Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find all $x\in\mathbb{R}$ such that $\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$. Find all $x\in\mathbb{R}$ such that:
$$
\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x\,.
$$
Immediately we notice that $x=2$ satisfies the equation.
Then we see that $LHS=a^x+b^x$, where $a<1$ and $b<2$, therefore $RHS$ grows faster (for larger $x$, $LHS\approx b^x<2^x$)
Hence $x=2$ is the only real solution.
Unfortunately I don't know whether this line of reasoning is correct. Moreover, if it is indeed correct, how to write this formally?
| It is not difficult (formula for the double angle) to show that $$\sin \left(\frac{ \pi }{8} \right)= \sqrt{ \frac{2- \sqrt{2} }{4} }$$ which in combination with the trigonometric one gives $$\cos^2\left(\frac{ \pi }{8} \right)=1-\sin^2\left(\frac{ \pi }{8} \right)=\frac{2+ \sqrt{2} }{4}\Rightarrow \cos \left( \frac{ \pi }{8} \right)= \sqrt{ \frac{2+ \sqrt{2} }{4} }$$ thus our equation can be expressed equivalently in the form $$\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$$ $$\left( \sqrt{\frac{2- \sqrt{2} }{4}}\right)^x+\left( \sqrt{\frac{2+ \sqrt{2} }{4}}\right)^x=1 $$ $$\sin^x\left( \frac{ \pi }{8} \right)+\cos^x \left( \frac{ \pi }{8} \right)=1$$ of course thanks to the trigonometric one $x=2$ is a trivial solution. Uniqueness of this solution is due to the fact $\sin \& \cos \le 1$. Formally, you can consider cases $x>2$ or $x<2$ and estimate the left side.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge a+b+c +\frac{3n}{a^2+b^2+c^2}$$
or
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge ab+bc+ca +\frac{3n}{a^2+b^2+c^2}$$
i dont know what to do next. Any ideas preferably using am-gm.
source Samin Riasat Basics in Olympiad ineq.
| We only need to prove this inequality for $n=3.$ Write the inequality as homogeneous form$:$
$${\dfrac {{a}^{2}}{b}}+{\dfrac {{b}^{2}}{c}}+{\dfrac {{c}^{2}}{a}}+{
\dfrac { 9\left( ab+bc+ac \right) ^{3}}{ \left( {a}^{2}+{b}^{2}+{c}^{2}
\right) \left( a+b+c \right) ^{3}}}\geqslant {\dfrac {6(ab+bc+ac)}{a+b+c}}\quad (\text{1})$$
Since the inequality $(\text{1})$ is homogeneous$,$ we may assume $a+b+c=1.$
Let $ab+bc+ca=\frac{1-t^2}{3} \quad (t\in [\,0,1\,]),abc=r.$
We can prove$:$ $${\dfrac {{a}^{2}}{b}}+{\dfrac {{b}^{2}}{c}}+{\dfrac {{c}^{2}}{a}}\geqslant {\dfrac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( 1+2\,t \right) \left( t+1 \right) }} (\text{2}),$$
You can prove this inequality by yourself, I will post my proof for it in another question.
Therefore we need to prove$:$ $${\frac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( 1+2\,t \right) \left( t+1 \right) }}+2\,({t}^{2}-1)-{ \frac { \left( t-1 \right) ^{3} \left( t+1 \right) ^{3}}{1+2\,{t}^{2}} }\geqslant 0.$$
I think now, you can prove it very easy! And the inequality $(\text{2})$ is strong!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $a,b,c,x,y,z$ satisfying $a+b+c=xyz$ and $x+y+z=abc$.
Find all positive integers $a,b,c,x,y,z$ satisfying $$a+b+c=xyz,\tag{1}$$and$$x+y+z=abc,\tag{2}$$ where $a\ge b\ge c\ge 1$ and $x\ge y\ge z\ge 1$.
My try: I think this problem is unique in a way that there are 6 variables and 2 equations , in that way, there may be lot of cases. Also unlike other diophantine equations factoring is not possible. Here is something i did.
Obviously $xyz\ge 3$ and similarly $abc\ge 3$ ,starting with the equality $xyz=3$ or $x=3,y=1,z=1$,and it happens when $a=b=c=1$. Naturally it does not satisfy equation $(2)$ .I tried randomly setting variables some values to see if some pattern popped up,but all efforts were futile.
Next i tried setting $y=1 ,z=1$ which implies $a+b+c=abc-2$ again one could get many triplets.
i am totally stuck.Could anyone nudge me to the right track
| If all the integers are greater than or equal to $2$, then
$$
a+b+c < abc = x+y+z < xyz = a+b+c.
$$
Contradiction. Therefore, assume WLOG that $z=1$. You get
$$
a+b+c = xy, \qquad x+y+1 = abc.
$$
Now assume again, $a, b, c, x, y\geq2$, you get
$$
a+b+c < abc = x+y+1 \leq xy + 1 = a+b+c+1.
$$
Therefore, $abc=a+b+c+1$ and $x+y=xy$. Since $x, y\geq 2$, this implies $x=y=2$. Easy to find $a, b, c$ from here (if there are any).
The remaining cases are when $c=1$ or $y=1$.
$(a, b, c) = (3, 2, 1)$ and $(x, y, z) = (3, 2, 1)$ is one example.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k. For $a,b,c\geqslant 0;ab+bc+ca>0.$ Find $k_\max$ and proving in that case$:$
$$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{kabc(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^3(ab+bc+ca)}.$$
I use computer and I found $k_\max =4.$ Then proving$:$
$$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{4abc(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^3(ab+bc+ca)}.$$
Let $p=a+b+c,q=ab+bc+ca, r=abc,$ need to prove$:$
$$f(r)=(48q -16p^2) r^3+pq ( 23{p}^{2}-96q
) {r}^{2}+6{p}^{2}{q}^{2} ( 3{p}^{2}+8q ) r+{p
}^{3}{q}^{2} \left( {p}^{2}-4\,q \right) \left( 4{p}^{2}-q
\right) \geqslant 0.$$
Since $$f'(r)=3 \left( 48q -16\,{p}^{2}\right) {r}^{2}+2pq ( 23\,{p}^{2}
-96\,q ) r+6\,{p}^{2}{q}^{2} \left( 3\,{p}^{2}+8\,q \right)
\geqslant 0$$
$$\because (a-b)^2 (b-c)^2 (c-a)^2 \geqslant 0.$$
Therefore we need to prove $f(r)\geqslant 0$ when $r$ get the minimum values.
Let $p=1,q=\dfrac{1-t^2}{3} (0 \leqslant t \leqslant 1),r=abc$ we have$:$
$$\dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2} \leqslant r$$
(see the proof here)
So $$f(r) \geqslant f\Big(\dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2}\Big)=$$
$$={\frac {8}{19683}}\,{t}^{2} \left( 4\,{t}^{3}-6\,{t}^{2}-15\,t+49
\right) \left( t-2 \right) ^{2} \left( 2\,t-1 \right) ^{2} \left( t+
1 \right) ^{2}
\geqslant 0,$$
which is true.
But I can't prove $f'(r) \geqslant 0$ with $(a-b)^2(b-c)^2(c-a)^2\geqslant 0.$
I only check it with computer and know it's true because $\prod (a-b)^2 \geqslant 0.$
In addition$,$ I wish to know how do you find $k_\max=4$ if without computer$?$
Source. I post it to found a proof for this inequality, there was a SOS's proof.
| Find $k_{\max}.$ Let $a=b=1$ then
$$(2c+1)\left[\frac 14+\frac{2}{(c+1)^2}\right] \geqslant \frac 94+\frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
equivalent to
$$\frac{c(c-1)^2}{2(c+1)^2}\geqslant \frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
or
$$c(c-1)^2\left[k - \frac{(2c+1)(c+2)^3}{2(c+1)^2}\right] \leqslant 0.$$
Therefore, we will to find $k$ satisfy
$$k \leqslant \frac{(2c+1)(c+2)^3}{2(c+1)^2}.$$
Let $c=0,$ we get $k \leqslant 4.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $ has exactly three solutions. Find the sum of all values of the parameter $ b $ for which the equation $$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $$ has exactly three solutions.
I tried to represent this equation in two axes and draw this graph in order to analyze the cases when there are exactly three solutions, but it comes out too cumbersome. I will be glad for any hint, thanks :)
| Just reminding some basics:
*
*To plot $f(|x|)$, draw $f(x)$, ignore the left hand side part of the y-axis, reflect the graph to the right of the y-axis in the y-axis.
*To plot $|f(x)|$, draw $f(x)$, reflect any part of graph that was below the x-axis in the x-axis.
Let's draw $4|4|x| - b^2| = |16|x| - 4b^2|$:
*
*$f(x) = 16x - 4b^2$.
*$f(|x|) = 16|x| - 4b^2$
*$|f(|x|)|$ = $|16|x| - 4b^2|$
Then we have a $y = x - b$, a line parallel to lines below (or one of the lines below):
In order for $x - b = 4|4|x| - b^2|$ to have exactly three answers, $x - b$ should have three intersections with $4|4|x| - b^2|$. So $x - b$ should be one of the blue lines below and can't be any line else:
So:
$$
4b^2=-b \Rightarrow
\begin{cases}
b=0\\
b=-4\\
\end{cases}
$$
$$
\frac{-b^2}{4}=b \Rightarrow
\begin{cases}
b=0\\
b=\frac{-1}{4}\\
\end{cases}
$$
But if $b=0$, then $\frac{b^2}{4}=\frac{-b^2}{4}$, so the graph we drew will change and the only answer will be $x=0$.
So
$
\begin{cases}
b=-4
\begin{cases}
x=-4\\
x=\frac{60}{17}\\
x=\frac{68}{15}\\
\end{cases} \\
b=\frac{-1}{4}
\begin{cases}
x=\frac{-1}{34}\\
x=0\\
x=\frac{1}{30}\\
\end{cases} \\
\end{cases}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821186",
"timestamp": "2023-03-29T00:00:00",
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In an acute angled triangle with angles $A,B$ and $C$, prove that $\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac6\pi$ In an Acute Angled Triangle Prove That : -
$$\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac{6}{\pi}$$
A , B , C represent the angles of the triangle.
Edit:
I wasn’t able to think of a direct approach so I assumed it to be a equilateral Triangle (we get some few inequalities by this method) but that didn’t work ...
| Let $$f(x) = \frac{\sin x}{x}, \quad x \in \left[0, \frac{\pi}{2}\right].$$
We have
$$f'(x) = \frac{x \cos x - \sin x}{x^2} = \frac{g(x)}{x^2},$$
and
$$g'(x)= - x\sin x < 0.$$
So $g(x) < g(0) = 0,$ deduced $f'(x) < 0,$ or
$$f(x) > f\left(\frac{\pi}{2}\right) = \frac{2}{\pi}.$$
Therefore
$$\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} > \frac{2}{\pi}+\frac{2}{\pi}+\frac{2}{\pi} = \frac{6}{\pi}.$$
| {
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Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.
I have tried the question like this.
\begin{align*}
\frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x+ 48y &= 7xy\\
48x &= 7xy-48y\\
48x &= y\cdot(7x-48)\\
y &= \frac{48\cdot x}{7x-48}\\
\end{align*}
As $y$ is a positive number the denominator have to be $<0$, so $x \ge 7$. Now If I put $x =7$
I get $y = \frac{48\cdot 7}{49-48} = 336$. But
lcm(7,336) is 336
and $\frac{7+336}{336} \neq \frac{7}{12}$. Where am I making mistake? the hcf is not 4 for (7,336)
second method I Tried was,
$x = 4a, y =4b.$
now ,
\begin{align*}
\frac{4a + 4b}{4\cdot a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{1}{a}+\frac{1}{b} &= \frac{7}{12}\\
\frac{1}{a} &= \frac{7}{12} - \frac{1}{b}\\
\frac{1}{a} &= \frac{7b-12}{12b}\\
or\\
a &= \frac{12b}{7b-12}
\end{align*}
so $b\ge 2$ to get an integer, now for $b=2, a= 12. \Rightarrow x=48, y=4$, also if I choose b=2, the hcf is not 4, so if I put $b=3, a = 4 \Rightarrow (12, 16)$ , here the hcf and lcm are,
$4$ and $48$ respectively.and $\frac{12+16}{48} = \frac{7}{12}$.
What is the mistake I am making. ?
| From the point you get to, $y = \frac{48\cdot x}{7x-48}$
As HCF of $x, y$ is $4$, try $x = 8, 12, ... \, (\ge 7)$. $8$ is not possible as that gives $y = 48$ and hence HCF of $8$. So next try with $x=12$ and you get $y=16$ and those are your numbers.
Your second method gives $b = 3, a = 4$ which is the same. In your second method, you have defined $x = 4a, y = 4b$ so HCF of $a$ and $b$ will be $1$. If you choose $b = 2$, you get $a = 12$ and HCF of $a, b$ is $2$ which is not what you are looking for. So, go with $b = 3$ and that works.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1 + \sum_{k=1}^{n-1} \frac{k}{n-k} { 2n-k-1 \choose n} = C_n$ where $C_n$ is the $n$'th Catalan number? Prove that $$1 + \sum_{k=1}^{n-1} \frac{k}{n-k} { 2n-k-1 \choose n} = C_n$$ where $C_n$ is the $n$'th Catalan number?
Wrote a program to validate it, appears to be correct.
This looks awfully like the formula $C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i}$
Both computational proof and combinatoric proof are welcome, but obviously combinatoric proof might be more beautiful. :)
| We seek to show that with $C_n$ the Catalan number we have
$$C_n = 1 + \sum_{k=1}^{n-1} \frac{k}{n-k} {2n-k-1\choose n}.$$
This holds by inspection when $n=0$ and $n=1$ when the sum is zero so we
may assume that $n\ge 2.$ Observe that
$$\frac{k}{n-k} {2n-k-1\choose n}
= k \times \frac{(2n-k-1)!}{n! \times (n-k)!}
= \frac{k}{n} {2n-k-1\choose n-k}.$$
We thus obtain
$$1+ \frac{1}{n} \sum_{k=1}^{n-1} k {2n-k-1\choose n-k}
= \frac{1}{n} \sum_{k=1}^n k {2n-k-1\choose n-k}
\\ = \frac{1}{n} [z^n] (1+z)^{2n-1} \sum_{k=1}^n k
\frac{z^k}{(1+z)^k}.$$
Here the coefficient extractor enforces the upper range and we find
$$\frac{1}{n} [z^n] (1+z)^{2n-1} \sum_{k\ge 1} k
\frac{z^k}{(1+z)^k}
\\ = \frac{1}{n} [z^n] (1+z)^{2n-1} \frac{z/(1+z)}{(1-z/(1+z))^2}
\\ = \frac{1}{n} [z^n] (1+z)^{2n-1} \frac{z(1+z)}{(1+z-z)^2}
= \frac{1}{n} [z^{n-1}] (1+z)^{2n}
= \frac{1}{n} {2n\choose n-1}.$$
To see that this is indeed the Catalan number sequence we write
$$\frac{1}{n} {2n\choose n-1} = \frac{(2n)!}{n! \times (n+1)!}
= \frac{1}{n+1} {2n\choose n}.$$
This is the claim (and it holds for $n\ge 0.$)
| {
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Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$ I am trying to compute the Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$. When I compute the Cesàro means, I get the following sequence
$$\left(1, 1, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \frac{6}{9}, \frac{7}{10}, \frac{8}{11}, \frac{8}{12}, \cdots\right)$$
Where does this sequence converge to? Is it $\frac{2}{3}$? I fail to see the pattern of this sequence. If I can just rewrite the sequence into a more general form, then I might be able to compute the limit.
| Let $a_n = (-1)^n$ for $n\ge 0$. The partial sum $$s_k = \sum_{n = 0}^k a_n$$ for $k\ge 0$ is $1 , 0 , 1 , 0,\dots$ which clearly shows $s_k$ is divergent. Let $$t_n = \frac{1}{n}\sum_{k = 0}^{n-1}s_n$$for $n\ge 1$. So we have $t_1 = \frac{1}{1} , t_2 = \frac{1}{2} , t_3 = \frac{2}{3} , t_4 = \frac{2}{4} , t_5 = \frac{3}{5} , t_6 = \frac{3}{6} , \dots$ and we can easily see the pattern $t_n = \frac{\lfloor\frac{n+1}{2}\rfloor}{n}$. The question is $\lim_{n \to \infty} t_n$ . We know that $x \ge \lfloor x\rfloor \gt x - 1$, so we have $$\frac{n+1}{2} \ge \lfloor \frac{n+1}{2}\rfloor \gt \frac{n+1}{2} - 1 \implies \frac{n+1}{2n} \ge \frac{\lfloor \frac{n+1}{2}\rfloor}{n} \gt \frac{n+1}{2n} - \frac{1}{n}$$ Because $\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{n+1}{2n} - \frac{1}{n} = \frac{1}{2}$, according to the squeeze theorem we have $$\lim_{n \to \infty} t_n = \frac{1}{2}$$
In the same manner you can construct your sequence and compute the limit. The answer is
$b_n = \frac{\lfloor\frac{2n+2}{3}\rfloor}{n} \implies \lim_{n \to \infty} b_n = \frac{2}{3}$
| {
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Finding rational $p$ and $q$ that satisfy $(p+3\sqrt 7)(5+q\sqrt 7)=9\sqrt 7-53$
How do I find rational $p$ and $q$ that satisfy this equation?
$$(p+3\sqrt 7)(5+q\sqrt 7)=9\sqrt 7-53$$
The equation is equal to:
$$5p+15\sqrt 7+pq\sqrt 7+3q\sqrt 7=9\sqrt 7-53$$
What can I do next to find what $p$ and $q$ are equal to?
I know that since there are three $\sqrt 7$'s here, and I know that they have something to do with the answer; so, what can I do next to find out the values of $p$ and $q$?
| Whoa! IMPORTANT $(3\sqrt 7)\cdot (q\sqrt 7) = 3q\sqrt 7^2 = 3q\cdot 7=21q\ne 3q\sqrt 7$.
So we have $5p + 15 \sqrt 7 +pq \sqrt 7 + 21q = 9\sqrt 7 -53$
Now rewrite that as $(15+pq)\sqrt 7 + (5p+21a) = 9\sqrt 7 - 53$
The trick is that no matter how you add up the rational multiples of $\sqrt 7$ you will always get a a rational multiple of $7$ and no matter how you add up the rationals you get a rational and you can never "breakup" or mixup the rational multiples of $\sqrt 7$ and rationals. (They are like oil and vinegar.)
So if $M\sqrt 7 + Q = A\sqrt 7 + B$ and $M,Q,A,B$ are all rational we must have $M=A$ and $Q = B$.
Pf: Suppose $M\ne A$ then $M\sqrt 7 - A\sqrt 7 = B-Q$ then $\sqrt 7(M-A) = B-Q$ and $M-A \ne 0$ so $\sqrt 7 =\frac {B-Q}{M-A}\in \mathbb Q$ which is a contradiction. So $M = A$ and so $M\sqrt 7 + Q = M\sqrt 7 + B$ so $Q=B$.
So we just have to solve $15 +pq = 9$ and $5p+21q = -53$.
$p = \frac {-53-21q}5$ and so
$15 - \frac {53+21q}5q = 9$
$75 - 53q -21q^2 = 45$
$21q^2 +53q -30 = 0$ so $q = \frac {-53 \pm \sqrt {53^2+4*30*21}}{42}=\frac {-53\pm 73}{42}= \frac {20}{21}$ or $-\frac {126}{42}=3$ and so
$p =\frac {-53 -20}5 = \frac {-73}5$ or $p=\frac {-53+63}5= 2$.
So $(p,q) =(\frac {-73}5,\frac {20}{21})$ or $(p,q) = (2,-3)$.
| {
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Show $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$ The problem :
*
*find the minimal polynomial of $\sqrt{5} + \sqrt{7}$
*what is the degree of $ [ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ] $
*conclude $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$
My question and my works
I found 1., I found that the minimal polynomial of $\sqrt{5} + \sqrt{7}$ is $P : X \mapsto X^4 - 24X^2 + 4 $
I know that $\deg P = [ \mathbb{Q}( \sqrt{5} + \sqrt{7} ): \mathbb{Q} ]$ where $P$ is the min. polynomial of $\sqrt{5} + \sqrt{7}$ over $\mathbb{Q}$. But, how can i calculate $ [ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ] $ with 1. ?
I found 3., by showing each inclusion, without using 2...
Thanks you
| Once you know that $[ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ]=4$, with basis $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$, you can proceed as follows, without finding the minimal polynomial of $\sqrt{5}+\sqrt{7}$.
Let $\alpha=\sqrt{5}+\sqrt{7}$. Then
$$
\begin{pmatrix} 1 \\ \alpha \\ \alpha^2 \\ \alpha^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 12 & 0 & 0 & 2 \\ 0 & 26 & 22 & 0
\end{pmatrix}
\begin{pmatrix} 1 \\ \sqrt{5} \\ \sqrt{7} \\ \sqrt{35} \end{pmatrix}
$$
The matrix has nonzero determinant and so is invertible. Therefore, $\{1,\alpha,\alpha^2,\alpha^3\}$ is also a basis and so generates the same space, that is, $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
| As an alternative
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )=$$
$$=n\left(\frac12 \frac1n+\frac12 \frac1n- \frac{1}{n+1} + \frac12\frac{1}{n+2}+\frac12\frac{1}{n+2}-\frac{1}{n+3}+\frac12\frac{1}{n+4}+\ldots\right)=$$
$$=\frac12+n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \to \frac12$$
indeed
$$n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \le n\sum_{k=0}^\infty \frac{1}{(n+2k)^3} =$$
$$=\frac1n\int_0^\infty \frac1{(1+2x)^3}dx=\frac 1{4n} \to 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Find limit at 0 of cosine function with embedded sine I'm getting stuck on the following exercise where I have to find the limit as x approaches zero, for this cosine function:
$$\lim_{x \to 0}\cos\left(\frac{\pi\sin^2(x)}{x^2}\right)$$
The graph shows that there should be a limit of $-1$ at $0$, but I can't find a nice trigonometric identity that allows me to rewrite this such that the $x^2$ in the denominator disappears.
