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How use Lagrange interpolation prove this inequality How can I use Lagrange interpolation to solve the following problem?
Let $a,b,c,d\in \mathbb{R}$ such that
$$|ax^3+bx^2+cx+d|\le 1 $$
for every $x\in[-1,1]$. Show that
$$|a|+|b|+|c|+|d|\le 7$$
|
Let $f(x)=ax^3+bx^2+cx+d$.
From Lagrange Interpolation (or solving a system of linear equations),
we find that $$f(x)=\frac{-2f(-1)+4f(-\frac{1}{2})-4f(\frac{1}{2})+2f(1)}{3}x^3+\frac{2f(-1)-2f(-\frac{1}{2})-2f(\frac{1}{2})+2f(1)}{3}x^2+\frac{\frac{1}{2}f(-1)-4f(-\frac{1}{2})+4f(\frac{1}{2})-\frac{1}{2}f(1)}{3}x+\frac{-\frac{1}{2}f(-1)+2f(-\frac{1}{2})+2f(\frac{1}{2})-\frac{1}{2}f(1)}{3}$$
Now, we divide cases. The four polynomials in the form of $\pm f(\pm x)$ satisfy the exact same conditions as $f(x)$.
It is safe to assume that $a,b \ge 0$.
Case 1. $c \ge 0$.
If $d \ge 0$, we are done because $|a|+|b|+|c|+|d| =a+b+c+d=f(1) \le 1$.
If $d \le 0$, we have $|a|+|b|+|c|+|d|=a+b+c-d=f(1)-2f(0) \le 3$.
Case 2. $c \le 0$.
If $d \ge 0$, we have $$|a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}f(1)-\frac{1}{3}f(-1)-\frac{8}{3}f(\frac{1}{2})+\frac{8}{3}f(-\frac{1}{2}) \le \frac{4}{3}+\frac{1}{3}+\frac{8}{3}+\frac{8}{3}=7$$
If $d \le 0$, we have $$|a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}f(1)-4f(\frac{1}{2})+\frac{4}{3}f(-\frac{1}{2}) \le \frac{5}{3}+4+\frac{4}{3} = 7$$
This completes the proof. The bound is tight since $f(x)=4x^3-3x$ reaches equality.
EDIT: Note how $4x^3-3x$ is a Chebyshev Polynomial, which appears a lot in Lagrange Interpolation problems.
|
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|
How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$
How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at
$a_{2n+1}$ and $a_{2n-1}$
$a_{2n}$ and $a_{2n-2}$
But I am unable to prove that they are bounded.
NOTE: Look at joeys answer for correct solution.
|
It could of course be that I made a mistake in my calculations so proceed with caution!
We'll show by induction that for any $n \geq 1$:
$$
a_{2n} = \frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$$
where a sum $\sum_{i=1}^0\frac{1}{4^n}$ is assumed to be zero.
The first step: $n=1$ is trivial. Now let $n$ be arbitrary we'll first prove the form of $a_{2(n+1)}$: by the induction hypethesis we may assume that:
$
a_{2n} =\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$ thus we get:
$$
a_{2(n+1)} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{\frac{1}{2} + \frac{a_{2n}}{2}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \overset{IH}{=} \frac{1}{4} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}}{4} \\= \frac{1}{4} + \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n-1} \frac{1}{4^{i+1}}= \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n} \frac{1}{4^{i}}
$$
which proves the claimed equality for $a_{2n}$ we now have for arbitrary $n$:
$$
a_{2n+1} = \frac{1}{2} + \frac{a_{2n}}{2} = \frac{1}{2} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^n}}{2} = \frac{a}{4^{n+1}} + \frac{1}{2} \sum_{i=0}^n \frac{1}{4^i}.$$
Can you deduce the result from this? I can work it out further if you want but I think it's clear how you should continue now (just calculating the sums yields that the liminf is 1/3 and the limsup is 2/3).
Now without liminf and limsup, first we show that $a_n \geq 1/3 $ always holds by induction. For $a_{2n+1}$ this is immediately clear since we have $a_{2n+1} \geq 1/2$. For $a_{2n}$ we proceed by induction so suppose we have $a_{2n} \geq 1/3$ and we show that then also $a_{2(n+1)} \geq 1/3$, we have:
$$
a_{2(n+1)} = a_{2n + 2} = \frac{a_{2n-1}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \geq \frac{1}{4} + \frac{\frac{1}{3}}{4} = 1/3,
$$
in the same way one shows that $a_n \leq 2/3$ for all $n$. Now it remains to show that $\lim_{n \rightarrow \infty} a_{2n} \leq 1/3$ and $\lim_{n\rightarrow \infty} a_{2n+1} \geq 2/3$ if you show this you are done.
|
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|
Exponential Diophantine equation $7^y + 2 = 3^x$
Find all positive integer solutions to $$7^y + 2 = 3^x.$$
ATTENTION: MY SOLUTION HAS A TERRIBLE MISTAKE WHICH I HAVE OVERLOOKED!
Obviously, $x > y$. Then, we have $3^x = 7^y + 2 \equiv 0 \pmod {3^y}$. Also, $$7^y = (6 + 1)^y = \sum_{k = 0}^{y} {y \choose k} 6^k \equiv \sum_{k = 0}^{y - 1} {y \choose k} 6^k \pmod {3^y}.$$ We claim that the highest power of $3$ that divides ${y \choose k}$ is at most $2$. Indeed, $$\sum_{i = 1}^{\infty} \left [\frac {y} {3^i} \right] - \left (\sum_{i = 1}^{\infty} \left [\frac {y - k} {3^i} \right] + \sum_{i = 1}^{\infty} \left [\frac {k} {3^i} \right] \right) \leqslant 2.$$ Hence, $$7^y \equiv \sum_{k = 0}^{y - 1} {y \choose k} 6^k \leqslant 2 \sum_{k = 0}^{y - 1} 6^k = \frac {2} {5} (6^y - 1).$$ Since $(5, 3^y) = 1$, we have by Euler's Theorem that $$5^{\phi (3^y)} = 5^{3^y - 3^{y - 1}} \equiv 1 \pmod {3^y}.$$ Then, $2 \cdot 3^{y - 1} \equiv \frac {2} {5} (6^y - 1) \pmod {3^y}$ and $$0 \equiv 7^y + 2 \leqslant 2 \cdot 3^{y - 1} + 2 \pmod {3^y}.$$ Take $s > 0$ an integer for which $$3^{y} s \leqslant 2 \cdot 3^{y - 1} + 2.$$ It follows from this that $0 \leqslant 3^{y} (s - 1) \leqslant 2 - 3^{y - 1}$. Hence, $y < 2$. So the solutions are $$(x, y) = (1, 0), (2, 1).$$
"Notes on Olympiad Problems", Nima Bavari, Tehran, 2006.
|
Okay so I got this solution after modular bashing for an hour. This better be right.
Since the cases $x,y \le 2$ are already investigated above easily, we look at $x,y \ge 3$.
Rewrite this equation to $$7(7^{y-1}-1)=9(3^{x-2}-1)$$
Now, since $7|3^{x-2}-1$, and the order of $3$ modulo $7$ is $6$, we have $6|x-2$.
This gives $13|3^6-1|3^{x-2}-1$, so $13|7^{y-1}-1$.
Now we have $12|y-1$ since the order of $7$ modulo $13$ is $12$. This gives $19|7^{12}-1|7^{y-1}-1$.
Now we have $19|3^{x-2}-1$, so $18|x-2$, since the order of $3$ modulo $19$ is $18$.
Now $37|3^{18}-1|3^{x-2}-1$. This gives $37|7^{y-1}-1$.
Now we have $9|y-1$. Now we have $27|7^9-1|7^{y-1}-1$, since the order of $7$ modulo $37$ is $9$.
However, $9(3^{x-2}-1) \equiv -9 \pmod{27}$, so it cannot be a multiple of $27$.
We now have a contradiction, so the answer is $(x,y)=(1,0),(2,1)$, as desired. GG!!
|
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|
Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
|
Apply $\ln$ to get
$$\tag 1 \frac{\ln (a^x-x\ln a) - \ln (b^x-x\ln b)}{x^2}.$$
Using $e^u = 1 + u + u^2/2 +O(u^3)$ as $u\to 0,$ we get
$$a^x = e^{x\ln a} = 1 + x\ln a + (x\ln a)^2/2 + O(x^3).$$
A similar result holds for $b^x.$ Therefore $(1)$ equals
$$\tag 2 \frac{\ln (1+(x\ln a)^2/2 + O(x^3)) - \ln (1+(x\ln b)^2/2 + O(x^3))}{x^2}.$$
Now $\ln (1+u) = u + O(u^2).$ Therefore $(2)$ equals
$$\frac{(x\ln a)^2/2 -(x\ln b)^2/2 + O(x^3)}{x^2} \to (\ln a)^2/2 -(\ln b)^2/2 .$$
Exponentiating back gives $e^{(\ln a)^2/2 -(\ln b)^2/2}.$
|
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|
Integrate $\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$ I need integrate:
$$\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$$
How can i solve it? is it good way substitute argument of arccos? $$t=\sqrt{\frac{x-4}{x+6}}$$
|
Hint. Here is a route.
By the change of variable
$$
t=\sqrt{\dfrac{x-4}{x+6}}\qquad x= 2\:\dfrac{ 3 t^2+2}{1-t^2} \qquad dx= 2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'dt
$$ one may obtain
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\int\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'\arccos(t)\:dt
$$ then one may integrate by parts to get
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)\arccos(t)+2\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt
$$ The latter integral may be evaluated by the change of variable $t:=\sin u$ yielding
$$
\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt=\frac{10 t}{\sqrt{1-t^2}}-6\: \text{arcsin}(t)+C, \quad |t|<1.
$$
|
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|
Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$ Prove the following equation ($|x|<1$)
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
I made this question and I have the following answer but I think it may be incomplete.
If anyone can point out a flaw in my proof or give a better proof then it would be appreciated.
My solution:
$$\begin{align}
&f(x)=\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right)\\
&N(p,q)=\{n\in\mathbb N \mid n\ne (2m-1)2^{k-1};m,k\in\mathbb N,2m-1\leq p,k\leq q\}\\
&f(x)(1+x+x^2)=(1+x+x^2)(1-x+x^2)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^2+x^4)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^2+x^4)(1-x^2+x^4)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^4+x^8)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^4+x^8)(1-x^4+x^8)\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^8+x^{16})\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)=\cdots\\
&=\lim_{k\to\infty}(1+x^{2^k}+x^{2^{k+1}})\prod_{n\in N(1,k)}^{} \left(1-x^n+x^{2n}\right)=\prod_{n\in N(1,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\text{Similarly,}\\
&f(x)(1+x+x^2)(1+x^3+x^6)=\prod_{n\in N(3,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&f(x)(1+x+x^2)(1+x^3+x^6)(1+x^5+x^{10})=\prod_{n\in N(5,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\cdots\\
&f(x)\prod_{m=1}^{\infty} \left(1+x^{2m-1}+x^{2(2m-1)}\right)=\lim_{p\to\infty} \prod_{n\in N(p,\infty)}^{} \left(1-x^n+x^{2n}\right)=1
\end{align}$$
(*) Is it obvious that $\{(2m-1)\cdot2^{k-1}\mid m,k\in \mathbb N\}$ is equivalent to $\mathbb N$, or should I also prove it?
Thanks.
|
We have
\begin{align}
\prod_{n = 1}^\infty (1 - x^n + x^{2n}) &= \prod_{n = 1}^\infty \frac{1+x^{3n}}{1+x^n} \\
&= \prod_{n = 1}^\infty \frac{1 - x^{6n}}{1-x^{3n}}\prod_{n=1}^\infty\frac{1-x^{n}}{1-x^{2n}}\\
& = \prod_{n = 1}^\infty \frac{1}{1-x^{3(2n-1)}}\prod_{n = 1}^\infty (1 - x^{2n-1})\\
&= \prod_{n = 1}^\infty \frac{1-x^{2n-1}}{1-(x^{2n-1})^3}\\
&= \prod_{n = 1}^\infty \frac{1}{1 + x^{2n-1} + x^{4n-2}}.
\end{align}
|
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|
limits of inverse trigonometric functions without L'hospital's rule How do I solve this without using L'Hospital's rule? $$\lim_{h\rightarrow0} \frac{\cos^{-1}(\frac{1}{2}-h) -\cos^{-1}(\frac{1}{2})}{h}$$
I already tried letting $\theta=\cos^{-1}(\frac{1}{2}-h)$ gets $\cos\theta=\frac{1}{2}-h$ then $h=\frac{1}{2}-\cos\theta$, replaced all $h$ with this and I'm lost. I think this doesn't help. I'm getting a zero as value or indefinite one which shouldn't be because the value must be $\frac{2\sqrt{3}}{3}$. Please help.
|
Your start is good.
$$\lim_{h\to0}\frac{\cos^{-1}\left(\frac{1}{2}-h\right)-\cos^{-1}\left(\frac{1}{2}\right)}{h}$$
Let $\theta=\cos^{-1}\left(\frac{1}{2}-h\right)$ so $h=\frac{1}{2}-\cos h$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\frac{1}{2}-\cos\theta}$$
Rewrite fraction as a trig value and apply sums to products.
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\cos\left(\frac{\pi}{3}\right)-\cos\theta}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{2\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)}\times\lim_{\theta\to\frac{\pi}{3}}\frac{1}{\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=1\times\frac{1}{\frac{\sqrt{3}}{2}}$$
$$=\frac{2}{\sqrt{3}}$$
$$=\frac{2\sqrt{3}}{3}$$
|
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A different approach in distributing $8$ distinct balls into $6$ distinct boxes such that each box has at least $1$ ball Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.
We are well acquainted with the traditional Inclusion-Exclusion principle in solving this and it will come up as $191520$ ways.
I tried in this way:
When we distribute $8$ balls into $6$ boxes we have two possible cases
Case $1.$ There are exactly $2$ boxes with $2$ balls each
Case $2.$ There is exactly $1$ Box with $3$ balls in it.
For Case $1$, the distribution is done by choosing $2$ boxes from $6$ boxes and $2$ balls from $8$ balls and those two balls can be arranged in $2!$ ways. But I require another two balls from remaining $6$ balls to be placed in each of these two boxes which can be done in $\binom{6}{2} \times 2!$ ways. Finally remaining $4$ balls can be distributed in remaining $4$ boxes in $4!$ ways. So total number of ways for Case $1.$ is
$$\binom{6}{2} \times \binom{8}{2} \times 2! \times \binom{6}{2} \times 2! \times 4!=604800$$
For Case $2$, the number of ways is
$$\binom{6}{1} \times \binom{8}{3} \times 5!=40320$$
I am pretty sure the value obtained for Case $2.$ is correct. but I could not find where I went wrong in Case $1.$
|
On the case 1 you must choose first $2$ boxes over $6$ i.e. $C(6,2)$ and after for the first box you choose $(8\cdot 7)/2!$ and after $(6\cdot 5)/2!$ for the second box, and in third place you count the permutations of the others boxs that is $4!$, so the total is $\binom{6}{2}^2\cdot \binom{8}{2}\cdot4!=15^2\cdot 28\cdot 24=151200$.
And $151200+40320=191520$ as stated.
|
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$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $1^{\infty}$.
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}=\exp \left[{\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}-1\right)\times \frac{3x^2+8}{x^2}}\right]=e^{-8}$$
But i want to know,cant it be found using Taylor/Maclaurin series?What are the limitations of Taylor/Maclaurin series?
When can we not use them?
Is it true that we can use them whenever L Hospital rule is applicable(in $\frac{0}{0}$ and $\frac{\infty}{\infty}$ cases.)Can we use them in $0^0,\infty^0,1^\infty$ cases.
Please help me.Thanks.
|
Notice, we know $\color{blue}{\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$
Now we have
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$$$=\lim_{x\to 0}\left(1-\frac{2x^2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$
let, $\large \frac{2x^2}{5x^2+2}=\frac{1}{t}$$\implies x^2=\large \frac{2}{2t-5}$, hence
$$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t-17}$$
$$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t}\left(1-\frac{1}{t}\right)^{-17}$$
$$=\left(\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^t\right)^{8}\cdot \lim_{t \to \infty}\left(1-\frac{1}{t}\right)^{-17}$$
$$=(e^{-1})^{8}\cdot (1)^{-17}=\color{red}{\large e^{-8}}$$
|
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|
The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
|
Here is a different way. Let $f(x)=(1+x)^3(2+x^2)^{10}$. Then \begin{align}
f'(x)&=3(1+x)^2(2+x^2)^{10}+20x(1+x)^3(2+x^2)^9\\
f''(x)&=...\\
f'''(x)&=6 \left(x^2+2\right)^{10}+360 x (x+1) \left(x^2+2\right)^9+9 (x+1)^2
\left(20 \left(x^2+2\right)^9+360 x^2
\left(x^2+2\right)^8\right)+(x+1)^3 \left(1080 x
\left(x^2+2\right)^8+5760 x^3 \left(x^2+2\right)^7\right)
\end{align}
Hence the coefficient you are after is $$\frac16f'''(0)=16384$$
|
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|
What's $\lim\limits_{n\to\infty} \frac{x^n}{y^n + 1}$ For $x,y \in \mathbb{R}$, x > 1, y >0 what's the limit of $\frac{x^n}{y^n + 1}$?
If $x = y \implies x^n = y^n: $
Let $\varepsilon > 0, y^N > \frac{1}{\varepsilon}, N \in \mathbb{N}$
$|\frac{x^n}{y^n + 1} - 1| = |\frac{x^n}{y^n + 1} - \frac{y^n+1}{y^n+1}| = |\frac{x^n - y^n + 1}{y^n + 1}| = |\frac{1}{y^n+1}| = \frac{1}{y^n+1} < \frac{1}{y^n} < \frac{1}{y^N} < \varepsilon\ \forall n>N \implies \lim\limits_{n\to\infty}\frac{x^n}{y^n + 1} = 1$
But what if $x<y$?
I know that the limit is 0, so I need to show that $|\frac{x^n}{y^n + 1} - 0| < \dots < \varepsilon$. I know $x < y \implies x^n < y^n \implies \frac{x^n}{y^n} < 1$ which is probably useful.
And how to prove that if $x > y$ the limit is $\infty$. I probably need to show that $\forall a \in R \exists\varepsilon > 0: |a_n-a|>\varepsilon$ with using $\frac{x^n}{y^n}>1$
Am I on the right track here?
|
Assume $y<1$. Then the nominator goes to infinity while the denominator goes to $1$.
Now assume $y>1$. Write the quotient as $\frac{x^n}{y^n} \frac{1}{1+\frac{1}{y^n}}$. Since $y>1$ whe have that $\frac{1}{1+\frac{1}{y^n}}$ goes to $1$. So we only have to worry about $\frac{x^n}{y^n}$. If $x>y$ we see that the limit diverges and if $y>x$ the limit is $0$. If $x=y$ then the limit is $1$.
|
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|
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?
|
$$(3X^2+4X+2X^2-X-2)/(X^3+2X^2 )=(3X^2+4X)/(X^3+2X^2 )+(2X+3)/X^2 -8/(X+2)$$
$ln(X^3+2X^2 )+lnX^2-3/x-8ln(X+2)+c$
|
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|
Show that if $n \equiv 4\pmod 9$, then $n$ cannot be written as the sum of three cubes. This is a repeat of a question asked here.
