Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$ Compute the given line integral:
*
*$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$ then $$\color{blue}{\frac{\partial P}{\partial y}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{green}{\frac{\partial P}{\partial z}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{blue}{\frac{\partial Q}{\partial x}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial Q}{\partial z}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{green}{\frac{\partial R}{\partial x}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial R}{\partial y}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$
So the vector field is conservative which means there is $f$ such that $\vec \nabla f=F$, I don't know how to continue.
| Consider
$$f(x,y,z)=\sqrt{x^2+y^2+z^2}+C$$
then $\vec \nabla f=F$ and
$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}=f(3,4,5)-f(0,0,0)=5\sqrt{2}.$$
| {
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"url": "https://math.stackexchange.com/questions/4147367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Binomial coefficient and series expansion I had to expand $y=\sqrt{1+x^2}$ into the Maclaurin series, which (prior to simplification) would give me: \begin{align} \sqrt{1+x^2}= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}+...+\frac{1/2(1/2-1)...(1/2-n+1)x^{2n}}{n!}+...\end{align}
What is the nice way to compactify the series using the $\sum$ notation? I need that because I will have to check the convergence of the series at $x=-1$ and $x=1$ (at which the series seems to converge).
From my lecture notes I can see that $(1+x)^m=\sum^{\infty}_{n=0} C^{n}_{m}x^n,|x|<1$, but because we've never used the combination notion, I'm confused as to what the correct way to implement it is.
It appears that I can multiply the numerator and the denominator of my fraction $\frac{1/2(1/2-1)...(1/2-n+1)x^{2n}}{n!}$ by $(1/2-n)!$ to turn it into $\frac {1/2!}{n!(1/2-n)!}$, which seems to be the normal expansion for the notion $C^n_m, m=1/2$.
However, would it be right of me to use the factorial of a fraction here or am I mixing some incompatible notations? Is there something I'm doing completely wrong? I would be grateful for a clarification.
| A convenient generalisation of the binomial coefficient is given for real (or even complex) values $\alpha$ by
\begin{align*}
\binom{\alpha}{n}:=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}&n\geq 0\tag{1}\\
0&n<0
\end{cases}
\end{align*}
Using (1) we can write
\begin{align*}
\sqrt{1+x^2}&= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}\\
&\qquad +\cdots+\frac{1/2(1/2-1)\cdots(1/2-n+1)x^{2n}}{n!}+\cdots\\
&= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}\\
&\qquad +\cdots+\color{blue}{\binom{\frac{1}{2}}{n}}x^{2n}+\cdots\\
&=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n}x^{2n}
\end{align*}
The formula (1) can be found for instance as formula (5.1) in Concrete Mathematics by R. L. Graham, D. Knuth and O. Patashnik.
| {
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"url": "https://math.stackexchange.com/questions/4147520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation $\tan x = \tan 2x \tan 4x \tan 8x$ In the question we have the equality
$$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$
which is equivalent to
$$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$
This means that the equation
$$\tan x = \tan 2x \tan 4x \tan 8x$$
has the solution $x =6^{\circ} = \frac{\pi}{30}$. How to find all the solution of this equation?
| $$\tan x=\tan 2x\tan 4x\tan 8x$$
$$\frac{\sin x}{\cos x}=\frac{2 \sin x \cos x}{\cos 2x}\left(\frac{2\sin 4x \sin 8x}{2 \cos 4x \cos 8x}\right) $$
If $\sin x\neq0$ then,
$$\frac{\cos 2x}{2\cos^2 x}=\frac{\cos 4x - \cos 12x}{\cos 4x + \cos 12x}$$
Now,using componendo-dividendo,
$$\frac{\cos 2x+ 2\cos^2x}{2\cos^2 x-\cos 2x}=\frac{\cos 4x}{\cos 12x}$$
$$ 4 \cos^2 x -1 =\frac{1}{4\cos ^2 4x-3}$$
$$(2\cos 2x+1)(4(2\cos^2 2x -1)^2-3)=1$$
These are the solutions of above equation :wolframalpha
can someone solve the last equation by hand and edit the answer$?$
also,note that I cancelled some terms like $\cos 4x$,because $\cos 4x=0$ is not in the domain of the equation
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve it to get the complete answer?
| Another way to tackle this problem:
\begin{align}\frac 1{x+1} - \frac 1{x}=-\frac 1{x(x+1)} &\le \frac 1{x-1} - \frac 1{x-2}= -\frac 1{(x-2)(x-1)}\\
\iff \frac{1}{(x-2)(x-1)}&\le \frac{1}{x(x+1)}
\end{align}
Now observe that:
*
*outside of $\,(-1,0)\cup(1,2)$, both denominators are positive, so
$$\frac{1}{(x-2)(x-1)}\le \frac{1}{x(x+1)}\iff x(x+1)\le(x-2)(x-1)\iff 4x\le 2.$$
*In $(-1,0)$ and in $(1,2)$, compare the signs of the denominators.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing I need to show that the sequence $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing.
I tried showing $a_{n+1}\le a_n$ but it was too messy.
I did manage to do it by showing that $f'(x)$ will be negative at some point by taking the limit of the nominator of the derivative to infinity, but this method was pretty exhausting. I wonder if there a simpler method I just didn't see, and i'd be happy to have it in my "toolbox".
thanks!
| $$a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}} \implies a_n^2=\frac {n^{2}+2021n+420}{n^{3}+2022n+420 }$$
Long division or Taylor series
$$a_n^2 =\frac{1}{n}+\frac{2021}{n^2}-\frac{1602}{n^3}+O\left(\frac{1}{n^4}\right)$$ Doing the same
$$a_{n+1}^2=\frac{1}{n}+\frac{2020}{n^2}-\frac{5643}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\frac{a_{n+1}^2 } {a_n^2}=\frac{\frac{1}{n}+\frac{2020}{n^2}-\frac{5643}{n^3}+O\left(\frac{1}{n^4}\right) } {\frac{1}{n}+\frac{2021}{n^2}-\frac{1602}{n^3}+O\left(\frac{1}{n^4}\right) }$$ Long division again
$$\frac{a_{n+1}^2 } {a_n^2}=1-\frac{1}{n}-\frac{2020}{n^2}+O\left(\frac{1}{n^3}\right)= 1-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{a_{n+1} } {a_n}=1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Lagrange multiplier does not find global minimum I want to minimize the function
$$f(r_1,r_2,r_3)=\frac{1}{6}(r_1^2+r_2^2+r_3^2)-(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})$$
subject to the conditions:
$$g(r_1,r_2,r_3)=\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})=0$$
$$r_1,r_2,r_3\geq0$$
The first thing I notice is that the inequality constraints are not active. So I set
$$\nabla f(r_1,r_2,r_3)=\lambda\nabla g(r_1,r_2,r_3)$$
which gives me $$\lambda=\frac{r_i^5+9}{2r_i^5+27}$$ for each $i=1,2,3$. Since $\lambda$ is the same for each equation, I start to inspect the new function $h(r)=\frac{r^5+9}{2r^5+27}$, which is strictly increasing on $(0,\infty)$. So in order to minimize $f$, $r_i$'s must be the same. Then I plug this into the equality constraint and find $f=(\frac{3}{2})^\frac{2}{5}3^\frac{-3}{5}$
However, this does not give me the global minimum, which is when $r_1+r_2=r_3$ and $r_1=r_2$. But in that case, $\lambda$ is not a constant for each $i$. I can prove that $f$ attains a global minimum on the set $\{
g=0,r_1,r_2,r_3>0\}$. What did I do wrong here?
| Calling
$$
L(r_k,\lambda) = \frac{1}{6}(r_1^2+r_2^2+r_3^2)-\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)-\lambda\left(\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)\right)
$$
the stationary points for this lagrangian are the solutions for
$$
0=\nabla L = \cases{9-27\lambda-(2\lambda-1)r_k^5, \ \ \{k=1,2,3\}\\
\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)
}
$$
now substituting the found $r_k^*(\lambda)$ into the restriction we get $\lambda^* = \frac 25$ and then $r_k = 3^{\frac 25}$ so apparently we have only a stationary point. This point seems to be a global minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show power series is solution to differential equation I am currently studying for my analysis 2 course, and I've run into a couple of questions from old exams regarding power series as solutions for differential equations which I have a trouble completing.