Any indication on how to solve this?
| \begin{align*}
\lim_{x \rightarrow 0} &{}\cos \left( \frac{\pi \sin^2 x}{x^2} \right) \\
&= \cos \left( \lim_{x \rightarrow 0}\frac{\pi \sin^2 x}{x^2} \right) & &\text{cosine is continuous} \\
&= \cos \left( \pi \lim_{x \rightarrow 0}\frac{\sin^2 x}{x^2} \right) & &\text{constant multiple} \\
&= \cos \left( \pi \left(\lim_{x \rightarrow 0} \frac{\sin x}{x}\right)^2 \right) & &\text{$x \mapsto x^2$ is continuous} \\
&= \cos \left( \pi \left(1\right)^2 \right) & &\text{trigonometric limit} \\
&= -1 \text{.}
\end{align*}
The trigonometric limit used is one of a pair of fundamental limits, $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0$, which are usually found by application of the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solving $x+\sqrt{a+\sqrt{x}}=a$ for $x$. I have to solve the following equation for $x$, for all values of $a$:
$$x+\sqrt{a+\sqrt{x}}=a$$
Clearly $x\gt0$
For $a\lt0,x\in\phi$ as RHS$\lt0$ but LHS$\gt0$.
$a=0\Rightarrow x=0$
Now for $a\gt0$
$$a-x=\sqrt{a+\sqrt{x}} ,x\in(0,a)$$
$$(a-x)^2=a+\sqrt{x}.....(1)$$
Now plotting the graphs for $$y=(a-x)^2...a\gt0,x\in(0,a)$$ $$y=a+\sqrt{x}$$
For $a\in(0,1)$ the graphs don't intersect$\Rightarrow x\in\phi$. Also, $a=1\Rightarrow x=0$.
For $a\gt1$ the graphs intersect so $$a^2-a+x^2-2ax=\sqrt{x}...from (1)$$Squaring and rearranging
$$x^4-4ax^3+6a^2x^2-2ax^2-4a^3x+4a^2x+a^4-2a^3+a^2=x$$ which I am not able to solve further.
| Let $x=(y^2-a)^2$ with $y^2>a$ then
$$x+\sqrt{a+\sqrt{x}}=a \iff (y^2-a)^2+y=a \iff y^4-2ay^2+y-a+a^2=0$$
and since the $y^3$ term is equal to zero we can guess
$$y^4-2ay^2+y-a+a^2=(y^2+y+A)(y^2-y+B)=$$
$$=y^4+(A+B-1)y^2+(B-A)y+AB$$
which leads to $A+B-1=-2a$, $B-A=1$, $AB=a^2-a$ that is
$$y^4-2ay^2+y-a+a^2=( y^2 + y-a) ( y^2 - y + 1-a) = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Please give your explanation.
I've tried this and got stuck
\begin{align}|f(x) - f(u)|
&= \Big|\frac{2x}{1+2x^2} - \frac{2u}{1+2u^2}\Big| \\
&=\Big|\frac{(-4xu+2)(x-u)}{(1+2x^2)(1+2u^2)}\Big| \\
& = \Big|\frac{-4xu+2}{(1+2x^2)(1+2u^2)}\Big| |x-u| \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{(1+2x^2)(1+2u^2)} \Big) |x-u| \\
& \leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{1+2u^2} \Big) |x-u| \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + 2\Big) |x-u|.\end{align}
So I confused to make $$\frac{|-4xu|}{(1+2x^2)(1+2u^2)}$$ to be less than some positive number. What should I do next?
| Actually any continuous function $f$ on $\mathbb R$ such that $\lim_{x\to \pm\infty}f(x)=0$ is uniformly continuous on $\mathbb R.$ We have that hypothesis satisfied with $f(x)=2x/(1+2x^2).$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Inverse z-transform by convolution method I've been asked to find the poles, impulse response of a system (in digital signal processing).
The transfer function was,
$$H(z) = \frac{z^2+z}{z^2-z+0.5}$$
I solved it by the partial fractions method as follows.
$$\color{blue}{H(z) = \frac{z(z+1)}{z^2-z+0.5}}$$
$$\begin{align} &\Rightarrow \frac{H(z)}{z} = \frac{z+1}{z^2-z+0.5}=\frac{z+1}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)}
\\ &\Rightarrow \frac{z+1}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)} = \frac{A}{z-(0.5+0.5i)} +\frac{A^*}{z-(0.5-0.5i)}\\ & \\
& A = 0.5 - 1.5i \text{ and } A^* = 0.5+1.5i \\ &
\\& \Rightarrow H(z) = (0.5-1.5i)\frac{z}{z-(0.5+0.5i)} + (0.5+1.5i)\frac{z}{z-(0.5-0.5i)} \\&
\\& \Rightarrow \color{blue}{h(n) = \left[(0.5-1.5)(0.5+0.1i)^n +(0.5+1.5i)(0.5-0.5i)^n\right]u(n)}\end{align}$$
(This can be further reduced to terms containing $\cos(kn)$ and/or $\sin(kn)$ but that wasn't required.)
Then as a continuum, I've been asked to find the impulse response (Inverse z-transform of $H(z)$) by convolution method.
We have,
$$H(z) = \frac{z(z+1)}{z^2-z+0.5}$$
If it were of the form, $\frac{z^2}{(z-a)(z-b)}$, we can consider $F(z) = \frac{z}{z-a}$ and $G(z) =\frac{z}{z-b}$, find $f(n) = a^nu(n)$ , $g(n) = b^nu(n)$. Then the inverse could be found by convoluting $f(n)$ and $g(n)$ easily.
But in this case, the term $z+1$ in the numerator makes it difficult to rewrite $H(z)$ as a product of 2 functions of $z$ whose inverse z-transform is known.
Can somebody help in finding $2$ such functions so that it could be evaluated by convolution method ? Or is it impossible to find the inverse z-transform of this function by convolution?
I also thought of using the fact $Z(a^n \sin(\omega n)) = \dfrac{az\sin\omega}{z^2-2az\cos\omega + a^2}$, but this left me behind with $z+1$.
| Thanks to my professor for providing me a hint (To split $\dfrac{z+1}{z-b}$ into $\dfrac{z}{z-b}$ and $\dfrac{1}{z-b}$ and then convolute)
$$ H(z) = \dfrac{z(z+1)}{z^2-z+0.5} = \dfrac{z(z+1)}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)}$$
Let $a= 0.5+0.5i$ and $b= 0.5-0.5i$.
$$H(z) = \dfrac{z}{z-a}\cdot\dfrac{z+1}{z-b}$$
Let
$$\begin{align} & H_1(z) = \dfrac{z}{z-a} \Rightarrow \color{blue}{h_1(n) = a^nu(n) }\\&H_2(z) = \dfrac{z+1}{z-b}=\dfrac{z}{z-b}+z^{-1}\cdot\dfrac{z}{z-b}\Rightarrow \color{green}{h_2(n) = b^nu(n)+b^{n-1}u(n-1)}\\&
\\& h(n) = \color{blue}{h_1(n)}*\color{green}{h_2(n)} = \sum_{k=-\infty}^{\infty}a^ku(k)b^{n-k}u(n-k) + \sum_{k=-\infty}^{\infty}a^ku(k)b^{n-k-1}u(n-k-1)
\\&\Rightarrow h(n) =\sum_{k=0}^na^kb^{n-k} + \sum_{k=0}^{n-1}a^kb^{n-k-1} = b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k + b^{n-1}\sum_{k=0}^{n-1}\left(\frac{a}{b}\right)^k
\\&\Rightarrow\color{red}{h(n) = \left[\frac{a^{n+1}-b^{n+1}}{a-b} +\frac{a^n - b^n}{a-b}\right]u(n) = \left[\frac{a^n(a+1)-b^n(b+1)}{a-b}\right]u(n)}\end{align} $$
Subbing $a$ and $b$,
$$\begin{align}& h(n) = \frac{(0.5+0.5i)^n(1.5+0.5i)-(0.5-0.5i)^n(1.5-0.5i)}{i}u(n)
\\& \Rightarrow\color{blue}{h(n) = \left[(0.5+0.5i)^n(0.5-1.5i)+(0.5-0.5i)^n(0.5+1.5i)\right]u(n)}\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/3840165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?
To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.
Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.
A family of solutions is then given by $x=m$, $y=m^{2}z - m$ where $z$ takes on any positive integer.
How do I go about proving the striking observation: There are no positive integer solutions $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?
| Let $a$ be a positive integer that is not a cube, and let $x$, $y$ and $z$ be positive integers such that
$$(xz+1)(yz+1)=az^3+1.$$
Expanding the left hand side and rearranging a bit then shows that
$$az^2-xyz-(x+y)=0,\tag{1}$$
so $z$ is an integral root of a quadratic with discriminant $x^2y^2+4a(x+y)$. In particular this discriminant is a perfect square, so there exists a positive integer $v$ such that
$$x^2y^2+4a(x+y)=(xy+2v)^2,$$
and with a bit of rearranging we find the curious identity
$$(a-xv)(a-yv)=a^2-v^3.$$
We see that $v<a$ as otherwise the right hand side is negative, whereas the left hand side is not. Applying the quadratic formula to $(1)$ shows that
$$z=\frac{xy+\sqrt{x^2y^2+4a(x+y)}}{2a}=\frac{xy+(xy+2v)}{2a}=\frac{xy+v}{a},$$
where we have the $+$-sign because $z$ is positive. It follows that
$$z<\frac{xy}{a}+1,$$
so now to prove that $z<a^2+2a$ it suffices to show that $xy<a(a+1)^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all functions $f:\mathbb N_0\to \mathbb N_0$ such that $f(a^2+b^2)=f(a)^2+f(b)^2$ I think the answer is that there are only 2 such functions: the zero function and the identity function, but I'm not able to prove it.
A few findings:
*
*$f(0)=0$ and thus $f(a^2)=f(a)^2$.
*If $f(1)=0$, then $f(2^n)=0$ for all $n\in \mathbb N_0$; if $f(1)=1$, then $f(2^n)=2^n$ for all $n\in \mathcal N_0$ (can be proven by PMI).
*For any functions satisfying the condition, say $f,g$, $f\circ g$ also satisfies the condition.
Source of this problem: https://www2.math.binghamton.edu/p/pow/problem2f20
| I found a much simpler proof, heavily based on and inspired by the ideas of Hagen von Eitzen's solution. For the sake of completeness, we first note that $f(0) = 2f(0)^2$ and because $1 \neq 2f(0)$, we must have $f(0) = 0$. Next we find $f(1)^2 = f(1)$ and so $f(1) \in \{ 0,1 \}$. For convenience, denote $u = f(1)$. More generally, we have that $f(a^2) = f(a)^2$.
We proceed by noting that $f(2) = 2u^2 = 2u$ and so $f(4) = f(2)^2 = 4u$. Further, observe that $f(5) = f(1)^2 + f(2)^2 = 5u$. Next we note that
$f(3)^2 + f(4)^2 = f(25) = f(5)^2.$ We conclude that $f(3) = 3u$ also. Lastly, we note that $f(8) = f(2)^2 + f(2)^2 = 8u$ and $f(10) = f(3)^2 + f(1)^2 = 10u,$ so that from
$f(6)^2 + f(8)^2 = f(100) = f(10)^2$ we find that $f(6) = 6u$ too.
We will now prove with induction that $f(n) = nu$ for all non-negative integers $n$. Suppose that for some integer $n > 6$ we have shown that $f(k) = ku$ for all $k < n$. If $n$ is odd, write $n = 2m+1$ for some $m > 2$. Then the equality
$$
(2m+1)^2 + (m-2)^2 = (2m-1)^2 + (m+2)^2
$$
implies via the functional equation that
$$
f(2m+1)^2 + f(m-2)^2 = f(2m-1)^2 + f(m+2)^2
$$
Since all of $m-2$, $2m-1$ and $m+2$ are smaller than $2m+1$, the result for $n = 2m+1$ follows by induction. Similarly, for even $n$, we write $n = 2m+2$ for some $m > 2$ to find that
$$
(2m+2)^2 + (m-4)^2 = (2m-2)^2 + (m+4)^2
$$
implies via the functional equation that
$$
f(2m+2)^2 + f(|m-4|)^2 = f(2m-2)^2 + f(m+4)^2.
$$
Therefore the induction hypothesis ensures that the result also holds for $n = 2m+2$, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Integral $\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt$ I am preparing for a master's degree entrance exam. One of the questions from the past exam asks to show the following formula. Could someone provide some hint?
$$\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt = \frac{\pi}{6} (e^{-x} - e^{-2x})$$ for $x \geq 0.$
Thank you!
| First $$I(x):=\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt$$ then applying a Laplace transform we get $$\mathscr{L}_{x\to s}\{I(x)\}=\int_0^{\infty} e^{-sx}\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dtdx$$ here by the $x\to s$ I just denote which vairables I change to which, to avoid confusion. Now since all integrals converge we change order of Integration getting. $$\int_0^{\infty} \int_0^{\infty} e^{-sx}\frac{t\sin(tx)}{(1+t^2)(4+t^2)}dxdt=\int_0^{\infty} \frac{t}{(1+t^2)(4+t^2)}\mathscr{L}_{x \to s} \{ \sin(tx) \}dt$$ Now using the fact that $\mathscr{L}_{x\to s} \{ \sin(tx)\}=\frac{t}{t^2+s^2}$ which is obtainable by using $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, we get $$\int_0^{\infty} \frac{t}{(1+t^2)(4+t^2)}\mathscr{L}_{x \to s} \{ \sin(tx) \}dt=\int_0^{\infty} \frac{t^2}{(1+t^2)(4+t^2)(s^2+t^2)}dt$$ Now using partial fraction decomposition $$\frac{t^2}{(1+t^2)(4+t^2)(s^2+t^2)}=-\frac{1}{3(s^2-1)(1+t^2)}+\frac{4}{3(s^2-4)(4+t^2)}-\frac{s^2}{(s^2-4)(s^2-1)(s^2+t^2)}$$ Plugging this back in we get $$\int_0^{\infty} -\frac{1}{3(s^2-1)(1+t^2)}+\frac{4}{3(s^2-4)(4+t^2)}-\frac{s^2}{(s^2-4)(s^2-1)(s^2+t^2)}dt=$$$$=-\frac{1}{3(s^2-1)}\int_0^{\infty} \frac{1}{1+t^2}dt+\frac{4}{3(s^2-4)}\int_0^{\infty} \frac{1}{4+t^2}dt-\frac{s^2}{(s^2-4)(s^2-1)}\int_0^{\infty} \frac{1}{s^2+t^2}dt$$ Which using the fact that for $z>0$ we have $\int_0^{\infty} \frac{1}{z^2+t^2}dt=\frac{\pi}{2z}$ is equal to $$\pi(-\frac{1}{6(s^2-1)}+\frac{1}{3(s^2-4)}-\frac{s}{2(s^2-4)(s^2-1)})$$ Now this is equal to $\mathscr{L}_{x\to s} \{I(x)\}$ thus the inverse Laplace transform of this is I(x). Which can be easy be found using partial fraction decompositon to be $$\boxed{\frac{\pi}{6}(e^{-x}-e^{-2x})}$$
Very neat solution!
| {
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How do I evaluate $\int_{-1}^1\frac{dx}{(1+x^2)(e^x+1)}$? How do I evaluate:
$$\int_{-1}^1\frac{dx}{(1+x^2)(e^x+1)}$$
The answer given in the book is written below.
$$\int_{-1}^1\frac{dx}{(1+x^2)(e^{-x}+1)} \tag{1}$$
$$=\int_{-1}^1\frac{e^x}{(1+x^2)(e^x+1)}dx \tag{2}$$
On adding $(1)$ and $(2)$, we get:
$$\Rightarrow 2I=\int_{-1}^1\frac{e^x+1}{(1+x^2)(e^x+1)}dx \tag{3}$$
$$=\int_{-1}^1\frac{dx}{1+x^2} =2\int_{0}^1\frac{dx}{1+x^2} \tag{4}$$
$$\Rightarrow I=\int_{0}^1\frac{dx}{1+x^2}=[ \tan^{-1}x]\:_{0}^{1}\:=\:\frac{\pi}{4} \tag{5}$$
$(2): \int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(a+b-x)dx}$
$(4): \frac{1}{1+x^2} \text{ is an even function}$
I understood how to solve the integral, but I'm not able to understand how the original integral changed to the integral in $(1)$.
It might be something basic, most probably related to exponents but please tell because I'm a complete beginner in calculus.
| Answer:
$I= \int\limits_{-1}^{1} \frac{dx}{(1+x^2)(e^x+1) } $=
We put $x=-y$
$I=\int\limits_{1}^{-1} - \frac{dy}{(1+y^2)(e^{-y}+1)} =\int\limits_{-1}^{1} \frac{e^{y}}{(1+y^2)(e^{y} +1)} =
\int\limits_{-1}^{1} \frac{e^{y}+1-1}{(1+y^2)(e^{y} +1)} $
$2I=\int\limits_{-1}^{1} \frac{1}{1+y^2 } $
$I=\frac{[arct(y) ]_{-1}^{1}}{2} =\frac{\pi}{4}$
| {
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} |
Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
| Use the fact that\begin{align}x^3-4x^2-4x+16=0&\iff x(x^2-4)-4(x^2-4)=0\\&\iff(x-4)(x^2-4)=0.\end{align}
| {
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"question_score": "6",
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reduce a differential equation $y^{'}=\dfrac{1+y}{1-x}$ I want to reduce a differential equation.
$$y^{'}=\dfrac{1+y}{1-x}$$
I reduce this, but my answer don't much "Wolfram alpha".
Please tell me what is wrong.
divide both sides of the equality by $(1+y)$
$$\dfrac{y'}{1+y}=\dfrac{1}{1-x}$$
integrate both sides by $x$
$$\int \dfrac{1}{1+y}dy=\int \dfrac{1}{1-x}dx$$
$$\log \left| 1+y\right| =-\log \left| 1-x\right| +C$$
$$1+y=\dfrac{C}{1-x}$$
Last
$$y=\dfrac{C}{1-x}-\dfrac{1}{1-x}$$
| From your calculations
$$1+y = \dfrac C {1-x} \implies y = \dfrac C{1-x} - 1$$
which is already the solution.
If you want to obtain the form given by WolframAlpha, notice that:
$$\frac C{1-x} - 1 = \frac C{1-x}-\frac 1{1-x} + \frac x{1-x} = \frac{C-1}{1-x}+\frac x{1-x}$$
and $C-1$ is absorbed into $C$ since $C$ can be any constant, giving
$$y = \frac {c_1}{1-x}+\frac x{1-x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $ I have been trying to derive
$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$
with $G$ being the the Catalan constant.
I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.
| Write the numerator of the integrand as
$$\ln(1-x)\ln(x+x^2)=\ln(1-x)\ln x+\ln(1+x)\ln\frac{1-x}{1+x}+\ln^2(1+x)
$$
and, correspondingly
$$I=\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx
= I_1+I_2+I_3
$$
where
\begin{align}
I_1 &= \int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx
= \frac12\int_0^1\frac{\ln^2(1-x)+\ln^2x-\ln^2\frac x{1-x}}{1+x^2}dx\\
I_2 &=\int_0^1\frac{\ln(1+x)\ln \frac{1-x}{1+x}}{1+x^2}dx
\overset{ \frac{1-x}{1+x} \to x} = -\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-G\ln2 \\
I_3 &= \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx
= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \int_1^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{x\to 1/x} {dx }\\
&= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \left(
I_3 + \int_0^1 \frac{\ln^2x}{1+x^2}dx-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx\right)\\
&=\frac12 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx
- \frac12 \int_0^1\frac{\ln^2 x}{1+x^2}dx
+\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
\end{align}
Then, add up the three integrals above to obtain
\begin{align}
I=& \ I_1+I_2+I_3\\
=& \ \frac12 \int_0^1\overset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx
+\frac12\int_0^\infty\overset{t=1+x}{ \frac{\ln^2(1+x)}{1+x^2}}dx
- \frac12\int_0^1\overset{t=x/(1-x)}{ \frac{\ln^2\frac x{1-x}}{1+x^2}}dx-G\ln2 \\
=& \ 2\int_0^\infty \underset{t^2\to 2t} {\frac{t\ln^2t}{t^4+4}dt } -G\ln2
=\frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2
\end{align}
| {
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Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$, and so $$\int\frac{2x+2}{x^2+2x+3}\,dx=\int\frac{du}{u}=\ln{|x^2+2x+3|}.$$
Next consider the other integral. Put $t\sqrt{2}=x+1$. Then $dx=\sqrt 2\,dt$. Now
\begin{align*}
\int\frac{x+2}{x^2+2x+3}\,dx&=\int\frac{(x+1)+1}{(x+1)^2+2}\,dx\\
&=\int\frac{t\sqrt 2+1}{2t^2+2}\sqrt 2\,dt\\
&=\frac{1}{2}\int\frac{2t+\sqrt 2}{t^2+1}\,dt\\
&=\frac{1}{2}\left(\int\frac{2t}{t^2+1}\,dt+\sqrt 2\int\frac{1}{t^2+1}\,dt\right)\\
&=\frac{1}{2}\left(\ln{|t^2+1|}+\sqrt 2\arctan t\right)
\end{align*}
and hence this is equal to
$$\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+1}{2}+1\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)=\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)$$
therefore
\begin{align*}
\int\frac{x}{x^2+2x+3}\,dx&=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx\\
&=\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C.
\end{align*}
apparently, the correct answer is $\frac{(\ln|x^2+2x+3|)}{2}-\frac{\sqrt{2}\arctan{\frac{(x+1)}{\sqrt{2}}}}{2}+C.$ what went wrong?
| Recall $\ln(a/b) = \ln(a) - \ln(b)$, so
$$\ln\left|\dfrac{x^2+2x+3}{2} \right| = \ln\dfrac{|x^2+2x+3|}{|2|} = \ln|x^2+2x+3|-\ln2$$
so, distributing the $-\dfrac{1}{2}$, we obtain
$$\ln|x^2+2x+3|-\dfrac{1}{2}\ln|x^2+2x+3|-\dfrac{1}{2}\ln2 - \dfrac{\sqrt{2}\arctan\frac{x+1}{\sqrt 2}}{2}+C$$
which is just
$$\dfrac{1}{2}\ln|x^2+2x+3| - \dfrac{\sqrt{2}\arctan\frac{x+1}{\sqrt 2}}{2}-\dfrac{1}{2}\ln 2 + C$$
and because $-\dfrac{1}{2}\ln 2$ is a constant, we can absorb that into $C$.
| {
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"url": "https://math.stackexchange.com/questions/3859541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to evaluate the infinite sum of $\frac{1}{ (n^2+a)(n^2+b)(n^2+c)}$ Playing around with Mathematica, I found a rather nice result for this summation
\begin{align}
\sum_{n=0}^\infty \frac{1}{(n^2 + a)(n^2 +b) (n^2 + c)}
&= \frac{1}{2 a b c}
- \frac{ \pi \left[
\sqrt{ab}(a-b) \coth(\pi\sqrt{c})
+
\sqrt{bc}(b-c) \coth(\pi \sqrt{a})
+
\sqrt{ca}(c-a) \coth( \pi \sqrt{b})
\right]}{2 \sqrt{a b c} \, (a-b)(b-c)(c-a)}.
