Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ holds.
My work.In fact, I want to solve another problem ( Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ ) This another problem has already been solved, but I want to solve it by the method that the author of the problem intended. The fact is that in the original problem there was another inequality ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). It seems to me that the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ is proved as the author of the problem wanted. I think that the author of the problem wanted us to prove inequality $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ on the basis of inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$. To this end, it suffices to prove the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ I checked this inequality by numerical methods on a computer. This inequality holds for all $n$. Interestingly, if $n \to \infty$ then this inequality (in the limit) is an equality.
Perhaps this will help to solve the problem: let $x_k=\frac{k+1}{k}$. Then $1<x_k \le 2$ and inequality takes the form $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$
| (update 2019/08/09)
Let $H_n = \sum_{k=1}^n \frac{1}{k}$. Since $\frac{1}{k} \le -\ln (1 - \frac{1}{k}) $ for $k\ge 2$,
we have $H_n - 1 \le -\sum_{k=2}^n \ln (1 - \frac{1}{k}) = \ln n$ or $H_n \le \ln n + 1$.
Rewrite the inequality as
$$(n^2+2n-H_n)\left(n^2 - 2n - 3 + \frac{3}{n+1} + 3H_n\right) \le n^2(n+ H_n)\left(n + 1 - \frac{1}{n+1} - H_n\right)$$
or
$$-(n^2-3)H_n^2 - \frac{n(n^2+10n+11)}{n+1}H_n + \frac{n^2(n+2)(n+5)}{n+1}\ge 0. \qquad (1)$$
When $1\le n\le 12$, the inequality in (1) is verified directly.
When $n\ge 13$, rewrite (1) as
$$-H_n^2 - \frac{n(n^2+10n+11)}{(n+1)(n^2-3)}H_n + \frac{n^2(n+2)(n+5)}{(n+1)(n^2-3)} \ge 0.$$
Since $\frac{n(n^2+10n+11)}{(n+1)(n^2-3)} \le \frac{7}{4}$ and $\frac{n^2(n+2)(n+5)}{(n+1)(n^2-3)} \ge n+6$, it suffices to prove that
$-H_n^2 - \frac{7}{4}H_n + n+6 \ge 0$ or
$$-\frac{7}{8} + \frac{1}{8}\sqrt{433+64n} \ge H_n.$$
Since $H_n \le \ln n + 1$, it suffices to prove that $-\frac{7}{8} + \frac{1}{8}\sqrt{433+64n} \ge \ln n + 1$.
Let $f(x) = -\frac{7}{8} + \frac{1}{8}\sqrt{433+64x} - \ln x - 1$.
We have $f(13) > 0$ and $f'(x) = \frac{4}{\sqrt{433+64x}} - \frac{1}{x} > 0$ for $x\ge 13$.
Thus, $f(n) \ge 0$ for $n\ge 13$. This completes the proof.
| {
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"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
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Solve the initial value problem below using the method of Laplace transform I am able to get up to a certian point this problem but get stuck when doing the Laplace transforms. The problem is
Solve the initial value problem below using the method of Laplace transform :
$$y''-4y=8t-20e^{-2t}$$
with $y(0)=0$ and $y'(0)=19$.
My try :
Taking Laplace to both sides we get
$$\mathcal{L} \{ y'' \} - 4 \mathcal{L} \{ y \} = 8 \mathcal{L} \{ t \} - 20 \mathcal{L} \{ e^{-2t} \}$$
$$s^2 y(s) - s\ y(0) - y'(0) - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$
$$s^2 y(s) - 19 - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$
$$y(s) (s^2-4) = \frac{8}{s^2} - \frac{20}{s+2} + 19$$
$$y(s) = \frac{8}{s^2(s^2-4)} - \frac{20}{(s+2)(s^2-4)} + \frac{19}{s^2-4}$$
| Since you already have the expression for $y(s)$, you can further simplify each term using partial fractions.
First, we have $\frac{8}{s^2 (s^2 - 4)} = \frac{A}{s^2} + \frac{B}{s^2 - 4}$, which gives us $8 = A (s^2 - 4) + Bs^2$. Thus, $A = -2$ and $B = 2$.
Then we have $\frac{20}{(s + 2)(s^2 - 4)} = \frac{A}{s + 2} + \frac{B}{(s + 2)^2} + \frac{C}{s - 2}$, which gives us $20 = A (s - 2)(s + 2) + B (s - 2) + C (s + 2)^2$. Set $s = -2$, and we get $B(-4) = 20$, or $B = -5$. Then set $s = 2$ to get $20 = 16C$, which gives us $C = 5/4$. Arbitrarily set $s = 1$, and we find $20 = -3A + 5 + 45/4$, which gives us $A = -5/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314720",
"timestamp": "2023-03-29T00:00:00",
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Showing $\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $
Show that
$$\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $$
I tried to find a polynomial that had its roots, but the degree grows too fast and I'm getting lost.
| We make use of the general equality that $$\sf a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\tag1$$ where $\sf a,b,c=\sqrt[3]{\cos\frac{2\pi}9},\sqrt[3]{\cos\frac{4\pi}9},\sqrt[3]{\cos\frac{8\pi}9}$ such that $\sf a^3+b^3+c^3=0$ and $\sf abc=-\frac12$.
We can also substitute in that $\sf a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ where simplifying gives: $$\sf \frac32=(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)$$ and so $$\sf \frac32=p(p^2-3q)\tag2$$ where $\sf p=a+b+c$ and $\sf q=ab+bc+ca$.
Next we can also substitute $\sf (a,b,c)\mapsto(ab,bc,ca)$ in $\sf(1)$ to get that $$\sf a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2=(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-a^2bc-ab^2c-abc^2)$$ such that $\sf a^3b^3+b^3c^3+c^3a^3=-\frac34$ and $\sf a^2b^2c^2=\frac14$.
Now substitute $\sf a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$ to get: $$\sf -\frac32=(ab+bc+ca)\left((ab+bc+ca)^2-3abc(a+b+c)\right)$$ and so since $\sf abc=-\frac12$, $$\sf -\frac32=q(q^2+\frac32p)\tag3$$
Thus we get a system of cubics $\sf(2)$ & $\sf(3)$.
We can solve by first canceling out the $\sf pq$ terms to get $$\sf q^3=-\frac12p^3-\frac34$$ so that $\sf(2)$ becomes $$\sf \left(p^3-\frac32\right)^3=27p^3\left(-\frac12p^3-\frac34\right)$$
This is a cubic in $\sf p^3$, so we can proceed through the usual way and get that $$\sf p=\sqrt[3]{\frac{3\sqrt[3]9-6}2}$$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:
$ n^2+n =p^2$
$ n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$ (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$ (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$
I am not able to proceed further from here. Any suggestion on what to do next?
| You are actually very close to the answer. All you really need to do is multiply both sides of $(n+{1\over2}-p)(n+{1\over2}-p)={1\over4}$ by $4$ and then note that the only integer solutions of $ab=1$ is $a=b=\pm1$. Here is a complete presentation:
Assuming only $n,p\in\mathbb{Z}$, we have
$$\begin{align}
n(n+1)=p^2&\iff4n^2+4n=4p^2\\
&\iff(2n+1)^2=(2p)^2+1\\
&\iff(2n+1)^2-(2p)^2=1\\
&\iff(2n+1-2p)(2n+1+2p)=1\\
&\iff2n+1-2p=2n+1+2p=\pm1\\
&\iff p=0\land n\in\{0,-1\}
\end{align}$$
The requirement $n\in\mathbb{N}$ eliminates $n=-1$, leaving $(n,p)=(0,0)$ as the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317993",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Complex root used to find $a$. Find $a$ if $z=2+i$ is a root of $2z^{2}+3z+2a-14+3i=0$
Workings:
$2(2+i)^{2}+3(2+i)+2a-14+3i=0$
$2(4+4i-1)+6+3i+2a-14+3i=0$
$8-8i-2+6+3i+2a-14+3i=0$
$-2+14i=-2a$
$1-7i=a$
Is this correct?
| Your answer is correct.
You could say $2a=14-3i-3z-2z^2=14-3i-3(2+i)-2(2+i)^2$
$=14-3i-3(2+i)-2(3+4i)=2-14i,$ so $a=1-7i$.
By Vieta's formulas, the sum of the roots is $-\dfrac32$, so the other root besides $2+i$ is $-\dfrac72-i$.
As noted in comments, these roots are not complex conjugates;
their product is $(2a-14+3i)/2,$ which is not real.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the limit $b_n = n - \sqrt{n^2 + 2n}$ if $(1/n)\rightarrow 0$. Consider the sequence given by $b_n = n - \sqrt{n^2 + 2n}$. Taking $(1/n)\rightarrow 0$ as given, and using both the Algebraic Limit Theorem, and the fact that if $(x_n)\rightarrow x$ then $(\sqrt{x_n})\rightarrow \sqrt{x}$, show $\lim b_n$ exists and find the value of the limit.
Well, because $(1/n) \rightarrow 0$, the sequence $(n)\rightarrow \infty$. Consider $n=\sqrt{n^2}>\sqrt{n^2 - 2n}$ which implies $n>n-\sqrt{n^2 - 2n}>0 $. So if $(n)\rightarrow \infty$ then so does $(n-\sqrt{n^2 - 2n})\rightarrow \infty$ $\blacksquare$
But after writing I wonder if it is okay for me to assume that $n\in \mathbb{N}$. If $n\notin \mathbb{N}$ then my first statement of $(n)\rightarrow \infty$ is wrong because $n$ could also diverge to $-\infty$.
| Observe that $n - (n^2 + 2n)^{\frac{1}{2}}=n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}$.
By the binomial theorem, it follows that
$$
\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=1+\frac{1}{n}-\frac{1}{2n^2}+o(n^{-2})
$$
So
$$
n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=n - \left(n + 1 - \frac{1}{2n} + o(n^{-1}) \right)=-1+\frac{1}{2n}+o(n^{-1})
$$
Therefore, $\lim n - (n^2 + 2n)^{\frac{1}{2}} = -1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x, y \in \mathbb R$: $(x^2 - 1)^2 + 3 = \dfrac{6x^5y}{x^2 + 2}, 3y - x = \sqrt{\dfrac{4x - 3x^2y - 9xy^2}{x + 3y}}$
Solve the following system of equations $$\large \left\{ \begin{align} (x^2 &- 1)^2 + 3 = \frac{6x^5y}{x^2 + 2}\\ 3y - x &= \sqrt{\frac{4x - 3x^2y - 9xy^2}{x + 3y}}\end{align} \right.$$ such that $(x, y)$ is a root but $(-x, -y)$ is not.
I have provided a solution below. It looks disordered and there might be better solutions out there, I don't know.
| $(x \le 3y \ne -x)$
Let $a = \dfrac{2}{x^2} > 0$ and $b = \dfrac{3y}{x}$. The system of equation becomes
$$\left\{ \begin{align} \left(\frac{2}{a} - 1\right)^2 + 3 = \frac{2b \cdot \left(\dfrac{2}{a}\right)^3}{\dfrac{2}{a} + 2}\\ b - 1 = \sqrt{\frac{2a - b - b^2}{1 + b}} \end{align} \right.$$
$$\iff \left\{ \begin{align} (2 - a)^2 + 3a^2 = \frac{8b}{a + 1}\\ (b - 1)^2 + b = \frac{2a}{b + 1} \end{align} \right.$$
$$\iff \left\{ \begin{align} a^2 - a + 1 = \frac{2b}{a + 1}\\ b^2 - b + 1 = \frac{2a}{b + 1} \end{align} \right.$$
$$\iff \left\{ \begin{align} a^3 + 1 = 2b\\ b^3 + 1 = 2a \end{align} \right.$$
$\iff a^3 - 2b = b^3 - 2a \iff (a - b)(a^2 + ab + b^2 + 2) = 0$
However $a^2 + ab + b^2 + 2 \ge 2 > 0, \forall a, b \in \mathbb R$
$\implies a^3 - 2a + 1 = b^3 - 2b + 1 = 0$ and $a = b$
$\iff (a - 1)(a^2 + a - 1) = (b - 1)(b^2 + b - 1) = 0$ and $a = b$
$\iff a = b = 1$ or $a = b = \dfrac{\sqrt 5 - 1}{2} (a > 0)$
$\implies \dfrac{2}{x^2} = \dfrac{3x}{y} = 1$ or $\dfrac{2}{x^2} = \dfrac{3x}{y} = \dfrac{\sqrt 5 - 1}{2}$
$\implies (x, y) = \left(\pm \sqrt 2, \pm \dfrac{\sqrt 2}{3}\right)$ or $(x, y) = \left(\pm \sqrt{\sqrt 5 + 1}, \pm \dfrac{\sqrt{\sqrt 5 + 1}\left(\sqrt 5 - 1\right)}{6}\right)$
Nevertheless $x \le 3y$, $\implies \left(- \sqrt{\sqrt 5 + 1}, - \dfrac{\sqrt{\sqrt 5 - 1}}{3}\right)$ is a solution but $\left(\sqrt{\sqrt 5 + 1}, \dfrac{\sqrt{\sqrt 5 - 1}}{3}\right)$ is not.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ How do I formally prove that $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ which can be seen graphically. Here $\epsilon$ is a small positive constant $\ll1$, $x\gg0$ and $c>0$.
| Suppose that $x$ is very large and use Taylor expansion
$$\log \left(1-\frac{1}{x-c}\right)=-\frac{1}{x}-\frac{c+\frac{1}{2}}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$(\epsilon x-c)\log \left(1-\frac{1}{x-c}\right)=-\epsilon +\frac{-c \epsilon +c-\frac{\epsilon }{2}}{x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}=\left(1-\frac{1}{x-c}\right)^{\epsilon x-c}=\exp\left((\epsilon x-c)\log \left(1-\frac{1}{x-c}\right) \right)$$
$$\text{lhs}=e^{-\epsilon }+\frac{e^{-\epsilon } \left(\left(-c-\frac{1}{2}\right) \epsilon
+c\right)}{x}+O\left(\frac{1}{x^2}\right)$$
Do the same for the rhs to get
$$\text{rhs}=e^{-\epsilon }-\frac{e^{-\epsilon } \epsilon }{2
x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}-\text{rhs}=\frac{c e^{-\epsilon } (1-\epsilon )}{x}+O\left(\frac{1}{x^2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry : what is the $\phi$ angle, if area of yellow rectangle is equal with area of red triangle? I have a right triangle and in it area of yellow rectangle is equal with area of red triangle. How could prove that $\phi=45^{\circ}$?
$$\text{Area of Yellow Rectangle}=\text{Area of Red Triangle}$$
This problem has no additional information.
I'm sorry for the inconvenience.
Update:
If we add some constraint to this problem like bellow, is it have a unique solution?
$$|FG|=x+y$$
| Thanks for all.
first equality: area of red and yellow are equal.
$$ 2xy=vw \label{eq:1} \tag{1}$$
Now by using Thales' theorem in this triangle:
$$ \frac{v+y}{c}=\frac{b-x}{b} \Rightarrow v=c(\frac{b-x}{b})-y \label{eq:2} \tag{2}$$
$$ \frac{x+w}{b}=\frac{c-y}{c} \Rightarrow w=b(\frac{c-y}{c})-x \label{eq:3} \tag{3}$$
by importing $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:1}$ and simplifying equations, I gained this equation:
$$ \frac{(x-b)^2}{b^2}+\frac{(y-c)^2}{c^2}=1 \label{eq:4} \tag{4}$$
It's means location of P, which area of red and yellow are equal, is an elliptic with canonical point $B$ and $C$. So we have $\infty$ points for $P$ in the $ABC$.
Now, How to find $\phi$?
By loking at this picture we have:
$$ AG=\sqrt{(x+w)^2+y^2} \label{eq:5} \tag{5}$$
$$ AF=\sqrt{(y+v)^2+x^2} \label{eq:6} \tag{6}$$
$$ FG=\sqrt{v^2+w^2} \label{eq:7} \tag{7}$$
Law of cosines
$$ FG^2=AG^2+AF^2-2AF.FG.\cos(\phi) \label{eq:8} \tag{8}$$
Now by using $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:5}$, $\ref{eq:6}$ and $\ref{eq:7}$ and finally in $\ref{eq:8}$, and simplifying that, I reach to this equation:
$$ \sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}\cos(\phi)= \left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]\label{eq:9} \tag{9}$$
So
$$ \cos(\phi)=\frac{\left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]}{\sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}}\label{eq:10} \tag{10}$$
$$for \: b > x > 0 \: and \: c > y > 0 \: $$
For finding a $\phi$ as a function on $x$ and $y$ we need to considering $\ref{eq:4}$ and $\ref{eq:9}$. If I find exact solution I post it here.
I have writed a python code for determine all possible solution of this equation, but this not have a unique solution. Since all $x$ and $y$ that for them we have $\cos(\phi)=\frac{\sqrt{2}}{2}$ are solutions.
| {
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Integration with $\ln(x)$ in the denominator
Find$$\displaystyle\int_1^\infty\frac{(x^2-1)(x^4-1)(x^6-1)}{\ln(x)(x^{14}-1)} dx$$
I tried simplifying the terms without logarithm
$x^2-1=(x-1)(x+1)\\x^{14}-1=(x^7-1)(x^7+1)$
to see if any substitution is possible but since the logarithm is in the denominator it is getting hard and I am not sure if there is any substitution that would work.
I am looking for hints to get started
| The following answer is similar to Song's answer except that it involves using a common integral representation of the digamma function:
$$\psi(z) = -\gamma + \int_{0}^{1} \frac{1-t^{z}}{1-t} \, \mathrm dt\, , \quad \mathcal \Re (z) >0$$
The integral representation of the tetragamma function used in Song's answer can be derived from this integral representation by differentiating under the integral sign twice.
$$ \begin{align} &\int_{1}^{\infty} \frac{(x^{2}-1)(x^{4}-1)(x^{6}-1)}{(x^{14}-1)\ln x} \, \mathrm dx \\ &= \int_{0}^{1}\frac{(1-u^{2})(1-u^{4})(1-u^{6})}{(1-u^{14}) \ln \frac{1}{u}} \, \mathrm du\\ & = \int_{0}^{1} \int_{0}^{6} \frac{(1-u^{2})(1-u^{4}) \, u^{a}}{(1-u^{14}) \ln u} \, \mathrm da \, \mathrm du\\ &= \int_{0}^{6} \int_{0}^{1} \frac{(1-u^{2})(1-u^{4}) \, u^{a}}{(1-u^{14}) } \, \mathrm du \,\mathrm da \\\ &= \frac{1}{14}\int_{0}^{6} \int_{0}^{1} \frac{(1-w^{2/14})(1-w^{4/14}) \, w^{a/14}}{1-w } \, w^{-13/14} \, \mathrm dw \, \mathrm da \\ &= \frac{1}{14}\int_{0}^{6} \int_{0}^{1} \left(\frac{w^{(a+1)/14-1}-w^{(a+5)/14-1}}{1-w} + \frac{w^{(a+7)/14-1} -w^{(a+3)/14-1}}{1-w} \right) \, \mathrm dw \,\mathrm da \\&= \frac{1}{14} \int_{0}^{6} \left[ \left(\psi \left(\frac{a+5}{14} \right) - \psi \left(\frac{a+1}{14} \right) \right) + \left(\psi \left(\frac{a+3}{14} \right) - \psi \left(\frac{a+7}{14} \right) \right) \right] \, \mathrm d a \\ &= \ln \left[\frac{\Gamma \left(\frac{11}{14} \right) \Gamma\left(\frac{9}{14} \right)}{\Gamma \left(\frac{1}{2} \right)\Gamma \left(\frac{13}{14} \right)} \right]\, - \ln \left[\frac{\Gamma \left(\frac{5}{14} \right)\Gamma \left(\frac{3}{14} \right)}{\Gamma \left(\frac{1}{14} \right)\Gamma \left(\frac{1}{2} \right)}\right] \\&= \ln \left[\frac{\Gamma \left(\frac{1}{14} \right)\Gamma \left(\frac{9}{14} \right) \Gamma\left(\frac{11}{14} \right)}{\Gamma \left(\frac{3}{14} \right)\Gamma \left(\frac{5}{14} \right)\Gamma \left(\frac{13}{14} \right)} \right] \end{align} $$
See Song's answer for how to use properties of the gamma function to show that the above result reduces to $\ln 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Interesting relation derived from the identity $\sin^2 x + \cos^2 x \equiv 1$ Every one knows that
$$\sin^2 x + \cos^2 x \equiv 1.$$
It is also well known that
$$\sin x \equiv \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n + 1}}{(2n+1)!},\quad\cos x \equiv \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.$$
One can rewrite the above mentioned identity as
$$\sum_{n, k = 0}^{\infty} \frac{(-1)^{n+k} x^{2(n+k+1)}}{(2n+1)! (2k+1)!} + \sum_{m, l = 0}^{\infty} \frac{(-1)^{m+l} x^{2(m+l)}}{(2m)! (2l)!} \equiv 1.$$
Terms of the $\sin^2 x$ series cancel out all but one (which is equal to 1) terms of the $\cos^2 x$ series. This fact can be expressed as
$$(-1)^{c-1} x^{2c} \sigma(c) + (-1)^c x^{2c} \gamma(c) = 0,$$
where $n+k+1 = m+l= c \geq 1$ and
$$\sigma(c) = \sum_{n=0}^{c-1} \frac{1}{(2n+1)! (2(c-n)-1)!},\quad\gamma(c) = \sum_{m=0}^{c} \frac{1}{(2m)! (2(c-m))!}.$$
At the moment it is not obvious for me that $$\sigma(c) \equiv \gamma(c).$$
Does any one have an idea how last identity can be proved?
| Here I provide explicit reasoning to make the identity in question obvious (note that I use the notation ${N \choose k} \equiv C(N,k)$).