Show that if $n \equiv 4 (\mod 9)$, then $n$ cannot be written as the
sum of three cubes.
Solution: Any integer has least residue as either $0,1,2,3,4,5,6,7,8 (\mod 9)$
Now, $$0^3\equiv0 \pmod 9\\
1^3\equiv1 \pmod 9\\
2^3\equiv8 \pmod 9\\
3^3\equiv0 \pmod 9\\
4^3\equiv 1\pmod 9\\
5^3\equiv 8\pmod 9\\
6^3\equiv 0\pmod 9\\
7^3\equiv 1\pmod 9\\
8^3\equiv 8\pmod 9$$.
So,any integer cube is congruent to either $0,1 \text{or} 8 \pmod 9$
It is not possible to produce $4$ with the combination of $0,1,8$.
Closest we can get is $1+1+1=3$ or $0+0+8=8$.
My question (1) is this method correct?
Question (2) is: Why the answer to the linked question says that $m^3 \equiv 0, \pm1 \pmod9$? How the user got "$-1$" and why "$8$" is missing?
|
HINT.- You have $8\equiv-1\pmod9$. Besides the same property goes for $n\equiv5\pmod9$, I mean
If $n\equiv5\pmod9$ then $n$ cannot be written as the sum of three cubes.
The proof is equal to the same as you have made for $n\equiv4\pmod9$
|
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|
How does this determinant calculation work? Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows:
$$
\text{det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_0 \\
-1 & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(x) \text{ det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_1 \\
-1 & x & 0 & ... & 0 & a_2 \\
0 & -1 & x & ... & 0 & a_3 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
+
\text{det}
\begin{bmatrix}
0 & 0 & ... & 0 & a_0 \\
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ = x(x^{n-1} + a_{n-1}x^{n-2}+...+a_1) + (-1)^{n-1}\text{det}
\begin{bmatrix}
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
0 & 0 & ... & 0 & a_0 \\
\end{bmatrix}
$$
I do not understand:
(1) how the determinant can be broken up into the sum of the determinants of the 2 smaller matrices and (2) how are the determinants of the 2 smaller matrices what they are?
|
The determinant was expanded along the first column, so you get
$$
\begin{bmatrix}
\color{red}{x} & 0 & 0 & ... & 0 & a_0 \\
\color{blue}{-1} & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(\color{red}{x}) \text{ det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_1 \\
-1 & x & 0 & ... & 0 & a_2 \\
0 & -1 & x & ... & 0 & a_3 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
- (\color{blue}{-1})^{1+2}
\text{det}
\begin{bmatrix}
0 & 0 & ... & 0 & a_0 \\
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
$$
After this just keep expanding along the first column.
|
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|
Definite integral problem involving trigonometric functions
$$\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+2\sin^2x}dx=?$$
Attempt:
$$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+(1-\cos 2x)}dx$$
$$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{2-\cos 2x}dx$$
$$=\sqrt{2}\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{1+\cos 2x}(2-\cos 2x)}dx$$
$$=\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{1+\cos 2x+2-\cos 2x }{\sqrt{1+\cos 2x}(2-\cos 2x)}dx$$
$$=\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\cos 2x}}{(2-\cos 2x)}dx+\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{1+\cos 2x}}dx$$
The second integrand is easy to calculate. How can I solve the first integrand?
(I would also like to know if there are easier methods)
|
Hint:
$$\begin{align}
I
&=\int_{0}^{\frac{\pi}{4}}\frac{\sec{\left(x\right)}}{1+2\sin^{2}{\left(x\right)}}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\cos^{2}{\left(x\right)}\left(1+2\sin^{2}{\left(x\right)}\right)}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\left(1-\sin^{2}{\left(x\right)}\right)\left(1+2\sin^{2}{\left(x\right)}\right)}\,\mathrm{d}x\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}y}{\left(1-y^{2}\right)\left(1+2y^{2}\right)};~~~\small{\left[x=\arcsin{\left(y\right)}\right]}\\
\end{align}$$
Presumably you can take it from there.
|
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|
Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$
|
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
$$(x-1)^2(x+2)\left[2(x-1)-3(x+2)\right]=0$$
$$(x-1)^2(x+2)(-x-8)=0$$
$$(x-1)^2(x+2)(x+8)=0$$
So this polynomial of degree $4$ has $4$ real roots i.e. $1,1,-2,-8$.
|
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|
A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac {\pi} {4} = \lim_{N \to \infty} s_N.$$
By Euler-MacLaurin summation formula,
$$\sum_{k = 0}^{N} \frac {1} {4k + 1} = \int_{0}^{N} \frac {dx} {4x + 1} + \frac {1} {2} \left(1 + \frac {1} {4N + 1} \right) + o \left (\frac {1} {N^2} \right)$$
and
$$\sum_{k = 0}^{N} \frac {1} {4k + 3} = \int_{0}^{N} \frac {dx} {4x + 3} + \frac {1} {2} \left(\frac {1} {3} + \frac {1} {4N + 3} \right) + o \left (\frac {1} {N^2} \right).$$
We then have
$$s_N = \frac {1} {4} \log \left(3 - \frac {6} {4N + 3} \right) + \frac {1} {3} + \frac {1} {(4N + 1) (4N + 3)} + o \left (\frac {1} {N^2} \right)$$
and
$$\lim_{N \to \infty} s_N = \frac {\log 3} {4} + \frac {1} {3}.$$
But $\frac {\log 3} {4} + \frac {1} {3} \ne \frac {\pi} {4}$. How come? Where have I done the mistake?
|
As pointed by others, you misused the Euler-Maclaurin formula.
Indeed, the next terms will involve some coefficients times powers of $1,\dfrac13,\dfrac1{4N+1}$ and $\dfrac1{4N+3}$. The constant terms do not vanish as $N\to\infty$.
|
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|
Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$ I need to find the sum of n terms of the series
$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$
And I've no idea how to move on. It doesn't look like an arithmetic progression or a geometric progression. As far as I can tell it's not telescoping. What do I do?
|
Clearly $$U_{r+1}=\frac{r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{2r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)-1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$
$$2U_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
Now let $$V_r=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$
Then $$V_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
Thus $$2U_{r+1}=V_r-V_{r+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\sum_{r=1}^{n} \{V_r-V_{r+1}\}=V_1-V_{n+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=V_1-V_{n+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\frac{1}{1}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}$$
|
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|
Hessian of a function Given this equation: $f(x,y,z) = \sqrt{1+x^2+y^2+z^2}$
I tried to calculate the Hessian --> for example $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ The second derivativ respect to x is hard to calculate for me: I tried the product rule: $x*(1+x^2+y^2+z^2)^{-1/2}$.
Then i get: $\frac{1}{1+x^2+y^2+z^2} - \frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$
But this is wrong.
Could anyone help?
|
Indeed $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ which you can write as $\frac{\partial f}{\partial x} = x(1+x^2+y^2+z^2)^{-\frac{1}{2}}$ so that $$\frac{\partial^2 f}{\partial x^2} = (1+x^2+y^2+z^2)^{-\frac{1}{2}} -\frac{1}{2} 2x^2 (1+x^2+y^2+z^2)^{-\frac{1}{2}-1} $$ $$ = (1+x^2+y^2+z^2)^{-\frac{1}{2}} - x^2 (1+x^2+y^2+z^2)^{-\frac{3}{2}}$$ $$ = (1+x^2+y^2) / \sqrt{1+x^2+y^2+z^2}^3$$ by putting $1/ \sqrt{1+x^2+y^2+z^2}$ in factor, and $$\frac{\partial^2 f}{\partial x \partial y} = xy (1+x^2+y^2+z^2)^{-\frac{3}{2}}.$$
Do you see now what you don't need to perform any other calculation to get all other partial derivatives needed to calculate the Hessian matrix ?
Note that I just used this : if $f(t)=g(t)^{\alpha}$ then $f'(t)=\alpha g'(t)g(t)^{\alpha-1}$ if $g$ is differentiable in one variable $t$
|
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|
Integrate $4x/(x^4-1)$ dx I'm having trouble solving the integral
$$\int_{5/4}^{13/12}\frac{4x}{x^4-1}\,dx$$
I have a feeling it is to do with the log integration identity but can't seem to manipulate it without involving complex numbers, something outside of this module (so I should be able to solve without them).
Any help appreciated, thanks in advance.
|
$$\int \frac{4x}{x^4-1}dx$$
Partial Fraction Decomposition on $\frac{4x}{x^4-1} = \frac{4x}{(x^2+1)(x^2-1)}$ gives $\frac{a}{x^2+1} + \frac{b}{x^2-1} = 4x \implies a(x^2-1)+b(x^2+1) = 4x$. Solving this gives $a=-2x, b=2x$
$$=\int\frac{2x}{x^2-1}-\int\frac{2 x}{x^2+1}dx$$
$$=\log|x^2-1| -\log(x^2+1) + C$$
|
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|
Minimal c satisfying $x+y-(xy)^c \geq 0$ for all $x,y\in [0,1]$ What is the minimal real $c$ satisfying $x+y-(xy)^c \geq 0$ for all $x,y \in [0,1]$?
Experimentally (though my experiments weren't necessarily accurate enough) I reached as low as $c=\tfrac{13}{32}$, where $c=\tfrac{12}{32}$ violates it.
|
Your answer is $c = \frac12$.
If $c = 1/2$,
$\begin{array}\\
x+y-(xy)^c
&=x+y-(xy)^{1/2}\\
&=x-(xy)^{1/2}+y/4-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2+3y/4\\
&\ge 0\\
\end{array}
$
so it is true for
$c \ge \frac12$
(since
$(xy)^{1/2+d}
\le (xy)^{1/2}
$
for
$d \ge 0$).
Suppose $c = \frac12-d$
where
$\frac12 > d > 0$.
Let $y = x$.
Then
$\begin{array}\\
x+y-(xy)^c
&=2x-(x^2)^{1/2-d}\\
&=2x-x^{1-2d}\\
&=x(2-x^{-2d})\\
\end{array}
$
and this is negative when
$2 < x^{-2d}$
or
$x^{2d} < \frac12$
or
$x <\frac1{2^{1/(2d)}}
$.
Therefore,
if $c < \frac12$,
there is $x$ and $y$
for which the inequality is false.
Therefore,
your answer is $c = \frac12$.
|
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|
Need help with $\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$ Please help me to evaluate this integral
$$\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$$
I tried a change of variable $x=\tanh z$, that transforms it into the form
$$\int_0^\infty\frac{4z}{\left(1+\log^2\tanh z\right)\sinh2z}\,dz,$$
but I do not know what to do next.
|
An alternative way to evaluate $$\frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-u}}{u} \, du ,$$ which is line $3d$ in robjohn's answer, is to add a parameter and then differentiate under the integral sign.
Specifically, let $$I(a) = \frac{\pi}{2}\int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-au}}{u} \, du.$$
Then $$ \begin{align} I'(a) &= - \frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) e^{-au} \, du \\ &= -\frac{\pi}{2} \int_{0}^{\infty} \left(\frac{1}{1+e^{- \pi u}}- \frac{e^{- \pi u}}{1+e^{-\pi u}} \right)e^{-au} \, du \\ &= -\frac{\pi}{2} \int_{0}^{\infty} \left(\sum_{n=0}^{\infty} (-1)^{n} e^{-n \pi u} + \sum_{n=1}^{\infty} (-1)^{n} e^{-n \pi u} \right)e^{-au} \, du \\ &= \frac{\pi}{2} \int_{0}^{\infty} \left(1- 2 \sum_{n=0}^{\infty} (-1)^{n}e^{-n \pi u} \right) e^{-au} \, du \\ &= \frac{\pi }{2a} -\pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{a+n \pi} \\ &= \frac{\pi }{2a} -\frac{1}{2} \psi \left(\frac{a+\pi}{2 \pi} \right) + \frac{1}{2} \psi \left(\frac{a}{2 \pi} \right) \tag{1}. \end{align}$$
Integrating back, we get $$ \begin{align} I(a) &= \frac{\pi}{2} \log(a) - \pi \log \Gamma \left(\frac{a+\pi}{2 \pi} \right) + \pi \log \Gamma\left(\frac{a}{2 \pi} \right) +C \\ &= \pi \log \left(\frac{\sqrt{a} \, \Gamma \left(\frac{a}{2 \pi } \right)}{\Gamma \left(\frac{a}{2 \pi} + \frac{1}{2} \right)} \right) + C,\end{align} $$
where $$\lim_{a \to \infty} I(a) =0 = \lim_{a \to \infty} \pi \log \left(\frac{\sqrt{a} \, \Gamma \left(\frac{a}{2 \pi } \right)}{\Gamma \left(\frac{a}{2 \pi} + \frac{1}{2} \right)} \right) +C $$
$$= \pi \log (\sqrt{2 \pi}) + C. \tag{2} $$
Therefore,
$$\frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-u}}{u} \, du = I(1) =\pi \log \left(\frac{ \Gamma \left(\frac{1}{2 \pi } \right)}{\sqrt{2 \pi} \, \Gamma \left(\frac{1}{2 \pi} + \frac{1}{2} \right)} \right).$$
$$ $$
$(1)$ http://mathworld.wolfram.com/DigammaFunction.html (6)
$(2)$ In general, for $x,y>0$, $\lim_{a \to \infty} \frac{a^{x} \Gamma(ya)}{\Gamma(ya+x)} = y^{-x}$. This can be proven using Stirling's approximation formula for the gamma function.
|
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|
Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather not $0$.
|
For
$\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}
$,
the ratio of consecutive terms is
$\dfrac{\dfrac{2^{n+1}(n+1)!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)(2n+5)}}{\dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}}
=\dfrac{2(n+1)}{(2n+5)}
=\dfrac{2n+2}{2n+5}
=1-\dfrac{3}{2n+5}
$
so that the terms decrease
in absolute value
and,
since
$\prod (1-\dfrac{3}{2n+5})
\to 0
$
(because
$\sum \dfrac{3}{2n+5}$
diverges,
the terms go to zero
so the series converges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I calculate $\int_0^ax^2\sqrt{a^2-x^2}dx$ via substitution? I have to calculate the integral $$\int_0^ax^2\sqrt{a^2-x^2}dx$$ using solely substitution (no integration by parts). $a$ is a positive constant.
I'm confused on how to do this?
|
Let $x=a\sin\theta\implies dx=a\cos \theta\ d\theta$
$$\int_0^a x^2\sqrt{a^2-x^2}\ dx=\int_{0}^{\pi/2} a^2\sin^2\theta(a\cos\theta)(a\cos \theta\ d\theta)$$
$$=a^4\int_{0}^{\pi/2} \sin^2\theta\cos^2\theta\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} (2\sin\theta\cos\theta)^2\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} \sin^22\theta\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} \frac{1-\cos4\theta}{2}\ d\theta$$
$$=\frac{a^4}{8}\int_{0}^{\pi/2} (1-\cos4\theta)\ d\theta$$
$$=\frac{a^4}{8}\left(\frac{\pi}{2}-0\right)=\frac{\pi a^4}{16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1584365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int \frac{1}{\sin x+\sec x}\,dx $ Evaluate $$\int \frac{1}{\sin x+\sec x}\,dx $$
Expressing $\sin x$ and $\cos x$ in terms of $\tan\frac{x}{2}$ i.e. putting $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ and hence $dx=\dfrac{2\,dt}{1+t^2}$
$$\int \frac{1-t^2}{1+t-t^3}\,dt $$
|
Probably a cleaner expression:
$$I=\int\dfrac{\cos x}{\sin x\cos x+1}dx =\int\dfrac{2\cos x}{2+2\sin x\cos x}dx$$
Now write $2\cos x=\cos x+\sin x+(\cos x-\sin x),$
As $\int(\cos x\pm\sin x)dx=\sin x\mp\cos x$
and $(\sin x\mp\cos x)^2=1\mp2\sin x\cos x,$
$$I=\int\dfrac{(\cos x+\sin x)}{3-(\sin x-\cos x)^2}dx+\int\dfrac{(\cos x-\sin x)}{1+(\sin x+\cos x)^2}dx$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Uniquely determining complex multiplication I encountered the following problem and it's not entirely clear to me exactly what I am supposed to do:
Show that the following rules uniquely determine complex multiplication on $\mathbb{C}=\mathbb{R^2}$:
*
*(a) $(z_1+z_2)w=z_1w+z_2w$
*(b) $z_1z_2=z_2z_1$
*(c) $i\cdot i=-1$
*(d) $z_1(z_2z_3)=(z_1z_2)z_3$
*(e) If $z_1$ and $z_2$ are real, $z_1\cdot z_2$ is the usual product of real numbers.
I get that (a) refers to distributivity, (b) refers to commutativity, (c) I'm not sure what the goal is there (is there a special name for this property?), (d) refers to associativity, and (e) refers to something else.
Looking at (b), for example (with $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$), I get that
\begin{align}
z_1z_2 &= (a_1+b_1i)(a_2+b_2i)\\[0.5em]
&= (a_1a_2+a_1b_2i+b_1ia_2+b_1ib_2i)\\[0.5em]
&= (a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
\end{align}
and
\begin{align}
z_2z_1 &= (a_2+b_2i)(a_1+b_1i)\\[0.5em]
&= (a_2a_1+a_2b_1i+b_2ia_1+b_2ib_1i)\\[0.5em]
&= (a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
\end{align}
Since $\operatorname{Re}(z_1z_2)=\operatorname{Re}(z_2z_1)$ and $\operatorname{Im}(z_1z_2)=\operatorname{Im}(z_2z_1)$, it is clear that $z_1z_2=z_2z_1$. But what does this really show? It seems like the question is asking for more (something about uniqueness, etc.).
Any ideas?
|
Well, using only the rules given, denoting the complex product with $\cdot$, we can say that for any complex numbers $z = a + i \cdot b$ and $w = c + i\cdot d$, we must have (using $(d)$ whenever there are terms involving at least two multiplications and writing $a \cdot b \cdot c$ for $a \cdot (b \cdot c) = (a \cdot b) \cdot c)$):
$$\begin{align*} z\cdot w &= (a+i\cdot b)\cdot (c+i\cdot d) \\ &= a \cdot (c+i\cdot d) + i\cdot b \cdot (c+i\cdot d) \quad \text{using (a)} \\ &= (c+i\cdot d) \cdot a + (c+i\cdot d)\cdot i\cdot b \quad \text{using (b)} \\ &= c\cdot a + i\cdot d \cdot a + c \cdot i\cdot b + i\cdot d \cdot i\cdot b \quad \text{using (a)} \\ &= c \cdot a + i \cdot d \cdot a + i \cdot c \cdot b + i \cdot i \cdot d \cdot b \quad \text{using (b)} \\ &= c \cdot a + i \cdot d \cdot a + i \cdot c \cdot b - d \cdot b \quad \text{using (c)} \\ &= c \cdot a - d \cdot b + i \cdot (d \cdot a + c \cdot b)\quad \text{using (a)} \\
&= c a - d b + i \cdot (d a + c b)\quad \text{using (e), denoting real multiplication by juxtaposition} \\
\end{align*}$$
So complex multiplication is uniquely determined by these rules.
|
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|
Sine of argument with large n approximation I have worked an integral and reduced the integral to
$$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$
I want to show that for $$n\rightarrow \infty$$ the above equation reduces to
$$\frac{1}{2}$$
Evidently, this means the $2$ sine functions must cancel each other. But what is a good way to do this? Large $n$ results in sine toggling between $-1$ and $1$. Notice that the either sine function will have a sign opposite to the other.
|
HINT:
$$\lim_{n\to\infty}\space\frac{n\pi+\sin\left(\frac{n\pi}{2}\right)-\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\lim_{n\to\infty}\space\left(\frac{1}{2}+\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}\right)=$$
$$\lim_{n\to\infty}\space\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\frac{1}{2}+\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{n}-\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{n}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$. Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$.