\end{align}
Does anybody know how to obtain this result analytically?
| Partial fraction decomposition gives
$$\frac1{(n^2+a)(n^2+b)(n^2+c)}=\sum_{\mathrm{cyc}}\frac1{(a-c)(b-c)}\cdot\frac1{n^2+c}$$
Now use the well-known sum
$$\sum_{n=0}^\infty\frac1{n^2+k}=\frac{1+\sqrt k\pi\coth\sqrt k\pi}{2k}$$
which should lead you to the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $x^{3}-3 x=\sqrt{x+2}$
Solve the equation $x^{3}-3 x=\sqrt{x+2}$
In the solution of this author wrote -
Let $-2 \leq x \leq 2 .$ Now Setting $x=2 \cos a, 0 \leq a \leq \pi$....
But here i did not understand the reason behind taking $ 0 \leq a \leq \pi$,i mean if we can take $a$ to be any real number since domain of $\cos x$ is real numbers then why we are restricting to
$ 0 \leq a \leq \pi$ ?
| In the answer by Dan Fulea, there is a cubic factor as "Just as a digression," namely $x^3 + x^2 - 2x - 1.$ This is well known, the roots are real numbers
$$ 2 \cos \left( \frac{2 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{4 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{8 \pi}{7} \right) \; \; . \; \; $$
This is verified easily enough, using nothing worse than the formula for $\cos 3 \theta. $
In the original problem $x^3 - 3 x = \sqrt{x+2},$ just one of the three works, namely $ x =2 \cos \left( \frac{4 \pi}{7} \right) \approx -0.445$
Note that the given roots are all strictly between $-2$ and $2.$
This page from Reuschle(1875) is missing from the scanned version online, I don't know what happened:
| {
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If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I concluded that $n^2+1=2a^2$ and $n+1=2b^2$ for some $a,b\in\mathbb{N}$ and it is here where I need some help.
Please only provide hints.
| Hint: If $\gcd(n+1,n^2+1)=2$ then $n$ is odd, say $n=2k+1$, and so you get
$$a^2=\frac{n^2+1}{2}=\frac{4k^2+4k+2}{2}=k^2+(k+1)^2=k^2+b^4,$$
which is a Pythagorean triple. Can you continue from here? [There is still quite some work to be done!]
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a way to rewrite this expression as a sum of the previous two terms?
The context of this question:
Consider for ${n \in \mathbb{N}, n\geq 0}$ a rectangular board divided into a ${2 \times n}$ square grid.
A ${2 \times 1}$ domino can cover any two horizontally or vertically adjacent squares. Prove by induction that there are exactly $F_{n+1}$ different ways to cover every square of the board with no two dominoes overlapping, where ${F_{n+1}
= F_{n} + F_{n-1}, F_{0}=0, F_{1}=1}$.
The (unfinished) answer so far:
A ${2 \times 2}$ "pair" in this answer is a pair of adjacent horizontal dominoes that take up a ${2 \times 2}$ space on the ${2 \times n}$ grid.
Take ${n}$ to be the length of the grid's rows. The number of ways of arranging non-overlapping ${2 \times 1}$ dominoes on a ${2 \times n}$ grid is equal to the number of ways some ${2 \times 2}$ "pairs" fit on a ${2 \times n}$ grid.
For example, for ${n=3}$, there is one way of fitting zero ${2 \times 2}$ "pairs" on a ${2 \times n}$ grid (all dominoes are placed vertically), and there are two ways of fitting one ${2 \times 2}$ "pair" on a ${2 \times n}$ grid i.e. the first arrangement has a ${2 \times 1}$ domino on its leftmost column with a ${2 \times 2}$ "pair" to its right and the second arrangement has a ${2 \times 1}$ domino on its rightmost column with a ${2 \times 2}$ "pair" to its left.
A ${2 \times 2 }$ "pair" takes up two of the ${n}$ columns. This means that the maximum number of "pairs" that can fit on a ${2 \times n}$ grid is ${\frac{n-1}{2}}$ if ${n}$ is odd and ${\frac{n}{2}}$ if n is even. Combining this into one formula gives:
${M}$, The maximum number of "pairs" for a given ${n}$, ${ = \frac{n+\frac{1}{2}((-1)^n-\frac{1}{2})}{2} = \frac{1}{4}{(2n+(-1)^{n}-1)}}$
The number of "pairs" ${k}$ on a ${2 \times n}$ grid is therefore ${0 \leq k \leq M}$. A "pair" takes up two columns, so the number of columns available after ${k}$ "pairs" have been placed for any given ${k}$ is equal to ${n-2k}$. When ${k}$ "pairs" have been placed, there are ${k}$ ${2 \times 2}$ "spaces" each taken up by a "pair" and ${n-2k}$ ${2 \times 1}$ "spaces" taken up by single vertical dominos, which means ${k+(n-2k)=n-k}$ total "spaces" filled. The number of ways of arranging these ${k}$ "pairs" on a ${2 \times n}$ grid is therefore given by ${n-k \choose k}$. For example, for ${n=5, k=2}$, placing ${k=2}$ "pairs" on the grid means there is ${n-2k=1}$ column left and so ${k+(n-2k)=n-k=3}$ "spaces" (two ${2 \times 2}$ spaces each taken up by a "pair" and one ${2 \times 1}$ space taken up by a single domino). Given that ${k}$ of these "spaces" are taken up by "pairs", the number of possible arrangements of these "pairs" within the "spaces" is ${{n-k \choose k} = {3 \choose 2} = 3}$.
All the ways of arranging these ${k}$ "pairs" (${\forall k}$ where ${0 \leq k \leq M}$) for a given ${n}$, is given by:
${A_n={\sum \limits_{k=0}^{k=M}} {n-k \choose k}}$
To get the ${M}$ values for ${n-1}$ and ${n-2}$, i.e. the maximum value for ${k}$ (number of ${2 \times 2}$ "pairs" that can fit on a ${2 \times n}$ grid) for ${n-1}$ and ${n-2}$, first consider when ${n}$ is odd. When ${n}$ is odd, the maximum number of "pairs" is the same as the maximum number of "pairs" for ${n-1}$. When ${n}$ is even, the maximum number of "pairs" for ${n-1}$ is one less than the maximum number of "pairs" for ${n}$. So this number, ${M_{n-1}}$, is given by
${M_{n-1}=M-(\frac{1}{2}(-1)^{n}+\frac{1}{2})}$.
Substituting in ${M}$, rearranging, and using the fact that ${(-1)=(-1)^{-1}}$ gives
${M_{n-1}=\frac{1}{4}{(2(n-1)+(-1)^{n-1}-1)}}$
which is the maximum value of ${k}$ for the case ${n-1}$. To get the ${M}$ value for ${n-2}$, ${M_{n-2}}$, the value ${M-1}$ is considered:
${M_{n-2}=M-1=\frac{1}{4}{(2n+(-1)^{n}-1)}-1}$
$=$ ...
${ = \frac{1}{4}{(2(n-2)+(-1)^{n-2}-1)}}$.
What's left to consider is the possible relation between
${A_n={\sum \limits_{k=0}^{k=M}} {n-k \choose k}}$ and the two formulas
${M_{n-1}=\frac{1}{4}{(2(n-1)+(-1)^{n-1}-1)}}$ and
${M_{n-2}=\frac{1}{4}{(2(n-2)+(-1)^{n-2}-1)}}$ to give the inductive step of this answer.
The question given the above context:
Is there a way to rewrite the expression ${{\sum \limits_{k=0}^{k=M}} {n-k \choose k}}$
as ${{\sum \limits_{k=0}^{k=M_{n-1}}} {n-1-k \choose k}} + {{\sum \limits_{k=0}^{k=M_{n-2}}} {n-2-k \choose k}}$, where
${M=\frac{1}{4}(2n + (-1)^n - 1)}$
${M_{n-1}=\frac{1}{4}(2(n-1) + (-1)^{n-1} - 1)}$
${M_{n-2}=}{\frac{1}{4}(2(n-2) + (-1)^{n-2} - 1)}$
$n \in \mathbb{N},\forall n \ge 0$?
| One thing to do in a situation like this (especially with such a very complicated chain of reasoning) is to check the results for some small input values.
For example, if $n = 6,$ then
$M = \frac14(2(6) + (-1)^6 - 1) = 3,$ and
\begin{align}
\sum_{k=0}^{k=M} \binom{n-k}{k} &= \sum_{k=0}^{k=3} \binom{6-k}{k}\\
&= \binom60 + \binom51 + \binom42 + \binom33 \\
&= 1 + 5 + 6 + 1 \\
&= 13.
\end{align}
For the first few values of $n$:
\begin{align}
A_0 &= \binom00 \\
A_1 &= \binom10 \\
A_2 &= \binom20 + \binom11 \\
A_3 &= \binom30 + \binom21 \\
A_4 &= \binom40 + \binom31 + \binom22 \\
A_5 &= \binom50 + \binom41 + \binom32 \\
A_6 &= \binom60 + \binom51 + \binom42 + \binom33 \\
A_6 &= \binom70 + \binom61 + \binom52 + \binom43 \\
\end{align}
For $n$ even, write the sum
\begin{align}
A_{n-2} + A_{n-1}
&= \left(\binom{n-2}0 + \binom{n-3}1 + \cdots
+ \binom{n/2}{(n/2) - 2} + \binom{(n/2) - 1}{(n/2) - 1}\right) \\
& \qquad + \left(\binom{n-1}0 + \binom{n-2}1 + \binom{n-3}2 + \cdots
+ \binom{n/2}{(n/2) - 1}\right) \\
&= \binom{n-1}0 + \left(\binom{n-2}0 + \binom{n-2}1\right)
+ \left(\binom{n-3}1 + \binom{n-3}2\right) \\
& \qquad + \cdots + \left(\binom{n/2}{(n/2) - 2} + \binom{n/2}{(n/2) - 1}\right)
+ \binom{(n/2) - 1}{(n/2) - 1} \\
&= \binom n0 + \binom{n-1}1 + \binom{n-2}2 + \cdots +
\binom{(n/2) + 1}{(n/2) - 1} + \binom{n/2}{n/2}
\end{align}
using the identity $$\binom{p - 1}{q - 1} + \binom{p - 1}q = \binom pq.$$
In short, we interleave the terms, carry the first and last term down individually,
and add the other terms in pairs.
The sum for odd $n$ is similar except that it has an odd number of terms in the intermediate sums, carrying down the first term and adding the rest in pairs.
This is still a somewhat informal proof due to all the "$\cdots$" parts.
To eliminate those I think you would need double induction.
On the other hand, as hinted in comments, there is a much simpler
inductive proof if you can find it.
It has been written up on this site before if you want to go looking for a spoiler,
but here's a hint: consider what you might find in the last column of two squares on the right.
| {
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| $ 0 \leq (b-c)^{2} =b^{2}+c^{2}-2bc=3-3bc$ so $bc \leq 1$. Now $(b+c)^{2}=b^{2}+c^{2}+2bc=3+bc \leq 4 $ Now take square root.
| {
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Evaluate the limit $\lim_{x\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$ Evaluate the limit:
$$\lim_{n\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$$
Using the fact that ${(1 + x)^{1/2} \approx 1 + x/2}$ for "small" x,
I have that $\sqrt{n+1}\approx\sqrt{n}(\frac{1}{2n}+1)$ then $n\rightarrow \infty$. However, following this procedure I end up with the following limit: $\lim_{n\rightarrow \infty}2n^2=\infty$, but the answer is $\frac{1}{2}$. I would be thankful for any help.
| Answer :
$\sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}) =\sqrt{n^3}\frac{(2\sqrt{n^2 +3n+2}-2\sqrt{n^2 +3n})}{\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3}}=\frac{4 \sqrt{n^3} }{(\sqrt{n^2 +3n+2}+\sqrt{n^2 +3n})(\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3})}=\frac{4 \sqrt{n^3} }{n \sqrt{n + 3} + \sqrt{n^3 + 4 n^2 + 5 n + 2} +\sqrt{n^3 + 5 n^2 + 8 n + 4} + \sqrt{n^3 + 3 n^2 + 2 n} +\sqrt{n^3 + 6 n^2 + 11 n + 6} + \sqrt{n^3 + 4 n^2 + 3 n} + \sqrt{n^3 + 5 n^2 + 6 n} + n^{3/2} + 3 \sqrt{ n}}$ $=\frac{4}{\sqrt{1+\frac{3}{n}}+\sqrt{1+\frac{4}{n}+\frac{5}{n^2 }+\frac{2}{n^3 }} +\sqrt{1+\frac{5}{n}+\frac{8}{n^2} +\frac{4}{n^3 }} +\sqrt{1 +\frac{3}{n}+\frac{2}{n^2 }} +\sqrt{1 +\frac{6}{n}+\frac{11}{n^2 }+\frac{6}{n^3 }} +\sqrt{1+\frac{4}{n^2 }+\frac{3}{n^3 }} +\sqrt{1+\frac{5}{n}+\frac{6}{n^2 }} +1+3\sqrt {\frac{1}{n^2 }}} $
So :
$\lim _{n\to+\infty} \sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})=\frac{4}{8} =\frac{1}{2}$
| {
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Are there restrictions to applying $\lim\limits_{x\to0}\frac {\tan x} {x}$? Question
Evaluate $$\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2}\ .$$
My working
So far, I have gotten to the step where
\begin{align}
\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & =
\exp\left(\lim\limits_{x\to0}\ \frac {\ln(\frac {\tan x} {x})} {x^2}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \left[\left(\frac {x\sec^2 x - \tan x} {x^2} \div \frac {\tan x} {x} \right) \div 2x \right]\right)
\end{align}
Here, I note that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$, so I continue with
\begin{align}
\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & =
\exp\left(\lim\limits_{x\to0}\ \left[\left(\frac {x\sec^2 x - \tan x} {x^2} \div 1 \right) \div 2x \right]\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {x\sec^2 x - \tan x} {2x^3}\right)
\end{align}
This is where I believe I have made a mistake. I tried to use the fact that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ again by factoring out an $x$ from the denominator, so I proceeded with
\begin{align}
\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & =
\exp\left(\lim\limits_{x\to0}\ \frac {x\sec^2 x - \tan x} {2x^3}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x - \frac {tan x} {x}} {2x^2}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x - 1} {2x^2}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {2\sec^2 x \tan x} {4x}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x \tan x} {2x}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {2\sec^2 x \tan^2 x + \sec^4 x} {2}\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to0}\ \frac {\sec^4 x} {2}\right)
\\[5 mm] & =
e^\frac {1} {2}
\end{align}
Answer
$$\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} = e^\frac {1} {3}$$
The solution provided by my professor also used the fact that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ at the same point where I used it for the first time, so I believe there is nothing wrong with my steps until that point. I am thinking that I have gone wrong when I applied it the second time, but I am not sure. If so, can anyone tell me why I cannot apply $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ for the second time? Or perhaps, did I go wrong somewhere else? I certainly hope it is not because of some careless mistake...
Any help/intuition/explanation will be greatly appreciated :)
Edit
Seeing how my professor worked out part of the limit to simplify the limit, I tried to be smart and extend his idea, but that did not work out too well for me, as pointed out by the answers! Really insightful comments from the community once again :)
| Indeed in the first step we have that
$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 \frac {\tan x} {x}} $$
and since $\frac {\tan x} {x} \to 1$ we reduce to study the following
$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 } $$
this one is a perfectly fine step but in the subsequent steps this is not allowed as for example for
$$\lim\limits_{x\to0} \frac{\sin x-x }{ x^3 } = \lim\limits_{x\to0} \frac{\frac{\sin x}x-1 }{ x^2 } \neq 0$$
Refer to the related
*
*Analyzing limits problem Calculus (tell me where I'm wrong).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove this geometric inequality Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$
My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have to proove $a+b+c\geq 3z$, i.e. $a+b\geq 2c$
Since $R(K,N;L,M)=R(B,C;D,E)$ so $R(K,N;L,M)= \frac{1}{2}:\frac{1}{2}$ thus $4ac=(a+b)(b+c)$.
I am not able to move further from here.
| Let $K$ and $N$ be placed on the sides $AB$ and $AC$ respectively.
Also, let $\frac{AK}{AB}>\frac{AN}{AC}$ and $G\in NC$ such that $KG||BC$, $KG\cap AD=\{P\}$ and $KG\cap AE=\{Q\}$.
Thus, $KP=PQ=QG$, which says that it's enough to solve our problem, when $K\equiv B$.
Now, let $AN=kAC$, where $0<k<1$ and $F\in EC$ such that $NF||AE.$
Thus, $$\frac{BM}{BN}=\frac{BE}{BF}=\frac{\frac{2}{3}BC}{\frac{2}{3}BC+\frac{k}{3}BC}=\frac{2}{2+k}.$$
Now, let $G\in NC$ such that $EG||BN.$
Thus, $$\frac{AM}{ME}=\frac{AN}{NG}=\frac{kAC}{\frac{2}{3}(1-k)AC}=\frac{3k}{2(1-k)}.$$
Now, let $I\in DE$ such that $MI||AD$.
Thus, $$\frac{LM}{BM}=\frac{DI}{BI}=\frac{DI}{2DI+EI}=\frac{\frac{DI}{EI}}{\frac{2DI}{EI}+1}=$$
$$=\frac{\frac{AM}{ME}}{2\cdot\frac{AM}{ME}+1}=\frac{\frac{3k}{2(1-k)}}{2\cdot\frac{3k}{2(1-k)}+1}=\frac{3k}{2(2k+1)}.$$
Id est, $$\frac{LM}{BN}=\frac{LM}{BM}\cdot\frac{BM}{BN}=\frac{3k}{2(2k+1)}\cdot\frac{2}{2+k}=\frac{3k}{(2k+1)(k+2)}$$ and it's enough to prove that:
$$\frac{3k}{(2k+1)(k+2)}\leq\frac{1}{3}$$ or $$(2k+1)(k+2)\geq9k,$$ which is true by AM-GM:
$$(2k+1)(k+2)\geq3\sqrt[3]{k^2}\cdot3\sqrt[3]k=9k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question on probability theory with 3 urns with return Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?
I think, this is $\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$ by Bayes'theorem and Law of total probability, but can't be sure.
Thanks for any help.
| I am unsure of where your math came from, and I am unfamiliar with the "Law of total probability". The following is how I would compute the probability.
Let $p(k)$ represent the probability that the 1st ball came from urn $k : k \in \{1,2,3\}.$
Then, the chance that the new drawing will also be a white ball, since the sampling is done with replacement is
$$\left[p(1) \times \frac{10}{20}\right]
~+~ \left[p(2) \times \frac{6}{20}\right]
~+~ \left[p(3) \times \frac{2}{20}\right].$$
Let $D$ (i.e. denominator) =
$$\left[\frac{1}{3} \times \frac{10}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{6}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{2}{20}\right].$$
Then:
$$p(1) = \frac{\frac{1}{3} \times \frac{10}{20}}{D}.$$
$$p(2) = \frac{\frac{1}{3} \times \frac{6}{20}}{D}.$$
$$p(3) = \frac{\frac{1}{3} \times \frac{2}{20}}{D}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.
Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.
Also, just some hints would be appreciated, thanks
| Since the expression is cyclic, we can, WLOG, reduce it into two cases:
Suppose $a\ge b\ge c > 0$. We have
$$\frac{a^2-c^2}{b+c} = \frac{a^2-b^2}{c+b} + \frac{b^2-c^2}{c+b} \ge \frac{a^2-b^2}{c+a} + \frac{b^2-c^2}{a+b}$$
Now suppose $a \ge c \ge b > 0$. We then have
$$\frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge \frac{c^2-b^2}{a+c} + \frac{a^2-c^2}{a+c} =\frac{a^2-b^2}{c+a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb N$
Solution:
Let $a+b+c=(x+1)^2=x^2+2x+1$
If $a+b=x^2$ then third number $c=2x+1$.
If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :
$a=4x$, so $b=x^2-4x$
Now due to statement we must have $a+c=6x+1=n^2$
$n^2=6x+1$ can have infinite solutions such as:
$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$
Which give:
$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$
I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:
1-Does this problem have solutions for N>3?
2-Any idea for better method?
| The solution given by "@ player3236" has $k=325$.
The integer $325$ can be written as sum of two squares
in three way's & is shown below:
$k= x_1^2+x_2^2=x_3^2+x_4^2=x_5^2+x_6^2$
($x_1,x_2,x_3,x_4,x_5,x_6$)=$(15,10,17,6,18,1)$
$325= 15^2+10^2=17^2+6^2=18^2+1^2$
Above can help in arriving at a parametric solution.
Also mathematician "Seiji Tomita has given parametric
solution in which $(a,b,c,d)$ taken two at a time is a
square & the link is given below.
http://www.maroon.dti.ne.jp/fermat/dioph115e.html
And click on (dioph115e)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $A = \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix} $. Find $A^n$ I am new here so please let me know if I must resentence the exercice. I considered it too short not to include it in the title too.
Let $ A= \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix}$
Find $A^n$
I tried finding the matrix's first 4 or 5 powers and it looks like a pettern can be noticed but I can't seem to find a form for the element on line 3 row 3. I haven't tried using a binomial expansion though, writing the matrix as $$ \begin{pmatrix} 0 & b & c \\ 0 & 0 & b \\ 0 & 0 & 0 \\ \end{pmatrix}+ a \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$
the second one being the identity matrix times $a.$
| Let $X=aI,Y=bJ$ and $Z=cJ^2$, where $J$ denotes the $3\times3$ nilpotent Jordan block. Then $X,Y,Z$ commute. Since $J^3=0$, among all degree-$n$ monomials of them, only $X^n,\,X^{n-1}Y,\,X^{n-1}Z$ and $X^{n-2}Y^2$ are nonzero. It follows that
\begin{aligned}
A^n
&=(X+Y+Z)^n\\
&=X^n+nX^{n-1}Y+nX^{n-1}Z+\binom{n}{2}X^{n-2}Y^2\\
&=\pmatrix{a^n&na^{n-1}b&na^{n-1}c+\binom{n}{2}a^{n-2}b^2\\ 0&a^n&na^{n-1}b\\ 0&0&a^n}.