We have a specific power series i.e.
$$f(x)=\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}x^{4n+3}$$
And we have to show this solves the differential equation,
$$x^2y''-3xy'+(3-4x^4)y=0$$
I have done some work to the left hand site, and gotten it to
$$\frac{-2x^{11}}{3}-7x^7+3x^3+\sum_{n=2}^{\infty}((\frac{16n^2+8n-4x^4}{(2n+1)!})x^{4n+3})$$
However I dont know how to continue from here in showing that the equation is equal to $0$. Of course we have the trivial solution for $x=0$, however how would I proceed for $x\neq0$?
| I will get you started. The goal is to get an expression
$$
x^2y''-3xy'+(3-4x^2)y=\sum_{n=0}^\infty a_n x^n
$$
where $a_n$ is a function of $n$ alone. You sort of got that in your last equation, but you have $x^4$ in your expression for the coefficient of $x^{4n+3}$. If all goes well, you should have $a_n=0$ for all $n$. It helps to work piece by piece:
\begin{align}
x^2y''
&=\sum_n \frac{(4n+3)(4n+2)}{(2n+1)!} x^{4n+3}
\\
-3xy' &=-3\sum_n \frac{(4n+3)}{(2n+1)!} x^{4n+3}
\\
(3-4x^4)y
&=3\sum_n\frac1{(2n+1)!}x^{4n+3}-4\sum_{n=0}^\infty\frac1{(2n+1)!} x^{4n+7} \\
&=3\sum_n\frac1{(2n+1)!}x^{4n+3}-4\sum_{n=1}^\infty\frac1{(2n-1)!} x^{\color{blue}{4n+3}}
\end{align}
Notice how I re-indexed the last series so that everything was a coefficient times $x^{4n+3}$. This lets you combine everything together, except for the $n=0$ of the first three series, which will have to be pulled out and dealt with separately. The result is
\begin{align}
&x^2y''-3xy'+(3-4x^2)y=\\
&\big(\frac{3\cdot 2}{1!}-3\cdot \frac{3}{1!}+3\frac{1}{1!}\big)x^{4\cdot 0+3}+
\\&\sum_{n=1}^\infty \underbrace{\Big(\frac{(4n+3)(4n+2)}{(2n+1)!}-3\cdot \frac{(4n+3)}{(2n+1)!}+3\cdot \frac1{(2n+1)!}-4\frac1{(2n-1)!}\Big)}_{a_n}x^{4n+3}
\end{align}
Now you have a big complicated expression for $a_n$ you need to show is zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy$. Q: Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \, dx\,dy$.
I imagined that $x$ should varie between $\dfrac{1}{9}$ and $\sqrt{\dfrac{2}{3}}$ and $y$ varie between $x^2$ and $\sqrt x$, then we get:
$$\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy = \int_{\frac{1}{9}}^{\sqrt{\frac{2}{3}}} \int_{x^2}^{\sqrt x} f(x,y) \, dy \, dx$$
But that is not the answer cuz i tested it with some $f(x,y)$ and got different values on the integral.
Can somebody help me? Thanks!
| The 'x' limit $y^2$ means $x=y^2$, so we have $y=\sqrt{x}$
The 'x' limit $\sqrt{y}$ means $x=\sqrt{y}$, so we have $y=x^2$
Notice you have exchanged lower and upper 'x' limits
For the 'y' limit $y=\frac{1}{3}$ we must get the $x=y^2 = \frac{1}{9}$ function because $\left(\frac{1}{3} \right)^2 < \sqrt{\frac{1}{3}}$.
Same reasoning for $y=\frac{2}{3}$ we use the other, $x=\sqrt{\frac{2}{3}}$
Thus, we arrive to
$$\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) dxdy=
\int_{\frac{1}{9}}^{\sqrt{\frac{2}{3}}} \int_{\sqrt x}^{x^2} f(x,y) dydx$$
Being that said, draw both functions (square and root), because perhaps you have to decompose the region into three pieces.
| {
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"timestamp": "2023-03-29T00:00:00",
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if $u = \arccos \left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)$ then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ Edit: this is a repost
if possible i need some sort of shorcut, 80 questions, 90 min to solve these type of questions i have only 1 min time solt for each question
Context: i've looked at PYQ's and this year mock tests and found out that this question was one of the most repeated ones and i'm having an exam next week and my teachers are not available
The answer i got
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{-x\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{x}}\right)} {\sin^2(u)(\sqrt{x}+\sqrt{y})^2}- \frac{y\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{y}}\right) } {\sin^2(u)\left(\sqrt{x}+\sqrt{y}\right)^2}$
The answer im getting is nowhere near the options that have been given
Options
1.$\frac{-1}{2}\sin(u)$
2. $\frac{-1}{2}\cot(u)$
3. $\frac{-1}{2}\tan(u)$
4. $\frac{-1}{2}\cos(u)$
| I got :
$$\tag{1} \boxed{ x \frac{\partial u}{\partial x} = -~ \frac{2x \sqrt{y} + x \sqrt{x} -y \sqrt{x}}{2~\sin(u) ~(\sqrt{x} + \sqrt{y})^2}} $$
$$\tag{2} \boxed{ y \frac{\partial u}{\partial y} = -~ \frac{2y \sqrt{x} + y\sqrt{y} -x \sqrt{y}}{ 2~\sin(u)~ (\sqrt{x} + \sqrt{y})^2}} $$
$$x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y} = - \frac{1}{2\sin(u)}~ \frac{\sqrt{y} (~ 2x+y-x)+\sqrt{x}(~x-y+2y)}{(\sqrt{x}+\sqrt{y})^2}$$
$$x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y} = - \frac{1}{2\sin(u)}~ \frac{ x+y}{(\sqrt{x}+\sqrt{y})} \tag{3}$$
From the equation : $$ u = \arccos( \frac{x+y}{(\sqrt{x}+\sqrt{y})})$$
$$ \cos(u) = \frac{x+y}{\sqrt{x}+\sqrt{y}}$$
replace in $(3)$ :
$$ \tag{4}\boxed{- \frac{\cos(u)}{2\sin(u)} = - \frac{\cot(u)}{2}}$$
Do confirm yourself the answer and partial derivatives
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to find $\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du $ How to find
$$\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du ?
$$
Where should I even start with this one? Can I pull out $\frac{1}{\sin(x^3)}$ outside the integral as $x$ is an constant and we're finding integral with respect to to $u$?, and then apply Leibniz integral rule?
