \cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
\end{alignat*}
Since there are $n$ columns in that sum, we obtain
\begin{align*}
2S_n & = n[2a_1 + (n - 1)d]\\
2S_n & = n[a_1 + a_1 + (n - 1)d]\\
2S_n & = n(a_1 + a_n)\\
S_n & = \frac{n(a_1 + a_n)}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_{20}$.
|
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|
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 the right rules to apply.
$$\lim_{x \to 0}\frac{\sin(x)\cos(a) + \sin(a)[\cos(x) - 1]}{2\sin(x)\cos(x)}$$
I need to simplify it down to apply $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}$.
|
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
$$\lim_{x\to0}\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}=\frac{\cos a}{2\times1}-0=\frac{\cos a}{2}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
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 $a,b,c,d >0$
$$ = \frac{a+b}2 \cdot \frac{c+d}2 \cdot \frac{a+b+c+d}4 = \Big( \frac{a+b+c+d}4 \Big)^3 $$
Here's the entire proof. The problem is from Arthur Engel's problem solving strategies in the section on inequalities. The question was to prove the inequality below
$3$. We have
$$ \frac{abc + abd + acd + bcd}4 = \frac12 \Big( ab \frac{c+d}2 + cd \frac{a+b}2 \Big) $$
$$ \le \frac12 \Big( \Big( \frac{a+b}2 \Big)^2 \frac{c+d}2 + \Big( \frac{c+d}2 \Big)^2 \frac{a+b}2 \Big) $$
$$ = \frac{a+b}2 \cdot \frac{c+d}2 \cdot \frac{a+b+c+d}4 \le \Big( \frac{a+b+c+d}4 \Big)^3. $$
Hence,
$$ \sqrt[3]{ \frac{abc + abd + acd + bcd}4 } \le \frac{a+b+c+d}4 \le \sqrt{ \frac{a^2+b^2+c^2+d^2}4}. $$
|
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 $x=y$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 $I_3+2\rho M \gt 0$ (definite positiveness)
with the conditions in the sub-determinants
$$
1-\rho^2 > 0\\
(\rho -1)^2 (2 \rho +1) > 0
$$
follows in $$\rho \ge -\frac 12$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
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|
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 questions, maybe expand $1=\cos^2(\theta)+\sin^2(\theta)$ or some different arguments? Still not working to me.
|
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
$$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{n=0}^{\infty} \frac{1}{2\pi n!}\int\limits_{0}^{2\pi}\cos^n\theta \ d\theta$$
But for even $n$ we have: $$\int\limits_{0}^{2\pi} \cos^n \theta = \frac{2 \pi}{2^{2k}}\frac{(2k)!}{(k!)^2}$$
and zero for odd $n$ (See here why),
Therefore:
$$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{2 \pi}{2^{2k}}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2^{2k}}\frac{(1}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{(2^k k!)^2}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
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",
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|
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 $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$$A^n=\begin{pmatrix} \frac12(3^n+1) & 0&\frac12(3^n-1)\\ 0 &( -3)^n& 0 \\ \frac12(3^n-1) & 0&\frac12(3^n+1)\end{pmatrix}$$ And is just induction left which is easy, but I found by check and try the n-th term, is it possible to find it in a proper way?
|
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} \sum_{k=0}^{n} C_n^k \begin{pmatrix} 2^{n-k} & 0&0\\ 0 & (-4)^{n-k} & 0 \\ 0 & 0&2^{n-k} \end{pmatrix} \begin{pmatrix} (-1)^k+1 & 0&(-1)^{k+1}+1\\ 0 & 2& 0 \\ (-1)^{k+1}+1 & 0&(-1)^k+1 \end{pmatrix} $
$=\begin{pmatrix} \frac12(3^n+1) & 0&\frac12(3^n-1)\\ 0 &( -3)^n& 0 \\ \frac12(3^n-1) & 0&\frac12(3^n+1)\end{pmatrix}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$ $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$
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 $5^x$ , I get a term $\frac{7^x}{5^x}$ ... which is unsolvable for the limit $x \rightarrow \infty $
However , the answer provided by book is $-7$ . I doubt there is mistake in the question.
|
$\dfrac{5^{x+1}+7^{x+1}}{5^{x}+7^{x}}=\dfrac{5(\frac{5}{7})^x+7}{(\frac{5}{7})^x+1}\to \dfrac{5\cdot0+7}{0+1}=7$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
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 $\phi(x) = \frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$ and $\Phi(-x) = \int_x^\infty \phi(t) \,\mathrm{d}t$ are the normal pdf and cdf respectively. Moreover, observe that setting $a = 0$ recovers the upper bound for the normal cdf on the right-hand side of the above chain. My question, therefore, is whether a tighter upper bound for the normal tail exists for some $a \in (0,1)$ when we restrict to $x \geq 1$. Specifically, this yields the following two questions
Prove or disprove: There exists $a \in (0, 1)$ such that for all $x \geq 1$
$$
\Phi(-x) \leq \Big(\frac{1}{x} - \frac{a}{x^3}\Big) \phi(x)
$$
Prove or disprove: There exists $a \in (0,1)$ such that for all $x \geq 1$,
$$
\Phi(-x) \leq \Big(\frac{x}{a + x^2}\Big) \phi(x)
$$
So far, I've exhausted most of my "painless" tricks, and thought I'd post a fun problem that looks to me like it might have a positive answer.
|
Let
$$ G(x) = \left(\frac {1} {x} - \frac {a} {x^3}\right)\phi(x) - \Phi(-x)$$
for $a \in (0, 1)$. Then
$$ \begin{align} G'(x)
&= \left(-\frac {1} {x^2} + \frac {3a} {x^4}\right)\phi(x)
+ \left(\frac {1} {x} - \frac {a} {x^3}\right)(-x)\phi(x) + \phi(-x) \\
&= \left(-\frac {1} {x^2} + \frac {3a} {x^4} - 1 + \frac {a} {x^2} + 1\right)\phi(x) \\
&= \frac {\phi(x)} {x^4}\left[-(1-a)x^2 + 3a\right] \\
&\begin{cases}
<0 & \text{when} & \displaystyle x < -\sqrt{\frac {3a} {1-a}} \\
=0 & \text{when} & \displaystyle x = -\sqrt{\frac {3a} {1-a}} \\
>0 & \text{when} & \displaystyle -\sqrt{\frac {3a} {1-a}} < x < \sqrt{\frac {3a} {1-a}}\\
=0 & \text{when} & \displaystyle x = \sqrt{\frac {3a} {1-a}} \\
<0 & \text{when} & \displaystyle x > \sqrt{\frac {3a} {1-a}} \\
\end{cases}
\end{align}$$
Since we are interested in $x > 1$ only, we can safely ignore the first two negative cases. Note that
$$ \sqrt{\frac {3a} {1-a}} > 1 \iff a > \frac {1} {4}$$
So we can conclude that when $\displaystyle 0 < a \leq \frac {1} {4}$, $G$ is strictly decreasing on $[1, +\infty)$. And it is trivial to check that $\displaystyle \lim_{x \to +\infty} G(x)= 0 $ which implies $G(x) \geq 0$ for all $x \in (1, +\infty)$. And this proves the first claim - it holds when $a$ is small enough.
We can also check the case when $\displaystyle \frac {1} {4} < a < 1$, $G$ first strictly increasing on $\displaystyle \left[1, \sqrt{\frac {3a} {1-a}}\right)$, attains the maximum at $\displaystyle x = \sqrt{\frac {3a} {1-a}}$, and then strictly decreasing on $\displaystyle \left(\sqrt{\frac {3a} {1-a}}, +\infty\right)$. So we also need to check the value of $G(1)$ as the boundary point. Note that
$$ G(1) = (1 - a)\phi(1) - \Phi(-1) \geq 0 \iff a \leq 1 - \frac {\Phi(-1)} {\phi(1)} \approx 0.3443205$$
So the inequality also holds when $\displaystyle \frac {1} {4} < a \leq 1 - \frac {\Phi(-1)} {\phi(1)} $
The another claim can also be checked similarly I think.
This is also worth a read:
https://mathoverflow.net/questions/19404/approximation-of-a-normal-distribution-function
|
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|
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:
$$v=-\dfrac{1}{1+x}, du=[2\dfrac{\arctan{x}\ln(x+1/x+2)}{1+x^2}-\dfrac{(1-x)}{x(1+x)}{(\arctan{x})^2}]dx$$.
$$\dfrac{(1-x}{x(1+x)^2}=-\dfrac{1}{1+x}-\dfrac{2}{(1+x)^2}+\dfrac{1}{x}$$
$$\int_0^1\dfrac{(\arctan{x})^2}{x}dx=\dfrac{\pi}{2}G-\dfrac{7}{8}\zeta(3),\int_0^1\frac{(\arctan{x})^2}{(1+x)^2}dx=-\dfrac{G}{2}+\dfrac{\pi}{4}\ln2,\int_0^1\dfrac{(\arctan{x})^2}{1+x}dx=\dfrac{\pi}{4}G-\dfrac{21}{32}\zeta(3)+\dfrac{\pi^2}{32}\ln2$$
$$ And \int_0^1\dfrac{(1-x)}{x(1+x)^2}{(\arctan{x})^2}dx=\dfrac{\pi}{4}G-\dfrac{7}{32}\zeta(3)-\dfrac{\pi^2}{32}\ln2+G-\dfrac{\pi}{2}\ln2$$
How to calculate$$\int_0^1\frac{\arctan{x}\ln({x+1/x+2})}{(1+x)(1+x^2)}dx??$$
|
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}\ln^22\right) + \frac12\left(\frac7{32}\zeta(3)+\frac{\pi^2}{32}\ln2 \right)\\
= &\ \frac7{64}\zeta(3)- \frac{\pi^3}{192} + \frac{\pi^2}{64}\ln2
+ \frac{5\pi}{32}\ln^22
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
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
$$P(\max\{X,Y \} = z) = \frac{2z - 1}{36},$$
but I can't find the other expression. I've tried to use Bayes' theorem
and it has lead me to nothing.
|
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:
$$\eqalign{
P(\mathbf X = x, \mathbf Y = y) &= P(\mathbf X = x) P(\mathbf Y = y) \\
&= \left( {1 \over 6}\right)^2 \\
&= {1 \over 36}}$$
for all $x, y \in \{1, 2, \ldots, 6\}$.
And all possible values for $max \{ \mathbf X, \mathbf Y \}$ given $\mathbf X$ and $\mathbf Y$ are:
X\Y || 1 2 3 4 5 6
====================================
1 || 1 2 3 4 5 6
2 || 2 2 3 4 5 6
3 || 3 3 3 4 5 6
4 || 4 4 4 4 5 6
5 || 5 5 5 5 5 6
6 || 6 6 6 6 6 6
And we know that by definition
$$P( \mathbf X = x \mid max \{ \mathbf X, \mathbf Y \} = z) = {{P(\mathbf X = x, max \{ \mathbf X ,\mathbf Y \} = z)} \over {P(max \{ \mathbf X , \mathbf Y \} ) = z}}$$
I'll show a solution with intuition and a purely analytical one.
Solution with intuition
From the table we can calculate the probability of an event $A$ by counting all the possible occurences of $(\mathbf X, \mathbf Y)$ in which the event $A$ occurs and divide this by 36 since each $(x, y)$ has a probability of ${1 \over 36}$ to occur.
If $max \{ \mathbf X , \mathbf Y \} = z$ we can see that if $\mathbf X = z$ then $\mathbf Y$ can have any of the $z$ values in $\{ 1, \ldots, z \}$ and if $\mathbf X$ has any of the $z-1$ values in $\{1, \ldots, z-1 \}$ then $\mathbf Y = z$. So these are $z + (z - 1) = 2z - 1$ combinations of $( \mathbf X , \mathbf Y )$ such that $max \{ \mathbf X , \mathbf Y \} = z$ and therefore $P(max \{ \mathbf X , \mathbf Y \} = z) = {2z-1 \over 36}$.
Now lets check $P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z )$.
Clearly if $x \geq z$ then $P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) = 0$.
From the table we can see that whenever $x \lt z$, then we will have just one combination where $max \{ \mathbf X , \mathbf Y \} = z$ because $\mathbf Y = z$ is the only possible value of $\mathbf Y$. So $P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) = {1 \over 36}$ for $x \lt z$.
And when $x = z$ then $\mathbf Y$ can take any of the $z$ values where $y \leq z$. So $P(\mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) = {z \over 36}$ for $x = z$.
In conclusion:
$$P(max \{ \mathbf X , \mathbf Y \} = z) = {2z - 1 \over 36}$$
and
$$P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) =
\begin{cases}
{1 \over 36} & x \lt z \\
{z \over 36} & x = z
\end{cases}
$$
therefore
$$P( \mathbf X = x \mid max \{ \mathbf X , \mathbf Y \} = z) =
\begin{cases}
{1 \over 2z - 1} & x \lt z \\
{z \over 2z - 1} & x = z
\end{cases}
$$
Purely analytical solution
Since $\mathbf X$ and $\mathbf Y$ are independent
$$P(max \{ \mathbf X , \mathbf Y \} = z) = P( \mathbf X = z, \mathbf Y \leq z) + P( \mathbf X \lt z, \mathbf Y = z)$$
and calculating each summand separately we can see that
$$\eqalign{
P( \mathbf X = z, \mathbf Y \leq z) &= P(\mathbf X = z, \mathbf Y = 1) + \cdots + P( \mathbf X = z, \mathbf Y = z) \\
&= {z \over 36}
}$$
and
$$\eqalign{
P( \mathbf X \lt z, \mathbf Y = z) &= P( \mathbf X = 1, \mathbf Y = z) + \cdots + P( \mathbf X = z-1, \mathbf Y = z) \\
&= {z-1 \over 36}
}$$
So
$$\eqalign{
P(max\{ \mathbf X , \mathbf Y \} = z) &= P( \mathbf X = z, \mathbf Y \leq z) + P( \mathbf \lt z, \mathbf Y = z) \\
&= {z \over 36} + {z-1 \over 36} \\
&= {2z - 1 \over 36}
}$$
For $P( \mathbf X = x, max \{\mathbf X, \mathbf Y\} = z)$ we can see different cases.
If $x \lt z$ then
$$\eqalign{
P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) &= P( \mathbf X = x, \mathbf Y = z) \\
&= {1 \over 36}
}$$
If $x = z$ then
$$\eqalign{
P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) &= P( \mathbf X = z, \mathbf Y = 1) + \cdots + P(\mathbf X = z, \mathbf Y = z) \\
&= {z \over 36}
}$$
So
$$ P( \mathbf X = x, max \{ \mathbf X , \mathbf Y \} = z) =
\begin{cases}
{1 \over 36} & x \lt z \\
{z \over 36} & x = z
\end{cases}
$$
therefore
$$P( \mathbf X = x \mid max \{ \mathbf X , \mathbf Y \} = z) =
\begin{cases}
{1 \over 2z - 1} & x \lt z \\
{z \over 2z - 1} & x = z
\end{cases}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
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
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
|
Direct way:
Factor out $\cos^2\theta$ from numerator and denominator of the l.h.s. fraction and do some trigonometry:
$$\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\dfrac1{\cos^2\theta}+2\tan\theta}{1-\tan^2\theta}=\frac{1+\tan^2\theta+2\tan\theta}{1-\tan^2\theta}=\frac{(1+\tan\theta)^2}{1-\tan^2\theta}$$
and simplify.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 of the problem. I used L'Hopital. But it became too tedious to calculate.
Can anyone please give me suggestion to solve it?
Edit: I used the hint given By lab bhattacharjee. I expand the inverse functions and I got $$\lim _{x \to 0} \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right] =1$$.