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum binomial coefficients Show that
$$\sum_{k=0}^{n}\binom{n}{k}\left(\binom{m+k+1}{m-n}+(-1)^{k+1}\binom{m+2(n-k)+1}{m}\right)=0$$
for all integers $m$ and $n$ with $m\ge n\ge 0$.
I tried induction on $n$, but there's not a very nice way to change the LHS from the $n$ case to $n+1$.
| We get for the first of the two pieces
$$\sum_{k=0}^n {n\choose k} {m+k+1\choose m-n}
= [z^{m-n}] (1+z)^{m+1} \sum_{k=0}^n {n\choose k} (1+z)^k
\\ = [z^{m-n}] (1+z)^{m+1} (2+z)^n.$$
The second piece is
$$\sum_{k=0}^n {n\choose k} (-1)^{k+1} {m+2(n-k)+1\choose m}
\\ = - [z^m] (1+z)^{m+2n+1}
\sum_{k=0}^n {n\choose k} (-1)^k (1+z)^{-2k}
\\ = - [z^m] (1+z)^{m+2n+1}
\left(1-\frac{1}{(1+z)^2}\right)^n
\\ = - [z^m] (1+z)^{m+1} (2z+z^2)^n
= -[z^m] z^n (1+z)^{m+1} (2+z)^n
\\ = -[z^{m-n}] (1+z)^{m+1} (2+z)^n.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Computing Joint Moment Generating Function Good morning, Stack Exchange. My problem is that I have a joint distrubtion of discrete random variables with the following PMF and support:
$f(x,y) = \frac{2^{x-y} e^{-3}}{x!(y-x)!}$ when $x = 0, 1, 2, ... y$ and $y = 0, 1, 2, ... \infty $ and $f(x,y) = 0$ otherwise
So, my problem here is that I know that to get the MGF, $M(t_1, t_2)$, we must evaluate the sum:
$$\sum_{y=0}^{\infty} \sum_{x=0}^{y} \frac{e^{x_1t_1 + yt_2} * e^{-3} * 2^{y-x}}{x!{y-x!}}$$
My strategy for evaluating the inner sum was to move all the terms that contain a $y$ to the outermost sum and to move the constant $e^{-3}$ out of the sum entirely, hence solve the inner sum as $$\sum_{x=0}^{y} \frac{e^{xt_1} * 2^{-x}}{x!{y-x!}}$$
The hint from the textbook tells me that this sum is easy to evaluate if the term $2^{-x}$ weren't there, but I am not sure how to deal with the sum once this term is present. Am I on the right track here, and how do I solve this innermost sum and help simplifying the sum in general would be much appreciated. Thanks for reading.
| You have PMF:
$$
f(x,y) =
\begin{cases}
\frac{2^{x-y} e^{-\frac{3}{2}}}{x!(y-x)!},& x = 0, 1, 2, \ldots y\text{ and } y = 0, 1, 2,\ldots\\
0& \text{otherwise}
\end{cases}
$$
And the MGF is:
\begin{align}
M_{X,Y}(t_1,t_2)
&=
E(e^{t_1x+t_2y})\\
&=
\sum\limits_{y=0}^\infty\sum\limits_{x=0}^y
e^{t_1x+t_2y}\frac{2^{x-y} e^{-\frac{3}{2}}}{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
e^{t_1x}\frac{2^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{\left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{y! \left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} \\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} 1^{y-x}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}(2e^{t_1}+1)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(2^{-1}e^{t_2}(2e^{t_1}+1)\right)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(e^{t_1+t_2}+\dfrac{1}{2}e^{t_2}\right)^y\\
&=
e^{-\frac{3}{2}} e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}}\\
&=
e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}-\frac{3}{2}}.
\end{align}
Note
I use the fact:
(1) Binomial theorem
$$\sum\limits_{x=0}^n \begin{pmatrix}n\\x\end{pmatrix}a^x b^{n-x} = (a+b)^n$$
(2) The Taylor series about $x=0$ of $e^x$,
$$\sum\limits_{n=0}^\infty \dfrac{x^n}{n!}=e^x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?
| $(x+y) ^2 =16$$\Rightarrow$$ x^2 +y^2 +2xy=16$$\Rightarrow$$ x^2 +y^2 =12$
Because $xy=2$
$x^2 +y^2 =12$ $\Rightarrow $ $ (x^2 +y^2)^3 =12^3$$\Rightarrow $$x^6+y^6+3(xy)^2(x^2 +y^2) =1728$$\Rightarrow $$x^6+y^6=1728 - 3(4)(12)=1584$
Finally :
$x^6+y^6 =1584$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:
$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$
Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$
When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.
| From
$$x\ne-\sqrt{x+1}$$
you are deducing
$$x^2\ne|x+1|$$
But this is invalid! You can't deduce $a^2\ne b^2$ from $a\ne -b$, because $a$ may be equal to $b$. (And in fact $x$ is equal to $\sqrt{x+1}$ when $x=\frac{1+\sqrt 5}{2}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Stuck solving this PDE problem I need to solve $$uu_x+yu_y=x, u(x,1)=2x$$ i tried with characteristics method but got stuck solving the ODEs $$\frac{dx}{dt}=u ~,~\frac{dy}{dt}=y~,~\frac{du}{dt}=x$$ with initial values $x(0)=s,y(0)=1,u(0)=2s$. From the second one i get $y=e^t$, but how to solve the first two ODEs? I tried $$\frac{dx}{du}=\frac{u}{x} \implies x^2-u^2=k_1$$ where $k_1=-3s^2$ which gives $u=\sqrt{x^2+3s^2}$. Similarly $$\frac{dy}{y}=\frac{dx}{\sqrt{x^2+3s^2}} \implies y=k_2(x+\sqrt{x^2+3s^2})$$ And $k_2$ can be found out as $k_2=\frac{1}{3s}$, giving me $y=\frac{x+\sqrt{x^2+3s^2}}{3s}$. I'm not sure how to proceed further and solve for $u$?
| According to your results we have
$$
\cases{
u^2-x^2-3s^2=0\\
y-\frac{x+\sqrt{x^2+3s^2}}{3s}=0
}
$$
or
$$
\cases{
u^2-x^2-3s^2=0\\
(3sy-x)^2-x^2-3s^2=0
}
$$
So solving the first for $s$ and substituting after choosing the feasible one into the second, we have
$$
u = x\frac{3y^2+1}{3y^2-1}
$$
| {
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"answer_id": 1
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Prove Cauchy Sequence and limit Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.
Let $\{x_n\}$ be recursively defined as $ \left\{
\begin{array}{ll}
x_{n+1} =\frac{1-x_n}{4} \\
x_1 = \frac{1}{2}
\end{array}
\right. $
$(a)$ Prove that $|x_{n+1}-x_n| \le \frac{1}{4}|x_n-x_{n-1}|$
$(b)$ Use $(a)$ to prove that ${x_n}$ is a cauchy sequence
$(c)$ Compute $\underset{n \rightarrow \infty}{\rm lim} x_n$
Answer:
$(a)$
$x_{n+1} - x_n=\frac{1-x_n}{4}-x_n$ and $ x_n = \frac{1-x_{n-1}}{4}$. Thus, $x_{n+1}-x_n = (\frac{1-x_n}{4})-(\frac{1-x_{n-1}}{4}) = \frac{1}{4} \{1-x_{n}-1+x_{n-1}\}$. Therefore, $|x_{n+1} -x_n| \le |\frac{1}{4}(1-x_n-1+x_{n-1})| \Rightarrow |x_{n+1} -x_n| \le \frac{1}{4} |x_{n-1}-x_n|$ and $\frac{1}{4} |x_{n-1}-x_n| = \frac{1}{4} |x_{n}-x_{n-1}|$
$(b)$ From $(a)$, it is observable that $|x_{n+1}-x_n| \le \frac{1}{4}|x_{n-1}-x_n|$. Now $|x_{n-1}-x_{n-2}| \le \frac{1}{4} |x_{n-2} - x_{n-3}|$. Continuing this way, we get $|x_{n+1}-x_n| \le \frac{1}{4^n} |x_2-x_1| \Rightarrow |x_{n+1}-x_n| \le \frac{1}{4^n}(\frac{+3}{8})$. As $x \rightarrow \infty , \frac{1}{4^n} \rightarrow 0$. Hence, $\{x_n\}$ becomes a Cauchy sequence. Since every sequence in $\mathbb{R}$ is convergent, $\{x_n\}$ is convergent.
$(c)$ In $(b)$ we have shown $\{x_n\}$ is convergent. Therefore, we assume $l \in \mathbb{R}$ and $ lim\hspace{3pt} x_n = l$. We know $x_{n+1} = \frac{1}{4} (1-x_n)$;
$lim \hspace{3pt} x_{n+1} = \frac{1}{4}(1-lim\hspace{3pt} x_n) \Rightarrow l=(\frac{1}{4})(1-l) \Rightarrow l=\frac{1}{5}$
| For (a) and (c) you proofs are correct But your argument for (b) is wrong. $x_{n+1}-x_n \to 0$ does not imply that $(x_n)$ is a Cauchy sequence. Use the fact that $|x_{n+m}-x_n| \leq |x_{n+m}-x_{n+m-1}|+|x_{n+m-1}-x_{n+m-2}|+\cdots +|x_{n+1}-x_{n}|$ to show that $x_{n+m}-x_n \to 0$ as $n,m \to \infty$. [You will have to compute a geometric sum for this].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$ I am working on this problem:
$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have
$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$, $b' =\sqrt[3]{b}$ and $c' =\sqrt[3]{c}$
So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.
| Note $X^3+Y^3+Z^3 - 3XYZ = (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)$.
Using $X^3=ab, Y^3=bc, Z^3 = ca$, as $X, Y, Z$ are distinct (why?), $X^2+Y^2+Z^2> XY+YZ+ZX$ and hence the sign of $X+Y+Z = \sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}$ is the same as the sign of $X^3+Y^3+Z^3-3XYZ = -24-3=-27$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
maximum of the function $S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi\sqrt{2+2\cos\phi}$ In a problem I have the function:
$$S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi \sqrt{2+2\cos\phi}$$
in which $\phi \in [0,\pi]$.
I have to find the point of maximum and calculate the maximum value of the function.
$\frac{dS}{d \phi}= 2 \pi*sin(\phi) cos(\phi)+\frac{\pi*((3cos(\phi))^2+2cos(\phi)-1))}{\sqrt{2+2cos(\phi)}}$
Putting this derivative equal to $0$ we have the equation:
$-8 (cos\phi)^5-17 (cos\phi)^4-4(cos\phi)^3+(cos\phi)^2+4cos\phi-1=0\Rightarrow (cos\phi+1)^3*(-8(cos\phi)^2+7cos\phi-1)=0$
The maximum of this function is for $\cos \varphi = \frac{7+ \sqrt{17}}{16} $
Then $$\sin\varphi=\frac{\sqrt{190-14 \sqrt{17}}}{16}$$
Now I'm trying to find the maximum value of the function but I'm lost in the calculation and I don't know what is the mistake.
The result should be $$S_{\max}= \frac{\pi(107+51 \sqrt{17})}{128}$$
| Note
\begin{align}
S(\phi) &= \pi \sin^2\phi+\pi\sin\phi \sqrt{2+2\cos \phi}\\
&= \pi \sin^2\phi+2\pi\sin\phi \cos \frac\phi2\\
&= 4\pi\left(\sin\frac\phi2 + \sin^2\frac\phi2-\sin^3\frac\phi2-\sin^4\frac\phi2\right)\tag1
\end{align}
Set $S’(\phi)= 0$ to get
\begin{align}
&2\pi \cos\frac\phi2\left(
4\sin^3\frac\phi2 +3 \sin^2\frac\phi2 - 2\sin\frac\phi2-1\right)\\
= &2\pi \cos\frac\phi2 \left(\sin\frac\phi2+1\right)\left( 4\sin^2\frac\phi2- \sin\frac\phi2-1\right)=0
\end{align}
and the valid solution $\sin\frac\phi2=\frac{1+\sqrt{17}}8$. Then, plug it into (1) to obtain the maximum value
$$S_{max}=\frac{107+51\sqrt{17}}{128}\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$
I've calculated several values of this function, and it seems to have a limit of $1$.
| For $x\in\left(0,\frac\pi2\right]$, the concavity of $\sin(x)$ says
$$
\frac2\pi\le\frac{\sin(x)}x\le1
$$
Therefore,
$$
\underbrace{\left(\frac4\pi\sqrt{x}-\sqrt{x}\right)^x}_{\left(\frac4\pi-1\right)^x\sqrt{x^x}}\le\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x\le\underbrace{\left(2\sqrt{x}+\sqrt{x}\right)^x}_{3^x\sqrt{x^x}}
$$
The Squeeze Theorem says
$$
\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .
What I Tried: Here is a picture:-
Let the perpendicular bisector of $BC$ pass through $BC$ at $E$ .
Then I first noticed that $\Delta BDE \cong \Delta CDE$ from $(SAS)$ congruency.
This gives the required information in the diagram, as well as we have $BD = 7$ .
Now :- $$\Delta ABD \sim \Delta ACB$$
$$\rightarrow \frac{AD}{AB} = \frac{7}{BC} = \frac{AB}{AC}$$
So let $AD = k$ , $AB = m$ , $BE = EC = n$ . We have :-
$$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$
| Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have
$$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$
Let $t=\tan \angle ACB$, then
$$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$
And we have
$$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t^2}\right),\quad |AB|=2\alpha\frac{1+t^2}{3-t^2}.$$
Thus, the area
\begin{align}
S_{\triangle ABD} &= \frac12|AB|\cdot|DE|\\
&=\alpha\beta\frac{1+t^2}{3-t^2}\\
&=\frac{49t}{3-t^2}\in(0,\infty).
\end{align}
If you have some other conditions such that $\frac{t}{3-t^2}=\sqrt5$, then $a=49$. If not, $a$ can be any positive real number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3906150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Legendre symbols So for odd prime $p$, I have Legendre symbol $(\frac{-1}{p}) = (-1)^{(p-1)/2}$, and let $n\geq 3$ be an odd positive integer.
I have to show that for odd positive integers $n_1$ and $n_2$, we have
$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$
And from this I have to deduce $(\frac{-1}{n}) = (-1)^{(n-1)/2}$.
So I have manage to show the first part:
$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$
$\frac{n_1n_2-n_1-n_2+1}{2} \equiv 0\pmod{2}$
$\frac{(n_1-1)(n_2-1)}{2} \equiv 0\pmod{2}$
$n_1,n_2$ odd $\implies (n_1-1),(n_2-1)$ even $\implies n_1-1=2a, n_2-1=2b; a,b\in \mathbb{Z}$
$\frac{4ab}{2} \equiv 0\pmod{2}$
$2ab \equiv 0\pmod{2}$
$0 \equiv 0\pmod{2}$
So now, I have to somehow use this to deduce $(\frac{-1}{n}) = (-1)^{(n-1)/2}$, but I have no idea what to do. Any suggestions?
| You already know the result when $n$ is prime, so assume that $n$ is composite and positive. Then there exist positive integers $n_1$ and $n_2$ such that $2\le n_1\le n_2<n$ and $n_1 n_2 = n.$ What determines whether a power of $-1$ is $1$ or $-1$ is the exponent modulo $2.$ By what you have already proven, $$(-1)^{\frac{n-1}{2}}=(-1)^{\frac{n_1 n_2 -1}{2}} = (-1)^{\frac{n_1-1}{2}+\frac{n_2-1}{2}}=(-1)^{\frac{n_1-1}{2}}\cdot (-1)^{\frac{n_2-1}{2}}.$$
You can use strong induction on $n,$ so we get $$(-1)^{\frac{n-1}{2}}=\left(\frac{-1}{n_1}\right)\left(\frac{-1}{n_2}\right).$$ This is equal to $\left(\frac{-1}{n}\right)$ by the complete multiplicativity of the Jacobi symbol (in either entry when the other entry is fixed), assuming that is what you are using for $\left(\frac{-1}{n}\right)$ since the Legendre symbol is defined only for primes in the lower entry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the eigenvalues of $x x^T - y y^T$?
If $x$ and $y$ are two linearly independent column $n$-vectors, where $n \geq 2$, find all the eigenvalues of $x x^{T} - y y^{T}$
Could you please tell me whether what I have written is correct or not?
Using the identity
\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}
for $A \in F^{m \times n}$ and $B \in F^{n \times m}$, we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \times n}$ directly as:
\begin{align*}
\varphi(\lambda) &= \det(\lambda I_{(n)} - (xx^T - yy^T)) =
\lambda^{n - 2}\det\left(\lambda I_{(2)} - \begin{pmatrix} x^T \\ -y^T \end{pmatrix}\begin{pmatrix} x & y \end{pmatrix}\right) \\
&=
\lambda^{n - 2}\begin{vmatrix}
\lambda - x^Tx & -x^Ty \\
y^Tx & \lambda + y^Ty
\end{vmatrix} \\
&= \lambda^{n - 2}[(\lambda - x^Tx)(\lambda + y^Ty) + (x^Ty)^2] \\
&=
\lambda^{n - 2}(\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2))
\end{align*}
Since $x$ and $y$ are linearly independent, by Cauchy-Schwarz inequality $(x^Tx)(y^Ty) > (x^Ty)^2$ (that is, the equality of C-S inequality cannot hold), whence the determinant $\Delta$ of the quadratic equation $\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2) = 0$ equals to
\begin{align*}
\Delta = (x^Tx - y^Ty)^2 + 4(\|x\|^2\|y\|^2 - (x^Ty)^2) > 0.
\end{align*}
Hence the two non-zero eigenvalues are two distinct real numbers
\begin{align*}
\lambda_1 = \frac{y^Ty - x^Tx + \sqrt{\Delta}}{2}, \quad
\lambda_2 = \frac{y^Ty - x^Tx - \sqrt{\Delta}}{2}.
\end{align*}
| Every vector simultaneously orthogonal to $x$ and $y$ is an eigenvector of the matrix $xx^T - yy^T$ of eigenvalue 0 (note that the matrix has rank 2, so his kernel must have dimension $n-2$ and this is always an eigenspace). So, the remaining eigenvectors must be linear combination of $x$ and $y$.
Solve
$$ (xx^T - yy^T)(\alpha x + \beta y) = \lambda (\alpha x + \beta y) \qquad \lambda,\alpha,\beta \in \mathbb{R}. $$
You will find two polinomial equation in $\lambda,\alpha,\beta$. It is easy to see that it can't be $\alpha=0$ and $\beta=0$. So, from the two equation you can find a polinomial equation of degree 2 in $\frac{\alpha}{\beta}$. Then you can compute the two eigenvalues and the corrisponding eigenvectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving linear in $\langle a_a, a_a \rangle =
\begin{cases}
\frac{1}{9}, & \text{if $a$ $\ne$ $a$}\
\end{cases}$
| To begin with, let us consider the linear combination
\begin{align*}
\alpha_{1}a_{1} + \alpha_{2}a_{2} + \ldots + \alpha_{k}a_{k} = 0
\end{align*}
If we take the inner product of such linear combination with $a_{1}$ we get the equation
\begin{align*}
2\alpha_{1}b + \alpha_{2} + \alpha_{3} + \ldots + \alpha_{k} = 0\\\\
\end{align*}
If we apply such inner product to $a_{j}$, where $2\leq j\leq k$, we get the following system of equations:
\begin{align*}
\begin{bmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{bmatrix}
\begin{bmatrix}
\alpha_{1}\\
\alpha_{2}\\
\vdots\\
\alpha_{k}
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
\vdots\\
0
\end{bmatrix}
\end{align*}
Such system of linear equations has only the trivial solution iff the determinant of the coefficient matrix is different from zero. Based on such result, we need to determine its value in terms of $b$ and $k$:
\begin{align*}
\begin{vmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix} =
\begin{vmatrix}
2b + k - 1 & 1 & \cdots & 1\\
2b + k - 1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
2b + k - 1 & 1 & \cdots & 2b
\end{vmatrix} & =
(2b + k - 1)
\begin{vmatrix}
1 & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix}\\\\
& =(2b + k - 1)
\begin{vmatrix}
1 & 0 & \cdots & 0\\
1 & 2b - 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
1 & 0 & \cdots & 2b - 1
\end{vmatrix}\\\\
& = (2b+k-1)(2b-1)^{k-1} \neq 0
\end{align*}
and we are done.
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
JEE Circles Doubt Can you derive a general formula for length of intercept made by a circle $x^2+y^2+2gx+2fy+c=0$ on the line $ax+by+c=0$? If no, then is there any other method to solve these type of questions?
| \begin{align}
x^2+y^2+2gx+2fy+d &= 0
\tag{1}\label{1}
,\\
ax+by+c &= 0
\tag{2}\label{2}
.
\end{align}
Expression \eqref{1} represent a circle
with radius $r=\sqrt{g^2+f^2-d}$
centered at $O=(-g,-f)$:
\begin{align}
(x+g)^2+(y+f)^2
&=g^2+f^2-d
\tag{3}\label{3}
.
\end{align}
Intersection points $X_+,\ X_-$
of the circle $(O,R)$
and the line through the point $B$
in direction $D=z_2-z_1$ can be found as follows:
\begin{align}
X_{\pm}&=B+t_{\pm}\cdot D
\tag{4}\label{4}
,\\
t_{\pm}&=\frac{-(B-O)\odot D\pm\sqrt{\left((B-O)\odot D\right)^2+|D|^2(R^2-|B-O|^2)}}{|D|^2}
\tag{5}\label{5}
,
\end{align}
where $\odot$ is
the dot product.
Given \eqref{2},\eqref{3}, we can apply \eqref{4}, \eqref{5}
by setting (assuming that $b\ne0$)
\begin{align}
O&=(-g,-f)
,\quad
z_1=(0,-\tfrac cb)
,\quad
z_2=(1,-\tfrac{a+c}b)
,\quad
B=z_1
,\quad
D=z_2-z_1=(1,-\tfrac ab)
\tag{6}\label{6}
.