One can easily note that
$$(2c)! \sigma(c) = \sum_{n = 0}^{c-1} \frac{(2c)!}{(2n+1)! (2c-(2n+1))!}
= \sum_{n = 0}^{c-1} \frac{r!} {b_{odd}! (r -b_{odd})!}
= \sum_{n = 0}^{c-1} C(r,b_{odd}), $$
where $r= 2c,$ $b_{odd} = 2n+1.$
$$(2c)! \gamma(c) = \sum_{m = 0}^{c} \frac{(2c)!}{(2m)! (2c-2m)!}
= \sum_{m = 0}^{c} \frac{r!} {b_{even}! (r - b_{even})!}
= \sum_{m = 0}^{c} C(r,b_{even}),$$
where $r = 2c,$ $b_{even} = 2m.$
From the trivial identity $0 \equiv (1 - 1)^{2c}$ one can derive the relation
$$0 \equiv (1 - 1)^{2c} = \sum_{d = 0}^{2c} C(2c, d) 1^{2c - d} (-1)^{d}
= \sum_{o = 0}^{c-1} C(2c,2o+1) 1^{2c-(2o+1)} (-1)^{2o+1}
+ \sum_{e = 0}^{c} C(2c,2e) 1^{2c-2e} (-1)^{2e}
= -\sum_{o = 0}^{c-1} C(2c,2o+1)
+ \sum_{e = 0}^{c} C(2c,2e).$$
So $$\sum_{o = 0}^{c-1} C(2c,2o+1) \equiv \sum_{e = 0}^{c} C(2c,2e),$$
and hence
$$\sigma(c) \equiv \gamma(c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
extracting a coefficient from formal power series multiplication The Questions is Compute the value of $[x^n] \frac{(1+x)^n}{1-x} $
The solution is as follows:
$[x^n] \frac{(1+x)^n}{1-x} $ = $[x^n](1+x)^n(1 + x + x^2 + x^3 + . . . ) $
= $\sum_{k=0}^n [x^k](1+x)^n * [x^{n-k}](1 + x + x^2 + x^3 + . . . ) $
= $\sum_{k=0}^n {n \choose k}(1)$
= $2^n $
Could someone please explain how this sum is formed?
| $\begin{array}\\
S(n)
&=[x^n] \frac{(1+x)^n}{1-x}
\qquad\text{want coefficient of }x^n\\
&=[x^n](1+x)^n\sum_{m=0}^{\infty} x^m
\qquad\text{expand } \dfrac1{1-x}\\
&=\sum_{k=0}^n [x^k](1+x)^n [x^{n-k}]\sum_{m=0}^{\infty} x^m
\qquad\text{for each }0\le k \le n
\text{ look at separate terms in the product}\\
&=\sum_{k=0}^n \binom{n}{k}\cdot 1
\qquad [x^k](1+x)^n=\binom{n}{k}, [x^{n-k}]\sum_{m=0}^{\infty} x^m=1\\
&=2^n
\qquad \sum_{k=0}^n \binom{n}{k}=2^n\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be able to cancel anything out. I am also struggling with other methods like AM-GM since $a,b$ are not necessarily positive. Any help would be appreciated.
| With variable subsitutions $a = x + y$, $b = x - y$, the inequality becomes $$(x^2 + y^2)^3 - 8 (x^3 + xy^2)(x^2 - y^2 - 2x) \geq 0$$ in the domain $x \geq 0$. Denote the LHS $f(x, y)$. As $f(x, y) = f(x, -y)$, it suffices to prove $f \geq 0$ in the closed first quadrant.
First note that $$f(x, 0) = x^6 - 8x^5 + 16 x^4 = x^4 (x-4)^2 \geq 0$$ so the inequality holds on the $x$-axis. Routine computation gives $$ \frac{\partial f}{\partial y} = 6x^4 y + 12x^2 y^3 + 6y^5 + 32 xy^3 + 32 x^2 y$$ which is clearly nonnegative in the first quadrant, so we have $f(x, y) \geq f(x, 0) \geq 0$ for all $x \in \mathbb{R}_{\geq 0}, y \in \mathbb{R}$, and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$ Given the circle $C: 2x^2 + 2y^2 + 2x + 2y - 13 = 0$. Line $L$, with slope $m$ and passing through the point $P(0,2)$, cuts the circle at points $A$ and $B$ such that $AB=\sqrt{2}$.
*
*Find the equation of $L$.
*Find the equation of the locus of the centers of the circles passing through $A$ and $B$.
The equation of the line is obviously $L:y=mx+2$
The length of the chord $AB$ is $\sqrt{2}$ so the following must be satisfied:
$(x_B-x_A)^2+(y_B-y_A)^2 = 2$
I tried a few things to calculate $m$, but unsuccessfully.
I guess that once I find $m$ the equation of the locus of the centers of the circles passing through $A$ and $B$ is calculated based on the condition that the locus is perpendicular to $AB$ (please correct me if I am wrong).
Any hint would be useful.
| Let $x$ the distance from the centre of the circle $(x+1/2)^2+(y+1/2)^2=7$, I have $x=\sqrt{7-\frac{1}{2}}=\frac{\sqrt{13}}{\sqrt{2}}$. The distance of the line from the centre $P(-1/2,-1/2)$ has to be equal to $x$, so I obtain: $$\frac{|\frac{m}{2}-\frac{1}{2}-2|}{\sqrt{m^2-4}}=\frac{\sqrt{13}}{\sqrt{2}}$$ Solving for $m$ I have that: $25m^2+10m+1=0$ that has only one solution when $m=-\frac{1}{5}$. The equation of the line is: $$y=-\frac{1}{5}x+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a simple pattern to memorize the sine of $0^\circ$, $15^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $75^\circ$, $90^\circ$? We know there is a nice pattern to memorize the sine of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ as follows.
\begin{align}
\sin 0^\circ &= \tfrac12\sqrt0\\
\sin 30^\circ &= \tfrac12\sqrt1\\
\sin 45^\circ &= \tfrac12\sqrt2\\
\sin 60^\circ &= \tfrac12\sqrt3\\
\sin 90^\circ &= \tfrac12\sqrt4
\end{align}
We also know that
\begin{align}
\sin 15^\circ &= \tfrac14(\sqrt6-\sqrt2)\\
\sin 75^\circ &= \tfrac14(\sqrt6+\sqrt2)
\end{align}
Question
If I want to combine these two groups, is there a simple nice pattern available for us to easily rote memorize them?
| Hmm, this pattern works:
$$
\sin 0^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{4}}, \\
\sin 15^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{3}}, \\
\color{gray}{\sin 22.5^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{2}},} \\
\sin 30^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{1}}, \\
\sin 45^{\circ} = \frac{1}{2}\sqrt{2\pm\sqrt{0}}, \\
\sin 60^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{1}}, \\
\color{gray}{\sin 67.5^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{2}},} \\
\sin 75^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{3}}, \\
\sin 90^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{4}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$? A friend asked me to prove
$$\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$$
and here is my approach:
\begin{align}
I&=\int_0^1\frac{1-x}{\ln x}\frac1{1+x}\ dx\\
&=\int_0^1\left(-\int_0^1x^y\ dy\right)\frac1{1+x}\ dx\\
&=\int_0^1\left(-\int_0^1\frac{x^y}{1+x}\ dx\right)\ dy\\
&=\int_0^1\left((-1)^n\sum_{n=1}^\infty\int_0^1x^{y+n-1}\ dx\right)\ dy\\
&=\int_0^1\left(\sum_{n=1}^\infty\frac{(-1)^n}{y+n}\right)\ dy\\
&=\sum_{n=1}^\infty(-1)^n\int_0^1\frac1{y+n}\ dy\\
&=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\tag{1}
\end{align}
Now how can we finish this alternating sum into $\ln\left(\frac2{\pi}\right)$?
My idea was to use
$$\operatorname{Li}_a(-1)=(1-2^{-a})\zeta(a)=\sum_{n=1}^\infty\frac{(-1)^n}{n^a}$$
and if we differentiate both sides with respect to $a$ we get
$$2^{1-a}(\zeta^{'}(a)-\ln2\zeta(a))-\zeta^{'}(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}\ln(n)}{n^a}\tag{2}$$
wolfram says that $\sum_{n=1}^\infty(-1)^n\ln(n)$ is divergent which means we can not take the limit for (2) where $a\mapsto 0$ which means we can not break the sum in (1) into two sums. So any idea how to do (1)?
Other question is, I tried to simplify the sum in (1) as follows
\begin{align}
S&=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\\
&=-\left[\ln(2)-\ln(1)\right]+\left[\ln(3)-\ln(2)\right]-\left[\ln(4)-\ln(3)\right]+\left[\ln(5)-\ln(4)\right]-...\\
&=-2\ln(2)+2\ln(3)-2\ln(4)+...\\
&=2\sum_{n=1}^\infty(-1)^n\ln(n+1)
\end{align}
which is divergent again. what went wrong in my steps?
Thanks.
| Note that $$\int_0^1 x^y dy=\frac{x-1}{\ln x}$$using this result our integral becomes $$\begin{aligned} I &=\int_0^1\frac{x-1}{(x+1) \ln x}\\&=\int_0^1\left(\frac{1}{x+1}\int_0^1x^ydy\right)dx \\&=\int_0^1\int_0^1\frac{x^y}{1+x}dy dx\\&=\int_0^1\underbrace{\int_0^1\frac{x^y}{1+x}dx }_{I_1}dy \end{aligned}$$ Since the integral $$I_1=\int_0^1\frac{x^y}{1+x}dx =\frac{1}{2}\left(H_{\frac{
y}{2}}-H_{\frac{y-1}{2}}\right)=\frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$$ is well know result which I have proved here by polynomial long division. Integrating $I_1$ we yield $$ \begin{aligned}I &= \int_0^1 I_1 dy \\&=\left[\ln\Gamma\left(\frac{y+2}{2}\right)-\ln\Gamma\left(\frac{y+1}{2}\right)\right]_0^1\\& =\ln\left[
\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\right)\cdots(1)\\&=\ln\left(\frac{\sqrt{\pi}}{2}\sqrt{\pi}\right)=\ln\left(\frac{\pi}{2}\right)\end{aligned}$$ we use the half gamma $\displaystyle \Gamma\left(\frac{1}{2}+n\right)=\frac{(2n)!}{4^n n!}\sqrt{\pi}$ argument in $(1)$ or using functional equation of gamma function we can write $(1)$ as $\displaystyle=\ln\left(\frac{1}{2}\Gamma^2\left(\frac{1}{2}\right)\right)=\ln\left(\frac{\pi}{2}\right)$
Alternatively$$\begin{aligned}I_1 & =\int_0^1\frac{x^y}{1+x}dx \\&=\int_0^1x^y \left(\sum_{r=0}^{\infty} (-1)^r x^r\right)dx\\&=\sum_{r=0}^{\infty} \frac{(-1)^r}{y+r+1}=\Phi\left(-1,1,y+1\right)\cdots(3)\end{aligned}$$ where $\Phi(z,s,\alpha )$ is Lerch Transcendent function using the equation 5 and 6 we obtain $$I= \frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$$ To prove tha relationship between $ (3)$ and final result we can directly use series formula of digamma function.
The above integral can be related as
$$\int_0^{\infty} \frac{\operatorname{tanh}x}{e^{2x}} \frac{dx}{x}=\int_0^1\frac{x-1}{(1+x)\ln x} =-\ln\left(\frac{2}{\pi}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$ Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
My attempt
$\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
$= \lim_{x \to \infty} (\ln(x+1) - \ln (x))(\ln (x+1) + ln(x))$
$=\lim_{x \to \infty} (\ln (\frac{x+1}{x})(\ln x(x+1))$
It's $0 \times \infty$ indeterminate form.
Then I'm stuck.
How to proceed$?$
| Write
$$\log(x+1)=\log \left(x\left(1+\frac{1}{x}\right) \right)=\log(x)+\log \left(1+\frac{1}{x}\right) $$ and, by Taylor,
$$\log \left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$
So,
$$A=\log ^2(x+1)-\log ^2(x)=\left(\log(x)+\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \right)^2-\log^2(x)$$ Expand to get
$$A=2 \frac{\log(x)}x+\frac{1-\log (x)}{x^2}+O\left(\frac{1}{x^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
} |
What is the solution $f:\mathbb{R^3}\to \mathbb{C}$ to the functional equation $f(A)f(B)=f(A+B+A\times B)$? Given $3$-vectors $A$ and $B$ and an unknown function $f:\mathbb{R^3}\to \mathbb{C}$, are there any solutions to this functional equation?
$$f(A)f(B)=f(A+B+A\times B)$$
apart from the trivial solutions $f(A)=1$ or $f(A)=0$.
Edit: As a simpler example a solution to $f(A)f(B)=f(A+B)$ is $f(A)=e^{k\cdot A}$ for some $3$-vector $k$.
| The functional equation has no non-trivial solution.
For any $a, b \in \mathbb{R}^3$ such that $|a| = |b|$ and $a \ne -b$. Let
$$A = \frac{a+b}{4} + \frac{2}{|a+b|^2}(a \times b)
\quad\text{ and }\quad
B = \frac{a+b}{4} - \frac{2}{|a+b|^2}(a \times b)
$$
Notice
$$\begin{align}A + B + A \times B
&= \frac{a+b}{2} - \frac12(A+B)\times(A-B)\\
&= \frac{a+b}{2} - \frac{1}{|a+b|^2} (a+b)\times(a \times b)\\
&= \frac{a+b}{2} - \frac{1}{|a+b|^2}\left((a\cdot b + |b|^2)a - (|a|^2+b\cdot a)b\right)\\
&= \frac{a+b}{2} - \frac{a-b}{2}\\
&= b
\end{align}
$$
and similary, $A +B - A \times B = a$. We have
$$\begin{align}
f(a) &= f(A+B-A\times B) = f(A+B+B\times A) = f(B)f(A)\\
f(b) &= f(A+B+A\times B) = f(A)f(B)\\
\implies f(a) &= f(b)
\end{align}
$$
For any $a \in \mathbb{R}^3$ such that $a \ne 0$, take a $b\in \mathbb{R}^3$ with $|b| = |a|$ and $b \ne a, -a$. This leads to
$$f(a) = f(b) = f(-a)$$
Combine these, we can deduce $f(a)$ depends only on $|a|$.
For any $a \in \mathbb{R}^3$, we have
$$f(a)^2 = f(a)f(-a) = f(a - a - a \times a) = f(0)$$
This forces $f(a)$ to be a constant function.
Since $f(0)^2 = f(0+0+0\times 0) = f(0)$, the constant can only be $0$ or $1$.
i.e. the functional equation has no non-trivial solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$
$$ = \frac{\frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} + \tan(c)}{1 - \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \tan(c)} $$
$$ = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a) \tan(b) \tan(c)}{ (1-\tan(a)\tan(b)) - \tan(a) \tan(c) -\tan(b) \tan(c) } $$
Then the denomenator must be $0$, so it is proven?
Another way:
$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1 $$
$$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$
$$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$
$$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$
| Writing $$\tan c=\tan (\pi/2 -(a+b))=\frac 1{tan(a+b)}=\frac {1-\tan a\tan b}{\tan a +\tan b}$$
So we have $$\tan a\tan b+\tan a\tan c+\tan b\tan c=\tan a\tan b+ \frac {\tan a-\tan^2 a\tan b}{\tan a +\tan b}+ \frac {\tan b-\tan a\tan^2 b}{\tan a +\tan b}= \frac {\tan^2 a\tan b+\tan a\tan^2 b+\tan a-\tan^2 a\tan b+\tan b-\tan a\tan^2 b}{\tan a +\tan b}= \frac {\tan a+\tan b}{\tan a +\tan b} =1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix} a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix} a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b \\ 0 & 1 \end{pmatrix}$
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time.
| I'm partial to the SVD for this. The eigenvalues are $a$ and $1$ associated with eigenvectors $(1,0)$ and $(b,1-a)$ respectively.
So, we get
$$\begin{bmatrix}
a & b \\
0 & 1
\end{bmatrix}^n= \begin{bmatrix}
1 & b \\
0 & 1-a
\end{bmatrix} \begin{bmatrix}
a^n & 0 \\
0 & 1
\end{bmatrix} \begin{bmatrix}
1 & \frac{-b}{1-a} \\
0 & \frac{1}{1-a}
\end{bmatrix} = \begin{bmatrix}
a^n & \frac{b(1-a^n)}{1-a} \\
0 & 1
\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to solve given differential equation $(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$? The equation to solve is:
$$(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$$
I tried putting $xy=t$ but that just gave me this:
$$\frac{t^3-t^2-t+1}{t^3+t^2+t+1}dt=\frac{dx}{x}$$
I suppose there must be some clever factoring involved somewhere but I can't see it so can someone guide me on how to advance or perhaps suggest an alternate method?
| Grouping terms, we get
$$(x^3y^3+1)(ydx+xdy) - (x^2y^2+xy)(xdy-ydx) = 0$$
Now substitute $t=xy$ and $s=\frac{y}{x}$
$$\implies dt = ydx+xdy \hspace{20 pt} x^2 ds = \frac{t}{s}ds = xdy - ydx$$
turning the differential equation into
$$s(t^3+1)dt - t^2(t+1)ds = 0$$
which is now separable and yields
$$\int \frac{ds}{s} = \int \frac{t^3+1}{t^2(t+1)}dt = \int 1 - \frac{1}{t} + \frac{1}{t^2}dt$$
$$\implies \log|s| = t - \log|t| - \frac{1}{t} + C$$
Substituting back in for $x$ and $y$
$$\log (y^2) - xy + \frac{1}{xy} = C$$
or rearranging terms we can have
$$y^2\exp\left(\frac{1}{xy} - xy \right) = C > 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something particularly interesting happens:
$$e^{i z} = 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots \\
=1 + iz - \frac{z^2}{2!} - \frac{i z^3}{3!} + \frac{z^4}{4!} + \frac{i z^5}{5!} + \cdots \\
= (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i (z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots),$$
so $$e^{i z} = \cos z + i \sin z.$$
But I think the author didn't use a rearrangement of the terms at all.
Am I right?