Now I think it remains to find the value of $\lim _{x \to 0}e^{f(x).g(x)}$ where $f(x) = \left[\frac{2}{x^3}(\sin^{-1}x - \tan^{-1}x )\right]$ and $g(x) = 2/x^2$
|
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 can choose some large $X$ such that $x > X$ implies $$\left|\frac{k_2}{x^\eta}\right| < \epsilon$$ and thus $$e^{k-\epsilon} = \lim_{x \to \infty} \left( 1 + \frac{k-\epsilon}{x}\right) \leq \lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}}\right) \leq \lim_{x \to \infty} \left( 1 + \frac{k+\epsilon}{x}\right) = e^{k+\epsilon}.$$ But since $\epsilon$ was arbitrary, this shows $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{k_2}{x^{1 + \eta}} \right) = e^k.$$
Now by Taylor expansions, for $x > 0$, $$\frac{x^3}{2} - \frac{x^5}{8} < \arcsin x - \arctan x < \frac{x^3}{2} - \frac{x^5}{8} + \frac{3x^7}{16} $$ so $$\lim_{x \to 0^+} \left(1 - \frac{x^2}{4} \right)^\frac{2}{x^2} \leq \lim_{x \to 0^+} \left[ \frac{2}{x^3} \left( \arcsin x - \arctan x\right) \right]^\frac{2}{x^2} \leq \lim_{x \to 0^+} \left(1 - \frac{x^2}{4} + \frac{3x^4}{16} \right)^\frac{2}{x^2}.$$ By substituting $y = 2/x^2$, the outer limits are made into $$\lim_{y \to \infty} \left( 1 - \frac{1}{2y} \right)^y$$ and $$\lim_{y \to \infty} \left( 1 - \frac{1}{2y} + \frac{3}{4y^2} \right)^y$$ both of which equal $\boxed{e^{-1/2}}$.
The $x \to 0^-$ limit can be treated very similarly, the only caveat being that the inequalities are reversed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 $$\left|\dfrac{\left( \dfrac{1}{x} - \dfrac{1}{3} \right)}{x-3} – L\right| < \epsilon \ \ \ \ $$ Or $$ \ \ \ \ \left|\dfrac{-(x-3)}{3x (x-3)}-L\right|< \epsilon $$ .
Now what ?
|
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 $|x-3|<t<1\implies x>2$.)
So,
$$\forall \epsilon>0 \exists \delta=\min\{1,18\epsilon\}|0<|x-3|<\delta \implies \dfrac{|x-3|}{9x}<\epsilon.$$
|
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"url": "https://math.stackexchange.com/questions/2890998",
"timestamp": "2023-03-29T00:00:00",
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|
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 is periodic.
My Attempt:
$$f(x+k)=1+(1+(C_0^5(1)^5(-f(x))^0+C_1^5(1)^4(-f(x))+C_2^5(1)^3(-f(x))^2+C_3^5(1)^2(-f(x))^3+C_4^5(1)^1(-f(x))^4+C_5^5(1)^0(-f(x))^5))^{\frac{1}{5}}$$
$$f(x+k)=1+(1+(1-f(x)^5)^{\frac{1}{5}}$$
The problem is that I am not able to establish $f(x)$ is periodic. How can I proceed ?
|
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 $2k$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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
$$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$
My Work:
$$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}$$
I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything.
I would appreciate it if someone could please take the time to explain this problem.
|
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\dfrac{k}{(k^2-1)^2}&=\dfrac{1}{4}\sum_{k=2}^n\left(\dfrac{1}{(k-1)^2}-\dfrac{1}{(k+1)^2}\right)\\
&=\dfrac{1}{4}\left(\sum_{k=2}^n\dfrac{1}{(k-1)^2}-\sum_{k=2}^n\frac{1}{(k+1)^2}\right);
\end{align*}
notice that by changing the variable $k-1$ into $k$, you get $$\sum_{k=2}^n\dfrac{1}{(k-1)^2}=\sum_{k=1}^{n-1}\dfrac{1}{k^2}.$$ Do the same for $\sum_{k=2}^n\frac{1}{(k+1)^2}$ and see what you'll get.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 positions in 15! ways. Then since there are 3 balls of 5 different colors each, we divide it by $(3!)^5$ and then divide by 3 as rotation is allowed.
But this is not correct as if 6 balls are needed to arrange like this where there are 3 balls each of 2 different colors, by this logic, the answer should be $6!/3!3!3$ which is not an integer.
|
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
$$[A^3 B^3 C^3 D^3 E^3] Z(G; A+B+C+D+E)
\\ = [A^3 B^3 C^3 D^3 E^3]
\frac{1}{3} (A+B+C+D+E)^{15}
\\ + [A^3 B^3 C^3 D^3 E^3]
\frac{2}{3} (A^3+B^3+C^3+D^3+E^3)^5
\\ = \frac{1}{3} \frac{15!}{3!^5}
+ [A B C D E]
\frac{2}{3} (A+B+C+D+E)^5
\\ = \frac{1}{3} \frac{15!}{3!^5} + \frac{2}{3} 5!
= 56056080.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
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.
Therefore, we have that $a^2 k_1 = b$ and $b^3 k_2 = c$ for some $k_1, k_2 \in \mathbb{Z}$.
I'm unsure of how to proceed from here. If I cube $a^2 k_1 = b$ to get $a^6 k_1 = b^3$, and then substitute this into $b^3 \mid c$, then I get $a^6k_1k_2 = c$, which isn't useful. The same happens if I do the other substitution.
I would greatly appreciate it if people could please take the time to clarify how I should be proceeding.
|
For some integers $k,l$, we have $a^2k =b, b^3l = c$.
Hence $c^3 = b^9l^3 = a^{18}k^9l^3 = (a^{10}k^5)a^8k^4l^3 = b^5a^4(a^4k^4l^3)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 am unsure where to start. Should I consider other series?
|
$$\sum\limits_{n=0}^{+\infty} \frac{n+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \frac{(2n+1)+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \bigg(\frac{1}{(2n)!}+\frac{1}{(2n+1)!}\bigg) = \frac{1}{2} \sum\limits_{k=0}^{+\infty} \frac{1}{k!} = \frac{e}{2}$$
Maybe the details are not explained well enough. If there is anything unclear just ask.
|
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|
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
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6} {\sqrt {3}}$
$y=\dfrac {1} {\sqrt {3}}x-6$
$m_{2}=\dfrac {1} {\sqrt {3}}$
the formula to find angle between two lines
$\tan \theta =\dfrac {m_{2-}m_{1}} {1+m_{1}m_{2}}$
$\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}.\frac{1}{\sqrt{3}}}$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 1 + 1 }$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 2 }$
$- \frac { 2 } { 2 \sqrt { 3 } }$
$- \frac { 1 } { \sqrt { 3 } }$
is this right till now and how to find angle after wards . should i use tan inverse of $- \frac { 1 } { \sqrt { 3 } }$ or my algebra calculation is wrong . help me thank you
|
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 the angle by dot product
$$\cos \theta = \frac{v_1\cdot v_2}{\|v_1\|\|v_2\|}=\frac{2\sqrt 3}{4}=\frac{\sqrt 3}{2}\implies \theta = 30°$$
As an alternative the angles between the two lines and the $x$ axis are
*
*for $y=\sqrt{3}x$ $$\tan \theta_1=\sqrt 3 \implies \theta_1=60°$$
*for $\sqrt{3}y=x$ $$\tan \theta_2=\frac1{\sqrt 3} \implies \theta_2=30°$$
$$\implies \theta =\theta_1-\theta_2= 30°$$
Then the acute angle between the two lines is equal to $30°=\frac{\pi}6$ and as a consequence the obtuse angle between the two lines is equal to $180°-30°=150°=\frac{5\pi}6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 2}{(\frac{\sqrt{3}}{2}u)^2+(\frac {\sqrt{3}}{2})^2} = \int \frac{u}{u^2+1}du + \frac{1}{\sqrt{3}}\int\frac 1 {u^2+1}du.$$
This gives $$\frac 1 2\log({u^2+1})+\frac{1}{\sqrt{3}}\arctan{u}.$$
Substituting back in x yields $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+\frac 1 {\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$
However, according to Wolfram Alpha, the integral should evaluate to $$\frac 1 2 \log(x^2-x+1)+\frac{1}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$After working some time on the integral, I know how to reach this solution, but I don't understand why my first attempt didn't arrive at the correct answer. Do you see where I went wrong?
Thank you for any help!
|
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 $\dfrac 12 \ln \left( x^2 - x +1\right)+\dfrac{1}{2} \ln \left(\dfrac 43\right)+\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C$ which is equal to $\dfrac 12 \ln \left( x^2 - x +1\right)+\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C'$.
So, this is basically it. The both answers are correct. Only their constants of integration are different.
|
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|
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
$$\int_1^2\frac{3x-5}{x^3}~dx = 9\int_{-2}^1 \frac{u}{(u-5)^3}~du$$
However, this doesn't solve the problem I have of the difference in degree between the numerator and the denominator. I thought that perhaps partial fractions would make a difference, but the quadratic I had to solve was untenable.
|
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 here ?
|
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"timestamp": "2023-03-29T00:00:00",
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|
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
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$
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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,
$$\int(x^{-1/2}-4x^{-3/2}+10x^{-5/2})\,dx=\frac21x^{1/2}-\frac21(-4)x^{-1/2}-\frac2310x^{-3/2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
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 JEE EXAM syllabus ,
i tried in this way..
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}$$
$$ =\lim_{x\to 1}{\left(\frac{(1-x^{n+1})(1-x^{n+2})\cdots(1-x^{2n})}{{(1-x)(1-x^2)\cdots(1-x^n)}}\right)}$$
then knowing that
$$\lim_{x\to 1}{(\frac{1-x^n}{1-x})}=n$$ hence
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}
= \binom{2n}{n}$$
Is this right?
Is there any other way to do it?
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We have that
$$(1-x^k)=(1-x)(\overbrace{1+x+\ldots x^{k-1}}^{k\,terms})$$
therefore
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}=$$$$=\lim_{x\to 1} {\left(\frac{(1)(1+x)\cdots(1+x+\ldots+x^{2n-1})}{({(1)(1+x)\cdots(1+x+\ldots+x^{n-1})})^2}\right)}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$$
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"timestamp": "2023-03-29T00:00:00",
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|
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 $$\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k\xi_i^2=0$$
After doing this transformation:$$\xi_k^2\le(a_k+b_k)^2-a_{k+1}^2$$
$\Rightarrow$ $$\sum_{i=1}^{k}\xi_k^2\le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+\cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.
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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
$$a_1 + \sum_{i=0}^{k+1} b_i \ge a_{k+1} + b_{k+1} \ge a_{k+1} + \big(\sqrt{a_{k+2}^2 + \xi_{k+1}^2} - a_{k+1} \big) \ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$\tag{2}\sum_{i=1}^k \xi_i^2 \leq \big(a_1 + \sum_{i=0}^k b_i \big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 \le (a_1+b_1)^2-\xi_1^2 \le (a_1+b_0+b_1)^2 - \xi_1^2.$$
Assume that (2) is true for $k$, then
\begin{align}
\big(a_1 + \sum_{i=0}^{k+1} b_i \big)^2-a_{k+2}^2 &\ge \sum_{i=1}^k \xi_i^2 + 2 \big( a_1 +\sum_{i=0}^k b_k \big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^k \xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^k \xi_i^2 + (a_{k+2}^2 +\xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^{k+1} \xi_i^2
\end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.
The previous result can be also applied with shifted-indices: The shifted sequence $(a_{n+j},b_{n+j},\xi_{n+j})_{n \in \mathbb{N}_0}$ satisfies the initial condition and therefore we can conclude (by using $(a+b)^2 \leq 2 a^2+2b^2$) that
\begin{align}
\frac{1}{k-j+1} \sum_{i=j+1}^k \xi_i^2 &\le \frac{2}{k-j+1} \Big(a_{j+1}^2 + (\sum_{i=j}^k b_i)^2\Big)\\
&\le \frac{2a_{j+1}^2}{k-j+1} + 2\sum_{i=j}^k b_i^2,
\end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Assuming now additionally that $\sum_{n=1}^\infty b_n^2 <\infty$, we find that
\begin{align}
\limsup_{k \rightarrow \infty} \frac{1}{k} \sum_{i=1}^k \xi_i^2 &\le \limsup_{k \rightarrow \infty} \frac{1}{k} \sum_{i=1}^{j} \xi_i^2 + \limsup_{k \rightarrow \infty} \frac{1}{k} \sum_{i=j+1}^k \xi_i^2 \\
&= \limsup_{k \rightarrow \infty} \frac{k-j+1}{k} \frac{1}{k-j+1} \sum_{i=j+1}^k \xi_i^2 \\
&\le \limsup_{k \rightarrow \infty} \frac{2a_{j+1}^2}{k-j+1} + 2\sum_{i=j}^k b_i^2 \\
&= 2\sum_{i=j}^\infty b_i^2 \rightarrow 0 \quad \text{for} \quad j \rightarrow \infty
\end{align}
proving the second part of the initial question.
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|
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 $ε>0$ and we have to find $x\in A$ such that $\frac{x}{x+1}<1-ε$
that is $x>\frac{1}{ε}-1$ ...
the proof continues but my problem is this segment, how does it go from here $\frac{x}{x+1}<1-ε$ to here $x>\frac{1}{ε}-1$ ?
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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 $x+1 > 1 > 0)$
$$ x+1 < \frac{1}{\epsilon} \quad \Leftrightarrow \quad x < \frac{1}{\epsilon}-1. $$
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"timestamp": "2023-03-29T00:00:00",
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|
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 point of $\mathbb R^2$?
Update: The interesting fact with this function is that we can not use the inverse function theorem to decide whether it is locally injective, since for all points $(x, y)$ we have $Jf (x, y) = 0$. Indeed,
\begin{align}
\det Jf(x,y) = &
\det \begin{pmatrix}
\frac{\partial}{\partial x} \frac{ (x^2-y^2)}{(x^2+y^2)}
&
\frac{\partial}{\partial y} \frac{ (x^2-y^2)}{(x^2+y^2)}
\\
\frac{\partial}{\partial x} \frac{ xy}{(x^2+y^2)}
&
\frac{\partial}{\partial y} \frac{xy}{(x^2+y^2)}
\end{pmatrix}
\\
=&
\det \begin{pmatrix}
\frac{ 2x(x^2+y^2)-(x^2-y^2)2x}{(x^2+y^2)^2}
&
\frac{-2y(x^2+y^2)-(x^2-y^2)2y}{(x^2+y^2)^2}
\\
\frac{y(x^2+y^2)-xy(2x) }{(x^2+y^2)^2}
&
\frac{ x(x^2+y^2)-xy(2y) }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\det \begin{pmatrix}
\frac{ 4xy^2}{(x^2+y^2)^2}
&
\frac{-4x^2y}{(x^2+y^2)^2}
\\
\frac{-x^2y+y^3 }{(x^2+y^2)^2}
&
\frac{ x^3-xy^2 }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2}
\det \begin{pmatrix}
y
&
-x
\\
\frac{-x^2y+y^3 }{(x^2+y^2)^2}
&
\frac{ x^3-xy^2 }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
\det \begin{pmatrix}
y
&
-x
\\
{-x^2y+y^3 }
&
{ x^3-xy^2 }
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
[ y({ x^3-xy^2 }) +x({-x^2y+y^3 })]
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
[ x^3y-xy^3-x^3y+xy^3]
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
\cdot 0
\\
=&0
\end{align}
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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.
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|
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 $x$ should be bigger then -1/2, let's say if we put x=-2
(as -2<-1/2) in the log we get $-3/-1=3$ which is good. I am not sure what is the domain of this function.
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$$\log_3\frac{x-1}{x+1}\ge 0$$
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
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|
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 also tried to flip the triangle UFO over side NO but again no success.
*I tried to find like triangles, but not enough.
Could you please give me a hint?
Thanks,
R. de Souza
|
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 $\angle NFU=\angle NEA_{34}=25^\circ$, so $\angle UFO= \angle NFO-\angle NFU=100^\circ-25^\circ=75^\circ$.
To prove $UF\parallel A_{34}E$, it is enough to prove $\frac{NU}{UT}=\frac{NF}{FE}$ ($T$ is as on the picture). For we can use the following two formulae:
*
*The length of a chord of a circle with inscribed angle $\alpha$ is $2R\sin\alpha$.
*If $E$ is on a side $BC$ of a $\triangle ABC$, then $\frac{BE}{EC}= \frac{AB\sin\angle BAE}{AC\sin\angle CAE}$.
Now, from $\triangle OEN$ we have: $$\frac{NF}{FE}= \frac{ON\sin\angle NOF}{OE\sin\angle EOF}=\frac{2R\sin 40^\circ\sin 20^\circ}{2R\sin 60^\circ\sin 60^\circ}.$$ From $\triangle NET$ we have: $$\frac{NU}{UT}= \frac{EN\sin\angle NEU}{ET\sin\angle TEU}= \frac{2R\sin 80^\circ\sin 5^\circ}{ET\sin 20^\circ}.$$ By the law of sines on $\triangle NET$, $\frac{ET}{NE}=\frac{\sin 60^\circ}{\sin 95^\circ}$, so $ET= NE\ \frac{\sin 60^\circ}{\sin 95^\circ}= 2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}$ and thus $$\frac{NU}{UT}= \frac{2R\sin 80^\circ\sin 5^\circ}{2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}\sin 20^\circ}= \frac{2R\sin 95^\circ\sin 5^\circ}{2R\sin 60^\circ\sin 20^\circ}.$$
So, for $\frac{NU}{UT}=\frac{NF}{FE}$ it is enough to check: $\sin 40^\circ\sin 20^\circ\sin 20^\circ= \sin 95^\circ\sin 5^\circ\sin 60^\circ$.