\end{align}
The number of intersections is defined by the discriminant
\begin{align}
\delta&=
(a\,f-b\,g)^2-d\,(a^2+b^2)+c\,(2a\,g+2b\,f-c)
\tag{7}\label{7}
.
\end{align}
If $\delta<0$, there are no intersections,
if $\delta>0$, there are two intersection points,
\begin{align}
X_{\pm}&=
\left(
\frac{-b\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-a\,c}{a^2+b^2},\
\frac{ a\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-b\,c}{a^2+b^2}
\right)
\tag{8}\label{8}
,
\end{align}
and if $\delta=0$, then the line is tangent to the circle at the point
\begin{align}
X_+=X_-&=
\left(
\frac{-b\,(b\,g-a\,f)-a\,c}{a^2+b^2},\
\frac{ a\,(b\,g-a\,f)-b\,c}{a^2+b^2}
\right)
\tag{9}\label{9}
.
\end{align}
The length of intercept is just
\begin{align}
|X_+-X_-|&=
2\sqrt{\frac{\delta}{a^2+b^2}}
\tag{10}\label{10}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $a^3 + b^3 \geq a^2b + ab^2$? Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--
For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$
I have no idea how to begin but this is what I did.
By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)
Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$
Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.
(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)
| Note
$$\frac{a^3+b^3}{a^2b+ab^2}=\frac{a^2-ab+b^2}{ab}=\frac{(a-b)^2}{ab}+1\ge 1
$$
Thus
$$a^3+b^3\ge a^2b+ab^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Domain of a function defined with another function Suppose $f(x)=\frac{1}{1+\frac{1}{x}}$, and $g(x)=\frac{1}{1+\frac{2}{f(x)}}$. To find the points not in the domain of $g(x)$, we obviously have to check the $x-$values where $f(x)$ becomes $0$ and $-2$, and count them first.
But where is the intuition behind considering the points for which $f(x)$ itself is undefined? On substituting $f(x)$ in $g(x)$, you get $g(x)=\frac{x}{3x+2}$. Why do we count the additional points where $f$ is discontinuous (in this case, the only extra one being $x=-1$, as $x=0$ is already necessary for $g(x)$'s denominator to be defined)?
| We want to find when $g$ is undefined and this is when the denominator, $1+ \frac {2}{f(x)} = 0 $ which is only true when $f(x) = -2$. So then we check $-2 = \frac {1}{1+\frac{1}{x}} \implies 1+\frac {1}{x} = -\frac{1}{2}\implies x=-\frac {2}{3}.$ So $g(-\frac{2}{3})$ is not defined and $x = -\frac {2}{3}$ is not in the domain of $g$. Notice that $g$ is also not defined when $f(x) = 0$, but $f(x)$ never equals $0$ so this is not a problem. We must check where $f(x)$ is undefined now. This is when $1+ \frac{1}{x} = 0 \implies x = -1$. so then $g(-1)$ is not defined either. Also $f(0)$ is undefined since $\frac {1}{0}$ is undefined which means $g(0)$ is undefined. Hence the points not in the domain of $g$ are $x = -1, -\frac {2}{3},0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Evaluating limits using a series I'm trying to use a Taylor series centered at $0$ to evaluate this limit:
$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$
I rewrote the function as its Maclaurin series:
$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$
In expanded form:
$$\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}=4x^{3-\frac{2}{x^3}}+\frac{4x^{3-\frac{4}{x^3}}}{2!}+\frac{4x^{3-\frac{6}{x^3}}}{3!}+...$$
As $x$ goes to $\infty$, $\frac{2}{x^3}$ goes to $0$. Thus, the limit of the first term is simply the limit of $4x^3$, which is $\infty$. Based on this fact alone, I would assume, the limit of the entire series is $\infty$, but apparently the answer is $-8$. What did I do wrong?
| As $x\to +\infty$, it should be
$$4x^3\left(e^{\color{blue}{\frac{-2}{x^3}}}-1\right)=4x^3\sum_{k=1}^{\infty}\frac{(\color{blue}{\frac{-2}{x^3}})^k}{k!}=
4x^3\left(-\frac{2}{x^3}+o(1/x^3)\right)=-8+o(1).$$
Then what is the limit?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$) The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.
This was question asked:
I don't know how to find eigenvalues of matrices ( larger than 3 $\times$ 3).
So, it's my humble request if you can tell some details or source of results and observations which are useful for dealing with such matrices.
I have studied linear algebra from David C lay.
Thanks a lot!!
| In addition to the existing answers I'd like to say that it's not hard to calculate the characteristic polynomial using Laplace expansion:
\begin{align}p(x)&=\det\begin{bmatrix} x & 0 & 0 & -1 & 0 & 0\\
0 & x & 0 & 0 & -1 & 0\\
0 & 0& x& 0 & 0& -1\\
-1 & 0 &0&x& 0 & 0\\
0 & -1 & 0 & 0&x &0\\
0 & 0& -1 & 0 & 0 &x
\end{bmatrix}\\
&=x\det\begin{bmatrix}
x & 0 & 0 & -1 & 0\\
0& x& 0 & 0& -1\\
0 &0&x& 0 & 0\\
-1 & 0 & 0&x &0\\
0& -1 & 0 & 0 &x
\end{bmatrix}+1\det\begin{bmatrix} 0 & 0 & -1 & 0 & 0\\
x & 0 & 0 & -1 & 0\\
0& x& 0 & 0& -1\\
-1 & 0 & 0&x &0\\
0& -1 & 0 & 0 &x
\end{bmatrix}\\
&=x\cdot x\det\begin{bmatrix}
x & 0 & -1 & 0\\
0& x& 0 & -1\\
-1 & 0 &x &0\\
0& -1 & 0 &x
\end{bmatrix}-1\det\begin{bmatrix}
x & 0 & -1 & 0\\
0& x & 0& -1\\
-1 & 0 &x &0\\
0& -1 & 0 &x
\end{bmatrix}\\
&=(x^2-1)\det\underbrace{\begin{bmatrix}
x & 0 & -1 & 0\\
0& x & 0& -1\\
-1 & 0 &x &0\\
0& -1 & 0 &x
\end{bmatrix}}_{=M}
\end{align}
Now it's your turn to prove $\det(M)=(x^2-1)^2$ using the same method (Laplace expansion) as above.
The final result is
$p(x)=(x^2-1)^3$
| {
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Is $s_n$ a multiple of $2^n$ for all $n \geq 0$? Let's define:
$$a_{0,0} = 1$$
$$\forall k \ne 0, a_{0,k} = 0$$
$$\forall n \geq 0, \forall k, a_{n+1,k} = -(4n-2k+3)a_{n,k-1} + (2k+1)a_{n,k}$$
| 0 1 2 3 4 ...
----+-----------------------------
0 | 1
1 | 1 -1
2 | 1 -8 3
3 | 1 -33 71 -15
4 | 1 -112 718 -744 105
... ... ... ... ... ... ...
Up to the alternated sign, this is OEIS' sequence number A214406.
Now let's define:
$$s_n = \sum_{k=0}^n a_{n,k}$$
$$ s_0 = 1, s_1 = 0, s_2 = -4, s_3 = 24, s_4=-32, \dots $$
It appears that for all $n$, $s_n$ is a multiple of $2^n$. Is it provable?
Context/own efforts: so far, I was able to prove connections with the sequence of functions $(R_n)_{n \in \mathbb{N}}$ defined by:
$$ R_0(x)=1 $$
$$ R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2} \right)' $$
The connection is:
$$ s_n = 2^n R_n(1) $$
We even have:
$$ R_n(x) = 2^n \frac{P_n(x^2)}{(1+x^2)^{2n}} $$
where
$$ P_n(x) = \sum_{k=0}^n a_{n,k} x^k $$
I don't know if it helps, but this also means that the exponential generating function
$$ \sum_{n \geq 1} \frac{s_{n-1}}{2^{n-1}} \frac{x^n}{n!} $$
is the series reversion of
$$ \frac{\ln(x+1)}{2} + \frac{x}{2} + \frac{x^2}{4} = 0 + 1x + 0 x^2
+ \frac{1}{6}x^3 - \frac{1}{8}x^4
+ \frac{1}{10}x^5 - \frac{1}{12}x^6
\dots $$
I also established that
$$ s_{n+1} = -4 \sum_{k=0}^n (n-k) a_{n,k} $$
but this appears just to push the problem forward, of proving that $ \sum_{k=0}^n k a_{n,k} $ is a multiple of $2^{n-1}$ ...
| Starting with $$s_n = 2^n R_n(1)$$
All we need to prove then is that $R_n(1)$ is an integer for all $n$
Clearly $s_n$ is a sequence of integers, implying that if $R_n(1)$ is not an integer, then it must be a fraction with denominator $2^m,m\leq n$.
Remember
$ R_0(x)=1, R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2} \right)' $
Note that since $R_n(x)$ can subsequently be written as $R_n(1)=\frac{a*2^b}{2^m}$ for integers $a,b$, we just need to prove $b\geq m$ to prove $R_n(1)$ is an integer
Proof by induction:
$R_0(1)$ clearly satisfies this condition.
Also note for $n>0$,
$R_{n+1}(x) = \cfrac{d}{dx} (R_n(x) \times (\frac{2x}{x^2+1}-1))=R^{'}_n(x)\times \frac{2x}{x^2+1}-R_n(x)\times \frac{2(x+1)(x-1)}{(x^2+1)^2}$
$R_{n+1}(1) = R^{'}_n(1)\times (1-0) = R^{'}_n(1)$
$R_{n+1}(1) = \frac{d^n}{dx^n}R_1(1) = \frac{d^{n+1}}{dx^{n+1}}\frac{2x}{x^2+1}$
By the quotient rule, if there are $m-b$ factors of $2$ in the simplified denominator of $R_n(1)$ (such that there are 0 factors of 2 in the numerator), then there is at most $2(m-b)$ factors of $2$ in the denominator of $R^{'}_n(1)$.
However, since for $R_0(1),m-b=0$, then for $R_1(1)$, the denominator can have at most $2(m-b) = 0$ factors of 2. Equivalently, $2(m-b)\leq 0,b\geq m$ By induction, this holds true for every $R_n(1)$, which is what we set out to prove.
| {
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Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$
Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;
in particular, $\left\lfloor\dfrac{\left\lfloor\dfrac {a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor\dfrac{a}{bc}\right\rfloor$
I don't know how to approach this problem, any hints to solve this would be greatly appreciated.
Thanks for your help!
| Let $m=\lfloor x\rfloor$. Then $m\leq x < m+1$. So $\dfrac{m}{n}\leq \dfrac{x}{n}< \dfrac{m+1}{n}$. By the division algorithm, $m =qn+r$ for some integers $q,r$ with $0\leq r\leq n-1$. Thus, $$q\leq\dfrac{m}{n}\leq\dfrac{x}{n}<\dfrac{m+1}{n}=\dfrac{qn+r+1}{n}\leq q+1.$$
So $\lfloor \tfrac{m}{n} \rfloor=\lfloor\tfrac{x}{n}\rfloor=q$.
Note: $\lfloor x\rfloor$ is the unique integer $N$ such that $N\leq x<N+1$.
| {
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prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.
When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac{x-1}{x} - \frac{1-x^2}{1+x^2} \\ \geq \frac{1-x}{x} - \frac{1-x^2}{1+x^2} \\ = \frac{(x-1)^2}{x(1+x^2)} \\ \geq 0 $$
So how to prove it when $0 < x <1 $?
| Disclaimer: Here is a solution which is not mine. I just slightly adapted it from the very same question, asked as Problem 5642, by R. Redheffer, American Mathematical Monthly 76, (1969), p.422, entitled "A delightful inequality"...
Let us fix $x$ with $0 < x < 1$:
Starting from the classical infinite product for the cardinal sine:
$$\frac{\sin \pi x}{\pi x}=\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2}\right)=(1-x^2)\underbrace{\prod_{k=2}^{\infty}\left(1-\frac{x^2}{k^2}\right)}_{\lim_{n \to \infty} P_n(x)}$$
where $P_n(x)$ is defined by:
$$P_n(x):=\prod_{k=2}^{n}\left(1-\frac{x^2}{k^2}\right),$$
it is enough to prove that $$\text{for any} \ \ n \ge 2, \ \ P_n(x) \ge \dfrac{1}{1+x^2} \ \ \iff \ \ (1+x^2)P_n(x) \ge 1 \ \ \tag{1}$$
Let us remark the evident relation:
$$P_{n+1}(x)=\left(1-\frac{x^2}{(n+1)^2}\right)P_n(x) \tag{2}$$
Actually we get, for all $n \ge 2$:
$$(1+x^2)P_n(x) \ge 1+\frac{x^2}{n}$$
by a simple induction argument based on (2). Therefore, (1) is established.
An interesting article where I have found the reference to this question can be found here.
Edit: I found the proof by River Li very interesting.
I noticed a possible alternative treatment beginning at the level where a third degree polynomial appears between square brackets.
Setting $a:=\pi^2$ and $Y:=y^2$,
$$\begin{align}
&\Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\
\ge & \ a(2Y+6a)-(a-Y)(a+Y)^2\\
= & \ Y^3 + aY^2 - a^2Y + 4aY + 12a^2- a^3\\
= & \ \underbrace{(Y+a/3)^3}_{>0}+4a\underbrace{\Big[Y(1-a/3)+a(3-7a/27)\Big]}_{p(Y)}
\end{align}$$
with first degree polynomial $p(Y)>0$ for $Y \in [0,1]$ because
$$p(0)=a(3-7a/27)>0 \ \ \text{and} \ \ p(1)=a(8/3-7a/27)+1>0$$
as a consequence of the fact that $a=\pi^2<10$.
| {
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"question_score": "3",
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| Let $t = \tan (x/2)$. Then $\sin x = 2t/(1+t^2)$ and $\cos x = (1-t^2)/(1+t^2)$. Your equation looks quartic after multiplying both sides by $(1+t^2)^2$.
$$12t(1-t^2)+4((1-t^2)^2-4t^2)-(2-2t^2-12t)(1+t^2)-6(1+t^2)^2$$
but it simplifies to $-4(3t-1)^2=0$, so $t = 1/3$, giving $x = 2(\tan^{-1} t + n\pi) = 2(\tan^{-1}{\frac13}+n\pi)$ for any integer $n$, since $\tan$ has period $\pi$.
The above $t$-substitution omits the possibility that $x = (2n+1) \pi$, which is actually a solution: cross out the $\sin$ in the equation and observe that $4(1)-2(-1)-6 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expected projected length of radial vectors of n-sphere Situation
In $n$-dimensional Euclidean space rests a unit $(n-1)$-dimensional sphere that is orthographically projected onto a $(n-1)$-dimensional plane. The topological definition of a sphere is used, i.e. only the surface points belong to a sphere.
Question
What is the expected projected length $\bar{x}_n$ of uniformly distributed radial unit vectors that start at the sphere center?
Solution strategy
One could calculate this by multiple integration using $n$-dimensional spherical coordinates but I hope there is a more elegant method. It seems there is a pattern that could be generalized to $n$ dimensions.
Solutions for $n=2$ and $n=3$
2-space (see drawing)
In 2-space (plane) a unit 1-sphere (circle) is projected onto a 1-plane (line).
The mean projected length from integration over a quadrant is $$\bar{x}_2=\frac{2}{\pi} =\frac{4}{2\pi}\int_{0}^{\frac{\pi}{2} } \sin(\alpha) \mathrm{d}\alpha$$
3-space
If we go 1 dimension higher, we arrive at the usual 2-sphere in 3-space. The radial vector is projected to a 2-plane. By integration over an octand we get for the mean projected length $$\bar{x}_3=\frac{\pi}{4} =\frac{8}{4\pi} \int_{0}^{\frac{\pi}{2} } \int_{0}^{\frac{\pi}{2} } \sin^2(\theta) \mathrm{d\theta} \mathrm{d}\phi$$
| Disclaimer: I used Mathematica for part of the problem.
As an expected value, this would be $$\mathbb{E}\left[ \sqrt{x_1^2+x_2^2 + ... + x_{n-1}^2} \right]$$
where $\left(x_1, x_2, ..., x_n\right)$ is a random point on the surface of the sphere. Since it is known that $x_1^2 + x_2^2 + ... + x_n^2 = 1$, this is equivalent to $$\mathbb{E}\left[ \sqrt{1 - x_n^2} \right] = \mathbb{E}\left[ \sqrt{1 - x_1^2} \right]$$
From here, $x_1$ can be chosen as $\frac{y_1}{\sqrt{\sum_{k=1}^n y_k^2}}$, where $y_k$ are chosen from $\mathcal{N}(0, 1)$. The expected value is then $$\mathbb{E}\left[ \sqrt{1 - \frac{y_1^2}{\sum_{k=1}^n y_k^2}} \right]$$
As an integral, this would be $$\int_0^{1} \left(1 - \mathbb{P}\left( \sqrt{1 - \frac{y_1^2}{\sum_{k=1}^n y_k^2}}<x \right)\right) dx$$
That inner probability is equal to $$\mathbb{P}\left( \frac{1-x^2}{x^2}\sum_{k=2}^n y_k^2 < y_1^2 \right)$$
Using the PDF of the chi-square distribution, this would be $$\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)}\frac{1}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{\frac{1-x^2}{x^2}z}^{\infty} z^{(n-1)/2-1}e^{-z/2}y^{1/2-1}e^{-y/2} dydz$$
Then the answer is (after some switching of bounds) $$1-\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{0}^{\infty}\int_{\sqrt{\frac{z}{y+z}}}^{1}\left(z^{\frac{n-3}{2}}e^{-\frac{z}{2}}y^{-\frac{1}{2}}e^{-\frac{y}{2}}\right)dxdydz$$
This simplifies to $$\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{0}^{\infty}\left(z^{\frac{n-2}{2}}e^{-\frac{y+z}{2}}y^{-\frac{1}{2}}\right)\sqrt{\frac{1}{y+z}}dydz$$
Mathematica then gives this as $$\frac{\pi^{\frac{3}{2}}}{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\frac{\cos\left(\frac{n\pi}{2}\right)\Gamma\left(1-\frac{n+1}{2}\right)}{\left(\sin\left(\frac{n\pi}{2}\right)\Gamma\left(1-\frac{n}{2}\right)\right)^{2}}$$ for non-integer $n$. Taking the limit as $n$ approaches an integer and using Euler's reflection formula, this simplifies to $$\frac{\Gamma\left(\frac{n}{2}\right)^2}{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)} \approx 1 - \frac{1}{2}n^{-1}-\frac{3}{8}n^{-2} + O(n^{-3})$$
Edit: Using that $\Gamma(k + \frac{1}{2}) = \frac{(2k-1)!!\sqrt{\pi}}{2^k} = \frac{(2k)!\sqrt{\pi}}{4^k}k!$ and $\Gamma(k) = (k-1)!$, splitting by odd and even cases, for odd $n$, it is $$\frac{\pi}{4^{n-1}}\cdot\frac{\left[\left(n-1\right)!\right]^{2}}{\left(\frac{n-3}{2}\right)!\left[\left(\frac{n-1}{2}\right)!\right]^{3}} = \frac{\pi(n-1)}{2^{2n-1}} \binom{n-1}{\frac{n-1}{2}}^2$$ while for even $n$, it is $$\frac{2^{2n-2}}{\pi}\frac{\left(\frac{n}{2}\right)!\left[\left(\frac{n}{2}-1\right)!\right]^{3}}{\left(n-2\right)!\left(n\right)!} = \frac{2^{2n+1}}{\pi}\frac{n-1}{n^{2}}\binom{n}{\frac{n}{2}}^{-2}$$
| {
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Evaluating Contour Integral of square Evaluating $\frac{1}{2i\pi} \oint_C \frac{z^2}{z^2+4} dz$ where $C$ is the square with vertices at $\pm 2$ and $\pm 2 + 4i$
So, I rewrote the contour integral:$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$
then used the Cauchy Integral Theorem:
$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$ = $\frac{1}{2i\pi} \oint_C \frac{\frac{z^2}{z+2i}}{z-2i} dz$ + $\frac{1}{2i\pi} \oint_C \frac{\frac{z^2}{z-2i}}{z+2i} dz$ and my answer for the two contour integrals is $0$.
Is it done correctly or I did incorrectly.
| You have done the integration incorrectly; the answer is $i$ (backed up by numerical evaluation using mpmath).
The contour encircles a simple pole of the integrand at $2i$. By Cauchy's integral formula applied to $f(z)=\frac {z^2}{z+2i}$
$$f(2i)=i=\frac1{2\pi i}\oint_C\frac{f(z)}{z-2i}\,dz=\frac1{2\pi i}\oint_C\frac{z^2}{z^2+4}\,dz$$
| {
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Primes of the form $x^2+31y^2$ and $x^2+xy+8y^2$ and quadratic forms While working on an exercise in algebraic number theory, I encountered the following problem:
A prime number $p$ can be represented as $x^2+31y^2$ if and only if it can be represented as $x^2+xy+8y^2$.
In this answer by Will Jagy, he claims that those quadratic forms represent the same odd numbers (where odd is in bold).
A while ago, I encountered a similar problem, with $x^2+3y^2$ and $x^2-xy+y^2$, which I managed to solve using some ad-hoc reasoning:
Claim. There exist $a,b\in \Bbb Z$ such that $p=a^2-ab+b^2$ if and only if there exist $x,y\in \Bbb Z$ such that $p=x^2+3y^2$. Proof. $\boxed{\Leftarrow}$ Let $a=x-y$ and $b=x+y$. $\boxed{\Rightarrow}$ If $a$ and $b$ have the same parity, let $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$. In the case where $a$ and $b$ have different parity, wlog $a$ is odd and $b$ is even. Write $a=2n+1$ and $b=2m$ with $n,m\in \Bbb Z$. Let $x=m-2n-1$ and $y=m$. $\blacksquare$
This is not very satisfying, I would like to know if there is some general method for tackling this. Here I found some explanation, pointing me into the direction of equivalent quadratic forms. It suggests to find a matrix $T\in GL_2(\mathbf{Z})$ with determinant $\pm 1$ such that $T^\top \begin{bmatrix}1 & \frac12 \\ \frac12 & 8 \end{bmatrix}T=\begin{bmatrix}1 & 0 \\ 0 & 31 \end{bmatrix}$. Letting $T=\begin{bmatrix} a & b \\ c& d \end{bmatrix}$, we get $1=\begin{bmatrix} a & b\end{bmatrix}\begin{bmatrix}1 & \frac12 \\ \frac12 & 8 \end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}=a^2+ab+8b^2$, which has unique integer solutions $(a,b)=(\pm 1,0)$. Similarly, $31=c^2+cd+8d^2$ has unique integer solutions $(c , d)=(\pm 1,\mp 2)$. Hence, such $T$ does not exist. I suspect this has to do with what Will Jagy is underlining.