Let
$$c_n := 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots + (-1)^n \frac{z^{2 n}}{(2 n)!},$$
$$s_n := z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots + (-1)^n \frac{z^{2 n + 1}}{(2 n + 1)!},$$
$$e_n := 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots + \frac{(iz)^n}{n!}.$$
Then, $$e_{2 n + 1} = c_n + i s_n,$$
$$\lim_{n \to \infty} e_{2 n + 1} = e^{i z},$$
$$\lim_{n \to \infty} c_n + i s_n = \lim_{n \to \infty} c_n + i \lim_{n \to \infty} s_n = \cos z + i \sin z,$$
so $$e^{i z} = \cos z + i \sin z.$$
| Technically a rearrangement of a series $\sum_{n=0}^\infty a_n$ is understood to be any series $\sum_{n=0}^\infty a_{f(n)}$ where $f : \mathbb N_0 \to \mathbb N_0$ is a bijection. Here $\mathbb N_0$ is the set of nonnegative integers.
In that sense no rearrangement of $e^{iz} = \sum_{n=0}^\infty \frac{(iz)^n}{n!}$ can produce the desired equation $e^{iz} = \cos z + i\sin z$. The "trick" is that for any strictly increasing $g : \mathbb N_0 \to \mathbb N_0$ we have
$$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a'_n \text{ with } a'_n =
\begin{cases}
a_m & n = g(m) \\
0 & n \notin g(\mathbb N_0)
\end{cases}$$
Let us write $\sum_{n=0}^\infty a'_n = [\sum_{n=0}^\infty a_n]*g$.
Applying this to $a_n = (-1)^n\frac{z^{2n}}{(2n)!}$ with $g(m) = 2m$ and $b_n = (-1)^n\frac{z^{2n+1}}{(2n+1)!}$ with $h(m) = 2m+1$ yields the result: We get
$$\left[\sum_{n=0}^\infty (-1)^n\frac{z^{2n}}{(2n)!}\right]*g + i\left[\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{(2n+1)!}\right]*h = e^{iz} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How am I supposed to find the area of the shaded quadrilateral? In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.
|
It can be deduced that the area $[FED] = 7\cdot 4/14=2$.
Furthermore, examine the ratios below,
$$\frac{[FDC]}{[BDC]}=\frac{9}{18}=\frac{\frac{a}{a+b} \frac{d}{c+d}[ABC] }{ \frac{b}{a+b}[ABC]} =\frac ab \frac{1}{\frac cd +1} \tag{1}$$
$$\frac{[FDB]}{[FCB]}=\frac{6}{21}=\frac{\frac{c}{c+d} \frac{b}{a+b} [ABC] }{ \frac{d}{c+d} [ABC]} =\frac cd \frac{1}{\frac ab +1} \tag{2}$$
From (1) and (2),
$$ \frac cd = \frac 12$$
Then, the area of the quadrilateral is
$$[AFED]= 2+ [AFD] = 2+\frac 12\cdot 9 = \frac {13}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$
If $a+b+c=2$ where $0 < {a,b,c} <1$ find the range of $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}.$$
I have tried using AM-GM but I was not able to solve it . I also assumed the expression equal to $k$ and tried to find the range of $k$ but I derived some absurd results. Please help me to solve it and guide me to the correct approach.
| By Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
$$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}\geq 8$$
where the equality holds for $a=b=c=2/3$.
Moreover, for $t\geq 3$, by letting $b(t)=c(t)=1-1/t\in (0,1)$ and $a(t)=2/t\in (0,1)$, then $a(t)+b(t)+c(t)=2$ and
$$g(t):=\frac{a(t)}{1-a(t)}\cdot\frac{b(t)}{1-b(t)}\cdot\frac{c(t)}{1-c(t)}= \frac{2(t-1)^2}{t-2}.$$
Hence $g(3)=8$ and $\lim_{t\to +\infty} g(t)=+\infty$.
Finally, by continuity, it follows that the required range is $g([3,+\infty))=[8,+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196$$
$$\alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 \implies 84 + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 $$
$$ \alpha \beta + \beta \gamma + \alpha \gamma = 56 $$
Also
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{3} + \beta^{3} + \gamma^{3} + 3(\alpha^{2} \beta + \alpha^{2} \gamma + \beta^{2} \alpha + \beta^{2} \gamma + \gamma^{2} \alpha + \gamma^{2} \beta) + 3 \alpha \beta \gamma = 2744$$
$$ 584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + 6 \alpha \beta \gamma$$
$$ \alpha \beta \gamma = 32$$
So that
$$ (\alpha + \beta + \gamma)^{4} = ( \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) )^{2} $$
$$ 38416 = (\alpha^{2} + \beta^{2} + \gamma^{2})^{2} + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2}) + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)^{2}$$
$$ 38416 = \left[ \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta)^{2} + (\beta \gamma)^{2} + (\alpha \gamma)^{2} \right) \right] + 4 (56)(84) + 4 (56^{2}) $$
$$ 7056 = \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta + \alpha \gamma + \beta \gamma)^{2} - 2\alpha \beta \gamma(\alpha + \beta + \gamma ) \right) $$
$$ 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(32)(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} = 2576$$
| Define $A,B,C$ such that $(x-\alpha)(x-\beta)(x-\gamma) = x^3 - Ax^2 + Bx - C$.
For any $n \ge 1$, let $p_n = \alpha^n + \beta^n + \gamma^n$. In terms of $p_n$, the question becomes:
Given $p_1 = 14, p_2 = 84, p_3 = 584$, what is $p_4$?
There is a bunch of Newton's identities connecting $p_n$ and the coefficients $A,B,C$ above.
Using these identities, the computation of $p_4 = \alpha^4 + \beta^4 + \gamma^4$ is mechanical.
*
*$p_1 = A \\ \quad \implies A = 14$
*$p_2 = A p_1 - 2B \\ \quad \implies B = \frac12(A^2 - p_2) = \frac12(14^2 - 84) = 56$
*$p_3 = A p_2 - B p_1 + 3C \\ \quad \implies C = \frac13 ( p_3 - Ap_2 + B p_1)
= \frac13(584 - 14\cdot 84 + 56\cdot 14) = 64$
Finally, we have
$$p^4 = Ap_3 - Bp_2 + Cp_1 = 14\cdot 584 - 56\cdot 84 + 64\cdot 14 = 4368$$
This is different from what you have got.
When you compute $\alpha\beta\gamma$, you have make a mistake in one of the coefficient. The correct step should be
$$584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + \color{red}{3} \alpha \beta \gamma\\
\implies \alpha\beta\gamma = \color{red}{64}
$$
Other than this, the rest of your derivation seems fine. In fact, if you substitute $32$ by $64$ in last two lines, you get the right answer.
$$\require{cancel}
7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(\color{red}{\cancelto{64}{\color{gray}{32}}})(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} =
\color{red}{\cancelto{4368}{\color{gray}{2576}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3\theta=1/2$ are
$$\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}$$
| Consider the diagram below.
When does $\sin\theta = \sin\varphi$?
The sine of a directed angle is the $y$-coordinate of the point where its terminal side intersects the unit circle. Clearly, one solution is when $\theta = \varphi$. By symmetry, $\sin\theta = \sin(\pi - \theta)$, so $\varphi = \pi - \theta$ is another solution. Moreover, any angle coterminal with $\theta$ or $\pi - \theta$ will also have the same sine. Since coterminal angles differ by integer multiples of $2\pi$, $\sin\theta = \sin\varphi$ if
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = \pi - \theta + 2m\pi, m \in \mathbb{Z}$$
Solve the equation $\sin(3\theta) = \dfrac{1}{2}$ in the interval $[0, 2\pi]$.
A particular solution of the equation
$$\sin\alpha = \frac{1}{2}$$
is
$$\alpha = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$
Substituting $3\theta$ for $\alpha$ yields
$$\sin(3\theta) = \sin\left(\frac{\pi}{6}\right)$$
Thus, we have the general solution
\begin{align*}
3\theta & = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z} & 3\theta & = \pi - \frac{\pi}{6} + 2m\pi, m \in \mathbb{Z}\\
\theta & = \frac{\pi}{18} + \frac{2k\pi}{3}, k \in \mathbb{Z} & 3\theta & = \frac{5\pi}{6} + 2m\pi, m \in \mathbb{Z}\\
& & \theta & = \frac{5\pi}{18} + \frac{2m\pi}{3}, m \in \mathbb{Z}
\end{align*}
Since we seek solutions in the interval $[0, 2\pi]$, $k = 0, 1, 2$ and $m = 0, 1, 2$, which yields
\begin{align*}
\theta & = \frac{\pi}{18}, \frac{13\pi}{18}, \frac{25\pi}{18} & \theta & = \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{29\pi}{18}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Confused about dealing with Big $\mathcal{O}$ in functions For the following defined functions
$$f(x) = \frac{1}{x}+\frac{n-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)$$
and
$$y(x) = c\left(\frac{1}{x}+\frac{m-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)$$
where $m > n > 0$ and $c$ is large.
I want to bound $f(x)-y(x))$.
I thought that $f(x)-y(x) \leq \frac{1}{x}+\frac{n-1}{x^2}- c\left(\frac{1}{x}+\frac{m-1}{x^2} \right)$, since we are talking about an approximation!
Is this true?
After that, I found this question where the answer states that
For every $x\geq0$, $f(x)-y(x) \leq f(x)$
Is this the tighter bound? if so, how can I get rid of the $\mathcal{O}$ in $f(x)$?
| Regarding your first question
I thought that $ f(x) − y(x) \leq \frac{1}{x} + \frac{n−1}{x^2} −c (\frac{1}{x} + \frac{m−1}{x^2}) $
, since we are talking about an approximation!
Is this true?
No. Consider for example $ f(x) = \frac{1}{x} + \frac{n-1}{x^2} + \frac{c+1}{x^3}, g(x) = c(\frac{1}{x} + \frac{m-1}{x^2} + \frac{1}{x^3}) $. Then you get $ f(x) - g(x) = \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) + \frac{1}{x^3} > \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\ln(x^2)=x$? I am struggling to solve the equation $\ln(x^2)=x.$ I can't figure out how to isolate the variable. Exponentiating both sides leaves me with $x=e^{\frac{x}{2}}.$ I would really appreciate any guidance, or first steps.
Also, isn't one of the properties of logs that $\ln(x^2)=2\ln(x)?$ How can these two things be equal if the term on the left hand side can input negative numbers, but the term on the right cannot?
| Consider that you look for the zero's of function
$$f(x)=\log(x^2)-x$$ for which
$$f'(x)=\frac 2 x-1 \qquad \text{and} \qquad f''(x)=-\frac 2 {x^2} < 0 \qquad \forall x$$
The first derivative cancels at $x=2$ and $f(2)=\log(4)-2 <0$. By the second derivative test $x=2$ corresponds to a maximum; then, no solution for $x >0$.
By inspection, $f(-1)=1$ making the solution to be between $-1$ and $0$.
Being lazy, build the Taylor series around $x=-1$ to get
$$f(x) \sim 1-3 (x+1)-(x+1)^2+O\left((x+1)^3\right)$$ Solving the quadratic and keeping the solution in the range, then $x=\frac{1}{2} \left(\sqrt{13}-5\right)\approx -0.697224$ which is not too bad.
For better approximations, instead of Taylor series, build $[2,n]$ Padé approximants and set the numerator equal to zero. For example, for $n=2$,
$$f(x)\sim \frac{1-\frac{23 }{6}(x+1)+\frac{14}{9} (x+1)^2 } {1-\frac{5 }{6}(x+1)+\frac{1}{18} (x+1)^2 }\implies x=-\frac{3 \sqrt{305}-13}{56}\approx -0.703442$$ while the exact solution given by Lambert function would be $\approx -0.703467$.
Using the $[2,4]$ Padé approximant would lead to the exact result (for six significant figures).
Edit
Using @Quanto's answer, we can make amazing approximations building the $[1,n]$ Padé approximant of function $(e^x-x^2)$ around $x=-1$. The problem reduces to a linear equation in $(x+1)$. Below are given the expressions and their decimal values
$$\left(
\begin{array}{ccc}
n & \text{formula} & \text{value} \\
0 & -\frac{2+e}{1+2 e} & -0.73304361 \\
1 & -\frac{3+13 e+2 e^2}{1+11 e+6 e^2} & -0.70599412 \\
2 & -\frac{8+140 e+164 e^2+12 e^3}{2+80 e+194 e^2+48 e^3} & -0.70361456 \\
3 & -\frac{30+1674 e+6426 e^2+3390 e^3+144 e^4}{6+738 e+5058 e^2+5142 e^3+720 e^4} & -0.70346790
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach :
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} $$
$$ \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
$$ \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
Applying L Hopital,
$$\lim_{x\to0}\frac{2\sin(x)\cos(x)}{2x}=1$$
But the actual answer is $\frac{2}{3}$. What am I doing wrong here?
| $$L=\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2[x] \right)= \frac{1}{x^2}-\frac{1}{(x+x^3/3+O(x^5))^2}$$ Let us use $(1+z)^{\nu} \approx 1+\nu z, ~if ~$z$ <<1.$
$$\Rightarrow L=\frac{1}{x^2}-\frac{1}{x^2} (1+x^2/3+O(x^4)^{-2}=\frac{1}{x^2}-\frac{1}{x^2} (1-2x^2/3+O(x^4))= \lim_{x \rightarrow 0} \frac{2}{3}+O(x^2)=\frac{2}{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the coefficient of $x^{19}$ in the generating function of $f(x) = \frac{(1-x^{5})^5}{(1-x)^{10}}$
Find the coefficient of $x^{19}$ in the generating function development of $f(x) = \dfrac{(1-x^{5})^5}{(1-x)^{10}}$
My Attempt:
Let $\mathbf{(I)} \ (1-x^5){^5}$, then $(1-x^5){^5} = \sum_{k=0}^{5} \binom{5}{k} \left((-1)^{k}\cdot x^{5k}\right)$
And let $\mathbf{(II)} \ \left(\frac{1}{1-x}\right)^{10} = \left(\sum_{l=0}^{\infty}{\left(x^{l}\right)^{}}\right)^{10} = \sum_{l=0}^{\infty} \binom{l+9}{9}\cdot{x^{l}}$
I'd like to find all values of $k,l$ for which we'll get $x^{19}$.
as $k=0,l=19:x^{5\cdot(0)}\cdot x^{19}=x^{19}$, and the coefficient of $x^{19}$ is $\binom{28}{9}$
as $k=1,l=14:x^{5\cdot(1)}\cdot x^{14}=x^{19}$,
and the coefficient of $x^{19}$ is $-5\cdot \binom{23}{9}$
for $k=2,l=9:x^{5\cdot(2)}\cdot x^{9}=x^{19}$,
and the coefficient of $x^{19}$ is $10\cdot\binom{18}{9}$
for $k=3,l=4:x^{5\cdot(3)}\cdot x^{4}=x^{19}$,
and the coefficient of $x^{19}$ is $-15\cdot \binom{13}{9}$
Finally, the coefficient of $x^{19} = \binom{28}{9} -5\binom{23}{9} +10\binom{18}{9} -15\binom{13}{9}$
Is that True?
| You are almost correct! Since $\binom{5}{3}=10$ (not $15$), it should be
$$[x^{19}]f(x) = \binom{28}{9} -5\binom{23}{9} +10\binom{18}{9} -\color{red}{10}\binom{13}{9}=3300000.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Roots of x inside square root In finding the roots of
$ \sqrt{6-4x-x^2}=x+4$
I get that the roots of $x$ are $-5$ and $-1$.
However first I need to take into account that:
*
*$x+4 \ge 0$
and
*$6-4x-x^2 \ge 0$.
Considering the second equation gives me:
$x \ge -2+ \sqrt{10}$,
would the roots of $x$ I found in the first equation be considered invalid?
| Let $x+4 \ge 0~~~(1)$ and for tje reality of the roots $$6-4x-x^2 \ge 0~~~~~~(2) \Rightarrow x^2+4x-6 \le 0 \Rightarrow -2-\sqrt{10} \le x \le -2+\sqrt{10}.$$ So that we can square the given equation, then we have
$$x^2+6x+5 \Rightarrow x=-5, -1$$. As per the conditiona (1,2), $x=-1$ is the correct root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to identify that a polynomial has factor $(x+y)^2$ About integer polynomial $f(x,y)$, conclusions such as "If $f(x,x)=0$, then $y-x$ is a factor" "If $f(x,wx)=0$, then $x^2+xy+y^2$" is a factor" are easy.
But what about $(x-y)^2$? How to verify, or identify that $f(x,y)$ has a factor $(x+y)^2$? Does differentiation work?
| Let $t:=x+y$.
Then we must have
$$P(x,t-x)=t^2Q(x,t-x).$$
E.g.
$$P(x,y)=x^3+x^2-xy^2+2xy-y^3+y^2+yx^2\to P(x,t-x)=2xt^2-t^3+t^2.$$
You can simply check that the polynomial has a double root at $t=0$,
$$\left.P(x,t-x)\right|_{t=0}=P(x,-x)=0,$$
$$\left.\frac{\partial P}{\partial y}(x,t-x)\right|_{t=0}=\frac{\partial P}{\partial y}(x,-x)=0.$$
Indeed
$$x^3+x^2-x(-x)^2+2x(-x)-(-x)^3+(-x)^2+(-x)x^2=0$$
and
$$x^2-2x(-x)+2x-3(-x)^2+2(-x)=0.$$
Another option is to divide by $(x+y)^2=x^2+2xy+y^2$:
$$x^3+x^2-xy^2+2xy-y^3+y^2+yx^2\\
=x(x^2+2xy+y^2)-x^2y+x^2+2xy-y^3+y^2\\
=(x-y)(x^2+2xy+y^2)+x^2+2xy+y^2\\
=(x-y+1)(x^2+2xy+y^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that the unequality is right for $n = k + 1$ as well, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$
I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
| Using the induction hypothesis:
$$\frac{2k+1}{2k+2} ≤ \frac{\sqrt{2k+1}}{\sqrt{2k+3}}$$
Now let $u = 2k+2$:
$$\frac{u-1}{u} ≤ \frac{\sqrt{u-1}}{\sqrt{u+1}}$$
For $u ≥ 1$ or $2k + 2 ≥ 1$ which implies $k ≥ -\frac{1}{2}$, both sides are positive. So we can square both sides:
$$\frac{u^2-2u+1}{u^2} ≤ \frac{u-1}{u+1}$$
$$u^3-u^2-u+1 ≤ u^3-u^2$$
$$u ≥ 1$$
Obviously this is not a formal proof. To do so, you need to invert all the steps starting from $u ≥ 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
| You may use the hockey-stick identity, like in the computation of $\sum_{n=1}^{N}n^m$. We have
$$ \binom{n}{2}^2 = 6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1} $$
hence
$$\begin{eqnarray*} \sum_{n=1}^{N}\binom{n}{2}^2 &=& 6\binom{N+1}{5}+12\binom{N+1}{4}+7\binom{N+1}{3}+\binom{N+1}{2}\\&=&\color{red}{\frac{N(N+1)(N+2)(3N^2+6N+1)}{60}}.\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$ Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$
First, we have
$$ n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 $$
for all positive integers $n$. (Since $\frac{\sin x}{x} < 1 \forall x > 0$.) Therefore,
$$ \sum_{n=1}^{\infty}\left| \log \left( n \sin \frac{1}{n} \right)\right| = \sum_{n=1}^{\infty}-\log \left( n \sin \frac{1}{n} \right) = \sum_{n=1}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right). $$
\begin{align*} \sin \frac{1}{n} &= \frac{1}{n} - \frac{1}{6n^3} + \cdots \\[9pt] \implies n \sin \frac{1}{n} &= 1 - \frac{1}{6n^2} + \cdots \\[9pt] &\geq 1 - \frac{1}{6n^2} \\[9pt] &\geq 1 - \frac{1}{n^2}. \end{align*}
Could someone please guide me on how to proceed from here?