We have: $$\sin 95^\circ\sin 5^\circ\sin 60^\circ=\frac{1}{2}(\cos 90^\circ-\cos 100^\circ)\sin 60^\circ= \frac{1}{2}\cos 80^\circ\sin 60^\circ= \frac{1}{4}(\sin 140^\circ-\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ),$$ and: $$\sin 40^\circ\sin 20^\circ\sin 20^\circ=\frac{1}{2}(\cos 20^\circ-\cos 60^\circ)\sin 20^\circ= \frac{1}{2}(\cos 20^\circ\sin 20^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{2}(\frac{1}{2}\sin 40^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ).$$
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|
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!
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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, so you cannot apply those rules here.
However, your observations that $32 = 2^5$, $8 = 2^{3}$, and $\sqrt{2} = 2^{1/2}$ would be useful if we were working with logarithms to the base $2$. To do so, we use the Change of Base Formula. Let $a, b, x > 0$, with $a, b \neq 1$. Then
$$\log_a x = \frac{\log_b x}{\log_b a}$$
By setting $a = 7$ and $b = 2$, we obtain
\begin{align*}
\frac{\log_7 32}{\log_7 8\sqrt{2}} & = \frac{\dfrac{\log_2 32}{\log_2 7}}{\dfrac{\log_2 8\sqrt{2}}{\log_2 7}}\\
& = \frac{\log_2 32}{\log_2 8\sqrt{2}}\\
& = \frac{\log_2 2^5}{\log_2 2^32^{1/2}}\\
& = \frac{5}{\log_2 2^{7/2}}\\
& = \frac{5}{\frac{7}{2}}\\
& = \frac{10}{7}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 $\arccos(x)=\frac{\pi}{2}-2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$?
|
$y = \arccos x\\
\cos y = x\\
\sec^2 y = \frac {1}{x^2}\\
\tan^2 y + 1 = \frac {1}{x^2}\\
\tan y = \sqrt {\frac {1}{x^2} - 1} = \frac {\sqrt {1-x^2}}{x}\\
\arccos x = \arctan \frac {\sqrt{1-x^2}}{x}$
Except the range of $\arccos x$ will be $[0, \pi]$ and the range of $\arctan x$ is $(-\frac {\pi}{2}, \frac{\pi}{2})$ if that is relavant you may need to play some games if $x<0$
|
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|
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 sum converges.
Now
$$ S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)$$
but I don't know how to go on with this. Now $ \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0$ but the right answer should be $1$.
|
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 $$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots\\=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\dots$$
|
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|
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.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
|
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 start with expanding only the term involving $\sin(x)$. Using the Taylor expansion:
$$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} +\mathcal{O}(x^7)$$
Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $\mathcal{O}(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $\dfrac{1}{\sin^2(x)}$:
$$\frac{1}{\sin^2(x)} = \frac{1}{x^2}\left[1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^{-2}$$
To expand the square brackets, we can use:
$$\frac{1}{(1+u)^2} = 1-2 u + 3 u^2 + \mathcal{O}(u^3)$$
This can be derived by differentiating the geometric series term by term. We can then substitute $u = - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)$. We have:
$$u^2 = \left[- \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^2 = \frac{x^4}{36} +\mathcal{O}(x^6)$$
Therefore:
$$\frac{1}{1+u}= 1-2 u + 3 u^2 +\mathcal{O}(u^3)= 1 + \frac{x^2}{3} + \frac{x^4}{15} +\mathcal{O}(x^6)$$
And we see that:
$$\frac{1}{\sin^2(x)} = \frac{1}{x^2} + \frac{1}{3} + \frac{x^2}{15} +\mathcal{O}(x^4)$$
The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $\dfrac{1}{\sin^4(x)}$ by squaring both sides of this expansion, like:
$$\lim_{x\to 0}\left[\frac{1}{\sin^4(x)}-\frac{1}{x^4} - \frac{2}{3 x^2}\right]= \frac{11}{45}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 differential anyways? Is it to multiply $e^\frac{y}{x}$ into $\frac{G}{x}$ and differentiate $e^\frac{y}{x}(3x^2+2xy)$ by $x$ and $e^\frac{y}{x}(2x^2+xy-2y^2)$ by $y$ and prove that both functions are the same?
I'm stuck on this question and would appreciate some help. Thanks!
|
Hint.
Note that
$$
\frac{\partial}{\partial y}\left(\frac{e^{\frac{y}{x}} \left(2 x^2+x y-2 y^2\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\right)}{x^2}\\
\frac{\partial}{\partial x}\left(\frac{e^{\frac{y}{x}} \left(3 x^2+2 x y\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\right)}{x^2}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
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 allowed to use graphs.
As for how many times they intersect, I think it's three because the polynomial is third degree
|
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 three distinct real roots.
Dividing \eqref{1} by $4$
and substituting $t+\tfrac16$ for $x$, we get
a reduced cubic equation
\begin{align}
t^3+pt+q=&0
\tag{2}\label{2}
,
\end{align}
\begin{align}
\text{where}\quad
p&=-\tfrac{25}{12}
,\quad
q=-\tfrac{16}{27}
.
\end{align}
Applying
Trigonometric solution for three real roots
to \eqref{2},
we can find the roots as
\begin{align}
t_k&=
2\sqrt{-\frac{p}3}
\cos\left(\frac1{3}\arccos
\left({\frac{3q}{2p}}\sqrt{-\frac3p}\right)
-\frac{2\pi k}3\right)
\quad \text{for}\quad k=0,1,2
,\\
\end{align}
\begin{align}
t_0&=\phantom{-}\tfrac53\cos(\tfrac13\arccos(\tfrac{64}{125}))
\approx 1.56878
,\\
t_1&=-\tfrac53\cos(\tfrac13\arccos(\tfrac{64}{125})+\tfrac\pi3)
\approx -0.29702
,\\
t_2&=-\tfrac53\sin(\tfrac13\arccos(\tfrac{64}{125})+\tfrac\pi6)
\approx -1.27176
.
\end{align}
Then the corresponding roots of \eqref{1} are
\begin{align}
x_0&=t_0+\tfrac16\approx 1.73545
,\\
x_1&=t_1+\tfrac16\approx -0.13036
,\\
x_2&=t_2+\tfrac16\approx -1.10509
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
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}\Big) \mathrm{d} y\\
&= \frac{1}{8}y + \frac{1}{8y} + C \\
&= \frac{1}{8}\sqrt{\frac{1-3x}{x+1}} + \frac{1}{8}\sqrt{\frac{x+1}{1-3x}} + C\\
&= \frac{1-x}{4\sqrt{(1-3x)(x+1)}} + C.
\end{align}
Euler's 3rd substitution, see: https://en.wikipedia.org/wiki/Euler_substitution
|
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|
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 equation,
$$
(x^2)p(p-1)x^{p-2}+xpx^{p-1}-n^2x^p=0
$$
So I have
$$
x^p(p^2-n^2)=0
$$
And there, i'm stuck.
My answer is
$$
C_1x^{p_1}+C_2x^{p_2}=y(x)
$$
But is this correct? :'(
|
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,
$$\begin{align}
\ \frac{dy}{dx} & =\frac{dt}{dx}\cdot\frac{dy}{dt} \\
& = \frac{d}{dx}\left(\ln(x)\right)\cdot\frac{dy}{dt} \\
& = \frac{1}{x}\cdot\frac{dy}{dt} \\
& = e^{-t}\cdot\frac{dy}{dt} \\
\end{align}$$
and
$$\begin{align}
\ \frac{d^2y}{dx^2} & =\frac{d}{dx}\left(\frac{dt}{dx}\cdot\frac{dy}{dt}\right) \\
& = \frac{d}{dx}\left(\frac{1}{x}\cdot\frac{dy}{dt}\right) \\
& = \frac{d}{dx}\left(\frac{1}{x}\right)\cdot\frac{dy}{dt}+\left(\frac{1}{x}\right)^2\cdot\frac{d^2y}{dt^2} \\
& = \frac{1}{x^2}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right) \\
& = e^{-2\cdot t}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)
\end{align}$$
Substitution into the differential equation gives
$$\begin{aligned} e^{2\cdot t}\cdot e^{-2\cdot t}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)+e^t\cdot e^{-t}\cdot\frac{dy}{dt}-n^2\cdot y&=0 \\
\frac{d^2y}{dt^2}-n^2\cdot y&=0 \\
\end{aligned}$$
Assume that $y\propto e^{\lambda\cdot t}$ for some constant $\lambda$.
Substitution into the differential equation gives
$$\begin{aligned} \frac{d^2}{dt^2}(e^{\lambda\cdot t})-n^2\cdot e^{\lambda\cdot t}&=0 \\
(\lambda^2-n^2)\cdot e^{\lambda\cdot t}&=0 \\
\end{aligned}$$
Because $e^{\lambda\cdot t}\ne 0$ for any finite $\lambda$, $\lambda^2-n^2=0\rightarrow \lambda_{1,2}=\pm n$.
Therefore,
$$y(t)=\text{c}_1\cdot e^{-n\cdot t}+\text{c}_2\cdot e^{n\cdot t}$$
Let $t=\ln(x)$:
$$\begin{align}
\ y(x) & =\text{c}_1\cdot e^{-n\cdot\ln(x)}+\text{c}_2\cdot e^{n\cdot\ln(x)} \\
& = \text{c}_1\cdot e^{\ln(x^{-n})}+\text{c}_2\cdot e^{\ln(x^{n})} \\
& = \text{c}_1\cdot x^{-n}+\text{c}_2\cdot x^{n} \\
\end{align}$$
where $n\in\mathbb{Z}$.
|
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|
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} \equiv 1.96\\
\sqrt 2 & = 1.41^{2} \equiv 1.9881\\
\sqrt 2 & = 1.414^{2} \equiv 1.999396\\
& \ldots
\end{align}
First I tell whether it passes such that $1.x^{2}$ would be not greater than 3.
If that passes, I will add a new decimal to it. Let's say $y.$ $1.xy^{2}$
If that y fails, I increment $y$ by 1 and square it again.
The process will keep repeating. Unfortunately, the process takes so much time.
|
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:
$$ \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \ddots}}}. $$
(This follows from the fact that if $x = \sqrt{2}-1$, then $x = \sqrt{2}-1 = \frac{1}{\sqrt{2}+1} = \frac{1}{2+x}$.)
Now, from this we can calculate subsequent rational approximations to $\sqrt{2}$:
$$
\begin{matrix}
& & 1 & 2 & 2 & 2 & 2 & 2 & \cdots \\
0 & 1 & 1 & 3 & 7 & 17 & 41 & 99 & \cdots \\
1 & 0 & 1 & 2 & 5 & 12 & 29 & 70 & \cdots
\end{matrix} $$
So, for example $\frac{99}{70} \approx 1.4142857$ whereas $\sqrt{2} \approx 1.4142136$.
(It also happens that this procedure generates solutions to Pell's equation $a^2 - 2 b^2 = \pm 1$; for example, $99^2 - 2 \cdot 70^2 = 1$. The connection is: if $a^2 - 2 b^2 = \pm 1$ then $a - b \sqrt{2} = \pm \frac{1}{a + b \sqrt{2}}$; so if $a$ and $b$ are large positive integers satisfying Pell's equation, then $a - b\sqrt{2} \approx \pm\frac{1}{2a}$ which implies $\frac{a}{b} - \sqrt{2} \approx \pm\frac{1}{2ab} \approx \pm\frac{1}{a^2\sqrt{2}}$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
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 not be at $1$st, $2$nd, and $3$rd place.
If we fix $6$ at fourth place then all three $1$'s are fixed at first three places and there are $4!$ such sequences possible.
How to calculate for $6$ be fixed at $5$th to $8$th place?
|
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 $1$s and the $6$, which can be done in $\binom{8}{4}$ ways. There is only one way to arrange them in these positions so that all the $1$s appear before the $6$. The remaining four distinct numbers may be placed in the remaining four positions in $4!$ ways. Hence, there are
$$\binom{8}{4}4! = \frac{8!}{4!}$$
admissible arrangements of the numbers.
Method 3: First, we count the number of distinguishable arrangements, then use symmetry to determine the number of admissible arrangements.
Number of distinguishable arrangements: Choose three of the eight positions for the $1$s, which can be done in $\binom{8}{3}$ ways. The remaining five distinct numbers can be placed in the remaining five positions in $5!$ ways. Hence, the number of distinguishable arrangements of $1, 1, 1, 2, 3, 4, 5, 6$ is
$$\binom{8}{3}5! = \frac{8!}{4!}$$
Number of admissible arrangements: By symmetry, in one fourth of these arrangements will the $6$ appear after all three $1$s. Hence, the number of admissible arrangements is
$$\frac{1}{4}\binom{8}{3}5! = \frac{8!}{4!}$$
|
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|
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 the shaded area as the solutions, but I still don't understand how the solution is $[-6,-3] \cup [-2, \infty]$. I understand $[-6,-3]$ but not the $\infty$ part. Yes, I can do this by plugging in values and checking if the solutions work but that is not efficient. What am I doing wrong? I appreciate anyone's help.
https://www.desmos.com/calculator/7j8yamvbzv
This is my graph I did to find the solutions.
|
$$|x + 5| = \begin{cases}
x + 5 & \text{if $x \geq -5$}\\
-x - 5 & \text{if $x < -5$}
\end{cases}
$$
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 $x = -2$ or $x = -3$. It is positive when the factors are both positive or both negative. Both factors are positive when $x > -2$. Both factors are negative when $x < -3$. Hence, if $x \geq -5$, the inequality is satisfied when $x \leq -3$ or $x \geq -2$.
