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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which arose from the deduction that $\frac{1}{4}\sqrt{\frac{256\sin^{4}40^{\circ}-80\sin^{2}40^{\circ}+12-\ 8\sqrt{3}\sin40^{\circ}}{\left(16\sin^{4}40^{\circ}-4\sin^{2}40^{\circ}+1\right)}}=\cos50^{\circ}$. Despite the apparent simplicity of the relationship, it seems quite tricky to prove. I managed to prove it by solving the equation as a quadratic in $(\sin\frac{\pi}{9})$ and then using the identity $\sqrt{\sec^2 x-1}=|\tan x|$, the double angle formulae and finally that $\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x$ can be written in the form $\sin\left(x+\frac{2\pi}{3}\right)$.
But it seems like quite a neat problem. So, does anyone have a better way of proving it?
| Let $s=\sin 20° \quad → \sin (3×20°) = \frac{\sqrt{3}}{2} = 3s-4s^3$
Let $c=\cos 20° \quad → \cos (3×20°) = \frac{1}{2} = -3c+4c^3$
$LHS = 4s^2 - 2\sqrt{3}\,s + 1 = 4s^2 - 4s\,(3s-4s^3) + 1 = (1-4s^2)^2$
$\displaystyle RHS = \left(\frac{1}{2c}\right)^2 = \left(-3+4c^2\right)^2 = (1-4s^2)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the rational function $f(x)/g(x)$ has a partial fraction decomposition in the case when $g(x)$ factors into distinct linear factors. To be honest i don't even know where to start. I thought about using diagonalizable matrices and characteristic polynomials, but the class hasn't gotten there yet so there should be a way to solve this with simpler stuff. Any ideas?
| We proceed by strong induction. Given that $n = 1,$ we have that $f(x) = a$ for some real number $a$ so that $$\frac{f(x)}{g(x)} = \frac a {x - c_1},$$ as desired. We will assume inductively that for any polynomial $f(x)$ of degree $\leq n - 1,$ we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_n}{x - c_n}$$ for some real numbers $a_1, \dots, a_n.$ Consider the polynomial $f(x) = b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0$ of degree $\leq n.$ We have that $$\frac{f(x)}{g(x)} = \frac{b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)(x - c_{n + 1})} = \frac 1 {x - c_{n + 1}} \biggl(\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} + \frac{b_{n - 1} x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)} \biggr).$$ Given that $b_n = 0,$ the expression in the parentheses is the ratio of a polynomial of degree $\leq n - 1$ and a polynomial of degree $n,$ hence by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(\frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr) = \frac{d_1}{(x - c_1)(x - c_{n + 1})} + \cdots \frac{d_n}{(x - c_n)(x - c_{n + 1})}$$ for some real numbers $d_1, \dots, d_n.$ Once again, by strong induction, we can have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $a_1, \dots, a_{n + 1}.$ (Each of the terms has its own partial fraction decomposition.)
Consider the case that $b_n \neq 0.$ Observe that $(x - c_1) \cdots (x - c_n)$ is a monic polynomial of degree $n$ and $b_n x^n$ is a polynomial of degree $n,$ hence by the Division Algorithm, we have that $$\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} = b_n + \frac{r(x)}{(x - c_1) \cdots (x - c_n)}$$ for some polynomial $r(x)$ of degree $\leq n - 1.$ Consequently, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(b_n + \frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr)$$ for some real numbers $d_1, \dots, d_n.$ Like before, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $a_1, \dots, a_{n + 1}.$ Our proof is complete by strong induction. QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 , y_0)$
My attempt : Here $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$
Now i put $x= e^z$ then $z= \log x$
So $y= Dy- D^2y$
$D^2y-Dy -y=0$
so $(D^2-D-1)y=0$
so auxiliary equation will be $m^2-m-1=0$ ,$m= \frac{1 +_{-}\sqrt - 3}{2}$
so $y= e^{\frac{1}{2}x} (c_1 \cos(\frac{\sqrt - 3}{2} ) + c_2\sin ({\frac{\sqrt - 3}{2}} )x)$
After that im not able to proceed further
| Note that $y=bx-b^2$ is a solution, where
$$
b=\frac{-x_0\pm\sqrt{x_0^2-4y_0}}{2}
$$
This gives us two distinct solutions whenever $x_0^2\neq 4y_0$. If instead $x_0^2=4y_0$, we can still find a second solution $y=x^2/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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how should Cayley's resolvent be used in order to see if a given quintic equation is solvable or not?
Hi. I chose a simple equation to work with: $x^5+x^4+x^3+x^2+x+1=0$ (which is solvable and $x=-1$).
the amounts of $p$, $q$, $r$ and $s$ where $\frac{3}{5}$, $\frac{14}{25}$, $\frac{87}{125}$ and $\frac{2604}{3125}$ respectively.
I calculated $P$ to be $P=z^3-15z^2-69z+1235$ and $\Delta$ to be $\Delta=1296$.
as the picture says, $P^2-1024z\Delta$ (Cayley's resolvent) should have rational root(s) in $z$ in order to the first qintic function be solvable.
but the problem is that $P^2-1024z\Delta=0$ is $z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=0$ which is even worse and impossible to solve!
what should I do now?!
| Hint:
We have the following factorization:
$$
z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=
(z^2 + 22z + 169)(z^2 - 26z + 361)(z - 1)(z - 25).
$$
Of course, the splitting field of
$$
x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}
$$
is $\Bbb Q(\zeta_6)$, so that the Galois group is solvable.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?
| Let $z=e^{ \frac{i\pi}{n}},$ for example. If we can show it for this point, it is true in general since this is and its conjugate are the closest to the point $z=1$.
$$|z-1|^2 = (\cos \frac{\pi}{n} -1)^2 + \sin^2 \frac{\pi}{n} = 2 \left( 1-\cos \frac{\pi}{n} \right).$$
$$|z-1|^2=2\left[ 1-\left(1-\frac{1}{2!}\left(\frac{\pi}{n} \right)^2 -\frac{1}{4!}\left( \frac{\pi}{n}\right )^4 + \frac{1}{6!}\left( \frac{\pi}{n}\right )^6 -\cdots\right) \right] $$
$$|z-1|^2=\left(\frac{\pi}{n} \right)^2 -\frac{2}{4!}\left( \frac{\pi}{n}\right)^4 + \frac{2}{6!}\left( \frac{\pi}{n}\right )^6 -\cdots$$
When $n=3, |z-1|=1.$
When $n>3$,
$$|z-1|^2 > \left(\frac{\pi}{n} \right)^2 -\frac{1}{12}\left( \frac{\pi}{n}\right)^4 > \frac{4}{(n-1)^2}$$
So
$$|z-1|\ge\frac{2}{n-1}, \quad n\ge3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
3(3)^2+3(3)+1 = ((3)+1)^3-(3)^3\\
3(4)^2+3(4)+1 = ((4)+1)^3-(4)^3\\
\vdots \\
3(n-1)^2+3(n-1)+1 = ((n-1)+1)^3-(n-1)^3\\
3n^2+3n+1 = (n+1)^3-(n-1)^3\\
$$
The result of adding these formulas is
$$
3[1^2+2^2 + ... + (n-1)^2+n^2] + 3[1 + 2 + ... + (n-1)+n] + n = (n+1)^3-1^3
$$
I am able to follow up-to this point easily. I don't understand how this last equation goes from that to this $ n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$ I know thee sum of arithmetic series to n is $n(n-1)/2$ and that replaces the second expression on the left side.
Can some one please show me the algebra step by step?
| Perhaps it's easier if we use the sum notation. Notice that your first equation can be used to show
$$(n+1)^3 - 1 = \sum_{k=1}^n(k+1)^3 - k^3 = \sum_{k=1}^n (3k^2 + 3k + 1) = 3\sum_{k=1}^n k^2 + \frac 32n(n+1) + n,$$
where the first equality comes from the telescopic property.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| x>0;
$f(x):=\sqrt{x^2+6}= x\sqrt{1+6/x^2}=$
$x(1+3/x^2+$
$(1/2)(-1/2)(1/2!)(6/x^2)^2+ O(1/x^6))=$
$x(1+3/x^2-(6^2/8)/x^4+O(1/x^6));$
$x^2(x^2-xf(x)+3)=$
$x^2(- 3+3+(9/2)/x^2+O(1/x^4))$;
Take the limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
| Using first approximation for $\sin x\approx x$ for $x$ near $0$, the limit can be rewritten without change of variables as $$\begin{aligned} &\lim_{x\to \infty}x^3\left(\frac{1}{x+2}-\frac{2}{x+1}+\frac{1}{x}\right)\\ = &\lim_{x\to \infty}x^3\left[\left(\frac{1}{x}-\frac{1}{x+1}\right)-\left(\frac{1}{x+1}-\frac{1}{x+2}\right)\right]\\=&\lim_{x\to \infty}x^3\left[\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}\right]\\=&\lim_{x\to \infty}\frac{2x^3}{x(x+1)(x+2)}\to 2\end{aligned}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int \sec x \, dx = \ln \left\vert\frac{1+\sin x}{1-\sin x}\right\vert^\frac{1}{2}+C = \ln \vert \sec x+\tan x\vert + C$ When our teachers give the rule
$$\int \sec x \, dx = \ln \vert \sec x+\tan x\vert + C$$
We ignore the other solution when we say
$$\sec x = \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} = \frac{\cos x}{1-\sin^2x} $$
and multiply it by $2$ then integrate it we get this equation:
$$\ln \left\lvert\frac{(1+\sin x)}{1-\sin x}\right\rvert^\frac{1}{2}+C$$
$$\int \sec x \, dx = \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert^\frac{1}{2}+C = \ln \left\lvert \sec x+ \tan x\right\rvert + C$$
How that is possible? I don't know much about mathematics please keep it range of calculus 1-1.5.
| You can also show they are the same using trigonometry:
$$\begin{aligned}\ln|\sec x+\tan x|
&=\ln\left\lvert\frac{1+\sin x}{\cos x}\right\rvert\\
&=\ln\left\lvert\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right\rvert\\
&=\ln\left\lvert\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right\rvert\\
&=\ln\left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert^{1/2}
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Big Oh and Little oh notation I have to prove or disprove the following:
$$2n^3 + 5n^2 -3 = o(n^3)$$
$$\frac{n+2}{\sqrt{n^2+4}} = O(1)$$
My attempt at the first question
$$2n^3 + 5n^2 - 3 \leq c |g(x)|$$
$$2n^3+5n^2-3 \leq c |2n^3+5n^3-3n^3|$$
$$2n^3+5n^2-3 \leq c |4n^3|$$
$$2n^3+5n^2-3 \leq 4 |n^3|$$
evaluating when n = 1
$$4 \leq 4$$
when n = 2
$$53 \leq 32$$
hence $$fn \neq o(n^3)$$
My attempt at the second question
$$f_n = \frac{n+2}{\sqrt{n^2+4}} $$
$$ \lim_{n \to -\infty} f_n = \lim_{n \to -\infty} \frac{n+2}{\sqrt{n^2+4}} $$
$$ \lim_{n \to -\infty} f_n = \lim_{n \to -\infty} \frac{n}{\sqrt{n^2}} $$
$$ \lim_{n \to -\infty} f_n = 1 $$
Therefore $$f_n = O(1)$$
can anyone just verify my solution and guide me as to whatever I did wrong.
| *
*$2n^3 + 5n^2 -3 = o(n^3)$ means $ \frac{2n^3+5n^2-3}{n^3} \to 0$ as $n \to \infty.$
But this is fals, since $ \frac{2n^3+5n^2-3}{n^3} \to 2$ as $n \to \infty.$
*$\frac{n+2}{\sqrt{n^2+4}} = O(1)$ means that the sequence $(\frac{n+2}{\sqrt{n^2+4}})$ is bounded.
This is true, since $0 \le \frac{n+2}{\sqrt{n^2+4}} \le 4$ for all $n.$
| {
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How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it.
A hint or a detailed answer would be much appreciated
| Step 1: Using extended Euclidean algorithm, find $g(x)$ and $h(x)$ such that $$g(x)(x^2+2x-4)+h(x)(x^2+x+1)=1$$.
\begin{align}
x^2+2x-4=x^2+x+1+(x-5)\implies& \begin{cases} q = 1\\
r=x-5\\
s=1\\
t=-1\end{cases}\\
x^2+x+1=(x+6)(x-5)+31\implies &\begin{cases} q = x+6\\
r=31\\
s=-(x+6)\\
t=1-(x+6)(-1)=x+7\end{cases}\\
x-5=\left(\frac{x-5}{31}\right)31\implies& \begin{cases} q = \frac{x-5}{31}\\
r=0\\
s= 1+\frac{x-5}{31}(x+6)=\frac{x^2+x+1}{31}\\
t=-1-\frac{x-5}{31}(x+7)=\frac{x^2+2x-4}{31}\end{cases}
\end{align}
Therefore, $g(x)=-\frac{x+6}{31}$ and $h(x)=\frac{x+7}{31}$.
Step 2: Using Chinese remainder theorem,
\begin{align}
f(x)&= (−3+3) + [(4−1)-(−3+3)]h(x)(x^2+x+1)\\&=(−3+3) +(7^3+3x^2+3x−4)\frac{(x+7)}{31}\\
&=\frac{7}{31}x^4+\frac{52}{31}x^3+\frac{24}{31}x^2-\frac{76}{31}x+\frac{65}{31}
\end{align}
| {
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How to factor $x^6-4x^4+2x^3+1$ by hand? I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which other numbers does this equation hold for?
The six possible answers are the roots of $x^6-4x^4+2x^3+1=0$. Note that I am not interested in solving for $x$ itself as much as I am interested in a method which would allow me to completely factor out this polynomial into lowest degree factors which still have real coefficients. Note that I am treating this equation as if I had no clue that the golden ratio is one of the solutions. In other words, I am trying to factor this equation as if I never saw it before, so I can't just immediately factor out $(x^2-x-1)$ without a justifiable process, even though it is indeed one of the factors.
I first observed that the equation holds for $x=1$, so I was able to divide out $(x-1)$ to get the factorization of:
$$(x-1)(x^5+x^4-3x^3-x^2-x-1)$$
I tried making an assumption that the quintic reduces to a product of $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$, multiplying out, and equalling coefficients, but I ended up with a system of two extremely convoluted equations which I had no idea how to solve. I also tried to turn the first five terms of the quintic into a palindromic polynomial and then perform the standard method of factoring palindromic polynomials, to no avail.
I am either missing something, or I don't know of a nice method that would let this expression be factored. I'm looking forward to being enlightened, thanks for any help.
| Here's a possible way to do it:
$x^6-4x^4+2x^3+1 = (x^6+2x^3+1)-4x^4 = (x^3+1)^2 - 4x^4$
$(x^3+1)^2-4x^4 = [x^3+1-2x^2][x^3+1+2x^2]$
$x^6-4x^4+2x^3+1= [(x^3-x^2)+(1-x^2)][x^3+2x^2+1]$
Then, we have:
$x^6-4x^4+2x^3+1 =[x^3+2x^2+1][x^2(x-1)+(1-x)(1+x)]$
$x^6-4x^4+2x^3+1 = (x-1)(x^2-x-1)[x^3+2x^2+1]$
So that gives you a decently nice factored form.
| {
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Maximum of $\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$ If $a>b>c>0$ such that $7a+8b=15c+24\sqrt[3]{abc}$, find the maximum value of
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$$
I tried to use $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3$ so that
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=(\frac{a}{c}+\frac{b}{a}+\frac{c}{a})-(\frac{a}{b}+\frac{b}{c}+\frac{c}{b})\le (\frac{a}{c}+\frac{b}{a}+\frac{c}{a}) -3$$
and here I am stuck.
I don't know how to find the maximum of $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$.
| Notice that:
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b} = \frac{(a-b)(b-c)(a-c)}{abc}$$
Now, let $x=a-b > 0$ and $y=b-c>0$. Then $a-c=x+y$ and from the condition
$$7a+8b=15c+24\sqrt[3]{abc}\Leftrightarrow \sqrt[3]{abc}=\frac{7x+15y}{24}$$
Therefore, using AM-GM:
$$
\begin{aligned}
\frac{(a-b)(b-c)(a-c)}{abc} &= \frac{24^3 xy(x+y)}{(7x+15y)^3}\\
&= \frac{24^3}{144}\cdot \frac{(4x)\cdot (12y)\cdot 3(x+y)}{(7x+15y)^3}\\
&\leq \frac{24^3}{144\cdot 27} \cdot \frac{[4x+12y+3(x+y)]^3}{(7x+15y)^3}\\
&= \frac{32}{9}
\end{aligned}$$
Equality occurs when $a : b : c=9 : 3 : 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}$ for $a+b+c=1$. Source: RMO 2019, question 3
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}.$$
I tried using Holder's inequality but couldn't get to the desired result.
| Hint: Use AM-GM to prove the following ineq: for all $a,b,c>0$,
$$\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq \frac{3a+b+c}{25abc}.$$
| {
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"url": "https://math.stackexchange.com/questions/3571711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi - c.$$ This follows that \begin{align*} \cos(a+b)=\cos(\pi - c) & \implies \cos a \cos b - \sin a \sin b = - \cos c \\ & \implies xy-\sqrt{1-x^2} \sqrt{1-y^2}=z.\end{align*} Now i am not able to proceed from here. Please help me to solve this.
| This is equivalent to the equation
of the sum of cosines of the tree angles of triangle,
\begin{align}
\cos \alpha+\cos\beta+\cos\gamma
&=\frac{3}{2}
,
\end{align}
it is well-known that this sum can be expressed
in terms of $r$ and $R$,
the radii or inscribed and circumscribed circle, respectively as
\begin{align}
\cos\alpha+\cos\beta+\cos\gamma
&=\frac rR+1=\frac{3}{2}
,
\end{align}
which gives
\begin{align}
R&=2\,r
,
\end{align}
and this can be true only for
the equilateral triangle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Any shortcuts for integrating $\frac{x^6}{(x-2)^2(1-x)^5}$ by partial fractions? Is there a faster way to get the partial fraction decomposition of this $\frac{x^6}{(x-2)^2(1-x)^5}$?
$\frac{x^6}{(x-2)^2(1-x)^5} = \frac{A_1}{x-2} + \frac{A_2}{(x-2)^2} + \frac{B_1}{1-x} + \frac{B_2}{(1-x)^2} + \frac{B_3}{(1-x)^3} + \frac{B_4}{(1-x)^4} + \frac{B_5}{(1-x)^5}$
It's fairly easy to get $A_2$ and $B_5$, just by "covering up" $x-2$ and substituting $x=2$ and "covering up" $1-x$ and substituting $x=1$. So $A_2 = 64$ and $B_5=1$. But to proceed from there is there no other faster way other than to mulitply both sides of the equation by $(x-2)^2(1-x)^5$ and compare coefficients, which would result in a system of 6 equations with 6 unknowns?
| set $$x=1-\frac{1}{u+1}$$ to get
$$\int{\frac{{{x}^{6}}}{{{(x-2)}^{2}}{{(1-x)}^{5}}}dx}=\int{\frac{{{u}^{6}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du}$$
long division for the integrand gives:
$${{u}^{3}}-5{{u}^{2}}+17u-49+\frac{129{{u}^{2}}+324u+196}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$. Question: Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$.
My approach: After trying some examples I can conjecture that, the minimum value of $x^2+y^2$ is attained at $$x=\left\lceil \frac{k}{2}\right\rceil \text{and } y=\left\lfloor\frac{k}{2}\right\rfloor \text{and equivalently at } x=\left\lfloor\frac{k}{2}\right\rfloor \text{and }y=\left\lceil \frac{k}{2}\right\rceil.$$ This also implies that the minimum value of $x^2+y^2=\left\lceil \frac{k}{2}\right\rceil^2+\left\lfloor\frac{k}{2}\right\rfloor^2.$
But, how to prove the same?
Also, since $x+y=k$, this implies that $(x+y)^2=k^2\implies x^2+y^2=k^2-2xy.$ Therefore, we need to maximize $xy$ in order to minimize $x^2+y^2$.
But, again this is leading me nowhere.
| Why not to write
$$y=k-x \implies x^2+y^2=2x^2-2kx+k^2=2 \left(x-\frac{k}{2}\right)^2+\frac{k^2}{2}$$
| {
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"source": "stackexchange",
"question_score": "2",
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How to compute $\lim\limits_{n\to\infty}\frac{1}{\sqrt{4n^2-1^2}}+\dots+\frac{1}{\sqrt{4n^2-n^2}}$. I want to compute $$\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4n^2-k^2}}.$$
I found it on this question and the exercise appears to be from previous years of a Latvian competition.
I tried writing $\frac1{\sqrt{4n^2-k^2}}=((2n-k)(2n+k))^\frac{-1}2$ and wanted to continue with partial fractions but the square root is annoying. Also, I thought about $$\frac1{\sqrt{4n^2-k^2}}=\frac{\sqrt{4n^2+k^2}}{\sqrt{16n^4-k^4}}$$ but I don't think that this can help me.
| You can rewrite it to become $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{n}{\sqrt{4n^2-k^2}} = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{1}{\sqrt{4-\left(\frac{k}{n}\right)^2}}$$
This is the Riemann sum for $\frac{1}{\sqrt{4-x^2}}$ from $0$ to $1$. Therefore, the limit of the sum is $$\int_0^1 \frac{1}{\sqrt{4-x^2}} dx$$
Integrating, we get $$\arcsin \left( \frac{x}{2} \right) \bigg\vert_0^1 = \arcsin\left(\frac{1}{2}\right) - \arcsin\left(\frac{0}{2} \right) = \frac{\pi}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Showing moderate decrease property in the real line for $f(z)=\frac{a}{a^2+z^2}$. Let $f(z) = \frac{a}{a^2 + z^2}$ for $a>0$. Then $f$ is holomorphic in the horizontal strip
$|\Im(z)| < a$. I would like to show that in if $|\Im (z)| < a/2$, we have some constant $A>0$ such that
$$|f(x+iy)| \le \frac{A}{1+x^2}$$ for all $x \in \mathbb{R}$ and $|y|<a/2$.
To get this bound, first note that $f(x+iy) = \frac{a}{a^2 + x^2 -y^2 + 2xy i}$.