I know how to do it with the $x$ intercepts but I can't solve this.
|
The vertex form of the equation of a parabola is
$$f(x) = a(x - h)^2 + k$$
where $(h, k)$ is the vertex of the parabola. In this case, we are given that $(h, k) = (5, 0)$. Hence,
\begin{align*}
f(x) & = a(x - 5)^2 + 0\\
& = a(x - 5)^2
\end{align*}
Since we also know the parabola passes through the point $(7, -2)$, we can solve for $a$ because we know that $f(7) = -2$.
\begin{align*}
a(7 - 5)^2 & = -2\\
a(2)^2 & = -2\\
4a & = -2\\
a & = -\frac{1}{2}
\end{align*}
Thus, the given parabola has equation
$$f(x) = -\frac{1}{2}(x - 5)^2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to get $a \cos\left( \omega t \right)+b\sin\left( \omega t \right)=A \cos\left( \omega t+\phi \right)$? So this sprung up in a lecture note on quantum harmonic oscillation.
The equation is
$a \cos\left( \omega t \right)+b\sin\left( \omega t \right)=A \cos\left( \omega t+\phi \right)$
I must have forgotten how the right hand side follows from the left hand side.
Some hints would be helpful
|
Let $A = \sqrt{a^2+b^2}$ and $\phi$ be an angle such that $\cos\phi = \dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi = \dfrac{-b}{\sqrt{a^2+b^2}}$. Then, by using the identity $\cos(x+y) = \cos x\cos y-\sin x\sin y$, we have:
\begin{align}a\cos(\omega t)+b\sin(\omega t) &= \sqrt{a^2+b^2}\left[\dfrac{a}{\sqrt{a^2+b^2}}\cos(\omega t)-\dfrac{-b}{\sqrt{a^2+b^2}}\sin(\omega t)\right] \\ &=A\left[\cos\phi\cos(\omega t)-\sin\phi\sin(\omega t)\right] \\ &= A\cos(\omega t + \phi).\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1588325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $\frac{AC \times BC}{AD \times BD}$
$AC$ is $2004$. $CD$ bisects angle $C$. If the perimeter of $ABC$ is $6012$, find $\dfrac{AC \times BC}{AD \times BD}$.
Attempt
Let $c = AD+BD$.
We have that $\dfrac{AC}{AD} = \dfrac{BC}{BD}$. Thus, $$\dfrac{BD}{AD}+1 = \dfrac{BC}{AC}+1 \implies \dfrac{BD+AD}{AD} = \dfrac{AC+BC}{AC} = \dfrac{c}{AD} \implies AD = \dfrac{AC \cdot c}{AC+BC}.$$ Likewise it is easy to see by the angle bisector theorem that $BD = \dfrac{BC \cdot c}{AC+BC}$. Thus, we have that $\dfrac{AC \times BC}{AD \times BD} = \dfrac{AC \times BC}{\dfrac{AC \cdot BC \cdot c^2}{(AC+BC)^2}} = \dfrac{(AC+BC)^2}{c^2} = \dfrac{(2004+BC)^2}{(4008-BC)^2}$. But this answer seems to depend on $BC$, which makes no sense as the question indicates it is constant. What have I done wrong?
|
The problem is not well-posed. In fact, $BC$ can be anything between $1002$ and $3006$ (not inclusive), and the quantity in question can be any number greater than $1$ and less than $25$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the minimum value of $\sqrt { \frac { a }{ b+c } } +\sqrt [ 3 ]{ \frac { b }{ c+a } } +\sqrt [ 4 ]{ \frac { c }{ a+b } }$ If $a, b, c\ge 0$ with $(a+b)(b+c)(c+a) > 0$,
find the minimum of $\sqrt { \frac { a }{ b+c } } +\sqrt [ 3 ]{ \frac { b }{ c+a } } +\sqrt [ 4 ]{ \frac { c }{ a+b } }$.
The minimum is $\frac{3}{\sqrt[3]{4}}$ achieved at $b = 0, \frac{a}{c} = 2^{-4/3}$.
I am not able to progress in this problem.I tried applying AM-GM,Cauchy,Weighted AM-GM,etc. but none seem to provide fruitful results. Please help.
Source: A collection of problems which couldn't be solved by any teacher of my school.
Thanks.
|
Remark: Here is an ugly solution. Hope to see nice solutions.
Problem: Let $a, b, c \ge 0$ with $(a+b)(b+c)(c+a)\ne 0$. Find the minimum of
$f(a,b,c) = \sqrt{\frac{a}{b+c}} + \sqrt[3]{\frac{b}{c+a}} + \sqrt[4]{\frac{c}{a+b}}$.
Solution:
If $b=0$, by AM-GM inequality, we have $f = \sqrt{\frac{a}{c}} + \sqrt[4]{\frac{c}{a}} \ge \frac{3}{\sqrt[3]{4}}$
with equality if $\frac{a}{c} = 2^{-4/3}$.
Let us prove that the minimum of $f$ is $\frac{3}{\sqrt[3]{4}}$. It suffices to prove that, for $a, c \ge 0$ and $a+c > 0$,
$$\sqrt{\frac{a}{1+c}} + \sqrt[3]{\frac{1}{c+a}} + \sqrt[4]{\frac{c}{a+1}} \ge \frac{3}{\sqrt[3]{4}}$$
or
$$\sqrt{\frac{2a\sqrt[3]{2}}{1+c}} + \sqrt[3]{\frac{4}{c+a}} + \sqrt[4]{\frac{8c}{(a+1)\sqrt[3]{2}}} \ge 3.$$
Let
$$\frac{2a\sqrt[3]{2}}{1+c} = y^2, \quad \frac{8c}{(a+1)\sqrt[3]{2}} = x^4.$$
Correspondingly, we have
$$a = \frac{y^2(x^4\sqrt[3]{2} + 8)}{(16-x^4y^2)\sqrt[3]{2}}, \quad
c = \frac{x^4(y^2 + 2\sqrt[3]{2})}{16-x^4y^2}.$$
The constraint is $x, y \ge 0; \ x^4y^2 < 16$. It suffices to prove that
$$y + \sqrt[3]{\frac{2\sqrt[3]{2}(16 - x^4y^2)}{x^4y^2\sqrt[3]{2} + x^4\sqrt[3]{4} + 4y^2}} + x \ge 3.$$
It suffices to prove that, for $x, y \ge 0; \ x^4y^2 < 16; \ x + y < 3$,
$$\frac{2\sqrt[3]{2}(16 - x^4y^2)}{x^4y^2\sqrt[3]{2} + x^4\sqrt[3]{4} + 4y^2} \ge (3-x-y)^3$$
which is written as (after clearing the denominators)
$$-x^4(3 - x - y)^3 \sqrt[3]{4} + g(x, y)\sqrt[3]{2} - 4y^2(3 - x - y)^3 \ge 0$$
where $g(x, y) = 32 - 2x^4y^2 - (3-x-y)^3x^4y^2$.
Consider
$$F(q) = -x^4(3 - x - y)^3 q^2 + g(x, y)q - 4y^2(3 - x - y)^3.$$
Since $F(q)$ is concave and $\frac{5}{4} < \sqrt[3]{2} < \frac{19}{15}$, to prove $F(\sqrt[3]{2}) \ge 0$, it suffices to prove that
$$F(\tfrac{5}{4}) \ge 0, \quad F(\tfrac{19}{15}) \ge 0.$$
It suffices to prove that, for $x, y\ge 0$ and $x+y \le 3$,
$$F(\tfrac{5}{4}) \ge 0, \quad F(\tfrac{19}{15}) \ge 0.$$
They are verified by Mathematica. There are ugly proofs. Omitted.
|
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|
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
|
The Inequality $$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{9}{a+b+c}$$ follows by Cauchy Schwarts. Indeed, multiply by $a+b+c$ and write $$[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq 9$$
|
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|
How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5$$
where at least $2$ of the variables are $\geq 2$.
I found the total number of candidate solutions using the $\binom{B+N-1}{B} = 21$
However, only $9$ of them have two variables $\geq 2$.
\begin{align*}
2+0+3& =5\\
2+1+2& =5\\
3+0+2& =5\\
1+2+2& =5\\
3+2+0& =5\\
0+2+3& =5\\
0+3+2& =5\\
2+3+0& =5\\
2+2+1& =5
\end{align*}
I feel there is a connection to the Associated Stirling numbers of the second kind. But I can't place it :(
EDIT:
Here is my code for enumerating them all to count the number of ways of select B elements from a set of N (uniformly with replacement), such that you have at least C copies of K elements - also shows the output for this question I'm asking here as it's the core piece. Obviously can't be run for very large values of the parameters - that's why I'm here :) Code is here
Here is another example for B = 6, N = 3, C = 2 and K = 2 there are 16 solutions:
\begin{align*}
0+2+4& = 6\\
0+3+3& = 6\\
0+4+2& = 6\\
1+2+3& = 6\\
1+3+2& = 6\\
2+0+4& = 6\\
2+1+3& = 6\\
2+2+2& = 6\\
2+3+1& = 6\\
2+4+0& = 6\\
3+0+3& = 6\\
3+1+2& = 6\\
3+2+1& = 6\\
3+3+0& = 6\\
4+0+2& = 6\\
4+2+0& = 6\\
\end{align*}
There are a number of different and correct solutions below. I don't know which to accept.
|
Let $A_i$ be the set of solutions in nonnegative integers to $x_1+\cdots+x_n=B$ with $x_i\ge C$, for $1\le i\le n$,
and let $T_l=\sum\big|A_{i_1}\cap\cdots\cap A_{i_l}\big|$, where the sum is taken over all $l$-subsets of $\{1,\cdots,n\}$, for $1\le l\le n$.
Using Inclusion-Exclusion, the number of elements in at least k of $A_1,\cdots,A_n$ is given by
$\displaystyle\sum_{i=k}^n(-1)^{i-k}\binom{i-1}{k-1}T_i=\color{blue}{\sum_{i=k}^n(-1)^{i-k}\binom{i-1}{k-1}\binom{n}{i}\binom{B-Ci+n-1}{n-1}}$
since $T_i$ is the sum of $\binom{n}{i}$ terms, each of which has the value $\binom{B-Ci+n-1}{n-1}$.
See also this answer https://math.stackexchange.com/a/362516
for a related version of Inclusion-Exclusion.
Here are a few examples, which have been verified by considering cases:
$\textbf{1)}$ $B=6, n=3, C=2, k=2$:
$\displaystyle\sum_{i=2}^3(-1)^{i-2}\binom{i-1}{1}\binom{3}{i}\binom{8-2i}{2}=\binom{1}{1}\binom{3}{2}\binom{4}{2}-\binom{2}{1}\binom{3}{3}\binom{2}{2}=18-2=\color{blue}{16}$
$\textbf{2)}$ $B=12, n=5, C=4, k=2$:
$\displaystyle\sum_{i=2}^5(-1)^{i-2}\binom{i-1}{1}\binom{5}{i}\binom{16-4i}{4}=\binom{1}{1}\binom{5}{2}\binom{8}{4}-\binom{2}{1}\binom{5}{3}\binom{4}{4}=700-20=\color{blue}{680}$
$\textbf{3)}$ $B=18, n=6, C=3, k=4$:
$\displaystyle\sum_{i=4}^6(-1)^{i-4}\binom{i-1}{3}\binom{6}{i}\binom{23-3i}{5}=\binom{3}{3}\binom{6}{4}\binom{11}{5}-\binom{4}{3}\binom{6}{5}\binom{8}{5}+\binom{5}{3}\binom{6}{6}\binom{5}{5}=\color{blue}{5,596}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of limit with confusing moment Evaluate the following limit:
$\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x\cdot\sin x}{2\sin x+x\cdot \cos x}\color {\blue}{=}\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x}{2++x\cdot \frac{\cos x}{\sin x}}$ but $$\lim \limits_{x\to 0}x\cdot \frac{\cos x}{\sin x}=1$$ then $\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x}{2++x\cdot \frac{\cos x}{\sin x}}=\frac{2}{3}.$ Right?
But I have one question. In the second equality I divide enumerator and denominator to $\sin x$. Does it legal? What about if $\sin x=0$? Can anyone explain it to me please?
|
Since, $x\ne 0$ hence, one can also cancel $x$ as factor in numerator & denominator $$\lim_{x\to 0}\frac{\cos^{-3}x\cdot \sin x}{2\sin x+x\cdot \cos x}$$
$$=\lim_{x\to 0}\frac{\cos^{-3}x\cdot \frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}$$
$$=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}\left(\frac{1}{\cos x}\right)^3$$
$$=\frac{1}{2\cdot 1+1}\left(1\right)^3=\color{red}{\frac{2}{3}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
$5^n+2*3^{n-1}+1 = 5+2*1+1=8$ (mod n) for n coprime with 3 and 5. $n=8$ is such a number.
|
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|
Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$
This is not the starting inequality.
Is there something wrong in this method?
|
METHOD 1: Non-Calculus Based
In This Answer, I showed using basic tools only that the logarithm function satisfies the inequalities
$$\frac{x}{x+1}\le \log(1+x) \le x \tag 1$$
for $x\ge -1$. Note that $\frac{a+1}{a}=1+\frac1a$. Then, setting $x=\frac1a$ in $(1)$ gives the inequalities
$$\frac{1}{a+1}\le \log \left(\frac{a+1}{a}\right)\le \frac1a$$
The strict inequalities follows since the equality in $(1)$ occurs only when $x=0$. Since $\frac1a>0$, we have
$$\frac{1}{a+1} < \log \left(\frac{a+1}{a}\right) < \frac1a$$
METHOD 2: Calculus Based
Form the functions $f(x)=\log \left(1+x\right)-x$ and $g(x)=
\log(1+x)-\frac{x}{x+1}$ for $0<x$.
Then, note that $f'(x)=\frac{-1}{1+x}<0$ and $g'(x)= \frac{x}{(x+1)^2}>0$.
Therefore, $f(0)=0$ and $f'(x)<0$ implies $f(x) > 0$ for $x>0$. Then set $x=1/a$ and we have
$$\log \left(\frac{a+1}{a}\right) < \frac1a$$
Finally, $g(0)=0$ and $g'(x)>0$ implies $g(x)>0$. Then, set $x=1/a$ and we have
$$\frac{1}{a+1}< \log \left(\frac{a+1}{a}\right) $$
And we are done.
|
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|
Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$ Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequence.
It is $\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
So the sequence becomes $\sum_{k=1}^{1999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
I tried telescoping but i could not split it into two partial fractions.And this raised to $\frac{1}{2}$ is also troubling me.What should i do to find the answer?
|
After taking LCM, we get the general term of the series as:
$$\sqrt{\frac{(k^{2}+k+1)^{2}}{k^{2}(k+1)^{2}}} $$
$$=> \frac{k^{2}+k+1}{k^{2}+k}$$
$$=> 1 + \frac{1}{k^{2}+k}$$
So we have
$$\sum_{k=1}^{1999} 1 + \frac{1}{k^{2}+k}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k(k+1)}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k} - \frac{1}{(k+1)}$$
$$=> 1999 + 1 - \frac{1}{2000}$$
$$=> 2000 - \frac{1}{2000}$$
|
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|
How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
|
This is the graph of $x(\sqrt{x^2-x}-\sqrt{x^2-1})$.It is tending to $-\infty$ as $x$ is tending to $-\infty$
After rationalising
$L=\lim_{x\to -\infty}\frac{x(1-x)}{\sqrt{x^2}(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{x(1-x)}{|x|(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}$
$L=\lim_{x\to -\infty}\frac{x(1-x)}{-x(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{(1-x)}{-(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=-\infty$
Because when $x$ is negative ,$|x|=-x$,not $x$
|
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|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
|
$$\int \frac{x}{2x^2+x+3}dx$$
$$=\frac{1}{4} \int \frac{4x+1}{2x^2+x+3} dx- \frac{1}{4} \int \frac{1}{2x^2+x+3}$$
I presume you know how to solve the first integral [Do so using the sub: $u=2x^2+x+3$].
Complete the square of the second and substitute $m=\sqrt{2}x+\frac{1}{2\sqrt{2}}$to get:
$$\frac{-1}{4}\int\frac{1}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}dx=\frac{-1}{4}\int\frac{1}{\sqrt{2}}\frac{1}{(m^2+\frac{23}{8})}dm$$
Factor out the 23/8 and the constants, to get $$\frac{-\sqrt{2}}{23}\int\frac{1}{\frac{8m^2}{23}+1}dm$$
Substitute $q=2\sqrt{\frac{2m}{23}}$, solve the integral (it should be easy knowing the derivative of arctan(q)). Substitute back everything.
You should eventually get the below as the answer:
$$\frac{1}{4}\ln(2x^2+x+3)-\frac{\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)}{2\sqrt{23}}+C$$
|
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|
Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix},
&
&\text{(B)}\;
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \\
&\text{(C)}\;
\begin{pmatrix}
\phantom{-}\cos^{30} \theta & \sin^{30} \theta \\
-\sin^{30} \theta & \cos^{30} \theta
\end{pmatrix},
&
&\text{(D)}\;
\begin{pmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{pmatrix}.
\end{align*}
I think answer is B. But I am not sure.
|
Note that $A$ is a rotation matrix that takes a vector and rotates it by $1/67$th of a revolution. Now notice that $2010 = 67 \times 30$, so the answer is indeed (B).
|
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|
Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}
Find the area of the region determined by the system: \begin{align} y
& \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}
My attempt
*
*Assuming $x>0$
I have the system
$$\begin{cases}\begin{align}
y &\ge x \\
y & \le -x+3 \\
\end{align}\end{cases}$$
*
*Assuming $x<0$
$$\begin{cases}\begin{align}
y &\ge -x \\
y & \le x+5 \\
\end{align}\end{cases}$$
A relevant interval is $-1<x<0$ because in this interval $y \le -|x+1| +4 $ is still positive while $y \ge |x|$ is negative.
How do I combine now this information to solve the problem ?