Thanks a lot for your help.
| We have $\sin \dfrac{1}{n} = \dfrac{1}{n}-\dfrac{1}{3! n^3} + o\left(\dfrac{1}{n^3}\right)$, then :
$$\log\left( n \sin \dfrac{1}{n} \right) = \log\left( 1-\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) \right) = -\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) + \mathcal{O}\left( \dfrac{1}{n^4} \right)$$
Then :
$$\log\left( n \sin \dfrac{1}{n} \right) = \left(-\dfrac{1}{6}+o(1)\right) \dfrac{1}{n^2}+ \mathcal{O}\left( \dfrac{1}{n^4}\right)$$
Then:
$$n \to +\infty, \log\left( n \sin \dfrac{1}{n} \right) \sim \dfrac{-1}{6n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| If $a \ge b$, then $a^3 \ge b^3$ and $a^5 \ge b^5$, which implies $a^3 - b^3 \ge 0$ and $a^5 - b^5 \ge 0$. Since a positive times a positive is positive (or $0$ times anything is $0$), $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
If instead $a < b$, then
$a^3 < b^3$ and $a^5 < b^5$, which implies $a^3 - b^3 < 0$ and $a^5 - b^5 < 0$. Since a negative times a negative is positive, $$(a^3 - b^3)(a^5 -b^5) > 0.$$
Therefore, for all $a,b \ge 0$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Calculating $\iiint_E\sqrt{3x^2+3z^2}\,dV$
Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.
I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $\theta$ I got $\tan^{-1}4\leq\theta\leq\pi-\tan^{-1}4$.
$$\begin{align}\iiint_E\sqrt{3x^2+3z^2}\,dV &=\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}\int_8^{2r^2}\sqrt{3}\cdot r\ dy \ r\ dr\ d\theta\\&=2\sqrt3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}r^2(r^2-4)dr\ d\theta\\&=2\sqrt3\ 8^3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\csc^3\theta\bigg(\dfrac{8^2\csc^2\theta}{5}-\dfrac{4}{3}\bigg)d\theta\end{align}$$
Now, this is ugly to get through, please help, is there any efficient way to carry out this problem?
| We do not have to convert to other coordinate system. Since for a fixed value of y, the cross section is a circle with radius of $\sqrt{\frac{y}{2}}$. Then the integral becomes
$$\int_{y=0}^{8} \int_{x=-\sqrt{\frac{y}{2}}}^{\sqrt{\frac{y}{2}}} \int_{z=-\sqrt{\frac{y}{2}-x^2}}^{\sqrt{\frac{y}{2}-x^2}} dz dx dy$$
$$= \int_{y=0}^{8} \frac{\pi}{2} \frac{\sqrt{3}}{\sqrt{2}} y^{\frac{3}{2}} dy$$
$$= \frac{128\sqrt{3}\pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But I don't see a pattern in the left side.
| Let $x$ be the given iterated radical. Render
$x^2=2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
Now multiply $x$ by $\sqrt{3/2}$:
$(\sqrt{3/2})x=\sqrt{3+(3/2)\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+(9/4)\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+(81/16)\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+\sqrt{(19683/128)+...}}}}$
The second and subsequent radicands in $(\sqrt{3/2})x$ are larger than those in $x^2-2$ forcing
$x^2-2<(\sqrt{3/2})x$
$x^2-(\sqrt{3/2})x-2<0$
$x<\sqrt{3/8}+\sqrt{19/8}$ from the quadratic formula
The sum of two square roots is less than twice the square root of the average of the radicands:
$x<2\sqrt{11/8}, x^2<11/2<9, \color{blue}{x<3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the coefficient of $ x^6 $in the given expansion. Find the coefficient of $x^6$ in $\left[(1+x)(1+x^2)^2(1+x^3)^3 \cdots (1+x^n)^n\right]$.
Expansion
$$\left[\left(1+\binom11x \right)\left(1+\binom21x + \binom22x^2 \right)\left(1 +\binom31x+\binom32x^2+\binom33x^3 \right)\cdots \left(1 + \binom n1 x + \binom n2x^2 + \cdots +\binom n nx^n \right) \right]$$
I expanded it as shown above but couldn't proceed further. Any Help would be appreciated.
| Here we give a supplement regarding the cases $1\leq n\leq 5$. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
Cases $n=1,2$:
\begin{align*}
[x^6](1+x)=[x^6](1+x)(1+x^2)^2\color{blue}{=0}
\end{align*}
Case $n=3$:
\begin{align*}
\color{blue}{[x^6]}&\color{blue}{(1+x^3)^3(1+x^2)^2(1+x)}\\
&=[x^6](1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\tag{1}\\
&=\left([x^6]+3[x^3]+3[x^0]\right)(1+2x^2+x^4)(1+x)\tag{2}\\
&=0+3\cdot2+3\cdot 1\\
&\,\,\color{blue}{=9}
\end{align*}
Case $n=4$:
\begin{align*}
\color{blue}{[x^6]}&\color{blue}{(1+x^4)^4(1+x^3)^3(1+x^2)^2(1+x)}\\
&=[x^6](1+4x^4)(1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\\
&=[x^6](1+3x^3+4x^4+3x^6)(1+2x^2+x^4)(1+x)\tag{3}\\
&=\left([x^6]+3[x^3]+4[x^2]+3[x^0]\right)(1+2x^2+x^4)(1+x)\\
&=0 + 3\cdot 2+4\cdot 2+3\cdot 1\\
&\,\,\color{blue}{=17}
\end{align*}
Case $n=5$:
\begin{align*}
\color{blue}{[x^6]}&\color{blue}{(1+x^5)^5(1+x^4)^4(1+x^3)^3(1+x^2)^2(1+x)}\\
&=[x^6](1+5x^5)(1+4x^4)(1+3x^3+3x^6)(1+2x^2+x^4)(1+x)\\
&=[x^6](1+4x^4+5x^5)(1+2x^2+3x^3+x^4+6x^5+3x^6)(1+x)\tag{4}\\
&=\left([x^6]+4[x^2]+5[x]\right)(1+2x^2+3x^3+x^4+6x^5+3x^6)(1+x)\\
&=(6+3)+4\cdot 2+5\cdot 1\\
&\,\,\color{blue}{=22}
\end{align*}
Comment:
*
*In (1) we expand the terms skipping summands with powers of $x$ greater than $6$.
*In (2) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (3) we multiply out the left-most two terms ignoring summands which do not contribute to $[x^6]$.
*In (4) we multiply out the left-most two terms and the next two terms ignoring summands which do not contribute to $[x^6]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
My attempt:-
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$
Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$
and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
| In your proof, after $6-2x\sqrt{6}+x^2=5+2\sqrt{6}$ we have that
$$x^2+1=2(x+1)\sqrt{6}$$
If $x=-1$ then, from the above equation, it follows that $2=0$. Therefore $x$ is a rational number different from $-1$. After dividing by $2(x+1)\not=0$ we get
$$\sqrt{6}=\frac{x^2+1}{2(x+1)}\in \mathbb{Q}.$$
Contradiction! Hence $x=\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not a rational number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Trigonometric systems of equations I seek to express $DE$ and $UI$ and $IE$ in terms of $\theta$:
$$\sec { \theta } =DE+IE+UI \tag{1.1}\label{myeqone}$$
$$\frac { 1 }{ 4 } =UI(UI+IE) \tag{2.1}\label{myeqtwo}$$
$$(\frac { 1 }{ 2 } -\tan { \theta } )^{ 2 }=DE(DE+IE) \tag{3.1}\label{myeqthree}$$
I rewrote $(2)$ and $(3)$ as
$$\frac { 1 }{ 4 } =UI(sec\theta -DE) \tag{2.2}\label{myeqtwo_one}$$
$$(\frac { 1 }{ 2 } -{ tan }^{ 2 }\theta )=DE(sec\theta -UI) \tag{3.2}\label{myeqthree_one}$$
This is where I'm stuck. I tried solving the system by replacing $DE$ in $(2.2)$ with $$\frac { (\frac { 1 }{ 2 } -tan^{ 2 }\theta ) }{ (sec\theta -UI) } $$but to no avail. I've also tried multiplying, adding, and subtracting the two systems in the hopes of something symmetric or meaningful, but once again to no avail. I even replaced $UI$, $DE$, and $IE$ with $x$,$y$, and $z$ in the hopes of simplifying some notation and hopefully solving the equation, but to no avail.
| For notational brevity, let $x=DE$, $y=IE$ and $z=UI$ and work with the set below,
$$x+y+z = \sec \theta \tag{1}$$
$$z(z+y) =\frac 14\tag{2}$$
$$x(x+y)=\left(\frac 12 -\tan\theta \right)^2\tag{3}$$
Take (3)-(2) and use (1),
$$x-z=\sin\theta(\tan\theta-1)\tag{4}$$
From (3)+(2),
$$x^2+z^2+y(x+z)=\left(\frac 12 -\tan\theta \right)^2+\frac14\tag{5}$$
Square (1) and rearrange,
$$y^2-(x-z)^2+2[x^2+z^2+y(x+z)]= \sec^2\theta\tag{6}$$
Substitute (4) and (5) into (6) to get the equation for $y$,
$$y^2-\sin^2\theta(\tan\theta-1)^2+2\left(\frac 12 -\tan\theta \right)^2+\frac12=\sec^2\theta$$
Then, simplify and obtain the solution for $y$,
$$y^2=\sin 2\theta,\>\>\>\>\>y=\pm\sqrt{\sin 2\theta}$$
Then, the solutions for $x$ and $z$ follow,
$$\begin{align}
x&= -\frac12\left(y+\cos\theta+\sin\theta\right) +\sec\theta\\
z&= -\frac12 \left(y-\cos\theta-\sin\theta\right)
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality involving Harmonic mean Let $a, b \ge 0 $ and $x,y > 1$ Show $\frac{1}{1/x+1/y}(a+b) \le \max(ax,by)$
If $a = b = 0$ then this is clear so assume not both are zero.
This seems to be related to the harmonic mean but I am not quite getting it.
$\frac{1}{1/x+1/y}(a+b) = \frac{xy}{x+y}(a+b) = \frac{1}{2}H(x,y)(a+b)$
Where $H(x,y) = \frac{2xy}{x+y}$ : the harmonic mean of $x,y$
And, $\max(ax,by) = \frac{1}{2}[ax+by+\mid ax - by \mid ] = A(ax,by) + \frac{1}{2}\mid ax - by \mid$
Where $A(ax,by) = \frac{ax+by}{2}$ : the arithmetic mean of $ax,by$
This is where I get stuck.
EDIT: I am closer but not quite:
$\displaystyle \frac{a+b}{x+y} = \frac{a}{x+y}+\frac{b}{x+y} \le \frac{a}{y} + \frac{b}{x} \implies xy\frac{a+b}{x+y} \le xy(\frac{a}{y} + \frac{b}{x} ) = ax + by \le 2\max(ax,by) $
| Cauchy Schwarz says for $a,b\in\mathbb{R}$ and $x,y\gt0$,
$$
(a+b)^2\le\left(a^2x+b^2y\right)\left(\frac1x+\frac1y\right)
$$
which gives, for $a+b\ge0$,
$$
\begin{align}
\frac1{\frac1x+\frac1y}(a+b)
&\le\frac{a(ax)+b(by)}{a+b}\\
&\le\max(ax,by)
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreciated.
| Like How to prove: $11=10^{12}+10^{7}-45\sum_{n=1}^{999}\csc^4\frac{n\pi}{1000}$
Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$
Observe that $n\left(\frac{\pi (2 k+1)}{2 n}+y\right)=ny+k\pi+\dfrac\pi2$
As $\tan nx=\dfrac{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}{\cot^nx-\binom n{n-2}\cot^{n-2}x+\binom n{n-4}\cot^{n-4}x-\cdots},$
if $\tan nx=\tan\left(ny+k\pi+\dfrac\pi2\right)=-\cot ny$
$$-\dfrac1{\tan ny}=\dfrac{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}{\cot^nx-\binom n{n-2}\cot^{n-2}x+\binom n{n-4}\cot^{n-4}x-\cdots}$$
$$\implies\cot^n x+n\tan ny\cot^{n-1}x-\binom n2\cot^{n-2}x-\binom n3\tan ny\cot^{n-3}x+\cdots=0$$
So, the roots of $$c^n+n(\tan ny)c^{n-1}-\binom n2c^{n-2}-\binom n3(\tan ny)c^{n-3}+\cdots=0$$
are $c_k=\cot\left(y+\dfrac{(2k+1)\pi}{2n}\right), 0\le k\le n-1$
$$\sum _{k=0}^{n-1} \csc ^2\left(\frac{\pi (2 k+1)}{2 n}+y\right)=\sum _{k=0}^{n-1}\left(1+ \cot^2\left(\frac{\pi (2 k+1)}{2 n}+y\right)\right)=n+(n\tan ny)^2-2\binom n2=?$$
Similarly,
$$\csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=\left(1+ \cot^2\left(\frac{\pi (2 k+1)}{2 n}+y\right)\right)^2=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Reducible polynomials similar to Sophie Germain's identity Are there known any other simple nontrivial polynomials in two or more variables that can be factored like the one in Sophie Germain's identity?
$$x^4+4y^4=(x^2-2 x y+2 y^2) (x^2+2 x y+2 y^2)$$
Besides the well known (for $n \in \mathbb {Z} ^{+}$):
$$x^n-y^n=(x-y) \sum _{i=1}^n x^{i-1} y^{n-i}$$
$$x^{2 n-1}+y^{2 n-1}=(x+y) \sum _{i=1}^{2 n-1} (-1)^{i-1} x^{i-1} y^{2 n-1-i}$$
And of course besides trivial ones like $5 x^2+x y^2=x (5 x+y^2)$, where all terms share a common factor.
By simple polynomial I mean a polynomial that has few terms. The less terms the better - ideally only two terms like in the Germain's polynomial or in the known ones.
| More formulas similar to Sophie Germain's identity can be derived from Aurifeuillean factorizations such as
$$x^6+27y^6=(x^2+3y^2)(x^2-3xy+3y^2)(x^2+3xy+3y^2),$$
$$x^{10}-3125y^{10}=(x^2-5y^2)(x^4-5x^3 y+15x^2 y^2-25x y^3+25y^4)(x^4+5x^3 y+15x^2 y^2+25x y^3+25y^4).$$
| {
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Proof that inequality $|a| < |b-c+d|, |b| < |a-c+d|, |c| < |a-b+d|, |d|< |a-b+c| $ does not have solution in $\mathbb{R}$
Proof that inequality $$|a| < |b-c+d|, \\|b| < |a-c+d|, \\|c| < |a-b+d|,\\ |d|< |a-b+c| $$ does not have solution in $\mathbb{R}$.
Solution:
$$|a| < |b-c+d| \leq |b|+|c|+|d|$$
$$|b| < |a-c+d| \leq |a|+|c|+|d|$$
$$|c| < |a-b+d| \leq |a|+|b|+|d|$$
$$|d| < |a-c+c| \leq |a|+|b|+|c|$$
and next
$$ |a| + |b| + |c| + |d| < 3|a| + 3|b| + 3|c| + 3|d|$$
$$ 0 < 2|a| + 2|b| + 2|c| + 2|d|$$
And I have problem,
but we have any $a,b,c,d > 0$
Where is mistake ?
| There is no mistake, but the inequality you obtained is not useful!
Note that
$$|a| < |b-c+d|\Leftrightarrow (b−c+d)^2−a^2>0$$
and similarly we have that $(a-c+d)^2-b^2>0$, $(a-b+d)^2-c^2>0$, and $(a-b+c)^2-d^2>0$. Then by taking the their product, and by recalling that
$x^2-y^2=(x+y)(x-y)$, we find
$$\begin{align}((&b-c+d)^2-a^2)((a-c+d)^2-b^2)((a-b+d)^2-c^2)((a-b+c)^2-d^2)\\
&=-(b-c+d+a)^2(b-c+d-a)^2(a-c+d-b)^2(a-b+c+d)^2
\end{align}$$
which is a contradiction because the left-hand side is $>0$, whereas the right-hand side is $\leq 0$.
| {
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Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
| Because of the special case
$$x\equiv -1\pmod {2,3,4,5,6}$$
we have a shortcut for handling these.
$$x+1\equiv 0\pmod {2,3,4,5,6}$$
It is way easier to work with $x+1$ here.
This gives us
$$x+1 \equiv 0 \pmod {LCM(2,3,4,5,6)=60}$$
Now we need to incorporate
$$x+1 \equiv 1 \pmod 7 $$
Which is easy, as $7$ is prime. List out the first 6 multiples of $60$.
$$\begin{align}
1 \cdot 60 \equiv 4 \pmod 7 \\\\
2 \cdot 60 \equiv \color{red}1 \pmod 7 \\\\
3 \cdot 60 \equiv 5 \pmod 7 \\\\
4 \cdot 60 \equiv 2 \pmod 7 \\\\
5 \cdot 60 \equiv 6 \pmod 7 \\\\
6 \cdot 60 \equiv 3 \pmod 7 \\\\
\end{align}$$
So our answer is
$$ x +1 \equiv 2 \cdot 60 \pmod{ 7 \cdot 60}$$
or
$$ x \equiv 2 \cdot 60 -1 \pmod {7 \cdot 60}$$
| {
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If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$ If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$
All angles are in degrees
I will be writing $\omega$ as w to make it easier to type
From the value of $w$ it is clear that $w^9=1$
Solving the arithmetic geometric progression we get the the sum inside the modulus as
$$\frac{9w}{1-w}$$
Taking the inverse of the modulus ie.
$$\left|\frac{9w}{1-w}\right|^{-1}$$
I don’t know how to proceed. Please help.
The answer is $\frac 29 \sin 20$
Thanks!!
| If $S=\omega + 2\omega ^2 + 3\omega ^3+\cdots+9\omega ^9$
$(1-\omega)S=\omega + \omega ^2 + \omega ^3+\cdots+\omega ^9-9\omega ^{10}$
$=1+\omega + \omega ^2 + \omega ^3+\cdots+\omega^8-9w$ as $\omega ^9=1$
$=\dfrac{1-\omega^9}{1-w}-9w=-9w$
$$S=\dfrac{-9\omega}{1-\omega}=\dfrac9{1-\omega^{-1}}$$
$1-\omega^{-1}=1-(\cos2x-i\sin2x)=2\sin x(\cos x+i\sin x)$
$\dfrac1{1-\omega^{-1}}=\dfrac1{2\sin x(\cos x+i\sin x)}=\dfrac{\cos x-i\sin x}{2\sin x}$
Here $x=20^\circ$
| {
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"timestamp": "2023-03-29T00:00:00",
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Triple integral for the volume between $z=2$ and the top of the sphere $x^2 + y^2 + z^2 = 9$ Apparently we do something like this:
$z=2$, so $x^2 + y^2 + z^2 = 9$ tells us that $x^2+y^2=5$, and so our integral is:
$$\int_{-\sqrt{5}}^{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1\,dz\,dy\,dx$$
That solution makes very little sense to me. Why do we substitute the formula for the plane $z=2$ into the formula for the sphere? What exactly does that accomplish? The innermost integral makes sense, but the limits on the first two do not.
Shouldn't the outer two integrals simply be something like $$\int_0^3\int_{-\sqrt{9-x^2}}^\sqrt{9-x^2}$$
since the radius of the sphere is $3?$
Any help is appreciated.
| You want to project your solid onto the $xy$-plane first (Hint: look at the last two variables where you integrate). What does it look like?
Essentially, when $$z=2,$$ the solid has a region $R$ bounded by $$x^2+y^2=5,$$
a circle, which you pointed out already. From this region, we start to look at $y$ first and where it tends to and from (Hint: look again at the last two variables where you integrate:$\,dy$ is the outermost, so we look at $y$ first).
From the equation of the region $R$ above, we can see that the radius is equal to $\sqrt{5}$. Here, we can finally say that $$-\sqrt{5}≤y≤\sqrt{5}$$
Now we look at $x$, where you just manipulate the equation of the same region
$$x^2+y^2=5,$$
$$\implies x=±\sqrt{5-y^2}$$
Hence, the bounds for $x$ is
$$-\sqrt{5-y^2}≤x≤\sqrt{5-y^2}$$
For $z$, since the solid is bounded below by the plane $z=2$ and above by the sphere, we have
$$2≤z≤\sqrt{9-x^2-y^2}$$
Finally, we can set up the iterated triple integral
$$\int_{-\sqrt{5}}^\sqrt{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1dzdydx$$
| {
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Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$. Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$.
So we have the (Macluarin) series
\begin{equation*}
\sinh{(x)} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}.
\end{equation*}
at $x = 0$ so
\begin{equation*}
\sinh{(x^4)} = \sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!}.