Where should I even get started with this one its kinda confusing.
| $$\lim_{x \to 0} \left(\frac{1}{\sin(x^3)} \int_0^x \sin^2 u \ du \right)$$
$$= \lim_{x \to 0} \left( \frac{1}{\sin(x^3)} \left[\frac{1}{2}u - \frac{1}{4} \sin2u \right]_0^x \right)$$
and when $x^3$ is small, $\sin x \approx x - \frac{x^3}{3!}$ and $\sin x^3 \approx x^3$, so:
$$= \lim_{x \to 0} \left(\frac{1}{x^3} \left[\frac{1}{2}x - \frac{1}{4} \sin(2x) \right] \right)$$
$$= \lim_{x \to 0} \left(\frac{1}{x^3} \left[\frac{1}{2}x - \frac{1}{4} (2x - 8x^3/6) \right] \right)$$
$$= \lim_{x \to 0} \left(\frac{1}{x^3} (2x^3/6) \right) =\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question from pathfinder for Olympiad mathematics 2 If $p$, $q$, $r$ are the real roots of equation $x^3-6x^2+3x+1=0$, determine the possible value of
$p^2q+q^2r+r^2p$.
My Attempt:
$p+q+r=6 (1)$
$pq+qr+pr=3 (2)$
$pqr=-1 (3)$
Multiplying (1) and (2) and substituting (3) in it...
I came close but still unable to find solution...
Please Help
| Since $v_1=p^2q+q^2r+r^2p$ is not a symmetric polynomial, there might be multiple values. Another possible value is $v_2=p^2r+q^2p+r^2q.$
Now $v_1+v_2$ and $v_1v_2$ are symmetric. So we can express those values in terms of $p+q+r,$ $pq+pr+qr,$ and $pqr.$
Then solve for the two values, $v_1,v_2.$
I’ll do the harder case. Writing $s_1=p+q+r,$ $s_2=pq+pr+qr,$ and $s_3=pqr,$ we have:
$$\begin{align}v_1v_2=&(p^3+q^3+r^3)s_3+3s_3^2\\&+\left((pq)^3+(qr)^3+(rs)^3\right)
\end{align}$$
And $$\begin{align}p^3+q^3+r^3=&s_1^3-3s_1s_2+3s_3\end{align}$$
So that leaves us the last term, $(pq)^3+(pr)^3+(qr)^3.$ One clever way is to write it as:
$$(pq)^3+(pr)^3+(qr)^3=s_3^3\left(\frac1{p^3}+\frac1{q^3}+\frac1{r^3}\right)$$
then reuse the formula for the sum of three cubes we used previously, with $s_1’=s_2/s_3,$ $s_2’=s_1/s_3,$ $s_3’=1/s_3.$
So we have:
$$(pq)^3+(pr)^3+(qr)^3=s_2^3-3s_1s_2s_3+3s_3^2$$
So:
$$\begin{align}v_1v_2&=\left(s_1^3-3s_1s_2+3s_3\right)s_3+3s_3^2+(s_2^3-3s_1s_2s_3+3s_3^2)\\
&=s_1^3s_3-6s_1s_2s_3+s_2^3+9s_3^2\end{align}$$
Since $s_1=6,s_2=3,s_3=-1,$ this means: $$v_1v_2=-6^3+6\cdot 6\cdot 3 +3^3+9=-72.$$
I also get $v_1+v_2=21.$ So:
$$v_i=\frac{21\pm \sqrt{729}}{2}=-3\text{ or }24.$$
A numerical check using the roots provided by Wolfram Alpha indicate this is probably right.
| {
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"timestamp": "2023-03-29T00:00:00",
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What values of $n$ is $A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ an integer?
Suppose $a_i\in \mathbb{N} $ such that $a_i\ne a_j$ and $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ for $i \ne j$ $($take $a_{n+1}=a_1)$ and $\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$.
Then for what values of $n\ge 2$ can $A$ be an integer?
I found that if $a_i=(n-1)^i$ then $A$ is a integer but I realised that this violates the condition $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$. It was easy to prove that $A$ is not an integer for $n=2$, but I was stuck for the cases where for $n>2$. I then I plugged in some values for $a_i$ but for $n>2$ but I found no solution.
Intuitively I think there is no solution but I am unable to prove it. Any help would be appreciated.
Edit-1: As pointed out by @WillJagy it is possible for $n=3,4,5$. So can we actually prove it is true for $n>6$ or are there some $n>6$ for which it isn't true?
Edit-2: I was able to find a solution for $n=5$ from a solution for $n=4$.
$$~~~~~~~~~~\frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6$$
$$\implies \frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 5$$
$$\implies \frac{6}{2} +\frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 8$$
But I could not find a solution for $n=6$ from a solution for $n=5$ because when I rewrite $\frac{4}{3} =1+\frac{1}{3}$, the $1$ in the numerator of $\frac{1}{3}$ is used by another $a_i$. I don't know whether we can prove by induction. Any suggestions is welcome.
| $$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{15}{6}+\frac{24}{15}+\frac{10}{24}+\frac{2}{10}=10$$
$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{30}{7}+\frac{14}{30}+\frac{24}{14}+\frac{2}{24}=13$$
$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{36}{9}+\frac{15}{36}+\frac{42}{15}+\frac{2}{42}=15$$
$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{10}{ 9}+\frac {28}{ 10}+\frac {27}{ 28}+\frac {36}{ 27}+\frac {2}{ 36}=14$$
| {
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Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the integer just above 7.3 . It is 8. Then , Q wants you to prove that 8 is divisible by $2^{n+1}$.
\begin{equation}
\begin{array}{l}
(\sqrt{3}+1)^{2 n}=(4+2 \sqrt{3})^{n}=2^{n}(2+\sqrt{3})^{n}\\
=2^{n}(2+\sqrt{3})^{n}=2^{n}\left[^n C _{0}2^{n}+^n C_{1} 2^{n-1} \sqrt{3}+^n C_{{2}} 2^{n-2} \sqrt{3}^{2}+\right.\\
\begin{array}{l}
(\sqrt{3}-1)^{2 n}=(4-2 \sqrt{3})^{n}=2^{n}(2-\sqrt{3})^{n} \\
=2^{n}(2-\sqrt{3})^{n}=2^{n}\left[{ }^{n} c_{0} 2^{n}-n_{C}, 2^{n-1} \sqrt{3}+{ }^{n} c_{2} 2^{n-1} \sqrt{1}^{2} \ldots\right.
\end{array}\\
I+f+f)=2^{n}\left[2(\text { Integer) }]=2^{n+1}\right. \text { . Integer }\\
I+1=2^{n+1} \text { . Integer }
\end{array}
\end{equation}
In the image is the way this question is solved.
My Q from this method of solving is that if we notice at the end , we somehow got $ 2^{n+1}$. If the Q has taken some other value like $3^{n+3}$ or Something else. Then , it was not possible to prove this question.
What is another method to prove this Q or can you help me justify that the above method can be used for all kinds of Q.