I don't see how I can get something of the form $y \ge a$ and $y \le b$ which give clear bounds about area... in this problem it seems more complicated.
|
For $x > 0$, your area is the area enclosed by the graphs of $y = x$ and $y = -x + 3$ between $0$ and their intersection point. Thus, determine $x = -x +3$ which gets you $2x = 3 \Leftrightarrow x = {3\over 2}$ . So the first area is
$$\int_{0}^{3 \over 2} 2x - 3 dx$$
Process for $x < 0$ analogously.
|
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|
How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
|
We have
$$
\sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)=\ln\left(\prod_{k=1}^n\Big(1+\frac{1}{k+a}\Big)\right)=\ln\left(\prod_{k=1}^n\frac{k+1+a}{k+a}\right)=\ln\left(\frac{n+1+a}{1+a}\right)
$$
and thus
$$
\sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)=
\ln\left(\frac{n+1+a}{1+a}\right)-\ln\left(\frac{n+1+b}{1+b}\right) \\=\ln\left(\frac{1+b}{1+a}\right)-\ln\left(\frac{n+1+b}{n+1+a}\right)\to \ln\left(\frac{1+b}{1+a}\right),
$$
as $n\to\infty$, since
$$
\ln\left(\frac{n+1+b}{n+1+a}\right)=\ln\left(1+\frac{b-a}{n+1+a}\right)\to 0.
$$
|
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|
Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$ Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$
Let $P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$
By using $AM-GM$ inequality on the first two terms give me
$8(27)^{\log_{6}x}+27(8)^{\log_{6}x}\geq2\sqrt{8\times 27\times(27\times 8)^{\log_6x}}\geq 12\sqrt{6x^3}$
$P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3\geq12\sqrt{6x^3}-x^3$
I found the maximum value of $12\sqrt{6x^3}-x^3$ by differentiating and equating it to zero.I found that $12\sqrt{6x^3}-x^3$ is maximum at $x=6$.And its maximum value is $216$.
Can the maximum value of $12\sqrt{6x^3}-x^3$ be the maximum value of $P(x)?$I am confused because P(x)is greater than or equal to $12\sqrt{6x^3}-x^3$.$P(x)$ may be greater than the maximum value of $12\sqrt{6x^3}-x^3$.
|
Hint:
Let $E=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3=8(3^3)^{\log_{6}x}+27(2^3)^{\log_{6}x}-x^3$
$$=8(3)^{\log_{6}x^3}+27(2)^{\log_{6}x^3}-x^3$$
Now put $\log_{6}x^3=t$
$ \Rightarrow 6^t=x^3$
So we have that $E=8(3)^{t}+27(2)^{t}-6^t$
|
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|
Question about cancellations in a series resulting in $\frac {m+1}{m-k+1}$ I have the following series:
$$\displaystyle\sum_{i=1}^{k+1}iP_i = 1\left(1-\frac k m\right)+2\frac k m \left(1-\frac{k-1}{m-1}\right)+3\frac k m \frac {k-1}{m-1}\left(1-\frac{k-2}{m-2}\right)\\+...+(k+1)\frac k m \frac {k-1}{m-1}...\left(\frac{k-(k-1)}{m-(k-1)}\right) =\frac {m+1}{m-k+1}$$
I don't get how was the cancellation done here to get to that expression? How were the increasing $i$s canceled?
Trying to see the pattern with numbers, plugging $k=3, m=7$:
$1-\frac 3 7 +2\frac 3 7(1-\frac 2 6)+3\frac 3 7 \frac 2 6 (1-\frac 1 5)+4\frac 3 7 \frac 2 6 \frac 1 5 = \frac 4 7 (1+ 1 +\frac 3 5+ \frac 1 5 )=1.6$ which is correct but I still don't see the pattern.
|
This looks harder than it is.
Notice that your series can be written as:
\begin{align}
1 - & \frac{k}{m} + 2\frac{k}{m} - 2\frac{k}{m} \cdot \frac{k-1}{m-1}+3\frac{k}{m}\cdot \frac{k-1}{m-1}-3\frac{k}{m}\cdot \frac{k-1}{m-1}\cdot \frac{k-2}{m-2}+\dots+ \\ & (k-1)\cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} - (k-1)\cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} \cdot \frac{2}{m-k+2} + \\ & \dots + k \cdot \frac{k}{m}\cdot \frac{k-1}{m-1}\cdot \dots \cdot \frac{2}{m-k+2} - k \cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} + \dots + \\ & (k+1) \cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} \text{(last element equals $0$ as you noticed)} = \\ & 1 + \frac{k}{m} + \frac{k}{m} \cdot \frac{k-1}{m-1} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \frac{k-2}{m-2} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3}+ \\ & \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} \cdot \frac{2}{m-k+2} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} = \\ & 1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot(1+\frac{2}{m-k+2} \cdot (1+\frac{1}{m-k+1}))\dots)))) = \\ & 1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot(1+\frac{2}{m-k+2} \cdot (\frac{m-k+2}{m-k+1}))\dots)))) = \\ & 1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot (\frac{3}{m-k+3}\cdot(\frac{m-k+3}{m-k+1} )\dots)))) = \dots = \\ & 1+ \frac{k}{m-k+1} = \frac{m+1}{m-k+1}
\end{align}
|
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|
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
|
$$\cos x=\frac {e^{ix}+e^{-ix}}2$$
so
$$32\cos^6x=\cos 6x+6\cos 4x+15\cos 2x+10=$$$$=(1-18x^2+54x^4+\dots)+(6-48x^2+64x^4-\dots)+(15-30x^2+10x^4-\dots)+10$$
|
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|
Evaluating the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$
Evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^3}.$$
I tried integration by partial fraction. My work is below:
$$\int_{0}^{\infty}\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^{\infty}\frac{1}{x+1}+\frac{1}{3} \int_{0}^{\infty}\frac{2-x}{x^2-x+1}.$$
It seems that the result would go to infinity. But the answer is $\dfrac{2\pi}{3\sqrt3}$.
Did I do something wrong?
|
\begin{align}
\int_{0}^{\infty}\frac{1}{1+x^3}dx&\overset{x\to\frac1x}=\int_{0}^{\infty}\frac{x}{1+x^3}dx
= \frac12\int_{0}^{\infty}\frac{1+x}{1+x^3}dx\\
& = \frac12\int_{0}^{\infty}\frac{1}{x^2-x+1}dx
= \frac12\int_{0}^{\infty}\frac{d(x-\frac12)}{(x-\frac12)^2+\frac34}\\
&=\frac1{\sqrt3}\tan^{-1}\frac{2x-1}{\sqrt3}\bigg|_0^\infty= \frac1{\sqrt3}\left(\frac\pi2- (-\frac\pi6)\right)=\frac{2\pi}{3\sqrt3}
\end{align}
|
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|
Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.
Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you solve this using Cauchy-Schwarz?
|
More generally, the following is called Titu's Lemma or Engel's form of Cauchy-Schwarz inequality:
For all $a_i\in\mathbb R$, $b_i\in\mathbb R^+$:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$
Proof: by Cauchy-Schwarz: $$(b_1+b_2+\cdots+b_n)\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\right)$$
$$\ge \left(\sqrt{b_1}\sqrt{\frac{a_1^2}{b_1}}+\sqrt{b_2}\sqrt{\frac{a_2^2}{b_2}}+\cdots+\sqrt{b_1}\sqrt{\frac{a_n^2}{b_n}}\right)^2$$
$$=(|a_1|+|a_2|+\cdots+|a_n|)^2\ge (a_1+a_2+\cdots+a_n)^2$$
with equality if and only if $\frac{a_1^2}{b_1^2}=\frac{a_2^2}{b_2^2}=\cdots=\frac{a_n^2}{b_n^2}$ and $(|a_1|+|a_2|+\cdots+|a_n|)^2= (a_1+a_2+\cdots+a_n)^2$, i.e. if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.
|
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|
$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$
How does one calculate this limit?
Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:
$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =\lim_{x \to 0} \frac{x^2-x^2}{\tan(3x^4)} =0$
|
$$L=\lim_{x \to 0}\frac{x^2-sin^2x}{3x^4} \times \lim_{x \to 0} \frac{3x^4}{tan(3x^4)}=\frac{1}{3}\lim_{x \to 0}\frac{x^2-sin^2x}{x^4}$$
$$L=\frac{1}{3}\lim_{x \to 0}\frac{x-sinx}{x^3} \times \lim_{x \to 0}\frac{x+sinx}{x}=\frac{2}{3}\lim_{x \to 0}\frac{x-sinx}{x^3}$$
Now you can use Lhopital's rule
|
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|
Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\
(a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\
\end{array}
However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?
I think I've made the problem harder than it needs to be .
|
By C-S $\sum\limits_{cyc}(1+1)(a^2+b^2)\geq\sum\limits_{cyc}(a+b)^2$
and we are done!
|
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|
Prove that $(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$
For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$
What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$
Then it suffices to prove that $$(abc+2\sqrt2)^2 \geq 3(a+b+c)^2 \\ \Rightarrow abc+2\sqrt2 \geq \sqrt3(a+b+c)$$
But I don't know how to proceed. I also think I applied Holder's Inequality incorrectly.
|
By C-S $ (a^{2}+2)\left(1+\frac{(b+c)^{2}}{2}\right)\geq(a+b+c)^{2}$.
Hence, it remains to prove that $ (b^{2}+2)(c^{2}+2)\geq3\left(1+\frac{(b+c)^{2}}{2}\right)$,
which is equivalent to $ (b-c)^{2}+2(bc-1)^{2}\geq0$.
|
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|
Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\dfrac{\pi}{6}$,
$$2b^2+b-4=\sqrt{12-3b^2}$$
then $b$ such
$$b^3-3b+1=0$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
|
Hint $cos50\approx 0.6$ so create $cos(x+π/18)$ in terms of b from second equation and create $cosx$ in terms of b from first equation creating them and simplifying you get a huge equation! Which is $$8b^4+4b^3-18b^2+4b^2\sqrt{4-b^2}+6.6b\sqrt{4-b^2}-26.4\sqrt{4-b^2}+4=0$$ solving it you get roots but only $1.15023$ satisfies this as $b$ belongs to $(1,2)$ so its $1.42$ then approximately due to numerical approximation you get $cos(x)=0.62$ thus $x=π/6$
|
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|
What's wrong with this use of Taylor's expansions? I'm trying to find the value of the following limit:
$$
\lim_{x \to 0} \frac{x^2\cos x - \sin(x\sin x)}{x^4}
$$
Which I know equals to $-\dfrac13$.
I tried to do the following:
$$
\lim_{x \to 0} \frac{x^2(1 - \frac{x^2}{2} + o(x^4)) - \sin(x(x + o(x^3)))}{x^4}\\
= \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - \sin(x^2 + o(x^4)))}{x^4}\\
= \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - x^2 + o(x^4)}{x^4}
= -\frac12
$$
The result is clearly wrong. I suspect the mistake to be in the expansion of $\sin (x \sin x)$ but I don't get it.
What's wrong?
|
Since you have $x^4$ in the denominator, you need to expand the numerator up to $x^4$. You did that for $x^2\cos x$ but not for $\sin (x \sin x)$.
So, you need to include terms up to $x^4$ in the expansion of $\sin (x \sin x)$:
$$\sin (x \sin x) = x^2-x^4/6+o(x^6)$$
This comes for
$$\sin (x) = x-x^3/6+o(x^5)$$
$$x\sin (x) = x^2-x^4/6+o(x^6)$$
which you then feed to
$$\sin (y) = y+o(y^3)$$
|
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|
Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and say the answer is for the positive intervals. Is this correct / is there easier way to do it?
|
Raise $\frac{1}{3}$ to the power of both sides, and flip the inequality because $\frac{1}{3}^x$ is a decreasing function:
$$
\begin{aligned}
\log_{1/3}(x^2 - 3x + 3 ) &≥ 0\\
x^2 - 3x + 3 &\le \left(\frac{1}{3}\right)^0\\
x^2 - 3x + 3 &\le 1\\
x^2 - 3x + 2 &\le 0\\
\end{aligned}
$$
Now factor to get $(x-1)(x-2) \le 0$. We need a negative and a positive sign to make a negative, hence
$$1 \le x \le 2$$
|
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|
How to calculate the expected return in this coin toss game? In a coin tossing game that is made of two rounds, there are two cases.
In first round, if heads comes up, case 1 will be played in second round, if tails comes up, case 2 will be played in second round.
In case 1, two coins are tossed. If at least one of them is heads, player wins.
In case 2, two coins are tossed:
HH: Player wins,
HT and TH: Player loses
TT: Coin gets tossed again with the rules of case 2.
Now the probabilities of the first round are easy to calculate:
$$P(case_1) = P(case_1) = \frac{1}{2}$$
Probabilities of each individual cases are also easy:
For case 1:
$$P(win) = \frac{3}{4}$$
For case 2:
$$P(win) = \frac{1}{3} $$
Winning makes a profit of 1 dollar, and losing makes a loss of 1 dollar. What is the expected return of the game?
|
If we land on case 1 in round two, our game is guaranteed to end with either a win or a loss. The probability of a win is simply $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}.$
However, if we land on case 2 in round two, our game could last for an infinite amount of time. To solve for the probability of a win, we sum the infinite geometric series $\frac{1}{2} \cdot \left((\frac{1}{4})^{1} + (\frac{1}{4})^{2} + (\frac{1}{4})^{3} + ... \right) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.$
Our total win probability is $\frac{3}{8} + \frac{1}{6} = \frac{13}{24}.$
The probability of loss is just one minus the probability of a win. This is just $1 - \frac{13}{24} = \frac{11}{24}.$
Our expected return is $\frac{13}{24} \cdot 1 - \frac{11}{24} \cdot 1 = \boxed{+\frac{1}{12}}.$
|
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|
How many integer solutions does the following system have? $$y_1 + y_2 + y_3 = 20$$
$$1 \le y_1 \le 5$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
I know how to solve it if it were:
$$y_1 + y_2 + y_3 = 20$$
$$y_1 \ge 1$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
then I would do:
$$\begin{align}
x_1 + x_2 + x_3 &= (y_1 - 1) + (y_2 - 5) + (y_3 - 5) \\[2ex]
&= (y_1 + y_2 + y_3) - 11 \\[2ex]
&= 20 - 11 \\[2ex]
&= 9
\end{align}$$
and then the answer is $\binom{9+3-1}{3-1}=\binom{11}{2}$
but from here I don't understand how I can solve it when $(1 \le y_1 \le 5)$.
|
You wish to solve the equation
$$y_1 + y_2 + y_3 = 20 \tag{1}$$
in the integers subject to the restrictions $1 \leq y_1 \leq 5$, $y_2 \geq 5$, $y_3 \geq 5$. By making the substitutions of $x_1 + 1$ for $y_1$, $x_2 + 5$ for $y_2$, and $x_3 + 5$ for $y_3$, you obtained
$$x_1 + x_2 + x_3 = 9 \tag{2}$$
and correctly noted that if there were no restrictions, then equation 2 would have $$\binom{9 + 2}{2} = \binom{11}{2}$$ solutions. Note that the substitutions you made impose the restrictions that
$0 \leq x_1 \leq 4$, $0 \leq x_2$, and $0 \leq x_3$. Thus, we must exclude those solutions of equation 2 in which $x_2 \geq 5$. To do so, let
\begin{align*}
w_1 & = x_1 - 5\\
w_2 & = x_2\\
w_3 & = x_3
\end{align*}
Substituting for $w_1 + 5$ for $x_1$, $w_2$ for $x_2$, and $w_3$ for $x_3$ in equation 1 yields
$$w_1 + w_2 + w_3 = 4$$
which is an equation in the nonnegative integers with
$$\binom{4 + 2}{2} = \binom{6}{2}$$
solutions. Hence, the number of solutions of equation 1 subject to the restrictions $1 \leq y_1 \leq 6$, $y_2 \geq 5$, and $y_3 \geq 5$ is $$\binom{11}{2} - \binom{6}{2}$$
|
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|
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have no experience with it.
The answer according to Mathematica is when $A=B=C=60$.
Any ideas?
|
Due to the inequality of means and the law of cosines, we know that:
\begin{equation*}
4a^4=((a^2+b^2-c^2)+(a^2+c^2-b^2))^2\geq 4(a^2+b^2-c^2)(a^2+c^2-b^2)=\\ 16a^2bc\cos\beta \cos\gamma \Rightarrow a^2\geq 4bc\cos\beta \cos\gamma
\end{equation*}
\begin{equation*}
4b^4=((a^2+b^2-c^2)+(b^2+c^2-a^2))^2\geq 4(a^2+b^2-c^2)(b^2+c^2-a^2)=\\ 16ab^2c\cos\alpha \cos\gamma \Rightarrow b^2\geq 4ac\cos\alpha \cos\gamma
\end{equation*}
\begin{equation*}
4c^4=((a^2+c^2-b^2)+(b^2+c^2-a^2))^2\geq 4(a^2+c^2-b^2)(b^2+c^2-a^2)=\\ 16abc^2\cos\alpha \cos\beta \Rightarrow c^2\geq 4ab\cos\alpha \cos\beta
\end{equation*}
Multiplying the last three inequalities and extracting the root (assuming, without loss of generality, that alpha, beta and gamma are in the first quadrant) we will have the required inequality.
|
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|
Is mathematical Induction possible in this situation? Is mathematical Induction possible with this sigma sign?
$\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
with $n = 2s+1 ; s \epsilon \mathbb{N}$
Statement: $\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
Assumption: $\sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n+2}+1}{b+1}$
I already checked the basis but I have problems with splitting up the assumption.
I tried it successfully with:
$\sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) \equiv \sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) + (-1)^{n+1}*b^{n+1} + (-1)^{n}*b^{n}$
$\Leftrightarrow \sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) \equiv \sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) + b^{n+1} - b^{n}$
It is kind of consequentially if you look at this:
$\frac{b^{n}+1}{b+1}$
$\frac{b^{3}+1}{b+1} = b^{2}-b+1$
$\frac{b^{5}+1}{b+1} = b^{4} - b^{3} + b^{2}-b+1$
$
\frac{b^{7}+1}{b+1} = b^{6}-b^{5}+ b^{4} - b^{3} + b^{2}-b+1$
but I cant find a mathematical correct way to split my sigma sign like above.
I am pretty sure that the Statement is correct for any possible natural uneven number.
Now Im not even sure if i used the right Assumption for n -> n+2.