\end{equation*}
The Maclaurin series for $\sinh{(x^4)}$ is
\begin{equation*}
\sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!} = \sum_{m=0}^{\infty} \frac{x^m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
We can see that the coefficient for $x^m$ is $\frac{1}{\left(\frac{m}{4}\right)!}$. On the other hand, this is the Maclaurin series for $g(x)$, and so the coefficient on $x^m$ is equal to $\frac{g^{(m)}(0)}{m}$. Equating these two, we have
\begin{equation*}
\frac{g^{(m)}(0)}{m} = \frac{1}{\left(\frac{m}{4}\right)!} \Longleftrightarrow g^{(m)}(0) = \frac{m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
Substituting $m = 30$ gives $g^{(30)}(0) = \frac{30}{7.5!}$. Am I on the right track here?
| Your second formula
$$g(x)=\sum_{n=0}^\infty{x^{8n+4}\over(2n+1)!}$$
is still correct. Now the $m^{\rm th}$ derivative of any power series $\sum_{k=0}^\infty a_kx^k$ at $0$ is equal to the coefficient $a_m$ belonging to $x^m$, times a combinatorial factor. In your case $m=30$. Since $8n+4\ne30$ for all $n\geq0$ it follows that $a_{30}=0$, hence $g^{(30)}(0)=0$.
| {
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Under what conditions is the following matrix diagonalisable Let
$$A=\begin{bmatrix} 0&1/2&1/2 \\ 0&0&1/2 \\ 1&1/2&0 \end{bmatrix}.$$
Could anyone advise me on how to show $M=(1-m)A+mS,$ where $S_{ij}=\frac{1}{3},$ is not diagonalisable for $0\leq m < 1 \ ?$ Do we compute the characteristic polynomial of $M$, show it has $2$ distinct eigenvalues and the dimension of eigenspace corresponding to each distinct eigenvalue is $1 \ ?$
| The matrix $M$ is
$$M=\frac{1}{6}\left(
\begin{array}{ccc}
2 m & 3-m & 3-m \\
2 m & 2 m & 3-m \\
6-4 m & 3-m & 2 m \\
\end{array}
\right).$$
Now compute the characteristic polynomial
$$f(x)=\frac14(m-1-2x)^2 (x-1).$$
This shows that $M$ has eigenvalues $1$ and $\frac{m-1}{2}$. Now check that $\dim\ker\left(M-\left(\frac{m-1}{2}\right)I\right)=1$, for example by row reducing the matrix $M-\left(\frac{m-1}{2}\right)I$ (here $I$ is the $3\times 3$ identity matrix). This shows that $\frac{m-1}{2}$ corresponds to a $2\times 2$ Jordan block.
I did this with a computer, so this sort of obscured the part where you use $m\neq 1$. If you look at the Jordan normal form for $M$, you see that it is similar to
$$\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{m-1}{2} & 1 \\
0 & 0 & \frac{m-1}{2} \\
\end{array}
\right)$$
via conjugation by
$$\left(
\begin{array}{ccc}
\frac{m-3}{2 (m-2)} & 0 & -\frac{2}{m-1} \\
\frac{1}{2-m} & -1 & \frac{2}{m-1} \\
1 & 1 & 0 \\
\end{array}
\right).$$
Of course, this matrix assumes $m\neq 1$. If $m=1$, then in fact we get the Jordan normal form
$$\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}
\right),$$
and so $M$ is diagonalizable if and only if $m=1$.
| {
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"question_score": "1",
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Showing continuity of $g(x, y)$ Let $g : \mathbb{R}^2 \to \mathbb{R}$. Let the domain be $y^2 \leq \frac{x}{3}$.
$g(x, y) = \begin{cases}
\frac{x^3}{y^2} \left (\sqrt{x} - \sqrt{y^2 + x} \right ) & y \neq 0\\
0 & y = 0
\end{cases} $
I want to show $g$ is continuous at $(0, 0)$.
We have $|g(x, y) - g(0,0)| = \frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )$.
Fix some $\epsilon > 0$.
Now, I want to find a $\delta > 0$ such that $x^2 + y^2 < \delta$ $\implies $ $\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right ) < \epsilon$
| Write your term as $$\frac{x^3}{y^2}\frac{x+y^2-x}{\sqrt{x}+\sqrt{y^2+x}}$$
| {
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Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25} \le A(a) < 2$$
Note that the bounds are rather tight. I encountered this issue in determining a narrow range of an angle in a geometry problem. I was only able to examine certain limits and identified the correct answer. But, I did not manage to fully prove the above inequalities.
Since I am not all that versed in dealing with such type of problems and would appreciate if anyone could suggest solutions for the proof.
| Here's a simple approach: let $b = 1 + a + 1/a$, so $b \geq 3$, and
$$\frac{1}{A(a)} = \frac{(ab - 1/2)(b/a - 1/2)}{b^2} = 1 - \frac{a + 1/a}{2b} + \frac{1}{4b^2} = 1 - \frac{b-1}{2b} + \frac{1}{4b^2} = \frac{1}{2} + \frac{1}{2b} + \frac{1}{4b^2}.$$
As $a$ varies over $(0, \infty)$, $b$ varies over $[3, \infty)$, and it's clear that $1/A(a)$ is strictly decreasing as a function of $b$, so since $1/A(a) = 25/36$ at $b = 3$, and $1/A(a)$ decreases to $1/2$ as $b \to \infty$, we have $1/2 < 1/A(a) \leq 25/36$ for all $a$.
| {
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Can anyone prove this function is concave? Can anyone help me prove the following function $$f(x) = \frac{ax}{1-x^a}-\frac{x}{1-x}$$ is concave for any parameter $a \geq 1$ when $0<x<1$ ?
I have tried to prove its second derivative is negative, but it becomes another difficult problem.
| AmerYR calculated $$f''(x)=\frac{a^2x^{a-1}\left(1+a+x^a(a-1)\right)}{(1-x^a)^3} - \frac{2}{(1-x)^3} \, .$$
As you noticed $f''(x)=0$ at $a=1$. So it suffices to show that $\partial_a f''(x) \leq 0$. In the comments I gave the derivative with respect to $a$ by $${\frac { \left( \left( \left( {a}^{2}-a \right) \ln \left( x \right) -3\,a+2 \right) {x}^{3\,a-1}+ \left( 4\,{a}^{2}\ln \left( x \right) -4 \right) {x}^{2\,a-1}+{x}^{a-1} \left( \left( {a}^{2}+a \right) \ln \left( x \right) +3\,a+2 \right) \right) a}{ \left( 1- {x}^{a} \right) ^{4}}} \, .$$ Since $\frac{(1-x^a)^4}{a \, x^{a-1}}>0$ we can multiply by this and obtain the objective $$\left( a\left( a-1 \right) {x}^{2\,a}+4\,{a}^{2}{x}^{a}+{a}^{2}+a \right) \ln \left( x \right) + \left( -3\,a+2 \right) {x}^{2\,a}-4 \,{x}^{a}+3\,a+2 \stackrel{!}{\leq} 0 \, .\tag{0}$$
We now need to get rid of the logarithm in order to get a manifest signature. I leave it to you to check that this expression goes to $-\infty$ as $x\rightarrow 0$, while it vanishes at $x=1$. Taking the derivative with respect to $x$ and dividing the resulting expression by $a \, x^{a-1}>0$ yields $$2\,a \left( \left( a-1 \right) {x}^{a}+2\,a \right) \ln \left( x \right) + \left( a+1 \right) {x}^{-a}+ \left( -5\,a+3 \right) {x}^{a} +4\,a-4 \stackrel{!}{\geq}0 \tag{1} \, .$$
Now rinse and repeat. Check the limiting cases (they are $+\infty$ and $0$), derive by $x$ and divide by $a\,x^{a-1}$ to obtain $$2a\left( a-1 \right) \ln \left( x \right) - \left( a+1 \right) {x}^{-2\,a}+4\,{x}^{-a}a-3\,a+1 \stackrel{!}{\leq} 0 \, .\tag{2}$$ Almost there; The limiting cases are manifestly $-\infty$ and $0$ respectively and so after deriving with respect to $x$ and multiplying by $\frac{x^{2a+1}}{2a}>0$ this becomes $$\left( a-1 \right) {x}^{2\,a}-2\,a\,{x}^{a}+a+1 \stackrel{!}{\geq} 0 \, .\tag{3}$$ One more time: At $x=0$ this is $a+1>0$ and at $x=1$ it vanishes again. Deriving with respect to $x$ and dividing by $2a\, x^{a-1}>0$ we have $$(a-1)x^a - a \stackrel{!}{\leq} 0 \, .\tag{4}$$
| {
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.
| $$ \begin{align}& a^2+b^2+c^2+ab+ac+bc\le1 \\
& (a+b+c)^2 -(ab+bc+ac) \le 1 \\
& (a+b+ c) \le \sqrt{1+(ab+bc+ac)} \quad \quad \color{red}{\text{ (1.)}} \end{align}$$
Also note that :
$$\frac {(ab+bc+ac)}{3} \ge \sqrt[3]{a^2b^2c^2}$$
Assuming you are correct :
$${(ab+bc+ac)} \ge 3 \cdot\frac16 \implies (ab+bc+ac) \ge \frac12$$
Hence from $\color{red}{\text{ (1.)}}$:
$$(a+b+c) \le \sqrt{1+\frac12} \implies \boxed {\color {blue}{(a+b+c) \le \sqrt {\frac32}}}$$
| {
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$\int \frac{dx}{1-\sin x+\cos x}$ Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$
I tried:
$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos x+\sin x}{2 \cos x(\cos x+1)}$$ $$=\int (\frac{1}{2 \cos x}+ \frac{\sin x}{2 \cos x(\cos x+1)})dx=\frac{1}{2} \ln(\sec x+\tan x)+ \int \frac{\sin x}{2\cos x(\cos x+1)}dx$$
$ \cos x=u$ , $du=-\sin xdx$
$$\int \frac{-du}{2u(u+1)}= \frac{-1}{2} \int (\frac{1}{u}-\frac{1}{u+1})= \frac{-1}{2}(\ln(\cos x)-\ln(\cos x+1))$$
final answer : $$\frac{1}{2} \ln(\sec x+\tan x)-\frac{1}{2} \ln(\frac{\cos x}{1+\cos x})+ c$$
First: Is my answer right?
Second: Is there another approach or easier approach to solve this integral?
| Start by multiplying the top and bottom by $\sec(x)$, which will result in the following:
$$\int{\frac{\sec x}{\sec x-\tan x+1}}$$
Now we can perform a $u-$substitution for the following:
$$u=(\sec x-\tan x)\\du=(\sec x)(-(\sec x-\tan x))dx$$
The resulting integral is then:
$$\int{\frac{-1}{(u+1)(u)}}$$
We can now make use of partial fraction in the same manner that you did, giving us the following:
$$\int{\frac{1}{u+1}}+\int{\frac{-1}{u}}$$
Completing the remaining integrals and plugging back in for our substitution gives the following solution:
$$\ln\left|{\frac{1-\sin x+\cos x}{1-\sin x}}\right|+c$$
| {
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The sum of the series
$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$
MY SOLUTION
We can write it as
$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$
$$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$
$$=\frac{1}{\log 4}\left [\log(2.4.8....2^n)\right]$$
$$\frac{1}{\log 4}[log(2^n.2^{\frac{(n)(n+1)}{2}}
)]$$
$$\frac{2n+n^2+n}{4}$$
But the answer is $\frac{n^2+n}{4}$
| You are almost right, we have that
$$\dots=\frac{1}{\log 4}\left [\log(2\cdot 4\cdot 8\cdot \ldots \cdot 2^n)\right]=\frac{1}{\log 4}\left [\log(2\cdot 2^2\cdot 2^3\cdot \ldots \cdot 2^n)\right]=$$
$$=\frac{1}{\log 4}\left [\log(2^{1+2+3+\dots+n}
)\right]=\frac{1}{\log 4}\left [\log(2^{\frac{n(n+1)}2})\right]=\frac{n(n+1)}2\frac{\log 2}{\log 4}=\frac{n(n+1)}4$$
indeed $\frac{\log 2}{\log 4}=\frac12 \frac{2\log 2}{\log 4}=\frac12 \frac{\log 4}{\log 4}=\frac12$.
| {
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Do the coefficients of these polynomials alternate in sign? Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=x-1\\
p_4(x)&=x-2\\
p_5(x)&=x^2-3x+1\\
p_6(x)&=x^2-4x+3\\
p_7(x)&=x^3-5x^2+6x-1\\
p_8(x)&=x^3-6x^2+10x-4
\end{align}
It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts?
| Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ...
\begin{eqnarray*}
p_{2i}(x) = \sum_{j=0}^{i-1} \binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\
p_{2i-1}(x) = \sum_{j=0}^{i-1} \binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\
\end{eqnarray*}
Proof of the even formula ...
\begin{eqnarray*}
&p_{2i-1}(x) -p_{2i-2}(x) =\\ & x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \sum_{j=0}^{i-1} \binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\
&= x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2}
= p_{2i}(x).
\end{eqnarray*}
And the answer to the question in the title is $\color{red}{\text{yes}}$.
| {
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Evaluate $\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}dydx$ How to show
$$\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}d\sigma=-\frac{128 a}{9}+\frac{16}{3} \sqrt{2 a} \tan ^{-1}\left(\sqrt{\frac{a}{2}}\right)+\frac{8}{3} \sqrt{2} \log (2)$$
Where $a=\sqrt{2}-1$? Any kind of help will be appreciated.
| One can exploit a symmetry and split the integral in half along the line $y=x$ then use the following modified polar coordinates:
$$x = s^{\frac{2}{3}}\cos\theta \hspace{10 pt} y = s^{\frac{2}{3}}\sin\theta$$
$$\implies 2\cdot \frac{2}{3}\int_0^{\frac{\pi}{4}} \int_0^{\sec^{\frac{3}{2}}\theta} \frac{\log\left(s^{\frac{4}{3}}\right)}{\sqrt{\cos\theta+\sin\theta}}dsd\theta = \frac{16}{9}\int_0^{\frac{\pi}{4}} \frac{\sec^{\frac{3}{2}}\theta}{\sqrt{\cos\theta+\sin\theta}}\left(\frac{3}{2}\log(\sec\theta)-1\right)d\theta$$
then let $x=\tan\theta$
$$ \implies \frac{4}{3}\int_0^1 \frac{\log(1+x^2)}{\sqrt{1+x}}dx - \frac{16}{9}\int_0^1 \frac{1}{\sqrt{1+x}}dx$$
The integral on the right evaluates to $\frac{32}{9}(\sqrt{2}-1) \equiv \frac{32}{9}a$. The integral on the left becomes
$$ = \frac{8}{3}\sqrt{2}\log 2 - \frac{16}{3}\int_0^1 \frac{x\sqrt{1+x}}{1+x^2}dx$$
We have one and a half of the three terms the user posted, but this last integral is tricky and isn't yielding to a variety of methods. I will try to finish later, but in the meantime if anyone has any clever suggestions for this last integral I will be happy try it.
| {
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How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}$
Does anyone know how to solve this?
| Let $a_n = \left(\dfrac{x}{x + y}\right)^{n} + \left(\dfrac{y}{x + y}\right)^{n}$.
Since
$$
\dfrac{x}{x + y}+\dfrac{y}{x + y}=1, \quad
\dfrac{x}{x + y}\cdot\dfrac{y}{x + y}=1
$$
they are the roots of $t^2=t-1$ and so
we get
$$
a_{n+2} = a_{n+1}-a_n,
\quad
a_0=2,
\quad a_1=1
$$
This sequence is periodic of period $6$:
$$
2,1,-1,-2,-1,1,
2,1,-1,-2,-1,1,
\dots
$$
Thus, $a_{2013} = a_{2013 \bmod 6} = a_3 = -2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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how to simplify $\frac{2+\sqrt{3}}{6+4\sqrt{3}}$ among solving a problem I need to simplify: $$\frac{2+\sqrt{3}}{6+4\sqrt{3}}$$
I know it is $\frac{1}{2\sqrt{3}}$ but I dont know how to show that in mathematic way for example I can write $$\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}$$ but still not obvious.Is there better way to prove/show that?
| We have that
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{6+4\sqrt{3}}\cdot \frac{6-4\sqrt{3}}{6-4\sqrt{3}}=\frac{-2\sqrt{3}}{-12}=\frac{\sqrt 3}6$$
or by your idea
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}=\frac1{2\sqrt{3}}=\frac1{2\sqrt{3}}\cdot\frac{2\sqrt{3}}{2\sqrt{3}}=\frac{\sqrt 3}6$$
| {
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"source": "stackexchange",
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$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{x}{e^x - \cos x}\cdot\cancelto{1}{\frac{\sin x}{x}}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2+x}{e^x - 1 + 1 - \cos x}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^{11} - 3x^2 + x} + \frac {e^x-1}{x^{11}-3x^2+x})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^2\cancelto{-\infty}{(x^9-3-\frac {1}{x})}} + \frac {e^x-1}{x\cancelto{1}{(x^{10}-3x+1)}})^{-1}=$$
$$ =\lim_{x \to 0} (\cancelto{\frac{1}{2}}{\frac{1-\cos x}{x^2}}\cdot\cancelto{0}{\frac{1}{-\infty}} + \cancelto{1}{\frac {e^x-1}{x}})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1}{2} \cdot0+ 1)^{-1}=$$
$$=(0+1)^{-1} = 1$$
Is this correct?
| L'Hopital's rule also works as the limit is in the form $\frac{0}{0}$:
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}
=\lim_{x \to 0} \frac{11x^{10}-6x+\cos x}{e^x + \sin x}
= \frac{0-0+1}{1 + 0}=1$$
| {
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Proving limits involving floor function.
Let $\theta $ be real, and $0<\theta<1$, define $g_n = 0$ if $\lfloor n\theta\rfloor=\lfloor(n-1)\theta \rfloor$ and $g_n=1$ otherwise. For example, take $\theta = \frac{3}{4}$, then $g_2 =1$, since $\left\lfloor \frac{3}{2}\right\rfloor = 1 \ne 0 =\left\lfloor \frac{3}{4} \right\rfloor$. Prove that $$
\lim_{n \to \infty} \dfrac{g_1+g_2+\cdots + g_n}{n} = \theta$$ Note $\lfloor x \rfloor$ denotes the greatest integer function/ floor function.