Since we require that $x \geq 5$ and $x \leq -3$ or $x \geq 5$ and $x \geq -2$, $-5 \leq x \leq -3$ or $x \geq -2$. In interval notation, we write
\begin{align*}
\{x \in \mathbb{R} \mid -5 \leq x \leq -3\} & = [-5, -3]\\
\{x \in \mathbb{R} \mid x \geq -2\} & = [-2, \infty)\\
\end{align*}
Therefore, if $x \geq -5$ and satisfies the inequality $x|x + 5| \geq -6$, then
$$x \in [-5, -3] \cup [-2, \infty)$$
Case 2: $x < -5$
\begin{align*}
x|x + 5| & \geq -6\\
x(-x - 5) & \geq -6\\
-x^2 - 5x & \geq -6\\
x^2 + 5x & \leq 6\\
x^2 + 5x - 6 & \leq 0\\
(x + 6)(x - 1) & \leq 0
\end{align*}
The expression on the left-hand side equals zero when $x = -6$ or $x = 1$. It is negative when the two factors have opposite signs, which occurs when $-6 < x < 1$. Hence, the inequality is satisfied if $x < -5$ and $-6 \leq x \leq 1$, so $-6 \leq x < -5$. In interval notation, we write
$$\{x \in \mathbb{R} \mid -6 \leq x < -5\} = [-6, -5)$$
Therefore, if $x < -5$ and satisfies the inequality $x|x + 5| \geq -6$, then
$$x \in [-6, 5)$$
Solution: The solution of the absolute value inequality $x|x + 5| \leq -6$ is the union of the solutions for the cases $x \geq 5$ and $x < 5$, so we obtain the solution set
$$S = [-5, -3] \cup [-2, \infty) \cup [-6, -5) = [-6, -3] \cup [-2, \infty)$$
|
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|
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 know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
|
Hint: the real roots are the roots of the first factor:
$$
\begin{align}
x^8\color{red}{+2x^4-2x^4}-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \\
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
\end{align}
$$
|
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|
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:
$$
P(x) = 3x^2(x+2) \implies 3(0)^2(0+2)= 0,
$$
$$
\lim_{x \to 0}x\frac{Q(x)}{P(x)} = \lim_{x \to 0}x\frac{7x}{3x^2(x+2)} = \frac{7}{6} \text{ and}
$$
$$
\lim_{x \to 0}x^2\frac{R(x)}{P(x)} = \lim_{x \to 0}x^2\frac{-2}{3x^2(x+2)} = -\frac{1}{3}
$$
are both finite. Thus we assume a solution of the form
$$
y = \sum_{n=0}^\infty a_n x^{n+r}, y' = \sum_{n=0}^\infty a_n(n+r) x^{n+r-1}, \\y''= \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r-2}
$$
Which, after some algebra, yields the indicial equation
$$
(2r+1)(3r+5)=0 \implies r_1 = -\frac{1}{2}, r_2 = -\frac{3}{5}
$$
And the recurrence relation
$$
a_n=-\frac{-3a_{n-1}(n-3/2)(n-5/2)}{(6n+1)(n-1)}, \text{for} \ n \geq 1
$$
and $a_0$ is arbitrary. However, I soon realized that $a_1$ is undetermined as the denominator becomes 0. I was wondering if I got my algebra wrong or the entire approach is not correct. Thanks
|
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
$$
a_n=-\frac{3(n+r-1)(n+r-2)}{(2n+2r-1)(3n+3r+2)}.
$$
The recursion for $r=\frac12$ reduces to
$$
a_n=-\frac{3(2n-1)(2n-3)}{4n(6n+7)}a_{n-1},
$$
and for $r=-\frac23$
$$
a_n=-\frac{(3n-5)(3n-8)}{3n(6n-7)}a_{n-1}.
$$
This gives totally problem free recursions for the coefficient sequences of the two basis solutions.
|
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|
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^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3)\, dy dx &= \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}( x^2 )\, dy dx \\
&=\int_{0}^{1} 2x^2\sqrt{1-x^2}\, dx\\
&= \int_{0}^{\frac{\pi}{2}}2\sin^2\theta\cos^2\theta \,d\theta\\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin^22\theta \,d\theta\\
&= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-\cos 4\theta) \,d\theta\\
&= \frac{\pi}{8}.\end{align*}
Am I right? I think it's a little bit too complicated than what it should be?
|
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 equal to$$\left(\int_0^1r^3\,\mathrm dr\right)\left(\int_{-\frac\pi2}^\frac\pi2\cos^2(\theta)\,\mathrm d\theta\right)=\frac\pi8.$$
|
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|
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 start other than:
$l = w + 6$
$84 = lw$
|
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*}
(6+W)W = 84 \Longleftrightarrow W^{2} + 6W - 84 = 0
\end{align*}
Can you proceed from here?
EDIT
Observe that $W^{2} + 6W = (W^{2} + 6W + 9) - 9 = (W+3)^{2} - 9$, from whence we obtain
\begin{align*}
W^{2} + 6W - 84 = 0 \Longleftrightarrow (W+3)^{2} - 93 = 0 \Longleftrightarrow (W+3)^{2} = 93 \Longleftrightarrow W = \pm\sqrt{93} - 3
\end{align*}
Since $W$ must be positive, we conclude that $W = \sqrt{93}-3$ and $L = \sqrt{93} + 3$.
Another possible approach is to apply the Bhaskara's formula, which is given by:
\begin{align*}
ax^{2} + bx + c = 0 \Longleftrightarrow x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
\end{align*}
|
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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 $a^2+b^2 \gt a+b$ we get $a^2-a \gt b-b^2$
Since $a \geq b$ we can get $a^3-a^2 \gt b^2-b^3$ $\Rightarrow$ $a^3+b^3 \gt a^2+b^2$
If this solution is incorrect, please explain why and attach the correct solution. Thank you.
|
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 inequality uses the assumption $a^2+b^2>a+b$. As a consequence, $a^3+b^3>a^2+b^2$.
|
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|
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 associative if $+$ and $\cdot$ are. Moreover, $0$ is an identity.
Naturally this then lead me to explore what would happen if we used this operation to create a semigroup structure on $\mathbb{Z}_{n}$, where all of the arithmetic is carried out modular $n$. Here are a few of the semigroup tables for some small values of $n$:
$$n=4 \quad \begin{array}{|c|c|c|c|c|}
\hline
\hline
\odot & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} \\ \hline
\mathbf{0} & 0 & 1 & 2 & 3 \\ \hline
\mathbf{1} & 1 & 3 & 1 & 3 \\ \hline
\mathbf{2} & 2 & 1 & 0 & 3 \\ \hline
\mathbf{3} & 3 & 3 & 3 & 3
\end{array} \quad\quad\quad n=5 \quad \begin{array}{|c|c|c|c|c|c|}
\hline
\hline
\odot & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline
\mathbf{0} & 0 & 1 & 2 & 3 & 4 \\ \hline
\mathbf{1} & 1 & 3 & 0 & 2 & 4 \\ \hline
\mathbf{2} & 2 & 0 & 3 & 1 & 4 \\ \hline
\mathbf{3} & 3 & 2 & 1 & 0 & 4 \\ \hline
\mathbf{4} & 4 & 4 & 4 & 4 & 4 \\
\end{array}$$
$$
n=6 \quad \begin{array}{|c|c|c|c|c|c|c|}
\hline
\hline
\odot & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline
\mathbf{0} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
\mathbf{1} & 1 & 3 & 5 & 1 & 3 & 5 \\ \hline
\mathbf{2} & 2 & 5 & 2 & 5 & 2 & 5 \\ \hline
\mathbf{3} & 3 & 1 & 5 & 3 & 1 & 5 \\ \hline
\mathbf{4} & 4 & 3 & 2 & 1 & 0 & 5 \\ \hline
\mathbf{5} & 5 & 5 & 5 & 5 & 5 & 5
\end{array} \quad\quad\quad n=7 \quad \begin{array}{|c|c|c|c|c|c|c|c|}
\hline
\hline
\odot & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} \\ \hline
\mathbf{0} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\mathbf{1} & 1 & 3 & 5 & 0 & 2 & 4 & 6 \\ \hline
\mathbf{2} & 2 & 5 & 1 & 4 & 0 & 3 & 6 \\ \hline
\mathbf{3} & 3 & 0 & 4 & 1 & 5 & 2 & 6 \\ \hline
\mathbf{4} & 4 & 2 & 0 & 5 & 3 & 1 & 6 \\ \hline
\mathbf{5} & 5 & 4 & 3 & 2 & 1 & 0 & 6 \\ \hline
\mathbf{6} & 6 & 6 & 6 & 6 & 6 & 6 & 6
\end{array}
$$
It's straightforward to prove that $\odot$ always fails to be a group operation on $\mathbb{Z}_n$ because there is no inverse element for $n-1$. However, after investigating a number of tables I started to notice a few things, and made the following conjectures:
*
*When $n$ is prime, suddenly $\odot$ seems to become a group operation, although not on the enitre set. Rather, it seems to become a group action on the subset $\mathbb{Z}_{n-1}$.
*Also, the above condition seems to be necessary. That is, when $n$ is composite $\odot$ fails to be a group operation (again, missing inverses) on $\mathbb{Z}_{n-1}$.
I'm pretty lost when trying to prove these. Moreover, should these conjectures be true, I would love to know what finite groups these structures are representing. Does anyone have any suggestions? My algebra is quite rusty.
|
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 corresponding semigroup, since $x\odot(-1) = x-1 + x(-1) = -1$; hence you definitely need to exclude that element).
Your conjecture is indeed true.
You already know that $\odot$ is commutative and associative, so it’s just a matter of determining when do elements have inverses.
Suppose first that $n$ is prime, and let $x$, $0 < x <n-1$. You are looking for $y$ such that $x+y+xy = 0$; this is equivalent to $y(1+x)\equiv x \pmod{n}$. Since $x\not\equiv -1\pmod{n}$, then $x+1\not\equiv 0\pmod{n}$, so it has a multiplicative inverse $z$ modulo $n$. Thus, $y \equiv xz\pmod{n}$ will be an inverse for $x$. Thus, in this case you get a group.
Suppose next that $n$ is not a prime, and let $d$ be a proper divisor of $n$, so that $1 < d < n - 1$. Let $x=d-1$. Then as above, an inverse $y$ for $x$ would have to satisfy $y(x+1)\equiv x\pmod{n}$, or $yd\equiv d-1\pmod{n}$. Since $d$ divides $n$, this implies $yd\equiv d-1\pmod{d}$, which requires $-1\equiv 0\pmod{d}$, which is only possible if $d=1$, which cannot hold. Thus, $x$ does not have a multiplicative inverse, and so what you have is not a group.
Now as to the structure, you are getting cyclic groups so far, generated by $2$ in both cases. I’d have to think about it to see if this is indeed the case.
|
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|
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:
$$\int\frac{\ln u}{2u}du$$
and then substituting $v=\ln u$, which means $dv = \frac{1}{u} du$, and therefore:
$$\frac{1}{2}\int dv = \frac{v}{2} + c = \frac{\ln u}{2} +c = \frac{\ln (x^2+1)}{2}+c$$
However verifying this online from various sources, they indicate that the solution is in fact:
$$\int\frac{x\ln (x^2+1)}{x^2+1}dx = \frac{\ln(x^2+1)^2}{4}+c$$
because they substitute $u = \ln(x^2+1)$ which gives $du = \frac{2x}{x^2+1} dx$, which is clairly a shorter path.
But my question is, why does both of these substitution techniques not yield the same result? Where lies my error?
|
$$I=\int\frac{x\ln(x^2+1)}{x^2+1}dx$$
$$u=x^2+1$$
$$dx=\frac{du}{2x}$$
$$I=\frac{1}{2}\int\frac{\ln(u)}{u}du$$
$$v=\ln(u)$$
$$du=udv$$
$$I=\frac{1}{2}\int vdv=\frac{v^2}{4}+C=\frac{\ln^2(u)}{4}+C=\frac{\ln^2(x^2+1)}{4}+C$$
|
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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 -
$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{n+1}$
How to approach to prove this summation equality?, I could owrite the LHS as $2n - 3n(n-1) + 2n(n-1)(n-2) +...$ after simplification.
|
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
$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} \frac{r+1}{n+1} {n+1\choose r+1}
= \frac{1}{n+1}
\sum_{r=1}^n \frac{(-1)^{r+1}}{r} {n+1\choose r+1}
\\ = \frac{1}{n+1}
\sum_{r=2}^{n+1} \frac{(-1)^r}{r-1} {n+1\choose r}.$$
Working with the sum term and introducing
$$f(z) = (-1)^{n+1}
\frac{(n+1)!}{z-1} \prod_{q=0}^{n+1} \frac{1}{z-q}$$
we have for $2\le r\le n+1$ that
$$\sum_{r=2}^{n+1} \mathrm{Res}_{z=r} f(z)
= \sum_{r=2}^{n+1} (-1)^{n+1} \frac{(n+1)!}{r-1}
\prod_{q=0}^{r-1} \frac{1}{r-q} \prod_{q=r+1}^{n+1} \frac{1}{r-q}
\\ = \sum_{r=2}^{n+1} (-1)^{n+1} \frac{(n+1)!}{r-1}
\frac{1}{r!} \frac{(-1)^{n-r+1}}{(n+1-r)!}
= \sum_{r=2}^{n+1} \frac{(-1)^{r}}{r-1} {n+1\choose r},$$
so this is the desired sum. Now residues sum to zero and the residue
at infinity is zero since $\lim_{R\to\infty} 2\pi R / R / R^{n+2} =
0.$ The remaining poles are at $z=0$ and $z=1.$ We write
$$f(z) = (-1)^{n+1} (n+1)! \frac{1}{z} \frac{1}{(z-1)^2}
\prod_{q=2}^{n+1} \frac{1}{z-q}$$
to get
$$\mathrm{Res}_{z=0} f(z)
= (-1)^{n+1} (n+1)! \frac{(-1)^n}{(n+1)!} = -1$$
as well as
$$\mathrm{Res}_{z=1} f(z)
= (-1)^{n+1} (n+1)!
\left. \left(-\frac{1}{z^2} \prod_{q=2}^{n+1} \frac{1}{z-q}
- \frac{1}{z} \prod_{q=2}^{n+1} \frac{1}{z-q}
\sum_{q=2}^{n+1} \frac{1}{z-q} \right) \right|_{z=1}
\\ = (-1)^{n+1} (n+1)!
\left(- \frac{(-1)^n}{n!}
+ \frac{(-1)^n}{n!} H_n \right)
= n+1 - (n+1) H_n.$$
Restoring the multiplier in front we thus obtain
$$\frac{1}{n+1}
(- \mathrm{Res}_{z=0} f(z) - \mathrm{Res}_{z=1} f(z) )
= \frac{1}{n+1} (1 + (n+1) H_n - (n+1))
\\ = H_n - \frac{n}{n+1} = H_{n+1} - 1.$$
|
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|
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 for each day and adding them together, but of no result.
|
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 $$K = \frac{\frac{7!}{2!2!3!}}{7^7}\cdot 7^6 = \frac{\frac{7!}{2!2!3!}}{7}$$
$$K = \frac{6!}{2!2!3!} = 30$$
|
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|
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:
$$\left |\sqrt{n^2 + 2} - \sqrt{n^2 - 2} - 0\right | = \left(\sqrt{n^2 + 2} - \sqrt{n^2 - 2}\right)\cdot \frac{\sqrt{n^2 + 2}+\sqrt{n^2 - 2}}{\sqrt{n^2 + 2}+ \sqrt{n^2 - 2}}$$
$$\ \ = \frac{n^2 + 2 - n^2 + 2}{\sqrt{n^2 + 2}+\sqrt{n^2 - 2}}$$
$$\ \ = \frac{4}{\sqrt{n^2+2}+ \sqrt{n^2-2}}$$ (Because $\sqrt{n^2 - 2} > 0$)
$$< \frac{4}{\sqrt{n^2+2}} < \frac{2}{n} \ \ \ \ \ \ \ \ $$
We must find $N \in \mathbb{N}$ s.t. $\frac{2}{N} < \varepsilon \iff N > \frac{2}{\varepsilon}$
Fix $\varepsilon > 0$.
By the Archimedean Property, find $N \in \mathbb{N}$ with $N > \frac{2}{\varepsilon}$
Then if $n \geq N$,
$$|a_n-L| = \left |\sqrt{n^2 + 2} - \sqrt{n^2 - 2} - 0\right | < \frac{2}{n} \leq \frac{2}{N} < \varepsilon \ \ \ \text{(From Above)}$$
Thus proving, $$\lim_{n \to \infty} \sqrt{n^2 + 2} - \sqrt{n^2 -2} =0 .$$
|
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 $N = 2 + \lceil \frac2{\epsilon}\rceil .$
|
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|
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 which method should I use
|
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:
$$
\begin{align}
\sqrt[3]{1+2x} + 1 &=
\frac{1 + 2x + 1}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} =
\frac{2(x + 1)}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} \\
\sqrt{2+x} + x &=
\frac{2 + x - x^2}{\sqrt{2+x} - x} =
-\frac{(x + 1)(x - 2)}{\sqrt{2+x} - x}
\end{align}
$$
Therefore, your limit is equal to
$$
\lim_{x\to{-1}} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x} =
\lim_{x\to{-1}} -\frac{2(\sqrt{2+x} - x)}{(x - 2)({\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1)},
$$
which is pretty straightforward to compute.
|
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|
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 use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
|
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 $a^2$.
|
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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 find a pattern, which is:
$$S_n = \frac{S_{n-1}}{1-x^{2^n}} + \frac{x^{2^n}}{1-x^{2^{n+1}}}$$
I am not sure how to proceed onwards to solve this recurrence relation. Is there a clever trick I can do to solve it?
|
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 identity $a^2-b^2=(a-b)(a+b)$ with $a=1$ and $b=x^{2^{j-1}}$.