Clearly, my knowledge of quadratic forms is too weak. Could someone shed a light on this problem?
| It turns out I was thinking too difficult. "Completing the square" gives $$x^2+xy+8y^2=(x+\tfrac12 y)^2+31(\tfrac12 y)^2.$$
Let $n$ be an odd integer such that $n=x^2+xy+8y^2$ for some integers $x,y$. Reducing modulo 2 gives $1\equiv x(x+y)\bmod{2}$, hence $x$ and $x+y$ are odd, so in particular $y=(x+y)-x$ is even.
Conversely, if $n=z^2+31w^2$ for some integers $z,w$, then we can $y=2w$ and $x=z-2w$ to see that $n=x^2+xy+8y^2$.
Hence, we see that an odd number can be represented by the quadratic form $x^2+xy+8y^2$ if and only if it can be represented by the quadratic form $x^2+31y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck at solving differential equation by using integrating factor This is the text:
Solve the following differential eqution: $$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$$
This particular differential equation has to be solved by finding an integrating factor. I first tried making this equation into a Darboux differential eqaution, and didn't find any luck. These are the integrating factors that I tried: $u(x,y)=u(x)$ , $u(x,y)=x^my^n$ , $u(x,y)=u(y)$ .
I don't have any idea what else to try. You don't have to answer it fully, just give me a hint or something. Thank you in advance.
| $(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$
$\dfrac{dy}{dx}=-\dfrac{2x^3+3x^2y+y^2-y^3}{2y^3+3x^2y+x^2-x^3}$
Let $y=xu$ ,
Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=-\dfrac{2x^3+3x^3u+x^2u^2-x^3u^3}{2x^3u^3+3x^3u+x^2-x^3}$
$x\dfrac{du}{dx}=\dfrac{(u^3-3u-2)x-u^2}{(2u^3+3u-1)x+1}-u$
$x\dfrac{du}{dx}=\dfrac{(-2u^4+u^3-3u^2-2u-2)x-u^2-u}{(2u^3+3u-1)x+1}$
$((2u^4-u^3+3u^2+2u+2)x+u^2+u)\dfrac{dx}{du}=-(2u^3+3u-1)x^2-x$
Let $x=\dfrac{1}{v}$ ,
Then $\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$
$\therefore-\left(\dfrac{2u^4-u^3+3u^2+2u+2}{v}+u^2+u\right)\dfrac{1}{v^2}\dfrac{dv}{du}=-\dfrac{2u^3+3u-1}{v^2}-\dfrac{1}{v}$
$((u^2+u)v+2u^4-u^3+3u^2+2u+2)\dfrac{dv}{du}=v^2+(2u^3+3u-1)v$
This belongs to an Abel equation of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that $\frac{1}{1-z}= \frac{1}{2}\left(1 + i\cot(\theta/2)\right)$
Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$
I have been trying to rearrange the right-hand-side in order to make it look like the required form, but have been struggling to do so and want to see if anyone is able to show me what steps need to be taken to solve this question.
| Note that $e^{i\theta}=\cos\theta+i\sin\theta$.
$$\frac{1}{1-e^{i\theta}}$$
$$=\frac{1}{2}\times \frac{-2}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \frac{e^{i\theta}-1-(e^{i\theta}+1)}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \left(1-\frac{e^{i\theta}+1}{e^{i\theta}-1}\right)$$
Note that $e^{i\theta}+1=e^{\frac{i\theta}{2}}\cdot 2\cos(\frac{\theta}{2})$ and $e^{i\theta}-1=e^{\frac{i\theta}{2}}\cdot 2i\sin (\frac{\theta}{2})$.
Plug in these results above,
$$\frac{1}{1-e^{i\theta}}=\frac{1}{2}\times \left(1-\frac{\cot(\frac{\theta}{2})}{i}\right)$$
$$=\frac{1}{2}\times \left(1-\frac{i\cot(\frac{\theta}{2})}{i^2}\right)$$
$$\fbox{$\huge{\dfrac{1}{1-e^{i\theta}}=\frac{1}{2}\times \left(1+i\cot\left(\frac{\theta}{2}\right)\right)}$}$$
I used some results above. See how I got it below.
$$e^{i\theta}+1=\cos \theta+i\sin \theta+1$$
$$=2\cos^2 \frac{\theta}{2}-1+2i\sin\frac{\theta}{2} \times \cos\frac{\theta}{2}+1$$
$$=2\cos\frac{\theta}{2}\cdot e^{\frac{i\theta}{2}}$$
The other result:
$$e^{i\theta}-1=\cos \theta+i\sin \theta-1$$
$$=1-2\sin^2 \frac{\theta}{2}+2i\cos\frac{\theta}{2}\cdot \sin\frac{\theta}{2}-1$$
$$=2i\sin\frac{\theta}{2}\left(\frac{-\sin\frac{\theta}{2}}{i}+\cos\frac{\theta}{2}\right)$$
$$=2i\sin\frac{\theta}{2}\cdot e^{\frac{i\theta}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help solving Laplace Transform Question For context: I am just starting to learn LaPlace Transform and am stumped with this question.
Find the PS of $x''+x'-12x=sin(3t) + e^{-4t}+e^{3t}$
Where $x(0)=0, x'(0)=0$ using LaPlace Transform
My Approach:
Let X(s)=$\mathscr{L}[x(t)]$
Apply the LaPlace Transform to both sides of the equation
$\mathscr{L}[x''] + \mathscr{L}[x'] - 12\mathscr{L}[x]$ = $\mathscr{L}[sin(3t)] +\mathscr{L}[e^{-4t}] +\mathscr{L}[e^{3t}]$
to get $s^2X(s)-sx(0)-x'(0)+sX(s)-x(0)-12X(s) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$
Plugging in the IC's $x(0) = 0$ and $x'(0)=0$ and then factoring $X(s)$ we get
$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$
You should then make the RHS have a common denominator so you get
$X(s)[s^2+s-12] = \cfrac{3(s+4)(s-3)}{(s^2+9)(s+4)(s-3)}+\cfrac{(s^2+9)(s-3)}{(s+4)(s^2+9)(s-3)}+\cfrac{(s+4)(s^2+9)}{(s+4)(s-3)(s^2+9)} = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)(s+4)}$
Dividing both sides by $[s^2+s-12]$ AKA $(s+4)(s-3)$ to get $X(s)$ by itself
We get
$X(s) = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2}$
Now I Believe I'm supposed to use partial fraction decomposition here to make taking the Inverse LaPlace Transform easier so...
$\cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2} = \cfrac{As+B}{s^2+9} + \cfrac{C}{s-3} + \cfrac{D}{(s-3)^2} + \cfrac{E}{(s+4)} + \cfrac{F}{(s+4)^2}$
Then multiplying both sides by the LHS denominator and simplifying we get
$2s^3+4s^2+21s-27 = (As+B)(s-3)^2(s-4)^2+C(s^2+9)(s-3)(s-4)^2+D(s^2+9)(s+4)^2+E(s+4)(s-3)^2(s^2+9)+F(s^2+9)(s-3)^2$
Here is where I get stuck. How do I solve for A,B,C,D,E,F? This seems ridiculously complex and seems like so much algebra that it makes me think that partial fraction decomposition isn't the right way. So what I've come to ask is partial fraction decomposition to right way to tackle this question? Is there an more efficient way besides partial fraction decomposition? If there's not and partial fraction decomposition is the correct way, How can I solve for A,B,C,D,E,F? Can someone help me determine the algebra, I'm not used to partial fractions being this long unless I've made an error somewhere...
Thank You For Any Help
| It's more simple to decompose fractions from this step:
$$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)(s+4)(s-3) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)=\dfrac 1{(s+4)(s-3)}\left ( \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}\right)$$
And you have:
$$\dfrac 1{(s+4)(s-3)}=\dfrac 1 {7}\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)$$
Now what we have is this;
$$X(s)=\color {red}{\dfrac 37\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)\left ( \cfrac{1}{s^2+9}\right)}+\color {blue}{\dfrac 17\left (\dfrac 1{(s-3)^2}-\dfrac 1{(s+4)^2}\right)}$$
For the term in red you still need to decompose a little bit but for the term in blue you can already apply inverse Laplace Transform.
Note that:
$$\dfrac 1{(s-3)^2}=-\dfrac {d}{ds}\dfrac 1{(s-3)}$$
The inverse Laplace transform is therefore:
$$te^{3t}$$
And:
$$\dfrac 1{(s+4)^2}=-\dfrac {d}{ds}\dfrac 1{(s+4)}$$
The inverse Laplace transform is:
$$te^{-4t}$$
How to decompose this term ?
$$\dfrac 1{(s-3)}\cfrac{1}{(s^2+9)}$$
$$\dfrac {s+3}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\dfrac {1}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\left (\dfrac {A}{(s^2-9)}+\cfrac{B}{(s^2+9)}\right)$$
That's easy now ....
$$\dfrac {(s+3)}{18}\left (\dfrac {1}{(s^2-9)}-\cfrac{1}{(s^2+9)}\right)$$
$$\dfrac {1}{18}\left (\dfrac {1}{(s-3)}-\cfrac{s+3}{(s^2+9)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$ I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$
I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.
So, I don't know how to continue.
| Just use the definition: you need to verify that
$$\Bigg|\frac{x}{x^2-2}-1\Bigg| < \epsilon$$
gives you a neighbourhood of $2$.
This means to solve
$$\begin{cases}
\dfrac{x}{x^2-2}-1 < \epsilon \\\\
\dfrac{x}{x^2-2}-1 > -\epsilon
\end{cases}
$$
Solve this and you will get
$$\sqrt{\dfrac{2\epsilon-4}{\epsilon-1}} < x < \sqrt{\dfrac{-2\epsilon-4}{-\epsilon-1}}$$
Which, being $\epsilon$ small ad libitum, gives you $2 < x < 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculus Integral Question
Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.
I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabola.
Here's what I've done so far.
Step One:
Let $$y = cx^2 + dx + e$$ be the equation for the parabola.
Because the parabola has a root at the origin, $$e = 0$$ The function for parabola can be written as $$y = cx^2 + dx$$
Step Two:
We are given that the parabola passes through the point (a, b) in the first quadrant. Plugging the numbers into the function above, $$b = c(a)^2 + d(a)$$ $$\frac{b-a^2c}{a}=d $$
Therefore, $$y=cx^2+\frac{b-a^2c}{a}x$$
Step Three:
Next, to find the area under the parabola, use the integral: $$\int_0^n(cx^2+\frac{b-a^2c}{a}x)dx$$where n is the second root of the parabola (first root is the origin).
For n, $$0 = cx^2+\frac{b-a^2c}{a}x$$ $$0 = cx+\frac{b-a^2c}{a}$$ $$x=-\frac{b-a^2c}{ac} = n$$
Now, substitute the expression for n, $$\int_0^{-\frac{b-a^2c}{ac}} (cx^2+\frac{b-a^2c}{a}x)dx$$ which equals $$\frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2$$ which equals $$\frac{b-a^2c}{6a^3c^2}$$
It can be concluded that the area under the parabola has the function $$A=\frac{b-a^2c}{6a^3c^2}$$
Step Four:
To find the minimum area -- as requested, take the derivative of A. $$A'=\frac{6a^3c^2(-a)^2-(b-a^2c)(12a^3c)}{(6a^3c^2)^2}$$ $$=\frac{a^2c-2b}{6a^3c^3}$$ Find its root: $$0=\frac{a^2c-2b}{6a^3c^3}$$ $$0=a^2c-2b$$ $$c=\frac{2b}{a^2}$$
Since (a, b) is in the first quadrant, c is a positive number.
Can you tell me where have I gone wrong?
Thanks!
| The reqyired parabola should be of the type $y=-kx^2+hx, k>0$ if it passes through $(a,b)$ then $h=(b+ka^2)/a$. So the eq. of the parabola is $$y=-kx^2+\frac{(b+ka^2)}{a}x$$
This cots $x$ axis at $x=0$, $x=x_0=a+b/(ak)$, so the area made by the parabola on x-axis is
$$A(k)=\int_{0}^{x_0}\left(-kx^2+\frac{(b+ka^2)x}{a}\right) dx, k>0$$
$$A(k)=\frac{ab}{2}+\frac{b^3}{6a^3k^2}+\frac{b^2}{2ak}+\frac{a^3k}{6}$$
$$\implies A'(k)=\frac{a^3}{6}-\frac{b^3}{2a^3k^3}-\frac{b^2}{2ak^2}=0$$
Check that have three roots: $$k=-\frac{b}{a^2},-\frac{b}{a^2},\frac{2b}{a^2}$$
Third one being positive is useful here. Also check that $A''(k)>0$. Thus,
$$A_{min}=A(2b/a^2)=\frac{9ab}{8}$$
| {
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"url": "https://math.stackexchange.com/questions/3939556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Finding a function $f:(0, \infty) \to(0, \infty)$ satisfying the differential equation ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$
Let $f:(0, \infty) \to(0, \infty)$ be a differentiable function such that ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$ where $a$ is positive constant and $f'(1) = 1$, $f'(2) = 2$, then find the value of $f(5)$.
Since the arguments of $f$ and $f'$ are different, it is not possible for me to solve the differential equation directly. By hit and trial, I got the function as $\frac{x^2}2$, but cannot solve it mathematically. Can someone please provide me some hints for this?
| Let $\varphi(x) := f(x) f\left(\frac{a}{x}\right)$. Differentiating we get, for every $x > 0$,
$$
\varphi'(x) = f'(x) f\left(\frac{a}{x}\right)+f(x) \left[f\left(\frac{a}{x}\right)\right]'
= \frac{a}{x f\left(\frac{a}{x}\right)} f\left(\frac{a}{x}\right)- \frac{a}{x^2} f(x) \frac{x}{f(x)} = 0.
$$
Hence there exists $c > 0$ such that $\varphi(x) = c$ for every $x > 0$.
Since
$$
f'(x) = \frac{a}{x f(a/x)} = \frac{a}{c} \cdot \frac{f(x)}{x}
$$
integrating we get
$$
f(x) = f(1) x^{a/c}.
$$
Substituting in $f'(x) = \frac{a}{c} f(1) x^{a/c - 1}$ the values $x=1$ and $x=2$ we finally get $a/c = 2$ and $f(1) = 1/2$, so that $a=2$ and
$f(x) = x^2/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$ I have to solve the following statements, finding the tangent planes to $f$ and its points of tangency:
that contains the line in $\mathbb{R}^3$ that passes through the points $(3,0,3)$ and $(0,-3,3)$
and this other statement:
that contains the point $(0,0,2)$
For the first statement, I tried first making the parametric equation and I found that $L(t) = (3-3t, -3t, 3)$, and I also know that the equation of the plane is given by $$p(x,y) = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) = -2x_0 (x-x_0) -2y_0 (y-y_0) + (1 -x_0^2 - y_0^2)$$ but after that, I don´t see what else to do. Could you give me any hint for both statements?
| Given
$$
\cases{
f(x,y,z)=x^2+y^2+z-1=0\\
\pi(x,y,z) = a x+b y + z c + d=0\\
p=(x,y,z)\\
p_1=(3,0,3)\\
p_2=(0,-3, 3)
}
$$
Construct $\pi(x,y,z)$ such that it contains the line $L \to p = p_1+\mu(p_2-p_1)$ and is tangent to $f(x,y,z)$.
At the tangency point $p_0$ we have
$$
\cases{
\nabla f(p_0) = \lambda \nabla \pi(x_0)\\
f(p_0) = 0\\
\pi(p_0) = 0\\
\pi(p_1) = 0\\
(p_2-p_1)\cdot\nabla \pi(p_0) = 0\\
\|\nabla \pi(p_0)\|=1
}
$$
Here
$$
\cases{
\nabla f(p_0) = (2x_0,2y_0,1)\\
\nabla\pi(p_0) = (a,b,c)
}
$$
so we have $8$ equations on the $8$ unknowns $x_0,y_0,z_0,a,b,c,d,\lambda$. Solving those equations we have two solutions:
$$
p_0 = \left\{
\begin{array}{ccc}
(\frac{1}{2} \left(3-\sqrt{13}\right), & \frac{1}{2} \left(\sqrt{13}-3\right), & 3 \sqrt{13}-10) \\
( \frac{1}{2} \left(3+\sqrt{13}\right), & -\frac{1}{2} \left(3+\sqrt{13}\right), & -10-3 \sqrt{13})\\
\end{array}
\right.
$$
and the planes are characterized by
$$
(a,b,c,d) = \left\{
\begin{array}{cccc}
(-\sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \frac{1}{68} \left(\sqrt{102}
\left(9-\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9-\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9-\sqrt{13}\right)}-\sqrt{102}
\left(9-\sqrt{13}\right)^{3/2}\right)) \\
(-\sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)} ,& \frac{1}{68} \left(\sqrt{102}
\left(9+\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9+\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9+\sqrt{13}\right)}-\sqrt{102}
\left(9+\sqrt{13}\right)^{3/2}\right)) \\
\end{array}
\right.
$$
Attached a plot showing in red the given line and in black the tangentcy points.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?
| Using @Bernard solution
$$y=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}$$ using partial fractions
$$y=-\frac{1+i}{2(u-i)}-\frac{1-i}{2 (u+ i)}+\frac{\sqrt{2}+2 i}{2 \left(\sqrt{2} u+\sqrt{2}-2 i\right)}+\frac{\sqrt{2}-2 i}{2 \left(\sqrt{2} u+\sqrt{2}+2 i\right)}$$
$$\int y\,du=\left(-\frac{1}{2}-\frac{i}{2}\right) \log (-2 u+2 i)\left(-\frac{1}{2}+\frac{i}{2}\right) \log (2 u+2 i)+$$
$$\frac{\left(\sqrt{2}+2 i\right) \log \left(\sqrt{2} u+\sqrt{2}-2 i\right)}{2
\sqrt{2}}+\frac{\left(\sqrt{2}-2 i\right) \log \left(\sqrt{2} u+\sqrt{2}+2 i\right)}{2
\sqrt{2}}$$
Now, recombining everything
$$\int y\,du=\tan ^{-1}(u)-\sqrt{2} \tan ^{-1}\left(\frac{u+1}{\sqrt{2}}\right)+\frac 12 \log \left(\frac{u^2+2 u+3}{u^2+1}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find asymptotic expansion of $\frac{1}{\cos(z)}$
Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.
My work:
I know that:
$$
\frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !}
$$
To find the poles we solve $\cos(z)=0$ where z is a complex number. Getting, $\chi_k=\pi/2+\pi k$
where $k \in Z$.
Now wrie $\frac{1}{\cos(z)} \sim \frac{1}{z-\chi_k}= \frac{-1}{\chi_k} \frac{1}{1-z/\chi_k}$.(I'm not sure if this what I am supposed to do).
Then
$$
E_{2n}=-\left( 2n \right) !\sum_{k\in Z}{\frac{1}{\chi _k}}\frac{1}{\chi _{k}^{n}} \\
E_{2n}=-\left( 2n \right) !\left( \frac{1}{\left( \frac{\pi}{2} \right) ^{n+1}}+\sum_{k\ge 1}{\left( \frac{1}{\chi _{k}^{n+1}}+\frac{1}{\chi _{k_-}^{n+1}} \right)} \right)
$$
Then I can look at the internal sum in the last equation and try to simplify it.
Edited:
$f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k}$ where $\chi_k=\frac{\pi}{2}+2\pi k$ the
poles of $f(z)$ and $Res(f(z),\chi_k)= lim_{z\to \chi_k} (z-\chi_k) f(z)$.
After calculations we get that:
$\chi_0= \pi /2$ with Res= -1, $\chi_{-1} = -\pi / 2$ with Res = 1 , $\chi_1=3\pi /2$ with Res = 1, $\chi_{-2}= -3\pi /2$ with Res = -1 etc.
So, $f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k} \sim \frac{-1}{z-1/2 \pi} +\frac{1}{z+1/2 \pi} +\frac{1}{z-3/2 \pi} + \frac{-1}{z+3/2 \pi}+....$
Then if we expand these geometric coulomns and look at their coefficients we get:
$\frac{E_{2n}}{n!} \sim (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 3\pi)^{n+1} +(2/ 3\pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$
Thus, $E_{2n} \sim (2n)! ( (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$
Thus $E_{2n} \sim (2n)! ( (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} +O(2/ 5\pi)^{n})$
What do you think.
Any guide will be so appreciated.
| You found that
$$
\frac{1}{{\cos z}} = \sum\limits_{k = 0}^\infty {\left( {\frac{{( - 1)^k }}{{z + \left( {k + \frac{1}{2}} \right)\pi }} - \frac{{( - 1)^k }}{{z - \left( {k + \frac{1}{2}} \right)\pi }}} \right)} = \sum\limits_{k = 0}^\infty {( - 1)^k \frac{{(2k + 1)\pi }}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 - z^2 }}} .
$$
This expansion is actually convergent for all $z \neq (k+1/2)\pi$, $k \in \mathbb{Z}$. If $|z|<\frac{\pi}{2}$, then, by the geometric series,
$$
\frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 - z^2 }} = \frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 }}\frac{1}{{1 - \left( {z/\left( {k + \frac{1}{2}} \right)\pi } \right)^2 }} = \sum\limits_{n = 0}^\infty {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} .
$$
Thus
\begin{align*}
\frac{1}{{\cos z}} & = \sum\limits_{k = 0}^\infty {( - 1)^k (2k + 1)\pi \sum\limits_{n = 0}^\infty {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} }
\\ &
= \sum\limits_{n = 0}^\infty {( - 1)^n \left[ {( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}} } \right]\frac{{z^{2n} }}{{(2n)!}}} ,
\end{align*}
provided $|z|<\frac{\pi}{2}$. Consequently,
$$
E_{2n} = ( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}}
$$
for all $n\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$ Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$
How would I go about rearranging this to deduce a value for x?
Thanks for the help.
| Here I propose an alternative way to solve it for the sake of curiosity.