I would be very happy with some help :)
|
Let's pose
$$A_s = \sum_{k=1}^{2s+1} (-b)^{2s+1-k} = \frac{b^{2s+1}+1}{b+1}.$$
We start from $A_1$:
$$A_1 = b^2-b+1 = \frac{b^3+1}{b+1} = \frac{b^{2s+1}+1}{b+1}.$$
Now we need to find the inductive rule.
$$A_{s+1} = \sum_{k=1}^{2(s+1)+1} (-b)^{2(s+1)+1-k} = \sum_{k=1}^{2s+3} (-b)^{2s+3-k} = \\
= \sum_{k=1}^{2s+1} (-b)^{2s+3-k} + (-b)^{2s+3-(2s+2)} + (-b)^{2s+3-(2s+3)}= \\
= (-b)^2\sum_{k=1}^{2s+1} (-b)^{2s+1-k} + (-b)^{1} + (-b)^{0}= \\
= b^2 A_s -b + 1.$$
Finally:
$$A_{s+1} = b^2 A_s -b + 1 = b^2 \frac{b^{2s+1}+1}{b+1} -b + 1 = \frac{b^{2s+3}+b^2+(-b+1)(b+1)}{b+1} = \\
= \frac{b^{2(s+1)+1}+b^2-b^2 + 1}{b+1} = \frac{b^{2(s+1)+1} + 1}{b+1}.$$
|
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|
Convexity of the natural exponential fuction - directly from the definition Without using the Second Derivative Test, can the convexity of the natural exponential function be shown directly from the definition of convexity? The expression
\begin{equation*}
e^{t} = \sum_{n=0}^{\infty} \frac{t^{n}}{n!}
\end{equation*}
can be used. Here is what I have.
The natural exponential function is convex if, and only if, for any pair of conjugate real numbers $s$ and $t$ and for any real numbers $x$ and $y$,
\begin{equation*}
se^{x} + te^{y} \geq e^{sx + ty} .
\end{equation*}
This latter inequality is equivalent to
\begin{equation*}
\sum_{n=0}^{\infty} \frac{sx^{n} + ty^{n}}{n!}
\geq \sum_{n=0}^{\infty} \frac{(sx + ty)^{n}}{n!}
\end{equation*}
and to
\begin{equation*}
\sum_{n=0}^{\infty} \frac{sx^{n} + ty^{n} - (sx + ty)^{n}}{n!} \geq 0 .
\end{equation*}
Since $s + t = 1$, this last inequality is equivalent to
\begin{equation*}
\sum_{n=2}^{\infty} \left( s(1 - s^{n-1})x^{n} + t(1 - t^{n-1})y^{n}
- \frac{1}{n!} \sum_{i=1}^{n-1} \binom{n}{i} (sx)^{i}(ty)^{n-i}\right)
\geq 0 .
\end{equation*}
I am not sure that this last inequality is useful.
|
Demonstration in the case that $x > 0$, $y > 0$, and $s, \, t \in \mathbb{Q}^{+}$
Since $s$ is a positive, proper, rational number, there are positive integers $j < k$ such that $s = j/k$, and since $s$ and $t$ are a pair of conjugate numbers, $t = (k - j)/k$. For any positive, real numbers $z_{1} , \, z_{2} , \, \ldots \, z_{n}$, and for any positive integer $n$,
\begin{equation*}
\left(\frac{z_{1} + z_{2} + \ldots + z_{k}}{k}\right)^{n}
\leq \frac{{z_{1}}^{n} + {z_{2}}^{n} + \ldots + {z_{k}}^{n}}{k} .
\end{equation*}
If $z_{1} = z_{2} = \ldots = z_{j} = x$ and $z_{j+1} = z_{j+2} = \ldots = z_{k} = y$,
\begin{align*}
(sx + ty)^{n}
&= \left(\left(\frac{j}{k} \right) x + \left(\frac{k - j}{k} \right) y\right)^{n} \\
&= \left(\frac{jx + (k-j)y}{k}\right)^{n} \\
&\leq \frac{jx^{n} + (k-j)y^{n}}{k} \\
&= \left(\frac{j}{k}\right)x^{n} + \left(\frac{k-j}{k}\right)y^{n} \\
&= sx^{n} + ty^{n} .
\end{align*}
Consequently,
\begin{equation*}
\frac{(sx + ty)^{n}}{n!} \leq \frac{sx^{n} + ty^{n}}{n!} .
\end{equation*}
|
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|
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing the given number by $9$. The quotient was like $123456789012\cdots$
So the quotients sum is $(45\times 8)+1=361$
Is this the way to proceed? Is there a shorter way?
|
We are looking for
$ x = 10^{80} + 10^{79} + \dots + 10^2 + 10 + 1 \mod{9^3}$
$ 10^k = (9 + 1)^k = \sum\limits_{i=0}^{k}{k \choose i}9^i $ (Binomial theorem)
Considering the second equality $\mod{9^3} $, we only need first three summands of the sum. So $ 10^k \equiv 1 + k\cdot 9 + \frac{k\cdot(k-1)}{2}\cdot 9^2\mod{9^3}$. So what we're looking for is
$$ x\mod{9^3} \equiv\sum\limits_{k=0}^{80}1 + 9k + 81\frac{k\cdot(k-1)}{2} \mod{9^3} = \sum\limits_{k = 0}^{80}1 - \frac{63}{2}k + \frac{81}{2}k^2$$
Using formulas $ \sum_{k=0}^n k = \frac{n(n+1)}{2}, \sum\limits_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6} $ we get that
$$ x \mod{9^3} = 81 + \frac{-63}{2}\cdot80\cdot81\cdot\frac{1}{2} + \frac{81}{2}\cdot80\cdot81\cdot161\cdot\frac{1}{6} \mod{9^3}$$
$$ x \mod{9^3} = 81 + 81((-63)\cdot20 + 20\cdot27\cdot161) \mod{9^3}$$
Notice the term in brackets is divisible by $ 9 $, hence the answer is $ 81 $.
I'm not sure that's the best way to approach this. Please correct me if any of the calculations were wrong
|
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|
Evaluate $ \lim_{x \rightarrow - \infty} \frac{\sqrt{9x^6-x}}{x^3+1} $ I have started learning limits in calculus and I came across this question:
Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ .
I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\dfrac{1}{x^5}}}{1+\dfrac{1}{x^3}} = \lim_{x \rightarrow \infty} \dfrac{\sqrt{9}}{1} = \boxed{3} $
But now I am asked to compute $ \displaystyle \lim_{x \rightarrow -\infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $
How to solve for minus infinity? Where to put minus sign , I am getting confused , please help. Thanks.
|
Dividing by $|x^3|$ we get
$$\frac{\sqrt{9x^6-x}}{x^3+1}=\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}$$
As $x\to-\infty$ we have, for $x<0$,
\begin{align}
\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}&=\frac{\sqrt{\frac{9x^6-x}{x^6}}}{\frac{x^3+1}{-x^3}}=\frac{\sqrt{9-\frac{1}{x^5}}}{-1-\frac{1}{x^3}}\to\frac{\sqrt{9+0}}{-1+0}=\color{blue}{-3}
\end{align}
|
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|
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
|
Divide the interval into 3 pieces.
X is between (-4,0)
X is greater than 0
X is less than -4
On each interval try to figure out whether the sign is positive or negative.
$$\left (\frac{|x|}{|x+4|}\right)$$
|
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|
What is the inverse of $f(x) = \dfrac{x}{x^2 - 1}$ I need to find a continuous inverse of $$f(x) = \dfrac{x}{x^2 - 1}$$
Let $y = f(x) = \dfrac{x}{x^2 - 1} \Rightarrow y(x^2-1) = x \Rightarrow yx^2-x = y$
How should I proceed from here?
|
A plot of the function looks like this
To be invertible, a function of x and y must pass the horizontal line test. That is, any horizontal line cannot pass through more than one point of the graph.
This function fails the horizontal line test. But if you break the graph up into the blue and red branches shown (where we consider the two red branches to really be just one branch) each branch does pass the horizontal line test.
We haven't done any math yet, but we should expect that, when we solve for x in terms of y, we will get two answers. So let's do the math
\begin{align}
y &= \dfrac{x}{x^2 - 1}\\
y x^2 - y &= x\\
y x^2 - x - y &= 0\\
x &= \dfrac{1 \pm \sqrt{1 + 4y^2}}{2y}
\end{align}
A plot of $ \color{red}{x = \dfrac{1 + \sqrt{1 + 4y^2}}{2y}}$ and
$ \color{blue}{x = \dfrac{1 - \sqrt{1 + 4y^2}}{2y}}$ is shown below.
and the correspondences with the brances of the original function should be obvious.
There are a few things that still bother me.
First note that $ \color{blue}{x = \dfrac{1 - \sqrt{1 + 4y^2}}{2y}}$ passes through the origin even though it is undefined when $y = 0$. We also see that x is bound between plus and minus one but it is hard to see that by looking at the equation. For these reasons, I think we should write this particular equation in a different, more useful, form.
\begin{align}
x
&= \dfrac{1 - \sqrt{1 + 4y^2}}{2y}\\
&= \dfrac{1 - \sqrt{1 + 4y^2}}{2y}
\cdot
\dfrac{1 + \sqrt{1 + 4y^2}}{1 + \sqrt{1 + 4y^2}}\\
&= \dfrac{1 - (1 + 4y^2)}{2y(1 + \sqrt{1 + 4y^2})}\\
&= \dfrac{-4y^2}{2y(1 + \sqrt{1 + 4y^2})}\\
x &= \dfrac{-2y}{1 + \sqrt{1 + 4y^2}}
\end{align}
Now it's fairly clear that $x = 0$ when $y = 0$ and, as $y$ approaches minus and plus infinity, $x$ approaches plus and minus one.
|
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|
Is $7^{101}+8^{101}$ divisible by 25? If not, what is $ 7^{101} + 8^{101} \bmod 25$ What I derived is: $$\begin{align}7^{101}+8^{101} &\equiv (5+2)^{101}+ (5+3)^{101} \\
&\equiv 2^{101}+101\cdot5\cdot2^{100}+3^{101}+101\cdot 5\cdot 3^{100} \\
&\equiv 2^{101}+(100+1)\cdot5\cdot 2^{100}+3^{101}+(100+1)\cdot5\cdot3\cdot2^{100}\\
& \equiv 2^{101}+5\cdot 2^{100}+3^{101}+5\cdot3^{100} \pmod {25}.
\end{align}$$
Then I don't know what's the next step.
|
Use the lifting the exponent lemma:
$v_5(7+8)^{101}=v_5(7+8)+v_5(101)=1+0=1$, so $5^2\not\mid (7+8)^{101}$
The link is for the pdf-file called Lifting The Exponent Lemma by Amir Hossein Parvardi.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Linear combinations of vectors that depend on a parameter
Let $s \in \mathbb{R}$ and
$$\mathbf{u} = \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}, \quad
\mathbf{v}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad
\mathbf{w}=\begin{bmatrix} s^2 \\ 9 \\ s^2 \end{bmatrix}$$
for which values $s \in \mathbb{R}$ is
$$\mathbf{b} = \begin{bmatrix} 4+2s \\ 10 \\ 2s \end{bmatrix}$$ a linear combination of $\mathbf{u},\mathbf{v}$ and $\mathbf{w}$?
Through elementary row operations I've reduced the matrix to
\begin{bmatrix}
1 & 0 & 0 & 4\\
0 & 1 & s^2 & 2s\\
0 & 0 & \frac{s^2-9}{9} & \frac{1}{9}(2s+6) \\
\end{bmatrix}
I'm not too sure how to progress from here? I think this may be the next step:
$$\begin{bmatrix} x_1 \\ x_2+x_3 \\ x_3 \end{bmatrix} = \mathbf{b} = \begin{bmatrix} 4+2s \\ 10 \\ 2s \end{bmatrix}$$
then say that a solution is $s=0$, but it asks for values as in more than one, which I can't seem so see.
|
Note that $$\mathrm{det}\left(\begin{array}{rrr} 1 & 1 & s^2 \\ 4 & 1 & 9 \\ 0 & 1 & s^2\end{array}\right)=s^2-9.$$ Thus, if $s\ne \pm 3,$ the determinant is different from zero. Thus, the columns form a basis and so it is possible to write the vector $\bf{b}$ as a linear combination of the columns.
Now, if $s=-3$ it is
$$\left(\begin{array}{rrr} -2 \\ 10 \\ -6 \end{array}\right)=4\left(\begin{array}{rrr} 1 \\ 4 \\ 0 \end{array}\right)-6\left(\begin{array}{rrr} 1 \\ 1 \\ 1 \end{array}\right).$$
Finally, if $s=3$ one has to look if there exist $\alpha$ and $\beta$ such that
$$\left(\begin{array}{rrr} 10 \\ 10 \\ 6 \end{array}\right)=\alpha\left(\begin{array}{rrr} 1 \\ 4 \\ 0 \end{array}\right)+\beta\left(\begin{array}{rrr} 1 \\ 1 \\ 1 \end{array}\right).$$
(Note that $(9,9,9)^T=9(1,1,1)$ and thus we can omit it). Show that there is no solution. So, the vector ${\bf b}$ cannot be written in terms of the columns. An alternative way to show this is to prove that
$$\mathrm{det}\left(\begin{array}{rrr} 10 & 1 & 4\\ 10 & 4 & 1 \\ 6 & 0& 1 \end{array}\right)\ne 0.$$ In such a case the columns are linearly independent.
|
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|
Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$
Let $ \triangle ABC$ be an acute-angled triangle. Prove that
$ \sum\limits_\text{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2 $
Attempt
Since $\triangle{ABC}$ is acute, we may say that $A,B,C \in (0, \frac{\pi}{2})$. Now, we have that by a result for triangles $\displaystyle \sum_{cyc} \cot{A} \cdot \cot{B} = 1$. Then see that $$ \sum_{cyc} \sqrt{\cot{A}+\cot{B}} = \sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}}.$$ Now see that by Cauchy-Schwarz $\sqrt{(\cot{A}+\cot{B})(2)} \geq (\sqrt{\cot{A}}+\sqrt{\cot{B}})$ and thus $\sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}} \geq \sqrt{2}(\sqrt{\cot{A}}+\sqrt{\cot{B}}+\sqrt{\cot{C}})$. Now since $$\cot{A}\cot{B}+\cot{B}\cot{C}+\cot{A}\cot{C} = 1.$$
I get stuck here.
|
hint:
$\sqrt{\cot{A}+\cot{B}}=\sqrt{\dfrac{sin(A+B)}{sinAsinB}}=\sqrt{\dfrac{sinC}{sinAsinB}}=\dfrac{sinC}{\sqrt{sinAsinBsinC}}$
$3\sqrt[3]{sinAsinBsinC} \le sinA+sinB+sinC \le \dfrac{3\sqrt{3}}{2}$
|
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|
The tangent at a point $P$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$ meets the $y-$axis at $T,$then find $PT^2.$ The tangent at a point $P$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$ meets the $y-$axis at $T,$then find $PT^2.$
Let the point of tangency be $P(x_0,y_0)$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$.
Equation of the tangent is $(y-y_0)=(\frac{dy}{dx})_{(x_0,y_0)}(x-x_0)$
I found $\frac{dy}{dx}=\frac{2\sqrt{4-x^2}+x^2}{x\sqrt{4-x^2}}$
Now the question has become difficult to simplify.If i put $x=0$ to find the coordinates of $T(0,\ln(\frac{2+\sqrt{4-x_0^2}}{2-\sqrt{4-x_0^2}})-\sqrt{4-x_0^2}-\frac{2\sqrt{4-x_0^2}+x_0^2}{\sqrt{4-x_0^2}})$ and the coordinates of $P$ are $(x_0,\ln(\frac{2+\sqrt{4-x_0^2}}{2-\sqrt{4-x_0^2}})-\sqrt{4-x_0^2})$.
If i use the distance formula now it is very difficult to simplify.Is there an efficient method to find $PT^2$?
|
I found $\frac{dy}{dx}=\frac{2\sqrt{4-x^2}+x^2}{x\sqrt{4-x^2}}$
This is not correct. $$\small\begin{align}\frac{dy}{dx}&=\frac{2-\sqrt{4-x^2}}{2+\sqrt{4-x^2}}\cdot\frac{(2+\sqrt{4-x^2})'(2-\sqrt{4-x^2})-(2+\sqrt{4-x^2})(2-\sqrt{4-x^2})'}{(2-\sqrt{4-x^2})^2}-\frac{-2x}{2\sqrt{4-x^2}}\\&=\frac{\frac{-4x}{\sqrt{4-x^2}}}{(2+\sqrt{4-x^2})(2-\sqrt{4-x^2})}+\frac{x}{\sqrt{4-x^2}}\\&=\frac{-(4-x^2)}{x\sqrt{4-x^2}}\\&=-\frac{\sqrt{4-x^2}}{x}\end{align}$$
Then, having $P(x_0,y_0), T\left(0,y_0+\sqrt{4-x_0^2}\right)$ gives
$$PT^2=x_0^2+(4-x_0^2)=\color{red}{4}.$$
|
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|
How to simplify this? What did I do wrong? $$(3-\sqrt3) (2-\sqrt3)-\sqrt3\cdot\sqrt{27}.$$
So I simplified the $\sqrt3\cdot\sqrt{27}$ part into
$$\sqrt 3\cdot\sqrt{9\cdot3} = \sqrt3\cdot3\cdot\sqrt3=4\cdot\sqrt3$$
Then I multiplied the brackets :
$$6+\sqrt3-2\cdot\sqrt3-3\cdot\sqrt3 =6-4\cdot\sqrt3.$$
Then minus the $4\cdot\sqrt3$ we got
$$6-8\cdot\sqrt3$$
However the answer should be $-5\cdot\sqrt3$
How can I get that answer, what did I do wrong?
|
Second line, second equal sign is wrong.
$\sqrt{3}\cdot 3 \cdot \sqrt{3} = 9$
Multiplying the brackets yelds:
$6 - 2\sqrt{3} - 3\sqrt{3} +3 = 9 - 5\sqrt{3}$
Putting it all together:
$9 - 5\sqrt{3} - 9 = -5\cdot \sqrt{3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$ I start with a integral in complex plane $$\oint_c \frac{e^{izx} e^{zy} dz}{z\cosh(za)}$$ where $c$ is a countour starting in $z = -R$ along the real axis and jumping the pole at origin and continuing to $z = R$ and closing above the real axis with a semicircle circulating all the poles. I got $$\int_{0}^{\infty} \frac{e^{ky}\sin(kx)}{k\cosh(ky)}dk = \frac{\pi}{2} + 2\pi\sum_{n=1}^{\infty} (-1)^{n+1} \exp\left\{\frac{-xn\pi}{2a}\right\} \frac{\cos{\frac{yn\pi}{2a}}}{n}$$ and now i have no idea.
|
Assuming your summation is correct we have
$$-\sum_{n=1}^\infty (-1)^ne^{-\frac{\pi n x}{2a}}\frac{e^{\frac{\pi n y}{2a}i}+e^{-\frac{\pi n y}{2a}i}}{2n}=-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}+\frac{\pi n y}{2a}i}}{n}-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}-\frac{\pi n y}{2a}i}}{n}.$$
This can be rewritten
$$-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{\left(e^{-\frac{\pi x}{2a}+\frac{\pi y}{2a}i}\right)^n}{n}-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{\left(e^{-\frac{\pi x}{2a}-\frac{\pi y}{2a}i}\right)^n}{n}.$$
Using the series $-\log(1+z)=\sum_{n=1}^\infty (-1)^n\frac{z^n}{n}$, $|z|<1$, we obtain
$$\frac{1}{2}\log\left(1+e^{-\frac{\pi x}{2a}+\frac{\pi y}{2a}i}\right)+\frac{1}{2}\log\left(1+e^{-\frac{\pi x}{2a}-\frac{\pi y}{2a}i}\right),$$ for $\left|e^{-\frac{\pi x}{2a}\pm\frac{\pi y}{2a}i}\right|=e^{-\frac{\pi x}{2a}}<1\implies \frac{\pi x}{2a}>0$.