*In (2) we apply the telescoping series.
|
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|
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 roots.)
a) $z^3-z^{-3}=i$
$\frac{z^6-1}{z^{3}}=i$
$z^6-1=iz^3$
$z^3(z^3-i)=1$
1. $z^3=1$
$z=cis(\frac{2π+2kπ}{3})$, k=0,1,2
2. $z^3-i=1$
$z^3=\sqrt{2}cis(\frac{π}{4})$
$z=2^{\frac{1}{6}}cis(\frac{\frac{π}{4}+2kπ}{3})$, k=0,1,2
b) $(z-i)^6=-64$
$\sqrt{((z-i)^3)^2}=\sqrt{-64}$ square root to both sides
$(z-i)^3=\sqrt{-1*64}$
$(z-i)^3=8i$
$z-i=2cis(\frac{\frac{π}{2}+2kπ}{3})$
$z=2cis(\frac{5kπ}{6})+i=2cis(\frac{5kπ}{6})+cis(\frac{π}{2})$, k=0,1,2 ?
c) $z^5(1+i)=\bar z $ multiplying both sides by z
*$z^6(1+i)=\mid{z}\mid^2$ using absolute value on both sides
$\mid{z}\mid^6\mid{1+i}\mid=\mid{z}\mid^2$ dividing both sides by $\mid{z}\mid^2$
$\mid{z}\mid^4=\frac{1}{\mid{1+i}\mid}=\frac{1}{\sqrt{2}}$
$\mid{z}\mid=\frac{1}{2^\frac{1}{8}}$ then using * again
$z^6(1+i)=\frac{1}{2^\frac{1}{4}}$
$z^6=\frac{1}{2^\frac{1}{4}(1+i)}$
Don't know how to proceed.
|
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.
|
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|
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 for a slightly stronger claim:
$$
{{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}
$$
Indeed,
$$
{{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}
<\frac{2(2n+1)}{n+1}\frac{4^n}{\sqrt{2n+1}}
$$
So, we only have to prove that
$$
\frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}}
$$
which is true for all $n\ge0$.
|
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|
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 $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
|
$$f(z)=\left|z-\frac{1}{z}\right|=\frac{\left|z^2-1\right|}{2}$$
since we know $|z|=2$
|
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|
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
$$
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 2 \\
0 & 0 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 5
\end{pmatrix} \cdot (-1)^2
$$
|
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, $\det(A) = \det(A^T)$.
|
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|
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 $\Delta = b^2-4ac$ method. Use Vieta's formulas: $$S=\alpha+\beta=-\frac{b}{a} \qquad P=\alpha\beta=\frac{c}{a}$$
Actually I want to solve the main entry using difference of squares! Such as (x-a)(x+a)
And I note everybody that I know basical rules of quadratic equations like delta and etc.
|
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 $x=1$.
|
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|
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
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}=
\sqrt[4]{n^4+2n^2}-q(n)=\frac{n^4+2n^2-q^4(n)}{(\sqrt[4]{n^4+2n^2}+q(n))\,(\sqrt{n^4+2n^2}+q^2(n))}$$
and I thought of getting $q(n)=n$, but in that case
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\sim \frac{1}{n}$$
and in that case we get divergence, by limit comparison test.
Thanks in advance.
|
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}}\\
&=n\left(1+\frac a{3n}+\frac b{3n^2}-\frac a{9n^2}+O\left(\frac 1{n^3}\right)\right)\\
&=n+\frac a3+\left(\frac b3-\frac a9\right)\frac 1n+O\left(\frac 1{n^2}\right)\\
\end{align}
Comparing we get $\frac a3=0$ and $\frac b3-\frac a9=\frac 12$ which gives $a=0$ and $b=\frac 32$, hence $p(n)=n^3+\frac 32n+c$ for all $c\in\Bbb R$.
|
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|
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 $x = 2+\epsilon$, $y \to \dfrac{3(2+\epsilon)^2+4}{(2+\epsilon)-2} = \dfrac{K}{\epsilon} \to \infty$
Therefore, $\text{ran}(g(x)) = (-\infty, \infty)$
have I done this right or am I missing something? For instance, I thought about the case when $x \to \infty$, but this looks like $y \to 1$ in this case, nothing remarkable. I don't even know if that is what we say when we have an $\frac{\infty^2}{\infty}$....Do we just say that is one? Or do we say that is $\infty$ still?
|
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 $y^2-24y+144-192 = (y-12)^2-192 \geq 0$, i.e. $\lvert y-12 \rvert \geq \sqrt{192}$. It is clear that this is not true for every $y$, since for example $y=0$ does not satisfy the inequality. So the range is not $\mathbb{R}$, but is given by those points $y$ satisfying $\lvert y-12 \rvert \geq \sqrt{192}$.
One more comment on your last observation. If $x \to \infty$ then $$g(x) = \frac{x^2\left(3+\frac{4}{x^2}\right)}{x\left(1-\frac{2}{x}\right)} = \frac{x\left(3+\frac{4}{x^2}\right)}{\left(1-\frac{2}{x}\right)} \to \frac{\infty \cdot 3}{1}=\infty.$$
|
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|
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 the number of ways.
No. of ways to choose one question from each section = (5C1)^3
No. of questions remaining = 12
No. of ways to pick 2 questions from remaining 12 questions = 12 * 11
Total number of ways = (5C1)^3 * 12 * 11
Can somebody tell me where I'm going wrong
|
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.
In total he can make $3\cdot \left(\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}+ \binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}\right)=2250$ choices.
For every section we have one path. We choose 3 question from each section. For every path we have to add ($\text{not multiply}$) the ways
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)$
Now we make a case decision.
a) We choose 2 questions from one (other) section and 2 questions from the remaining section.
$ \binom{5}{2}\cdot \binom{5}{2}$
b) We choose 1 questions from one (other) section and 3 questions from the remaining section.
$ \binom{5}{1}\cdot \binom{5}{3}$
Therefore in total we have
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)\cdot \left(\binom{5}{2}\cdot \binom{5}{2}+ \binom{5}{1}\cdot \binom{5}{3}\right)=2250$
|
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|
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
$S_{n+1}$: $\frac{2^{4(n+1)}-(-1)^{n+1}}{17} = \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} = \frac{2^{4n} \cdot (2^4+1) - 2^{4n} +(-1)^{n}}{17}$.
Now, we are left with $2^{4n} - \frac{2^{4n}-(-1)^n}{17}$.
We know $\frac{2^{4n}-(-1)^n}{17}$, $2^{4n} \in \mathbb{N}$ and $2^{4n} > \frac{2^{4n}-(-1)^n}{17} \implies 2^{4n} - \frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
$$\tag*{$\blacksquare$}$$
Is my proof correct? I also feel there is a more elegant/smoother proof (assuming mine is correct).
|
Alternatively, assuming $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$:
$$\begin{align}\frac{2^{4(n+1)}-(-1)^{n+1}}{17} &= \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} =\\
&=\frac{2^{4} \cdot (2^{4n}-(-1)^n) + 2^{4}(-1)^n +(-1)^{n}}{17}=\\
&=2^4\cdot \underbrace{\frac{2^{4n}-(-1)^n}{17}}_{\ge 1}+(-1)^{n}\in \mathbb N.
\end{align}$$
|
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|
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? The digits are from $0-9$.
My solution:
_A_A_A_ thus $x_1 + x_2 + x_3 + x_4 = 7$, where $x_2 \geq 1$ and $x_3 \geq 1$
Therefore, there are $7-1-1 = 5$ (identical) digits to put in $4$ distinct spots.
Using combination with repetition formula: $(5+4-1)C5 = 8C5$
Then since there are $10$ digits, with repetition allowed, there are $10^7$ ways.
Answer: $8C5 \cdot 10^7$
But apparently this answer is wrong. Can someone please provide the correct step-by-step solution for this problem and where I went wrong?
|
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 must subtract those arrangements in which at least one pair of $A$s are adjacent.
A pair of $A$s are adjacent: We have to arrange nine objects: $A$, $AA$, and seven digits. There are nine ways to choose the position of the $A$ and eight ways to choose the position of the $AA$. Each of the remaining seven positions can be filled with a digit in $10$ ways. Hence, there are
$$9 \cdot 8 \cdot 10^7$$
such arrangements.
However, if we subtract the number of arrangements with a pair of $A$s from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of consecutive $A$s twice, once for each way of designating one of the pairs as the pair of consecutive $A$s. We only want to subtract such arrangements once, so we must add them back.
Two pairs of adjacent $A$s: Since we only have three $A$s, this can only occur if we have the block $AAA$. We have eight objects to arrange: $AAA$ and the seven digits. There are eight ways to place the block $AAA$. Each of the remaining seven places can be filled with a digits in $10$ ways. Hence, there are
$$8 \cdot 10^7$$
such arrangements.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$\binom{10}{3}10^7 - 9 \cdot 8 \cdot 10^7 + 8 \cdot 10^7 = \left[\binom{10}{3} - 9 \cdot 8 + 8\right]10^7 = (120 - 72 + 8)10^7 = 56 \cdot 10^7 = \binom{8}{5}10^7$$
Method 2: We arrange the seven digits, then insert the $A$s in the spaces between them or at the ends of the row.
Since we have $10$ choices for each of the seven digits, we can arrange seven digits in a row in $10^7$ ways. This creates eight spaces, six between successive digits and two at the ends of the row.
$$\square d_1 \square d_2 \square d_3 \square d_4 \square d_5 \square d_6 \square d_7 \square$$
To separate the $A$s, we must choose three of these eight spaces in which to place an $A$, which can be done in $\binom{8}{3}$ ways. Hence, there are
$$\binom{8}{3}10^7 = \binom{8}{8 - 3}10^7 = \binom{8}{5}10^7$$
admissible arrangements.
|
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|
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 was given but I don't know how to prove the lemma.
Lemma : Let, $a_1,a_2,a_3, \ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+\cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)\cdots(1+a_n)$$ $$\le 1+s+\dfrac{s^2}{2!}+\dfrac{s^3}{3!}+\cdots+\dfrac{s^n}{n!}$$
|
It can be easily shown that for any $x\in(0,1)$ we have
$$ \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9765}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9765}} <1+x < \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9450}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9450}} $$
and in general, by setting $D_m=\prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+\sum_{k=1}^{n}\frac{x^k}{D_k}$,
$$ \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+\frac{x^{n+1}}{D_{n+1}}}<1+x< \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+\frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
By telescoping it follows that
$$ p_n(1)+\frac{1}{D_{n+1}}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<p_n(1)+\frac{1}{D_n(2^{n+1}-2)} $$
and by picking $n=4$ we have
$$ \frac{3326}{1395}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<\frac{22531}{9450} $$
such that the first figures of the middle term are $\color{green}{2.3842}$. By picking $n=10$ we get that the middle term is $\color{green}{2.384231029\ldots}$.
As a continued fraction
$$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)=\left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,\ldots\right]$$
while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots]$.
|
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|
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 $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.
Since $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.
$$
\begin{aligned}
\lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\
&= \frac{5}{6}
\end{aligned}
$$
And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?
|
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
$$S_{3n}=\ln 4n-\frac{\ln 2n}{2}-\frac{\ln n}{2}+O(1/n)=\frac{3\ln2}2
+O(1/n)$$
so the limit is $\frac32\ln2$.
|
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|
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}
$$
\cos18^\circ=\sqrt{1-\frac{6-2\sqrt{5}}{16}}=\frac{\sqrt{10+2\sqrt{5}}}{4}
$$
$$
\sin81^\circ=\cos9^\circ=\sqrt{\frac{1+\cos18^\circ}{2}}=\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}=\frac{\sqrt{8+2\sqrt{10+2\sqrt{5}}}}{4}
$$
How do I proceed further and find the solution ?
Or from here, at least can I just identify the solution as $\sin81^\circ=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$ from the options given ?
|
$\sin81^\circ = \sin(45 + 36)\\
\sin45\cos 36 + \cos 45\sin 36$
$\cos 36 = 1 - 2\sin^2 18 = \frac {1+\sqrt 5}{4}$
$\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 {\sqrt{5-\sqrt5}}{4}$
But they have pulled a dirty trick on you.
$3+\sqrt{5} = \frac 12 (1+\sqrt{5})^2\\
\frac {\sqrt{3+\sqrt{5}}}{4} = \frac {1}{\sqrt 2} \frac {1+\sqrt{5}}{4} = \sin 45\cos 36$
|
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|
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(a+1) = 3k$
I'm just not sure where to go from here. Can anyone help?
|
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 on and then number $m + (k-i) \equiv i+(k-i) \equiv 0 \pmod k$.
(More generally, for any $j; 0 \le j < k-1$ then one of the consecutive integers is equivalent to $j \pmod k$.)
So if $(a-1), a, (a+1)$ are three consecutive numbers then one of them must be divisible by $3$.
So the product of all of them is divisible by $3$ so $(a-1)a(a+1) \equiv 0 \pmod 3$.
....
Another less elegant but more direct one would be to note that either $a \equiv 0 \pmod 3$ or $a \equiv 1 \pmod 3$ or $a \equiv 2 \equiv -1 \pmod 3$.
So either $a^3 \equiv 0^3 = 0\equiv a \pmod 3$ or $a^3 \equiv 1^3 =1 \equiv a\pmod 3$ or $a^3 \equiv 2^3 = 8\equiv 2 \equiv a \pmod 3$.
Those are the only three options.
....
Or if you really want a basic argument. Let $a = 3k +i$ for some $k$ and $i = 0, 1, $ or $2$.
Then $a^3 = (3k + i)^3 = 27k^3 + 27k^2i + 9ki^2 + i^3 \equiv i^3 \pmod 3$. And it's easy to verify $i^3 \equiv i \pmod 3$ if $i=0,1,2$.
|
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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}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
|
Claim:
For all positive integers $n$, we have
$$\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{*}$$
holds for all positive integers $n$.
To prove $(*)$, proceed by induction on $n$.
By direct evaluation, $(*)$ holds for the base case $n=1$.
Suppose $(*)$ holds for some positive integer $n$.
\begin{align*}
\text{Then}\;\;&\sum_{k=1}^{n+1} \ln\left(1+\frac{1}{k^3}\right)\\[4pt]
=\;&
\left(
\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right)
\right)
+
\ln\left(1+\frac{1}{(n+1)^3}\right)
\\[4pt]
<\;&
\left( 1-\frac{1}{(n+1)^2}\right)
+
\ln\left(1+\frac{1}{(n+1)^3}\right)
&&\text{[by the inductive hypothesis]}
\\[4pt]
<\;&
\left( 1-\frac{1}{(n+1)^2}\right)
+
\frac{1}{(n+1)^3}
&&\text{[since $\ln(1+x) < x$, for all $x > 0$]}
\\[4pt]
=\;&\left(1-\frac{1}{(n+2)^2}\right)-\frac{n^2+n-1}{(n+1)^3(n+2)^2}
\\[4pt]
<\;&1-\frac{1}{(n+2)^2}
&&\text{[since $n^2+n-1 > 0$]}
\\[4pt]
\end{align*}
which completes the induction, and thus proves the claim.
|
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|
I would really appreciate some help solving this induction problem!!
$3.$ for all $n\ge1$, $\displaystyle\sum_{i=1}^n(2i)^2=\frac{2n(2n+1)(2n+2)}{6}$
I have
$$\frac{2n(2n+1)(2n+2)(12(n+1)^2)}6= \frac{2n(2n+1)2(n+1)12(n+1)(n+1)}6.$$
I think I need to do something with the $(n+1)$.
However, I'm not sure where to go from here. I know the end goal is:
$$\frac{2(n+1)(2(n+1)+1)(2(n+1)+2)}6.$$
Any help is appreciated!
|
So to prove this we must use induction and since the LHS is a sum, we add $(2(n+1))^2$.
We do this because if $n$ and $n+1$ every natural number works as $n$ is arbitrary. Therefore, we only need to prove it works for $n+1$.
$$\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}(2i)^2=\frac{(4n^2+14n+12)(2n+2)}{6}$$
$4n^2+14n+12$ factors to give:
$$\sum_{i=1}^{n+1}(2i)^2=\frac{(2n+3)(2n+4)(2n+2)}{6}$$
Therefore:
$$\sum_{i=1}^{n+1}(2i)^2=\frac{(2(n+1)+2)(2(n+1)+1)(2(n+1))}{6}$$
All that remains is to check the "base-case" when $n=1$.
This will complete the proof.
|
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|
How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 + 250 + 1, which is why the answer is also 1?
|
$$342342^{1001} \equiv2^{1001} \space \bmod5$$
$$2^{1001}=2 \times (2^4)^{250}\equiv2\times1^{250}\bmod 5=2$$
Case $342343^2 \bmod 3$ is even simpler. You can easily prove that $a^2\equiv 0 \bmod 3$ iff $a\equiv0\bmod3$. In all other cases $a^2\equiv 1 \bmod3$.