Thank you.
| Part of the problem is that the image was a bit obscured and so the transcription was a bit off. Here is a more accurate transcription with tags added for reference:
$$\require{cancel}
\left(\sqrt3+1\right)^{2n}=\left(4+2\sqrt3\right)^n=2^n\left(2+\sqrt3\right)^n\tag1
$$
$$
\!\!\!\!I+f=2^n\left(2+\sqrt3\right)^n=2^n\left[{}^nC_02^n+\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2+\dots\right]\tag2
$$
$$
\left(\sqrt3-1\right)^{2n}=\left(4-2\sqrt3\right)^n=2^n\left(2-\sqrt3\right)^n\tag3
$$
$$
f'=2^n\left(2-\sqrt3\right)^n=2^n\left[{}^nC_02^n-\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2-\dots\right]\tag4
$$
$$
I+\bbox[3px,border:1px solid black]{f+f'}=2^n[2(\text{Integer})]=2^{n+1}\cdot\text{Integer}\tag5
$$
$$
I+1=2^{n+1}\cdot\text{Integer}\quad\checkmark\tag6
$$
$(1)$ is just simple algebraic manipulation
$(2)$ is expanding via the Binomial Theorem ($I$ and $f$ are the integer and fraction parts)
$(3)$ is $(1)$ with the substitution $\sqrt3\mapsto-\sqrt3$; note that $\left(\sqrt3-1\right)^{2n}=\left(-\sqrt3+1\right)^{2n}$
$(4)$ is expanding via the Binomial Theorem
$(5)$ adding $(2)$ and $(4)$ cancels all of the terms with $\sqrt3$ to an odd power
$\phantom{\text{(5)}}$ and doubles all of the terms with $\sqrt3$ to an even power (these terms are integers)
$(6)$ since $0\lt\left(4-2\sqrt3\right)\lt1$, $f'\in(0,1)$ and by definition, $f\in[0,1)$
$\phantom{\text{(6)}}$ since $I+f+f'\in\mathbb{Z}$, we must have $f+f'=1$
Here is, in my experience, a more usual proof of this.
$\left(1\pm\sqrt3\right)^2=4\pm2\sqrt3$ are roots of $x^2-8x+4$. Therefore, the solution to the linear recurrence equation
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[3pt]
&=4(2u_{n-1}-u_{n-2})\tag7
\end{align}
$$
is
$$
\begin{align}
u_n
&=a\left(1+\sqrt3\right)^{2n}+b\left(1-\sqrt3\right)^{2n}\\[3pt]
&=a\left(4+2\sqrt3\right)^n+b\left(4-2\sqrt3\right)^n\tag8
\end{align}
$$
In particular, the sequence with $a=b=1$ starts out with
$$
u_0=2,u_1=8,u_2=56\tag9
$$
Note that $\left.2^{n+1}\,\middle|\,u_n\right.$ for $n=1$ and $n=2$. Then $(7)$ guarantees that
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[6pt]
&=8\cdot2^n\frac{u_{n-1}}{2^n}-4\cdot2^{n-1}\frac{u_{n-2}}{2^{n-1}}\\
&=2^{n+1}\left(4\frac{u_{n-1}}{2^n}-\frac{u_{n-2}}{2^{n-1}}\right)\tag{10}
\end{align}
$$
By induction, $(9)$ and $(10)$ show that $\left.2^{n+1}\,\middle|\,u_n\right.$ for all $n\ge1$.
Since $u_n\in\mathbb{Z}$,
$$
\begin{align}
u_n
&=\left(1+\sqrt3\right)^{2n}+\overbrace{\left(1-\sqrt3\right)^{2n}}^{\text{in }(0,1)}\\[3pt]
&=\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\tag{11}
\end{align}
$$
Therefore, $(10)$ and $(11)$ show that
$$
\left.2^{n+1}\,\middle|\,\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\right.\tag{12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157841",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
How to compute this determinant
I think adding some rows or column to each other will get a triangular matrix which is easy to compute but i cant see how
| \begin{align}
\Delta_n&= \left|\begin{matrix}
1 & 2 & 3 & \cdots & n-1 & n\\
-1 & x & 0 & \cdots & 0 & 0 \\
0 & -1 & x & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & 0 \\
0 & 0 & 0 & \cdots & -1 & x
\end{matrix}\right|
\tag{1}\label{1}
\end{align}
By expanding the determinant $\Delta_n$
with respect to the $n$-th column,
we get a recurrence relation
\begin{align}
\Delta_n&=x\cdot\Delta_{n-1}+(-1)^{n-1} n\cdot(-1)^{n-1}
=x\cdot\Delta_{n-1}+n
\end{align}
with initial condition $\Delta_1=1$, which has a solution
in a compact form
\begin{align}
\Delta_n&=\frac{x(x^n-1)-n(x-1)}{(x-1)^2}
\quad \text{for }\quad x\ne 1
\tag{2}\label{2}
.
\end{align}
In the limiting case for $x=1$ we have
\begin{align}
\Delta_n&=\lim_{x=1}
\frac{x(x^n-1)-n(x-1)}{(x-1)^2}
=\tfrac12n(n+1)
\tag{3}\label{3}
.
\end{align}
Expression \eqref{2} can be also expressed in an expanded form
as follows
\begin{align}
\Delta_n
&=\frac{(x\sum_{k=0}^{n-1} x^k)-n}{x-1}
\tag{4}\label{4}
\\
&=
\frac1{x-1}\sum_{k=1}^n (x^k-1)
\tag{5}\label{5}
\\
&=
\sum_{k=1}^n \sum_{j=0}^{k-1} x^j
=
\sum_{k=1}^n k x^{n-k}
\tag{6}\label{6}
.
\end{align}
| {
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"answer_count": 3,
"answer_id": 2
} |
How many possible Jordan decompasitions for an Endomorphism? Let V be an vector space and $\phi \in End_\mathbb{C}(V)$ a linear map with the characteristic polynomial $P_\phi(X) = (x-2)^3(x-5)^2$. How many possible Jordan normal form are there for $\phi$?
| Hint: You should also provide the minimal polynomial that could be one of
*
*$(x-2)(x-5)$,
*$(x-2)(x-5)^2$,
*$(x-2)^2(x-5)$,
*$(x-2)^2(x-5)^2$,
*$(x-2)^3(x-5)$,
*$(x-2)^3(x-5)^2$.
With that in mind the posibles respectively are
$\left[\begin{array}{ccccc}
2&0&0&0&0\\
0&2&0&0&0\\
0&0&2&0&0\\
0&0&0&5&0\\
0&0&0&0&5\\
\end{array}
\right]$
,
$\left[\begin{array}{ccccc}
2&0&0&0&0\\
0&2&0&0&0\\
0&0&2&0&0\\
0&0&0&5&1\\
0&0&0&0&5\\
\end{array}
\right]$
,
$\left[\begin{array}{ccccc}
2&1&0&0&0\\
0&2&0&0&0\\
0&0&2&0&0\\
0&0&0&5&0\\
0&0&0&0&5\\
\end{array}
\right]$
,
$\left[\begin{array}{ccccc}
2&1&0&0&0\\
0&2&0&0&0\\
0&0&2&0&0\\
0&0&0&5&1\\
0&0&0&0&5\\
\end{array}
\right]$
,
$\left[\begin{array}{ccccc}
2&1&0&0&0\\
0&2&1&0&0\\
0&0&2&0&0\\
0&0&0&5&0\\
0&0&0&0&5\\
\end{array}
\right]$
,
$\left[\begin{array}{ccccc}
2&1&0&0&0\\
0&2&1&0&0\\
0&0&2&0&0\\
0&0&0&5&1\\
0&0&0&0&5\\
\end{array}
\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Combinatorial proof of ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$
Prove that ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$
I found the algebraic proof of the above statement.