So when we take the absolute value, we need a bound like $|a^2+x^2 - y^2+2xy i | \ge 1+x^2$.
I can see that we have $|a^2+x^2-y^2 + 2xyi| \ge |a^2+x^2-y^2| \ge (a^2+x^2) - y^2 > \frac{3a^2}{4} + x^2$.
So we have $|f(x+iy)| \le \frac{a}{\frac{3a^2}{4} + x^2}$. But how can I bound this right fraction by some $\frac{A}{1+x^2}$?
| Note that the follow must hold,
$$\frac{A}{1+x^2}- \frac{a}{\frac{3a^2}{4} + x^2} \ge 0$$
for all $x$, which is equivalent to
$$(A-a)x^2+ a(\frac34 aA-1) \ge 0$$
Thus,
$$A = \max( a, \frac4{3a})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1}{\frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)}(5x^2)$ to $\frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)x^2$ I was following the steps from the book Ordinary Differential Equations (Lesson 25B page 274) to find a particular solution of
$4y'' - 3y' + 9y = 5x^2, \quad (4D^2 -3D +9)y = 5x^2$
And theses were the steps
$ y_p = \frac{1}{9\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)}(5x^2) $
$ y_p = \frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)x^2 $
My problem is: I don't know how they had
$ y_p = \frac{5}{9}\bigr(1 + \frac{D}{3} - \frac{D^2}{3} \bigr)x^2$
| Use $$\dfrac 1 {1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+\dots$$
$$4y'' - 3y' + 9y = 5x^2, \quad (4D^2 -3D +9)y = 5x^2\implies y_p=\dfrac 5 {4D^2 -3D +9} (x^2).$$
Hence
$$\begin{align}
y_p &= \frac{5}{{9}\bigr(1 - \frac{D}{3} + \frac{4D^2}{9} \bigr)}(x^2) =\frac 5 9 \frac{1}{\bigr(1 -( \frac{D}{3} - \frac{4D^2}{9}) \bigr)}(x^2)\\
&=\frac 5 9 \left(1 +\left( \frac{D}{3} - \frac{4D^2}{9}\right) +\left( \frac{D}{3} - \frac{4D^2}{9}\right) ^2+\dots\right)(x^2)\\
&=\frac 5 9 \left(1 +\left( \frac{D}{3} - \frac{4D^2}{9}\right) +\frac{D^2}{9} \right)(x^2)\\
&=\frac 5 9 \left(1 + \frac{D}{3} - \frac{D^2}{3} \right)(x^2).
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$ Let be $a,b,c,x,y,z>0$ such that $ax\ge \sqrt{(b^2+c^2)(y^2+z^2)}$. Prove that
$$(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$$
I tried to expand
$$a^2(y^2+z^2)+x^2(b^2+c^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
Here my idea was to use the condition after the means inequality:
$$a^2(y^2+z^2)+x^2(b^2+c^2) \ge 2ax\sqrt{(b^2+c^2)(y^2+z^2)}$$
$$\ge 2(b^2+c^2)(y^2+z^2)$$
but it's not good enough to prove the question
$$2(b^2+c^2)(y^2+z^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
is not true when $a$ and $x$ can be very big.
Thank you for your help.
| Let $\frac{b}{a}=p$, $\frac{c}{a}=q$, $\frac{y}{x}=u$ and $\frac{z}{x}=v$.
Thus, the condition it's $$(p^2+q^2)(u^2+v^2)\leq1$$ and we need to prove that:
$$(u-p)^2+(v-q)^2\geq(pv-qu)^2,$$ which is true by C-S twice:
$$(u-p)^2+(v-q)^2=\sqrt{\left((u-p)^2+(v-q)^2\right)^2}\geq$$
$$\geq\sqrt{\left((v-q)^2+(p-u)^2\right)(p^2+q^2)\cdot\left((v-q)^2+(p-u)^2\right)(u^2+v^2)}\geq$$
$$\geq\sqrt{(vp-qp+pq-uq)^2(vu-qu+pv-uv)^2}=(pv-qu)^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^6-6x^4+12x^2-11$ is irreducible over $\mathbb{Q}$ Extracted from Pinter's Abstract Algebra, Chapter 27, Exercise B1:
Let $p(x) = x^6-6x^4+12x^2-11$, which we can transform into a polynomial in $\Bbb{Z}_3[x]$:
\begin{align*}
x^6+1
\end{align*}
Since none of the three elements $0,1,2$ in $\Bbb{Z}_3$ is a root of the polynomial, the polynomial has no factor of degree 1 in $\Bbb{Z}_3[x]$. So the only possible factorings into non constant polynomials are
\begin{align*}
x^6+1 &= (x^3+ax^2+bx+c)(x^3+dx^2+ex+f)
\end{align*}
or
\begin{align*}
x^6+1 &= (x^4+ax^3+bx^2+cx+d)(x^2+ex+f)
\end{align*}
From the first equation, since corresponding coefficients are equal, we have
\begin{align}
x^0:\qquad & cf &= 1 \tag{1} \\
x^1:\qquad & bf + ce &= 0 \tag{2} \\
x^2:\qquad & af + be + cd &= 0 \tag{3} \\
x^3:\qquad & c + f + bd + ae &= 0 \tag{4} \\
x^5:\qquad & a + d &= 0 \tag{5} \\
\end{align}
From (1), $c = f = \pm1$, and from (5), $a + d = 0$. Consequently, $af + cd = c(a + d) = 0$, and by (3), $eb = 0$. But from (2) (since $c = f$), $b + e = 0$, and therefore $b = e = 0$. It follows from (4) that $c + f = 0$, which is impossible since $c = f = \pm1$. We have just shown that $x^6 + 1$ cannot be factored into two polynomials each of degree 3.
For the second equation, however, $x^6+1=(x^2+1)^3$ in $\Bbb{Z}_3[x]$. So we cannot say $p(x)$ is irreducible over $\Bbb{Q}$ because $x^6+1$ is irreducible over $\Bbb{Z}_3$. What am I missing here?
| Let $p(x) = x^6 - 6x^4 + 12x^2 - 11$
Substitute $x^2 = y$
$h(y) = y^3 - 6y^2 + 12y - 11$
Letting $g(y) = h(y+2)$
$$g(y) = (y+2)^3 -6(y+2)^2 + 12(y+2) - 11$$
$$g(y) = y^3 - 3$$
$g(y)$ is obviously irreducible, thus so is $h(y)$ and $p(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Second-order difference equation solution Say we have a second-order difference equation:
$$x_n = x_{n-1} + x_{n-2} $$
Many of the notes that I have found online regarding how to solve this type of equation will have a step such as "guess" $x_n=Ar^n$.
What is the intuition behind this procedure? How would one know to "guess" that $x_n=Ar^n$?
| We can transform the recurrence relation into a linear algebraic problem.
Let $\vec{a}_n=
\begin{pmatrix}
x_n\\
x_{n+1}\\
\end{pmatrix}$,
$\vec{a}_0=
\begin{pmatrix}
0\\
1
\end{pmatrix}$.
Then $\vec{a}_{n+1}=
\begin{pmatrix}
x_{n+1}\\
x_{n}+x_{n+1}
\end{pmatrix}=A\vec{a}_n
$
where $A
=\begin{pmatrix}
0 & 1\\
1 & 1\\
\end{pmatrix}$
In this way, we can easily see that $\vec{a}_n=A^n\vec{a}_0$ by induction.
(Note that if you know $\vec{a}_n$, then you can recover $x_n$ by looking at its first coordinate)
So it remains to compute $A^n\vec{a}_0$ in a more efficient way.
It turns out that if we can find a number $\lambda$ such that
$A\vec{x}=\lambda \vec{x}$ for special values of $\vec{x}$ then the problem can be easily solved.
The value $\lambda$ is called eigenvalue and it's corresponding value $\vec{x}$ is called eigenvector.
The eigenvalue $\lambda$ can be found by solving the
characteristic equation $\det(\lambda I-A)=0$
In this case, we have $\det(\lambda I-A)=\lambda^2-\lambda-1=0$
Hence the eignvalues are $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$
For the eigenvalue $\frac{1+\sqrt{5}}{2}$, to find its eignevector, we
have to solve $A\vec{x}=\frac{1+\sqrt{5}}{2}\vec{x}$
This is the same as finding $Ker(A-\frac{1+\sqrt{5}}{2}I)$
Since $A-\frac{1+\sqrt{5}}{2}I=
\begin{pmatrix}
-\frac{1+\sqrt{5}}{2} & 1\\
1 & \frac{1-\sqrt{5}}{2}
\end{pmatrix}$,
we see that $\displaystyle Ker(A-\frac{1+\sqrt{5}}{2}I)=\mathbb{R}(1,\frac{1+\sqrt{5}}{2})$
Similarly, we have $Ker(A-\frac{1-\sqrt{5}}{2}I)=\mathbb{R}(1,\frac{1-\sqrt{5}}{2})$
Denote $\vec{v}_1=(1,\frac{1+\sqrt{5}}{2})$, $\vec{v}_2=(1,\frac{1-\sqrt{5}}{2})$.
When $\vec{x}$ is a multiple of $\vec{v}_1$,
then $A\vec{x}=\frac{1+\sqrt{5}}{2}\vec{x}$ and so
$A^n\vec{x}=\left(\frac{1+\sqrt{5}}{2}\right)^n\vec{x}$
Similar for the $\frac{1-\sqrt{5}}{2}$ case.
This gives us intuition to express $\vec{a}_0$ as a linear combination
of $\vec{v}_1$ and $\vec{v}_2$:
$$\vec{a}_0=
\frac{1}{\sqrt{5}}\vec{v}_1-\frac{1}{\sqrt{5}}\vec{v}_2
$$
Thus,
$$\begin{align*}
\vec{a}_n&=A^n(\frac{1}{\sqrt{5}}\vec{v}_1-\frac{1}
{\sqrt{5}}\vec{v}_2)\\
&=\frac{1}{\sqrt{5}}(A^n\vec{v}_1-A^n\vec{v}_2)\\
&=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n \vec{v}_1-\left(\frac{1-\sqrt{5}}{2}\right)^n\vec{v}_n\right)
\end{align*}$$
Finally, we pick the first coordinate and get:
$$x_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n \right) $$
| {
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"url": "https://math.stackexchange.com/questions/3580370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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please solve quadratic simultaneous equations. $x,y,z$ are real variables. $l,m,n$ are real positive constants.
Solve for real $x,y,z$:
$$\begin{align}
x^2+xy+y^2 &= l\\
y^2+yz+z^2 &= m\\
z^2+zx+x^2 &= n
\end{align}
$$
This problem has a relevance in Electrical Engineering.
| $$x^2+xy+y^2=l~~(1),~ y^2+yz+z^2=m ~~~(2),~ z^2+zx+x^2=n~~~(3)$$
Subtract (1) from (2) and (1) from (3) to get
$$(x-y)(x+y+z)=m-l~~~(4),~ (x-z)(x+y+z)=n-l~~~(5)$$
Let $(x+y+z)=w$, then (4) and (5) give
$$(x-y)=\frac{m-l}{w}~~~(6), ~(x-z)=\frac{n-l}{w}~~~(7)$$
From these two we get $$[3x-(x+y+z)]w=m+n-2l \implies x=\frac{w^2+m+n-2l}{3w}~~~(8).$$
Similarly, $$y=\frac{w^2+l+b-2m}{3w}, ~~z=\frac{w^2+l+m-2n}{3w}~~~(9)$$
Putting these in Eq. (4), we get
$$w^4-(l+m+n)w^2+(l^2+m^2+n^2)-lm-mn-nl=0$$
This gives
$$w^2=\frac{(l+m+n)\pm\sqrt{(l+m+n)^2-4[l^2+m^2+n^2-lm-mn-nl]}}{2}$$
$$w^2=\frac{(l+m+n) \pm \sqrt{3[2(lm+mn+nl)-(l^2+m^2+n^2)}}{2}~~~~(10)$$
Lastly, Eqs. (8), (9) and (10) give the values of $x,y,z$ in terms of$l,m,n.$. The roots will be real if $$l^2+m^2+n^2 \le 2(lm+mn+nl)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$ How to prove that
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$
$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$
where $\text{Li}_r$ is the polylogarithm function and $\zeta$ is Riemann zeta function.
I managed to prove the equality above using the following harmonic series,
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n}, \ \sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}H_n}{n},\ \sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\ \text{and }\ \sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$
so definitely this approach is pretty boring. Is it possible to solve it in a different way? Thank you.
| I proved here
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$
Divide both sides by $1+a$ the integrate from $a=0$ to $a=1$ we get
$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da-\text{Li}_3\left(\frac{1}{2}\right)\underbrace{\int_0^1\frac{da}{1+a}}_{\ln(2)}$$
$$-\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da+\frac12\ln2\underbrace{\int_0^1\frac{\ln^2(1+a)}{1+a}\ da}_{1/3 \ln^3(2)}$$
where
$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x} \left(\int_x^1\frac{da}{1+a}\right)\ dx$$
$$=-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx=-I$$
$$\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da=\text{Li}_4\left(\frac{1+a}{2}\right)\bigg|_0^1=\zeta(4)-\text{Li}_4\left(\frac{1}{2}\right)$$
$$\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da\overset{IBP}{=}\ln(1+a)\text{Li}_3\left(\frac{1+a}{2}\right)\bigg|_0^1-\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da$$
$$=\ln(2)\zeta(3)-\zeta(4)+\text{Li}_4\left(\frac{1}{2}\right)$$
Combine all results and use $\text{Li}_3\left(\frac{1}{2}\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$, we get the closed form of $I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$ show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$
Here is the algebric proof:
We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
So, when $n\to\infty$, we get $\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.
Question: I am wondering does episilon-delta method work here as an alternative proof?
| You want to prove that for all $\varepsilon > 0$ there is $N > 0$ such that
$\sqrt{n+1}-\sqrt{n}<\varepsilon$ when $n>N$. Your algebraic manipulation shows that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
so you would like to solve the inequality $\frac{1}{\sqrt{n+1}+\sqrt{n}} < \varepsilon$.
To find $N$ it is enough to note that $\sqrt{n+1}+\sqrt{n} > 2\sqrt{n} > \varepsilon^{-1}$, and the last gives $n > 1/4\varepsilon^2$. Set $N$ to be the least integer greater than $1/4\varepsilon^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What are the number of ordered m-tuples of integers,such that sum of square of elements is a given integer? Given a non-zero integer $n$,the problem is to find a m-tuple of integers,$(x_1,x_2,x_3,...,x_m)$such that the following equation is satisfied---$$\sum_{i=1}^mx_i^2=n$$
I have no idea how to approach the problem,neither can I find exactly a solution to this,but I hope maybe counting and a bit of number theory can help.Is there any analytic solution at all?
| Given $$\sum_{i=1}^m x_i^2=n$$
We can use Euclid's formula for generating Pythagorean triples
$$A=x^2-y^2\qquad B=2xy\qquad C=x^2+y^2$$
We can find a case for $m=2$ easily if we solve the $C$-function for $(y)$ and test a defined range of $m$-values to see which yield integers. For example
$$C=x^2+y^2\implies y=\sqrt{C-x^2}\qquad\text{where}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le x \le \lfloor\sqrt{C-1}\rfloor$$
The lower limit ensures $x>y$ and the upper limit ensures $y\in\mathbb{N}$.
$$C=85\Rightarrow \bigg\lfloor\frac{ 1+\sqrt{170-1}}{2}\bigg\rfloor=6 \le x \le \lfloor\sqrt{85-1}\rfloor=9\quad\land \quad x\in\{7,9\}\Rightarrow y\in\{6,2\}$$
$$f(7,6)=(13,84,85)\quad f(9,2)=(77,36,85)$$
Side $C$ always takes the form $(C=4n+1)$ and we see that side $A$ of $F(7,6)=13=4(3)+1$ so we can try the process again. (Note: not all values of $4n+1$ are valid.)
$$C=13\implies \bigg\lfloor\frac{ 1+\sqrt{26-1}}{2}\bigg\rfloor=3 \le x \le \lfloor\sqrt{13-1}\rfloor=3\quad\text{ and we find} \quad x\in\{3\}\Rightarrow y\in\{2\}$$
$$f(3,2)=(5,12,13)\quad \text{and repeating the process for } A=5\quad f(2,1)=(3,4,5)$$
So we get a $4$-dimensional m-tupple where $3^2+4^2+12^2+84^2=85^2=7225$
There are an infinite number of these where $2\le m \le\infty$. It is easier to build them by finding a side $A$ to match a given side $C$ and, we can always find such since all odd numbers greater than $1$ are valid $A$-values.
$$A=x^2-y^2\implies y=\sqrt{x^2-A}\qquad\text{where}\qquad \sqrt{A+1} \le x \le \frac{A+1}{2}$$
The lower limit ensures $y\in\mathbb{N}$ and the upper limit ensures $x> y$.
$$A=15\implies \sqrt{15+1}=4\le x \le \frac{15+1}{2} =8\quad\text{ and we find} \quad x\in\{4,8\}\implies y \in\{1,7\} $$
$$fF4,1)=(15,8,17)\qquad \qquad f(8,7)=(15,112,113)$$
The example repeated yields $f(9,8)=(17,144,145)$ or $f(57,56)=(113,6384,6385)$ and this process will always work and for an unlimited number of iterations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Easy way to find the partial fraction I always have trouble trying to find the partial fraction, especially for complicated ones.
For example, this is what I will do to find the partial fraction of $\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$
*
*$\displaystyle \frac{A}{x}+\frac{B}{2x+3}+\frac{C}{(2x+3)^2}+\frac{D}{(2x+3)^3}$
*$\displaystyle A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $
*When x = $0$ -> $\displaystyle 27A = 27, A=1$
*$\displaystyle (2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27$
After what I can do is to expand all the elements and group them based on their exponent, and solve the system of equation.
However, I remember seeing there exists an easier solution. Also given that there will be no calculator available on the exam, doing this way will take a long time and results in possible errors.
Does anyone have an easier way to solve this question or similar ones?
Thanks!
| $A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $
With $ x = 0, 27 A = 27, A = 1 $.
This gives us $Bx(2x+3)^2+Cx(2x+3)+Dx = -x^2 - 12x $.
Dividing by $x$, this gives us $B(2x+3)^2+C(2x+3)+D = -x - 12 $.
With $x = - \frac{3}{2}$, $D = - 10 \frac{1}{2}$.
This gives us $ B(2x+3)^2 + C (2x+3) = -x - \frac{3}{2}$
Dividing by $(2x+3)$, this gives us $ B ( 2x+3) + C = - \frac{1}{2}$.
With $ x = - \frac{3}{2}$, this gives us $ C = - \frac{1}{2}$.
This gives us $ B ( 2x+3) = 0 $.
Dividing by $(2x+3) $, this gives us $ B = 0 $.
With this process, you basically get rid of terms sequentially. Use the cover up rule to determine one variable, move that over, divide by the common terms, rinse and repeat.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factorize this determinant?
The question is to factorize $$\det\begin{pmatrix}(x^2+1)^2 & (xy+1)^2 & (xz+1)^2 \\ (xy+1)^2 & (y^2+1)^2 & (yz+1)^2 \\ (xz+1)^2 & (yz+1)^2 & (z^2+1)^2 \end{pmatrix}.$$
I have a hint which is considering the factorization of $\det\begin{pmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix}$ where all entries are real numbers.
I don't have any idea how to use the hint. So, I factorize $\det\begin{pmatrix}(x^2+1)^2 & (xy+1)^2 & (xz+1)^2 \\ (xy+1)^2 & (y^2+1)^2 & (yz+1)^2 \\ (xz+1)^2 & (yz+1)^2 & (z^2+1)^2 \end{pmatrix}$ directly and I get the answer which is $2(z-y)^2(z-x)^2(y-x)^2$.
My question is how to use the hint to factorize the given determinant? It is because my method seems very tedious.
| By calculation (I got this slightly mixed up in my comment),
$$\begin{pmatrix} 1 & 2x & x^2 \\ 1 & 2y & y^2 \\ 1 & 2z & z^2 \end{pmatrix} \begin{pmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{pmatrix}^\top = \begin{pmatrix} 1 + 2x^2 + x^4 & 1 + 2xy + (xy)^2 & 1 + 2xz + (xz)^2 \\ 1 + 2xy + (xy)^2 & 1 + 2y^2 + y^4 & 1 + 2yz + (yz)^2 \\ 1 + 2xz + (xz)^2 & 1 + 2yz + (yz)^2 & 1 + 2z^2 + z^4 \end{pmatrix}.$$
Using the hint, the first matrix has determinant $2(x - y)(y - z)(x - z)$ and the second matrix has the determinant $(x - y)(y - z)(x - z)$. We can now just multiply these determinants together to get the determinant of the right hand side, which is the matrix in question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$
Question: Given that $X$ and $Y$ are two random variables satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$
Calculate
$$\mathbb{E}(X-Y \mid 2X+Y).$$
I tried to use the fact that if $A$ and $B$ are independent, then $\mathbb{E}(A\mid B) = \mathbb{E}(A)$ and uncorrelated implies independence in jointly normal distribution.
So, I attempted to express $X-Y$ as a linear combination of $2X+Y$ and $Z$ where $\operatorname{Cov}(2X+Y,Z) = 0.$
But I am not able to do so.