Converting to trigonometric functions,
$$\frac{1}{2}\log 2+\frac{1}{2}\log\left(\cosh \left(\frac{\pi x}{2 a}\right)-\sinh \left(\frac{\pi x}{2 a}\right)\right)+\frac{1}{2}\log\left( \cosh \left(\frac{\pi x}{2 a}\right)+\cos \left(\frac{\pi y}{2 a}\right)\right).$$
Hence, your sum is
$$\frac{\pi}{2}+\pi\log 2+\pi\log\left(\cosh \left(\frac{\pi x}{2 a}\right)-\sinh \left(\frac{\pi x}{2 a}\right)\right)+\pi\log\left( \cosh \left(\frac{\pi x}{2 a}\right)+\cos \left(\frac{\pi y}{2 a}\right)\right).$$
Simplifying gives
$$\frac{\pi}{2}+\pi\log 2-\frac{\pi^2 x}{2a}+\log\left(\cos\left(\frac{\pi x}{2a}\right)+\cosh\left(\frac{\pi x}{2a}\right)\right).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
|
$2 = 3-1$
$2^k = (3 - 1)^k = 3^k - k*3^{k - 1} ..... = \sum_{n = 0}^k {k \choose n}3^{k - n}(-1)^n = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n + {k \choose 3}3^{k - n}(-1)^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n \pm 1$
Since $k$ is odd:
$2^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n - 1$
$2^k + 1 = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n = 3\sum_{n = 0}^{k - 1} {k \choose n}3^{k - n - 1}(-1)^n$
which is a multiple of 3.
|
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"timestamp": "2023-03-29T00:00:00",
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|
is this correct $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$? I think i might have seen this result somewhere $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$ but cant remember for sure.
Is there a name for sums having limits in both lower and upper bound?
is there anything similar for other numbers e.g. $\lim_{ n \to \infty} \sum_{k=3^n}^{3^{n+1}} \frac{1}{k}= \ln 3 ?$
|
Let $\alpha=\lim\limits_{n\to\infty}\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$
The following is taken from this answer:
In this answer, It is shown that $\left(1+\frac1n\right)^n$ increases and $\left(1+\frac1n\right)^{n+1}$ decreases to $e$. Thus,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}\le\dots\le\left(1+\frac1{n+k}\right)^{n+k}\tag{5}
$$
and
$$
\left(1+\frac1n\right)^{n+1}\ge\left(1+\frac1{n+1}\right)^{n+2}\ge\dots\ge\left(1+\frac1{n+k}\right)^{n+k+1}\tag{6}
$$
Putting $(5)$ and $(6)$ together yields, for $0\le k\le n$,
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1{n+k}\right)^{\frac{n}{n+k}\frac{n+k+1}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)\tag{7}
$$
Therefore,
$$
\begin{align}
e^\alpha
&=\lim_{n\to\infty}\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{8}
\end{align}
$$
and by $(7)$
$$
\begin{align}
\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
&\le\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\dots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{9}
\end{align}
$$
Using the other direction of $(7)$, we get
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{10}
$$
By the Squeeze Theorem, $(8)$, $(9)$, and $(10)$ say
$$
e^\alpha=2\tag{11}
$$
which, by definition, means
$$
\alpha=\log(2)\tag{12}
$$
|
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|
Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$
Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$
Cauchy-Schwarz doesn't really help here and neither does rearrangement so I see nothing better to do than substitute: $$(1-2a)(1-2b)(1-2c)(a^2+b^2+c^2) = (2(b+c)-1)(2(a+c)-1)(2(a+b)-1)(a^2+b^2+c^2).$$ I get stuck here.
|
We need to prove that
$$9a^2b^2c^2\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)(a+b+c)$$
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Since $\prod\limits_{cyc}(a+b-c)>0$, we can assume that $x$, $y$ and $z$ are non-negatives.
Indeed, let $x<0$ and $y<0$. Hence, $x+y<0$, which gives $c<0$. Contradiction.
Thus, we need to prove that
$$9(x+y)^2(x+z)^2(y+z)^2\geq32xyz(x+y+z)\sum\limits_{cyc}(x^2+xy)$$
which is $\sum\limits_{sym}(9x^4y^2+9x^3y^3-7x^4yz-10x^3y^3z-x^2y^2z^2)\geq0$,
which is obvious by Muirhead.
|
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|
Solving a differential equation with Bernoulli's Method What approach do you take to solve the differential equation $ y' + (6y/x) = (y^3)/ x^5\ $ through the use of Bernoulli's method?
I've assumed u = y^(-2) for substitution, but I don't know where to go from there. The answer is y $ ((C\x^2\) + (1/8x^4))^(-.5) \ $
|
$\frac{dy}{dx}+6\frac{y}{x}=\frac{y^3}{x^5}$
Multiply by $\frac{1}{y^3}$
$\frac{1}{y^3}\frac{dy}{dx}+6\frac{1}{y^2x}=\frac{1}{x^5}$
Let $\frac{1}{y^2}=z$
$\frac{-2}{y^3}\frac{dy}{dx}=\frac{dz}{dx}$
$-\frac{dz}{dx}-12\frac{z}{x}=-2\frac{1}{x^5}$
$P(x)=-\frac{12}{x}$
$\int e^{-\frac{12}{x}}dx=x^{-12}$
${z}{x^{-12}}=\int \frac{1}{x^{12}}\frac{1}{x^5}dx$
${z}{x^{-12}}=-\frac{1}{16 x^{16}}+C$
$z=-\frac{1}{16x^4}+Cx^{12}$
I hope tou can continue by replacing $z=\frac {1}{y^2}$
|
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|
Solution of $ (dy/dx)^2 = (x^2+y^2) $ How do I find the solution of:
$$ (dy/dx)^2 = (x^2+y^2). $$
I tried $ y=tx$ and $x=r\cos(t) ,y=r\sin(t)$ substitutions but they did not help. I am not able to change it into any variable separable form and the only meaning I can make out is that the slope at any point is equal to the distance of it from origin.
|
$\left(\dfrac{dy}{dx}\right)^2=x^2+y^2$
$\dfrac{dy}{dx}=\pm\sqrt{x^2+y^2}$
Apply the Euler substitution:
Let $u=y\pm\sqrt{x^2+y^2}$ ,
Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$
$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=u-\left(\dfrac{u}{2}-\dfrac{x^2}{2u}\right)$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=\dfrac{u}{2}+\dfrac{x^2}{2u}$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}=\dfrac{u}{2}+\dfrac{x^2+2x}{2u}$
$(u^2+x^2)\dfrac{du}{dx}=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $\dfrac{dv}{dx}=2u\dfrac{du}{dx}$
$\therefore\dfrac{u^2+x^2}{2u}\dfrac{dv}{dx}=u^3+(x^2+2x)u$
$(u^2+x^2)\dfrac{dv}{dx}=2u^4+(2x^2+4x)u^2$
$(v+x^2)\dfrac{dv}{dx}=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$\dfrac{dv}{dx}=\dfrac{dw}{dx}-2x$
$\therefore w\left(\dfrac{dw}{dx}-2x\right)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$w\dfrac{dw}{dx}-2xw=2w^2+(4x-2x^2)w-4x^3$
$w\dfrac{dw}{dx}=2w^2+(6x-2x^2)w-4x^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{dx}=-\dfrac{1}{z^2}\dfrac{dz}{dx}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dx}=\dfrac{2}{z^2}+\dfrac{6x-2x^2}{z}-4x^3$
$\dfrac{dz}{dx}=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
|
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|
Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?
|
While working on a non-variational approach, I present a variational apporach.
By the given condition, $ab+bc+ca=3$, we get
$$
(b+c)\,\delta a+(c+a)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
For interior critical points maximizing $a+b+c$, we want
$$
\delta a+\delta b+\delta c=0\tag{2}
$$
To get $(2)$ for all variations that satisfy $(1)$, we get $b+c=c+a=a+b$, which means $a=b=c=1$. That is
$$
a+b+c=3\tag{3}
$$
The edge critical points will come when either $a=b+c$ or $b=c+a$ or $c=a+b$. Without loss of generality, assume $a=b+c$. $(1)$ becomes
$$
(2b+3c)\,\delta b+(3b+2c)\,\delta c=0\tag{4}
$$
and $(2)$ becomes
$$
2\delta b+2\delta c=0\tag{5}
$$
To get $(5)$ for all variations that satisfy $(4)$, we get $2b+3c=3b+2c$, which means $b=c=\sqrt{\frac35}$. That is
$$
a+b+c=4\sqrt{\frac35}\tag{6}
$$
The corner critical points come from $a=0$ or $b=0$ or $c=0$. Without loss of generality, assume $c=0$, which means $a=b=\sqrt3$. That is
$$
a+b+c=2\sqrt3\tag{7}
$$
Combining $(3)$, $(6)$, and $(7)$ gives
$$
3\le a+b+c\le2\sqrt3\tag{8}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $(a+\sqrt b)^n+(a-\sqrt b)^n$ where $n$ is natural For the expression $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$ where $n \in \mathbb{N}$, and $a,b, \in \mathbb{Q}$, the radical is always ends up cancelled, and the result is always in $\mathbb{Q}$. Is there any way that this could be reexpressed with the assumption that n is always a positive integer without the use of a square root operation, such as an alternative closed form for even $n$, and another closed form for odd $n$, or am I always stuck having to calculate a square root?
|
Note: $(x + y)^n = x^n + {n \choose 1} x^{n-1}y + \dotsc + y^n$
While $(x - y)^n = x^n - {n \choose 1} x^{n-1}y + \dotsc \pm y^n$
The even terms are opposite parity while the odd terms are the same so
$(x + y)^n + (x - y)^n = 2x^n + 2{n \choose 2}x^{n-2}y^2 + \dotsc$
So $(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2(a^{n} + {n \choose 2}a^{n -2}\sqrt{b}^{2} + \dotsc)$
So we are only taking $\sqrt{b} $ to even powers, and $\sqrt{b}^{2k} = b^k$.
So $\displaystyle(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2 \sum_{i = 0; \,i \text{ even}}^n {n \choose i}a^{n - i}b^{i/2}$.
...
Or if you don't like taking a sum with only even summands:
$\displaystyle(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2\sum_{k = 0}^{\lfloor n/2 \rfloor} {n \choose 2k}a^{n - 2k}b^{k}$
|
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|
Compute all roots of $(-8)^{\frac{1}{3}}$ $$(-8)^{\frac{1}{3}}$$
The problem states to compute all roots of the complex number above. Below is my attempt, but my inquiries are if I did it right and why it doesn't match Wolfram. Wolfram only returns $1$ result - was I only supposed to have $1$ as well??
$$\text{modulus} = 8$$
$$\theta = \arctan \left(\frac{0}{-8}\right) = 0$$
Because we are working in the left half of the complex plane, $\theta=\pi$.
$$z=\left(8e^{i(2\pi k+\pi)}\right)^{\frac{1}{3}}$$
This complex number has $3$ roots, so $k=0,1,2$ :
$$z_0=2e^{i\frac{\pi}{3}}$$
$$z_1=2e^{i\pi}$$
$$z_2=2e^{i\frac{5\pi}{3}}$$
|
We can use elementary factorization to find all roots. Let $x = \left(-8\right)^{\frac{1}{3}}\Rightarrow x^3 = -8\Rightarrow x^3 + 8 = 0\Rightarrow (x+2)(x^2-2x+4) = 0\Rightarrow (x+2)((x-1)^2 + 3) = 0\Rightarrow x = -2, (x-1)^2 = -3\Rightarrow x = -2, x-1 = \pm i\sqrt{3}\Rightarrow x = -2, 1 \pm i\sqrt{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Path needed for solving these linear equations in Zn (my example Z105) So these are two equations :
$$49x \equiv 21 \pmod {105}$$
$$64x \equiv 21 \pmod {105}$$
I should find the multiplicative inverse of $64$ and then $49$ that gets the result of $1$
so....
In the first equation i know that the $\gcd(49, 21)$ is $7$
So reducing them, I have $7x \equiv 7 \pmod {105}$
and $105 = 15 \times 7$
here I get lost!!
In the second equation
I know that $64$ and $21$ are coprime with each other, so euclidean algorithm, bezout identity, smth like 64v + 21w = 1 right.. and if I found V I basically found the x... but I get lost somewhere...
first I do the $\gcd(105, 64)$ then with bezout identity I want to find v and w such that = $1$
and I have $25$ for the bigger term and $-41$ for the smaller term...
but $64 x 25 + 21(-41)$ doesn't equal to $1$
again,, lost... connection not found!! :D
please help... maybe i chose the wrong direction, and math doesn't do for me!
|
The inverse of $64$ modulo $105$ can be found with the Euclidean algorithm:
\begin{align}
105&=64\cdot 1+41\\
64&=41\cdot 1+23\\
41&=23\cdot 1+18\\
23&=18\cdot 1+5\\
18&=5\cdot 3+3\\
5&=3\cdot 1+2\\
3&=2\cdot 1+1
\end{align}
Thus
\begin{align}\def\c#1{\color{red}{#1}}
1&=\c{3}-\c{2}\\
&=\c{3}-(\c{5}-\c{3})=(-1)\cdot\c{5}+2\cdot\c{3}\\
&=(-1)\cdot\c{5}+2(\c{18}-3\cdot\c{5})=(-7)\cdot\c{5}+2\cdot\c{18}\\
&=-7(\c{23}-\c{18})+2\cdot\c{18}=(-7)\cdot\c{23}+(-5)\cdot\c{18}\\
&=(-7)\cdot\c{23}+(-5)(\c{41}-\c{23})=(-2)\cdot\c{23}+(-5)\cdot\c{41}\\
&=(-2)(\c{64}-\c{41})+(-5)\cdot\c{41}=(-2)\cdot\c{64}+(-3)\cdot\c{41}\\
&=(-2)\cdot\c{64}+(-3)(\c{105}-\c{64})=\c{64}+(-3)\cdot\c{105}
\end{align}
and it turns out that the inverse is $64$.
Therefore, from the second equation, we get that
$$
x\equiv 64\cdot21\equiv1344\equiv84\pmod{105}
$$
Now
$$
49\cdot 84=4116\equiv21\pmod{105}
$$
and you're done.
|
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|
A sequence to achieve $\frac{1}{a_{2016}}$ It is given that $a_ka_{k-1} + a_{k-1}a_{k-2} = 2a_k a_{k-2}$ , $k\geq3$ and $a_1=1$.
We have $S_q= \sum_{k=1}^{q} \frac{1}{a_k} $ and given that $\frac{S_{2q}}{S_{q}}$ is independent of q then $\frac{1}{a_{2016}}$ is = ?
I think no information is given to find $a_2$. How should we approach this problem?
|
You can find $a_2$ by solving three simultaneous equations ($a_1$ fixed at $1$):
$$
a_3 a_2 + a_2 = 2_a3 \implies a_2 = \frac{2a_3}{a_3+1} \\
a_4a_3 + a_3a_2 = 2a_4a_2\implies a_4a_3 + a_3\frac{2a_3}{a_3+1} = 2a_4\frac{2a_3}{a_3+1}\implies a_4 = \frac{2a_3}{3-a_3} \\
\frac{S_4}{S_2} = \frac{S_2}{S_1} \implies \frac{1+\frac{a_3+1}{2a_3}+\frac{1}{a_3}+\frac{3-a_3}{2a_3}}{1+\frac{a_3+1}{2a_3}} = 1+\frac{a_3+1}{2a_3}
$$
That last equation simplifies to $$5a_3^2-6a_3+1=0$$. This has two solutions.
If we take $a_3 = 1$ then $a_2 = 1$ and $a_i = 1$ for all $i$, and in that case
$$
\frac{1}{a_{2016}} = 1$$
However, the answer your problem poser intended was probably the solution taking $a_3=\frac{1}{5}$. Then $a_2 = \frac{1}{3}$ and it is easy to show that in general
$$a_i = \frac{1}{2i-1}$$
Thus the answer you are probably looking for is
\frac{1}{a_{2016}} = 2015$$
|
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"url": "https://math.stackexchange.com/questions/1631077",
"timestamp": "2023-03-29T00:00:00",
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|
Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where to start. It’s trivial to construct some solution of the form $\int u'v dx = uv - \int uv' dx$ to the integral on the left, but I can’t see how to get at this exact one.
EDIT:
I have tried to solve it by splitting it up,
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx
\end{equation}
but as far as I can tell this results in something rather different from where I am supposed to end up:
\begin{equation}
\int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx = \frac{1}{a}arctan \Big(\frac{x}{a}\Big) \cdot \frac{1}{(x^2 + a^2)^{n-1}} + \frac{2(n-1)}{a} \int{\frac{x^2 arctan \big(\frac{x}{a}\big)}{(x^2 + a^2)^{n}}}dx
\end{equation}
It is quite possible that I have made a very obvious mistake, so apologies in advance.
I have also tried this:
\begin{equation}
\int{(x^2 + a^2)^{-n}dx} = \int{\Big(1 \cdot (x^2 + a^2)^{-n}\Big) dx}
= x(x^2 + a^2)^{-n} + n\int{\frac{2x^2}{(x^2 + a^2)^{n+1}} dx}
\end{equation}
Again, it doesn’t seem to lead me nearer the specific solution I need.
|
Let $I_{n}=\int \frac{dx}{(x^{2}+a^{2})^{n}}$, then
\begin{align*}
I_{n} &=
\frac{x}{(x^{2}+a^{2})^{n}}-
\int x \, d\left[\frac{1}{(x^{2}+a^{2})^{n}} \right] \\
&=\frac{x}{(x^{2}+a^{2})^{n}}+2n\int \frac{x^{2}dx}{(x^{2}+a^{2})^{n+1}} \\
&=\frac{x}{(x^{2}+a^{2})^{n}}+
2n\int \left[ \frac{1}{(x^{2}+a^{2})^{n}}-\frac{a^{2}}{(x^{2}+a^{2})^{n+1}}
\right] dx \\
&=\frac{x}{(x^{2}+a^{2})^{n}}+
2nI_{n}-2a^{2}nI_{n+1} \\
2a^{2}nI_{n+1} &=\frac{x}{(x^{2}+a^{2})^{n}}+(2n-1)I_{n} \\
I_{n+1} &=
\frac{x}{2a^{2}n(x^{2}+a^{2})^{n}}+\frac{2n-1}{2a^{2}n}I_{n} \\
I_{n} &=
\frac{x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}}+\frac{2n-3}{2a^{2}(n-1)}I_{n-1}
\end{align*}
|
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|
Integral of a trig function divided by the square root of a polynomial: $\int_a^b\frac{\sin x}{\sqrt{(x-a)(b-x)}}dx$? I was trying to help some physics students with an integral on their homework and they've presented me with something that has me stumped.