This is what I did: $g_1$ is always $0$, since $\lfloor\theta\rfloor=0$, and $g_2 = 0$ if $\theta \in \left(0,\frac{1}{2}\right)$. Now consier $g_4$, $$
\begin{array}{cccc}
& \left( 0 ,\frac{1}{4}\right) & \left[ \frac{1}{4} , \frac{1}{3} \right) & \left[ \frac{1}{3} , \frac{1}{2} \right) & \left[ \frac{1}{2} , \frac{2}{3} \right) & \left[ \frac{2}{3} , \frac{3}{4} \right) & \left[ \frac{3}{4} , 1 \right) \\ \hline
\lfloor 4\theta \rfloor & 0 & 1 & 1 & 2 & 2 & 3\\
\lfloor 3\theta \rfloor & 0 & 0 & 1 & 1 & 2 & 2\\
\end{array}$$
Hence $g_4 = 0$ when $\theta \in\left( 0 ,\frac{1}{4}\right) \cup \left[ \frac{1}{3} , \frac{1}{2} \right) \cup \left[ \frac{2}{3} , \frac{3}{4} \right) $ and $g_4 = 1$ otherwise. Similarly, $g_n = 0$ when $\theta \in \left( 0 ,\frac{1}{n}\right) \cup \left[ \frac{1}{n-1} , \frac{2}{n} \right) \cup \left[ \frac{2}{n-1} , \frac{3}{n} \right) \cup \cdots \cup \left[\frac{n-2}{n-1}, \frac{n-1}{n} \right)$ and $g_n=1$ otherwise. Consider the case when $\theta = \frac{p}{q}$ is a rational number with $ p,q>0 \in \mathbb{Z}$ and $p,q$ are copime. Then we see a periodic behaviour, $g_1 = g_{mq+1} = 0$ where $m \in \mathbb{Z_{\ge 0}}$ and $g_i = g_{mq+i}$ for $i \in \{ 1,2,\cdots,q-1\}$. But for irrational $\theta$ I don't even know how to start. Any hints on how to further progess in this problem ?
| Hint: Show that $g_n=\lfloor n\theta \rfloor-\lfloor (n-1)\theta \rfloor$, so that $\sum_{i=1}^n g_n=\lfloor n\theta\rfloor$ by a telescoping sum argument. So what you need to do is show that
$$\lim_{n\to\infty}\frac{\lfloor n\theta\rfloor}{n}=\theta$$
which shouldn't be too difficult for you since that is a series of rational approximations that you can show that $|\theta-a_n|<\frac1n$ for all $n$
| {
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$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty.
| HINT:
$x=2\text{sec}(\theta)$, $dx=2\text{sec}(\theta)\text{tan}(\theta)d\theta$ then:
$\int \frac{1}{(x^2-4)^2}dx=\int \frac{2\text{sec}(\theta)\text{tan}(\theta)}{(4\text{sec}^2(\theta)-4)^2}d\theta=\frac{1}{8}\int \frac{\text{sec}(\theta)\text{tan}(\theta)}{\text{tan}^4(\theta)}d\theta$
$$=\frac{1}{8}\int \text{ctg}^2(\theta)\text{csc}(\theta) d\theta= \frac{1}{8}(\int \text{csc}^3(\theta)d\theta-\int\text{csc}(\theta) d\theta)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the limit as $x$ approaches infinity of $x-\sqrt{x^2-x+2}$ $$
\lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)
$$
I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.
| Yes rationalize is a good idea, indeed we obtain
$$x - \sqrt{x^2 - x +2 }=x - \sqrt{x^2 - x +2 }\cdot\frac{x + \sqrt{x^2 - x +2 }}{x + \sqrt{x^2 + x +2 }}=\frac{x^2-x^2+x-2}{x+ \sqrt{x^2 + x +2 }}$$
$$=\frac{x-2}{x+ \sqrt{x^2 + x +2 }}=\frac x x\frac{1-\frac 2x}{1+ \sqrt{1 + \frac1x +\frac2{x^2} }}\to \frac12$$
| {
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Complex numbers such that $(z-4)^4=-1$
Find all complex numbers $z$ in Cartesian form such that $(z-4)^4=-1$.
This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$ and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.
|
This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$
This is on the right track. Except, when you square root both sides, you need to do $\pm$. So instead of getting $i$ you get $\pm i$:
$$
(z - 4)^2 = \pm i
$$
and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.
The next step is to take the square root again (not squaring both sides), is that what you meant to do?
This part is a little tricky: when we take square root of both sides we need to know what are all the square roots of $i$ and $-i$.
The square roots of $i$ are
$$
\pm \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right)
$$
and the square roots of $-i$ are
$$
\pm \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \right)
$$
So, when we take the square root of the equation $(z - 4)^2 = \pm i$, we get four possibilities (the two square roots of $i$, and the two square roots of $-i$):
\begin{align*}
z - 4 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\
z - 4 &= \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \\
z - 4 &= -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\
z - 4 &= -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i.
\end{align*}
Can you find the answer in all of the four cases?
| {
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Is $(n + 1)^n = n^n + (n + 1)$? I'm trying to solve my homework, but I don't know if I'm on the right way.
I have
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}$$
and I want to cancel $(n+1)^n$ with $n^n$.
So I have a sequence of series and try to check the convergence with Ratio Test.
$$\lim_{n\to \infty}\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)} $$
| Always try with trivial cases at first, notably for $n=1$ we have
$$(n+1)^n=n^n+(n+1) \implies 2=1+2$$
Edit
Since the original problem seems related to the ratio test for convergence, we have that
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}=\frac{(n+1)^n}{n^n}\frac{n+1}{4n+2}=\left(1+\frac1n\right)^n\frac{n+1}{4n+2}$$
then take the limit.
Edit 2
For the given limit
$$\lim_{n\to \infty}\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)}$$
by ratio test we obtain
$$\frac{(n+2)^{n+3}}{(n+1)^{n+1}(2n+3)(2n+4)}\frac{n^n(2n+1)(2n+2)}{(n+1)^{n+2}}=$$
$$=\frac{n^n(n+2)^{n+3}}{(n+1)^{n+1}(n+1)^{n+2}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\frac{n^n(n+2)^{n}}{(n+1)^{n}(n+1)^{n}}\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\left(\frac{n^2+2n}{n^2+2n+1}\right)^n\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\left(1-\frac{1}{n^2+2n+1}\right)^n\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)} \to 1$$
which is not conclusive.
Anyway we don't need ratio test since
$$\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)}=\frac{(n+1)^n}{n^n}\frac{(n+1)^{2}}{(2n+1)(2n+2)}=$$
$$=\left(1+\frac1n\right)^n\frac{n^2+2n+1}{4n^2+6n+2} \to e \cdot \frac14 = \frac e 4$$
| {
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Probability Counting Problem Confusion Im having difficulty understanding how to do the following problem using Combinations.
Q: An urn contains 2 red balls, 4 blue balls and 6 green balls. If three balls are drawn at random one at a time, without replacement, what is the probability that the first and last are the same colour?
The book answer solves this by considering that the third choice of a ball does not affect the experiment and treats this experiment as of choosing two balls instead of three and thus solving the problem as: $\dfrac{{2\choose2}+{4\choose2}+{6\choose2}}{12\choose2}=\dfrac{1}{3}$
Now I was trying to solve this question without disregarding the third ball. My attempt to solve this was: $\dfrac{{10\choose1}[{2\choose2}+{4\choose2}+{6\choose2}]}{12\choose3}=1$ which is obv wrong. My motivation was the following $P(RR \cup BB \cup GG)=P(RR)+P(BB)+P(GG)$ where for example $P(BB)=\dfrac{{4\choose2}{10\choose1}}{{12\choose3}}$ since I have to fill two spots with blue balls and I have 4 choose 2 ways of doing that, however for each of those outcomes I have to fill the last spot with an arbitrary ball, now since I chose 2 blue balls, I am left with 10 balls In the urn and thus 10 choose 1 options to fill the last spot, so therefore the number of ways I can choose 2 blue balls and an arbitrary third ball is 4 choose 2 multiplied by 10 choose 1 ways. Now I divide that by the total number of ways of choosing 3 balls from 12. I carry the same analogy for the Red and Green situations.
I am just confused where Im making a mistake in my thinking and how to solve this qs without disregarding the last third ball.
Thank You in advance
| The book's answer can ignore order. Your answer cannot. You need specifically for the first and last ball to be the same. For the book's answer, they are only looking at the first and last ball, and they do not care which order they pull them so long as they are the same color. Once you add in the third ball, order matters. So, you have RAR, BAB, GAG with A being any color. Choose the first ball, then the last ball, then the middle ball, but order matters, so you are assigning balls to positions, and therefore cannot use combinations.
$$\require{cancel} \begin{align*}\dfrac{2\cdot 1\cdot 10 + 4\cdot 3\cdot 10 + 6\cdot 5\cdot 10}{12\cdot 11\cdot 10} & = \dfrac{2\cdot 1\cdot 10 + 4\cdot 3\cdot 10 + 6\cdot 5\cdot 10}{12\cdot 11\cdot 10}\cdot \dfrac{\tfrac{1}{2\cdot 10}}{\tfrac{1}{2\cdot 10}} \\ & = \dfrac{2\cdot 1\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}} + 4\cdot 3\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}} + 6\cdot 5\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}}}{12\cdot 11\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}}} \\ & = \dfrac{\dbinom{2}{2} + \dbinom{4}{2} + \dbinom{6}{2}}{\dbinom{12}{2}}\end{align*}$$
So these two methods give the same answer, which is apparent after some algebra to show how they are the same.
| {
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"url": "https://math.stackexchange.com/questions/3425014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Alternative proof of reduction formula for $\int {dx \over (a\sin x + b\cos x)^n }$
The problem i was working on is to prove:
$$
\begin{align}
J_n &= \int {dx \over (a\sin x + b\cos x)^n } \\
&={1\over (n-1)(a^2 + b^2)}\left({b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right)
\end{align}
$$
Where $a^2+b^2 \ne 0$ and $n>1\in\Bbb N$
It is indeed true, but the method I've used requires proving additional reduction formula and now I'm trying to find a simpler method. Below is what I've used to prove the statement.
I have started with a simpler case, namely $J_1$:
$$
J_1 = \int {dx\over a\sin x + b\cos x}
$$
Using a bit of trigonometry one may show that:
$$
J_1 = \int {dx\over \sqrt{a^2+b^2}\left({a\sin x\over \sqrt{a^2+b^2}} + {b\cos x\over \sqrt{a^2+b^2}}\right)}
$$
Where for some $\phi$:
$$
\begin{align*}
\sin \phi &= {a\over \sqrt{a^2+b^2}}\tag 1\\
\cos \phi &= {b\over \sqrt{a^2+b^2}}\tag 2
\end{align*}
$$
So:
$$
\begin{align}
J_1 &= \int {dx\over \sqrt{a^2+b^2}\left(\sin x\sin\phi + \cos x\cos \phi\right)}\\
&=\int {dx\over \sqrt{a^2+b^2}\cos(x-\phi)}
\end{align}
$$
So going back to $J_n$ we obtain:
$$
\begin{align}
J_n &= \int {dx\over (\sqrt{a^2+b^2})^n\cos^n(x-\phi)} \\
&= \int {1\over (\sqrt{a^2+b^2})^n}{dx\over \cos^n(x-\phi)}\\
&\stackrel{t=x-\phi}{=} {1\over (\sqrt{a^2+b^2})^n} \int {dt\over \cos^n(t)}\\
\end{align}
$$
Some time ago I've proven a reduction formula for integrals of ${1\over \cos^n x}$, so I'm just going to use it without a proof:
$$
J_n = {1\over (\sqrt{a^2+b^2})^n}\left({\sin t\over (n-1)\cos^{n-1}t} + (n-2)\int{dt\over \cos^{n-2}t}\right)
$$
The tricky part here is to arrive to the expression in the problem statement:
$$
J_n = {1\over (n-1)(a^2 + b^2)}{1\over (\sqrt{a^2 + b^2})^{n-2}}\left({\sin(x-\phi)\over \cos^{n-1}(x-\phi)} + (n-2)\int {dt\over cos^{n-2}t}\right)\\
= {1\over (n-1)(a^2 + b^2)}\left({\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} + {(n-2)\over (\sqrt{a^2 + b^2})^{n-2}}\int {dt\over cos^{n-2}t}\right)\\
={1\over (n-1)(a^2 + b^2)}\left({\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} + J_{n-2}\right)
$$
Consider the following part of the expression:
$$
\begin{align}
E &= {\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} \\
&= {\sin x\cos \phi - \cos x\sin \phi\over (\sqrt{a^2 + b^2})^{n-2}(\cos x\cos \phi + \sin x\sin \phi)^{n-1}}
\end{align}
$$
Now replace back using $(1)$ and $(2)$:
$$
E = {b\sin x - a\cos x\over (\sqrt{a^2 - b^2})^{n-1}\left({b\cos x\over \sqrt{a^2+b^2}} + {a\sin x\over \sqrt{a^2+b^2}}\right)^{n-1}} \\
= {b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}}
$$
Finally:
$$
J_n = {1\over (n-1)(a^2 + b^2)}\left({b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right)
$$
I'm wondering if there are alternative ways to prove the statement? Also, I would appreciate it if someone verifies my writings. Thank you!
| Note that,
$$\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right)'
=-{a^2+b^2\over (a\sin x + b\cos x)^2}
$$
and apply the integration-by-parts,
$$J_n = \int {1 \over (a\sin x + b\cos x)^n } $$
$$ = -\frac{1}{a^2+b^2}\int {1 \over (a\sin x + b\cos x)^{n-2} } d\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right)
$$
$$ = -\frac{1}{a^2+b^2}\left[{a\cos x - b\sin x\over (a\sin x + b\cos x)^{n-1}}
+(n-2)\int {(a\cos x - b\sin x)^2\over (a\sin x + b\cos x)^{n} } dx\right]
$$
Recognize,
$$(a\cos x - b\sin x)^2 = (a^2+b^2) - (a\cos x + b\sin x)^2 $$
to express the integral as,
$$J_n = -\frac{1}{a^2+b^2}\left[{a\cos x - b\sin x\over (a\sin x + b\cos x)^{n-1}}
+(n-2)(a^2+b^2)J_n - (n-2)J_{n-2} \right]
$$
Rearrange to obtain,
$$
J_n={1\over (n-1)(a^2 + b^2)}\left[ {b\sin x-a\cos x \over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Centroid of area enclosed by $x^n$ and $x^{1/n}$. I am interested in finding the centroid of the surface enclosed by the lines $y=x^n$ and $y=x^{1/n}$ where $n \in \mathbb N, n >1$. The exercise, gives the following sketch.
I know that in order to find the centroid, we need to calculate its $x$ and $y$ coordinates, namingly :
$$x_c = \frac{\iint_A x \mathrm{d}A}{A} , \; y_c = \frac{\iint_A y \mathrm{d}A}{A}$$
Now, since $n>1$ I think that the only two points that the two lines meet are at $0$ and $1$. Thus, the area should be :
$$A = \int_0^1 ( x^{1/n} - x^n)\mathrm{d}x = \left[ \frac{x^{1/n + 1}}{1/n + 1}- \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{\frac{1}{n}+1} - \frac{1}{n+1} $$
I am having a bit of a struggle now, though, calculating the numerator quantities. Please, correct me if I am wrong in my approach for them as follows :
$$\iint_A x\mathrm{d}A = \int_0^1 \left( \int_0^{y^{1/n}} x \mathrm{d}x \right) \mathrm{d}y, \; \iint_A y\mathrm{d}A = \int_0^1 \left( \int_0^{x^{1/n}}y\mathrm{d}y\right)\mathrm{d}x$$
Thanks in advance.
| $$\iint_A x\mathrm{d}A = \int_0^1 \left( \int_{y^n}^{y^{1/n}} x \mathrm{d}x \right) \mathrm{d}y$$
$$= \int_0^1 \left[\frac {x^2}2 \right]_{y^n}^{y^{1/n}} \mathrm{d}y$$
$$= \frac 12 \int_0^1 \left[y^{\frac 2n}-y^{2n} \right] \mathrm{d}y$$
$$= \frac 12 \left[\frac{y^{\frac 2n+1}}{\frac 2n+1}-\frac{y^{2n+1}}{2n+1} \right]_0^1 $$
$$= \frac 12 \left[\frac{y^{\frac {2+n}{n}}}{\frac {2+n}{n}}-\frac{y^{2n+1}}{2n+1} \right]_0^1 $$
$$= \frac 12 \left[\frac{1}{\frac {2+n}{n}}-\frac{1}{2n+1} \right] $$
$$= \frac{n}{2(2+n)}-\frac{1}{2(2n+1)}$$
$$= \frac{n(2n+1)-(2+n)}{2(2+n)(2n+1)}$$
$$= \frac{2n^2-2}{2(2+n)(2n+1)}$$
$$= \frac{n^2-1}{(2+n)(2n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the set of values of $a$ ($a \in \mathbb{R}$) for which $\log_a(x^2 + 4) \ge 2$, $\forall x \in \mathbb{R}$? How can I find the set of values of $a$, with $a \in \mathbb{R}$ such that:
$$\log_a(x^2+4) \ge 2, \forall x \in \mathbb{R}$$
How should I approach this?
| Note if $0 < a < 1$ then $\log_a M$ is decreasing; If $M > 1$ then $\log_a M < 0$, $\log_a 1 = 0$ and if $0< M < 1$ then $\log_a M > 0$. And $M < N\iff M> N$.
So $\log_a (x^2 +4) \ge 2 \iff a^{\log_a x^2 +4 } = x^2 + 4 \le a^2 \iff x^2 \le a^2 -4$. But because $a^2 < 1$ then $a^2 -4 < -3$ and $x^3 < -3$ is impossible. So $\log_a(x^2 + 4) < 2$ for all $0< a < 1$.
And if $a > 1$ then $\log_a M$ is increasing; If $M > 1$ then $\log_a M > 0$, $\log_a 1 = 0$ and if $0< M < 1$ then $\log_a M < 0$. And $M < N\iff M< N$.
So $\log_a (x^2 +4) \ge 2 \iff a^{\log_a x^2 +4 } = x^2 + 4 \ge a^2 \iff x^2 \ge a^2 -4$.
For this be be true for all $x$ then $x^2 \ge 0$ and $x^2 =0$ is possible. So this is only possible and is always possible if $x^2 \ge 0 \ge a^2 -4$.
Or in other words if $a^2 \le 4$ or $|a| \le 2$. But as $a >1$ we have:
$1 < a \le 2$.
=== badly worded earlier answer below ===
If $a > 1$ then $\log_a(x^2 + 4) \ge 2\iff a^{\log_a (x^2 + 4)}=x^2 +4 \ge a^2\iff \sqrt{x^2 + 4} \ge a$. Now $\sqrt{x^2 + 4}$ can be as small, but no smaller than $2$ and can take on all values higher than $2$ so this can only be true for all $x$ (assuming $a > 1$) and will be true for all $x$ if $a \le 2$.
Or to put it another way. If $a > 2$ then if $0 \le x < \sqrt{a^2 -4}$ we have $x^2 + 4 < a^2$ and $\log_a (x^2 + 4) < 2$. But if $1 < a \le 2$ then $a^2 \le 4 \le x^2 + 4$ for all $x$ and $\log_2 (x^2 + 4) \ge 2$.
And if $0< a < 1$ then $x^2 + 4 \ge 4 > 1$ so $\log_a x^2 + 4 < 0$ for all $x$.
And $\log_a$ is undefined for $a\le 0$ or $a = 0$.
so $\log_a (x^2 + 4) \ge 2 \iff 1 < a^2 < x^2 + 4$ and that is only true for all $x$ if $1< a \le 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Combinatorics: Selecting pebbles making up a 9-sided shape Find the number of ways to simultaneously select $3$ pebbles from $9$ pebbles that are arranged in a circle such that no two of the three selected pebbles are adjacent to each other, and one selection is the same as another rotation if the former can be rotated to form the latter.
My Solution: At first, we can pick any $9$ points. Then, any $6$ points. And at last, $4$ points. This turns out be $6\cdot4\cdot9$. However, we have over-counted. We divide this by $3$! because we do not care about the order. Lastly, we divide by $9$ to account for the last part of the problem. Answer: $4$
Is this correct? And is there an easier way using combinatorics?
| We may also use the Polya Enumeration Theorem. Selecting three
non-adjacent pebbles leaves a total of six non-selected ones that must
be placed in three available slots with the cyclic group $C_3$ acting
on them and at least one pebble in every slot. The cycle index is
$$Z(C_3) = \frac{1}{3} a_1^3 + \frac{2}{3} a_3.$$
We then have
$$[z^6] Z\left(C_3; \frac{z}{1-z}\right)
= [z^6] \left(\frac{1}{3} \frac{z^3}{(1-z)^3}
+ \frac{2}{3} \frac{z^3}{1-z^3} \right)
\\ = \frac{1}{3} [z^3] \frac{1}{(1-z)^3}
+ \frac{2}{3} [z^3] \frac{1}{1-z^3}
= \frac{1}{3} {3+2\choose 2} + \frac{2}{3} [z^1] \frac{1}{1-z}
\\ = \frac{1}{3} \times 10 + \frac{2}{3} = 4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real part of a complex number
Compute $\operatorname{Re}\left(\frac{1}{z+1}\right)$ when when $|z| = 1$.
The only way I could think to go about this is to simply go by definitions. If $z\in \Bbb C$, then $z\bar z$ = $|z|^2$. Now $$z=\frac{|z|^2}{\bar z}$$and$$z=\frac{1^2}{\bar z}.$$ So $z$ must be the inverse of the conjugate of $z$ which can be written as $$z=\frac{z}{\bar z z}$$ I don't know how to proceed from here. Is this even a good approach?
| Let $z = a+bi$
$$\begin{align}Re\left(\frac{1}{z+1}\right) &= Re\left(\frac{1}{a+1+bi}\right) = Re\left(\frac{1}{a+1+bi}\cdot\frac{a+1-bi}{a+1-bi}\right) = \\ &= Re\left(\frac{a+1-bi}{(a+1)^2-(bi)^2}\right) = Re\left(\frac{a+1-bi}{(a+1)^2+b^2}\right) = \\ &= Re\left(\frac{a+1}{(a+1)^2+b^2} - \frac{b}{(a+1)^2+b^2}i\right) = \frac{a+1}{(a+1)^2+b^2} = \\ &= \frac{Re(z)+1}{(Re(z)+1)^2+Im(z)^2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Proof $a_{n}$ by induction This topic is a continuation of Prove that $a_{n} = a_{n-1} + a_{n-2}$.