Number 342343 is not divisible by 3 so the result must be 1.
|
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|
How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$? Wolfram Alpha shows that $$\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}.$$
I am not sure how to continue from this point. I need some help.
|
HINT
By using the basic trigonometric identity
$$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\right)\right)~dx\\
&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx
\end{align}$$
where within the second integral the substitution $x=\frac x2$ was used. Now use the Fourier series expansion
$$\ln(\sin x)=-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$
to further get
$$\begin{align}
\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\left[-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right]~dx\\
&=\frac{8\pi^3}{3}\ln(2)-16\int_0^{\pi}x^2\ln(2)~dx-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx
\end{align}$$
The second integral can be evaluated by applying integration by parts. Can you finish it from hereon?
|
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If $x_1, x_2$ are roots of $P(x)=x^2-6x+1$, prove that $5 \nmid x_1^n+x_2^n$
Given $x_1$ and $x_2$ are roots of $P(x)=x^2-6x+1$ prove using Binomial expansion that $x_1^n+x_2^n$ is not divisible by 5 if $n\in \Bbb N\setminus\{0\}$.
Source: list of problems for the preparation for math contests.
Notice: this problem is different from other SE question, as it requires a specific proof that was suggested by @DeepSea in a comment in that post, but not developed.
My attempt: By solving the roots of $P(x)$ it is easy to see that
$$x_1=3+2\sqrt{2}\ \ \text{and}\ \ \ x_2=3-2\sqrt{2}$$
and, using the Binomial expansion,
$$x_1^n+x_2^n=(x_1+x_2)^n-\sum_{k=1}^{n-1}{{n-1}\choose {k}} x_1^{n-k}x_2^k=6^n-\sum_{k=1}^{n-1} {{n-1}\choose {k}}x_1^{n-k}x_2^k$$
It is easy to see that $6^n\equiv 1\pmod{5}$ for all possible $n$. Therefore the problem is showing that for $$A(n)=\sum_{k=1}^{n-1}{{n-1}\choose {k}}x_1^{n-k}x_2^k,\ \ A(n)\not\equiv 1\pmod{5},\ \forall n\in \Bbb N\setminus\{0\}.$$
My problem is on how to proceed with the proof of this last statement. Hints and answers are welcomed.
|
Here is a solution inspired by Pell's equations, which tangentially is related to Binomial expansion. $(3,2)$ is a solution for
$$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\sqrt{2}\right)^n-\left(3-2\sqrt{2}\right)^n}{2\sqrt{2}}$$
where
$$x+y\sqrt{2}=\left(3+2\sqrt{2}\right)^n \space\space\space\text{ and }\space\space\space
x-y\sqrt{2}=\left(3-2\sqrt{2}\right)^n$$
is also a solution. Using binomial expansion it is easy to check $x,y \in \mathbb{Z}$ and
$$2x=x_1^n+x_2^n \tag{2}$$
Now, proof by contradiction. If $5 \mid x_1^n+x_2^n \overset{\color{red}{(2)}}{\Rightarrow} 5 \mid x$, this is from Euclid's lemma. But then
$$5\mid x^2 \overset{\color{red}{(1)}}{\Rightarrow} 5 \mid 2y^2+1$$
However $2y^2+1$ is never divisible by $5$. This is easy to see from $y^4 \equiv 1 \pmod{5}$ (LFT) or $5 \mid \left(y^2-1\right)\left(y^2+1\right)$ (applying the same Euclid's lemma)
*
*if $5 \mid y^2+1 \Rightarrow 5 \nmid 2y^2+1=y^2+1+y^2$, otherwise $5 \mid y^2$ leading to $5 \mid 1$.
*if $5 \mid y^2-1 \Rightarrow 5 \nmid 2y^2+1=2(y^2-1)+3$, otherwise $5 \mid 3$.
Alternatively, it's easy to check that $2y^2+1$ never has $0$ or $5$ as the last digit.
$$2\cdot \color{blue}{0}^2+1=1 \text{, }\space\space
2\cdot \color{blue}{1}^2+1=3 \text{, }\space\space
2\cdot \color{blue}{2}^2+1=9 \text{, }\space\space
2\cdot \color{blue}{3}^2+1=19 \text{,}\\
2\cdot \color{blue}{4}^2+1=33 \text{, }\space\space
2\cdot \color{blue}{5}^2+1=51 \text{, }\space\space
2\cdot \color{blue}{6}^2+1=73 \text{, }\space\space
2\cdot \color{blue}{7}^2+1=99 \text{,}\\
2\cdot \color{blue}{8}^2+1=129 \text{, }\space\space
2\cdot \color{blue}{9}^2+1=163$$
So we have a contradiction.
|
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How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\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\cos x(\cos x-\sqrt{3}\sin x)=0$$
$2\cos x=0\Rightarrow x\in \{\frac{\pi }2(2n-1):n\in\Bbb Z\}$
But how do I solve $$\cos x-\sqrt{3}\sin x=0$$
|
Multiply by the conjugate: $(\cos(x) - \sqrt{3} \sin(x))(\cos(x) + \sqrt{3} \sin(x)) = 0$. Then we have $\cos^2(x)-3\sin^2(x)=0$. This is the same thing as $1-4\sin^2(x)=0$ or $\sin(x)=\pm \frac{1}{2}$.
*
*NOTE OF CAUTION: This gives you the answers to both the question and its conjugate. You'd have to plug in and check which ones are the answers you're looking for.
|
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|
Maximum length perimeter of a box whose diagonal is 10 unit long. Suppose the diagonal of a three-dimentional box has length $10$, what is its maximum perimeter length?
Here is my solution:
Let the three edges of the box adjacent to a vertex be labelled $a,b,c$. Then by Cauchy-Schwarz inequality, we have the size of the perimeters of the box is $$4 \cdot \langle(1,1,1),(a,b,c)\rangle \le 4 \sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}=4 \cdot 10 \cdot \sqrt3.$$ Hence we know that the size of the perimeter of the box is bounded by $40\sqrt3$. Then I still need to find an actual box with perimeter that value and with the size of its diagonal equals to $10$.
|
You want to maximise $a+b+c$ subject to the constraint $a^2+b^2+c^2=100$. This constraint is the surface of a sphere with radius $10$, so it is fairly clear from geometrical considerations that the maximum occurs when $a=b=c=\sqrt{\frac{100}{3}}$.
To show this formally, we can use Lagrange multipliers: let $f(a,b,c)=a+b+c$ and $g(a,b,c)=a^2+b^2+c^2-100$. Then we have to find $\lambda,a,b,c$ such that the Lagrange function
$$\left(\frac{\partial f}{\partial a},\frac{\partial f}{\partial b},\frac{\partial f}{\partial c}\right)-\lambda\left(\frac{\partial g}{\partial a},\frac{\partial g}{\partial b},\frac{\partial g}{\partial c}\right)$$
is zero. This gives
$$(1,1,1)-\lambda(2a,2b,2c)=0$$
So $$a=b=c=\frac{1}{2\lambda}$$
The constraint $g(a,b,c)=0$ gives us $\lambda=\pm\sqrt{\frac{3}{400}}$, and we get the solutions
$$(a,b,c)=\left(\sqrt{\frac{100}{3}},\sqrt{\frac{100}{3}},\sqrt{\frac{100}{3}}\right)$$
and
$$(a,b,c)=\left(-\sqrt{\frac{100}{3}},-\sqrt{\frac{100}{3}},-\sqrt{\frac{100}{3}}\right)$$
corresponding to the maximum and minimum values of $a+b+c$.
|
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Apostol Calculus, Method of Exhaustion In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 \ to\ b$. Using the fact that,
\begin{align}
&1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2\label{1} \\
&\Rightarrow \frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < \frac{b^3}{3} < \frac{b^3}{n^3}(1^2+2^2+...+n^2) \nonumber \\
&\Rightarrow s_{n} < \frac{b^3}{3} < S_{n} \nonumber
\end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=\frac{b^3}{3}$ is the only number that satisfies
\begin{equation}
s_{n}<A<S_{n} \label{2}
\end{equation}
for every $n\geq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $\frac{b^3}{n^3}$,
\begin{align*}
&\frac{b^3}{n^3}(1^2+2^2+...+n^2)<\frac{b^3}{3}+\frac{b^3}{n^2} \\
&\Rightarrow S_{n}<\frac{b^3}{3}+\frac{b^3}{n^2}
\end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $\frac{b^3}{n^3}$,
\begin{align*}
&\frac{b^3}{3}-\frac{b^3}{n^2}<\frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\\
&\Rightarrow \frac{b^3}{3}-\frac{b^3}{n^3}<s_{n}
\end{align*}
this implies,
\begin{align*}
\frac{b^3}{3}-\frac{b^3}{n^2}<A<\frac{b^3}{3}+\frac{b^3}{n^2}
\end{align*}
This is where it gets confusing. He says the only possibilities are:
$$\begin{array}{ccc}
A>\frac{b^3}{3},& A<\frac{b^3}{3},& A=\frac{b^3}{3}
\end{array}$$
and proceeds to show that $A=\frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $\frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $\frac{b^3}{n^2}$?
|
First, there is a typo in your question. The inequality from the book is
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$
Picking it up from here. Note that the Apostol's ultimate goal in this proof is to somehow show that $A = \frac{b^3}{3}$. He basically just applied the law of trichotomy which says (I write informally) that given any two arbitrary numbers $x$ and $y$, exactly one of the following is true: $x < y$ or $x > y$ or $x = y$.
Applying that, we have $A$ which is an arbitrary number and we have $\frac{b^3}{3}$ also a number. We do not have any idea about the relationship between $A$ and $\frac{b^3}{3}$. However, we know from the law of Trichotomy that exactly one of the following (relations) is true: $A < \frac{b^3}{3}$ or $A > \frac{b^3}{3}$ or $A = \frac{b^3}{3}$.
If we're able to show that the first two cases are false, then the third case must be true which is exactly what he did.
|
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|
Global extrema for $\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$? Does the function
$$\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$$
have either a global minimum or a global maximum for $q \geq 5$ and $k \geq 1$?
WolframAlpha is unable to find any.
WolframAlpha computation for the global minimum
WolframAlpha computation for the global maximum
|
can we do better than this?
No, we cannot.
Let $f(k)$ be your function.
Then, we have
$$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)},\qquad \lim_{k\to\infty}f(k)=\lim_{k\to\infty}\frac{(1-\frac{1}{q^k})(1-\frac{2}{q}+\frac{1}{q^{k+1}})}{(q-1)(1-\frac{1}{q^{k+1}})}=\frac{q-2}{q(q-1)}$$
we have
$$\frac{q-1}{{q}(q+1)}\le f(k)\lt \frac{q-2}{q(q-1)}$$
Added :
Let $f(q)$ be your function.
We get, with some help of WA,
$$f'(q)=\frac{- (q^{k + 1} - k q + k - q) (q^{2k+1}(q-4)+2q^{k+1}+2q^{2k}-1)}{q^{k+1}(q-1)^2(q^{k+1}-1)^2}$$
Here, let
$$g(q):=q^{k + 1} - k q + k - q$$Then,$$g'(q)=(q^k-1)(k+1)\gt 0$$So, $g(q)$ is increasing and $$g(q)\ge g(5)=5^{k + 1} - 4k - 5\gt 0$$
It follows that $f(q)$ is strictly decreasing for $q\ge 5$.
Since $\displaystyle\lim_{q\to\infty}f(q)=0$, we have
$$0\lt f(q)\le f(5)=\frac{(5^k-1)(5^{k+1}-2\cdot 5^k+1)}{4\cdot 5^k(5^{k+1}-1)}\lt \frac{3}{20}$$
|
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"url": "https://math.stackexchange.com/questions/2998091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$x$ intercept problem How would I find the $x$ intercept of $x^5-x^3+2=0$? I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.
|
$$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(-\sqrt{\frac{3}{5}} \right)=2+\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 2.1859 \qquad f''\left(-\sqrt{\frac{3}{5}} \right)=-6 \sqrt{\frac{3}{5}}$$ So, ther is only one real root which is negative. By inspection, $f(-1)=2$, $f(-2)=-22$; so the root is close to $-1$.
To approximate the root, use Taylor series to get
$$f(x)=2+2 (x+1)-7 (x+1)^2+O\left((x+1)^3\right)$$ and the root to be considered is given by
$$x=\frac{1}{7} \left(\sqrt{15}-6\right)\approx -1.41043$$ Now, use Newton method which will update the guess according to
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{-4 x_n^5+2 x_n^3+2}{3 x_n^2-5 x_n^4}$$ to get the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & -1.410426192 \\
1 & -1.354288685 \\
2 & -1.347943175 \\
3 & -1.347867906 \\
4 & -1.347867896
\end{array}
\right)$$ which is the solution for ten significant figures.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate the indefinite integral $\int {(1-x^2)^{-3/2}}dx$ Evaluate the indefinite integral
$$\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} +\frac38\cdot\left[\frac1{1-x}+\frac1{1+x}\right]^2 + \frac{1}{4(1+x)^3}=\\
\frac{1}{4(1-x)^3} +\frac{3}{8(1-x)^2}+\frac38\cdot \left[\frac1{1-x}+\frac1{1+x}\right] + \frac{3}{8(1+x)^2}+ \frac{1}{4(1+x)^3}.$$
After edit: $\int \frac {dx}{(1-x^2)^{3/2}}$.
Make the change $x=\sin t \Rightarrow dx=\cos tdt$:
$$\int \frac {dx}{(1-x^2)^{3/2}}=\int \frac {\cos tdt}{(\cos^2t)^{3/2}}=\int \frac{dt}{\cos^2t}=\\
=\tan t+C=\frac{\sin t}{\sqrt{1-\sin^2t}}+C=\frac{x}{\sqrt{1-x^2}}+C.$$
|
{
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|
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 I don't know how to continue..
|
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,
using $k$ instead of $n$,
$d(k+1)-d(k)
=b(k)/c^{k+1}$.
Summing from $0$ to $n-1$,
$\begin{array}\\
d(n)-d(0)
&=\sum_{k=0}^{n-1}(d(k+1)-d(k))\\
&=\sum_{k=0}^{n-1}\dfrac{b(k)}{c^{k+1}}\\
&=\sum_{k=0}^{n-1}\dfrac{2^{k-1}-1}{c^{k+1}}
\qquad\text{since }b(n) = 2^{n-1}-1\\
&=\sum_{k=0}^{n-1}\dfrac{2^{k-1}}{c^{k+1}}-\sum_{k=0}^{n-1}\dfrac{1}{c^{k+1}}\\
&=\dfrac{1}{2c}\sum_{k=0}^{n-1}\dfrac{2^{k}}{c^{k}}-\dfrac1{c}\sum_{k=0}^{n-1}\dfrac{1}{c^{k}}\\
&=\dfrac{1}{2c}\dfrac{1-(2/c)^n}{1-2/c}-\dfrac1{c}\dfrac{1-(1/c)^n}{1-1/c}\\
&=\dfrac{1-(2/c)^n}{2c-2}-\dfrac{1-(1/c)^n}{c-1}\\
&=\dfrac{c^n-2^n}{2(c-1)c^n}-\dfrac{c^n-1}{(c-1)c^n}\\
&=\dfrac{c^n-2^n-2(c^n-1)}{2(c-1)c^n}\\
&=\dfrac{-c^n-2^n+2}{2(c-1)c^n}\\
\text{so}\\
\dfrac{a(n)}{c^n}-a(0)
&=\dfrac{-c^n-2^n+2}{2(c-1)c^n}\\
\text{or}\\
a(n)-a(0)c^n
&=\dfrac{-c^n-2^n+2}{2(c-1)}\\
\end{array}
$
or
$\begin{array}\\
a(n)
&=a(0)c^n+\dfrac{-c^n-2^n+2}{2(c-1)}\\
&=c^n(a(0)-\dfrac1{2(c-1)}-\dfrac{2^n-2}{2(c-1)}\\
&=c^n(a(0)-\dfrac1{2(c-1)}-\dfrac{2^{n-1}-1}{c-1}\\
&=3^n(a(0)-\dfrac1{4}-\dfrac{2^{n-1}-1}{2}
\qquad\text{since }c = 3\\
\end{array}
$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
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 another (over the equal sign).
|
\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.$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
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}} = r+\sqrt{8}.$$
|
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|
Seeking methods to solve $ \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx $ As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:
$$ \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx $$
I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.