So we have to show that $$ \frac{(n+1)(n)(n-1)}{3\times 2}-\frac{(n-1)(n-2)(n-3)}{3\times 2} \stackrel{?}{=} (n-1)(n-1)$$ $$\implies \frac{(n+1)(n)}{3\times 2}-\frac{(n-2)(n-3)}{3\times 2} \stackrel{?}{=} (n-1) $$$$ \implies (n+1)n-(n-2)(n-3)\stackrel{?}{=}3 \times 2
(n-1), $$ which is true and we are done!
But I couldn't get the combinatorial proof. Any hints?
| I would use Pascal's identity to get
$$\binom{n+1}{3}-\binom{n-1}{3} = \left[\binom{n+1}{3} - \binom{n}{3}\right] + \left[\binom{n}{3}-\binom{n-1}{3}\right] = \binom{n}{2} + \binom{n-1}{2}.$$
This can be solved algebraically:
$$\binom{n}{2} + \binom{n-1}{2} = \frac{n(n-1)}{2} + \frac{(n-1)(n-2)}{2} = \frac{(n-1)[(n-2)+n]}{2} = (n-1)^{2}.$$
I'm shocked I figured this out so fast. Sometimes you just see the solutions when you read the problem and sometimes you get stuck on a wrong path. It's just the story of math :)
| {
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Show that $2 ^{n(n+1)} \mid 32 · φ(2^{2^n} − 1) $
Show that for every $n \in \mathbb{N}$
$$2 ^{n(n+1)}\mid 32 · \varphi(2^{2^n}− 1)$$
where $\varphi$ denotes the Euler-Phi function.
So the right hand side can be broken down into:
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1)\cdots(2^1 + 1)(2-1)$$
Since $φ$ is multiplicative, all that's left is to figure out the value of $\varphi(2^{2^k}+1)$ for $k$ from $1$ to $n-1$. This is where I've gotten stuck.
| We could use your idea of breaking $2^{2^n} - 1$ up into
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1) \ldots (2^1 + 1)(2-1)$$
Next, we could determine appropriate lower limits of each $\varphi\left(2^{2^k} + 1\right)$ and then add them together.
However, another similar (but I think a bit easier) way, with your idea being implicitly used, to prove your statement of
$$2^{n(n+1)}\, \mid \, 32 · \varphi(2^{2^n}− 1) \tag{1}\label{eq1A}$$
is to use induction. First, using the $p$-adic order function, \eqref{eq1A} is equivalent to proving
$$n(n + 1) \le 5 + \operatorname{ord}_2\left(\varphi(2^{2^n}− 1)\right) \tag{2}\label{eq2A}$$
Start with $3$ base cases. First, with $n = 1$, we have $2 \le 5 + 1 = 6$. Next, with $n = 2$, we get $6 \le 5 + 1 + 2 = 8$. Finally, with $n = 3$, we have $12 \le 5 + 1 + 2 + 4 = 12$.
Next, assume for some $k \ge 3$ that \eqref{eq2A} is true for $n = k$. To prove \eqref{eq2A} is true for $n = k + 1$, it's sufficient to show the value of the right side increases by at least as much as the left side. To do this, first note the factoring of a difference of squares gives
$$2^{2^{k+1}} - 1 = \left(2^{2^{k}} - 1\right)\left(2^{2^{k}} + 1\right) \tag{3}\label{eq3A}$$
Since $2^{2^{k}} - 1$ and $2^{2^{k}} + 1$ differ by just $2$ and are both odd, they are also relatively prime to each other. Thus, since $\varphi$ is multiplicative, as you noted, then
$$\begin{equation}\begin{aligned}
& \varphi\left(2^{2^{k+1}} - 1\right) = \varphi\left(2^{2^{k}} - 1\right)\varphi\left(2^{2^{k}} + 1\right) \implies \\
& \operatorname{ord}_2\left(\varphi\left(2^{2^{k+1}} - 1\right)\right) = \operatorname{ord}_2\left(\varphi\left(2^{2^{k}} - 1\right)\right) + \operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This means it's sufficient to prove
$$\operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right) \ge (k+1)(k+2) - k(k+1) = 2(k+1) \tag{5}\label{eq5A}$$
If $2^{2^{k}} + 1$ is prime, then $\varphi\left(2^{2^{k}} + 1\right) = 2^{2^{k}}$. It's easy to prove that $2^{k} \ge 2(k+1) \implies 2^{k-1} \ge k + 1$ for $k \ge 3$ (e.g., for $k = 3$, we get an equality, and for all larger $k$ the left side doubles but the right side increases by just $1$).
If $2^{2^{k}} + 1$ is not prime, then it must have $2$ or more prime factors. Similar to Mindlack's question comment, the Other theorems about Fermat numbers section of Wikipedia's article states and outlines a proof of a theorem by Édouard Lucas, i.e.,
Any prime divisor $p$ of $F_{n}=2^{2^{n}}+1$ is of the form $k2^{n+2}+1$ whenever $n \gt 1$
Note the How to prove that if a prime divides a Fermat Number then $p=k\cdot 2^{n+2}+1$? post also gives a proof of this result. Using this, then for each such prime $p$ factor of $2^{2^{k}} + 1$, we have
$$\operatorname{ord}_2(\varphi(p)) \ge k + 2 \tag{6}\label{eq6A}$$
Thus, for at least $2$ such prime factors, we get
$$\operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right) \ge 2(k + 2) \gt 2(k + 1) \tag{7}\label{eq7A}$$
This shows that, in either case, \eqref{eq5A} is true, which means that \eqref{eq2A} is true for $n = k + 1$. This proves the inductive step and, therefore, that \eqref{eq1A} is true for all $n \in \mathbb{N}$.
| {
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Calculate $\displaystyle \iiint_V x^2 \,dx\, dy\, dz$ where $V=\{(x, y, z)|z^2\le \frac{x^2}{4}+\frac{y^2}{9}\le 2z\}$ Calculate $\displaystyle \iiint_V x^2 \,dx\, dy \,dz$, where $V=\{(x, y, z)\mid z^2\le \frac{x^2} {4} + \frac{y^2}{9}\le 2z\}$.
I made the substitution $x=2r\cos\theta$, $y=3r\sin \theta$ and $z=z$. From here I got that $z^2\le r^2\le 2z$, which means that $z\le r \le \sqrt{2z}$ and also implies that $0\le z \le 2$. Furthermore, $0\le \theta \le 2\pi$ (this may also be seen if we draw a picture), so
$$\iiint_V x^2 \, dx \, dy \, dz = \int_0^2 \left(\int_0^{2\pi}\left(\int_z^{\sqrt{2z}}24r^3 \cos^2 \theta \,dr\right)d\theta\right)dz$$and this is not hard to compute. Is my substitution correct? Did I get this one right?
| Your working is absolutely correct. The region is bound below by the paraboloid and above by the cone and we could have alternatively set up the integral wrt $z$ first. The surfaces intersect at $z = 2$ and at intersection, $r \leq 2$.
We have $z^2 \leq r^2 \leq 2z \implies \dfrac{r^2}{2} \leq z \leq r$. So the integral becomes,
$\displaystyle \int_0^{2\pi} \int_0^2 \int_{r^2/2}^r 24r^3 \cos^2 \theta \ dz \ dr \ d\theta = \dfrac{128 \pi}{5}$
| {
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If $x+2y+3z=10$ find the maximum value of $x^2y^2z^6$. If $x+2y+3z=10$ where x,y,z are postive real numbers, find the maximum value of $x^2y^2z^6$.