The integral they are working on is:
$$\int_a^b\frac{\sin x}{\sqrt{(x-a)(b-x)}}dx$$
They're definitely going to need a change of variable but any changes to simplify the bottom portion will greatly complicate the top. Any suggestions on how to tackle such a beast?
|
Rewrite as
$$\begin{align}\int_0^{b-a} dx \frac{\sin{(x+a)}}{\sqrt{x (b-a-x)}} &= \int_0^1 du \frac{\sin{[(b-a) u+a]}}{\sqrt{u (1-u)}}\\ &= \cos{a}\int_0^1 du \frac{\sin{[(b-a) u]}}{\sqrt{u (1-u)}} + \sin{a} \int_0^1 du \frac{\cos{[(b-a) u]}}{\sqrt{u (1-u)}}\\ &= 2 \cos{a} \int_0^1 du \frac{\sin{[ (b-a) u^2 ]}}{\sqrt{1-u^2}} + 2 \sin{a} \int_0^1 du \frac{\cos{[ (b-a) u^2 ]}}{\sqrt{1-u^2}}\end{align}$$
Now, consider
$$\begin{align}\int_0^1 du \, (1-u^2)^{-1/2} e^{i (b-a) u^2} &= \int_0^{\pi/2} dt \, e^{i (b-a) \sin^2{t}}\\ &= e^{i (b-a)/2} \int_0^{\pi/2} dt \, e^{-i [(b-a)/2] \cos{2 t}}\\ &= \frac12 e^{i (b-a)/2} \int_0^{\pi} dt \, e^{-i [(b-a)/2] \cos{t}}\\ &= \frac{\pi}{2} e^{i (b-a)/2} J_0 \left ( \frac{b-a}{2}\right ) \end{align}$$
Putting this altogether, the integral is
$$\pi J_0 \left ( \frac{b-a}{2}\right ) \left [ \cos{a} \sin{ \left ( \frac{b-a}{2}\right )} + \sin{a} \cos{ \left ( \frac{b-a}{2}\right )}\right ] $$
or
$$\int_a^b dx \frac{\sin{x}}{\sqrt{(x-a)(b-x)}} = \pi J_0 \left ( \frac{b-a}{2}\right ) \sin{ \left ( \frac{a+b}{2}\right )}$$
|
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|
How to find the real values of the parameter a, so that the inequality doesn't have positive solutions? The inequality is the following:
$$
\frac{x^2 - 6ax + 2x - 5a - 1}{x+a+1} < 0
$$
If we write this inequality like $\frac{a}{b} < 0$, I may say that for $a=0$, I found that
$X_1=3a - 1 - \sqrt{9a^2-a+2}$
$X_2=3a - 1 + \sqrt{9a^2-a+2}$
What should I do next? The equation should have only negative solutions.
|
For a fraction to be negative we need both the numerator and the denominator to have the opposite sign. For example, $(-2)/(+4) = -1/2$ and $(+3)/(-4) = -3/4$.
You need to solve the simultaneous inequalities:
Either $x^2+2(1-3a)x-(1+5a) > 0$ and $x+a+1<0$,
or $x^2+2(1-3a)x-(1+5a) < 0$ and $x+a+1>0$.
|
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|
Prove that if $|x| < 1$ then $\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}$. By multiplying power series, show that if $|x| < 1$ then
$$\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}.$$
So I know that:
$$\ln(1+x)=-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n},$$
when $|x|<1$, and:
$$\frac{1}{1+x}=\sum\limits_{n=0}^\infty (-1)^nx^n=1+\sum\limits_{n=1}^\infty (-1)^nx^n,$$
when $|x|<1$. So:
\begin{align*}
\frac{ln(1+x)}{1+x}={}&\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(1+\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}+\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\left(\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k} (-1)^n\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty\left[\frac{-(-1)^nx^n}{n}+ \sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k} (-1)^n\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}\left[\frac{-(-1)^nx^n}{n(-1)^{n-1}}+ \sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k} \frac{(-1)^n}{(-1)^{n-1}}\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}\left[\frac{1}{n}+ \sum\limits_{k=1}^n\frac{(-1)^{n-k}}{n-k} \right]x^n.
\end{align*}
Now I am stuck and not sure how to continue
|
The following is not an answer but cannot be a comment either.
Let me take from your attempt:
\begin{align*}
\frac{\ln(1+x)}{1+x}={}&\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(1+\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}+\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\left(\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-(-1)^kx^k}{k}(-1)^{n-k}x^{n-k}\right]={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-1}{k}(-1)^{n}x^{n}\right]={} \\
{}={}&\sum\limits_{n=1}^\infty\left[\frac{-1}{n}+\sum\limits_{k=1}^n\frac{-1}{k} \right](-1)^nx^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}x^n\left[\frac{1}{n}+\sum\limits_{k=1}^n\frac{1}{k}\right]x^n={}
\end{align*}
So the inside sum is indeed $s_n$, but we have a spurious $\frac1n$… If only the index of the inside sum stopped at $n-1$…
|
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|
Find a Mobius Transformation that carries the points $ -1, i, 1+i$ to the following: My goal is to find a Mobius transformation that transforms $-1, i, 1+i$ onto the points
a) $0, 2i, 1-i$
b) $i, \infty, 1$
For part a, I know that the Mobius transformation $M$ will be such that $M(-1) = 0$. So, this means that it should have the form
$$M(z) = \frac {z+1} {az+b}$$
Therefore, what should happen is that $M(i) = 2i$. That is,
$$M(i) = \frac {i+1} {ai+b}$$
Multiplying by the conjugate of the denominator, we obtain
$$M(i) = \frac {-(a+b)+(a-b)i} {-(a^2+b^2)} = 2i $$
Now, this means that $a = -b$, because the real part of this has to be $0$. Therefore, with this substitution, we can say that
$$M(i) = \frac {2ai} {-2a^2}= \frac {-i} {a} = 2i$$
Therefore, $ a = -\frac {1}{2}, b = \frac {1}{2}$.
However, when I use these values for the third point I do not find that $M(1+i) = 1-i$ is true. Is there something I'm missing here?
Also, for part b, I know the Mobius transformation has to be of the form
$$M(z) = \frac {az+b}{z-i} $$
because $0$ is the denominator whenever $z=i$, yielding a value of $\infty$ for $M$. However, I eventually run into similar problems like I'm having for part a). Can somebody tell why this isn't working, whether it's an algebraic mistake, or if I'm missing something more important? Thank you!
|
A Mobius transformation is an automorphism of the Riemann sphere $S = \Bbb C \cup \{ \infty \}$. It comes from a rational fraction of degree $1$ on $\Bbb C$, which is a map of the form $z \mapsto \frac {az+b}{cz+d}$ where $a,b,c,d$ are complex numbers.
In order for the map to be a bijection you need a condition on $a,b,c,d$, namely that $ad-bc \neq 0$ (if it is zero then $f$ is a constant map instead).
Even though you need $4$ numbers, multiplying all of them by a nonzero complex number doesn't change the map itself, so really the "space" of Mobius transformations has dimension $3$. This means that usually, you need $3$ data points to determine $f$. In our case, we have the image of $3$ distinct points, so this should be enough to determine $a,b,c,d$ up to a multiplicative constant.
Given a Mobius transformation $f$, the graph of $f$ is $\{(z,w) \mid w = f(z)\} = \{(z,w) \mid czw+dw-az-b = 0\}$, that is, the graph of $f$ in $\Bbb C^2$ is given by a linear relationship between $1,z,w,$ and $zw$.
If you think about this long enough, this shows that if you have three points $(z_i,w_i)$ on the graph of $f$, then the relationship between $z$ and $w$ is described by the equation :
$$ 0 = \begin{vmatrix} z_1w_1 & z_1 & w_1 & 1 \\ z_2w_2 & z_2 & w_2 & 1 \\ z_3w_3 & z_3 & w_3 & 1 \\ zw & z & w & 1 \end{vmatrix}$$
For case a), plugging in the values of $(z_i,w_i)$ you get
$$ 0 = \begin{vmatrix} 0 & -1 & 0 & 1 \\ -2 & i & 2i & 1 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix}$$
Next we do some row manipulations on the top $3$ rows to simplify it
$$\begin{vmatrix} 0 & -1 & 0 & 1 \\ -2 & i & 2i & 1 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 1+2i & 1+i & 2 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 0 & 1+i & 3+2i \\ 2 & 0 & 1-i & 2+i \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 0 & 1+i & 3+2i \\ 2 & 0 & 0 & 4i \\ zw & z & w & 1 \end{vmatrix}
$$
Now we divide row 3 by $2$, then we can develop this on the bottom row, to get the equation (getting the signs right is the hardest part)
$((-1)(1+i)(2i))zw + (1.(1+i).1)z + (1.(-1)(3+2i))w - (1.(-1).(1+i)) = 0$
This is equivalent to $2(1-i)zw + (1+i)z + (-3-2i)w + (1+i) = 0$
and to $M(z) = w = \frac {(1+i)(1+z)} {(-2+2i)z+(3+2i)}$
As for case b), since we got an $\infty$ in there, I can't exactly plug $w_2 = \infty$.
However, this is still equivlent to a linear relationship between $a,b,c,d$ : $f(z_2)=\infty \iff cz_2+d = 0$.
Or we can replace $z$ and $w$ with projective coordinates. Replace $z$ with $z/t$ and $w$ with $w/v$. While before we had a row $(zw : z : w : 1)$, now doing the replacement and multiplying it by $vt$ we get a row $(zw : zv : tw : tv)$, and now we can plug $\infty$ be putting $w/v = 1/0$
Anyway, you get :
$$ 0 = \begin{vmatrix} -i & -1 & i & 1 \\ i & 0 & 1 & 0 \\ 1+i & 1+i & 1 & 1 \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} 1-i & -1 & 0 & 1 \\ i & 0 & 1 & 0 \\ 1 & 1+i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} -i & -2-i & 0 & 0 \\ i & 0 & 1 & 0 \\ 1 & 1+i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix}
= \begin{vmatrix} -i & -2-i & 0 & 0 \\ i & 0 & 1 & 0 \\ 0 & 3i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix}
$$
We develop the bottom row and we get
$-(-2-i)zw + (-i)z + i(-2-i)w - (-i)(3i) = 0$
then $(2+i)zw - iz + (1-2i)w -3 = 0$
Which means $M(z) = w = \frac {iz+3}{(2+i)(z-i)}$
|
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|
Find the Jordan normal form of a nilpotent matrix $N$ given the dimensions of the kernels of $N, N^2, N^3$
Let $N\in \text{Mat}(10 \times 10,\mathbb{C})$ be nilpotent. Furthermore let $\text{dim} \ker N =3 $, $\text{dim} \ker N^2=6$ and $\text{dim} \ker N^3=7$. What is the Jordan Normal Form?
The only thing I know is that there have to be three blocks, since $\text{dim} \ker N = 3$.
Thank you very much in advance for your help.
|
Since $N$ is nilpotent, it has $0$ as unique eigenvalue. Since $\dim\ker N=3$, by the rank nullity theorem, the rank of $N$ is $7$. Hence the JNF of $N$ has seven 1's. Assume that there are $k$ blocks, and denote by $n_i$ the size of the $i$-th block. Of course, $n_1+\cdots+n_k=10$. Now, the rank of $N$ is $(n_1-1)+(n_2-1)+\cdots+(n_k-1)=10-k$, and we know that this rank is 7 (by rank nullity theorem). Hence k=3. You already got to that point.
Now, let $p$ be the number of blocks of size $1$. Then, the rank of $N^2$ is $(n_1-2)+\cdots+(n_k-2)+p=10-2k+p$. Since $\dim\ker N^2=6$, we can conclude (by the rank–nullity theorem again) that the rank of $N^2$ is $4$. Hence (since $k=3$) we conclude that $p=0$. So all blocks have a size at least 2.
Finally, let $q$ be the number of blocks of size $2$. Then, the rank of $N^3$ is $(n_1-3)+\cdots+(n_k-3)+q=10-3k+q$. Since $\dim\ker N^3=7$, we conclude that the rank of $N^3$ is $3$. Since $k=3$, we conclude that $q=2$. Hence $N$ has 2 blocks of size $2$ and one block of size $6$: we hence conclude that the JNF is (up to reordering of the blocks):
$$\left(\begin{array}{cc|cc|cccccc}0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&0&0&1&0&0&0&0\\0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&0&0&0\end{array}\right).$$
|
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|
integrate $\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx=\int \frac{3\left(\frac{16}{9}-x^2\right)^{\frac{3}{2}}}{x^6}dx$$
$x=\frac{4}{3}\sin\theta$
$dx=\frac{4}{3}\cos\theta d\theta$
$$\int \frac{3\left(\frac{16}{9}-\frac{16}{9}\sin^2\theta\right)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(1-\sin^2\theta)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(\cos\theta)^3}{x^6}d\theta=4\int \left(\frac{\cos\theta}{\frac{16}{9}\sin\theta}\right)^3d\theta$$
How should I continue?
|
Notice, you have made mistake while substituting
let $3x=4\sin\theta\implies dx=\frac{4}{3}\cos\theta\ d\theta$
$$\int \frac{(16-9x^2)^{3/2}}{x^6}\ dx$$$$=\int\frac{(16-16\sin^2\theta)^{3/2}}{\left(\frac 43\sin\theta\right)^6} \left(\frac{4}{3}\cos\theta\ d\theta\right)$$
$$=\frac{243}{16}\int \frac{\cos^4\theta}{\sin^6\theta}\ d\theta$$
$$=\frac{243}{16}\int \cot^4\theta\csc^2\theta\ d\theta$$
$$=-\frac{243}{16}\int \cot^4\theta(-\csc^2\theta\ d\theta)$$
$$=-\frac{243}{16}\int \cot^4\theta d(\cot\theta)$$
$$=-\frac{243}{16}\cdot \frac{\cot^5\theta}{5}+C$$
$$=-\frac{243}{80}\cot^5\theta+C$$
|
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|
Calculate $I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$ without using complex analysis
Calculate $$I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$$
without using complex analysis.
How to calculate without using the residue theorem?
The correct answer is
$$-\frac{1}{8}\ln^23+\frac{\pi^2}{72}-\frac{\pi\sqrt{3}}{36}\ln3$$
This integral is solved using complex analysis in the French book Gilbert Demengel "Balades sur les chemins du champs complexe"page 228 (exemple 4.53.)
The answer in the book $$\frac{1}{8}\ln^23-\frac{\pi^2}{72}-\frac{\pi\sqrt{3}}{36}\ln3$$ is wrong.
|
\begin{align}
I=&\int_0^{1}\frac{(1+x)\ln{\frac{x}{1-x}}}{x^2+x+1}dx\\
=&\ \frac12\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}
+ \frac{(2x+1)\ln x}{x^2+x+1}-\frac{(2x+1)\ln (1-x)}{x^2+x+1} \ dx\\
=&\ \frac12(I_1+I_2-I_3)
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}\overset{\frac {\sqrt3x}{1-x}\to x}{dx}
=\frac1{2\sqrt3}\int_0^\infty \frac{-\ln3}{x^2+\sqrt3x+1}dx
=-\frac{\pi\ln3}{6\sqrt3}\\
I_2&=\int_0^1 \frac{(2x+1)\ln x}{x^2+x+1}{dx}
= 2\Re \text{Li}_2\left(e^{i\frac{2\pi}3}\right)=-\frac{\pi^2}9\\
I_3&=\int_0^1 \frac{(2x+1)\ln (1-x)}{x^2+x+1}\overset{x\to 1-x}{dx}
= \int_0^1 \frac{(1-\frac23x)\ln x}{\frac13x^2-x+1}{dx}\\
&=-2\Re \text{Li}_2\left(\frac{1+i/\sqrt3}2\right)=\frac14\ln^23-\frac{5\pi^2}{36}
\end{align}
The identity $\Re\text{Li}_2(\frac{1+ia}2)=\frac{\pi^2}{12}-\frac12(\tan^{-1}a)^2-\frac18\ln^2\frac{1+a^2}4$ is used in evaluating $I_3$ and $\Re\text{Li}_2(e^{ia})=-\frac{\pi^2}{12}+\frac14(\pi-a)^2 $ in $I_2$. Substitute $I_1$, $I_2$ and $I_3$ into $I$ to obtain
$$I=\frac{\pi^2}{72}-\frac{\pi}{12\sqrt3}\ln3 -\frac{1}{8}\ln^23$$
|
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|
Number of ways of selecting 3 numbers from $\{1,2,3,\cdots,3n\}$ such that the sum is divisible by 3
Find the Number of ways of selecting 3 numbers from $\{1,2,3,\cdots,3n\}$ such that the sum is divisible by 3. (Numbers are selected without replacement).
I made a list like this:
The sum of all elements along such diagonals are divisible by 3. Number of ways of selecting such numbers is $3n-2$
Then I shifted the second row by one element and third row by one element behind second row:
Again the sum of elements are divisible by 3. Total number of ways is $3n-4$.
In the next list, I will shift the second row by two elements and third row by 2 elements behind second row. The elements along a diagonal will be $\{1,4,7\}\cdots\{3n-6,3n-3,3n\}$. Sum is divisible by 3. Number of such possibilities is $3n-6$.
The total number of ways is $3n-2+3n-4+3n-6+\cdots$
what will be the last case?
|
Let $S = \{1, 2, 3, \ldots, 3n - 2, 3n - 1, 3n\}$. Let
\begin{align*}
A & = \{k \in S \mid k \equiv 0 \pmod{3}\}\\
B & = \{k \in S \mid k \equiv 1 \pmod{3}\}\\
C & = \{k \in S \mid k \equiv 2 \pmod{3}\}
\end{align*}
Observe that $|A| = |B| = |C| = n$. We can choose three numbers from $S$ in the following ways:
*
*Choose $3$ elements of $A$, which can be done in $\binom{n}{3}$ ways.
*Choose $3$ elements of $B$, which can be done in $\binom{n}{3}$ ways.
*Choose $3$ elements of $C$, which can be done in $\binom{n}{3}$ ways.
*Choose $1$ element from each subset, which can be done in $\binom{n}{1}^3$ ways.
Hence, the number of ways of selecting a subset of three numbers in $S$ whose sum is divisible by $3$ is
\begin{align*}
3\binom{n}{3} + \binom{n}{1}^3 & = 3 \frac{n(n - 1)(n - 2)}{3!} + n^3\\
& = \frac{n^3 - 3n^2 + 2n}{2} + n^3\\
& = \frac{3n^3 - 3n^2 + 2n}{2}
\end{align*}
|
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|
Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\right) $$
Now no terms cancel. We get 0 in numerator and denominator too.
Ans: My teacher told me that the limit is $+\infty$, but didn't tell how.
|
The numerator is positive for both LHL and RHL, but the denominator is $\rm +ve$ for one and $\rm -ve$ for other:
$$\begin{align}\text{Since }\\
&\text{as }x\to-3^-,\ \ \sqrt{x^2+16}-5 >0 \\
&\lim_{x\to-3^-}f(x)=\infty
\end{align}$$
$$\begin{align}\text{also,}\\
&\text{as }x\to-3^+,\ \ \sqrt{x^2+16}-5 <0 \\
&\lim_{x\to-3^+}f(x)=-\infty\\
\end{align}$$
$$\rm RHL\neq LHL \implies \text{lim D.N.E}$$
|
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|
Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.