More precisely, what about polar coordinates? If we set $x = r\cos(\theta)$ and $y = r\sin(\theta)$, one gets that
\begin{align*}
\sqrt{x^{2} + ay^{2}} - \sqrt{x^{2}+y^{2}} = z & \Longleftrightarrow r\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - r = z\\\\
& \Longleftrightarrow r = \frac{z}{\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - 1}
\end{align*}
Consequently, one has that
\begin{align*}
y = \frac{z\sin(\theta)}{\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - 1} & \Rightarrow y\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} = y + z\sin(\theta)\\\\
& \Rightarrow y^{2}(\cos^{2}(\theta) + a\sin^{2}(\theta)) = y^{2} + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow y^{2}(1 + a\sin^{2}(\theta)) = y^{2}(1 + \sin^{2}(\theta)) + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow a\sin^{2}(\theta)y^{2} = y^{2}\sin^{2}(\theta) + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow a\sin(\theta)y^{2} = y^{2}\sin(\theta) + 2yz + z^{2}\sin(\theta)\\\\
& \Rightarrow \sin(\theta)(ay^{2} - y^{2} - z^{2}) = 2yz\\\\
& \Rightarrow \sin(\theta) = \frac{2yz}{y^{2}(a-1) - z^{2}}
\end{align*}
Once you know $\sin(\theta)$, you do also know $\cos(\theta)$. Consequently, we obtain the value of $x$.
It is worth emphasizing that one must check for each values of $\theta$ and $a$ the above-mentioned expressions make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
alternative way of proving $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ converges to $\ln 2$ without using integrals I found a lot of answered to this problem using Riemann - sums, I myself solved it rewriting the logarithm sum, but there is another way where you should use the following hints.
Show that
$$\exp \left( \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n} \right) =
2$$
using the following identities
(1). There is a nonnegative sequence $x_k$, converging to zero, and
$$\exp \left(\frac{1}{k} \right) = \left(1+\frac{1}{k} \right)\times
\exp\left(\frac{x_k}{k} \right)$$
(2). $$\lim_{x \to 0} \frac{\exp(x)-1}{x}=1$$
To prove $(1)$, I wanted to rewrite
$$\exp\left(\frac{1}{k} \right)\times \frac{1}{1+\frac{1}{k}}$$ and expand $$\frac{1}{1+\frac{1}{k}} = \frac{1}{1-\frac{-1}{k}}$$ as geometric sequence to find $x_k$, but I don’t get anything useful.
For $(2)$ I thought may one should “stretch”
$$\sum_{k=1}^n \left(\exp\left(\frac{1}{k+n}\right)-1 \right)$$
which has the same limit (I proved that) as $\sum_{k=1}^n \frac{1}{k+n}$ and rewrite it as
$$\sum_{k=1}^n \frac{\exp\left(\frac{1}{k+n}\right)-1}{\frac{1}{k+n}}\frac{1}{k+n}$$
I’m thankful for any tips or answers!
| Using the inequality$$\frac{1}{(n+1)}<\ln(\frac{n+1}{n})<\frac{1}{n}$$( Since $\frac{1}{(n+1)}*1<\ \int_{n}^{n+1}\frac{1}{x} \,dx <\frac{1}{n}*1$ (think area graphically) ),we can write $\ln(\frac{n+1}{n})<\frac{1}{n}$$<$$\frac{1}{n-1}<ln(\frac{n-1}{n-2})$. Then sum from $n$ to $2n$ and take $\lim_{n\to\infty}$.
$ \sum_{i=0}^{n}ln(\frac{n+i+1}{n+i})$$<$ $\sum_{i=0}^{n} \frac{1}{n+i} $$<$$ \sum_{i=0}^{n} ln(\frac{n+i-1}{n+i-2})$
=$\ln(\frac{n+1}{n}*\frac{n+2}{n+1}*...*\frac{2n+1}{2n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{n-1}{n-2}*\frac{n}{n-1}*...*\frac{2n-1}{2n-2})$
= $\ln(\frac{2n+1}{n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{2n-1}{n-2})$
$\lim_{n\to\infty}$$\ln(\frac{2n+1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2n-1}{n-2})$
$\lim_{n\to\infty}$$\ln(2+\frac{1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2-\frac{1}{n}}{1-\frac{2}{n}})$
You will see by the sandwich theorem the value comes to be $\ln2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$ Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1)+C,
$$
so
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx=\frac{\pi}{4}.
$$
Is there an alternative way to calculate this integral?
| A trick: use $\int_0^1f(x)dx=\int_0^1f(1-x)dx$ to average two versions of your integral, reducing it to$$\tfrac12\left(\int_0^1\tfrac{xdx}{x^2+(1-x)^2}+\int_0^1\tfrac{(1-x)dx}{x^2+(1-x)^2}\right)=\tfrac12\int_0^1\tfrac{dx}{x^2+(1-x)^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.
Is my reasoning fully acceptable throughout my solution?
| If $\arg(f(z))=\pi /2$, that means that $\Re(f(z))=0$ and as well $0 < \Im (f(z))$: that's a more "sure" way to go.
The steps in detail are
$$
\eqalign{
& \arg \left( {{{z - 8} \over {z - 2}}} \right) = {\pi \over 2}\quad \Rightarrow \quad
\left\{ \matrix{
{\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) = 0 \hfill \cr
0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
z = x + i\,y \hfill \cr
{{z - 8} \over {z - 2}}\quad \left| {\;z \ne 2} \right.\quad = \hfill \cr
= {{x - 8 + i\,y} \over {x - 2 + i\,y}}
= {{\left( {x - 8 + i\,y} \right)\left( {x - 2 - i\,y} \right)}
\over {\left( {x - 2} \right)^{\,2} + \,y^{\,2} }} = \hfill \cr
= {{\left( {\left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} } \right) + i\,y\left( {x - 2 - x + 8} \right)}
\over {\left( {x - 2} \right)^{\,2} + \,y^{\,2} }}\quad \left| \matrix{
\;x \ne 2 \hfill \cr
\;y \ne 0 \hfill \cr} \right. \hfill \cr
0 = {\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
= \left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} \quad \left| \matrix{
\;x \ne 2 \hfill \cr
\;y \ne 0 \hfill \cr} \right. \hfill \cr
0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
= 6y\quad \left| \matrix{
\;x \ne 2 \hfill \cr
\;y \ne 0 \hfill \cr} \right. \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
z = x + i\,y \hfill \cr
0 = \left( {x - 5 - 3} \right)\left( {x - 5 + 3} \right) + \,y^{\,2} = \hfill \cr
\left( {x - 5} \right)^{\,2} + \,y^{\,2} - 9\quad \left| {\;0 < y} \right. \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
z = x + i\,y \hfill \cr
(x,y) \in circle(center\,(5,0),\,radius\,3)\; \wedge \;0 < y \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
\left| {z - 5} \right| = 3 \hfill \cr
0 < {\mathop{\Im}\nolimits} z = {\mathop{\Im}\nolimits} \left( {z - 5} \right) \hfill \cr} \right. \cr}
$$
The circle is the same as you found, but (specially in the complex field) you must be careful in justifying every step that you do
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can't find the inverse of Laplace Transform. Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}
I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.
Anyone can give me hint to give me hint for this problem?
| $$F(s)=\dfrac{2}{s(s^2+4)}$$
$$F(s)=\dfrac{1}s\dfrac 2{(s^2+4)}= \mathcal {L}(1) \mathcal{L} ( \sin (2t)$$
You can also use the theorem of convolution :
$$f(t)=1* \sin (2t)=\int_0^t 1 \times \sin(2\tau) d\tau$$
$$f(t)= \dfrac {-\cos (2 \tau)}{2} \bigg |_0^t$$
$$f(t)= \dfrac 12 -\dfrac {\cos (2 t)}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If A is an $m\times n$ matrix, show that $||A||_2 \le \sqrt{||A||_1*||A||_\infty}$
If $A$ is an $m\times n$ matrix, show that $$\| A \|_2 \le \sqrt{\|A\|_1 \, \| A \|_\infty}$$
I reduced this to:
$$\rho(A^TA) \le||A||1*||A||_\infty$$
I created a matrix for experimenting:
$$A = \begin{bmatrix}a&b\\c&d\end{bmatrix} \\
A^T = \begin{bmatrix}a&c\\b&d\end{bmatrix} \\
A^tA = \begin{bmatrix}a^2+b^2&ac+bd\\ac+bc&c^2+d^2\end{bmatrix}$$
I tried to calculate the eigenvalues but that got me nowhere.
$$-\lambda^2-\lambda(a^2+b^2+c^2+d^2)+(ad-bc)^2 = 0 \Rightarrow(ad-bc)^2 = \lambda(a^2+b^2+c^2+d^2)+\lambda^2$$
As for $||A||_1$ it can be either $a+c$ or $b+d$, and $||A|||_\infty$ can be either $a+b$ or $c+d$. This gives me 4 possible combinations:
$$c^2+cd+a(c+d) \\
a^2+a(b+c)+bc \\
b^2+a(b+d)+db \\
d^2+d(b+c)+bc$$
I suppose I could do:
$$\lambda = \frac{(a^2+b^2+c^2+d^2)\pm \sqrt{(a^2+b^2+c^2+d^2)^2+4*(ad-bc)^2}}{-2}$$
Since that root will always be bigger than $(a^2+b^2+c^2+d^2)$, $\lambda$ will be either negative (because of the -2), in which case it's "proven" because the right hand side of the initial equation is always positive, or... it will be so big that the denominator will approach zero from the negative side, when the sign is negative, because you'll have $(a^2+b^2+c^2+d^2)-((a^2+b^2+c^2+d^2)+ something)$. In that case, the result will be a very small negative value, divided by -2, which is a very small positive value.
Going from there it's "intuitive" that the equation is correct but not rigorous, I suppose.
Help?
| The conventional proof is
$$
\|A\|_2^2=\rho(A^\ast A)\le\|A^\ast A\|_\infty\le\|A^\ast\|_\infty\|A\|_\infty=\|A\|_1\|A\|_\infty,
$$
where the first inequality is due to the fact that the spectral radius is the infimum of all submulticative norms, and the second inequality is due to the submultiplicativity of the induced maximum norm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Write $x$ in terms of $y$ in inequality ${x}^{2}+x\le y$ How can I write the inequality ${x}^{2}+x\le y$ such that $?\le x\le ?$ (in terms of $y$) where $-2\le y\le 1$ ?
I tried $x(x+1)\le y\Rightarrow x\le \frac{y}{x+1}$ then became stuck.
Any help will be appreciated.
| Rewrite the inequality $$x^2 + x \le y $$ as follows: $$ x^2 + x - y \le 0 $$ Then, by discriminant formula we get $$x_{1,2} = \frac{-1 \pm \sqrt{1+4y}}{2} $$ and therefore $$\left(x - \frac{-1 + \sqrt{1+4y}}{2}\right)\left(x - \frac{-1 - \sqrt{1+4y}}{2}\right) \le 0$$ Thus, $$\frac{-1 - \sqrt{1+4y}}{2} \le x \le \frac{-1 + \sqrt{1+4y}}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to read and execute $\sum_{1 \leq \ell
How to read and execute this sum?
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$$
I am having trouble to understand where is my error.
The question does not say, but I am assuming that $\ell$ starts at $1$, $m$ at $2$, and so $n$ at $3$.
This is essential a product of pg:
$$\sum_{1 \leq \ell<m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}\frac{1}{5^{\ell}}\sum_{m=2}\frac{1}{3^{m}}\sum_{n=3}\frac{1}{2^{n}} = \frac{1/5}{1-1/5}\frac{1/9}{1-1/3}\frac{1/8}{1-1/2}$$
But this does not agree with the answer :/
| The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq \ell <m<n}}\frac{1}{5^{\ell}3^m2^n}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq \ell <m<n
\end{align*}
which has $1$ as lower limit and $n$ as upper limit. We have two indices $\ell$ and $m$, which means we can write it as double sum as shown in the evaluation below.
We obtain
\begin{align*}
\color{blue}{\sum_{1\leq \ell <m<n}\frac{1}{5^{\ell}3^m2^n}}
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^{\ell}}\sum_{m=\ell+1}^{n-1}\frac{1}{3^m}\tag{2}\\
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^{\ell}}\left(\frac{\left(\frac{1}{3}\right)^{l+1}-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\right)\tag{3}\\
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^l}\,\frac{1}{2}\left(\frac{1}{3^l}-\frac{1}{3^{n-1}}\right)\tag{4}\\
&= \frac{1}{2^{n+1}}\sum_{l=1}^{n-2}\frac{1}{15^l}- \frac{1}{2^{n+1}\,3^{n-1}}\sum_{l=1}^{n-2}\frac{1}{5^l}\\
&=\frac{1}{2^{n+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{n-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{n+1}\,3^{n-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{n-1}}{1-\frac{1}{5}}\right)\tag{5}\\
&=\frac{1}{2^{n+2}\cdot7}\left(1-\frac{1}{15^{n-2}}\right)-\frac{1}{2^{n+3 }\,3^{n-1}}\left(1-\frac{1}{5^{n-2}}\right)\\
&\,\,\color{blue}{=\frac{1}{2^{n+2}\cdot7}-\frac{1}{2^{n+3}\,3^{n-1}}+\frac{1}{2^{n+3}\,3^{n-1}\,5^{n-2}\cdot7}}
\end{align*}
Comment:
*
*In (2) we factor out $\frac{1}{2^n}$ and reorder the double sum using another common style.
*In (3) we evaluate the inner sum using the finite geometric summation formula.
*In (4) we do a simplification and multiply out in the next line.
*In (5) we apply the finite geometric summation formula twice and do a simplification in the following lines.
Note: A varying upper limit $n$ is not admissible in your case. Here $n$ is a free variable whereas the indices $\ell$ and $m$ are bound variables. This is different to the situation
\begin{align*}
\sum_{1\leq \ell <m<n\color{blue}{<\infty}}\frac{1}{5^{\ell}3^m2^n}
\end{align*}
where $n$ is bound by the upper limit $\infty$ and where $n$ varies between $m$ and $\infty$.
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
| {
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"url": "https://math.stackexchange.com/questions/3955907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Proof by induction, $2^{n} > n^{2} - 2$ So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.
I start of with the "basecase" for n = 3:
RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$
Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.
$2^{k} > k^{2} - 2$
We multiply both sides with 2:
$2^{k+1} > 2(k^{2} - 2)$
$2^{k+1} > 2k^{2} - 4$
$2^{k+1} > k^{2} + k^{2} - 2 - 2$
According to our assumption, $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $
Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:
$2^{k+1} > k^{2} + 2k + 1 - 2$
Thus:
$2^{k+1} > (k+1)^{2} - 2$
QED.
How does my proof look, does it have any deficits, and how would you prove it?
Thank you!
| I think this is a little clearer.
$2^{k+1}\ge2k^2-4$
$=k^2+2k+1-2+k^2-2k-3$
$=(k+1)^2-2+(k-3)(k+1)$
$\ge(k+1)^2-2$ for $k\ge3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956531",
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"source": "stackexchange",
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If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is Below are the steps I have done so far.
Find $a$ in terms of $c$
$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$
Find $b$ in terms of $c$
$$
ab = 432 \\
\implies 4.5c \cdot b = 432 \\
\implies b = \frac{96}c.
$$
Substituting, we get
$$
a + b + c = 4.5c + \frac{96}c + c = 5.5c + \frac{96}c.
$$
How should I proceed from here?
| $bc=96$ and $c<9$ constrains $c$ to be one of $1,2,3,4,6,8$. Since $\gcd(432,96)=48$, $c$ cannot be $1$ or $3$, leaving four cases: $(a,b,c)=(9,48,2),(18,24,4),(27,16,6),(36,12,8)$. Of these, $(18,24,4)$ achieves the smallest $a+b+c$ of $46$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{m+n}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
m +\frac{1}{2} -2 < \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \leq m +\frac{1}{2}.
$$
Which implies
$$
-1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2} \Bigr\rfloor -m \leq .5
$$
This results in
$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$
How to decide that the result is $0$?
Any help is welcome.
| If $m+n$ is even, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n}{2}$ and as $m-n=m+n -2m$ is also even, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n-1}{2}$.
So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$
If $m+n$ is odd, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n-1}{2}$ and as $m-n=m+n -2m$ is also odd, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n}{2}$.
So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$
| {
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Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $
I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.
I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.
| Since $\frac{x}{10}, \frac{y}{10}, \frac{z}{10}\in[0; 1]$, we have that $\left(\frac{x}{10}\right)^2\leqslant\frac{x}{10}, \left(\frac{y}{10}\right)^2\leqslant\frac{y}{10}, \left(\frac{z}{10}\right)^2\leqslant\frac{z}{10}$. Thus \begin{align*}2x^2+y^2+z^2&=100\cdot \left(2\cdot\left(\frac{x}{10}\right)^2+\left(\frac{y}{10}\right)^2+\left(\frac{z}{10}\right)^2\right)\\&\leqslant 100\cdot \left(2\cdot \frac{x}{10}+\frac{y}{10} +\frac{z}{10}\right)\\&=100\cdot \left(\frac{x}{10}+1\right)\\&\leqslant 100\cdot (1+1)=200\end{align*} With equality if and only if $x=10, y=0, z=0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x-3)(x+2)\\(x^2+2x-3)=(x-1)(x+3)\\
(x+2)\\
f(x)=(x^3+4x^2+x-6)g(x)+(Ax+B)\\
f(1)=A+B=x+2\\
f(-3)=-3A+B=x+2\\
f(-2)=-2A+B=1\\
\left\{\begin{matrix}
A+B=x+2& x=-1\\
-3A+B=x+2& B=1\\
-2A+B=1& A=0
\end{matrix}\right.\\
r=(-1)(0)+1=1$$
Can the remainder really be $r=1$? My intuition says it's wrong.
EDIT:
$$f(x)=(x^3+4x^2+x-6)g(x)+(Ax^2+Bx+C)\\
f(1)=A+B+C=x+2\\
f(-3)=9A-3B+C=x+2\\
f(-2)=4A-2B+C=1\\
\left\{\begin{matrix}
A+B+C=x+2& A=\frac{x+3}{6}\\
9A-3B+C=x+2& B=\frac{3x+5}{6}\\
4A-2B+C=1 &C=\frac{4-2x}{6}
\end{matrix}\right.\\
r=\frac{x+3}{6}x^2+ \frac{3x+5}{6}x+\frac{4-2x}{6}=\frac{x^3+6x^2+3x+9}{6}$$
| Since the divisor polynomial is of degree $3$, the remainder will be of atmost degree $2$. Hence, instead of $Ax+B$ that you have taken, take $Ax^2+Bx+C$ as the remainder and you also have $3$ equations to plug the values.
Note that $f(1)=A+B+C=3$, $f(-3)=9A-3B+C=-1$ and $f(-2)=4A-2B+C=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| After you substitute $(x,y)=(\cos{\theta},\sin{\theta})$, you just need to prove the followings:
$$
\begin{aligned}
\sin{\theta}+\cos{\theta}&\geq\sqrt{2}\\
\\
1-\sin{\theta}\cos{\theta}&\geq\sin{\theta}\cos{\theta}
\end{aligned}
$$
Which are easy. First one is well known, second one is just double angle sine identity
| {
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"source": "stackexchange",
"question_score": "2",
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Is there a reason why this technique is invalid? What is $ \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$? A simple way to evaluate this limit is to substitute $0$ for $x$ in the numerator to obtain
$ \displaystyle \lim_{x \rightarrow 0} \frac{1 - 1}{x} = \lim_{x \rightarrow 0} ( \frac{1}{x} - \frac{1}{x} )
= \lim_{x \rightarrow 0} (0) = 0 $
since $ \frac{1}{x} - \frac{1}{x} = 0$ since one quantity subtracted from the same quantity is 0. This technique circumvents the problem of division by zero while utilizing the fact that $\cos(0)$ is known.
| A counterexample:
$$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac12,\quad\enspace\text{not }0.$$
Indeed $\;1-\cos x=2\sin^2\tfrac x2$, so
$$\frac{1-\cos x} {x^2}= \frac{2\sin^2\frac x2}{4\bigl(\frac x2\bigr)^2}=\frac12\biggl(\underbrace{\frac{\sin\frac x2}{\frac x2}}_{\underset{\textstyle 1}{\downarrow}}\biggr)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Convergence/Divergence proof for Infinite products Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.
A close relative of the infinite series is the infinite product.
\begin{align*}
\prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdots
\end{align*}
which is understood in terms of its sequence of partial products.
\begin{align*}
p_m = \prod_{n=1}^{m} b_n = b_1 b_2 b_3 \cdots b_m
\end{align*}
Consider the special class of infinite products of the form
\begin{align*}
\prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots
\end{align*}
where $a_n > 0$.
(a) Find an explicit formula for the sequence of the partial products in the case where $a_n = 1/n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_n = 1/n^2$ and make a conjecture about the convergence of this sequence.
(b) Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty}a_n$ converges. (The inequality $1+x \le 3^x$ for positive $x$ will be useful in one direction.)
Proof.
(a) If $a_n = 1/n$, the partial product
\begin{align*}
p_m &:= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{m}\right)\\
&= 1 + \sum_{m} \frac{1}{m} + \sum_{m < n} \frac{1}{m \cdot n} + \sum_{m < n < p} \frac{1}{m \cdot n \cdot p} + \ldots
\end{align*}
Since $\sum 1/m$ is the harmonic series, which is well-known to be divergent, this sequence of partial products is divergent.
Consider the case of $a_n = 1/n^2$. We are going to use the inequality $1 + x^2 \le e^{x^2}$. We have:
\begin{align*}
p_m &= \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{m^2}\right)\\
&\le e^{\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{m^2}}\\
&\le e^{\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{m\cdot (m+1)}}\\
&\le e^{\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{m} - \frac{1}{m+1}\right)}\\
&= e^{1 - \frac{1}{m+1}}\\
&\le e
\end{align*}
Moreover, $p_1 = 2$, $p_2 = (2)(5/4)$, $p_3 = (2)(5/4)(10/9)$, $p_4 = (2)(5/4)(10/9)(17/16)$. So, $(p_m)$ is monotonic increasing and bounded by $e$. Consequently, by the Montone Convergence Theorem, $(p_m)$ is convergent.
(b) ($\longrightarrow$) direction.