I need to prove that $a_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$ for all $n \geq 1$.
This is how I tried to prove this.
Basic step: $a_{1} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{1} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{1}$ which is equal to $2$ and is correct.
Inductive step: Let $k \geq 1$ and let the statement apply to each $k$ where $1 \geq j \geq k$. From the previous topic we know that $a_{k} = a_{k-1} + a_{k-2}$ and so that $a_{k+1} = a_{k} + a_{k-1}$. Because of the hypothesis we know that the statement applies for $a_{k}$ and $a_{k-1}$. As $a_{k+1}$ exists of the sum of $a_{k}$ and $a_{k-1}$, we can conclude that the statement also applies for $a_{k+1}$.
Is this proof by induction valid?
| We have $a_{n} = a_{n-1} + a_{n-2}$ with $a_{0} = 1$ and $a_{1}=2$.
Let $b_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$.
To prove that $a_n=b_n$, it is enough to prove that $b_{n} = b_{n-1} + b_{n-2}$ with $b_{0} = 1$ and $b_{1}=2$.
*
*$b_{0} = 1$ and $b_{1}=2$ is immediate.
*$b_{n} = b_{n-1} + b_{n-2}$ follows from the fact that $\frac{1\pm\sqrt{5}}{2} $ are roots of $x^2=x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ How to use Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ to show that if $a+b+c=1$ then
$\displaystyle \left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right) \geq 8$?
Any idea how to take the vectors? Thanks a lot.
| With a little algebra, you can write $$\bigg{(}\frac{1}{a}-1\bigg{)}\bigg{(}\frac{1}{b}-1\bigg{)}\bigg{(}\frac{1}{c}-1\bigg{)} = \frac{1}{abc}(1-c-b+bc-a+ac+ab-abc) = \frac{1}{abc}(bc+ac+ab-abc) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 $$
Now, take $x = \bigg{(}\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}\bigg{)}$ and $y=(\sqrt{a},\sqrt{b},\sqrt{c})$. Then $$\langle x,y \rangle^{2} = 3^{2} = 9 \le \bigg{(}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg{)}(a+b+c) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
Thus $$8 \le \frac{1}{a}+\frac {1}{b}+\frac {1}{c}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Systems of equations over reals
Solve over reals
$$2x^2+xy+x-3y+3=0$$
$$3xy+y^2-2x+6y-6=0$$
Adding or subtracting these equations doesn't help, and completing the square also doesn't work. I'm considering substituting $x+y$ and $xy$, but it also doesn't show much promise. Thanks!
| Another way to do it.
From the first equation, extract $y$
$$y=-\frac{2 x^2+x+3}{x-3}$$
Plug $y$ in the second equation to get
$$\frac{-2 x^4+5 x^3+49 x^2+51 x+9}{(x-3)^2}=0$$ By inspection $x=-1$ and $x=-3$ are roots of the numerator. So, we are left with
$$\frac{(x+1) (x+3) \left(2 x^2-13 x-3\right)}{(x-3)^2}=0$$ So, the four roots for $x$ and the corresponding $y$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$ I'm trying to solve the following problem:
Find the coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$.
Can I use the multinomial theorem to solve it? I'm unsure how to start..
Thanks!
| Well it's simple
we are multiplying every term by $x^3$ so coefficient of $x^{10}$ should be coeffiecent of $x^7$ in the expansion of $(x^2-3x^3-1)^6$
Now, divide three terms into two groups $((x^2-3x^3)+(-1))^6$
To get $x^7$, the only combination possible is when $x^2$ is raised to the power two and $-3x^3$ is rased to the power one and multiplied.So we get $6C3*(x^2-3x^3)^{(2+1)}*(-1)^{(6-(2+1))}$. I took 3 (2+1) because thats when i get the terms $x^7$.. for $x^7$ we get $6C3*3C2*(x^2)^2*(-3x^3)*(-1)^3$.Now multiply all the coefficent we get $20*3*-3*-1=180$
And you got your answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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cauchy's first theorem on limits of sequences Cauchy's first theorem on limits goes like this
If$\ <f_n> $ be a sequence of positive terms and $$ \lim_{\ n\to\infty}\ f_n=l$$ Then
$$ \lim_{ n\to\infty}\ [\ \frac{f_1+f_2+\dots+f_n}{n}]=l$$
Now this is an example of its application.
$Q)$ Find the value of $$ \lim_{\ n \to \infty}\frac{1}{n}[1+\frac{1}{2}+\frac{1}{3}\dots+\frac{1}{n}] $$
$A)$ By cauchy's theorem if $$ f_n=\frac{1}{n} \ and \lim_{\ n\to\infty}\frac{1}{n}=0 $$ Then $$ \lim_{\ n\to\infty}\frac{1}{n}[1+\frac{1}{2}+\frac{1}{3}\dots+\frac{1}{n}]=0$$ I understand this example. Now here is another example and its solution which I found in various texts but I don't understand how can we apply cauchy's theorem to it
$Q)$ Find the value of $$ \lim_{\ n\to\infty}[\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2}] $$
$A)$ Multiply and divide by n
$$ \lim_{\ n\to\infty}\frac{1}{n}[\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\dots+\frac{n}{2n^2}] $$
Let $$ <f_n>=\frac{n}{(n+n)^2}=\frac{n}{4n^2}$$
Then, $$ \lim_{\ n\to\infty}f_n=\lim_{n\to\infty}\frac{1}{4n}=0$$
By cauchy's theorem
$$ \lim_{\ n\to\infty}[\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2}]=0 $$
As you can see there is clearly a distinction between the first and second question
In th first we have $ 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$
while in the second one we have $\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\dots+\frac{n}{(2n)^2} $
I see how the first one can be written as the sum of sequence $\frac{1}{n}$
But how can the second one written as the sum of the sequence $\frac{n}{(2n)^2}$
Please help....
| Answer to the second question.
Observe that for all $n \ge k$,
$$ \frac{1}{2n} \le \frac{1}{n+k} \le \frac{1}{n}$$
Thus,
$$\left( {\frac{1}{{{{(2n)}^2}}} + \frac{1}{{{{(2n)}^2}}} + \ldots + n{\mkern 1mu} {\mkern 1mu} terms} \right) \le \left( {\frac{1}{{{{(n + 1)}^2}}} + \frac{1}{{{{(n + 2)}^2}}} + \ldots + \frac{1}{{2{n^2}}}} \right) \le \left( {\frac{1}{{{n^2}}} + \frac{1}{{{n^2}}} + \ldots + n{\mkern 1mu} {\mkern 1mu} terms} \right)$$
$$\implies \frac{1}{4n} \le \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2} \le \frac{1}{n}$$
The answer follows by the Sandwich theorem.
| {
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"source": "stackexchange",
"question_score": "6",
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Prime-free sequence
Prove that there exist infinitely many positive integers $k$ such that
the sequence $\{x_n\}$ satisfying
$$ x_1=1, x_2=k+2, x_{n+2}-(k+1)x_{n+1}+x_n=0(n \ge 0)$$ does not
contain any prime number.
I found a similar question: The sequence ${x_n}$ satisfies $x_{n+2}=(k+1)x_{n+1}-x_n, x_0=1, x_1=k+2$ and every term of the sequence is either $1$ or composite. Prove that the set of such $k$ is infinite.
Solution: Let $\alpha=\frac{\sqrt{k+3}+\sqrt{k-1}}{2}$, $\beta=\frac{\sqrt{k+3}-\sqrt{k-1}}{2}$, then$$x_n=\frac{\alpha^{2n+1}-\beta^{2n+1}}{\alpha-\beta}.$$We consider$$x_{n}^2=\frac{(\alpha^{2n+1}+\beta^{2n+1})^2-4}{k-1},$$let $a_n=\alpha^{2n+1}+\beta^{2n+1}$, $a_0=\sqrt{k+3}$, $a_1=\frac{(k+3)^{\frac{3}{2}}+3\sqrt{k+3}(k-1)}{4}$. We let $k+3=u^2$, and $2|u$, then $a_n$ is a positive integer, we have$$(k-1)x_{n}^2=(a_{n}+2)(a_{n}-2).$$If $x_n=p$ is a prime, $p$ is lagre enough when $n$ is lagre enough, $gcd(a_n-2, a_n+2)=gcd(a_n-2, 4)\leq 4$, it follows that $p^2\mid a_n+2$, or $p^2\mid a_n-2$, a contradiction.
Is a different solution like this?
| I tried using the Z transform. From the question, we conclude that $x_0 = -1$. And taking the Z transform we get:
$$ x_{n+2} - (k+1)x_{n+1}+x_n = 0 $$
$$ z^2X(z)-x_0z^2-x_1z-(k+1)(zX(z)-x_0z) + X(z) = 0 $$
$$ X(z) = \frac{(k+1)z+z-z^2}{z^2-(k+1)z+1} $$
if we let $a=k+1$ then after taking the inverse Z transform we get:
$$ x_n = 2^{-n-1}\frac{ (-\sqrt{a^2-4}+a+2)(\sqrt{a^2-4}+a)^n - (\sqrt{a^2-4}+a+2)(a-\sqrt{a^2-4})^n }{\sqrt{a^2-4}}$$
This might be helpful for further solving.
For $a=2$ which is $k=1$ the sequence becomes:
$$ x_n = 2n - 1 $$ and it contains infinitely many prime numbers since it is a sequence of odd numbers and we have a contradiction.
| {
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"url": "https://math.stackexchange.com/questions/3441436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$ Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$
My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side)
$$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$
However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is:
$$\sqrt a+\sqrt b+\sqrt c\le\frac35$$
However, as $a,b,c<1$, $$\sqrt a +\sqrt b + \sqrt c> a+b+c=1>\frac35.$$ So my upper bound is always larger than the given one.
| Multiply both sides with $abc$ to get that your inequality is equivalent to
\begin{equation}\tag 1\label 1\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}\le\frac15.\end{equation}
Now note that, since $a+b+c=1$, we have $$a^2+b^3+c^3=a^2\cdot(a+b+c)+b^3+c^3=a^3+b^3+c^3+a^2b+a^2c.$$
Hence, by AM-GM inequality for 5 variables,
$$a^2+b^3+c^3\geq5\sqrt[5]{a^7b^4c^4}.$$
Hence, \begin{split}\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}&\le\frac15\sum_{\text{cyc}} \frac{a^2bc}{\sqrt[5]{a^7b^4c^4}}\\&=\frac15\sum_{\text{cyc}} \sqrt[5]{a^3bc}\\&\overset{\text{AM-GM}}\le\frac1{25}\sum_{\text{cyc}}3a+b+c\\&=\frac{5(a+b+c)}{25}\\&\overset{a+b+c=1}=\frac15. \end{split}
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$
As $3>0$ and $\dfrac{1}{y}>0$
$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$
$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$
$$x=\dfrac{3y-1}{y+3}$$
$y+3\in[4,\infty)$ as $y\ge1$, $3y-1\in [2,\infty)$ as $y\ge1$
For $x$ to be positive integer, $3y-1$ should be multiple of $y+3$
$$3y-1=m(y+3) \text { where } m\in Z^{+}$$
$$3y-my=3m-1$$
$$(3-m)y=3m-1$$
Here R.H.S is positive, so L.H.S should also be positive.
So $3-m>0$, hence $m<3$
So possible values of $m$ are {$1$,$2$}.
For $m=1$, $$3y-1=y+3$$
$$2y=4$$
$$y=2$$
$$x=\dfrac{3\cdot2-1}{2+3}$$
$$x=1$$
For $m=2$, $$3y-1=2(y+3)$$
$$3y-1=2y+6$$
$$y=7$$
$$x=\dfrac{3\cdot7-1}{7+3}$$
$$x=\dfrac{20}{10}$$
$$x=2$$
Is there some other nicer way to solve this problem.
| Another way
As we need $x,y>0$
If $x/y<1$
$$\tan^{-1}x+\tan^{-1}(1/y)=\tan^{-1}\dfrac{xy+1}{y-x}$$
$$xy+1=3(y-x)$$
$$\iff x=\dfrac{3y-1}{y+3}=3-\dfrac{10}{y+3}$$
So, $y+3(>3), $ must divide $10$ and must honor $x<y$
If $x/y >1,x>y$ as $x,y>0$
$$\tan^{-1}x+\tan^{-1}\dfrac1y>\tan^{-1}y+\cot^{-1}y=\dfrac\pi2>\tan^{-1}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$ I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.
| You can extract the factor $7$ as follows using
*
*$(\star)$: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
Hence,
$$3^{2n-1} + 2^{n+1}= 3\cdot 9^{n-1} + 4\cdot 2^{n-1}$$
$$= 3\cdot 9^{n-1} + (7-3)\cdot 2^{n-1} = 3(\underbrace{9^{n-1} - 2^{n-1}}_{\stackrel{ (\star)}{=}7\cdot m}) + 7\cdot 2^{n-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$ which step is wrong? So I'm trying to evaluate $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$
These are my steps:
I first rationalize the expression (square root trick) -
$$\lim \limits_{x\to -\infty} \frac{-2x}{x - \sqrt{x^2+2x}}$$
Then I simply divide by $x$ so
$$\lim \limits_{x\to -\infty} \frac{-2}{1 - \sqrt{1+\frac{1}{2x}}}$$
Then I get the following by evaluating the limit
$$\frac{-2}{1 - \sqrt{1}}$$
which then evaluates to $0$ in the denominator.
Would really appreciate some help in understanding what I'm doing wrong here.
Thanks in advance!
| What you got wrong is in the step "I simply divide by $x$", where you should get
$$
\frac{-2x}{x - \sqrt{x^2+2x}}=
\frac{-2}{1-\frac{1}{x}\sqrt{x^2+2x}}
=\frac{-2}{1{\color{red}+}\sqrt{\frac{x^2+2x}{x^2}}}
=\frac{-2}{1{\color{red}+}\sqrt{1+\frac{2}{x}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the closed form for the following power series I wish to find the closed form for this power series:
$1+\dfrac{1}{6}x+\dfrac{3}{40}x^2+\dfrac{5}{112}x^3+\dfrac{35}{1152}x^4...$
I have been able to spot the that the second term is the first term multiply with $\dfrac{1 \cdot 1}{2 \cdot 3}$, the third term is the second term multiply with $\dfrac{3 \cdot 3}{4 \cdot 5}$, the fourth term is the third term multiply with
$\dfrac{5 \cdot 5}{6 \cdot 7}$, the fifth term is the fourth term multiply with
$\dfrac{7 \cdot 7}{8 \cdot 9}$ and so on.
How do I start to find the closed form for this series? It is taken from James Stirling's Methodus Differentialis. The author doesn't present a closed form
| As noted in the comments we have for $|x|\lt1$
$$\arcsin{(x)}=x+\frac16x^3+\frac3{40}x^5+\frac5{112}x^7+\dots$$
Hence the provided power series is just
$$1+\frac16x+\frac3{40}x^2+\frac5{112}x^3+\dots=\begin{cases}1&x=0\\\frac{\arcsin{(\sqrt{x})}}{\sqrt{x}}&x\ne0\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $.
*
*For $\int_0^{1}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I prove $\ln(1+x^2)< x^2$, using that I can prove $\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ converges, but it take too long so is there shorter way to do this problem?
*For $\int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I have no idea how to do this problem.
| Your approach on $[0,1]$ seems fine. For the other part, you can verify that:
$$\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}} \le \frac{\ln (1 + x^2)}{\sqrt {x^6}}=\frac{\ln (1 + x^2)}{x^3} \quad \mbox{and} \quad \int_1^{\infty}\frac{\ln (1 + x^2)}{x^3}\,\mbox{d}x = \ln 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What's the length of EC? Consider the following construction (the lenght of segment EC in the image is just an approximation):
I wanted to calculate the exact value of the length of EC.
Using the law of cosines, law of sines, the angle bisector theorem and some manipulation I arrived at:
$$\sqrt{16+\left(5-\frac{35 \csc \left(\frac{\pi }{4}+\cos ^{-1}\left(\frac{29}{35}\right)\right)}{9 \sqrt{2}}\right)^2+\frac{8}{5} \left(5-\frac{35 \csc \left(\frac{\pi }{4}+\cos ^{-1}\left(\frac{29}{35}\right)\right)}{9 \sqrt{2}}\right)}$$ which simplifies to $\displaystyle \frac{28 \sqrt{1240129-291740 \sqrt{6}}}{4113}$.
While I believe this is correct I also believe there should be a easier way to solve this. Is there a simpler argument?
| Calling $A = \{x_a,y_a\}, B = \{0,0\}, C = \{x_c,0\}$ and equating
$$
\cases{
|A-B|^2 = 5^2\\
|B-C|^2 = 7^2\\
|C-A|^2 = 4^2
}
$$
we obtain
$$
\cases{
A = \{\frac{29}{7},-\frac{8\sqrt 6}{7}\}\\
C = \{7,0\}
}
$$
Calling now
$$
\vec v_{AC} = \frac{C-A}{|C-A|} =\left \{\frac{5}{7},\frac{2\sqrt 6}{7}\right\}\\
\vec v_{AB} = \frac{B-A}{|B-A|} =\left \{-\frac{29}{35},\frac{8\sqrt 6}{35}\right\}
$$
we have for the point $D$ determination
$$
A + \lambda (\vec v_{AC}+\vec v_{AB}) = B + \mu(C-B)
$$
which solved for $\lambda,\mu$ gives
$$
D = \left\{\frac{35}{9},0\right\}
$$
and finally the point $E$ determination
$$
A + \lambda(B-A) = D + \mu\{-1,-1\}
$$
which solved for $\lambda,\mu$ gives
$$
E = \left\{\frac{1015 \left(29-8 \sqrt{6}\right)}{4113},-\frac{560}{144+87 \sqrt{6}}\right\}
$$
hence
$$
|E-C| = \frac{28 \sqrt{1240129-291740 \sqrt{6}}}{4113}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| Use lagrange multipliers. $\nabla f=\langle 2x,2y,2z\rangle$ and $\nabla g=\langle x^{-2},y^{-2},z^{-2}\rangle$ where $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)= x^{-1}+y^{-1}+z^{-1}$. Solving $\nabla f=\lambda\nabla g$ yields $\lambda=2x^3=2y^3=2z^3$ and therefore $x=y=z$. Plugging this into $g(x,y,z)=1$ we have $x=y=z=3$ which we plug into $f(x,y,z)$ to obtain $3^2+3^2+3^2=27$ which is obviously a minimum as $x^2+y^2+z^2$ is unbounded above on the surface $g(x,y,z)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| I got a new way to solve this question and its just beautiful:
Multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$
$$\dfrac{1}{abc}\begin{vmatrix}
a^3+a\lambda&a^2b&a^2c\\
ab^2&b^3+b\lambda&b^2c\\
ac^2&bc^2&c^3+c\lambda
\end{vmatrix}$$
Taking a common from first column, b from second column, c from third column
$$\begin{vmatrix}
a^2+\lambda&a^2&a^2\\
b^2&b^2+\lambda&b^2\\
c^2&c^2&c^2+\lambda
\end{vmatrix}$$
Now its simple , just do $$C_1\rightarrow C_1-C_2, C_2\rightarrow C_2-C_3$$
$$\begin{vmatrix}
\lambda&0&a^2\\
-\lambda&\lambda&b^2\\
0&-\lambda&c^2+\lambda
\end{vmatrix}$$
Now it's simple to solve. Hope it will be useful to somebody.