|
My approach
Let
\begin{align}
\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\
&= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\
&= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right| \:dx \\
&= \int_{0}^{\frac{\pi}{2}} \left[ \ln\left|\cos^2(x) + 1 \right| - \ln\left|\cos^2(x)\right| \right]\:dx \\
&= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx
\end{align}
Now
$$ \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx = 2\int_{0}^{\frac{\pi}{2}} \ln\left|\cos(x)\right|\:dx = 2\cdot-\frac{\pi}{2}\ln(2) = -\pi \ln(2)$$
For detail on this definite integral, see guidance here
We now need to solve
$$ \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx $$
Here, Let
$$ I(t) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + t \right|\:dx $$
Thus,
$$
\frac{dI}{dt} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos^2(x) + t}\:dx
= \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{\cos(2x) + 1}{2} + t}\:dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x) + 2t + 1}\:dx
$$
Employ a change of variable $u = 2x$:
$$\frac{dI}{dt} = \int_{0}^{\pi} \frac{1}{\cos(u) + 2t + 1}\:du $$
Employ the Weierstrass substitution $\omega = \tan\left(\frac{u}{2} \right)$:
\begin{align}
\frac{dI}{dt} &= \int_{0}^{\infty} \frac{1}{\frac{1 - \omega^2}{1 + \omega^2} + 2t + 1}\:\frac{2}{1 + \omega^2}\cdot d\omega \\
&= \int_{0}^{\infty} \frac{1}{t\omega^2 + t + 1} \:d\omega \\
&= \frac{1}{t}\int_{0}^{\infty} \frac{1}{\omega^2 + \frac{t + 1}{t}} \:d\omega \\
&= \frac{1}{t}\left[\frac{1}{\sqrt{\frac{t+1}{t}}}\arctan\left( \frac{\omega}{\sqrt{\frac{t+1}{t}}}\right)\right]_{0}^{\infty} \\
&= \frac{1}{t}\frac{1}{\sqrt{\frac{t+1}{t}}}\frac{\pi}{2} \\
&= \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}
\end{align}
And so,
$$I(t) = \int \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}\:dt = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| + C$$
Now
$$I(0) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 0 \right|\:dx = -\pi \ln(2) = \pi\ln\left|\sqrt{0} + \sqrt{0 + 1} \right| + C \rightarrow C = -\pi \ln(2)$$
And so,
$$I(t) = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| -\pi \ln(2) = \pi\ln\left|\frac{\sqrt{t} + \sqrt{t + 1}}{2} \right|$$
Thus,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx = \pi\ln\left|\frac{\sqrt{1} + \sqrt{1 + 1}}{2} \right| = \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| $$
And Finally
\begin{align}
\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \\
&= \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| - \left(-\pi \ln(2)\right) \\
&= \pi\ln\left|1 + \sqrt{2} \right|
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.
|
This is just another way of saying what the others told you.
$$\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}$$
The theorem is
IF $\displaystyle \lim_{x\to 0}f(x) = L$
and $\displaystyle \lim_{x\to 0}g(x)=M$, where $M, N \in \mathbb R$,
THEN $\displaystyle \lim_{x\to 0}(f(x)-g(x))=L-M$
But, since $\displaystyle \lim_{x\to 0} \frac{\tan x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x^3} = \infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
\begin{align}
\frac{\tan x - \sin x}{x^3}
&= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \\
&= \frac{\sin x - \sin x \cos x}{x^3 \cos x} \\
&= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \\
&= \frac{1}{\cos x} \cdot\frac{\sin x}{x}
\cdot \frac{2\sin^2(\frac 12x)}{x^2} \\
&= \frac{1}{\cos x} \cdot\frac{\sin x}{x}
\cdot \frac 12 \cdot \left(\frac{\sin \frac x2}{\frac x2}\right)^2 \\
\end{align}
which approaches $\dfrac 12$ as $x$ approaches $0$.
|
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|
What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$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})+(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-x)(\frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=\frac{2x^2-x-1}{{2\sqrt{1-2x-x^2}}}$$
$$B'=2\bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=2 \bigg( \frac{1+x}{\sqrt{2}} \bigg)' \bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=\sqrt{2} \bigg( \frac{1}{\sqrt{1- \frac{1+2x+x^2}{2}} } \bigg)$$
$$B'=\sqrt{\frac{4}{1-2x-x^2}}$$
$$B'=\frac{4}{2\sqrt{1-2x-x^2}}$$
$$y'=A'+B'=\frac{2x^2-x+3}{2\sqrt{1-2x-x^2}}$$
The answer in the book is
$$y'=\frac{x^2}{\sqrt{1-2x-x^2}}$$
|
In first $A'$ instead of $(-2x)$ you have $(-2-2x)$
|
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|
Factoring cubic polynomial over R $z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4\cos^2(\frac{\pi}{14})$, $4\cos^2(\frac{3\pi}{14})$,$4\cos^2(\frac{9\pi}{14})$, but I don't know how to get them.
Thanks for help.
|
The roots of $$ 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} \; . \; \; $$
This is pretty easy if we take $\omega $ a 7th root of unity, then take $x = \omega + \frac{1}{\omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
|
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|
Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\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$$
which solves to $\sin x=(6\pm\sqrt{36-30})/10=(6\pm\sqrt6)/10$. Both are valid solutions, since $|\sin x|=|6\pm\sqrt6|/10$ and $|\cos x|=|2-3\sin x|=|2\mp3\sqrt6|/10$ are less than or equal to $1$ for both signs.
We now have
$${3\sin x\over4\sin x+3\cos x}={3\sin x\over4\sin x+3(2-3\sin x)}={3\sin x\over6-5\sin x}={3(6\pm\sqrt6)\over60-5(6\pm\sqrt6)}={3(6\pm\sqrt6)\over5(6\mp\sqrt6)}={3(6\pm\sqrt6)^2\over5\cdot30}\\={21\pm6\sqrt6\over25}$$
|
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|
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.
$$
= \int_0^\frac{\pi}{4} \frac{\sec^4(x)}{3\tan^4(x)+3-\sec^4(x)} \text{d}x
= \frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t
$$
Which looks a little tough but reminiscent of the well-known integral:
$$ \int_0^\infty \frac{1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac12\int_0^\infty \frac{x^2+1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac{\pi}{4\cos(a)} $$
Where we choose $a=\frac\pi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?
|
$$\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}$$
Observe that the integrand has period $\frac\pi2$, the problem can be easily done by using induction.
|
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}
|
Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\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}\left(\frac{1}{n^2}\cdot\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}\right)$$
Evaluate the following series:
$\sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$
$\frac{(k+1)^{k}}{k^{k-1}}=k\cdot\frac{(k+1)^{k}}{k^{k}}=k\cdot\left(1+\frac{k+1}{k}-1\right)^k=k\cdot\left(1+\frac{1}{k}\right)^k$
Then:
$\lim_{k\to\infty}\frac{(k+1)^{k}}{k^{k-1}}=\lim_{k\to\infty}k\cdot\left(1+\frac{1}{k}\right)^k=e\cdot\infty\neq0 \Longrightarrow \sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$ diverges.
Therefore:
$$\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)=0\cdot\infty$$
What to do next?
|
Hint: $$\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-x^2}}}=0$$
but its very tedious to find critical points here.
any other approach?
|
Let $(x,y)$ be a point on the unit circle $x^2+y^2=1$. We have to minimize the function:
$$
\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)\ ,\ (0,10)\ )
=\sqrt{3^2+10^2}\ ,
\end{aligned}
$$
with equality in the $\ge$ above only for the point of intersection of the segment with the above distance with the circle of radius $7$ centered in the origin.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3028968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Find close expression for the sum $S_{n,k}=\sum\limits_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$ Find close expression for the sum $$S_{n,k}=\sum_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$$
For small $k$ I have got following
\begin{align}
&S_{n,0}=1,\\
&S_{n,1}=2,\\
&S_{n,2}=-n+2,\\
&S_{n,3}=-2n+2,\\
&S_{n,4}=\frac{(n-1)(n-4)}{2},\\
&S_{n,5}=(n-1)(n-2),\\
&S_{n,6}=-\frac{(n-1)(n-2)(n-6)}{6}
\end{align}
What is the expression for $S_{n,k}$ for arbitrary $n,k$?
|
$$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$$$$\large=[x^k]\ \bigg(\sum_{j=0}^{\infty}(-1)^j \binom{n-1}{j}x^j\bigg)\bigg(\sum_{j=0}^{\infty}\binom{n+1}{j}x^j\bigg)$$$$\large=[x^k]\ (1-x)^{n-1}(1+x)^{n+1}=[x^k]\ (1-x^2)^{n-1}(x^2+2x+1)$$
$$\LARGE \therefore S_{n,k}=\begin{cases}
(-1)^{\frac{k}{2}}\Bigg(\binom{n-1}{\frac{k}{2}} - \binom{n-1}{\frac{k}{2}-1}\Bigg)& k\text{ is even} \\
2(-1)^{\frac{k-1}{2}}\huge\binom{n-1}{\frac{k-1}{2}} & k\text{ is odd}
\end{cases}$$
$\blacksquare$
Also see Cauchy product.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3030820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
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 $\cos \vartheta$ is concave on $[0, \frac{\pi}{2}]$.
When I plotted the graph of $1 - \frac{2\vartheta}{\pi} \sin \vartheta$ and the graph of $\cos \vartheta$ I noticed that indeed the stronger inequality
$$1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq \cos \vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
*
*Trying to find the zeros of $h(\vartheta)=\cos \vartheta - 1 + \frac{2 \vartheta}{\pi} \sin \vartheta$.
*Writing down the Taylor-expansion of $h(\vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
|
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\frac{θ}{2}\\
&\Longleftrightarrow \frac{\tan \dfrac{θ}{2}}{\dfrac{θ}{2}} \leqslant \frac{4}{π}.
\end{align*}
Define $f(t) = \dfrac{\tan t}{t}$ for $t \in \left(0, \dfrac{π}{4}\right]$, then $f'(t) = \dfrac{2t - \sin 2t}{2t^2 \cos^2 t} \geqslant 0$, which implies that$$
\frac{\tan \dfrac{θ}{2}}{\dfrac{θ}{2}} \leqslant f\left( \frac{π}{4} \right) = \frac{4}{π}.
$$
|
{
"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
$$
\alpha=2\tan^{-1}(2\sqrt{2}-1)=2\tan^{-1}(1.82)>2\tan^{-1}\sqrt{3}=2.\frac{\pi}{3}\\
\implies \boxed{\alpha>\frac{2\pi}{3}=0.67\pi}\\
\beta=3\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{3}{5}=3\sin^{-1}(0.33)+\sin^{-1}(0.6)\\
=\sin^{-1}(0.8)+\sin^{-1}(0.6)<\sin^{-1}(1)+\sin^{-1}(0.7=1/\sqrt{2})\\
\boxed{\beta<\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}=0.75\pi}
$$
I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to solve this ?
|
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 $2\tan^{-1}(2\sqrt{2}-1)>\dfrac{2\pi}{3}$, because the inverse of the tangent is an increasing function.
So, $\alpha>\dfrac{2\pi}{3}$
Note that $\sin3\theta=3\sin\theta-4\sin^2\theta$, then $3\theta=\sin^{-1}[3\sin\theta-4\sin^3\theta]$.
We have that $\theta=\sin^{-1}\dfrac13$ and $\sin\theta=\dfrac13$
So, $$\beta=3\sin^{-1}\dfrac13+\sin^{-1}\dfrac35$$
$$=\sin^{-1}\left[3\left(\dfrac13\right)-4\left(\dfrac13\right)^3\right]+\sin^{-1}\dfrac35$$
$$=\sin^{-1}\dfrac{23}{27}+\sin^{-1}\dfrac{3}{5}$$
Note that $\dfrac{23}{27}<\dfrac{\sqrt{3}}{2}$ and $\dfrac35<\dfrac{\sqrt{3}}{2}$, because $(23\cdot2)^2<(27\sqrt{3})^2$ and $(3\cdot2)^2<(5\sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.
So, $\sin^{-1}\dfrac{23}{27}+\sin^{-1}\dfrac35<\sin^{-1}\dfrac{\sqrt{3}}{2}+\sin^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$, because the inverse of the sine function is strictly increasing.
Therefore, $$\alpha>\dfrac{2\pi}{3},\ \beta<\dfrac{2\pi}{3}$$
$$\mbox{So, }\alpha>\beta$$
|
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"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
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|
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:
$$1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0$$
$$\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})$$
and the fact that $|A_i| = 1$ for $i =1,2,3,\cdots,n-1$.
|
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
$$
\sum_{k=1}^{n-1}\frac1{2-a_k}=\frac{(n-2)2^{n-1}+1}{2^n-1}
$$
|
{
"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\\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
\end{align}
More steps upon request.
I'll be grateful for any hints!
|
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 $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$\frac{a^3b^4c^5}{3^3\cdot 4^4\cdot 5^5}\leq 1$.
As $a^3b^4c^5 = 0.1\cdot 600^3 = 3^3\cdot 4^4\cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $\frac{a}{3}= \frac{b}{4}= \frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3046473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
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.]
$$
\binom{m+n}{s+1} - \binom{n}{s+1} = \sum_{i=0}^s \frac{m}{s-i+1}\binom{m+1+2(s-i)}{s-i}\binom{n-2(s-i+1)}{i}.
$$
$$
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \frac{m}{q+1}
{m+1+2q\choose q} {n-2-2q\choose s-q}
$$
$$
\binom{m+n}{s} = \sum_{i=0}^{s} \frac{m}{m+2i}
\binom{m+2i}{i}\binom{n-2i}{s-i}
$$
[Please ignore my attempt. It only explains one of equivalent forms.]
My attempt is to use the combinatorial argument.
The lefthand side could be understood as follow.
Suppose we have a box containing $m$ black balls and $n$ white balls.
We randomly draw $s+1$ balls out of it.
Then the LHS represents the number of ways that the selected $s+1$ balls.
However, not sure how to make a combinatorial argument on the RHS.
Based on the RHS, the sum of all cases of drawing $s-i$ balls from somewhere and $i$ balls from white balls.
But it is unclear to me to show the above identity.
Any suggestions/answers would be very appreciated. Thanks.
|
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
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \frac{m}{q+1}
{m+1+2q\choose q} {n-2-2q\choose s-q}}$$
we observe that
$${m+1+2q\choose q+1} - {m+1+2q\choose q}
\\ = \frac{m+1+q}{q+1} {m+1+2q\choose q} - {m+1+2q\choose q}
\\ = \frac{m}{q+1} {m+1+2q\choose q}.$$
Therefore we have two sums,
$$\sum_{q=0}^s
{m+1+2q\choose q+1} {n-2-2q\choose s-q}
- \sum_{q=0}^s
{m+1+2q\choose q} {n-2-2q\choose s-q}.$$
For the first one we write
$$\sum_{q=0}^s
[w^{q+1}] (1+w)^{m+1+2q} [z^{s-q}] (1+z)^{n-2-2q}
\\ = \underset{w}{\mathrm{res}}\;
(1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q=0}^s \frac{1}{w^{q+2}} z^q (1+w)^{2q} (1+z)^{-2q}.$$
We may extend $q$ beyond $s$ because of the coefficient extractor
$[z^s]$ in front, getting
$$\underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q\ge 0} z^q w^{-q} (1+w)^{2q} (1+z)^{-2q}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n-2}
\frac{1}{w^2} \frac{1}{1-z(1+w)^2/w/(1+z)^2}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w} \frac{1}{w(1+z)^2-z(1+w)^2}.$$
Repeat the calculation for the second one to get
$$\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w(1+z)^2-z(1+w)^2}.$$
Now we have
$$\left(\frac{1}{w}-1\right)\frac{1}{w(1+z)^2-z(1+w)^2}
= \frac{1}{w-z} \frac{1}{w(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}
\\ = \frac{1}{1-z/w} \frac{1}{w^2(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}.$$
We thus obtain two components, the first is
$$\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-z/w} \frac{1}{w^2(1+w)}
\\ = \underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m}
[z^s] (1+z)^{n} \frac{1}{1-z/w}
\\ = \underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m}
\sum_{q=0}^s {n\choose q} \frac{1}{w^{s-q}}
= \sum_{q=0}^s {n\choose q}
\underset{w}{\mathrm{res}}\; \frac{1}{w^{s-q+2}} (1+w)^{m}
\\ = \sum_{q=0}^s {n\choose q} [w^{s-q+1}] (1+w)^m
= [w^{s+1}] (1+w)^m \sum_{q=0}^s {n\choose q} w^q
\\ = - {n\choose s+1}
+ [w^{s+1}] (1+w)^m \sum_{q=0}^{s+1} {n\choose q} w^q.$$
We may extend $q$ beyond $s+1$ due to the coefficient extractor in
front, to get
$$- {n\choose s+1}
+ [w^{s+1}] (1+w)^m \sum_{q\ge 0} {n\choose q} w^q
= - {n\choose s+1}
+ [w^{s+1}] (1+w)^{m+n}$$
This is
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}.}$$
We have the claim, so we just need to prove that the second component
will produce zero. We obtain
$$\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-wz} \frac{1}{1+w}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m} [z^s] (1+z)^{n}
\frac{1}{1-wz}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m}
\sum_{q=0}^s {n\choose q} w^{s-q}
= \sum_{q=0}^s {n\choose q} \underset{w}{\mathrm{res}}\; w^{s-q} (1+w)^{m}
= 0.$$
This concludes the argument.
|
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"url": "https://math.stackexchange.com/questions/3049572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
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 on another approach which here is a combination of Feynman's Trick, Laplace Transforms and coupled ODE Systems. I would love for a qualified eye to have a look over to see if what I've done is correct and/or another method (Not using Complex Analysis) to solve.