Any hint is appreciated.
| We use two property:
First: $E(2X+Y|2X+Y)=2X+Y$
Second: $(X-dY,2X+Y)$ is bi-variate normal(for $d\neq - \frac{1}{2}$), if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent(by set $\rho=0$ in bivarite distribution of joint $(X-dY,2X+Y)$
Correlations_and_independence). so $E(X-dY|2X+Y)=E(X-dY)=0$.
$$E(2X+Y|2X+Y)=2X+Y$$
so
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y) \hspace{1cm} (1)$$
For first step let $\rho=0$
$$cou(X-2\frac{a}{b} Y,2X+Y)=2Var(X)-2\frac{a}{b} Var(Y)=2a-2\frac{a}{b}b=0$$
so since $X-2\frac{a}{b} Y$ and $2X+Y$$ are normal so they are independent.
in hence
$$E(X-2\frac{a}{b} Y|2X+Y)=E(X-2\frac{a}{b} Y)=0$$
$$E(X|2X+Y)=2\frac{a}{b} E(Y|2X+Y)\hspace{1cm} (2)$$
combine (1) and (2)
$$E(X|2X+Y)=\frac{2\frac{a}{b}}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$
$$E(Y|2X+Y)=\frac{1}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$
so
$$E(X -Y|2X+Y)=(\frac{2\frac{a}{b}}{1+4\frac{a}{b}}-\frac{1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)=(\frac{2\frac{a}{b}-1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)$$
**Now for general case ** $\rho \in[-1,1]$
if $$d=\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}} \hspace{1cm} (3)$$
$$cou(X-dY,2X+Y)=2a-db+(1-2d)\rho \sqrt{a} \sqrt{b}$$
$$=2a+\rho \sqrt{a} \sqrt{b}-d(b+2\rho \sqrt{a} \sqrt{b})=0$$
so $$E(X-dY|2X+Y)=E(X-dY)=0$$ and in hence
$$E(X|2X+Y)=dE(Y|2X+Y) \hspace{1cm} (4)$$
Combine (4) and (1)
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y)=2X+Y-2dE(Y|2X+Y)$$ so
$$E(Y|2X+Y)=\frac{1}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (5)$$ and
$$E(X|2X+Y)=dE(Y|2X+Y)=\frac{d}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (6)$$
(5) and (6)
$$E(X-Y|2X+Y)=\frac{d-1}{1+2d}\bigg(2X+Y\bigg)$$
$$=\frac{\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}-1}{1+2\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}}\bigg(2X+Y\bigg)$$
$$=\frac{2a-b-\rho \sqrt{a} \sqrt{b}}{b+4a+4\rho \sqrt{a} \sqrt{b}}\bigg(2X+Y\bigg)$$
detail for "@Student"
I explain now why I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent.
1)$(X-dY,2X+Y)$ is bi-variate normal for $d\neq \frac{-1}{2}$
I can write
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y
\end{bmatrix}
=\begin{bmatrix}
1 & -d \\
2 & 1
\end{bmatrix}
\begin{bmatrix}
X \\
Y
\end{bmatrix}
\end{eqnarray}
By linear-transformation-of-gaussian-random-variable I think
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y
\end{bmatrix}
\end{eqnarray}
is bi-variate normal.
2) Now by Correlations_and_independence I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent. wikipedia: "In general, random variables may be uncorrelated but statistically dependent. But if a random vector has a multivariate normal distribution then any two or more of its components that are uncorrelated are independent".
| {
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"url": "https://math.stackexchange.com/questions/3590203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Evaluating $\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln(1+x)-x^2}$ using L'Hôpital Considering this limit, assigned to my high school students,
$$\lim_{x\to 0}\frac{x\sin x-2+2\cos x}{x\ln \left(1+x\right)-x^2}=\left(\frac00\right)=\lim_{x\to 0}\frac{\frac{d}{dx}\left(x\sin \left(x\right)-2+2\cos \left(x\right)\right)}{\frac{d}{dx}\left(x\ln \left(1+x\right)-x^2\right)} \tag 1$$
After some steps, using L'Hôpital, I find:
$$\lim_{x\to 0}\frac{\left(x\cos \left(x\right)-\sin \left(x\right)\right)\left(1+x\right)}{-2x^2-x+x\ln \left(x+1\right)+\ln \left(x+1\right)}=\left(\frac00\right)$$
Should I continue to apply L'Hôpital? :-(
| The derivative of the denominator is
$$
\ln(1+x)+\frac{x}{1+x}-2x=\frac{(1+x)\ln(1+x)-x-2x^2}{1+x}
$$
Then after the first step you find
$$
\frac{x\cos x-\sin x}{(1+x)\ln(1+x)-x-2x^2}(x+1)
$$
The factor $x+1$ can be disregarded, because its limit is $1$.
At the next application, the numerator will become
$$
\cos x-x\sin x-\cos x=-x\sin x
$$
while the denominator will be
$$
\ln(1+x)-4x
$$
Thus the limit has become much simpler:
$$
\lim_{x\to0}\frac{x\sin x}{4x-\ln(1+x)}=
\lim_{x\to0}\frac{x^2}{4x-\ln(1+x)}\frac{\sin x}{x}
$$
and the part $(\sin x)/x$ can be disregarded. You can now apply l'Hôpital once more to get
$$
\lim_{x\to0}\frac{2x}{4-\dfrac{1}{1+x}}=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a positive integer $k$, describe all positive integers $n$ such that $\langle n\rangle=k$.
Question: For any positive integer $n$, let $\langle n \rangle$ denote the integer nearest to $\sqrt{n}$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n\rangle=k$.
(b) Show that $$\sum_{n=1}^\infty \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}=3.$$
My approach: It is clear that there are exactly $2n$ integers between $(n+1)^2$ and $n^2$, $\forall n\in\mathbb{N}$. Now, by the pattern that we observe we can conclude that out of these $2n$ integers the first $n$ integers, let us call them $k$, have $\langle k \rangle =n$ and the other $n$ integers, let us call them $l$, have $\langle l\rangle=n+1.$
Therefore, for any $k\in\mathbb{N}$, the positive integers $n$ that have $\langle n\rangle=k$ are $k^2-(k-1), k^2-(k-2),\cdots, k^2, k^2+1, k^2+2, \cdots, k^2+k.$
Therefore, we are done with part (a) of the question.
Now $\forall k\in\mathbb{N}$ we have, $$\sum_{n=k^2-(k-1)}^{k^2+k}\frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}\\=\sum_{n=k^2-(k-1)}^{k^2+k}\frac{2^{k}+2^{-k}}{2^n}\\=(2^k+2^{-k})\sum_{n=k^2-(k-1)}^{k^2+k}\frac{1}{2^n}\\=(2^k+2^{-k})\frac{2^{-(k^2-(k-1))}((2^{-1})^{2k}-1)}{2^{-1}-1}\\=(2^k+2^{-k})\frac{2^{k-1-k^2}(2^{-2k}-1)}{2^{-1}-1}\\=(2^k+2^{-k})\frac{2^{-k^2-k-1}-2^{k-1-k^2}}{2^{-1}-1}\\=(2^k+2^{-k})(2^{k-k^2}-2^{-k^2-k})\\=2^{2k-k^2}-2^{-k^2-2k}=2^{1-(k-1)^2}-2^{1-(k+1)^2}=(2^{1-(k-1)^2}+2^{1-k^2})-(2^{1-k^2}+2^{1-(k+1)^2}).$$
Define the sequence $\{a_n\}_n\ge1$, such that $$a_n=2^{1-(n-1)^2}+2^{1-n^2}, \forall n\in\mathbb{N}.$$
Therefore, $\forall k\in\mathbb{N}$ we have, $$\sum_{n=k^2-(k-1)}^{k^2+k}\frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}=a_k-a_{k+1}.$$
This implies that $$\sum_{n=1}^\infty\frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}=\sum_{i=1}^\infty a_i-a_{i+1}=\lim_{l\to\infty}(a_1-a_{l+1})=3.$$
Hence, $$\sum_{n=1}^\infty\frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}=3.$$
As one can see part (a) of the question is done based on observation and not rigorously. Can someone provide a rigorous solution to part (a) of the question and check if part (b) is done correctly and rigorously or not.
| The condition states that $ k- \frac{1}{2} \leq \sqrt{n} \leq k + \frac{1}{2}$
Squaring, we get
$ k^2 -k + \frac{1}{4} \leq n \leq k^2 + k + \frac{1}{4}$
Now, use the fact that $n $ is integer to conclude
$ k^2 - k + 1 \leq n \leq k^2 +k$
Your part b) is correct.
A slightly nicer solution to b) is to look at what the multiset $\{<n>-n\} \cup \{-<n>-n\}$ is.
Hence conclude that
$$\sum_n \frac{ 2^{<n> } + 2^{-<n> }} { 2^n} = \sum_{i=0}^{\infty} \frac{1}{2^i} + \sum_{i=1}^\infty \frac{1}{2^i} = 2 + 1 = 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)^{20}$ and then working with binomial expansion.
| $x^{40}(1+x^5+x^7)^{20}=x^{40}(1+x^5(1+x^2))^{20}=$
$x^{40}(1+20x^5(1+x^2)+$
$(20)(19)/2![(x^5)(1+x^2)]^2+$
$(20)(19)(18)/3![x^5(1+x^2)]^3+..).$
Need the coefficient of $x^{17}$ in the above expansion.
Which term is relevant?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx$ $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx$$
which is W3, Jozsef Wildt International Mathematical Competition 2019.
| A glance at the following equality
$$\frac{\cos x+1}{x\sin x+1}\mathrm{d}x=\frac{\mathrm{d}(x+\sin x)}{x\sin x+1}$$
inspires us to calculate that
$$\frac{\mathrm{d}}{\mathrm{d}x}\frac{x+\sin x}{x\sin x+1}=\frac{\cos x(\cos x+1-x^2)}{\left(x\sin x+1\right)^2}.$$
Let $y:=\frac{x+\sin x}{x\sin x+1}$, and compared with the integrand, we have
\begin{align*}
I& :=\int\frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx\\
& =\int\frac{x\sin x+1}{\cos x\sqrt{1-x^2}}\mathrm{d}y\\
& =\int\frac{x\sin x+1}{\sqrt{\left(1-x^2\right)\left(1-\sin^2 x\right)}}\mathrm{d}y.
\end{align*}
It's important to note that
\begin{align*}
\left(1-x^2\right)& \left(1-\sin^2 x\right)=\\
& \left(x\sin x+1\right)^2-\left(x+\sin x\right)^2.
\end{align*}
It follows that
\begin{align*}
I& =\int\left(1-\frac{\left(x+\sin x\right)^2}{\left(x\sin x+1\right)^2}\right)^{-\frac{1}{2}}\mathrm{d}y\\
& =\int\frac{\mathrm{d}y}{\sqrt{1-y^2}}=\arcsin y+C.
\end{align*}
Hence
\begin{align*}
I_1& :=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x+1-x^2}{(1+x\sin x)\sqrt{1-x^2}}dx\\
& =2\arcsin\left(\frac{4+\pi\sqrt{2}}{\pi+4\sqrt{2}}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
| We'll prove that $\max(x^2+y^2)=4.$
Indeed, let $xy\geq0$.
Thus, by C-S $$2=x^4+y^4+3xy\geq x^4+y^4=\frac{(1+1)(x^4+y^4)}{2}\geq\frac{(x^2+y^2)^2}{2},$$
which gives $$x^2+y^2\leq2<4.$$
Now, let $xy\leq0.$
Thus, by C-S again and AM-GM we obtain:
$$2=x^4+y^4+3xy\geq\frac{(x^2+y^2)^2}{2}-\frac{3(x^2+y^2)}{2},$$
which gives $$(x^2+y^2)^2-3(x^2+y^2)-4\leq0$$ or
$$(x^2+y^2-4)(x^2+y^2+1)\leq0$$ or
$$x^2+y^2\leq4.$$
The equality occurs for $(x^2,y^2)||(1,1)$ and $x=-y,$ which with our condition gives:
$$(x,y)=(\sqrt2,-\sqrt2)$$ or
$$(x,y)=(-\sqrt2,\sqrt2).$$
By the similar way we can prove that $\min(x^2+y^2)=1,$ where the equality occurs for $$(x,y)=\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$$ or
$$(x,y)=\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Evaluating Series Integral How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$
I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.
| \begin{align*}
J&=\int_0^\infty \frac{\ln^2 x}{1+y^2}\,dx\\
A&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{\ln^2(x)+\ln^2(y)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{2\ln^2(x)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=2J\Big[\arctan x\Big]_0^\infty\\
&=\pi J
\end{align*}
On the other hand,
\begin{align*}
A&\overset{u(x)=xy}=\int_0^\infty \left(\int_0^\infty \frac{\ln^2 u}{y(1+\left(\frac{u}{y}\right)^2)(1+y^2)}\,du\right)\,dy\\
&=\int_0^\infty \left(\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}\,du\right)\,dy\\
&=\frac{1}{2}\int_0^\infty \left[\frac{\ln\left(\frac{u^2+y^2}{1+y^2}\right)}{1-u^2}\right]_{y=0}^{y=\infty}\ln^2 u\,du\\
&=\int_0^\infty \frac{\ln^3 u}{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^3 u}{u^2-1}\,du+\int_1^\infty \frac{\ln^3 x}{x^2-1}\,dx\\
&\overset{u=\frac{1}{x}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}\,du\\
&=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\int_0^1 \frac{2x\ln^3 x}{x^2-1}\,dx\\
&\overset{u=x^2}=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du\\
&=\frac{15}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du
\end{align*}
Moreover,
\begin{align*}
\int_0^1 \frac{\ln^3 u}{1-u}\,du&=\int_0^1 \left(\sum_{n=0}^\infty u^n\right)\ln^3 u\,du\\
&=\sum_{n=0}^\infty\left(\int_0^1 u^n\ln^3 u\,du\right)\\
&=-6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\\
&=-6\zeta(4)
\end{align*}
Therefore,
\begin{align*}\
A&=\frac{15}{8}\times 6\zeta(4)\\
&=\frac{45}{4}\zeta(4)\\
J&=\frac{A}{\pi}\\
&=\frac{45\zeta(4)}{4\pi}\\
&=\frac{45\times\frac{\pi^4 }{90}}{4\pi}\\
&=\boxed{\frac{\pi^3}{8}}
\end{align*}
NB:
Observe that $\displaystyle \int_0^1 \frac{\ln x}{1+x^2}\,dx\overset{u=\frac{1}{x}}=-\int_0^1 \frac{\ln u}{1+u^2}\,du$
and, i assume $\displaystyle \zeta(4)=\dfrac{\pi^4}{90}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$ Find the limit:
$$\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$$
I applied the identity:
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$
by multiplying the numerator and denominator by the complementary part.
$$\lim\limits_{n\to\infty} \frac{(\sqrt[3]{n^6-6n^4+1} - n^2)(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}{(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}=\lim\limits_{n\to\infty} \frac{n^6-6n^4+1 - n^6}{(\sqrt[3]{(n^6(1-6/n^2+1/n^6))^2} + n^2\sqrt[3]{n^6(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6n^4+1}{(n^4\sqrt[3]{(1-6/n^2+1/n^6)^2} + n^4\sqrt[3]{(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6+1/n^4}{(\sqrt[3]{(1-6/n^2+1/n^6)^2} + \sqrt[3]{(1-6/n^2+1/n^6)}+ 1)}=\frac{-6}{3}=-2$$
Is there any more elegant way to approach the problem?
| With binomial series, it's $$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac6{n^2}+\dfrac1{n^6}\right)^{1/3}-n^2\right]$$
$$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac13\dfrac6{n^2}+O\left(\dfrac1{n^4}\right)\right) -n^2\right]$$
$$=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the values of $x$ in terms of $a$ in $x^2+\frac{(ax)^2} {(x+a)^2} =3a^2$ The question is as follows.
Find the values of $x$ in terms of $a$ in
$x^2+\dfrac{(ax)^2}{(x+a)^2} =3a^2 $
My solution:
Multiply both sides by $(x+a)^2$ and expand. On rearranging we get
$x^4+2ax^3-a^2x^2-6a^3x-3a^4=0$
Now dividing by $a^4$ we and taking $\dfrac{x}{a}=y$, we get
$y^4+2y^3-y^2-6y^3-3=0$
Now we can write the above equation as
$(y^2+ay+b)(y^2+cy+d)=0$
Now expanding it and comparing coefficient with the equation we get
$a=-1;b=-1;c=3;d=3$
So we get
$(y^2-y-1)(y^2 +3y+3)=0$
The second bracket has no real roots and by solving 1st bracket we get
$y=\dfrac{x}{a} =\dfrac{1\pm\sqrt5}{2}$
So $x=\dfrac{a(1\pm\sqrt5) }{2}$
My question is
1: As you can see the method is long and tedious, is there any other or elegant way to solve it?
2:Can any step in my solution or the solution itself be improved upon or be replaced by another easier step (like but not limited to factorising the equation we got in $y$ directly into two quadratics or solving the quatic directly)
Thanks!
I got the question from a preparatory book for olympiads.
| It's $$x^2+\frac{a^2x^2}{(x+a)^2}-\frac{2ax^2}{a+x}+\frac{2ax^2}{a+x}=3a^2$$ or
$$\left(x-\frac{ax}{a+x}\right)^2+\frac{2ax^2}{a+x}=3a^2$$ or
$$\left(\frac{x^2}{a+x}\right)^2+\frac{2ax^2}{a+x}-3a^2=0$$ or
$$\left(\frac{x^2}{a+x}\right)^2+\frac{2ax^2}{a+x}+a^2-a^2-3a^2=0$$
or
$$\left(\frac{x^2}{a+x}+a\right)^2-(2a)^2=0$$
or
$$\left(\frac{x^2}{a+x}-a\right)\left(\frac{x^2}{a+x}+3a\right)=0$$ and the rest is smooth:
The domain gives $x\neq-a$, which gives $a\neq0.$
Thus, $$x^2+3ax+3a^2=\left(x+\frac{3a}{2}\right)^2+\frac{3a^2}{4}>0,$$ which says that the right polynomial does not give roots.
Also, $$x^2-ax-a^2=0$$ gives
$$\left\{\frac{a(1+\sqrt5)}{2},\frac{a(1-\sqrt5)}{2}\right\}.$$
To solve an equation with parameter says to solve this equation for any values of the parameter.
Id est, we got the following answer.
If $a=0$, so $\oslash$;
if $a\neq0$, so $\left\{\frac{a(1+\sqrt5)}{2},\frac{a(1-\sqrt5)}{2}\right\}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes:
If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?
So what I tried was I know that since the line is normal to the 2nd circle, so it must pass through the center of the second circle which is $(-1,2)$.
So from that I got that $$-a+2b+c=0$$
But I cant really find any other equations here that would help, I tried differentiation the curves but I dont have the point of contact so can't really do anything there.
I also know that the tangent to the circle $x^2+y^2+2gx+2fy+c = 0$ at $(a,b)$ is $ax+by+(a+x)g+(b+y)f +c = 0$
I don't know how to proceed, can someone help?
| Another way.
Let $y=mx+n$ be an equation of the tangent.
Thus, $$mx-y+n=0,$$ which since equations of our circles they are
$$(x+1)^2+(y-2)^2=4$$ and $$(x-1)^2+y^2=\frac{8}{5},$$ we obtain that the point $(-1,2)$ is placed on the line $y=mx+n$
and the distance from $(1,0)$ to the line is equal to $\sqrt{\frac{8}{5}}.$
Thus,
$$-m-2+n=0$$ and
$$\frac{|m\cdot1-1\cdot0+n|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}},$$ which gives
$$\frac{|2m+2|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}}$$ or
$$5(m+1)^2=2(m^2+1).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find without L'Hospital's rule: $\lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}}$ $A = \sqrt{17 - 2x^{2}} \sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} \\
\Rightarrow \lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}} \\
= \lim\limits_{x \to 2} \frac{A^{6} - 9^{6}}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\
= \lim\limits_{x \to 2} \frac{(x - 2)^{2}(-72x^{10} - 192x^{9} + 940x^{8} + 2576x^{7} + 874x^{6} + 1992x^{5} - 24543x^{4} - 73908x^{3} - 82540x^{2} - 200414x - 102154)}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\
= \frac{-2.11.3^{10}}{6.9^{5}} = \frac{-11}{3}.$
Am I right? Is there a simple way?
| Yes, that works.
You can also use the fact that, if $f(x)=\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}$, then $f(2)=9$, $f'(2)=0$, and $f''(2)=-\frac{22}3$. So, near $0$,$$f(x)=9-\frac{11}3(x-2)^2+O\bigl((x-2)^3\bigr)$$ and therefore,\begin{align}\lim_{x\to2}\frac{\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}}{(x-2)^2}&=\lim_{x\to2}\frac{f(x)-f(2)}{(x-2)^2}\\&=-\frac{11}3.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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approaching $\int \sqrt{x^5+2}\; dx$ I recently came across this integral $\int\sqrt{x^5+2}\; dx$. From Wolframalpha i can see that it has a closed form. how does one get to that closed form? what techniques should i approach?
| As said in comments, using Taylor or the binomial theorem, we have
$$\sqrt{x^5+2}=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{
2^n\, n!} x^{5 n}$$
$$\int \sqrt{x^5+2}\,dx=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{
2^n\,(5n+1)\, n!} x^{5 n+1}$$ Computing the infinite summation
$$S=x\sqrt{2} \,\, _2F_1\left(-\frac{1}{2},\frac{1}{5};\frac{6}{5};-\frac{x^5}{2}\right)$$ which looks simpler that the result from Wolfram Alpha but which numerically does not agree with it. However, the formula given here matches the results obtained by numerical integration.
What happens ? That is the question !
Edit
If I use the result given by Wolfram Alpha,
$$\int \sqrt{x^5+2}\,dx=\frac{1}{7} x \left(5 \sqrt{2} \,
_2F_1\left(\frac{1}{5},\frac{1}{2};\frac{6}{5};-\frac{x^5}{2}\right)+2 \sqrt{x^5+2}\right)$$ and differentiate is, hoping that I am not mistaken, the result is
$$\frac{5 x^5}{2 \sqrt{x^5+2}}$$
On Wolfram Cloud, I obtained the result I gave.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $\cos C(\sin A +\sin B)=\sin C \cos(A-B)$, prove that triangle $ABC$ is equilateral.
Given $\cos C(\sin A +\sin B)=\sin C\cos(A-B)$, prove that triangle $ABC$ is equilateral.
My attempt:
By applying some trigonometric identities, I have:
$$\cos C(\sin A +\sin B)=\sin C \cos(A-B)\\
\iff \cos C \cdot2\sin \frac{A+B}{2} \cos \frac{A-B}{2}=\sin \frac{C}{2}\cos \frac{C}{2}\cos(A-B)\\
\iff \cos C \cos \frac{A-B}{2}=\sin \frac{C}{2}\cos(A-B)\\
\iff \left( 1-2\sin^2 \frac{C}{2}\right)\cos \frac{A-B}{2}=\sin \frac{C}{2}\left( 2\cos^2 \frac{A-B}{2}-1\right)\\ \iff \cos \frac{A-B}{2}+\sin \frac{C}{2}=2\sin \frac{C}{2}\cos \frac{A-B}{2}\left(\sin \frac{C}{2}+\cos \frac{A-B}{2} \right)\\ \implies 2\sin \frac{C}{2}\cos \frac{A-B}{2}=1$$
or $$2\cos \frac{A+B}{2}\cos \frac{A-B}{2}=1$$
or $$\cos A +\cos B=1.$$
And that's where I'm stuck.
| I got the same result!