| {
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"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$ twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom{18}{1k}\cdot \left ( \binom{19}{k}+\binom{19}{k-1} \right )=\sum_{k=0}^{19}\binom{18}{k}\cdot\left ( \binom{18}{k}+2\binom{18}{k-1}+\binom{18}{k-2} \right )$
Now have no idea what to do with that
| Your sum is equal to $c_n = \sum_{k=0}^n \binom{18}{k} \binom{20}{n-k}$ for $n=20.$ Note that $c_n$ is the coefficient of $x^n$ in $(1+x)^{18} (1+x)^{20}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(\cos x-1)^2(4\cos^2x+2\cos x-1)^2=16(1-\cos^2x)(2\cos^3x-\cos x)^2$
We obviously get $\cos x=1$ (which indeed is a solution of original equation) and it remains
$(1-\cos x)(4\cos^2x+2\cos x-1)^2=16(1+\cos x)(2\cos^3x-\cos x)^2$
Setting $\cos x=y$, this is :
$(1-y)(4y^2+2y-1)^2=16(1+y)(2y^3-y)^2$
Which is $64y^7+64y^6-48y^5-64y^4-4y^3+16y^2+5y-1=0$
I found no clever way to solve this degree-7 polynomial
| Factorirzing your given equation we get
$$2 \sin \left(\frac{x}{2}\right) \left(\sqrt{2} \sin
\left(\frac{5 x}{2}+\frac{\pi }{4}\right)+\cos
\left(\frac{x}{2}\right)+\cos \left(\frac{3
x}{2}\right)+\cos \left(\frac{7 x}{2}\right)\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Minimize $x + y + z$ subject to $x^2 + y^2 + z^2 \le 1$ and $x \ge 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + y + z \\
\text{s.t} & \quad x^2 + y^2 + z^2 &&\le 1 \\
& \quad x &&\ge 0
\end{align*}$$
The linear independence constraint constraint qualification - LICQ is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
The Mangasarian-Fromovitz constraint qualification - MFCQ is
The gradients of the equality constraints are linearly independent at $x^{*}$ and there exists a vector $v \in \mathbb{R}^{n}$ such that $\langle \nabla g_{i}\left(x^{*}\right), v \rangle<0$ for all active inequality constraints and $\langle \nabla h_{j}\left(x^{*}\right), v\rangle=0$ for all equality constraints.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
$\textbf{My attempt:}$
Let $f = x + y + z$, $g_1 = x^2+y^2 + z^2-1$, $g_2 = -x$, and $$\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid g_1(x,y,z) \le 0 \text{ and } g_2(x,y,z) \le 0\}$$
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $\nabla g_1 (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$, and $\nabla g_2 (x,y,z) = \begin{pmatrix} -1 \\ 0 \\ 0 \\ \end{pmatrix}$. Consider the system $$\begin{cases}
\mu_1 \nabla g_1 (x,y,z) + \mu_2 \nabla g_2 (x,y,z) &=0 \\
g_1(x,y,z) &= 0\\
g_2(x,y,z) &=0
\end{cases}
\iff
\begin{cases}
\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} -1\\ 0 \\ 0 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\
x^2 + y^2 + z^2 &= 1\\
-x &= 0
\end{cases}$$
$$\iff \begin{cases}
x &= 0\\
\mu_2 &= 0 \\
\mu_1 y &= 0 \\
\mu_1 z &= 0\\
y^2+z^2 &= 1
\end{cases}
\implies
\mu_1 = \mu_2 = 0
$$
Hence LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases}
\mu_1,\mu_2 &\ge 0\\
\mu_1 g_1(x,y,z) &= 0\\
\mu_2 g_2(x,y,z) &=0 \\
\nabla f (x,y,z) + \mu_{1} \nabla g_1(x,y,z) + \mu_{2} \nabla g_2(x,y,z)&=0
\end{cases}
\iff
\begin{cases}
\mu_1, \mu_2 &\ge 0\\
\mu_1 (x^2+y^2 + z^2-1) &=0\\
\mu_2 x &= 0 \\
\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} +\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} -1\\ 0 \\ 0 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix}
\end{cases}$$
$$\iff
\begin{cases}
\mu_1, \mu_2 &\ge 0\\
\mu_1 (x^2+y^2 + z^2-1) &=0\\
\mu_2 x &=0 \\
1+ 2 \mu_1 x &= \mu_2 \\
1+ 2 \mu_1 y &=0 \\
1+ 2 \mu_1 z &=0
\end{cases}
\iff
\begin{cases}
\mu_1 &> 0\\
\mu_2 &\ge 0\\
\mu_2 x &=0 \\
x^2+y^2 + z^2 &= 1 \\
1+ 2 \mu_1 x &= \mu_2 \\
1+ 2 \mu_1 y &=0 \\
1+ 2 \mu_1 z &=0
\end{cases}$$
$$\iff\begin{cases}
\mu_1 &= 1/\sqrt{2}\\
\mu_2 &= 1 \\
x &=0 \\
y &=-1/\sqrt{2} \\
z &=-1/\sqrt{2}
\end{cases} \qquad
\text{or} \qquad\begin{cases}
\mu_1 &= \sqrt{3}/2 \\
\mu_2 &= 0 \\
x &= -1/\sqrt{3} \\
y &= -1/\sqrt{3} \\
z &= -1/\sqrt{3}
\end{cases}$$
Comparing the values at these points, we get the solution is $(x,y,z) = (-1/\sqrt{3},-1/\sqrt{3},-1/\sqrt{3})$.
| Alternate Solution :
$x^2 + y^2 + z^2 \le 1 $ represents a solid sphere of radius 1
Now, we need to find a point on sphere such that the value of $x+y+z$ is minimum
It is also given that, $x \ge 0 $ , So we need to only consider the 4 octants out of 8
Now, out of these 4 octants, a minimum occurs when both $y$ and $z$ are negative, so we need to consider only 5th octant. Since $y$ and $z$ are negative a minimum occurs on the surface of the sphere i.e $y^2 + z^2 = 1$
Since $y$ and $z$ are already chosen to be negative, all it remains is what values of x make the result minimum, it is simply at x = $0$
All we need to find is a point on the sphere such that $x = 0$, $y$ and $z$ are negative. It doesn't take a lot of work to think that it occurs simply at $y = z$
$\implies y^2 + y^2 = 1 \implies y = z = \frac{-1}{\sqrt{2}} $
So, we can conclude that a minimum occurs at $(0,\frac{-1}{\sqrt{2}}\frac{-1}{\sqrt{2}})$
and hence the minimum value is $-\sqrt{2}$
Hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/3458496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find perfect number which in this form: $p^2q$ I want to find a perfect number in the form $p^2q$ where $p$ and $q$ are both prime numbers.
I get the diophantine equation :$p^2q=p^2+pq+p+q+1$ from the definition of the perfect number.
So I assume that there is one positive integer solution which is $(p,q)$ and $c=p+q$ which is a constant.
Rearranging the equation, I get $p^3-cp^2+cp+c+1=0$
So $p^3+1=(p+1)(p^2-p+1)=c(p^2-p-1)$
Thus $c=\frac{(p+1)(p^2-p+1)}{p^2-p-1}=p+1+\frac{2p+2}{p^2-p-1}$
Obviously $c-p-1$ is a positive integer since $p,q$ are prime numbers which are greater than $1$.
So assume positive integer $k=\frac{2p+2}{p^2-p-1}$
Thus $kp^2-(k+2)p-(k+2)=0$
Apply the root formula, we get $p=\frac{k+2+\sqrt{(k+2)^2+4k(k+2)}}{2k}$
so $(k+2)^2+4k(k+2)=n^2$ where $n$ is a positive integer.
Thus $n^2-(k+2)^2=(n-k-2)(n+k+2)=4k(k+2)$
Then what I can do next? I try to list all of the situations but $k$ is not a prime number.
| Solving $p^2q=p^2+pq+p+q+1$ for $q$ we get
$$
q = \frac{p^2 + p + 1}{p^2 - p - 1} = 1 + \frac{2p+2}{p^2 - p - 1}
$$
The last fraction needs to be an integer and so we need $2p+2\ge p^2 - p - 1$. This gives $p \in \{0,1,2,3\}$. The only solution is $p=2$ and $q=7$, which gives $p^2q=28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating polynomial is $x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1).$ Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
So, Possible CHaracteristic polynomials are $(x-1)(x+1),(x^2+x+1),(x^2-x+1)$.We can not take any 2-degree polynomial factor from the annihilating polynomial. Since complex roots occur in pairs. But the answer given is $5$. Can you help me where is my mistake?
| This statement is false:
Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
$$f(t) = (t-1)(t^4+7t^3+105t-999)$$
Annihilates $A=I$. The characteristic polynomial of $A$ does not divide $f(t)$.
It seems like you may be confusing characteristic polynomial with minimal polynomial.
Without loss of generality, you may assume $A$ is diagonal (it is diagonalizable, and similar matrices have the same characteristic polynomial). Your matrix, up to similarity will be
$$\pmatrix{1 & 0 \\ 0 & 1}, \pmatrix{1 & 0 \\ 0 & -1}, \pmatrix{-1 & 0 \\ 0 & -1},$$
$$ \pmatrix{-\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & -\dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}, \pmatrix{\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & \dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}$$
And so it's characteristic polynomial (respectively) will be one of
$$(x-1)^2, (x-1)(x+1), (x+1)^2, x^2+x+1, x^2-x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?
| Let $x=\dfrac{1}{a}, y=\dfrac{1}{b}, z=\dfrac{1}{c}$. It is easy to get
$$x^2+y^2+z^2\ge xy+yz+zx$$ which is $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}$$
We try to prove $a+b+c\ge\sqrt{3}$. As $a^2+b^2+c^2\ge ab+bc+ca\ge1$,
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\ge3\\ a+b+c\ge\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $2xyy^\prime=x^2+2y^2$
Solve $2xyy^\prime=x^2+2y^2$
I tried to solve with substitution $v=y^2$
then $\frac{dy}{dx}=\frac{1}{2}v^{-1/2}\frac{dv}{dx}$
and substituting the given equation I get $$\frac{x^2+v}{2x\sqrt{v}}=\frac{1}{2\sqrt{v}}\frac{dv}{dx}$$
$$\frac{x^2+v}{x}=\frac{dv}{dx}$$
$$\frac{dv}{dx}=x+\frac{v}{x}$$
But now it seems like I just need to make another substitution to solve this.
| Note that you have a linear equation
\begin{eqnarray*}
\frac{dv}{dx} +P v =Q
\end{eqnarray*}
where $P$ and $Q$ are functions of $x$ only. We multiply through by a function $R$ in order to make the LHS an exact diffeential of a product
\begin{eqnarray*}
R\frac{dv}{dx} +RP v =RQ \\
\frac{d}{dx} (Rv) = R \frac{dv}{dx} +R'v.
\end{eqnarray*}
So we need $R'=RP$ or $R=e^{\int P dx }$ ... This should shed some light on how we get the second line below.
You lost a factor of $2$ somewhere. We have
\begin{eqnarray*}
x\frac{dv}{dx}=x^2+2v \\
\frac{1}{ x^2}\frac{dv}{dx} - \frac{2v}{x^3}=\frac{1}{x} \\
\frac{d}{dx} \left( \frac{v}{x^2} \right) =\frac{1}{x} \\
\frac{v}{x^2} =\ln(x)+\ln(A) \\
v=x^2 \ln(Ax).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$ using polynomial I find the identity, $$\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$$
by counting the number of ways to distribute $K$ candies to $N$ people ($N>K)$.
For right-hand eq., I count the number of ways to distribute candies when there is at least $i$ people who cannot get candies.
Can we prove this identity using polynomials?
For left-hand eq., it corresponds to the coefficient of $x^K$ of $\frac{1}{(1-x)^N}$. Can we get the right-hand eq. by manipulating this polynomial?
| We can re-write the identity as
$$\sum_{q=0}^{N-1} (-1)^{q+1} {N\choose q} {N-q+K-1\choose K} = 0.$$
Equivalently (divide by $-1$),
$$\sum_{q=0}^{N-1} (-1)^q {N\choose q}
{N-q+K-1\choose N-1-q}
\\ = \sum_{q=0}^{N-1} (-1)^q {N\choose q}
[z^{N-1-q}] (1+z)^{N-q+K-1}
\\ = [z^{N-1}] (1+z)^{N+K-1}
\sum_{q=0}^{N-1} (-1)^q {N\choose q}
z^q (1+z)^{-q}.$$
Now we may add in the term for $q=N$ because it does not contribute to
the coefficient extractor:
$$[z^{N-1}] (1+z)^{N+K-1}
\sum_{q=0}^{N} (-1)^q {N\choose q}
z^q (1+z)^{-q}
\\ = [z^{N-1}] (1+z)^{N+K-1}
\left(1-\frac{z}{1+z}\right)^N
\\ = [z^{N-1}] (1+z)^{K-1}.$$
This is zero by inspection when $N\gt K$ as claimed. We obtain
${K-1\choose N-1}$ when $N\le K.$
Remark. If an inverse binomial is insisted upon we may start
from
$$\sum_{q=0}^{N-1} (-1)^q {N\choose q} {N-q+K-1\choose K}
\\ = \sum_{q=0}^{N-1} (-1)^q {N\choose q}
[z^{N-1-q}] \frac{1}{(1-z)^{K+1}}
\\ = [z^{N-1}] \frac{1}{(1-z)^{K+1}}
\sum_{q=0}^{N-1} (-1)^q {N\choose q} z^q.$$
Same action of the coefficient extractor as before:
$$[z^{N-1}] \frac{1}{(1-z)^{K+1}}
\sum_{q=0}^{N} (-1)^q {N\choose q} z^q
= [z^{N-1}] \frac{1}{(1-z)^{K+1}} (1-z)^N
\\ = [z^{N-1}] \frac{1}{(1-z)^{K-N+1}}.$$
Now when $N\gt K$ this becomes $[z^{N-1}] (1-z)^{N-K-1}$ which is zero
by inspection. With $N\le K$ we find ${N-1+K-N\choose K-N} =
{K-1\choose K-N} = {K-1\choose N-1}$ in agreement with the previous
result. We suppose in both versions that $N,K\ge 1.$
| {
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"url": "https://math.stackexchange.com/questions/3467933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show, that $ I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$
Show, that for $m,n\in\mathbb{N}$ holds: $$\displaystyle I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$$
I tried to do integration by parts and got $$\frac { ( 1 - x ) ^ { n + 1 } } { n + 1 } \cdot x ^ { m } - \int \frac { ( 1 - x ) ^ { n + 1 } } { n +1 }\cdot m\cdot x ^ { m - 1 } d x = \frac{(1-x)^{n+1}}{n+1}\cdot x^m-\left(\frac{1}{n+1}\cdot m\underbrace{\int (1-x)^{n+1}\cdot x^{m-1}dx}_{*}\right)\\
*=\left(\frac{(1-x)^{n+2}}{n+2}\cdot x^{m-1}-\left(\frac{1}{n+2}\cdot (m-1)\int (1-x)^{n+2}\cdot x^{m-2}dx\right)\right)=\frac{(1-x)^{n+1}}{n+1}\cdot x^m-\left(\frac{1}{n+1}\cdot m\left(\frac{(1-x)^{n+2}}{n+2}\cdot x^{m-1}-\left(\frac{1}{n+2}\cdot (m-1)\int (1-x)^{n+2}\cdot x^{m-2}dx\right)\right)\right)
$$
Do I see a pattern after integrating this a few times, because I don't really come to the conclusion, that this would equal $I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$. I've integrated two times now.
| $I_{m+1,n} = \frac{m+1}{n+1}\int_0^1 x^m(1-x)^{n+1}dx$ by IBP.
$= \frac{m+1}{n+1}\int_0^1 x^m(1-x)^{n}(1-x)dx = \frac{m+1}{n+1}(I_{m,n}-I_{m+1,n})$.
Thus $I_{m+1,n} = \frac{m+1}{m+n+2}I_{m,n}$. We may proceed by induction on $m$.
By symmetry we apply the same argument in $n$ and thus get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$ never a perfect square? In this question, the questioner asked to prove that $$f(n)=\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$$ is a perfect square for all $n\in\mathbb N$. However, I was not able to find any $n$ for which this is true. So I have the following "counter-conjecture": For any $n\in\mathbb N=\{1,2,3,4,\dots\}$, $f(n)$ is not a perfect square.
How can I prove this?
Notice that by simply evaluating the geometric series we get $$f(n)=\frac{8}{9} \left(10^n-1\right)+\frac{4}{9} \left(10^{2 n}-1\right)+4=\frac{4}{9} \left(2^{n+1} 5^n+100^n+6\right),$$
so it would be enough to prove that $$\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$$ is never an integer.
But I don't know how to do this.
| Whenever I see something of the form $A + 2A\times\text{something of magnitude} + A\times \times\text{something of geometrically more magnitude}$, I figure it fits roughly into the form of $A(\text{something of mangitude} + 1)^2$.
So so $4_{2n}= \frac 49\times (10^{2n} -1)$ is roughly of magnitude $(10^n)^2$ and $8_{n} = 49\times (10^n -1)$ is roughly of magnitude $10^n$ then we should figure $4_{2n}+ 8_n + 4$ is fairly close so $\frac 49(10^n + 1)^2$.
And it turns it is VERY close (but slightly more) that it is too large to be a square of $\frac 23(10^n+1)$ but too small to be the next integer square.
(We might pause and consider that $\frac 23(10^n+1)$ need not be an integer. That turns out not to be relevant.)
=====
To continue along your lines:
Express as $\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$
$=\frac 23\sqrt{(10^n)^2 + 2*10^n + 6}$
Notice that $n \ge 1$ then $(10^n + 1)^2 = 10^{2n} + 2*10^n + 1 < (10^n)^2 + 2*10^n + 6 < 10^{2n} + 4*10^n + 4 = (10^n + 2)^2$
So $(10^n)^2 + 2*10^n + 6$ is not a perfect square (but is an integer) so $\sqrt{2^{n+1} 5^n+100^n+6}$ is irrational and $\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$ is not an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$ Prove:
$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$
My solution:
$( x+1>x)$
$$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx< \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{1-x}}=0.5
$$
Is there a closed formula for the integrals
| Multiply numerator and denominator by $$\sqrt[3]{(1+x)}^2
-\sqrt[3]{1+x}\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and subtracting $1$ from the original limit. However, since $3-2x$ $\rightarrow - \infty$ and not $+\infty$, I got nowhere. Then I tried finding the logarithm of this limit. It resulted in a $\dfrac{0}{0}$ indeterminate form, I tried L'Hospital, but again, it led me nowhere. Either I made some mistakes in the calculations, or I should use a different approach.
| \begin{align*}
&\left(\dfrac{\sqrt{x^{2}+2x-3}}{x+2}\right)^{3-2x}\\
&=\left(\dfrac{\sqrt{x^{2}+2x-3}}{\sqrt{x^{2}+2x}}\right)^{3-2x}\left(\dfrac{\sqrt{x^{2}+2x}}{x+2}\right)^{3-2x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{3/2-x}\left(\dfrac{x}{x+2}\right)^{3/2-x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{-(x^{2}+2x)(x-3/2)/(x^{2}+2x)}\left(1-\dfrac{2}{x+2}\right)^{-(x+2)(x-3/2)/(x+2)}\\
&\rightarrow 1\cdot e^{2}\\
&=e^{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine $\lim_{x\to\infty} f(x)$, where $f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}$
Determine $\lim_{x\to\infty} f(x)$ where
$$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$
I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos and check value of this function at very large values like $10^9$ it shows function is equal to $1$.
I am more concerned about why this happens rather than answer.
| Take $$\ln f(x)=({x^3+3x})\ln\left(\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}\right)$$
and note
$$\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}
=\frac{1+\frac1{x^3}\sqrt{1+\frac1{x^2}}}{1-\frac1{x^3}+\frac2{x^4}}
=\frac{1+\frac1{x^3}+O(\frac1{x^5})}{1-\frac1{x^3}+\frac2{x^4}}
=1+ \frac2{x^3}+O(\frac1{x^4})$$
Then,
$$\lim_{x\to\infty} \ln f(x)=\lim_{x\to\infty}(x^3+3x)\ln(1+ \frac2{x^3}+O(\frac1{x^4}))$$
$$=\lim_{x\to\infty} (x^3+3x)(\frac2{x^3}+O(\frac1{x^4}))=\lim_{x\to\infty} (2+O(\frac1{x}))=2$$
Thus,
$$\lim_{x\to\infty} f(x)=e^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.