I started by trying to find some integer solutions. The first thing I found was $(x,y,z)=(1,3,1)$. But the value of $x^2y^2z^6$ was way too small (it was a multiple choice question in the paper I was solving).
The next thing I tried was setting $x=y=z$ because I thought the product of numbers would be highest if they were equal. And I got $x=y=z=\frac{5}{3}$. And so I got $x^2y^2z^6 = (\frac{5}{3})^{10}$
But this answer seems very weird and even my logic is not right, I merely went by intuition.
Please help me solve this.
| I presume $x,y,z$ are non-negative integers. Otherwise, if they are allowed to be negative, we may choose their magnitude to be arbitrarily large.
Write : \begin{align*}
10 &= x + 2y + 3z\\
&= \frac{x}{2} + \frac{x}{2} + y + y + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2}\\
&\geqslant 10\cdot\sqrt[10]{\frac{x}{2} \cdot \frac{x}{2} \cdot y \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2}}\\
&= 10 \cdot \sqrt[10]{\frac{x^2y^2z^6}{2^8}}\\
\end{align*}
We have used simple AM-GM in the first inequality step. Thus, $x^2y^2z^6 \leqslant 2^8 = 256$ , i.e. its maximum possible value is $256$.
The maximum value is attained when : $$\frac{x}{2}=y=\frac{z}{2}\quad\implies\quad x=2y=z\quad\implies\quad (x,y,z) = (2,1,2)$$
| {
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what is the residue of $\sin(z^2)/z^7$ what is the residue of $$\frac{\sin(z^2)}{z^7} \text{ at } z=0.$$ I'm finding trouble at the limit part at the end.
I am using the formula:
$$\operatorname{res}(f,z_0)=\lim_{z\to z_0}\frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} ((z-z_0)^n f(z))$$
| Recall that
$$\sin(z)= \sum_{n=0}^\infty \frac{(-1)^{n} z^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$$
Which also means that
$$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots$$
And thus
\begin{align}
\frac{1}{z^7}\sin{(z^2)}&=\frac{1}{z^7}\left( z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots\right)\\
&=\frac{1}{z^5}-\frac{1}{3!}\frac{1}{z}+\cdots
\end{align}
And so in the Laurent series expansion, the coefficient of the $1/z$ term is the residue at $z=0$. And thus
$$\mathrm{Res}_{z=0}\left( \frac{\sin(z^2)}{z^7}\right)=-\frac{1}{3!}=-\frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177996",
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"source": "stackexchange",
"question_score": "1",
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How do I solve this ominous integral? Let $ n\ge 1 $ be a positive integer. How do I prove the generalization: $$ \int_0^1\frac{\arctan(x)\log^{2n}(x)}{1+x} \, dx =\frac{\pi}{4}\left(1-2^{-2 n}\right) \zeta(2 n+1)(2 n)!+\frac{1}{2} \beta(2 n+2)(2 n) !-\frac{\pi}{16} \lim _{s \rightarrow 0}\left(\frac{d^{2 n}}{d s^{2 n}}\left(\csc \left(\frac{\pi s}{2}\right)\left(\psi\left(\frac{3}{4}-\frac{s}{4}\right)-\psi\left(\frac{1}{4}-\frac{s}{4}\right)\right)\right.\right.$$
The integral was offered to me by my good friend and it looks very difficult, I managed to solve only for $n=1$ and $n=2$.
$$\int_0^1 \frac{\arctan(x)\log^2(x)}{1+x} \, dx=\frac{21}{64}\pi\zeta(3)-\frac{\pi^3}{32}\log(2)-\frac{\pi^2}{24}G;$$
and
$$\int_0^1 \frac{\arctan(x)\log^4(x)}{1+x} \, dx=\frac{1395}{256}\pi\zeta(5)-\frac{9}{128}\pi^3\zeta(3)-\frac{7}{480}\pi^4G-\frac{5}{128}\pi^5\log(2)+\frac{\pi^6}{192}-\frac{\pi^2}{1536}\psi^{(3)}\left( \frac{1}{4} \right).$$
where $G$ represents the Catalan’s constant.
| The general "closed-form" is:
$$\int _0^1\frac{\arctan \left(x\right)\ln ^{2n}\left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{4}\left(1-2^{-2n}\right)\zeta \left(2n+1\right)\left(2n\right)!+\frac{1}{2}\beta \left(2n+2\right)\left(2n\right)!-\frac{\pi }{16}$$
$$\lim _{s\to 0}\left(\frac{d^{2n}}{ds^{2n}}\left(\csc \left(\frac{\pi s}{2}\right)\left(\psi \left(\frac{3-s}{4}\right)-\psi \left(\frac{1-s}{4}\right)\right)+\sec \left(\frac{\pi s}{2}\right)\left(\psi \left(1-\frac{s}{4}\right)-\psi \left(\frac{1}{2}-\frac{s}{4}\right)\right)-2\pi \csc \left(\pi s\right)\right)\right)$$
And you may find its evaluation in the book (Almost) Impossible Integrals, Sums, and Series through pages $140-145$ where double integration, symmetry and the following result are heavily exploited:
$$\int _0^{\infty }\frac{x^s}{\left(1+x\right)\left(1+y^2x^2\right)}\:dx=\frac{\pi }{2}\csc \left(\frac{\pi s}{2}\right)\frac{y^{-s}}{1+y^2}+\frac{\pi }{2}\sec \left(\frac{\pi s}{2}\right)\frac{y^{1-s}}{1+y^2}-\frac{\pi \csc \left(\pi s\right)}{1+y^2}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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What is $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y}$? I want to understand this multivariate limit.
WolframAlpha says $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} = 0$
But what if I take a curve $y = -x$, then the limit doesn't exist, right? Wouldn't this make the limit inexistent?
I tried to prove using sandwich theorem and definition, but didn't get anything good.
Any help would be appreciated.
| $$
\begin{aligned}
\lim\limits_{(x, y) \rightarrow (0, 0)}\frac{1-\cos(x)}{x + y} &=
\left|
\begin{array}{c}
\text{changing the variables:} \\
x = r\cos(\theta) \\
y = r\sin(\theta) \\
(r \in (0, +\infty), \text{ }\theta \in [0, 2\pi]) \\
\Downarrow \\
(x, y) \rightarrow (0, 0) \Leftrightarrow r \rightarrow 0
\end{array}
\right| =
\lim\limits_{r \rightarrow 0}\frac{1-\cos\left(r\cos(\theta)\right)}{r(\cos(\theta) + \sin(\theta))} = \\
=\lim\limits_{r \rightarrow 0}\frac{2\sin^2\left(\frac{r}{2}\cos(\theta)\right)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} &=
\left|
\begin{array}{c}
\sin^2\left(\frac{r}{2}\cos(\theta)\right) \underset{r \rightarrow 0}{\sim} \frac{r^2}{4}\cos^2(\theta)
\end{array}
\right| =
\lim\limits_{r \rightarrow 0}\frac{\frac{r^2}{2}\cos^2(\theta)}{r\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)} = \\
\frac{\cos^2(\theta)}{2\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)}\lim\limits_{r \rightarrow 0}r = 0, \forall \theta\text{ }\left(\theta \ne \frac{3\pi}{4}\right).
\end{aligned}
$$
So the limit equals to zero for all $\theta$ except $\theta = \frac{3\pi}{4}$.
Now, let us check the case when $\theta \rightarrow \pm\frac{3\pi}{4} \Leftrightarrow y \rightarrow -x$.