Now take $A=30^{\circ}$, for example, and we see that our triangle is not equilateral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Estimate expected payoff of rolling a dice, with choice of rolling up to $50$ times. This is an extended question of the classical rolling dice and give face value question.
You roll a dice, and you'll be paid by face value. If you're not satisfied, you can roll again. You are allowed $k$ rolls.
In the old question, if you are allowed two rolls, then the expected payoff is $E[\text{payoff}] = 4.25$.
If you are allowed $3$ rolls, the expected payoff is $E[\text{payoff}] = 4.67$.
If you can roll up to $50$ times, you can calculate the payoff using the formula and get $E = 5.999762$, notice that after $5^\text{th}$ roll, your expected payoff will be greater than $5$, so you'll only stop once you roll $6$.
So my question here is, without exact calculation(using geometric process), how would you estimate how many $9$s are there in the answer? Or another way to ask will be, is the expected payoff bigger than $5.9$? bigger than $5.99$? etc.
| Let $E_k$ be the expected payoff, if you're allowed to roll $k$ times, with the rules as you've described them. We can compute $E_k$ recursively.
With just $1$ roll, you must take what you get, since there are no more rolls. The expected value is therefore $$E_1 = \frac{1+2+3+4+5+6}{6} = 3.5$$
With $2$ rolls, if your first roll is $4$, $5$, or $6$, you will keep it, otherwise you will reroll and get $E_1$ from your next (and last) roll. Therefore, \begin{align*}E_2 &= \frac{4+5+6}{6}+\frac{1}{2}E_1 \\
&= 2.5+\frac{1}{2}(3.5) = 4.25\end{align*}
With $3$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_2$ from your next two rolls. Therefore, \begin{align*}
E_3 &= \frac{5+6}{6}+\frac{2}{3}E_2\\
&= \frac{11}{6}+\frac{2}{3}(4.25) = 4.\overline{6}
\end{align*}
With $4$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_3$ from your next three rolls. Therefore, \begin{align*}
E_4 &= \frac{5+6}{6}+\frac{2}{3}E_3\\
&= \frac{11}{6}+\frac{2}{3}(4.\overline{6}) = 4.9\overline{4}
\end{align*}
With $5$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_4$ from your next three rolls. Therefore, \begin{align*}
E_5 &= \frac{5+6}{6}+\frac{2}{3}E_4\\
&= \frac{11}{6}+\frac{2}{3}(4.9\overline{4}) = 5.1\overline{296} = \frac{277}{54}
\end{align*}
Now, we have reached the point at which the recursion relation is stable. With more than $5$ rolls, you will always only keep the first roll if it is a $6$.
With $k$ rolls, $k>5$ if your first roll is $6$, you will keep it, otherwise you will reroll and get $E_{k-1}$ from the next $k-1$ rolls. Therefore,\begin{align*}
E_k &= \frac{6}{6}+\frac{5}{6}E_{k-1}\\
E_k &= 1+\frac{5}{6}E_{k-1}\tag{1}\\\
\end{align*}
Notice that $$E_5 = \frac{277}{54} = 6 - \frac{47}{54}$$
The solution to the recurrence relation in $(1)$, with initial value $E_5 = 6- 47/54$, is:
$$E_k = 6 - \left(\frac{47 \cdot 144}{5^5}\left(\frac{5}{6}\right)^k\right)$$
Therefore, in general, the maximum expected payoff that you can achieve, when allowed $k$ rolls of a six-sided die, for any $k$, is $$\boxed{\,\,E_k \,=\,\begin{cases}7/2 \qquad &\text{if}\,\,\,k=1\phantom{l^{l^{l^{\overline{l}}}}}\\ 17/4 \qquad &\text{if}\,\,\,k=2\phantom{l^{l^{l^{\overline{l}}}}}\\ 14/3 \qquad &\text{if}\,\,\,k=3\phantom{l^{l^{l^{\overline{l}}}}}\\ 89/18 \qquad &\text{if}\,\,\,k=4\phantom{l^{l^{l^{\overline{l}}}}}\\\\6-\displaystyle\frac{6768}{3125}\left(\displaystyle\frac{5}{6}\right)^k \qquad &\text{if}\,\,\,k\geq 5\phantom{l_{l_{l_{l_l}}}}\\ \end{cases}\,\,\,}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Calculate $\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$ I want to calculate $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$$
where $|z|=2$ is run counterclockwise.
What I did: $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}=\int_{|z|=2}\frac{1}{(z+1)(z^2+1)} $$
Using Cauchy's integral formula
$$f(-1)=\frac{1}{2\pi i}\int_{|z|=2} \frac{f(z)}{z+1}dz$$
where $f(z)=\frac{1}{z^2+1}$
Therefore $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}=\pi i$$
However I realized that $f$ is not holomorphic in $1,-i,i$ (the roots of $z^3+z^2+z+1$) and this made me doubt my result. What is the correct answer?
| The poles $-1,\,\pm i$ all have modulus $1<2$, so we get contributions from all of them. The poles are also all first-order. The $z=-1$ residue is $\frac{1}{(-1)^2+1}=\frac12$. The $z=\pm i$ residue is $\frac{1}{(1\pm i)(\pm 2i)}=\frac{-1\mp i}{4}$. So the integral is $2\pi i(\frac12+\frac{-1}{2})=0$.
| {
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} |
Let the reduced row echelon form of A = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$. Determine A if ... Let the reduced row echelon form of $A$ be $R$ = $\begin{bmatrix}1&-3&0&4&0&5\\ 0&0&1&3&0&2\\ 0&0&0&0&1&-1\\ 0&0&0&0&0&0\end{bmatrix}$.
Determine A if the first, third, and sixth columns of A are $\begin{bmatrix} 1\\-2\\-1\\3\end{bmatrix}$, $\begin{bmatrix} -1\\1\\2\\-4\end{bmatrix}$, and $\begin{bmatrix} 3\\-9\\2\\5\end{bmatrix}$, respectively.
Given this information, I know second column of $A$ is $-3$ times the first column (by looking at the first two columns of $R$). How can I solve for the remaining two columns?
$A$ = $\begin{bmatrix}
1 & -3 & -1 & ? & ? & 3 \\
-2 & 6 & 1 & ? & ? & -9 \\
-1 & 3 & 2 & ? & ? & 2 \\
3 & -9 & -4 & ? & ? & 5
\end{bmatrix}$.
| Let $C_{i}$ be the $i$th column. By observing $R$ we notice that $C_{4} = 3C_{3} + 4C_{1}$ and $C_{5} = -C_{6} + 2C_{3} + 5C_{1}$.
The idea is to find a linear combination of the other columns which each
have only one non-zero entry ,to then express the values in the column that we're evaluating. You did right for column $2$.
| {
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} |
Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?
| The squared distance from $(2,1)$ to an arbitrary point $(x,y)=(x,4x/3+20/3)$ of the straight line is
$$(x-2)^2+\left(\frac43x+\frac{20}{3}-1\right)^2=\frac{25}{9}x^2+\frac{100}{9}x+\frac{35}{9}.$$
As the vertex of that parabola occurs at $x=-2$, the square of the minimal distance is $25$, hence the distance is $5$. Now subtract $2$.
| {
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Find x that satisfies the equality (matrix determinant): This is the exercise.
Find the value of $x\in\mathbb{R}$ that satisfies
$$
\begin{vmatrix}
x & -1\\
3 & 1-x
\end{vmatrix} =
\begin{vmatrix}
1 & 0 & -3 \\
2 & x & -6 \\
1 & 3 & x-5
\end{vmatrix}
$$
This is what I've done.
\begin{align*}
\det (A) &= \det (B)\\
\Rightarrow\det(x-x^{2}+3) &= \det(x^{2}-2x)\\
\Rightarrow x-x^{2}+3&=x^{2}-2x\\
\Rightarrow -2x^{2}+3x+3 &=0
\end{align*}
I've also tried to find $x$ with the quadratic formula but the above equation of determinants isn't equal when I plug in the $x$ I found.
What's wrong? What step do I need to follow?
| There's nothing wrong – just that $x=\frac{3\pm\sqrt{33}}4$, so simplifying the determinants after substituting $x$ in might take some work. The determinants on both sides should evaluate to $\frac{9\mp\sqrt{33}}8$.
| {
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Checking the argument for calculation of the limit $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
In computing this limit, i have used following steps:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}(\sin(x)-\frac{1}{2})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}\sin(x)-\frac{\sqrt3}{2}\frac{1}{2}} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(\frac{\pi}{6})\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$ , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x-\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
from here by the use of $\lim_{\theta\to 0} \frac{\sin(\theta)}{\theta}$ and theorem for limits of composite functions, it can be proved that:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $2 \over \sqrt3$, which is incorrect.
Can anyone tell me, what i did wrong in this computation?
It is likely that the assumption in this step is wrong:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$ , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
If so, then can anyone tell me, what is wrong with this assumption?
| We can not replace $\cos\dfrac\pi6$ with $\cos x$ just because $x\to\dfrac\pi6$
Better use Prosthaphaeresis Formula $$\sin x-\sin\dfrac\pi6=2\sin\dfrac{x-\dfrac\pi6}2\cos\dfrac{x+\dfrac\pi6}2$$
Then apply $\lim_{h\to0}\dfrac{\sin h}h=1$
| {
"language": "en",
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Finding a polynomial whose roots are connected to the roots of a different polynomial Suppose we have a polynomial function $$f(x) =x^5-4x^4+3x^3-2x^2+5x+1$$ Function $f$ will have 5 roots which can be denoted by $a, b, c, d, e$. I was interested in trying to find a degree 10 polynomial whose roots are given by $abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde$. My idea was that we can relate the coefficients of the degree 10 polynomial to the coefficients of the degree 5 polynomial using Vieta's relations. However, I soon realised that this led to expressions that were extremely difficult to simplify and the method in general, was time-consuming. I was interested in knowing if general techniques exist to solve such problems or if brute is the only way to go about it.
Thanks
| $f(x) = x^5-4x^4+3x^3-2x^2+5x+1$
$f$ has $5$ roots donated by $a$, $b$, $c$, $d$ and $e$
The elementary symmetric functions of the roots are
$a+b+c+d+e = 4$
$de+ce+be+ae+cd+bd+ad+bc+ac+ab = 3$
$cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$bcde+acde+abde+abce+abcd = 5$
$abcde = -1$
Let $z = abc$, Computing the elementary symmetric functions of $z$ which are symmetric functions in $a,b,c,d,e$ and expressing them in terms of the elementary symmetric functions of $x$
Writing out the conjugates of $z$ shows it's a polynomial of degree $10$
$(z-abc)(z-abd)(z-acd)(z-bcd)(z-abe)(z-ace)(z-bce)(z-ade)(z-bde)(z-cde)$
Expand to express the elementary symmetric functions of $z$
$z^{10}-s_1z^9+s_2z^8-s_3z^7+s_4z^6-s_5z^5+s_6z^4-s_7z^3+s_8z^2-s_9z+s_{10} = 0$
$s_1 = cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$s_2 = {.............}$
This process is large, requires tremendous calculations so I'll skip the details
$s_8 =
(abcde)^4(cde^2+bde^2+ade^2+bce^2+ace^2+abe^2+cd^2e+bd^2e+ad^2e+c^2de+b^2de+a^2de+bc^2e+ac^2e+b^2ce+a^2ce+ab^2e+a^2be+bcd^2+acd^2+abd^2+bc^2d+ac^2d+b^2cd+a^2cd+ab^2d+a^2bd+abc^2+ab^2c+a^2bc +3( bcde+acde+abde+abce+abcd ) )$
$s_9 = (abcde)^5(de+ce+be+ae+cd+bd+ad+bc+ac+ab) = (-1)^53 = -3$
$s_{10} = (abcde)^6 = 1$
Therefore our polynomial in $z$ is
$z^{10}-2z^9+19z^8-112z^7+82z^6+97z^5-15z^4+58z^3+3z^2+3z+1 = 0$
| {
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the range of the vector dot product $\overrightarrow{P M} \cdot \overrightarrow{P N}$
$ABCDEF$ is a equilateral regular hexagon with $AB=2 \sqrt2$. Let $P$ be an arbitrarily point of the hexagon and $MN$ be a movable chord of the circumcircle of the hexagon with $MN=4$. What would the range of the vector dot product $\overrightarrow{P M} \cdot \overrightarrow{P N}$ be?
My attemption:
Set up an coordinate system as following:
Let the coordinate of $P$ be $(x,-\sqrt6)$ with $-\sqrt2 \leq x \leq \sqrt2$, and set the coordinate values of the points $M, N$ to be $(2\sqrt2 \cos \alpha,2\sqrt2 \sin \alpha), (-2\sqrt2 \sin\alpha,2\sqrt2 \cos \alpha)$ respectively, where $0 \leq \alpha \leq 2\pi$. Thereby, we obtain
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=x^2+4x\sin(\alpha-\frac{\pi}{4})+4\sqrt6\sin(\alpha+\frac{\pi}{4})+6.$$
Set $t=\alpha-\frac{\pi}{4}$, then
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=x^2+4x\sin t+4\sqrt6\cos t+6,$$
for $0 \leq t \leq 2\pi$ and $-\sqrt2 \leq x \leq \sqrt2$.
On one hand,
\begin{align*}
x^2+4x\sin t+4\sqrt6\cos t+6 \leq& 2+4(\sqrt2|\sin t|+\sqrt6 |\cos t|)+6\\
&=8+4(\sqrt2|\sin t|+\sqrt6 |\cos t|)\\
&\leq 8+8\sqrt2.
\end{align*}
Obviously, the maximum value is $8+8\sqrt2$.
How do I proceed ahead now? Any help would be appreciated.
| Continue with
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=
f(x, t)=x^2+4x\sin t+4\sqrt6\cos t+6,$$
Note that $f_x’=f_t’=0$ yields $(x,t)=(0,0),(0,\pi),(0,2\pi)$, which leads to extreme values $6\pm 4\sqrt6$.
Moreover, check the boundary values at $x=\pm\sqrt2$,
$$f(\pm\sqrt2,t)= 2\pm4\sqrt2 \sin t+4\sqrt6 \cos t+6
=8\pm8\sqrt2\sin(t+\theta)$$
Together, the range is
$$6-4\sqrt6 \le \overrightarrow{P M} \cdot \overrightarrow{P N}\le 8+8\sqrt2$$
| {
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Matrix, and Differential Equation Solution Verification I given the following system:\begin{equation}\mathbf{X'}=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\mathbf{X}\end{equation}
Every variable in the system is a matrix. I am then given that $\mathbf{X}$ is a column matrix.\begin{align}\frac{d}{dt}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}&=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}\end{align}
The excerise involves in verifing that $\mathbf{X}$ is a solution to this linear system. I am wondering if my steps are correct in assessing the problem that is given at hand.
My Steps
\begin{equation}\begin{pmatrix}\frac32e^{-\frac{3t}2}\\ -3e^{-\frac{3t}{2}}\end{pmatrix}=\begin{pmatrix}e^{-\frac{3t}2}+\frac12e^{-\frac{3t}2}\\ -e^{-\frac{3t}2}-2e^{-\frac{3t}2}\end{pmatrix}=\begin{pmatrix}\frac32e^{-\frac{3t}2}\\ -3e^{-\frac{3t}{2}}\end{pmatrix}\end{equation}
$\because$ the LHS equals the RHS $\therefore$ the solution proposed by $\mathbf{X}$ is a valid one.
| Your solution looks correct to me.
$$\begin{align}\frac{d}{dt}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}&=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}\end{align}$$
You can also write it this way:
$$\begin{align}\begin{pmatrix}-1 \\ 2\end{pmatrix} \frac{d}{dt}e^{-\frac{3t}2}=&\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}- 1\\2\end{pmatrix}e^{-\frac{3t}2}\end{align}$$
$$\begin{align}-\frac 32\begin{pmatrix}-1 \\ 2\end{pmatrix}=&\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}- 1\\2\end{pmatrix}\end{align}$$
So that you only have matrices with numbers !!
| {
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}} = -6$.
I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.
| Starting from J.C.Santos result
$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$
I'd multiply num and den by $\left(2x+\sqrt{x^2+3}\right)$ to get
$$\lim_{x\to1}\frac{\left(4x^2-x^2-3\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$
that is
$$\lim_{x\to1}\frac{3(x+1)(x-1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$
and then
$$-\lim_{x\to1}\frac{3(x+1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{\left(2x+\sqrt{x^2+3}\right)}=-\frac{24}{4}=-6$$
| {
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Doubt in solution of APMO 1998 Inequality problem Question -
Let $a, b, c$ be positive real numbers. Prove that
$$
\begin{array}{c}
\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt[3]{x y z}} \\
(\text { APMO } 1998)
\end{array}
$$
My doubt -
in pham kim hung secrets they proved like this -
Solution. Certainly, the problem follows the inequality
$$
\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}
$$
which is true by AM-GM because
$$
3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\left(\frac{2 x}{y}+\frac{y}{z}\right)+\left(\frac{2 y}{z}+\frac{z}{x}\right)+\left(\frac{2 z}{x}+\frac{x}{y}\right) \geq \frac{3 x}{\sqrt[3]{x y z}}+\frac{3 y}{\sqrt[3]{x y z}}+\frac{3 z}{\sqrt[3]{x y z}}
$$
now i did not understand how they got to this
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}$
in starting not in end???
when i expand LHS i get total 6 reciprocal terms and 2 cancelled from both sides but i did not understand how they cancel other 2 on RHS and remaining 3 terms on LHS.......
thankyou
| $$\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)
=2+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).$$
And each term in parentheses satisfies the inequality you have proved.
| {
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Real solutions for $\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|$ Do there exist real numbers $a,b,c,d$ such that
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|?$$
| Consider the points $M=(a,b)$ and $N=(c,d)$. By the triangle inequality
$$OM-ON\leq MN \leq MP+NP $$
which gives us
$$\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\leq MN\leq |a-c|+|b-d| $$
Since $$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<\sqrt{a^2+b^2}-\sqrt{c^2+d^2} $$ we have the reverse inequality:
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<|a-c|+|b-d| $$
| {
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$.
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values.
Thank you very much for helping.
| Consider a linear function $ f(x)=ax+b$
$$ f(x^2+x) = ax^2+ax+b$$
$$ f(x^2-3x+2)= ax^2-3ax +2a+b$$
$$ f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$
$$a=3, b=-4$$
$$ f(x) = 3x-4$$
$$f(2016)=6044$$
| {
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Show that $\frac{1}{x+2}$ is continuous for $(-2,0]$. I must use an $ε-δ$ proof to prove the function $\frac{1}{x+2}$ is continuous for $(-2,0]$.
I set $x \in (-2,0]$, and $y\in (-2,0]$ s.t $|x-y|<δ$.
I am having a hard time determining what precisely I should set $y$ to, as
$|\frac{1}{x+2}-\frac{1}{y+2}|=|\frac{(y+2)-(x+2)}{(x+2)(y+2)}|=\frac{|y-x|}{(x+2)(y+2)}$
and at this point I am stuck on how to proceed.
| Hint
For $\vert y-x \vert \le \frac{x +2}{2}$ you have
$$-\frac{x +2}{2} \le y-x \le \frac{x +2}{2}, \text{ hence } 2 + y \ge \frac{x}{2} + 1 = \frac{x+2}{2}$$
and therefore
$$0 \le \frac{1}{2+y} \le \frac{2}{2+x}$$
Finally
$$\left\vert\frac{1}{x+2}-\frac{1}{y+2}\right\vert=\left\vert\frac{(y+2)-(x+2)}{(x+2)(y+2)}\right\vert=\frac{|y-x|}{(x+2)(y+2)} \le \frac{2}{(x+2)^2} \vert y-x\vert$$
For $\epsilon >0$ given, $\delta = \min\{\frac{x+2}{2}, \frac{\epsilon (x+2)^2}{2}\}$ should do the job.
| {
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Showing values of x fall into a range or domain, when y is also part of the equation I am told to consider the relation:
$$\frac{x^2}{\:20}+\frac{y^2}{5} = 1$$
(i) Show that the values of x for which the relationship is defined are given by:$-\sqrt{20}<=\:x\:<=\:\sqrt{20}$
(ii) Similarly, find the values of y for which the relation is defined.
Do I need to figure out what y is algebraically first and then try values of x that fall into the domain provided? And the next part it is saying.....I have no idea.
| Your equation is
$$\frac{x^2}{20} + \frac{y^2}{5} = 1 \implies \frac{x^2}{20} = 1 - \frac{y^2}{5} \tag{1}\label{eq1A}$$
You have $\frac{y^2}{5} \ge 0 \implies -\frac{y^2}{5} \le 0 \implies 1 - \frac{y^2}{5} \le 1$. Thus, $\frac{x^2}{20} \le 1 \implies x^2 \le 20 \implies -\sqrt{20} \le x \le \sqrt{20}$. Similarly, since $\frac{x^2}{20} \ge 0 \implies 1 - \frac{x^2}{20} \le 1$, you have that $\frac{y^2}{5} \le 1 \implies y^2 \le 5 \implies -\sqrt{5} \le y \le \sqrt{5}$.
| {
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Fixed points of $2$-dimensional linear system
Consider $$\begin{aligned} \dot{x} &= ax + by\\ \dot{y} &= bx + ay\end{aligned}$$ where $a>0$ and $b<0$. Find the fixed points and classify them.