The sequence of partial products,
\begin{align*}
p_n &= (1+a_1)(1+a_2)\cdots(1+a_n)\\
&\le e^{a_1}\cdot e^{a_2}\cdots e^{a_n}\\
&\le \exp \left({\sum_{k=1}^{n}a_k}\right)
\end{align*}
If the sequence of partial products is convergent and therefore bounded, the sequence of partial sums $\sum_{k=1}^{n} a_k$ must be bounded and convergent.
| For (a) it says "explicit formula". Yours is not. Here it is:
$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)
\cdots\left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)
\\
=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{n}{n-1}\cdot\frac{n+1}{n}
= \frac{n+1}{1} = n+1
$$
Each numerator cancels the following denominator.
| {
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Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}
Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!
| Hint: Rewrite as follows:
$$\begin{align*}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n+1-1}{n+1}-1-\frac{1}{n} \\
&=1-\frac{1}{n+1}-1-\frac{1}{n} \\
&=-\left(\frac{1}{n+1}+\frac{1}{n}\right).
\end{align*}$$
As $n\to\infty$, what do $1/n$ and $1/(n+1)$ approach?
| {
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"timestamp": "2023-03-29T00:00:00",
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Die and coin variance of random variable question. Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.
My work:
$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} + 3 \cdot \cfrac{1}{6} + 4 \cdot \cfrac{1}{6} + 5 \cdot \cfrac{1}{6} + 6 \cdot \cfrac{1}{6} = \cfrac{7}{2}$
$E(Z) = 0 \cdot \cfrac{1}{2} + 1 \cdot \cfrac{1}{2} = \cfrac{1}{2}$
So $E(X) = E(Y + Z) = E(Y) + E(Z) = \cfrac{7}{2} + \cfrac{1}{2} = 4$
$E(Y^2) = 1^2 \cdot \cfrac{1}{6} + 2^2 \cdot \cfrac{1}{6} + 3^2 \cdot \cfrac{1}{6} + 4^2 \cdot \cfrac{1}{6} + 5^2 \cdot \cfrac{1}{6} + 6^2 \cdot \cfrac{1}{6} = \cfrac{91}{6}$
$E(Z^2) = 0^2 \cdot \cfrac{1}{2} + 1^2 \cdot \cfrac{1}{2} = \cfrac{1}{2}$
so $E(X^2) = E(Y^2 + Z^2) = E(Y^2) + E(Z^2) = \cfrac{91}{6} + \cfrac{1}{2} = \cfrac{47}{3}$
$Var(X) = E(X^2) - (E(X))^2 = \cfrac{47}{3} - 4^2 = -\cfrac{1}{3}???$
Where did I go wrong, variance can't be negative so clearly my work is wrong, but I have no idea where I went wrong? Can someone point me in the right direction of how to do this question?
| $Y\perp\!\!\!\!\!\!\perp Z$ thus $V(X)=V(Y)+V(Z)$
Your error is this
$$E(X^2)=E[(Y+Z)^2] \ne E(Y^2)+E(Z^2)$$
| {
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Please explain how to derive the PGF of a 6 sided dice I understand the PGF is
$$ P(s) = \frac{s}{6} \frac{1-s^6}{1-s} $$
Can someone explain all the steps to get there
| We want
$$f(x)=x+x^2+x^3+\cdots +x^6.$$
By long division we can prove that
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$
If we multiply the above expression by $1-x^7$ we are going to see a pattern where all the $x^n$ exponents will cancel out after $x^6$:
$$\begin{align}&(1-\color{red}{x^7})+(x-\color{blue}{x^8}) +(x^2-x^9)+\cdots+(\color{red}{x^7}-x^{14})+(\color{blue}{x^8}-x^{15})+\cdots\\[2ex]&=1+x+x^2+\cdots+x^6
\end{align}$$
Now we have to get rid of the $1$ in front by multiplying the series by $x.$ That will require reducing the exponent in the numerator exponent from $1-x^7$ to $1-x^6.$
This leaves us with $x\frac{1-x^6}{1-x}.$ And since the probability of each outcome is $1/6,$ the expression is derived:
$$f(x)=\frac 1 6 \frac{x(1-x^6)}{1-x}$$
| {
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Find a closed form for the recurrence relation: $a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1$ Find a closed form for the following recurrence relation:
$$a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1\tag{$n \ge 0$}$$
This is what I've done so far:
Define a function $f(x):=\sum_{n\ge0}^{ }a_{n}x^{n}$,then:
$$\sum_{n\ge0}^{ }a_{n+2}x^{n+2}-x^{2}\sum_{n\ge0}^{ }a_{n}x^{n}=x^{2}\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}$$
The RHS is $\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}=\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{2n+1}$
$$f\left(x\right)-a_{0}-a_{1}x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$
Considering the initial values the equality transforms to:
$$f\left(x\right)-1-x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$
\begin{align}
a_n=[x^n]f\left(x\right)
&=[x^n]\frac{2x^{3}+x^{2}+x+1}{\left(1+x^{2}\right)\left(1-x^{2}\right)}\\
&=\left[x^{n}\right]\left(\frac{5}{4}\frac{1}{1-x}-\frac{1}{4}\frac{1}{1+x}-\frac{1}{2}\frac{x}{1+x^{2}}\right)\\
&=\left[x^{n}\right]\left(\frac{5}{4}\sum_{n\ge0}^{ }x^{n}-\frac{1}{4}\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{n}-\frac{1}{2}x\sum_{n\ge0}^{ }\underbrace{\binom{-1}{n}}_\text{=$(-1)^n$}x^{2n}\right)\\
&=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}\end{align}
But the problem is that the answer is $$a_n=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}-\frac{1}{2}\sin\frac{n\pi}{2}$$
And I don't understand why.
| Since a generating function answer has already been provided, I'll instead provide the direct proof. Note that
$$a_{n+2}=a_n+\sin(n\pi /2)=a_{n-2}+\sin((n-2)\pi/2)+\sin(n\pi/2).$$
But $\sin(n\pi/2-\pi)=-\sin (n\pi/2)$, so $a_{n+2}=a_{n-2}$. Hence the sequence is 4-periodic, and it suffices to know the first four entries. We have $a_0=a_1=1$ by assumption, and therefore also $$a_2=a_0+\sin(0\pi/2)=a_0=1,$$ $$a_{3}=a_1+\sin(\pi/2)=a_1+1=2.$$
We conclude that $a_{n}=2$ if $n\bmod 4=3$ and $a_n=1$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| Integration by parts
$$\int x \sqrt{1-x^2} \arcsin x\, dx$$
take
$$u'(x)=x \sqrt{1-x^2};\;v(x)=\arcsin x$$
so
$$u(x)=\int x \sqrt{1-x^2} \, dx=-\frac{1}{2}\int (-2x)(1-x^2)^{-1/2}\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}+C$$
and
$$v'(x)=\frac{1}{\sqrt{1-x^2}}$$
apply IBP formula
$$\int u'(x)v(x)\,dx=u(x)v(x)-\int u(x)v'(x)\,dx$$
$$\int x \sqrt{1-x^2} \arcsin x\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x-\int\left(-\frac{1}{3} \left(1-x^2\right)^{3/2}\right)\frac{1}{\sqrt{1-x^2}}\,dx=$$
$$=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x+\int \frac{1}{3} \left(1-x^2\right)\,dx=$$
$$=-\frac{1}{3} \left(1-x^2\right)^{3/2} \arcsin x+\frac{1}{3} \left(x-\frac{x^3}{3}\right)+C$$
which can be simplified to
$$=\frac{1}{9} \left(3x-x^3-3 \left(1-x^2\right)^{3/2} \arcsin x\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$ The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.
The easy way to solve this is by taking the last digit of each number:
*
*$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.
*$4^{2n+1}$ ends in $4$ as $2n + 1$ is odd.
So the sum will end in $4+1=5$ and because $81^{3^{n}}+4^{2n+1}>5$, it is $5|81^{3^{n}}+4^{2n+1}$, thus the number is not prime.
Are/is there any other proof(s) which do not reuse the same idea?
| \begin{align}
81^{3n} + 4^{2n+1} &= (3^4)^{3n} + (2^2)^{2n+1} \\
&= (3^{6n})^2 + (2^{2n+1})^2 \\
&= (3^{6n})^2 + (2^{2n+1})^2 + 2 \times 3^{6n} 2^{2n+1} - 2 \times 3^{6n} 2^{2n+1} \\
&= (3^{6n} + 2^{2n+1})^2 - 3^{6n} 2^{2n+2} \\
&= (3^{6n} + 2^{2n+1})^2 - (3^{3n} 2^{n+1})^2 \\
&= (3^{6n} + 2^{2n+1} - 3^{3n} 2^{n+1}) (3^{6n} + 2^{2n+1} + 3^{3n} 2^{n+1})
\end{align}
As you see $81^{3n} + 4^{2n+1}$ can be written as a product of two integers and none of them are $1$ for positive integers1, so it is not prime.
1. Because both of them are increasing functions and above $1$ for $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding Pythagorean triple where the squares of sides are weighted Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:
$$a^2 + 3 b^2 = 4 c^2$$
Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how to approach this problem.
| After discovering Find all integers satisfying $m^2=n_1^2+n_1n_2+n_2^2$, the paper Abdelalim, S., & Dyani, H. (2014). The Solution of the Diophantine Equation x2+ 3y2= z2. International Journal of Algebra, 8(15), 729–732. gives an easy answer:
Theorem 2.1 Let $E : x^2 + 3 y^2 = z^2$ diophantine equation and $(x, y, z) \in \mathbb{Z}^3$ with $x \wedge y = 1$, $y$ is even and $x z \wedge 3 = 1$. Then the following properties are equivalent:
*
*$(x, y, z)$ is the solution of $E$.
*$| z | = 3 y_1^2 + y_2^2, | x | = 3 y_1^2 - y_2^2, | y | = 2 y_1 y_2$ with $y_1 \wedge y_2 = 1$.
Theorem 2.2 Let $E : x^2 + 3 y^2 = z^2$ diophantine equation and $(x, y, z) \in \mathbb{Z}^3$ with $x \wedge y = 1$, $y$ is odd and $x z \wedge 3 = 1$. Then the following properties are equivalent:
*
*$(x, y, z)$ is the solution of $E$.
*$| z | = \frac{3 y_1^2 + y_2^2}{2}, | x | = \frac{3 y_1^2 - y_2^2}{2},
| y | = y_1 y_2$ with $y_1 \wedge y_2 = 1$.
Scaling the above primitive solutions $(x, y, z)$, and in our case restricting to $2 \mid z$, gives all solutions.
The following is a method that is less explicit, as it requires dividing by a $\gcd$.
Thanks to Daniel Hast's comment, I arrived at the following:
Finding such $(a, b)$ is equivalent to finding a single rational point on the ellipse $x^2 + 3 y^2 = 4$, then finding all rational lines through it. A rational point is $(1, 1)$, and a rational line through that is $t y = s x - s + t$ where $\gcd (s, t) = 1$. The other intersection with the ellipse is at:
$$ (x, y) = \left( \frac{3 s^2 - 6 s t - t^2}{3 s^2 + t^2}, - \frac{3 s^2 + 2 s t - t^2}{3 s^2 + t^2} \right) $$
So:
$$ \begin{array}{rll}
\left( \frac{3 s^2 - 6 s t - t^2}{3 s^2 + t^2} \right)^2 + 3 \left(
\frac{3 s^2 + 2 s t - t^2}{3 s^2 + t^2} \right)^2 & = & 4\\
(3 s^2 - 6 s t - t^2)^2 + 3 (3 s^2 + 2 s t - t^2)^2 & = & 4 (3 s^2 +
t^2)^2\\
4 (3 s^2 + t^2)^2 - 3 (3 s^2 + 2 s t - t^2)^2 & = & (3 s^2 - 6 s t -
t^2)^2
\end{array} $$
Note that the numerator and denominators of $x, y$ can be scaled while the
rational point stays the same. Let $d
= \gcd (3 s^2 + t^2, 3 s^2 + 2 s t - t^2)$. For $k \in \mathbb{Z}$, our
solutions are then:
$$ a = \frac{k}{d} \cdot (3 s^2 + 2 s t - t^2) \quad b = \frac{k}{d} \cdot (3 s^2 + t^2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$ Find all $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$.
I got this interesting question from a friend. I made an assumption and was able to get $n$ as $3$ and $7$ but am not able to get more solutions.
Any help to solve this question would be appreciated. Thanks in advance.
| Let
$$u(n)={n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}=\frac{(n+1)(n^2-n+6)}6$$
First, we must have $n\ge0$, otherwise the left hand side is zero.
If $u(n)$ is a factor of $2^{2018}$, that means $u(n)$ must be a power of two. Since $3$ must divide either $n+1$ or $n^2-n+6$, this means that either:
$$\begin{eqnarray}
n+1&=&3\cdot 2^p\\
n^2-n+6&=&2^q
\end{eqnarray}
$$
or
$$\begin{eqnarray}
n+1&=&2^p\\
n^2-n+6&=&3\cdot2^q
\end{eqnarray}
$$
In the first case, since $n^2-n+6>\dfrac{n+1}{3}$, we must have $2^p|2^q$, but
$$n^2-n+6=(n-2)(n+1)+8=3(n-2)\frac{n+1}{3}+8$$
Hence $\frac{n+1}{3}$ divides $8$. The possible values of $n$ are then $\{2,5,11,23\}$, and it's easy to check that only $2$ and $23$ are valid solutions.
In the second case, we have either $2^p|2^q$ or $2^q|2^p$ (or both).
If $2^p|2^q$, then $n^2-n+6=(n-2)(n+1)+8$, hence $n+1$ divides $8$, and the possible values of $n$ are $\{0,1,3,7\}$, which are all valid solutions.
If $2^q|2^p$, then $\frac{3(n+1)}{n^2-n+6}$ is an integer, but it's $\ge1$ for only finitely many values of $n$, namely for $n\le3$, but we already have all values of $n\le3$ as valid solutions (but it's easy to check that $\frac{3(n+1)}{n^2-n+6}$ is an integer only for $n\in\{1,3\}$).
All in all, the set of solutions is
$$n\in\{0,1,2,3,7,23\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$n-$section formulas in goniometry without the calculus We know that the formulas of bisection of an angle $\alpha$ it is:
$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$
In particular
$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}, \qquad \cos\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$$
My question is:
For the formulae of $n-$section of an angle $\alpha$, i.e.
$$\sin\left(\frac{\alpha}n\right), \qquad \cos\left(\frac{\alpha}n\right)$$
Do I must necessarily use the development in series of Taylor or the general formulas can be obtained with methods that do not use the derivatives, or the numerical methods?
My doubts arise for the students of an high school that not use the derivate.
| I doubt that there are such elementary general formulas. I will derive a possible (restricted) formula as an infinite series using De Moivre's formula, though this may not satisfy your requirement of calculus free derivation. One value of the multivalued expression $\cos\left(\dfrac{\theta}{n}\right)+i\sin\left(\dfrac{\theta}{n}\right),\quad n\ge2$ is given by
\begin{align}
\left(e^{i\theta}\right)^{1/n} &
= \left(\cos\theta+i\sin\theta\right)^{1/n} \\
& = \left(\cos\theta\right)^{1/n}\left(1+i\tan\theta\right)^{1/n}
\end{align}
Now, assuming $0\lt\theta\lt\pi/4,$ we can expand $\left(1+i\tan\theta\right)^{1/n}$ via binomial series as $$1+\dfrac{1}{n}i\tan\theta +\dfrac{(n-1)}{2!n^2}\tan^2\theta -\dfrac{(n-1)(2n-1)}{3!n^3}i\tan^3\theta -\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$
Hence we have two infinite series $$\dfrac{\sin\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=\dfrac{1}{n}\tan\theta - \dfrac{(n-1)(2n-1)}{3!n^3}\tan^3\theta+ \dfrac{(n-1)(2n-1)(3n-1)(4n-1)}{5!n^5}\tan^5\theta+$$ and
$$\dfrac{\cos\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=1+\dfrac{(n-1)}{2!n^2}\tan^2\theta-\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about generating functions about an exercise I have a question about generating functions,
I've understood that generating functions are also representing some series of numbers, but is this series could be formed by 2 different functions?
For example, an exercise that I've encountered goes like this:
Find the generating function that represents the number of solutions to the equation:
$$ x_1 + x_2 + \dots + x_k=n$$
Where the variables are even numbers that are not divisible by 3,
So I assume we should find a function that represents the series:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...
0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 ..
The zero's and one's stands for the coefficient of $x^n$
So to get to that series we have few options the first one i've thought about is to take the generating function: $1 + x + x^2 + x^3 + \dots$
$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x} $$
Which represents the series of 1 1 1 1 1 ...
and then to subtract those (subtract odd numbers and multiples of 6):
$$ f(x) = ((1 + x + x^2 + x^3 + ...) - x(1 + x^2 + x^4 + ...) - (1 + x^6 + x^{12} +...))^k $$
Which gives us the series we wanted.
But the solution to this exercise shows different answer and the generating function goes like this:
$$ g(x) = ((x^2 + x^4) + (x^8 + x^{10}) + (x^{14} + x^{16}) + ... )^k $$
which represents the same series as well but looks differently, does those 2 functions coefficients represent the same number of solutions to the equation?
Thanks alot !
| Your approach is fine. We can show the generating functions are equal: $f=g$.
We have
\begin{align*}
\color{blue}{g(x)}&=\left(\left(x^2 + x^4\right) + \left(x^8 + x^{10}\right) + \left(x^{14} + x^{16}\right) + ... \right)^k \\
&=\left(x^2\left(1+x^2\right)+x^8\left(1+x^2\right)+x^{14}\left(1+x^2\right)+\cdots\right)^k\\
&=\left(1+x^2\right)^k\left(x^2+x^8+x^{14}+\cdots\right)^k\\
&=\left(1+x^2\right)^kx^{2k}\left(1+x^6+x^{12}+\cdots\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}
On the other hand we obtain
\begin{align*}
\color{blue}{f(x)}&= \left(\left(1 + x + x^2 + x^3 +\cdots\right) - x\left(1 + x^2 + x^4 + \cdots\right)
- \left(1 + x^6 + x^{12} +\cdots\right)\right)^k\\
&=\left(\frac{1}{1-x}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1+x}{1-x^2}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{\left(1-x^6\right)-\left(1-x^2\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&=\left(\frac{x^2\left(1-x^4\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}
and the claim follows.
| {
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"source": "stackexchange",
"question_score": "1",
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Find $ \lim_{n \to \infty}n \log\left(1+ \left(\frac{f(x)}{n}\right)^p\right)$ Find $$\lim_{n \to \infty}n \log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)$$ where $0<p<1$
my attempt :By using L-Hospital Rule i got
$$\lim_{n\to \infty} \dfrac {\dfrac{1}{1+\left( \dfrac{f(x)}{n}\right)^p}
\cdot\dfrac {f(x)^p}{-n^{2p}} }{\dfrac{-1}{n^2}}$$
After that I’m not able to proceed further
| Let $y = f(x)$. Consider the essential case $y \neq 0$.
\begin{align*}
n\log\left(1 + \left(\frac{y}{n}\right)^p\right) &
= y^pn^{1-p}\log \left(\left( 1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right).
\end{align*}
Since $\left\{\left(\frac{n}{y}\right)^p \right\}_{n \in \mathbb{N}}$ is monotonic,
\begin{align*}
\log \left(\left( 1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right) \to 1.
\end{align*}
But
\begin{align*}
y^pn^{1-p} \to +\infty \ or \ -\infty.
\end{align*}
Thus the limit is $+\infty$ or $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| Denote $y = \tan x \cot 2x = \frac{1-\tan^2x}2$ and express $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$ as a system of equations in $x,y$
$$\tan x+\frac{2\tan x}{1-\tan^2x}=\left(1+\frac1y \right)\tan x=\frac{2}{\sqrt{3}}
$$
Then, eliminate $\tan x$ to get
$$\frac3{y^3} -\frac{13}{y}-6=0$$
which is a depressed cubic equation in $\frac1y$, yielding
$$\tan x \cot 2x=y= \left( \frac{2\sqrt{13}}3 \cos \left( \frac13\cos^{-1} \frac{27}{13\sqrt{13}}\right) \right)^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?
| Let $$x = \sum_{m=2}^{n}2^{m-2},$$ and note that $$1 + x = 1 + 1 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{1}.$$ Then, multiplying $(1)$ by $2$, we have $$2+2x = 2 + 2 + 4 + \dotsb + 2^{n-2} + 2^{n-1}\tag{2}.$$ We can rewrite $(1)$ as $$1 + x = 2 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{3}.$$ Subtracting $(3)$ from $(2)$ gives us $$1 + x = 2^{n-1},$$ which is the same as $$2^{n-1} = 1 + 1 + 2 + 4 + \dotsb + 2^{n-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Simplifying a trig relation My physics teacher wanted to prove Young's equation (my problem is pure mathematics), and in the derivation, he got the following expression:
$$\tag{1}
\begin{align}
\left(\gamma_{s l}-\gamma_{s o}\right) \cdot\left(\sin ^{2} \theta+\cos \theta\left(-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)\right) \\[6pt]
\quad+\;\gamma \cdot\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)=0
\end{align}$$
Which he says leads to:
$$\tag{2}
\gamma_{s l}+\gamma \cos \theta=\gamma_{s o}$$
It is easy to see from eq. (1) and (2) that we must have:
$$\tag{3}
\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)\left(\sin ^{2}(\theta)-\cos \theta \frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)^{-1}=\cos \theta
$$
But I have a hard time showing this (my trig-skills might not be up to par). Will someone help me show this?
| The coefficient of $\gamma$ can be simplified as,
$\begin{align}(1-\cos\theta)\frac{2+2\cos\theta - 2- \cos\theta}{1+\cos\theta} = \boxed{\cos\theta\ \left[\frac{1-\cos\theta}{1+\cos\theta}\right]} \end{align}$
Now, the coefficient of $\gamma_{sl} - \gamma_{so}$ can be simplified as,
$\begin{align}\sin^2\theta -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) &= (1-\cos\theta)(1+\cos\theta) -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) \\
&= (1-\cos\theta)\left[\frac{(1+\cos^2\theta +2\cos\theta) - (2\cos\theta+\cos^2\theta)}{1+\cos\theta}\right] \\& = \boxed{\frac{1-\cos\theta}{1+\cos\theta}}\end{align}$
Now the coefficients differ by a factor $\cos\theta$.
Assuming the coefficient $\frac{1-\cos\theta}{1+\cos\theta}$ is non-zero,
$\gamma_{sl} - \gamma_{so} + \gamma \cos\theta = 0$
or
$$\boxed{\gamma_{so} = \gamma_{sl} + \gamma \cos\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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Fractions in Questions and Answers
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