In this case,
$$
\lim\limits_{\theta \rightarrow \frac{3\pi}{4}}\lim\limits_{r \rightarrow 0}\left(\cdot \right) = 0 \ne \pm\infty = \lim\limits_{r \rightarrow 0}\lim\limits_{\theta \rightarrow \pm\frac{3\pi}{4}}\left(\cdot \right)
$$
This means that the limit doesn't exist.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx =\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3})$
Prove that
$$
I
=3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx
=\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3}).
$$
where, $G$ is catalan's constant
Above two Integrals are a part of a integral which I was trying to solve.
Let $I=3I_{1}-I_{2}$
Attempt:-1
$$I_{1}=\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx$$
$$\implies I_{1}=\int_{0}^{1}\frac{x}{x^2+3}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} x^{2n-1}\,\mathrm dx$$
$$\implies I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{x^2+3}\,\mathrm dx.$$
From my previous question 1 we have
$$
\int_{0}^{1}\frac{x^{2n}}{x^{2}+3}\,\mathrm dx
=(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}.
$$
Therefore,
$$
I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\bigg[(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}\bigg],
$$
$$
I_{1}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{(-1)^{2n-k}\space 3^{n-1-k}}{(2k+1)(2n-1)}-\frac{\pi}{6\sqrt{3}}\sum_{n=1}^{\infty}\frac{3^{n}}{2n-1}.
$$
Using Desmos Both of the series diverges so $I_{1}$ is of the form $\infty -\infty$ which have a finite answer. Same thing goes with $I_{2}$.
Attempt:- 2:
Try to convert one integral into another. Substitute $x=\frac{1}{x}$ in $I_{1}$, we get
$$I_{1}=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx
+3\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx $$
$$\implies I=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx -\int_{0}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx +4\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx$$
Surprisingly all three integrals diverges and convergence of $I$ is maintained by negative and positive sign.
How can I prove the original result?
Thank you for your help!
| Integrate by parts and then substitute $x=\tan \frac t2$
\begin{align}
I=& \>\int_{0}^{1}\frac{x\arctan x}{x^2+\frac13}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}dx
=\frac14 \int_0^{\frac\pi2} \ln\frac{1+\frac12 \cos t}{1-\frac12 \cos t}dt
\end{align}
Let $J(a) = \int_0^{\frac\pi2}\ln(1+\cos a\cos t)dt$. Then
$$J’(a)
=-\tan a\left( \frac\pi2-\int_0^{\frac\pi2}\frac{1}{1+\cos a\cos t}dt\right)
=a\sec a-\frac\pi2\tan a
$$
and
\begin{align}
I&= \frac14\left(J(\frac\pi3)-J(\frac{2\pi}3)\right)=-\frac14 \int^{\frac{\pi}2}_{\frac\pi3} J’(a)da - \frac14\int^{\frac{2\pi}3}_{\frac\pi2} \overset{a\to \pi -a}{J’(a)da }\\
&=\frac12 \int_{\frac\pi3}^{\frac\pi2} (\frac\pi2-a)\sec a \>da
\overset{a=\frac\pi2 - 2t}=2\int_0^{\frac\pi{12}} t\csc (2t) dt\\
&=- t\ln(\cot t)\bigg|_0^{\frac\pi{12}} + \int_{0} ^{\frac\pi{12}}
{\ln(\cot t) dt}\\
&= -\frac\pi{12} \ln(2+\sqrt3)+\frac23 G
\end{align}
where $\int^{ \frac\pi{12}}_{0}
\ln(\cot t)dt= \frac23G$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Can you check my proof? It would be really grateful.
First, we can define function $f(x)$ as following:
$$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$
$$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$
$$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$
Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0.
$$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$
Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get:
$$f(2) > f(3)$$
$$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.
I wish you a beautiful day, and stay safe.
| Your solution is fine. If you want a non-calculus approach, note that
\begin{align*}
f(x) &= (x+1)^{1/3}-x^{1/3}
\\
&= \dfrac{\left((x+1)^{1/3}-x^{1/3}\right)\left((x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}\right)}{(x+1)^{2/3}+2(x+1)^{1/3}x^{1/3}+x^{2/3}}
\\
&= \dfrac{(x+1)-x}{(x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}}
\\
&= \dfrac{1}{(x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}}
\end{align*}
is clearly decreasing since the numerator is constant and positive while the denominator is increasing and positive. Hence, $f(3) < f(2)$.
| {
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"url": "https://math.stackexchange.com/questions/4187958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{10}} = \frac{a}{b} \\
\frac{3+\sqrt{5}}{12 + 4\sqrt{5}} = \frac{a^2}{b^2} \\
\frac{3+\sqrt{5}}{4(3+\sqrt{5})} = \frac{a^2}{b^2} \\
\frac{1}{4} = \frac{a^2}{b^2}
\end{align*}
And because we get $\frac{1}{4}$ which is rational, we can conclude that the proof is right and there aren't any contradictions. Hence $\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{5}}$ is a rational number.
I guess my proof lacks something and I don't feel it's complete yet. I would appreciate any recommendations on how to improve my answer. Thanks in advance.
| $\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}\!=\!\dfrac{\sqrt{\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)^2}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac{\sqrt{\dfrac12}+\sqrt{\dfrac52}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac12.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
The easiest way to solve the following integral : $\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$ Today I passed the test to access to medicine, and I found some really cool questions on the mathematics part, anyways I just went to see the difficulty of the test, and sorry for being far from my main question :
$$\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$$
I used (obviously) the king property of integration which gives $\pi/12$ as an answer.
My question is how can we solve this using some basic properties of integration ?
Thanks in advance. and please wish me some luck for the next tests to access engineering universities.
| HINT
Make the change of variable $y = \pi/2 - x$. Then we get that
\begin{align*}
I & =
\int_{\pi/6}^{\pi/3}\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \int_{\pi/6}^{\pi/3}\frac{\cos(y)}{\cos(y) + \sin(y)}\mathrm{d}y
\end{align*}
Then add both expressions together and see what happens.
Can you take it from there?
EDIT
To begin with, notice that
\begin{align*}
\frac{\sin(x)}{\sin(x) + \cos(x)} & = \frac{1}{2}\left(\frac{2\sin(x)}{\sin(x) + \cos(x)}\right)\\\\
& = \frac{1}{2}\left[\frac{(\sin(x) + \cos(x)) + (\sin(x) - \cos(x))}{\sin(x) + \cos(x)}\right]\\\\
& = \frac{1}{2} - \frac{1}{2}\left(\frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)}\right)
\end{align*}
Moreover, we do also have that $(\sin(x) + \cos(x))' = \cos(x) - \sin(x)$.
Having said that, according to the substitution $u = \sin(x) + \cos(x)$, we get that
\begin{align*}
\int\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \frac{x}{2} - \frac{\ln(\sin(x) + \cos(x))}{2} + c
\end{align*}
Now it remains to apply the integration limits.
Hopefully this is helpful!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$ Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$
Thoughts: the "standard" method of starting by squaring (twice) to get rid of the square roots, because I don't have a nice way of showing the resulting polynomial is irreducible. So..