I haven't been able to find any fixed points other than the trivial $(x,y)=(0,0)$ solution. Any help would be appreciated!
| Fixed points are going to be whatever coordinates in the plane that make the derivative zero. If you set this up as a matrix differential equation you get
$$
\frac{d}{dt} \left [ \begin{array}{c}
x\\ y\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
b & a \\
\end{array} \right ]\left [ \begin{array}{c}
x\\ y\\
\end{array} \right ]
$$
and you can diagonalize this matrix as
$$
\left [ \begin{array}{cc}
a & b \\
b & a \\
\end{array} \right ] \;\; =\;\; \underbrace{\left [ \begin{array}{cc}
1/\sqrt{2} & -1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2} \\
\end{array} \right ]}_{ = P} \underbrace{\left [ \begin{array}{cc}
a + b & 0 \\
0 & a-b \\
\end{array} \right ]}_{=D} \underbrace{\left [ \begin{array}{cc}
1/\sqrt{2} & 1/\sqrt{2} \\
-1/\sqrt{2} & 1/\sqrt{2} \\
\end{array} \right ]}_{=P^T}.
$$
Notice that since $a>0$ and $b<0$ then the only way for this matrix to be singular is if $b = -a$. This would make the eigenvalue $a+b = 0$, and the corresponding eigenvector to this is $\left [ \begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array} \right ]^T$. Therefore we get two cases for fixed points:
$$
\textbf{Fixed Points} \;\; =\;\; \begin{cases}
\text{Span} \left \{\left [ \begin{array}{c}
1 \\ 1 \\
\end{array} \right ] \right \}, & \text{if} \; b = -a \\
\left [ \begin{array}{c}
0 \\ 0 \\
\end{array} \right ], & \text{otherwise} \\
\end{cases}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$
multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have
$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$
For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.
What I tried is subbing $1-2x=y$ which gives
$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$
$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$
I think I made it more complicated. Any help would be appriciated.
| $$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy=2\operatorname{Li}_4\left(\frac12\right)+\frac{133}{64}\ln2\zeta(3)-\frac{37}{768} \pi ^2 \log ^2(2)+\frac{77}{384}\log^42-\frac{3197\pi^4}{92160}-\frac{C^2}{2}-\frac{1}{8} \pi C \log (2)+\frac{3}{2}\text{Li}_4(2)+\frac{1}{2}\text{Li}_2(2)\log^22-\frac{3}{2}\text{Li}_3(2)\log2$$
$$\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\frac{47}{16}\operatorname{Li}_4\left(\frac12\right)+\frac{133}{64}\ln2\zeta(3)-\frac{61}{768} \pi ^2 \log ^2(2)+\frac{23}{96}\log^42-\frac{4367\pi^4}{92160}+\frac{1}{8} \pi C \log (2)+\frac{3}{2}\text{Li}_4(2)+\frac{1}{2}\text{Li}_2(2)\log^22-\frac{3}{2}\text{Li}_3(2)\log2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y)^2}{2(x^2+y^2)+5z^2}$$
following I want use C-S,But I don't Success
| Also, uvw helps.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition gives $$3u^2-2v^2=1,$$ which does not depend on $w^3$.
In another hand, we need to prove that $f(w^3)\geq0$, where $f$ is a concave function:
$f(w^3)\geq0$ is a quadratic inequality of $w^3$ with a coefficient before $w^6$ is equal to $-8$.
But the concave function gets a minimal value for an extreme value of $w^3$,
which happens in the following cases.
*
*$w^3=0$.
Let $z=0$.
Thus, after homogenization we can assume that $y=1$ and we need to prove that
$$\sum_{cyc}\frac{1}{5(x^2+y^2+z^2)-6xy}\leq\frac{1}{x^2+y^2+z^2}$$ or
$$\frac{1}{5x^2-6x+5}+\frac{2}{5x^2+5}\leq\frac{1}{x^2+1}$$ or
$$\frac{1}{5x^2-6x+5}\leq\frac{3}{5(x^2+1)}$$ or
$$5x^2-9x+5\geq0,$$ which is obvious;
*Two variables are equal.
Thus, after homogenization we can assume $y=z=1$ (for $y=z=0$ the inequality is obviously true),
which gives
$$\frac{2}{5x^2-6x+10}+\frac{1}{5x^2+4}\leq\frac{1}{x^2+2}$$ or
$$(x-1)^2(5x^2-2x+2)\geq0$$ and we are done!
Now we see that the starting inequality is true for any reals $x$, $y$ and $z$ such that $x^2+y^2+z^2=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $ Question -
Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$
\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4
$
My work -
first i multiply both numerator and denominator by $\sqrt{ab+1}$ and i apply CS in numerator but in end it does not work..
now i multiply both numerator and denominator by $\sqrt{a+1}$ and apply holder but it also fails..
i also try some substitutions but none of them work
In solution to this problem author apply am-gm and we need to prove
$(a+1)(b+1)(c+1)(d+1) \geq(a b+1)(b c+1)(c d+1)(d a+1)$
and he proves it by expanding , i understand his proof but can someone solve this problem using classic inequalities without using such a boring expansion ???
| Yes, your second inequality is also true.
Indeed, after expanding we need to prove that:
$$a+b+c+d+ac+bd+\sum_{cyc}abc\geq\sum_{cyc}a^2bd+abcd\sum_{cyc}ab+a^2b^2c^2d^2+abcd.$$
Now, by AM-GM $$4=a+b+c+d\geq4\sqrt[4]{abcd},$$ which gives $$abcd\leq1.$$
Thus, by AM-GM again we obtain:
$$ac+bd\geq2\sqrt{abcd}\geq abcd+a^2b^2c^2d^2.$$
Also, by AM-GM again $$\sum_{cyc}ab=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2=4.$$
Thus, by AM-GM again:
$$\sum_{cyc}abc\geq4\sqrt[4]{a^3b^3c^3d^3}\geq 4abcd\geq abcd\sum_{cyc}ab.$$
Id est, it's enough to prove that
$$\sum_{cyc}a^2bd\leq4,$$ which is true by AM-GM twice:
$$\sum_{cyc}a^2bd=(ab+cd)(ad+bc)\leq\left(\frac{ab+bc+cd+da}{2}\right)^2\leq4.$$
Done!
We can write this solution in one line:
$$\prod_{cyc}(1+ab)=(1+ab)(1+cd)(1+bc)(1+da)=$$
$$=(1+abcd+ab+cd)(1+abcd+bc+da)=$$
$$=(1+abcd)^2+(1+abcd)(ab+bc+cd+da)+(ab+cd)(bc+da)\leq$$
$$\leq1+2abcd+a^2b^2c^2d^2+\sum_{cyc}ab+abcd\sum_{cyc}ab+4=$$
$$=5+abcd+(abcd+a^2b^2c^2d^2)+\sum_{cyc}ab+abcd\sum_{cyc}ab\leq$$
$$\leq5+abcd+ac+bd+\sum_{cyc}ab+\sum_{cyc}abc=\prod_{cyc}(1+a).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Linear independence of complex basis of vectors. I understand that for a $\mathbb{C}^n$ as a real vector space, we choose $$\left\{\pmatrix{1\\0\\\vdots\\0},\pmatrix{\mathrm i\\0\\\vdots\\0},\pmatrix{0\\1\\\vdots\\0},\pmatrix{0\\\mathrm i\\\vdots\\0},\dots,\pmatrix{0\\0\\\vdots\\1},\pmatrix{0\\0\\\vdots\\\mathrm i}\right\}$$ as a standard basis. Now, how would I show linear independence of the basis? My first thought was to show that the determinant of the basis matrix is non-zero, but when looking at this basis I realise that there is no determinant because it is not a square matrix.
| See a little case for this, for example, to show the independence of
$$S = \left\{
\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} i \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ i \end{pmatrix}
\right\}$$
suppose that $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ can be written as
$$
\begin{pmatrix} 0 \\ 0 \end{pmatrix} =
a_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + b_1\begin{pmatrix} i \\ 0 \end{pmatrix} +
a_2\begin{pmatrix} 0 \\ 1 \end{pmatrix} + b_2\begin{pmatrix} 0 \\ i \end{pmatrix}
$$
for some choice of real numbers $a_1,b_1,a_2$ and $b_2$. Observe that the right hand side of this is
$$\begin{pmatrix} a_1+ib_1 \\ a_2 + ib_2 \end{pmatrix}$$
hence $a_1+ib_1 = 0$ and $a_2 + ib_2 = 0$. Of course, the latter implies that $a_1=b_1=a_2=b_2=0$, which means that the only way of writting $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ as a linear combination of the vectors in $S$ is the trivial combination. Thus, $S$ is linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Precalculus limits as x approaches negative infinity Example 1 $\displaystyle \lim _{x\rightarrow -\infty }\frac{\sqrt{9x^{2} +2}}{2x-9}$
Example 2
$\displaystyle \lim _{x\rightarrow -\infty }\frac{6x^{2} -x}{\sqrt{9x^{4} +7x^{3}}}$
I'm having difficulty understanding why in Example 1, we would divide the numerator and denominator by $-\sqrt{x^2}$, but in Example 2, we would divide the numerator and denominator by $+\sqrt{x^4}$. I know it has something to do with whether or not the exponent is even or odd, but I just need a little more clarification to get there. Thanks in advance.
| It would be far more convenient if, in the limit, almost every term were zero except for the "most important" term. For polynomials, the most important term is the leading term (highest power of $x$), so let's factor that one out of the two polynomials we can see. Recall that $\sqrt{a^2} = |a|$ (because the square root function cannot produce negative numbers) and as $x \rightarrow -\infty$, $|x| = -x$.
\begin{align*}
\lim _{x\rightarrow -\infty } \frac{\sqrt{9x^{2} +2}}{2x-9}
&= \lim _{x\rightarrow -\infty } \frac{\sqrt{9x^{2}(1 + \frac{2}{9 x^2})}}{2x(1-\frac{9}{2x})} \\
&= \lim _{x\rightarrow -\infty } \frac{|3x|\sqrt{1 + \frac{2}{9 x^2}}}{2x(1-\frac{9}{2x})} \\
&= \lim _{x\rightarrow -\infty } \frac{-3x\sqrt{1 + \frac{2}{9 x^2}}}{2x(1-\frac{9}{2x})} \\
&= \lim _{x\rightarrow -\infty } \frac{-3\sqrt{1 + \frac{2}{9 x^2}}}{2(1-\frac{9}{2x})} \\
&= \frac{-3\sqrt{1 + 0}}{2(1+0)} \\
&= \frac{-3}{2} \text{.}
\end{align*}
Notice that what we have done is equivalent to dividing by $-\sqrt{x^2}$ : the minus sign because $|x| = -x$ and the $\sqrt{x^2}$ to represent that the polynomial under the radical has degree $2$.
If you are ever working mechanically and happen to factor out a non-leading term, you will discover this in the step where you evaluate the limit at every term. You should only get one nonzero term in each of the numerator and denominator, but you will find that (at least) the leading term goes to infinity in this step. So back up and factor out the correct leading term.
In your second example, recall that $3x^2 \geq 0$ for all $x$, so $|3x^2| = 3x^2$ for any $x$, even those approaching $-\infty$.
\begin{align*}
\lim_{x\rightarrow -\infty} \frac{6x^{2} -x}{\sqrt{9x^{4} +7x^{3}}}
&= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{\sqrt{9x^{4}(1 + \frac{7x^{3}}{9x^4})}} \\
&= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{|3x^2| \sqrt{1 + \frac{7x^{3}}{9x^4}}} \\
&= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{3x^2 \sqrt{1 + \frac{7x^{3}}{9x^4}}} \\
&= \lim_{x\rightarrow -\infty} \frac{2(1 -\frac{x}{6x^2})}{\sqrt{1 + \frac{7x^{3}}{9x^4}}} \\
&= \frac{2(1 -0)}{\sqrt{1 - 0}} \\
&= 2 \text{.}
\end{align*}
What we have done is equivalent to dividing by $+\sqrt{x^4}$: the plus sign because $|x^2| = x^2$ and $\sqrt{x^4}$ to represent that the polynomial under the radical has degree $4$.
Notice that, really, once you have factored the leading terms and (if needed) migrated them outside the radicals (taking care about signs), you are interested in how much those terms cancel and what single power of $x$ remains. That power could be positive (larger degree in the numerator), zero (same degrees), or negative (larger degree in the denominator). The remaining power tells you what the function does in the limit and the "bunch of constants" that were left behind may scale that behaviour by a constant multiple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $p$ necessarily greater than $r$ in this number theory problem? From 1998 St. Petersburg City Mathematical Olympiad, presented in Andreescu & Andrica NT: SEP:
Let $n$ be a positive integer. Show that any number greater than $n^4/16$ can be written in at most one way as the product of two of its divisors having difference not exceeding $n$.
The presented solution is this:
Suppose, on the contrary, that there exist $a > c \ge d > b$ with $a-b \le n$ and $ab=cd>n^4/16$. Put $p=a+b, q=a-b, r=c+d,s=c-d.$ Now $$p^2-q^2=4ab=4cd=r^2-s^2>n^4/4.$$ Thus $p^2-r^2=q^2-s^2 \le q^2 \le n^2.$ But $r^2>n^4/4$ (so $r>n^2/2$) and $p>r\dots$
There is more to the solution, but that is irrelevant to my question. Why is $p>r$? It seems that this should be obvious, the way that it is presented. I notice that $p > r \Leftrightarrow p^2-r^2 > 0$, but I cannot prove that this is true. Manipulating the chain inequality $a>c\ge d > b$ has also done nothing for me.
| We have $a>c\ge d>b$, and $p=a+b$, $r=c+d$, $q=a-b$, $s=c-d$. These are all positive (as the question deals with only one sign for the divisors for uniqueness of solution, so we take the $+$ve route), although with the one exception of $s$ which could equal zero if $c=d$.
We cannot have $p<r$, since then $p^2-r^2<0$ which contradicts
$$(p^2-r^2)-(q^2-s^2)=4ab-4cd=0\tag{1}$$
If $p=r$ then $p^2-r^2=0$, which by $(1)$ means $q^2-s^2=0$, and so $q=s$ also. Hence
\begin{align*}
a+b&=c+d\tag{2}\\
a-b&=c-d\tag{3}
\end{align*}
or $a+b-c-d=a-b-c+d=0$. Adding/subtracting $(2)$ and $(3)$ gives either $a=c$ or $b=d$ respectively, contradicting $a>c\ge d>b$, and so we must have $p>r$.
| {
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"url": "https://math.stackexchange.com/questions/3667451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the all positive integer solutions $(a,b)$ to $\frac{a^3+b^3}{ab+4}=2020$.
Find the all positive integer solutions of given equation
$$\frac{a^3+b^3}{ab+4}=2020.$$
I find two possible solutions, namely $(1011,1009)$ and $(1009,1011)$, but the way I solve the equation was messy and I don't know if there are any other solutions.
Source: Turkey $1.$ TST for IMO $2020$
| Note that if $p\mid a^2-ab+b^2$, where $p$ is a prime natural number s.t. $p\equiv 2\pmod{3}$, then $p\mid a$ and $p\mid b$. For $p=2$, the claim is easily seen by inspection. Let now $p>2$. We prove by contradiction. Suppose that $p\nmid a$ or $p\nmid b$. It follows immediately that $p\nmid a$ and $p\nmid b$. Since
$$4(a^2-ab+b^2)=(2a-b)^2+3b^2\equiv0 \pmod{p},$$
we have $x^2\equiv-3\pmod{p}$, where $x=(2a-b)c$ if $c$ is an inverse of $b$ modulo $p$. Consequently,
$$\left(\frac{-3}{p}\right)=1.$$
By quadratic reciprocity,
$$\left(\frac{p}{3}\right)=\left(\frac{p}{-3}\right)=1.$$
Hence $p\equiv 1\pmod{3}$, which is a contradiction.
From $\frac{a^3+b^3}{ab+4}=2020$, we get
$$2020(ab+4)=a^3+b^3=(a+b)(a^2-ab+b^2).$$
If $101\mid a^2-ab+b^2$, then $101\mid a$ and $101\mid b$ by the paragraph above. Thus $101^3\mid (a+b)(a^2-ab+b^2)$, but clearly $101^3\nmid 2020(ab+4)$. Hence, $101\mid a+b$.
Similarly, $5\mid a+b$.
If $2\mid a^2-ab+b^2$, then $2\mid a$ and $2\mid b$. Write $a=2u$ and $b=2v$, then
$$1010(uv+1)=(u+v)(u^2-uv+v^2).$$
If $2\mid u^2-uv+v^2$, then $2\mid u$ and $2\mid v$. Therefore $8\mid (u+v)(u^2-uv+v^2)$, but clearly $8\nmid 1010(uv+1)$. Hence $2\mid u+v$. Consequently $4\mid a+b$. On the other hand, if $2\nmid a^2-ab+b^2$, then $4\mid a+b$ as $4\mid 2020$. In any case, $4\mid a+b$. This means $$a+b=4\cdot 5\cdot 101 \cdot k=2020 k$$ for some positive integer $k$.
Hence
$$k=\frac{a+b}{2020}=\frac{ab+4}{a^2-ab+b^2}.$$
Clearly $a\ne b$. Since $a+b=2020k$, $a\equiv b\pmod{2}$ so $(a-b)^2\ge 4$. If $(a-b)^2>4$, then $a^2-2ab+b^2>4$, making $ab+4<a^2-ab+b^2$, so
$$k=\frac{ab+4}{a^2-ab+b^2}<1,$$
which is a contradiction. Hence, $(a-b)^2=4$, so that $ab+4=a^2-ab+b^2$, making $k=1$. Thus we have $a+b=2020k=2020$ and $a-b=\pm\sqrt4=\pm2$. This gives
$$(a,b)=(1009,1011)\vee (a,b)=(1011,1009),$$
and both are solutions. In fact, these two solutions are the only integer solutions (positive or negative) to the required equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669889",
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"source": "stackexchange",
"question_score": "5",
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Is it possible to find an expression for $\frac{d^n}{dx^n}e^{-x^2}$? I am trying to find a general form to derivatives of the function $e^{-x^2}$. I tried to do it by finding a pattern for the first derivatives but with no success.
Any tip is welcome.
| Set $u(x)=e^{x}$, $v(x)=-x^2$ and use the famous Faà di Bruno's Formula.
$$
\frac{d^{n}}{d x^{n}} u(v(x))=
\sum_{m_1+m_2+\ldots+m_n=n}
\frac{n !}{m_{1} ! m_{2} ! \cdots m_{n} !}
\cdot
u^{\left(m_{1}+\cdots+m_{n}\right)}(v(x))
\cdot
\prod_{j=1}^{n}\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
$$
For $n=2$ we have
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!}
\cdot
u^{\left(m_{1}+m_{2}\right)}
\cdot
(v(x))
\cdot
\prod_{j=1}^{2}
\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
\\
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!}
\cdot
u^{\left(m_{1}+m_{2}\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{m_{1}}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{m_{2}}
\end{align}
The partitions of $m_1+m_2=2$ are $(m_1,m_2)=(2,0),(1,1),(0,2)$. Then
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\frac{2!}{2 ! 0!}
\cdot
u^{\left(2+0\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{2}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{0}
\\
+&
\frac{2!}{1 ! 1!}
\cdot
u^{\left(1+1\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{1}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{1}
\\
+&
\frac{2!}{0 ! 2!}
\cdot
u^{\left(0+2\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{0}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{2}
\end{align}
A most interessant formulation is
Faà di Bruno's determinant formula. If $u$ and $v$ are functions with a sufficient ber of derivatives, and $m \geq 1,$ then
$$
\frac{d^{m}}{d x^{m}} u(v(x))
=
\det
\left(
\begin{array}{cccccc}
\left(\begin{array}{c}
m-1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-1 \\
1
\end{array}\right) u^{\prime \prime} v & \left(\begin{array}{c}
m-1 \\
2
\end{array}\right) u^{\prime \prime \prime} v & \cdots & \left(\begin{array}{c}
m-1 \\
m-2
\end{array}\right) u^{(m-1)} v & \left(\begin{array}{c}
m-1 \\
m-1
\end{array}\right) u^{(m)} v \\
-1 & \left(\begin{array}{c}
m-2 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-2 \\
1
\end{array}\right) u^{\prime \prime} v & \cdots & \left(\begin{array}{c}
m-2 \\
m-3
\end{array}\right) u^{(m-2)} v & \left(\begin{array}{c}
m-2 \\
m-2
\end{array}\right) u^{(m-1)} v \\
0 & -1 & \left(\begin{array}{cc}
m-3 \\
0
\end{array}\right) u^{\prime} v & \cdots & \left(\begin{array}{c}
m-3 \\
m-4
\end{array}\right) u^{(m-3)} v & \left(\begin{array}{c}
m-3 \\
m-3
\end{array}\right) u^{(m-2)} v \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \left(\begin{array}{c}
1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
1 \\
1
\end{array}\right) v^{\prime \prime} v \\
0 & 0 & 0 & \cdots & -1 & \left(\begin{array}{c}
0 \\
0
\end{array}\right) u^{\prime} v
\end{array}
\right)
$$
where $u^{(i)}$ denotes $u^{(i)}(x)$ and $v^{k}$ is to be interpreted as $v^{(k)}(u(x))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ $(a>0)$
I just can analyse $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}=a\left(\frac{1}{1}-\frac{1}{1+a}+\frac{1}{2}-\frac{1}{2+a}+\frac{1}{3}-\frac{1}{3+a}...+\frac{1}{n}-\frac{1}{n+a}\right)$
Can anyone help me? Thanks
| Let's call the original sum $\lim_{n \to \infty} V_n$.
Asymptotic solution for $V_n$ with $a>0$: the first sum is Harmonic, so it is $\log n + O(1)$. The second sum is
$$
\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k+a} = \lim_{n \to \infty} S_n
$$
Each value in the (argument, value) tuple in this sum, $(1, \frac{1}{1+a}), (2, \frac{1}{2+a}) \ldots (n, \frac{1}{n+a})$ is in fact an area of a rectangle: $r_1 = (2-1) \times \frac{1}{1+a}, r_2= (3-2) \times \frac{1}{2+a} , \ldots r_n = (n+1-n) \times \frac{1}{n+a}$, so the sum $S_n$ is equa; to the sum of areas of these rectangles.