Can you please explain?
|
It's basically the number of ways you can write $12$ as a sum of three integers all greater than or equal to $3$, where order matters. So $12=3+3+6$, $12=3+4+5$, $12=4+4+4$. The first can be rearranged three ways, the second can be rearranged six ways and the last can only be arranged one way. So yes, I believe the answer is $10$.
|
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|
Integration of $\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx$ How can we integrate:
$$
\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx
$$
Using simple algebraic identities I deduced it to
$$
\int\frac{1-2\sin^2x\cdot\cos^2x}{(\sin x+\cos x)(1-\sin x\cdot\cos x)}dx
$$ but can't proceed further. Please provide some directions?
|
$\displaystyle\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{\sin^3 x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}$,
and $\;\;\displaystyle\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$
where $\sin x\cos x=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.
Then $A=\frac{1}{3}$ and $B=-\frac{1}{3}$,
so $\displaystyle\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx=\int\left(\sin x+\cos x-\frac{1}{3}\cdot\frac{\sin x+\cos x}{1-\sin x\cos x}+\frac{1}{3}\cdot\frac{1}{\sin x+\cos x}\right)dx$
$\displaystyle=-\cos x+\sin x-\frac{1}{3}\int\frac{2\sin x+2\cos x}{2-2\sin x\cos x}dx+\frac{1}{3}\int\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\int\frac{\sin x+\cos x}{1+(\sin x-\cos x)^2}dx+\frac{1}{3\sqrt{2}}\int\csc\left(x+\frac{\pi}{4}\right)dx$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\arctan(\sin x-\cos x)+\frac{1}{3\sqrt{2}}\ln\big|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\big|+C$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\arctan(\sin x-\cos x)+\frac{1}{3\sqrt{2}}\ln\left\vert\frac{\sqrt{2}-\cos x+\sin x}{\sin x+\cos x}\right\vert+C$
|
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|
Distribution of the number of times a player wins a game with random numbers Five distinct numbers are randomly distributed to players numbered
$1$ through $5$. Whenever two players compare their numbers, the one
with higher one is declared the winner. Initially, players $1$ and $2$
compare their numbers; the winner then compares with player $3$,
and so on. Let $X$ denote the number of times player $1$ is winner. Find $$P(X = i), \quad \text{ for }\ i = 0, 1, 2, 3, 4$$
|
In the following I denote the numbers that are distributed with $1,2,3,4$ and $5$ since the only thing that matters is their order and not their exact values. In total, there are $5!=120$ ways for these numbers to be distributed among the $5$ Players, each of these ways being equally probable. So, $$P(X=i)=\frac{\text{ways in which Player 1 wins exactly $i$ games}}{120}$$ for $i=0,1,2,3,4$.
*
*$i=4$. This is the most straightforward case, since Player $1$ can win all $4$ comparisons if and only if he has the highest number and the other numbers are distributed randomly among the others, i.e. $$P(X=4)=\frac{1\cdot4!}{120}=\frac{24}{120}$$
*$i=3$. For Player $1$ to win exactly $3$ comparisons (and hence lose the $4$-th and last comparison) he must have the second highest number, and the Player with the highest number must be Player $5$. This gives only $1$ choice for the numbers of Players $1$ and $5$ and the other $3$ numbers may be distributed randomly among the $3$ remaining Players, hence $$P(X=3)=\frac{1\cdot3!\cdot1}{120}=\frac{6}{120}$$
*$i=2$. I will spare me this one, because it is the most cumbersome. I will skip to $i=1$ and $i=0$ and use the fact that all probabilities must add up to $1$ and return to find this by complementarity.
*$i=1$. For Player $1$ to win exactly $1$ comparison (and hence lose the second comparison) he must have a higher number than Player $2$ and a lower number than Player $3$. This can happen in following ways (notation (no of Player $1$/ no of Player $2$ / no of Player $3$): $(2/1/3-4 \text{ or } 5)=3$ ways, $(3/1 \text{ or }2/ 4 \text{ or }5)=4$ ways, $(4/ 1,2 \text{ or }3/ 5)= 3$ ways. For any of these ways we must multiply with $2!$ for the ways to fill the last two places. So, in total, this gives
$$P(X=1)=\frac{10\cdot2!}{120}=\frac{20}{120}$$
*$i=0$. Player $1$ will win exactly $0$ comparisons if he has a lower number than Player $2$. So, the possible ways for this to happen are: $(1/ \text{ any })= 4$ ways, $(2/3,4 \text{ or } 5)=3$ ways, $(3/ 4 \text{ or }5)=2$ ways and $(4/ 5)=1$ way. For each of these ways we must multiply with $3!$ for the ways to distribute the rest of the numbers to Players $3$ to $5$. So, this gives $$P(X=0)=\frac{10\cdot3!}{120}=\frac{60}{120}$$
So, by complementarity $$P(X=2)=1-P(X\neq 2)=\frac{120-24-6-20-60}{120}=\frac{10}{120}$$
|
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|
Convergence of the series $\sum_n (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$ Let's consider the series $\sum a_n$, with $$a_n = (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$$
It looks absolutely like an example for the Leibniz test but here the signs don't interchange one for one. How can we prove the convergence of this series?
|
You can reduce the question to a question about Liebniz series by inserting parenthesis around the terms that have the same sign. The original series $\sum_{n=1}^{\infty} a_n$ is:
$$ \frac{1}{1} - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{9} - \frac{1}{10} - \dots. $$
Denote the partial sums of the series $\sum_{n=1}^{\infty} a_n$ by $S_n = \sum_{k=1}^n a_k$. Inserting parenthesis around the terms with the same sign, we obtain a new series $\sum_{n=1}^{\infty} b_n$:
$$ \frac{1}{1} - \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \dots + \frac{1}{9} \right) - \dots $$
Formally, what we do is we define $b_n$ by letting
$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}, \\
b_n = (-1)^{n+1} c_n = (-1)^{n+1} \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}. $$
Denote by $T_n$ the partial sums of the series $\sum_{n=1}^{\infty} b_n$. The fact that the new series $\sum_{n=1}^{\infty} b_n$ is obtained by inserting parenthesis around the terms corresponds to the fact that the sequence of partial sums of $\sum_{n=1}^{\infty} b_n$ is a subsequence of the sequence of partial sums of $\sum_{n=1}^{\infty} a_n$. Namely, $T_n = S_{n^2}$. If we can show that $c_n$ are monotonically decreasing to zero, we will obtain that $\sum_{n=1}^{\infty} b_n$ is a Liebniz series and hence converges. This shows that $S_{n^2}$ has a limit. Since the series $\sum_{n=1}^{\infty} b_n$ was obtained from the series $\sum_{n=1}^{\infty} a_n$ by inserting parenthesis around terms that have the same sign, this implies that $S_n$ also has a limit (the same limit).
Estimating $c_n$, we have
$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} \leq \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{(n-1)^2 + 1} = \frac{n^2 - (n-1)^2}{(n-1)^2 + 1} = \frac{2n - 1}{(n-1)^2 + 1} \xrightarrow[n \to \infty]{} 0 $$
so $c_n$ tends to zero.
Finally,
$$ c_n - c_{n+1} = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} - \sum_{k=n^2 + 1}^{(n+1)^2} \frac{1}{k} \\
= \sum_{i = 0}^{2n - 2} \left( \frac{1}{(n-1)^2 + 1 + i} - \frac{1}{n^2 + 1 + i} \right) - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
= \sum_{i = 0}^{2n - 2} \frac{2n - 1}{\left( (n-1)^2 + 1 + i \right) \left( n^2 + 1 + i \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
\geq \frac{(2n-1)^2}{\left( (n-1)^2 + 1 + (2n-2) \right) \left( n^2 + 1 + (2n-2) \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
\geq \frac{(2n-1)^2}{n^2(n^2 + 2n - 1)} - \frac{2}{(n^2 + 2n - 1)^2} \\
= \frac{2(n-1)^2 - 1}{(n^2 + 2n + 1)^2} > 0 $$
for $n > 1$ which shows that $c_n$ are monotonically decreasing.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but can not proceed further :(
|
Let $\displaystyle b+c-a=x,c+a-b=y,a+b-c=z\implies x+y+z=a+b+c$
We need $\displaystyle(x+y+z)^3-x^3-y^3-z^3=\{x+(y+z)\}^3-x^3-y^3-z^3$
$\displaystyle=x^3+(y+z)^3+3x(y+z)\{x+(y+z)\}-x^3-y^3-z^3$
$\displaystyle=3yz(y+z)+3x(y+z)\{x+(y+z)\}$
$\displaystyle=3(y+z)\{yz+x(x+y+z)\}$
$$\displaystyle\implies(x+y+z)^3-x^3-y^3-z^3=3(y+z)(z+x)(x+y)$$
Now $y+z=2a$ etc.
Can you take it from here?
|
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|
Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
$\bf{My\; Try::}$ Using the fact that the distance between two curve is independent of shifting.
So put $x=u+v$ and $y=u-v$ in thsese two curves, We get $\displaystyle u^2+v^2=\frac{9}{2}$
and $\displaystyle 2(u^2+v^2-2uv)+10(u^2+v^2-2uv)+6(u^2-v^2) = 1\Rightarrow 18u^2+6v^2-20uv=1$
But here again i get terms of $uv,$ So i did not understand how can i solve it,
Help me, Thanks
|
The original ellipse is congruent with the ellipse whose equation is $x^2+11y^2=1$.
And all the ellipse which is similar to it has the equation form:$x^2+11y^2=\lambda$ Make it and $x^2+y^2=9$ as a equation group and solve it: $ x = \sqrt \frac{99-\lambda}{10} or -\sqrt \frac{99-\lambda}{10}, y = \sqrt \frac{\lambda -9}{10} or - \sqrt \frac{\lambda -9}{10}$.That means when $ 9<=\lambda <= 99$, the ellipse has intersection points with circle.
Therefore, when $\lambda = 9$ the ellipse tangent with the circle.
The words are so abstract possibly.
Therefore the minimal distance becomes distance between the long axis vertexes. Since one of the long vertexes of $x^2+11y^2=1$ is (-1,0) and one of the long vertexes of $x^2+11y^2=9$ is (-3,0).
Hence, the answer should be 2.
Complement:
1) why elipse: $x^2+11y^2=1$ is congruent with elipse:$2x^2+10y^2+6xy=1$.
Designate $2x^2+10y^2+6xy=1$ as ellipse II.
Designate $x^2+11y^2=1$ as ellipse I.
Think one of point$(x,y)$ on the ellipse II. Make it rotate about original point$(0,0)$ and it should fall into the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. Find the rotation matrix:
$M = \left(\begin{matrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{matrix}\right)$
where $tan(\theta) = \frac{1}{3}$.
Using rotation matrix multiply the column matrix $ \left(\begin{matrix}x\\y\end{matrix}\right)$every point(x,y) on the ellipse II can get every point(u,v) on the ellipse I.
Now find the linear combination of $u^2, v^2$. Since $u^2 = f(x,y), v^2 = g(x,y)$ and the maximal degree of f(x,y) about x is 2. Similarly, the maximal degree of g(x,y) about x is 2. Through observing the coefficient, its equation should be $u^2 + 11 v ^2 = 1$.
2) Why choose $theta$ satisfying $tan(\theta) = \frac{1}{3}$?
Because our target is to make the ellipse formalized. Luckily, the ellipse's reflecting point is $(0,0)$ because there is no monomial item(i.e. x y) in the original equation. Therefore, assume a line: $y = kx$ passing through the ellipse II. Make $ y=kx$ and $ 2x^2+10y^2+6xy=1$ as a equation group. Solve it. Get $ x = \sqrt \frac{1}{10k^2+6k+2} y = \sqrt \frac{k^2}{10k^2+6k+2} $
Since the long vertex point of the ellipse has maximal distance on the ellipse about $(0,0)$, just find the $ \max x^2+y^2$.
*
*$x^2+y^2 = \frac{k^2+1}{10k^2+6k+2}$ By the division law in differentiation, $(x^2+y^2)' = \frac{-18k-5}{(10k^2+6k+2)^2} = 0$. So when $ k = -\frac{1}{3} $, $x^2 + y^2$ has maximal.
*Since the rotation angle should be positive, and then $ tan(\theta) = \frac{1}{3}$
In a nutshell, the answer is 2.
If my answer has some errors or short points, please leave a comment.
|
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|
Remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$. I am currently preparing for a certain quiz show when I encountered this question:
What is the remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$?
I know for a fact that $x^3-1=(x-1)(x^2+x+1)$.
And I got
$$x^{2016}+x^{2013}+\cdots+x^6+x^3=x^3(x^{2013}+x^{2010}+\cdots+x^3+1)$$
$$=[(x-1)(x^2+x+1)+1](x^{2013}+x^{2010}+\cdots+x^3+1)$$
$$=(x-1)(x^2+x+1)
(x^{2013}+x^{2010}+\cdots+x^3+1)+(x^{2013}+x^{2010}+\cdots+x^3+1)$$
I know that the remainder comes from the term not containing $x^2+x+1$ but I do not know to get it since it involved so many terms.
Any help please? Thanks in advanced!
|
As $x^3=1\implies x^{3n}=(x^3)^n=1\implies x^{3n}\equiv1\pmod{x^2+x+1}$
and $2016=3\cdot672$
$$\sum_{r=1}^{672}x^{3r}\equiv\sum_{r=1}^{672}1\pmod{x^2+x+1}$$
|
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|
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \rightarrow \infty} (x_n-n)=\frac{1}{4}$
|
My answer is wrong, but I would like to know why.
$$n(x_n-n-\frac{1}{4})=n \sum (\sqrt{(1+\frac{k}{n^2})}-(1+\frac{1}{4n}))$$
$$=n \sum \frac{(1+\frac{k}{n^2})-(1+\frac{1}{4n} )^2}{
\sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}
}$$
$$=n \sum \frac{
\frac{k}{n^2} - \frac{1}{16n^2} -\frac{1}{2n}}{
\sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}
}
$$
Since $$
\sqrt{(1+\frac{1}{n^2})}+1+\frac{1}{4n} \leq \sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n} \leq \sqrt{(1+\frac{n}{n^2})}+1+\frac{1}{4n}
$$,
and they have the same limit as $n$ tends to infinity, we will apply sandwich theorem to the sum later.
Now,
$$n \sum (\frac{k}{n^2} - \frac{1}{16n^2} -\frac{1}{2n} )$$
$$=n (\frac{n(n+1)}{2n^2} - \frac{1}{16n} -\frac{1}{2} )$$
$$=(\frac{n+1}{2} - \frac{1}{16} -\frac{n}{2} )$$
$$=(\frac{1}{2} - \frac{1}{16})=7/16$$
So even after dividing two, because of the denominator, it is $7/32$...
|
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|
Factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$ How do you go about factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$
Which factoring method would be the best?
|
$$\left({16+4t^2+\frac{t^4}{4}}\right)^{1/4}=\left({4^2+2\cdot4\cdot\frac{t^2}{2}+\left(\frac{t^2}{2}\right)^2}\right)^{1/4}=$$
$$=\left(\left(4+\frac{t^2}{2}\right)^2\right)^{1/4}=\left(4+\frac{t^2}{2}\right)^{2\cdot{1/4}}=\left(4+\frac{t^2}{2}\right)^{{1/2}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.
Prove that for positive integer $n$ we have $169| 3^{3n+3}-26n-27$.
It is easy to show that if we take the expression modulo $13$ we have $3^{n+3}-26n-27 \equiv 1^{n+1}-1 \equiv 1-1 \mod 13 = 0 \mod 13$. How do I prove it is divisible by $13^2$?
|
Use congruences.
$3$ has order $3$ modulo $13$, hence
$$3^{3n+3}-26n-27\equiv 1^{n+1}-0-1=0.$$
This proves it is divisible by $13$. More precisely, we have
\begin{align*}
3^{3n+3}-26n-27&=27(27^n-1)-26n=27\cdot26(27^{n-1}+27^{n-2}+\dots+27+1)-26n\\
&=2\cdot 13\bigl(27(27^{n-1}+27^{n-2}+\dots+27+1)-n\bigr).
\end{align*}
Modulo $13$, the second factor is congruent to
$$1\bigl(1(1+1+\dots+1)-n\bigr),$$
which is congruent to $0$, since there are $n$ $1$s in the sum. Hence the second factor is also divisible by 13.
|
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|
Find the number of distinct integers with non-decreasing digits formed from one or more of the digits $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$ Suppose integers are formed by taking one or more digits from the
following: $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$. For example, $355$ is a possible choice while $44$ is not. Find the number of distinct integers that can be formed in which the digits are non-decreasing.
|
An integer with non-decreasing digits selected from the digits $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$ is a selection of one of more digits of the string $2233455567$. For instance, the selection $22\color{blue}{33}4\color{blue}{5}55\color{blue}{67}$ corresponds to the integer $33567$.
Since the digits of the integer are non-decreasing, a particular integer is completely determined by the number of times each digit is selected. We can select a particular digit up to the number of times it occurs in the string. For instance, $2$ appears twice in the string $2233455567$, so we can select the digit $2$ either $0$, $1$, or $2$ times, giving us three choices for the number of occurrences of the digit $2$ in the integer. By similar reasoning, we have three choices for the number of occurrences of the digit $3$ in the integer, four choices for the number of occurrences of the digit $5$ in the integer, and two choices for the number of occurrences of each of the digits $4$, $6$, and $7$. Hence, it would appear that the number of integers we can form with the given digits is
$$3 \cdot 3 \cdot 2 \cdot 4 \cdot 2 \cdot 2 = 288$$
However, the integer must be formed using at least one digit, and we have counted the empty string. Therefore, the number of distinct integers with non-decreasing digits that can be formed from permutations of $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$ is $288 - 1 = 287$.
|
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|
Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I have managed to get that if $x$ is not divisible by $5$, then it is one of the following cases, where $n$ represents a positive integer:
\begin{align*}
x & = 5n + 1\\
x & = 5n + 2\\
x & = 5n + 3\\
x & = 5n + 4
\end{align*}
I have tried substituting for $x$, but with two different variables $n$ and $t$ and everything, I'm not sure how I would prove they can be equal. Any help would be much appreciated. Thank you!
|
Once you get that far, just check every case:
$x = 5n+1$:
$$x^2 = 25n^2 - 10n + 1 = 5(5n^2 + 2n) + 1 = 5k + 1$$
$x = 5n + 2$:
$$x^2 = 25n^2 + 20n + 4 = 25n^2 + 20n + 5 - 1 = 5(5n^2 + 4n + 1) - 1 = 5t - 1$$
$x = 5n + 3$:
$$x^2 = 25n^2 + 30n + 9 = 25n^2 + 30n + 10 - 1 = 5(5n^2 + 6n + 2) -1 = 5l - 1$$
$x = 5n + 4$:
$$x^2 = 25n^2 + 40n + 16 = 25n^2 + 40n + 15 + 1 = 5(5n^2 + 8n + 3) + 1 = 5m + 1$$
$k,t,l,m$ are all positive integers.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.