Here I've included more of my algebra to aid in those who wish to go over.
Consider the following two definite integrals
\begin{align}
I_{n,a,k} &= \int_{0}^{\infty} \frac{\sin\left(kx^n\right)}{x^n + a}\:dx \\
J_{n,a,k} &= \int_{0}^{\infty} \frac{\cos\left(kx^n\right)}{x^n + a}:dx
\end{align}
For $a,k \in \mathbb{R}^+$ and $n \in \mathbb{R}^{+}, n > 1$. Here we define:
\begin{align}
I_{n,a,k}(t) &= \int_{0}^{\infty} \frac{\sin\left(tkx^n\right)}{x^n + a}:dx \\
J_{n,a,k}(t) &= \int_{0}^{\infty} \frac{\cos\left(tkx^n\right)}{x^n + a}:dx
\end{align}
Here we observe that:
\begin{align}
I_{n,a,k}(1) &= I_{n,a,k} & I_{n,a,k}(0) &= 0 \\
J_{n,a,k}(1) &= J_{n,a,k} & J_{n,a,k}(0) &= a^{\frac{1}{n} - 1} \frac{\Gamma\left(1 -\frac{1}{n}\right)\Gamma\left(\frac{1}{n} \right)}{n} = \theta_{a,n}
\end{align}
Here we will address each integral individually. For $I_{n,a,k}$ we take the derivative with respect to '$t$':
\begin{align}
I_{n,a,k}'(t) &= \int_{0}^{\infty} \frac{kx^n\cos\left(tkx^n\right)}{x^n + a}\:dx = k\left[\int_{0}^{\infty} \cos\left(tkx^n\right)\:dx - a\int_{0}^{\infty} \frac{\cos\left(tkx^n\right)}{x^n + a}\:dx \right] \\
\frac{1}{k}I_{n,a,k}'(t) &= \frac{1}{k^{\frac{1}{n}}t^{\frac{1}{n}}}\int_{0}^{\infty} \cos\left(u^n\right)\:du - aJ_{n,a,k}(t)
\end{align}
Thus,
\begin{equation}
\frac{1}{k}I_{n,a,k}'(t) + aJ_{n,a,k}(t) = \frac{1}{k^{\frac{1}{n}}t^{\frac{1}{n}}}\int_{0}^{\infty} \cos\left(u^n\right)\:du
\end{equation}
From Section X, we arrive at:
\begin{equation}
\frac{1}{k}I_{n,a,k}'(t) + aJ_{n,a,k}(t) = \frac{\Gamma\left(\frac{1}{n}\right)\cos\left(\frac{\pi}{2n} \right)}{nk^{\frac{1}{n}}t^{\frac{1}{n}}}
\end{equation}
Applying the same method to $J_{n,a,k}\left(t\right)$ we arrive at:
\begin{equation}
-\frac{1}{k}J_{n,a,k}'(t) + aI_{n,a,k}(t) = \ \frac{\Gamma\left(\frac{1}{n}\right)\sin\left(\frac{\pi}{2n} \right)}{nk^{\frac{1}{n}}t^{\frac{1}{n}}}
\end{equation}
And thus, we arrive at the couple ordinary differential equation system:
\begin{align}
\frac{1}{k}I_{n,a,k}'(t) + aJ_{n,a,k}(t) &= \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)t^{-\frac{1}{n}}\\
aI_{n,a,k}(t) -\frac{1}{k}J_{n,a,k}'(t) &= \Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)t^{-\frac{1}{n}}
\end{align}
Where $\Psi_{k,n} = \frac{\Gamma\left(\frac{1}{n}\right)}{n}k^{-\frac{1}{n}}$. Although there are many approaches to solving this system, here I will employ Laplace Transforms:
\begin{align}
\frac{1}{k}\mathscr{L}\left[I_{n,a,k}'(t)\right] + a\mathscr{L}\left[J_{n,a,k}(t)\right] &= \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\mathscr{L}\left[t^{-\frac{1}{n}}\right]\\
a\mathscr{L}\left[I_{n,a,k}(t)\right] -\frac{1}{k}\mathscr{L}\left[J_{n,a,k}'(t)\right] &= \Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\mathscr{L}\left[t^{-\frac{1}{n}}\right]
\end{align}
Which becomes:
\begin{align}
\frac{s}{k}\bar{I}_{n,a,k}(s) + a\bar{J}_{n,a,k}(s) &= \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\kappa(s)\\
a\bar{I}_{n,a,k}(s) -\frac{s}{k}\bar{J}_{n,a,k}(s) &= \Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\kappa(s) + \frac{1}{k}\theta_{a,n}
\end{align}
Where
\begin{equation}
\kappa(s) = \mathscr{L}\left[t^{-\frac{1}{n}}\right] = \Gamma\left(1 - \frac{1}{n}\right)s^{1 - \frac{1}{n}}
\end{equation}
Solving for $\bar{J}_{n,a,k}(s)$ we find:
\begin{align}
\bar{J}_{n,a,k}(s) &= \frac{1}{s^2 + a^2k^2}\left[ak^2 \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\kappa(s) -k\Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)s\kappa(s) - s\theta_{a,n}\right] \\
&=ak^2 \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\frac{1}{s^2 + a^2k^2}\kappa(s)-k\Psi_{k,n}\frac{s}{s^2 + a^2k^2}\kappa(s)\\
&\qquad- \theta_{a,n}\frac{s}{s^2 + a^2k^2}
\end{align}
Taking the Inverse Laplace Transform, we arrive at:
\begin{align}
&J_{n,a,k}(t) = \mathscr{L}^{-1}\left[ \bar{J}_{n,a,k}(s) \right] = ak^2 \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\mathscr{L}^{-1}\left[\frac{1}{s^2 + a^2k^2}\kappa(s)\right]\\
&\qquad-ak\Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\mathscr{L}^{-1}\left[\frac{s}{s^2 + a^2k^2}\kappa(s)\right]- \theta_{a,n}\mathscr{L}^{-1}\left[\frac{s}{s^2 + a^2k^2}\right] \\
&= ak^2 \Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\int_{0}^{t} \frac{1}{ak}\sin\left(ak\left(t - \tau\right)\right) \tau^{-\frac{1}{n}}\:d
tau\\
&\qquad-ak\Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\int_{0}^{t} \cos\left(ak\left(t - \tau\right)\right) \tau^{-\frac{1}{n}}\:d
tau -\theta_{a,n}\cos\left(akt \right) \\
&= k\Psi_{k,n}\cos\left(\frac{\pi}{2n} \right)\int_{0}^{t} \left[\sin\left(akt\right)\cos\left(ak\tau\right) - \sin\left(ak\tau\right)\cos\left(akt\right) \right]\tau^{-\frac{1}{n}}\:d\tau\\
&\qquad-k\Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\int_{0}^{t} \left[\cos\left(akt\right)\cos\left(ak\tau\right) +\sin\left(ak\tau\right)\sin\left(akt\right) \right] \tau^{-\frac{1}{n}}\:d\tau \\
&\qquad-\theta_{a,n}\cos\left(akt \right) \\
&= k\Psi_{k,n}\cos\left(\frac{\pi}{2n} \right) \left[\sin\left(akt\right)\int_{0}^{t} \frac{\cos\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau - \cos\left(akt\right)\int_{0}^{t} \frac{\sin\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau \right] \\
&\qquad-k\Psi_{k,n}\sin\left(\frac{\pi}{2n} \right)\left[\cos\left(akt\right)\int_{0}^{t} \frac{\cos\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau+ \sin\left(akt\right)\int_{0}^{t} \frac{\sin\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau \right] \\
&\qquad-\theta_{a,n}\cos\left(akt \right) \\
&= k \Psi_{k,n}\left[\sin\left(akt + \frac{\pi}{2n}\right) \int_{0}^{t} \frac{\cos\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau-\cos\left(akt + \frac{\pi}{2n}\right) \int_{0}^{t} \frac{\sin\left(ak\tau\right) }{\tau^{\frac{1}{n}}}\:d\tau \right] \\
&\qquad-\theta_{a,n}\cos\left(akt \right) \\
&= k \Psi_{k,n}\left[\sin\left(akt + \frac{\pi}{2n}\right) k^{\frac{1}{n} - 1} a^{\frac{1}{n} - 1}\int_{0}^{akt} \frac{\cos\left(u\right) }{u^{\frac{1}{n}}}\:du-\cos\left(akt + \frac{\pi}{2n}\right)k^{\frac{1}{n} - 1} a^{\frac{1}{n} - 1} \int_{0}^{t} \frac{\sin\left(u\right) }{u^{\frac{1}{n}}}\:du \right] \\
&\qquad-\theta_{a,n}\cos\left(akt \right) \\
&= k k^{\frac{1}{n} - 1} a^{\frac{1}{n} - 1} \Psi_{k,n}\left[\sin\left(akt + \frac{\pi}{2n}\right) \int_{0}^{akt} \frac{\cos\left(u\right) }{u^{\frac{1}{n}}}\:du-\cos\left(akt + \frac{\pi}{2n}\right)\int_{0}^{akt} \frac{\sin\left(u\right) }{u^{\frac{1}{n}}}\:du \right] \\
&\qquad-\theta_{a,n}\cos\left(akt \right)
\end{align}
Hence,
\begin{align}
J_{n,a,k}(t) &= \int_{0}^{\infty} \frac{\cos\left(tkx^n\right)}{x^n + a}\:dx \\
&=a^{\frac{1}{n} - 1}\frac{\Gamma\left(\frac{1}{n} \right)}{n} \left[\sin\left(akt + \frac{\pi}{2n}\right) \int_{0}^{akt} \frac{\cos\left(u\right) }{u^{\frac{1}{n}}}\:du-\cos\left(akt + \frac{\pi}{2n}\right)\int_{0}^{akt} \frac{\sin\left(u\right) }{u^{\frac{1}{n}}}\:du \right] -\theta_{a,n}\cos\left(akt \right)
\end{align}
And finally,
\begin{align}
J_{n,a,k} &= J_{n,a,k}(1) = \int_{0}^{\infty} \frac{\cos\left(kx^n\right)}{x^n + a}\:dx \\
&=a^{\frac{1}{n} - 1}\frac{\Gamma\left(\frac{1}{n} \right)}{n} \left[\sin\left(ak + \frac{\pi}{2n}\right) \int_{0}^{akt} \frac{\cos\left(u\right) }{u^{\frac{1}{n}}}\:du-\cos\left(ak + \frac{\pi}{2n}\right)\int_{0}^{akt} \frac{\sin\left(u\right) }{u^{\frac{1}{n}}}\:d\tau \right] \\
&\qquad-\cos\left(ak \right) a^{\frac{1}{n} - 1} \frac{\Gamma\left(1 -\frac{1}{n}\right)\Gamma\left(\frac{1}{n} \right)}{n}
\end{align}
|
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 substituion and rearrangements: Let $\ \displaystyle x^nt = y \implies \mathrm{d}x = \frac{y^{1/n - 1}}{n t^{1/n}} \mathrm{d}y.$
Hence our Integral becomes:
$$\displaystyle I = \frac{1}{n} \sum_{r=0}^{\infty} (-1)^r \frac{k^{2r}}{(2r)!} \int_{0}^{\infty} t^{-2r - 1/n}e^{-at} \int_{0}^{\infty} y^{2r + 1/n -1} e^{-y} \ \mathrm{d}y \ \mathrm{d}t $$
And, $ \displaystyle \int_{0}^{\infty} y^{2r + 1/n -1} e^{-y} \ \mathrm{d}y =
\Gamma \left (2r + \frac{1}{n} \right ) $.
Similarly substitute $ \displaystyle at = z ,$ and get $ \ \displaystyle a^{2r + 1/n-1} \int_{0}^{\infty} z^{-2r - 1/n } e^{-z} \ \mathrm{d}z = \Gamma \left (1 - 2r - \frac{1}{n} \right ) \ $ for second integral.
Thus finally we have
$$\displaystyle I = \frac{a^{1/n-1}}{n} \sum_{r=0}^{\infty} (-1)^r \frac{(ka)^{2r}}{(2r)!} \Gamma \left (2r + \frac{1}{n} \right ) \Gamma \left (1 - 2r - \frac{1}{n} \right ) $$
And as $\displaystyle \Gamma(z)\Gamma(1-z) = \pi \csc (\pi z) \implies \Gamma \left (2r + \frac{1}{n} \right ) \Gamma \left (1 - 2r - \frac{1}{n} \right ) = \pi \csc \left ( 2\pi r + \frac{\pi}{n} \right) $.
Also, here $r \in \mathbb{Z}^+ $ thus $\displaystyle \csc \left ( 2\pi r + \frac{\pi}{n} \right) = \csc \left ( \frac{\pi}{n} \right) $
Thus $\displaystyle I = \frac{\pi}{n} \csc \left ( \frac{\pi}{n} \right) a^{\frac{1}{n}-1} \sum_{r=0}^{\infty} (-1)^r \frac{(ka)^{2r}}{(2r)!} $.
Finally we conclude:
$$\displaystyle I = \frac{\pi}{n} \csc \left ( \frac{\pi}{n} \right) a^{\frac{1}{n}-1} \cos (ka)$$
|
{
"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 $$\det\binom{ v+ku }{ w
}=\det\binom{v }{ w
}+k\det \binom{ u }{ w
}$$
$v,u,w$ are vectors $k$ is scalar.
I have tried to divide it into $n$ parts and tried to compose with sense but didn't acomplish.
Second I tried to make use of "Row Reduction" i.e. adding scalar multiple of some row to another does not change the determinant, so I added the all different rows into other i.e. adding $2, 3,4,\dots,n$th row to first row and similarly doing for all rows we got
$$\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=\det\begin{vmatrix}
n-1 & \cdots & n-1& n-1 & n-1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
n-1 & \cdots & n-1 & n-1 & n-1
\end{vmatrix}=0$$
The last determinant is zero (I guess) so the given determinant is zero?
I don't have the answer this question, so I am not sure. How to calculate this determinant?
|
Notice that
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix} = (n-1)\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0
\end{bmatrix} = -\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0
\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0
\end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0
\end{bmatrix}$$
$$\vdots $$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1
\end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1
\end{bmatrix}$$
All eigenvectors here are linearly independent so the eigenvalues are $-1$ with multiplicity $n-1$ and $n-1$ with multiplicity one.
Therefore the determinant is $(-1)^{n-1}(n-1)$.
|
{
"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{\pi}{2n+1} &=x
\end{align}
then
\begin{align}
f &= \dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \\
&=\dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\\
&=\dfrac{1}{(2w)^{2m+1}}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{2i}\sum_{k=1}^{n}w^{(4m-4i+2)k}
\end{align}
where $w=e^{ix}$.
|
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 $\infty$.
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\sum_{k=1}^n\cos^{2m+1}\left(\pi\,\frac{2k-1}{2n+1}\right)
&=\frac12-\frac12\sum_{k=0}^{2n}\cos^{2m+1}\left(2\pi\,\frac{k}{2n+1}\right)\\
&=\frac12+\frac12\Res_{z=0}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\
&-\frac12\Res_{z=\infty}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\
&=\frac12-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=0}^\infty z^{(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\
&-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=1}^\infty z^{-(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\
&=\frac12-\frac{2n+1}{2^{2m+1}}\sum_{k=0}^m\binom{2m+1}{k}[\,2n+1\,|\,2m+1-2k\,]
\end{align}
$$
where $[\dots]$ are Iverson Brackets.
Note that for $m\lt n$, the sum is $\frac12$ because $2n+1$ cannot divide $2m+1-2k$.
|
{
"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 continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
|
$x^2+1/4=x;$
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.