Next step is to compare each $r_j$ to the function $f(x) = \frac{1}{x+a}$. For each interval $[1,2], (2,3], \ldots (n, n+1)$ area of $r_j$ upper-bounds integral of $f(x)$:
$$
r_j > \int_{j}^{j+1} f(x)dx = \log \frac{j+1+a}{j+a}
$$
If we sum LHS and RHS of this inequality, we get the lower-bound on S_n:
$$
S_n > \sum_{j=1}^{n} > \sum_{j=1}^{n} \log \frac{j+1+a}{j+a} = \log (n+a+1) - \log (a+1)
$$
As a result, you get an upper bound on the original sum:
$$
V_n < H_n - \log (n+a+1) + \log (a+1) = \log (a+1) + \gamma + \log (\frac{n}{n+a+1}) = \log (a+1) + \gamma + O(\frac{1}{n})
$$
EDIT: got the sign wrong the first time. Also $\log \frac{n}{n+a+1} = -\log (1+\frac{a+1}{n}) \sim - \frac{a+1}{n} = O(\frac{1}{n})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n } = 1$? Question:
Find the limit
\begin{equation}
A = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }
\end{equation}
The series originated from the asymptotic analysis in this question. I can show that it converges. Numerical evaluation in Mathematica suggests that it is close to $1$.
I am just curious, is it possible to prove one of the following?
*
*$A > 1$
*$A = 1$
*$A < 1$
Perhaps one can think about the integral
\begin{equation}
\int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx
\end{equation}
Update 1:
Let $A_N = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }$, numerical evaluations up to $10,000$ shows $A_N < 1$
We can also plot the difference $1 - A_N$ as a function of $N$. To check the estimation $1- A_N \sim 0.3 / \ln N $ by one of the comments, we plot $( 1- A_N) \ln N$
update 2
Following Crostul's answer, I did an exercise to prove that the integral is also equal to $1$ in the limit $N \rightarrow \infty$
Relation between $A_N$ and the integral:
\begin{equation}
A_N = \sum_{n=1}^{N} \frac{1}{N} \frac{1}{ (1 + \frac{1}{N}) \ln ( N+ 1 ) - \frac{n}{N} \ln( \frac{n}{N} N ) }
\end{equation}
so to find $A_{N\rightarrow \infty}$, we may look at the limit
\begin{equation}
\lim_{N \rightarrow \infty} \int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx
\end{equation}
We can simplify the denominator
\begin{equation}
( 1 + \frac{1}{N} ) \ln ( N+1 ) - x \ln ( xN ) = \ln N ( 1 + \frac{b_N}{N} -x - \frac{x \ln x}{\ln N} )
\end{equation}
where
\begin{equation}
b_N = 1 + ( N + 1 )\frac{\ln (1 + \frac{1}{N})}{\ln N } = 1 + \frac{1}{\ln N} + o(\frac{1}{N} )
\end{equation}
So we study
\begin{equation}
I_N = \frac{1}{\ln N} \int_0^1 \frac{1}{ 1 + \frac{b_N}{N} - x - \frac{x \ln x}{\ln N} } dx
\end{equation}
Now use Crostul's relaxing trick
Define $f(x) = x + \frac{x \ln x}{ \ln N}$, the denominator is $f(1 + \frac{1}{N} ) - f(x)$. On one hand we have
\begin{equation}
f(1 + \frac{1}{N} ) - f(x) \ge f(1 + \frac{1}{N} ) - x \ge 1 + \frac{1}{N} - x
\end{equation}
where the second inequality holds for large enough $N$.
On the other hand, by mean value theorem
\begin{equation}
\frac{f( 1 + \frac{1}{N} ) - f(x) }{1 + \frac{1}{N} - x} = f'( y) \le f'( 1 + \frac{1}{N} ) = 1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }
\end{equation}
Hence
\begin{equation}
\frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }} \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx \le I_N \ln N \le \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx
\end{equation}
Both integrals on the left and right are $\ln N$. Therefore
\begin{equation}
\frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }} \le I_N \le 1
\end{equation}
Taking the limit $N\rightarrow$, we have $\lim_{N\rightarrow \infty } I_N = 1$ .
| I think if you set $b_{k+1} = a_{k+1} - a_k$, where $a_k = k \log k$ you could rewrite the denominator as $b_{k+1} = \log (1+\frac{1}{k})^k + \log (k+1) < 1+ \log (k+1) < 2 \log (k+1) < 2(k+1)$ hence the sum you have is $S_n = \sum_{k=1}^{n}\frac{1}{b_k} > \frac{1}{2}\sum_{k=1}^{n} \frac{1}{k+1}$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation which is of the form :
$$ \frac{1}{ {n \choose 2} } + \frac{1}{{n \choose 3} }+ \dotsb + \frac{1}{{n \choose n-2} }. $$
Notice that we may write $\frac{1}{ {n \choose 2} } = \frac{2! }{n(n-1)}$ and if keep doing the same for the other terms we obtain the following:
$$ \frac{ (n-2)! + (n-3)! (n-(n-2)) + (n-4)!(n-(n-2))(n-(n-3)) + \dotsb + 2! (n-3)! }{n!}, $$
which equals
$$ \frac{ (n-2)! + 2!(n-3)! + 3! (n-4)! + \dotsb + (n-3)! 2! }{n!} $$
and so this equals:
$$ \frac{1}{n(n-1)} + \frac{2}{n(n-1)(n-2)} + \dfrac{6}{n(n-1)(n-2)} + \dotsb + \dfrac{2}{n(n-1)(n-2) }. $$
But half of this terms are identical. Therefore, we are trying to sum up series of the form
$$\sum_{n \geq k} \frac{1}{(n-1)(n-2)(n-3)\dotso(n-k)} ,$$
which can be done by a telescoping trick, but it seems very formidable. Am I approaching this problem the right way? Any hints/suggestions?
| Partial Fractions and Telescoping Sums
$$
\begin{align}
\sum_{n=4}^\infty\sum_{k=2}^{n-2}\frac1{\binom{n}{k}}
&=\sum_{k=2}^\infty\sum_{n=k+2}^\infty\frac1{\binom{n}{k}}\tag1\\
&=\sum_{k=2}^\infty\frac{k}{k-1}\sum_{n=k+2}^\infty\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}\right)\tag2\\
&=\sum_{k=2}^\infty\frac{k}{k-1}\frac1{\binom{k+1}{k-1}}\tag3\\
&=\sum_{k=2}^\infty\frac2{(k-1)(k+1)}\tag4\\
&=\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)\tag5\\[3pt]
&=1+\frac12\tag6\\[9pt]
&=\frac32\tag7
\end{align}
$$
Explanation:
$(1)$: change order of summation
$(2)$: $\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}=\frac{\frac nk}{\binom{n}{k}}-\frac{\frac{n-k+1}k}{\binom{n}{k}}=\frac{\frac{k-1}k}{\binom{n}{k}}$
$(3)$: telescoping sum
$(4)$: $\binom{k+1}{k-1}=\binom{k+1}{2}=\frac{(k+1)k}2$
$(5)$: partial fractions
$(6)$: telescoping sum
$(7)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$
Let $x$ and $y$ be positive real numbers such that
$$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$
I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I should use it. Can anyone help?
Thanks!
| Cauchy-Schwarz implies $$((x+2)+2(y+2))\left(\frac 1{x+2}+\frac 1{y+2}\right)\geq (1+\sqrt{2})^2 $$ $$\Rightarrow x+2y+6\geq 3(1+\sqrt{2})^2,$$where equality is achieved when $$x+2=3(1+\sqrt{2}),y+2=\frac 3{\sqrt{2}}(1+\sqrt{2}).$$ This shows that the minimum of $x+2y$ is $3+6\sqrt{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An interesting property of a particular set of triples - multiplying two and adding the other always gives 1
Find all triples of real numbers such that multiplying any two in a
triple and adding the third always gives $1$.
When will this be the case? How can we find all such triples?
So far, I've let the numbers be $a$, $b$ and $c$.
Therefore, $ab+c$ etc must $= 1$ but how can I restrict the possibilities to find all possible triples?
I seem to think this has something to do with 1s and 0s, eg. $0,0,1$ or $0,1,1$
Many thanks guys!!
| You have three equations and three variables to solve for. The three equations are $$ab + c=1$$ $$ac +b=1$$ $$bc + a=1$$
Using the first equation and solving for $c$, we get $$c = 1-ab$$
This now reduces to two equations $$a(1-ab)+b=a+b-a^2b=1$$ $$b(1-ab)+a=a+b-ab^2=1$$
Subtracting the second equation from the first, I get $$-a^2b+ab^2=0$$ This can be factored as $$ab(b-a)=0$$
Which means that there are three possible cases $a=0, b=0,$ and $a=b$.
If $a=0$, it must also be true that $b=c=1$ in order to satisfy the original three equations. Similarly, if $b=0$, it must be true that $a=c=1$. If $a = b \not = 0$, there is the cubic $$2a-a^3=1$$ that must be solved. There are three solutions $$a= \frac{-1 \pm \sqrt{5}}{2}, 1$$
This means that all solutions to the problem are $(0, 1, 1), (1, 0, 1), (1, 1, 0), \left(\frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A tricky Inequality problem $ \dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1;\ a_1 , a_2 , \ldots , a_n > 0
$ show that $ \sqrt{a_1} + \sqrt{a_2} + \cdots + \sqrt{a_n} \ge (n-1) \left(\dfrac{1}{\sqrt{a_1}}+ \cdots + \dfrac{1}{\sqrt{a_n}}\right)$
I have tried AM - GM and this problem is from Inequalities a mathematical Olympiad approach.
| Here's a method that only involves algebraic manipulations and does not require the AM-GM inequality or any other more advanced inequalities. Not the most elegant of solutions however, but I believe it works.
$\dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1 \Rightarrow \left(1-\frac{a_1}{1+a_1}\right)+\left(1-\frac{a_2}{1+a_2}\right)+...+\left(1-\frac{a_n}{1+a_n}\right)=1$
Hence, we have: $\dfrac{a_1}{1+a_1}+\dfrac{a_2}{1+a_2}+...+\dfrac{a_n}{1+a_n}=n-1$.
Denote the inequality that is to be proven by $(1)$.
Now, $$(1) \iff \sum_{i=1}^{n}\sqrt{a_i}-(n-1)\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}\ge0 $$
We use a small trick and write the above inequality as:
$$(1) \iff \sum_{j=1}^{n}\frac{1}{1+a_j} \sum_{i=1}^{n}\sqrt{a_i}-\sum_{j=1}^{n}\frac{a_j}{1+a_j}\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}\ge0$$
$$ \iff \sum_{j=1}^{n}\sum_{i=1}^{n}\left(\frac{1}{1+a_j}\sqrt{a_i}-\frac{a_j}{1+a_j}\frac{1}{\sqrt{a_i}}\right)\ge 0$$
$$ \iff \sum_{j=1}^{n}\sum_{i=1}^{n}\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}} \ge 0 \hspace{70pt}$$
Note that, when $i=j$, $a_i-a_j=0$. Hence:
$$(1) \iff \sum_{i>j} \left( \frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}+\frac{a_j-a_i}{(1+a_i)\sqrt{a_j}} \right) \ge 0$$
$$\hspace{70pt} \iff \sum_{i>j} \frac{(a_i-a_j)(1+a_i)\sqrt{a_j}+(a_j-a_i)(1+a_j)\sqrt{a_i}}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$ \iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right) \left(\sqrt{a_i}+\sqrt{a_j}\right) (1+a_i) \sqrt{a_j} - \left(\sqrt{a_i}-\sqrt{a_j}\right) \left(\sqrt{a_i}+\sqrt{a_j}\right) (1+a_j) \sqrt{a_i}}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)\left( \sqrt{a_i}+\sqrt{a_j} \right) \left[(1+a_i)\sqrt{a_j}-\sqrt{a_i}(1+a_j)\right]}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)\left( \sqrt{a_i}+\sqrt{a_j} \right) \left[ - \left(\sqrt{a_i}-\sqrt{a_j}\right) + \sqrt{a_i}\sqrt{a_j}\left(\sqrt{a_i}-\sqrt{a_j}\right) \right]}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
$$\iff \sum_{i>j} \frac{\left(\sqrt{a_i}-\sqrt{a_j}\right)^2\left(\sqrt{a_i}+\sqrt{a_j}\right)\left(\sqrt{a_i}\sqrt{a_j}-1\right)}{(1+a_i)\left(1+a_j\right)\sqrt{a_i}\sqrt{a_j}} \ge 0$$
Indeed, it suffices for us to prove that $\sqrt{a_i}\sqrt{a_j} \ge 1$. We argue as follows:
$$1 \ge \frac{1}{1+a_i}+\frac{1}{1+a_j} \Rightarrow 1 \ge \frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j} \Rightarrow 1+a_i+a_j+a_ia_j \ge 2+a_i+a_j$$
$$\Rightarrow a_ia_j \ge 1 \Rightarrow \sqrt{a_i}\sqrt{a_j} \ge 1, \hspace{50pt}$$ and we are done, since all the terms in our sum are shown to be non-negative.
| {
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"url": "https://math.stackexchange.com/questions/3700783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$. I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:
The minimum distance $r$ (which will be the radius of the circle) between the origin and $y=e^{-x^2}$ can be found by finding the critical numbers of the derivative of the distance function.
\begin{align}
r&=\sqrt{x^2+(e^{-x^2})^2}\\
\frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\
0&=2x(1-2e^{-2x^2})\\
x&=0\quad\text{(obviously inadmissible, or)}\\
x&=\sqrt{\frac12\log2}\\
\\
r&=\sqrt{\frac12\log2+e^{-\log2}}\\
r^2&=\frac12\log2+\frac12\\
\implies\quad x^2+y^2&=\frac12\left(1+\log2\right)
\end{align}
as desired.
The relationship between a circle and an ellipse is like that of a square and a rectangle: given a set perimeter, the more square the rectangle, the larger the area. But since this question is based on the curve $y=e^{-x^2}$ and not on any set perimeter, it does not seem like the same conclusion can be drawn. I would need something stronger (perhaps based on concavity?) to show whether $x^2+y^2=\frac12(1+\log2)$ is the largest ellipse or not.
| We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $y = e^{-x^2}$ as well as a single point of intersection in the first quadrant. That is to say, $$b \sin \theta = e^{-(a \cos \theta)^2}$$ has a unique solution for $\theta \in (0, \pi/2)$, and at this point, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{b}{a} \cot \theta = -2x(\theta)e^{-x(\theta)^2} = -2a (\cos \theta )e^{-(a \cos \theta)^2}.$$ Consequently, $$-\frac{b}{a} \cot \theta = -2ab \cos \theta \sin \theta,$$ or $$\sin \theta = \frac{1}{a \sqrt{2}}.$$ Note if $a < 1/\sqrt{2}$, no such angle exists. The ellipse is "too narrow"--this is due to the fact that the point of tangency is at $(x,y) = (0,1)$. At the point of tangency, we also have $$\cos \theta = \sqrt{1 - (2a^2)^{-1}}$$ so that we now have $$\frac{b}{a \sqrt{2}} = e^{1/2 - a^2}$$ or $$b = a e^{1/2 - a^2} \sqrt{2}.$$ Finally, we seek to maximize the area of this family of ellipses parametrized by $a$. Since the area is proportional to $ab$, we need to maximize $$f(a) = a^2 e^{1/2-a^2}.$$ Computing the derivative with respect to $a$ and solving for critical values, we get $$0 = \frac{df}{da} = 2(a-1)a(a+1)e^{1/2-a^2},$$ hence $$a = 1$$ is the unique solution, with $b = \sqrt{2/e}$ and the ellipse has equation $$\frac{x^2}{1} + \frac{y^2}{2/e} = 1.$$ The area of this ellipse is simply $$\pi a b = \pi \sqrt{\frac{2}{e}}.$$ This is obviously not a circle.
For your enjoyment, I have included an animation of the family of ellipses $$\frac{x^2}{a^2} + \frac{y^2}{2a^2 e^{1-2a^2}} = 1,$$ for $a \in [1/\sqrt{2},2]$:
| {
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"url": "https://math.stackexchange.com/questions/3703410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$ For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$$
My proof$:$
$$4(a+b+c) \prod (a+b)^2 (\text{LHS}-\text{RHS})$$
$$=\prod (a+b) \Big[\sum\limits_{cyc} (ab+bc-2ca)^2+(ab+bc+ca)\sum\limits_{cyc} (a-b)^2 \Big]$$
$$+(a+b+c)(a-b)^2 (b-c)^2 (c-a)^2 \geqq 0$$
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\frac{\sum\limits_{cyc}(a^3+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b)}+\frac{3}{4}\geq\frac{(ab+ac+bc)\sum\limits_{cyc}(a+b)^2(a+c)^2}{\prod\limits_{cyc}(a+b)^2}$$ or
$$\frac{27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3}{9uv^2-w^3}+\frac{3}{4}\geq\frac{3v^2\sum\limits_{cyc}(a^2+3v^2)^2}{(9uv^2-w^3)^2}$$ or
$$4(9u^3-6uv^2+w^3)(9uv^2-w^3)+(9uv^2-w^3)^2\geq$$
$$\geq4v^2((9u^2-6v^2)^2-2(9v^4-6uw^3)+6(9u^2-6v^2)v^2+27v^4)$$ and since $$-4+1<0,$$ we see that our inequality it's $f(w^3)\geq0,$ where $$f(w^3)=-w^6+A(u,v^2)w^3+B(u,v^2),$$ which says that $f$ is a concave function.
Thus, by $uvw$ (see here https://artofproblemsolving.com/community/c6h278791 ) it's enough to prove our inequality in two following cases:
*
*$w^3\rightarrow0^+$;
*Two variables are equal (in this case it's enough to assume $b=c=1$).
But in the last case we obtain:
$$(a-1)^2\geq0,$$ which says that our inequality is true even for any reals $a$, $b$ and $c$ such that $\prod\limits_{cyc}(a+b)\neq0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.
| $$y'+\frac{xy}{1+x^2} =x$$
$$2x\dfrac {dy}{dx^2}+\frac{xy}{1+x^2} =x$$
Substitute $u=x^2$
$$2\dfrac {dy}{du}+\frac{y}{1+u} =1$$
$$2\sqrt {1+u}{y'}+\frac{y}{\sqrt {1+u}}=\sqrt {1+u}$$
$$(2\sqrt {1+u}{y})'=\sqrt {1+u}$$
Integrate:
$$2\sqrt {1+u}{y}=\int \sqrt {1+u}du$$
$$\sqrt {1+u}{y}=\frac 13({1+u})^{3/2}+K$$
$${y}=\frac 13({1+x^2})+\dfrac K {\sqrt {1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| $12(8^x+27^x)=14(12^x+18^x)$
$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$
$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$
We put :$t=(\frac{3}{2}) ^x$
$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$
Multiply by $t$
$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$
We can see $-1$ is a solution of the equation
So:after division by $t+1$ we see
$ t^2 - \frac{13}{6}x+1=0$
$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $
So :
$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $
$\Rightarrow $ $x_1=1$
And
$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $
$\Rightarrow $ $x_2=-1$
Finally
$x=-1$ or $ x= 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$ Let $m$ and $n$ be natural numbers. Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$
I interpreted this statement as $$\frac{m}{n} < \sqrt{2} \rightarrow \exists \frac{m'}{n'} \in \mathbb{Q} \left( \frac{m}{n} < \frac{m'}{n'} \wedge \frac{m'}{n'} < \sqrt{2} \right)$$ The contrapositive would then be $$\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right) \rightarrow m/n \geq \sqrt{2}$$
Partial Proof: Suppose that $\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right)$. Now suppose that $\frac{m}{n} < \sqrt{2}$. Since $\frac{m}{n} \in \mathbb{Q}$ and $\frac{m}{n} < \sqrt{2}$, we have $\frac{m}{n} \leq \frac{m}{n}$...
I'm just wondering why I was not able to draw a contradiction at the end. Is there something wrong with how I interpreted the statement?
| Here is a somewhat different approach: assume that $\dfrac mn < \sqrt 2$. Then
\begin{align*}
m^2 &< 2n^2\\
2m^2 &< 4n^2 \\
2m^2 + 3mn &< 3mn + 4n^2 \\
m(2m + 3n) &< n (3m + 4n) \\
\frac mn &< \frac{3m + 4n}{2m + 3n}
\end{align*}
and
\begin{align*}
m^2 &< 2n^2\\
9 m^2 - 8m^2 &< 18 n^2 - 16n^2 \\
9m^2 + 16n^2 &< 8m^2 + 18n^2 \\
9m^2 + 24 mn + 16 n^2 &< 8m^2 + 24mn + 18n^2 \\
(3m + 4n)^2 &< 2 (2m + 3n)^2 \\
\frac{(3m+4n)^2}{(2m+3n)^2} &< 2 \\
\frac{3m+4n}{2m+3n} &< \sqrt{2}
\end{align*}
so that
$$ \frac mn < \frac{3m+4n}{2m+3n} < \sqrt 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
To find $A$ given that $2I + A +A^2 = B$ where $B$ is given. How to find a matrix $A$ such that the following holds:
$$2I + A +A^2 = B,$$ where the matrix $B$ is given. I tried with char poly of $B$ but not getting any idea.
Note that it is also given that $B$ is invertible.
P.S. $B = \begin{pmatrix}-2&-7&-4\\ \:12&22&12\\ \:-12&-20&-10\end{pmatrix}$.
| According to the other posts, we may assume that $B=\begin{pmatrix}4&1&0\\0&4&0\\0&0&2\end{pmatrix}$. Note that $B$ is cyclic and that $AB=BA$; thus $A$ is a polynomial in $B$: $A=aI_3+bB+cB^2$.
Then $A$ is in the form -considered by copper.hat-
$A=\begin{pmatrix}u&p&0\\0&u&0\\0&0&v\end{pmatrix}$, where $2+u+u^2=4,2+v+v^2=2$. The sequel is easy; we obtain $4$ solutions
EDIT. We can also use the Mostafa's trick.
Since $B-7/4I_3$ is cyclic invertible, it admits $2^k=4$ square roots (where $k$ is the number of its distinct eigenvalues). In the same way as above, $\sqrt{B-7/4I_3}$ is in the form
$C=\begin{pmatrix}u&p&0\\0&u&0\\0&0&v\end{pmatrix}$, where $u^2=4-7/4,v^2=2-7/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Asymptotic behavior of recursive sequence Suppose, the real sequence
$x_{n+1}=\frac{1}{2}(x_n+\sqrt{x_n^2+c})$ with c>0 is given.
Find the asymptotic behavior of this sequence.
I have shown, that this sequence goes to infinity as $n\to\infty$ per contradiction. I guess, that it holds
$x_n \approx \frac{1}{2}\sqrt{cn}$
which is motivated by
$x_{n+1}^2=\frac{1}{2}x_n^2(1+\sqrt{1+\frac{c}{x_n}})+\frac{1}{4}c \approx x_n^2 + \frac{1}{4}c$
where I used the approximation $\sqrt{1+\frac{c}{x_n}} \approx 1$ as $n \to \infty$.
I have problems, to turn this into a formal proof, since it holds for the error term for my approximation
$\frac{1}{2}x_n^2(1-\sqrt{1+\frac{c}{x_n}}) \to \infty$
| You've already derived that
\begin{align}
x_{n+1}^2 - x_n^2 &= \frac 12 \left(\sqrt{1 + \frac{c}{x_n^2}} -1 \right) + \frac c4 \\
&= \frac{c}{2x_n^2} \frac{1}{\sqrt{1 + \frac{c}{x_n^2}} +1} + \frac c4.
\end{align}
Taking $n \to \infty$ we obtain
\begin{equation}
\lim\limits_{n\to\infty}\left[x_{n+1}^2 - x_n^2 \right] = \frac c4.
\end{equation}
Let $\epsilon > 0$ be arbitrary. Then there exists $N \in \mathbb{N}$ such that for all $n \geq N$ it holds that
\begin{equation}
\left| x_{n+1}^2 - x_n^2 - \frac c4\right| < \epsilon.
\end{equation}
We can then write
\begin{align}
\left|\frac{x_n^2}{n} - \frac c4\right| &= \left|\frac{(x_n^2 - x_{n-1}^2 - \frac c4) + ... + (x_{N+1}^2 - x_N^2 - \frac c4)}{n} + \frac{x_N^2}{n} - \frac{cN}{4n}\right| \\
&\leq \frac{(n-N)\epsilon}{n} + \frac{x_N^2}{n} + \frac{cN}{4n}.
\end{align}
Taking $n \to \infty$ yields
\begin{equation}
\limsup\limits_{n \to \infty} \left|\frac{x_n^2}{n} - \frac c4 \right| \leq \epsilon.
\end{equation}
As $\epsilon$ was arbitrary we get
\begin{equation}
\lim\limits_{n \to \infty} \frac{x_n^2}{n} = \frac c4
\end{equation}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?
| A combinatorial proof is in order!
The given expression is the number of ways to choose an even-sized subset of $n$ people, and promote one of them to be the leader. Notice that $n\cdot 2^{n-2}$ is the number of ways to choose the leader first, then choose an odd-sized subset from the remaining $n-1$ people (as the number of odd-sized subsets is the same as the number of even-sized ones). Hence the two sides are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
| $$f(x)=\frac{x+2}{(x+1)^2}$$
$$f'(x)=\frac{(x+1)-2(x+2)}{(x+1)^3}$$
$$=\frac{-x-3}{(x+1)^3}$$
so
$$f((-\infty,-1))=[f(-3),+\infty)$$
and
$$f((-1,+\infty))=(0,+\infty)$$
thus, the range is $$[-\frac 14,+\infty).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital Evaluating
$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.
| Yet another approach: as it's an even function, assume $x>0$. Cut a sector in a radius-$\sqrt{2}$ angle subtending $\pi x$ radians at the centre so $\pi x-\sin\pi x$ is the area in the sector outside the triangle with the same vertices. We'll approximate the arc as a parabola, in Cartesian coordinates with the line segment of the same endpoints a part of the $X$-axis, with extrema at $X=\pm\sqrt{2}\sin\frac{\pi x}{2}\sim\pm\frac{\pi x}{\sqrt{2}}$. The peak is at$$X=0,\,Y=\sqrt{2}(1-\cos\frac{\pi x}{2})=2\sqrt{2}\sin^2\frac{\pi x}{4}\sim\frac{\pi^2x^2\sqrt{2}}{8}.$$At leading order, the parabola is $Y=\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)$, so the area below it is$$\int_{-\pi x/\sqrt{2}}^{\pi x/\sqrt{2}}\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)dX=\frac{\pi^3x^3}{6}.$$So for small $x$,$$\pi x-\sin\pi x\sim\frac{\pi^3x^3}{6}\implies\frac{1}{\sin\pi x}-\frac{1}{\pi x}\sim\frac{\pi^2x^2}{6\sin\pi x}\sim\frac{\pi x}{6}\implies\frac{1}{\sin^2\pi x}-\frac{1}{\pi^2x^2}\sim\frac{\pi x}{2}\cdot\frac{2}{\pi x}=\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \sin ^{2} x+\cos ^{2} x=1$
$t^{2}+2-2 t^{2}+3 t \sqrt{1-t^{2}}=0$
$\left(t^{2}+2\right)^{2}=9 t^{2}\left(1-t^{2}\right)$
$t^{4}+4 t^{2}+h=9 t^{2}-9 t^{4}$
$10 t^{4}-5 t^{2}+4=0$
So the number of solution must be 4
P.s- Any other approach will be greatly appreciated!
correct me if I am wrong
| Here is another way to reach the goal,
The equation reduces to:
$$\begin{array}{l}
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+\cos ^{2} x+\sin x \cos x=0 \\
(\sin x+\cos x)^{2}+\cos x(\sin x+\cos x)=0
\end{array}$$
\begin{array}{l}
(\sin x+\cos x)(\sin x+2 \cos x)=0 \\
\Longrightarrow \tan x=-1 \text { and } \tan x=-2
\end{array}
$\tan x$ has a period of $\pi,$ Hence, it takes each value twice in an interval of $2 \pi .$ So the answer is 4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c=0, ab+bc+ca=1$ and $abc=1,$ then find the value of $\frac ab+\frac bc+\frac ca.$ Clearly $a, b, c$ are the roots of the cubic equation: $x^3+x-1=0\tag{1}.$ We have to find:
\begin{align}
\frac ab+\frac bc+\frac ca&=\frac{a^2c+b^2a+c^2b}{abc}\\\\
&=a^2c+b^2a+c^2b\\\\
&=p,\text{ say}.
\end{align} ($p$ is not a symmetric function of the roots.)
Now let: $q=ac^2+ba^2+cb^2.$ Then we have:
$0=\left(\sum ab\right)\left(\sum a\right)=p+s+3abc.$ This gives
$p+q=-3abc=-3.$
To find $p$ I multiplied $p$ and$q$ and obtained:
$pq=\sum a^3b^3+abc\left(\sum a^3\right)+3a^2b^2c^2.$
Now since $a, b, c$ are the roots of the Eq. $(1),$ so we can write:
$pq=\sum(1-a)(1-b)+abc\left[\sum(1-a)\right]+3(abc)^2=3-2\sum a+\sum ab+abc\left[3-\sum a\right]+3(abc)^2=3-0+1+1×(3-0)+3×1^2=10.$
This implies $p, q$ are the roots of the quadratic: $\color{green}{t^2+3t+10=0.}$ Which on solving gives
$\color{green}{t=\frac{-3\pm i\sqrt{31}}2\tag*{}.}$ Now my actual question is: between these two values of $t$ which one is $p$ and which one is $q$ ?
Please suggest. Thanks in advance.
| Both, $p$ and $q$, are solutions. For instance if you change $a,b,c$ for $a,c,b$, the problem is the same, but $p$ goes to $q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Question regarding the Quotient Rule, How does the textbook reach this intermediate step? $$\begin{align*}
\left(\frac{f(x)}{g(x)}\right)’ &= \frac{(x-3)^{1/3}\frac{1}{2}(x+2)^{-1/2}}{(x-3)^{2/3}} - \frac{(x+2)^{1/2}\frac{1}{3}(x-3)^{-2/3}}{(x-3)^{2/3}}\\
&= \frac{(x-3)^{-2/3}(x+2)^{-1/2}}{(x-3)^{2/3}}\cdot\left[\frac{1}{2}(x-3) - \frac{1}{3}(x+2)\right]
\end{align*}$$
Can someone advise how they got to the step after the second equality sign? Step by Step would be most helpful.
| $$\dfrac{(x-3)^{1/3}\frac12(x+2)^{-1/2}}{(x-3)^{2/3}}-\dfrac{(x+2)^{1/2}\frac13(x-3)^{-2/3}}{(x-3)^{2/3}}$$
$$=\dfrac{(x-3)\color{blue}{(x-3)^{-2/3}}{\frac12\color{blue}{(x+2)^{-1/2}}}-(x+2)\color{blue}{(x+2)^{-1/2}}\frac13\color{blue}{(x-3)^{-2/3}}}{(x-3)^{2/3}}$$
$$=\dfrac{\color{blue}{(x-3)^{-2/3}(x+2)^{-1/2}}}{(x-3)^{2/3}}\left[\frac12(x-3)-\frac13(x+2)\right].$$
Let me know if you need further details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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integral calculation mistake I try to solve this:
$
\int_{0}^{\pi /2}\sin x \cos x\sqrt{1+\cos^{2}x } dx
$
This is what I do:
$ \cos x = t; -\sin x dx = dt; -\sqrt{1-u^{2}} dt $
$ \frac{-\sqrt{1-t^{2}}*t*\sqrt{1+t^{2}}}{-\sqrt{1-t^{2}}} dt $
$-\int_{0}^{1} t * \sqrt{1+t^{2}} dt$
$1+t^2 = a; 2tdt = da; tdt = da/2$
$-\int_{0}^{2}\sqrt{a}da = - \frac{2\sqrt{2}}{3}$
But the answer is $\frac{2\sqrt{2}}{3} - \frac{1}{3}$
Where do I make the mistake
| After your first substitution you should have $-\int_{1}^{0}t\sqrt{1+t^{2}}dt$= $\int_{0}^{1}t\sqrt{1+t^{2}}dt$.
The limits for your second substitution should be $a=1$ at ($t=0$) and $a=2$ (at $t=1$) so we have $\frac{1}{2}\int_{1}^{2} \sqrt{a}da=\frac{1}{3}a^{\frac{3}{2}}|^{a=2}_{a=1}=\frac{2\sqrt{2}}{3}-\frac{1}{3}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
-8 \\
1
\end{bmatrix} = \begin{bmatrix}
3 \\
8
\end{bmatrix}$$
and
$$M\begin{bmatrix}
1 \\
5
\end{bmatrix} = \begin{bmatrix}
-8 \\
7
\end{bmatrix}.$$
Evaluate $$M\begin{bmatrix}
1 \\
1
\end{bmatrix}.$$
What is the question essentially asking? Isn't this just a matrix equation?
| Note that the given input vectors form a basis of $\Bbb{R}^2$. So we can express any vector in $\Bbb{R}^2$ in terms of the given vectors such as:
$$\begin{bmatrix}a\\b\end{bmatrix}={\small\left(\frac{b-5a}{41}\right)}\begin{bmatrix}-8\\1\end{bmatrix}+{\small \left(\frac{a+8b}{41}\right)}\begin{bmatrix}1\\5\end{bmatrix}. \tag{1}$$
In particular,
$$\begin{bmatrix}1\\1\end{bmatrix}={\small\frac{-4}{41}}\begin{bmatrix}-8\\1\end{bmatrix}+{\small\frac{9}{41}}\begin{bmatrix}1\\5\end{bmatrix}.$$
Thus
\begin{align*}
M\begin{bmatrix}1\\1\end{bmatrix}& ={\small \frac{-4}{41}}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \frac{9}{41}}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\
&={\small \frac{-4}{41}}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small\frac{9}{41}}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\
&=\begin{bmatrix}\frac{-84}{41}\\\frac{31}{41}\end{bmatrix}
\end{align*}
Generalization:
In fact, we can answer more generally as to what will $M$ do to any vector in $\Bbb{R}^2$. From equation (1)
\begin{align*}
M\begin{bmatrix}a\\b\end{bmatrix}&={\small\left(\frac{b-5a}{41}\right)}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\
&={\small\left(\frac{b-5a}{41}\right)}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\
&={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix}
\end{align*}
This also helps us find $M$ as
$$M\begin{bmatrix}a\\b\end{bmatrix}={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix}=\begin{bmatrix}\frac{-23}{41}&\frac{-61}{41}\\ \frac{-33}{41}&\frac{64}{41}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with proof of Euler's criterion Problem
This problem is about finding square roots modulo a prime $p$.
(a) Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.
An integer $x$ is called a square root of $n$ mod $p$ when $x^2 \equiv n \pmod p$. An integer with a square root is called a square mod $p$. Assume that $p$ is an odd prime and $n \not \equiv 0 \pmod p$. It turns out there is a simple test we can perform to see if $n$ is a square mod $p$:
Euler's Criterion
i. If $n$ is a square modulo $p$, then $n^{(p-1)/2} \equiv 1 \pmod p$.
ii. If $n$ is not a square modulo $p$, then $n^{(p-1)/2} \equiv -1 \pmod p$.
(b) Prove Case (i) of Euler's Criterion.
(c) Prove Case (ii) of Euler's Criterion.
Solution
(a) Prove both directions of the if and only if:
*
*Assume that $x^2 \equiv y^2 \pmod p$. Then:
$$\begin{align}
&x^2 - y^2 \equiv 0 \pmod p \\
&\Rightarrow(x+y)(x-y) \equiv 0 \pmod p \\
&\Rightarrow p\text{ | }(x+y)(x-y) \\
&\Rightarrow [p\text{ | }(x+y)]\text{ or }[p\text{ | }(x-y)] &\text{(since }p\text{ is prime)} \\
&\Rightarrow [x+y \equiv 0 \pmod p]\text{ or }[x-y \equiv 0 \pmod p] \\
&\Rightarrow [x \equiv -y \pmod p]\text{ or }[x \equiv y \pmod p]
\end{align}$$
*Assume that $x\equiv y \pmod p$ or $x\equiv -y \pmod p$. In both cases, squaring both sides gives $x^2\equiv y^2 \pmod p$.
(b) Assume that $x^2 \equiv n \pmod p$ for some $x, n, p$.
Since $n \not \equiv 0 \pmod p$ and $p$ is prime, $n$ and $p$ are relatively prime. Then, $x^2$ and $p$ are also relatively prime. Therefore, $x$ must be relatively prime to $p$. Then, by Fermat's theorem, $x^{p - 1} \equiv 1 \pmod p$. So:
$$\textrm{rem}(n^{(p-1)/2}, p) = \mathrm{rem}((x^2)^{(p-1)/2}, p) = \mathrm{rem}(x^{p-1}, p) = 1,$$
which proves (i).
(c) Assume that $n$ is not a square modulo $p$.
Since $n$ and $p$ are relatively prime, by Fermat's theorem, $n^{p - 1} \equiv 1 \pmod p$. Since $p$ is odd, $p - 1$ is even, so this can be rewritten as:
$$(n^{(p - 1)/2})^2 \equiv 1 \pmod p.$$
Then, from (a), it must be the case that either $n^{(p - 1)/2} \equiv 1 \pmod p$ or $n^{(p - 1)/2} \equiv -1 \pmod p$.
I got stuck at this point. I don't know how to infer, from the above facts, that $n^{(p-1)/2} \equiv -1 \pmod p$. Any hints on how to proceed?
| Let $g$ be a primitive root in modulo $p$. Then for an integer $x$ relatively prime to $n$ we have, $x\equiv g^i$(mod $p$), for some $i=1,2,\dots,p-1$. Similarly since, $gcd(n,p)=1$, so there are some $j=1,2,\dots,p-1$ such that $n\equiv g^j$(mod $p$).
Now, there is an integer $x$ relatively prime to $p$ satisfying $x^2\equiv n$(mod $p$) if and only if,
$g^{2i}\equiv g^j$ (mod $p$)
$\Leftrightarrow$ $2i\equiv j$ (mod $p-1$)$\dots$ (1)
Hence $n$ is a quadratic residue in module $p$
$\Leftrightarrow$ (1) has a solution in $i$.
$\Leftrightarrow$ $gcd(2,p-1)=2$ divides $j$.
Now if $2|j$ then $j=2t$ for some integer t. Then
$n^{\frac{p-1}{2}}\equiv (g^t)^{p-1} \equiv 1$ (mod $p$).
Next, if 2 does not divide $j$ then $\frac{p-1}{2}j$ is not a multiple of $p-1$. So
$n^{\frac{p-1}{2}}\equiv g^{\frac{p-1}{2}j} \not\equiv 1$ (mod $p$). [ Since $g$ is primitive in module $p$ so $g^s\equiv 1$ (mod $p$) for $s=p-1$ but not for any $s=1,\dots,p-2$.]
Now from (a) then we have, $n^{\frac{p-1}{2}}\equiv -1$ (mod $p$).
Hence $n$ is quadratic residue module $p$ or not according to, $n^{\frac{p-1}{2}}\equiv 1$ or, $n^{\frac{p-1}{2}}\equiv -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find real numbers $s$ and $t$ such that $\Gamma$ is the graph of $r = 2s \cos (\theta + t)$ The question: Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of
$r = 2s \cos (\theta + t).$
I did this so far:
We can use the general circle equation, $$(x-h)^{2}+(y-k)^{2} = h^2 + k^2.$$
We take this equation and expand it: $$x^2 -hx-hx+h^2 + y^2 -ky-ky+k^2 = h^2 + k^2.$$
Simplify: $$x^2 -2hx + y^2 -2ky = 0.$$
Using what we know, we can get that: $$x = r \cos \theta$$ and $$y = r \sin \theta.$$
Plugging it in, we get: $$r^{2} \cos^{2} \theta - 2hr \cos \theta + r^{2} \sin^{2} \theta - 2kr \sin \theta = 0.$$
This looks like the Pythagorean identity, just "squared"! We will keep $r$ on one side, while we move the other over to the right: $$r(\sin^{2}\theta+\cos^{2}\theta)=2h\cos\theta+2k\cos\theta.$$
Which will give us: $$r = 2h\cos\theta+2k\sin\theta.$$
I'm not sure how to continue, I'm stuck at this point.
| From
$$r=2(h \cos \theta + k \sin \theta)$$
$$r=2\sqrt{h^2+k^2}\left( \frac{h}{\sqrt{h^2+k^2}} \cos \theta + \frac{k}{\sqrt{h^2+k^2}}\sin \theta \right)$$
We can pick $s=\sqrt{h^2+k^2}$
and we can pick $t$ to satisfy $\cos t = \frac{h}{\sqrt{h^2+k^2}}$ and $\sin t = -\frac{k}{\sqrt{h^2+k^2}}$.
Remark: handle the case when $h=k=0$ separately.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the size of the Turan graph $T_r(n)$ is at least $(1 - \frac{1}{r}) \binom{n}{2}$. A Turan graph $T_r(n)$ is defined as the complete $r$-partite graph of order $n$ such that the number of vertices in each of the $r$ classes is either $\lfloor \frac{n}{r}\rfloor$ or $\lceil \frac{n}{r} \rceil$. For fixed $n$ and $r$, $T_r(n)$ is unique up-to isomorphism. The size of $T_r(n)$ can be simply counted as: $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$.
Here is what I have: assume $n = kr + s,\ 0 \leq s \leq r-1$. Note that at least one class must have exactly $\lfloor \frac{n}{r} \rfloor$ vertices. Then, $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$
$\geq$ $\binom{n}{2} -(r-1) \binom{\lceil \frac{n}{r} \rceil}{2} - \binom{\lfloor \frac{n}{r} \rfloor}{2}$
$=\binom{n}{2} - (r-1) \binom{k+1}{2} - \binom{k}{2}$
$= \frac{n(n-1)}{2} - \frac{(r-1)k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)}{2} - \frac{rk(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$\geq \frac{n(n-1)}{2} - \frac{n(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + k(k+1) - k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + 2k}{2}$
$= \binom{n}{2}(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$.
But clearly, $(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$ may be smaller than $1 - \frac{1}{r}$, as seen by taking $n=31$ and $r=5$.
| The size $S$ of the Turán graph is $\frac 12\left(n^2-\frac {(n^2-s^2)}{r}-s\right)$, see my answer. It follows $$S- \left(1 - \frac{1}{r}\right) \binom{n}{2}=\frac{(n-s)(r-1)+s(s-1)}{2r}\ge 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the values of $a$ and $b$ as
$$\frac{4x^2}{144}+\frac{9y^2}{144}=1$$
$$\frac{x^2}{6^2}+\frac{y^2}{4^2}=1$$
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, gives
$a=6$ & $b=4$. From here I have no idea how to solve further?
| Consider the four corner/vertex points $(\pm 6\cos\theta, \pm 4\sin\theta)$ of rectangle lying on given ellipse: $4x^2+9y^2=144$
Now, the sides of the rectangle are length: $(2\cdot 6\cos\theta)$ & width $(2\cdot 4\sin\theta)$ which are in ratio $3:2$ as given in question therefore we have $$\frac{12\cos\theta}{8\sin\theta}=\frac32\implies \tan\theta=1\iff \theta=\frac{\pi}{4}$$
Now, the area of rectangle inscribed in given ellipse $$\text{Length}\times \text{Width}=12\cos\theta\cdot 8\sin\theta$$
$$=48\sin2\theta$$
$$=48\sin\frac{\pi}{2}=\color{blue}{48\ \text{unit}^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Analogue of $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ for vectors The Brahmagupta–Fibonacci identity $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ allows us to write a product of squares as a sum of squares. Is there an analogue of this identity when $a, b, c, d$ are vectors in $\mathbb{R}^n$ and the multiplication is the ordinary scalar product?
One of the main hurdles is of course $(a \cdot c)(b \cdot d) \neq (a \cdot b)(c \cdot d)$. The Cauchy-Binet identity gives an expression for $(a \cdot c)(b \cdot d) - (a \cdot b)(c \cdot d)$, however I do not see a way to turn this into a simple expression.
EDIT: I do not mind having extra terms in the end: something like $(a\cdot a + b \cdot b)(c \cdot c + d \cdot) = (a\cdot c - b\cdot d)^2 + (a\cdot d + b\cdot c)^2 + \mbox{other terms}$ where the remaining terms can be written explicitly.
| I assume (perhaps unreasonably closed-mindedly) that you want an analogue of the form
\begin{align}
\left(\left<a,a\right>^2 + \left<b,b\right>^2\right) \left(\left<c,c\right>^2 + \left<d,d\right>^2\right) = \left<v,v\right>^2 + \left<w,w\right>^2
\end{align}
that holds for any vectors $a, b, c, d \in \mathbb{Q}^n$, where $v$ and $w$ are two vectors in $\mathbb{Q}^n$ whose entries are fixed degree-$2$ homogeneous polynomials in the entries of $a, b, c, d$. Such an analogue exists only if $n \in \left\{1,2,4\right\}$. Indeed, if we number the entries of the vectors $a, b, c, d$ such that $a = \left(x_1, x_2, \ldots, x_n\right)^T$ and $b = \left(x_{n+1}, x_{n+2}, \ldots, x_{2n}\right)^T$ and $c = \left(y_1, y_2, \ldots, y_n\right)^T$ and $d = \left(y_{n+1}, y_{n+2}, \ldots, y_{2n}\right)^T$, then the above analogue would imply that the polynomial $\left(x_1^2 + x_2^2 + \cdots + x_{2n}^2\right) \left(y_1^2 + y_2^2 + \cdots + y_{2n}^2\right)$ can be written as $u_1^2 + u_2^2 + \cdots + u_{2n}^2$ for some quadratic forms $u_1, u_2, \ldots, u_{2n}$ in the $x_i$ and $y_i$. But such quadratic forms only exist if $n \in \left\{1,2,4\right\}$, according to Theorem 1.1 in Keith Conrad, The Hurwitz theorem on sums of squares by representation theory. (Note that my $2n$ is his $n$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Consider $v=v_1+v_2$, where $v_1 \in M$ and $v_2 \in M^{\perp}$ $M = span\{\begin{pmatrix}8\\0\\-6\end{pmatrix}, \begin{pmatrix}8\\6\\-6 \end{pmatrix}\}$
I am trying to calculate $v_1$ and $v_2$ when $v=\begin{pmatrix}2 \\ 4 \\ 6 \end{pmatrix}$.
I know that since $v_1 \,\in \, M$ and $v_2 \, \in \, M^{\perp}$, $v_1\cdot v_2 = 0$.
Let $v_1 = a_1\begin{pmatrix}8\\0\\-6\end{pmatrix} + a_2\begin{pmatrix}8\\6\\-6 \end{pmatrix}$. The orthogonal basis of $M = \begin{pmatrix}8\\0\\-6\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix}$. From this, we can say that $v=a_1\begin{pmatrix}8\\0\\-6\end{pmatrix} + a_2\begin{pmatrix}8\\6\\6 \end{pmatrix}+a_3\begin{pmatrix}8\\0\\-6\end{pmatrix}+a_4\begin{pmatrix}0\\1\\0\end{pmatrix}$
But if we put this into an augmented matrix, it gives an inconsistent system (which is clearly wrong). What is the correct way to solve this?
| It is quite easy to see that $w=\begin{bmatrix}6\\0\\8\end{bmatrix}$ is orthogonal to the basis vectors given for $M$. Thus $w \in M^{\perp}$. Moreover $\text{dim}(M)=2$ and $M \subset \Bbb{R}^3$, so $\text{dim}(M^{\perp})=1$. This means we can say that $\{w\}$ is a basis for $M^{\perp}$.
So we want to solve for $a,b,c$ such that
$$\begin{bmatrix}2\\4\\6\end{bmatrix}=\underbrace{a\begin{bmatrix}8\\0\\-6\end{bmatrix}+b\begin{bmatrix}8\\6\\-6\end{bmatrix}}_{v_1 \in M}+\underbrace{c\begin{bmatrix}6\\0\\8\end{bmatrix}}_{v_2 \in M^{\perp}}.$$
This yields the system
\begin{align*}
4a+4b+3c&=1\\
3b&=2\\
-3a-3b+4c&=3
\end{align*}
Upon solving this, we get $a=-\frac{13}{15}, b=\frac{2}{3}$ and $c=\frac{3}{5}$.
Thus
$$v_1=\color{red}{\frac{-13}{15}}\begin{bmatrix}8\\0\\-6\end{bmatrix}+\color{red}{\frac{2}{3}}\begin{bmatrix}8\\6\\-6\end{bmatrix},$$
and
$$v_2=\color{red}{\frac{3}{5}}\begin{bmatrix}6\\0\\8\end{bmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my lights go off.
Any leads? Thanks in advance.
| Note
\begin{align}
&\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}} \\
=& \sqrt{\frac{(2a+b)+b +2\sqrt{(2a+b)b}}{2}}\\
= &\sqrt{\frac{(\sqrt{2a+b}+\sqrt b )^2}{2}}\\
= &\sqrt{\frac{2a+b}2}+\sqrt {\frac b{2}}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim_{n\to \infty} a_n=\frac{\sqrt{5}-1}{2}$ if $a_{n+1}=\sqrt{1-a_n}$ and $0If $0<a_0<1$ and $a_{n+1}=\sqrt{1-a_n}$, prove that:
$$\lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$$
Here what I do is that when $n\to \infty$, $a_{n+1}=a_n$
$\therefore a_n=\sqrt{1-a_n}\Rightarrow a_{n}^2+a_n-1=0$
which implies that when $n\to \infty$, $a_n\to \dfrac{\sqrt{5}-1}{2}$
$\therefore \lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$
My doubt here is, was I correct in simply assuming that $a_{n+1}=a_n$ when $n\to \infty$? (I'm sorry for I may be weak in basics.)
If not, then please suggest alternative method.
| $$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{\frac{\sqrt5-1}{2}-a_n}{\sqrt{1-a_n}+\frac{\sqrt5-1}{2}}$$ and
$$a_{n+2}-a_n=\sqrt{1-\sqrt{1-a_n}}-a_n=\frac{a_n\sqrt{1-a_n}\left(\frac{\sqrt5-1}{2}-a_n\right)\left(\frac{\sqrt5+1}{2}+a_n\right)}{(\sqrt{1-\sqrt{1-a_n}}+a_n)((1+a_n)\sqrt{1-a_n}+1)}.$$
Now, let $a_1<\frac{\sqrt5-1}{2}.$
Thus, for any odd $n$ we have $a_n<\frac{\sqrt5-1}{2}$ and $a_{n+2}>a_n$,
which says that for odd $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n$ and by your work
(we just need to solve the following equation: $a=\sqrt{1-\sqrt{1-a}}.$) it's equal to $\frac{\sqrt5-1}{2}.$
Now, $$a_2=\sqrt{1-a_1}>\sqrt{1-\frac{\sqrt5-1}{2}}=\frac{\sqrt5-1}{2},$$ which says that for even $n$ we have $a_n>\frac{\sqrt5-1}{2}$ and $a_{n+2}<a_n$,
which gives that for even $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n.$
Let this limit is equal $b$.
Thus, since $$a_{n+2}=\sqrt{1-\sqrt{1-a_n}}$$ and $f(x)=\sqrt{x}$ is a continuous function, we obtain:
$$b=\sqrt{1-\sqrt{1-b}},$$ which gives $$b\in\left\{0,1,\frac{\sqrt5-1}{2}\right\}.$$
Since, $a_{n+2}<a_n$, we see that $b\neq1$ and since, $a_n>\frac{\sqrt5-1}{2},$ we see that $b\neq0$.
Thus, $b=\frac{\sqrt5-1}{2}.$
Can you end it now?
| {
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System of equations and recurrence relation I am trying to find the general solution for $N$ of the following system of equations
$$
\begin{cases}
(x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\
{x_n}^2 + {y_n}^2 = 1
\end{cases}
$$
with the initial values $x_0 = 1$ and $y_0 = 0$ and the following
*
*$\theta$ is a constant and $0 \leqslant \theta \leqslant 2$
*$N$ is a constant and we want to find the terms $(x_N, y_N)$
Using substitution with respect to $N$, we have
$$
\begin{align}
x_0 = 1 \quad & ; \quad y_0 = 0 \\
x_1 = -\frac{\theta^2 - 2N^2}{2N^2} \quad & ; \quad y_1 = -\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2} \\
x_2 = \frac{\theta^4 - 4N^2\theta^2 + 2N^4}{2N^4} \quad & ; \quad y_2 = \frac{(\theta^3 - 2N^2\theta) \sqrt{4N^2 - \theta^2}}{2N^4} \\
x_3 = -\frac{\theta^6 - 6N^2\theta^4 + 9N^4\theta^2 - 2N^6}{2N^6} \quad & ; \quad y_3 = -\frac{(\theta^5 - 4N^2\theta^3 + 3N^4\theta) \sqrt{4N^2 - \theta^2}}{2N^6}
\end{align}
$$
By using substitution, it becomes very difficult with $N \geqslant 2$.
| Because it is very difficult to solve this relation, we can use complex numbers. Let
$$(x_n, y_n) = x_n + iy_n \quad \text{and} \quad z_n = x_n + iy_n$$
then we have
$$z_1 = -\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}$$
and the term $z_1$ is always given by the same expression with respect to $N$.
To compute the product of two complex numbers, we multiply the magnitudes and add the arguments. Since $\lvert z_1 \rvert = 1$, we have
$$\forall k \in \mathbb{N}^*, \lvert (z_1)^k \rvert = 1$$
and it rotates $z_1$ by $\arg(z_1)$ for $k$ times. Since the magnitude is always equal to 1, we stay in orbit around the unit circle.
Consider the following picture:
In particular, we notice that
$$\forall n,N \in \mathbb{N}^*, \arg(z_n) \leqslant \frac{\theta}{N} \quad , \quad n \leqslant N$$
because the length of the chord between two terms is given by
$$\frac{\theta}{N} = \lvert z_n - z_{n-1} \rvert = \sqrt{(x_n - x_{n-1})^2 + (y_n - y_{n-1})^2} \quad , \quad n \leqslant N$$
and we can reduce the length of each chord such that we obtain the term
$$z_N = \left(-\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}\right)^N$$
and by taking its limit, we eventually have
$$\cos(\theta) + i\sin(\theta) = \lim\limits_{N \to \infty} \left(-\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}\right)^N$$
as each chord becomes smaller and smaller and closer to the circle, we denote by $\theta$ the arc length measure around the unit circle in radians.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the probablity that the sum of two dice is 4 or 6?
What is the probablity that the sum of two dice is 4 or 6?
The explanation I found is as follows:
Total number of outcomes $6 \times6 = 36$
Number of outcomes where the event occurs: $1+3, 2+2, 3+1, 1+5, 2+4, 3+3, 4+2$ and $5+1$ (Total $ 8$)
The probability that the event occurs is $8/36$ or $2/9$
My doubt is why $1+3$ and $3+1$ are considered as separate outcomes. Isn't it the same as the dices are identical and we cannot really find out which dice had which number on it?
| In developing an understanding of the sample space, you should think of the two dice as being distinct. Yes, in principle, the two dice may be indistinguishable. However, from the point of view of probability theory, the behaviour of each die is described by a different random variable. Hence the two dice are not, really, indistinguishable. Maybe one of them is blue and the other red, or one is rolled before the other, or one has a cute little smiley face drawn in place of the $3$.
From this point of view, the outcome $1+3$ represent rolling a $1$ and then a $3$, while the outcome $3+1$ represents rolling a $3$ and then a $1$. In terms of the ultimate nature of the events being modeled, these two outcomes are the same (both yield a $4$), but as outcomes (or elementary events) in the underlying sample space, they are distinct. To make this more concrete, imagine that the dice are rolled one after the other, and suppose that they are colored differently. For example, if one of the dice is blue and the other red, then the sample space looks like
$$ \begin{array}{c|cccccc}
& \color{red}{1}
& \color{red}{2}
& \color{red}{3}
& \color{red}{4}
& \color{red}{5}
& \color{red}{6} \\\hline
\color{blue}{1}
& (\color{blue}{1}, \color{red}{1})
& (\color{blue}{1}, \color{red}{2})
& (\color{blue}{1}, \color{red}{3})
& (\color{blue}{1}, \color{red}{4})
& (\color{blue}{1}, \color{red}{5})
& (\color{blue}{1}, \color{red}{6}) \\
\color{blue}{2}
& (\color{blue}{2}, \color{red}{1})
& (\color{blue}{2}, \color{red}{2})
& (\color{blue}{2}, \color{red}{3})
& (\color{blue}{2}, \color{red}{4})
& (\color{blue}{2}, \color{red}{5})
& (\color{blue}{2}, \color{red}{6}) \\
\color{blue}{3}
& (\color{blue}{3}, \color{red}{1})
& (\color{blue}{3}, \color{red}{2})
& (\color{blue}{3}, \color{red}{3})
& (\color{blue}{3}, \color{red}{4})
& (\color{blue}{3}, \color{red}{5})
& (\color{blue}{3}, \color{red}{6}) \\
\color{blue}{4}
& (\color{blue}{4}, \color{red}{1})
& (\color{blue}{4}, \color{red}{2})
& (\color{blue}{4}, \color{red}{3})
& (\color{blue}{4}, \color{red}{4})
& (\color{blue}{4}, \color{red}{5})
& (\color{blue}{4}, \color{red}{6}) \\
\color{blue}{5}
& (\color{blue}{5}, \color{red}{1})
& (\color{blue}{5}, \color{red}{2})
& (\color{blue}{5}, \color{red}{3})
& (\color{blue}{5}, \color{red}{4})
& (\color{blue}{5}, \color{red}{5})
& (\color{blue}{5}, \color{red}{6}) \\
\color{blue}{6}
& (\color{blue}{6}, \color{red}{1})
& (\color{blue}{6}, \color{red}{2})
& (\color{blue}{6}, \color{red}{3})
& (\color{blue}{6}, \color{red}{4})
& (\color{blue}{6}, \color{red}{5})
& (\color{blue}{6}, \color{red}{6}) \\
\end{array}
$$
This sample space gives all of the possible outcomes, of which there are $36$. Moreover, and of crucial importance, is that every one of these outcomes is equally likely. Showing that all of these outcomes are equally likely requires a little bit of work, but the essential ideas are that
*
*each die is modeled by a uniform variable on the set $\{1,2,3,4,5,6\}$,
which means that the probability of rolling any particular number on either die is $1/6$, and
*the two die rolls are independent, which means (more or less) that the number rolled on one of the two dice does not depend on the result of the other.
These two observations are sufficient to show that all $36$ outcomes are equally probable.
However, we are only interested in the event which describes a sum of either $4$ or $6$. Replacing the specific die rolls with the corresponding sums in the table above gives
$$ \begin{array}{c|cccccc}
& \color{red}{1}
& \color{red}{2}
& \color{red}{3}
& \color{red}{4}
& \color{red}{5}
& \color{red}{6} \\\hline
\color{blue}{1}
& 2 & 3 & \boxed{4} & 5 & \boxed{6} & 7 \\
\color{blue}{2}
& 3 & \boxed{4} & 5 & \boxed{6} & 7 & 8 \\
\color{blue}{3}
& \boxed{4} & 5 & \boxed{6} & 7 & 8 & 9 \\
\color{blue}{4}
& 5 & \boxed{6} & 7 & 8 & 9 & 10 \\
\color{blue}{5}
& \boxed{6} & 7 & 8 & 9 & 10 & 11 \\
\color{blue}{6}
& 7 & 8 & 9 & 10 & 11 & 12\\
\end{array}
$$
From this presentation, it can be seen that there are $8$ "good" outcomes (i.e. there are $8$ elementary events giving a sum of either $4$ or $6$), out of a total of $36$ equiprobable outcomes. Thus
$$ P(\text{two dice sum to either $4$ or $6$})
= \frac{8}{36} = \frac{2}{9}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Deriving $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$ I tried but I could not get $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$ from $\psi_1(x) = \cos^3 x + \sin^2 x + \cos x + 1$ using $\phi_0, \phi_1...$
Can any help? Thanks.
| To obtain the required decomposition it suffices to sum the left-hand and right-hand sides of the following trigonometric identities:
$$1=\sqrt{\pi}\varphi_0(x),$$
$$\cos x=\frac{\sqrt{\pi}}{\sqrt{2}}\varphi_1(x),$$
$$\sin^2 x =\frac 12(1-\cos 2x)=\frac{\sqrt{\pi}}{2}\varphi_0(x)-\frac {\sqrt{\pi}}{2\sqrt{2}}\varphi_2(x),$$
$$\cos^3 x= \frac 14(\cos 3x+3\cos x)=\frac {\sqrt{\pi}}{4\sqrt{2}}\varphi_3(x)+\frac {3\sqrt{\pi}}{4\sqrt{2}}\varphi_1(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist. I turned that into one fraction, then I factored the polynomial on the numerator as $-(x-1)(x+2)$ and the one on the denominator as $(x+1)(-x^2-x-1)$. What did I do wrong?
| I believe you messed up by looking at $$\lim_{x \to 1}\left(\frac{1}{x-1} \color{red}{+} \frac{3}{1-x^3}\right)$$ instead of $$\lim_{x \to 1} \left( \frac{1}{x-1}\color{red}{-}\frac{3}{1-x^3} \right)$$
The value of the first one is $1$, but the second one does not exist.
We can factor $1-x^3$ as $(x-1)(-x^{2}-x-1)$ so that the limit becomes $$\lim_{x \to 1}\frac{(-x^2-x-1)-3}{(x-1)(-x^2-x-1)}$$
Then it is clear that the limit does not exist since the numerator when plugging in $x = 1$ is $-6$ and the denominator is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How find this nice minmum of this value $\frac{\prod_{i=1}^{n-1}(a_{i}+a_{i+1})\sum a_{i}}{\prod_{i=1}^{n}a_{i}}$ Let $n$ be give postive integer number. For any $a_{i}>0 (i=1,2,\cdots,n)$, find the minimum of the value
$$F_{n}(a_{1},a_{2},\cdots,a_{n})=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})\cdots (a_{n-1}+a_{n})(a_{1}+a_{2}+\cdots+a_{n})}{a_{1}a_{2}a_{3}\cdots a_{n}}.$$(by wang yong xi)
I try when $n=2$,then
$$F_{2}(a_{1},a_{2})=\dfrac{(a_{1}+a_{2})(a_{1}+a_{2})}{a_{1}a_{2}}=\dfrac{(a_{1}+a_{2})^2}{a_{1}a_{2}}\ge 4$$when $a_{1}=a_{2}$ is minimum
(2):when $n=3$,
$$F_{3}=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})(a_{1}+a_{2}+a_{3})}{a_{1}a_{2}a_{3}}$$WLOG $a_{3}=1$,so
$$F=\dfrac{(a_{1}+a_{2})(a_{2}+1)(a_{1}+a_{2}+1)}{a_{1}a_{2}}$$
I use this find this minimum is $$(F_{3})_{min}=\dfrac{1}{2}(11+5\sqrt{5})$$
when $a_{1}=1,a_{2}=\dfrac{\sqrt{5}-1}{2}$ seelinks
| Here's some partial progress. Maybe someone will have further ideas.
By homogeneity, we can set $a_1 + \cdots + a_n = 1$ throughout. First dispense with boundary behavior. Fix $1 \leq k \leq n$ and note that
$$\frac{\prod_{i=1}^k (a_i+a_{i+1})}{\prod_{i=1}^n a_i} = \prod_{i=1}^{k-1} \frac{a_i + a_{i-1}}{a_i} \cdot \prod_{i=k}^{n-1} \frac{a_i + a_{i+1}}{a_{i+1}} \cdot \frac{1}{a_k} \geq \frac{1}{a_k}.$$
Hence $F_n(a_1, \ldots, a_n) \geq \frac{1}{\min_i a_i}$, so $F_n \to \infty$ uniformly as we tend to the boundary.
Let $G_n(a_1, \ldots, a_n) = \log F_n(a_1, \ldots, a_n)$, which we may as well minimize. Recalling that $a_1+\cdots+a_n=1$ throughout, consider when $\nabla G_n = 0$. We find
$$\begin{align*}a_1 \partial_1 G_n &= \frac{a_1}{a_1+a_2} + a_1 - 1 \\
a_i \partial_i G_n &= \frac{a_i}{a_{i-1} + a_i} + \frac{a_i}{a_i + a_{i+1}} + a_i - 1 \qquad \text{for }1 < i < n \tag{*}\label{*}\\
a_n \partial_n G_n &= \frac{a_n}{a_{n-1}+a_n} + a_n - 1.
\end{align*}$$
In particular we have the relation
$$\sum_{i=1}^n a_i \partial_i G_n = 0$$
which seems important, though I haven't found a use for it.
Using $\eqref{*}$, we can rewrite the condition $\nabla G_n = 0$ as either
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= a_1 + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= a_1 + a_2 + \cdots + a_{n-1} \\
1 &= a_1+\cdots+a_n
\end{align*}$$
or
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= \frac{a_2}{a_1+a_2} + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= \frac{a_{n-1}}{a_{n-2}+a_{n-1}} + a_{n-1} \\
a_1+\cdots+a_n &= 1.
\end{align*}$$
Either of these forms is suitable for computer algebra systems. I threw the first 12 into Mathematica; here's the output. Using the first $n-1$ relations, we may iteratively solve for $a_2, a_3, \ldots, a_n$ as rational functions of $a_1$, and then we may impose the final condition to solve for $a_1$. This shows that $a_1$ (and hence all the others) are algebraic, so the algebraic minimum math110 suggests in a comment above is at least plausible on its face. Edit: indeed, I checked through $n=12$ that the values of $F_n$ when $\nabla G_n = 0$ are precisely the Galois conjugates of math110's guess of $\left(\sin\left(\frac{2\pi}{n+2}\right)/\sin\left(\frac{\pi}{n+2}\right)\right)^{n+2}$.
Based on that data, it seems that when $n$ is odd, $a_1$ satisfies $p_n(t) := \sum_{k=0}^{(n+1)/2} \binom{n+1-k}{k} (-t)^k = 0$. Intriguingly, this family of polynomials appears to be tending pointwise to 0 on $[0, 1]$ as $n \to \infty$, and they seem to be real-rooted. Moreover, it appears that the unique value of $a_1$ solving the above system subject to $a_i > 0$ and $a_1 + \cdots + a_n = 1$ is the smallest root of $p_n(t)$.
I get the sense that I'm missing some sort of physical meaning in this problem. I also can't see why $\eqref{*}$ implies the symmetry $a_i = a_{n-i+1}$, though it should. There's a bizarre amount of structure throughout this problem that I'm at a loss to explain!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
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