Attempt: It would be great if I could get our $\alpha$ in the form $$(a+b)^2=a^2+2ab+b^2=\sqrt{3+2\sqrt{2}}.$$ So, $$3+2\sqrt{2}=(\sqrt{2})^2+2\sqrt{2}+1=(\sqrt{2}+1)^2.$$So,
$$\alpha=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$$So,
$$(\alpha-1)^2=2\\
\alpha^2-2\alpha+1=2\\
\alpha^2-2\alpha-1=0.$$So let $f(x)=x^2-2x-1$. Since $f(x)$ is irreducible over $\mathbb{Q}$ by the Rational Roots Test (since it has degree $2$), $f(x)$ is monic, and $f(\alpha)=0$, we conclude that $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
Does this look okay?
| Since the comment has answered the OP's stated question, it is open season.
Alternative approach
If $a$ is rational and $b$ is irrational, then $(a+b)$ is irrational.
Further, if $a$ is irrational, then $\sqrt{a}$ is also irrational.
The proof to the 2nd assertion above is that if $\sqrt{a}$ can be rationally expressed as $\frac{P}{Q}$, this implies that $a$ has the rational expression $\frac{P^2}{Q^2}$, which contradicts the premise that $a$ is irrational.
Using the two assertions, you have that since $\sqrt{2}$ is known to be irrational, so is $\left[3 + 2\sqrt{2}\right]$, and therefore, so is $\sqrt{3 + 2\sqrt{2}}.$
This implies that there can not be any polynomial equation of degree 1, with form $x + b = 0,$ with $(b)$ rational, whose root is $\sqrt{3 + 2\sqrt{2}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Comparing sets $\{8^n-7n-1: n \in \Bbb{N}\}$ and $\{49(n-1): n \in \Bbb{N}\}$
If
$$\begin{align}
X&=\{8^n-7n-1: n \in \Bbb{N}\} \\
Y&=\{49(n-1): n \in \Bbb{N}\}
\end{align}$$
then,
a) $X\subset Y \qquad$ b) $Y\subset X\qquad$ c) $X=Y\qquad$d) none of these
I know this question can be solved by taking $X=8^n-7n-1$ and splitting $8^n$ into $(7+1)^n$ and then apply binomial theorem as follows:
Given,
$$\begin{align}X &=8^n−7n−1 \\[4pt]
&=(1+7)^n−7n−1 \\[4pt]
&=1+7n+\frac{n(n-1)}{2}+\cdots+7^n-7n-1 \\[4pt]
&=\frac{n(n-1)}{2}7^2+\cdots+7^n \\[4pt]
&=49\left[\frac{n(n-1)}{2}+\cdots+7^{n-2}\right]
\end{align}$$
Hence, the set $X$ will be some specific multiples of $49$.
$Y=49(n-1)$. Hence, the set $Y$ will be all multiples of $49$. So, it will contain the elements of $X$ too.
So, $$X\subset Y$$
But is there any alternate/simple method to solve this question without using binomial theorem?
| $8^n - 7n - 1 = (8^n - 1) - 7n = (8-1)(8^{n-1} + ..... + 8 + 1) - 7n = 7[(8^{n-1} + .... + 8 + 1) - n]$.
So $7|8^n - 7n - 1$.
And $8^n \equiv 1^n \equiv 1 \pmod 7$ so $(8^{n-1} + .... + 8 + 1) - n\equiv (1+1+1+....+1) -n \equiv n - n \equiv 8 \pmod 7$
So $7|(8^{n-1} + .... + 8 + 1) - n$ and $49|8^n - 7n - 1=7[(8^{n-1} + .... + 8 + 1) - n]$.
As $Y$ is the set of all non negative multiple of $49$ and $8^n-7n - 1$ is a non-negative multiple of $49$ we know $X\subset Y$.
But as $n$ gets large $8^n -7n -1$ will increase by larger amounts than $49$ and not all multiples of $49$ can possibly be covered by $8^n-7n-1$.
Indeed for $n = 2$ we have $8^2 - 7\cdot 2 -1 = 49$ but for $n =3$ we have $8^2 -3\cdot 7 -1 = 490$ and we never have any multiples of $49$ between $49=49\cdot 1$ and $490 = 49\cdot 10$.
And furthermore If $A_k = 8^k - 7k - 1$ then $D=A_{k+1} - A_k = 7\cdot 8^{k+1} - 7$. and $k> 1$ then $D > 49$ and $A_n$ is increasing and by more than $49$.
So $Y\not \subset X$.
| {
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"url": "https://math.stackexchange.com/questions/4200557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find locus of mid-point of chord of circle I am working through a pure maths book as a hobby and am struggling with the very last part of the question below. I give the whole of the question just in case it proves relevant to the answer.
A circle passes through the points A, B and C which have coordinates $(0,3),(\sqrt3,0)$ and $(-\sqrt3,0$ respectively.
Find (i) the equation of the circle , (ii) the length of the minor arc BC, (iii) the equation of the circle on AB as diameter.
And then follows this part which I cannot do:
A line $y=mx-3$ of variable gradient m cuts the circle ABC in two points L and M. Find in cartesian form the equation of the locus of the mid-point of LM.
I have worked out the first parts to get:
(i) $x^2+y^2-2y-3=0$
(ii)$\frac{4\pi}{3}$
(iii)$x^2+y^2-3\sqrt x-3y=0; x^2+y^2+2y-3=0$
But I cannot get the very last part of (iii). I thought if I could find the coordinates of intersection I could then take the coordinates of the mid-point.
I have tried as follows:
Circle equation $x^2+y^2-2y-3=0 \rightarrow y=\frac{x^2+y^2-3}{2}$
At intersection $\frac{x^2+y^2-3}{2}=mx-3$
I then try to solve for this but in no way can I get the answer which the book says is $x^2+y^2+2y-3=0$
| Equation of circle is $x^2+y^2-2y-3=0$
Equation of line is $y = mx - 3$
Substituting $y$ from equation of line into equation of circle, we get
$x^2 + (mx-3)^2 - 2 (mx-3) - 3= 0$
Simplifying, $(1+m^2) x^2 - 8 mx + 12 = 0$
As the line intersects the circle at two points, the sum of roots of the above equation is the sum of x-coordinates of the intersection.
So, $x_1 + x_2 = \cfrac {8m}{1+m^2}$
If the midpoint is $(h, k)$,
$h = \cfrac{x_1 + x_2}{2} = \cfrac{4m}{1+m^2} \tag1$
Now we know that, $k = mh - 3$ as $(h, k)$ is on the line $y = mx - 3$.
So, $ \ k = \cfrac{4m^2}{1+m^2} - 3 = \cfrac{m^2-3}{1+m^2} \tag2$
From ($1$) and ($2$), $h^2 + k^2 = \cfrac{16m^2 + m^4 - 6m^2 + 9}{(1+m^2)^2}$
$ = \cfrac{(1+m^2)(m^2 + 9)}{(1+m^2)^2} = \cfrac{m^2 + 9}{1+m^2} = \cfrac{(3m^2 + 3) - (2m^2 - 6)}{1+m^2}$
So, $h^2 + k^2 = 3 - 2k \implies h^2 + k^2 + 2k - 3 = 0$
Alternatively from ($2$), $m = \pm \sqrt{\cfrac{k+3}{1-k}}$. Plug $m$ into ($1$), square both sides and resolve. You get the locus.
| {
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"url": "https://math.stackexchange.com/questions/4201422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive solution $x_n$.
*The sequence $(x_n)$ has a finite limit with $n \to \infty+$. Find that limit.
Here's all I did:
$\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} = 1.$
$\Leftrightarrow 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) = n! (x+1)(x+2)...(x+n) $
Consider the polynomial $P(x) = 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) - n! (x+1)(x+2)...(x+n) $
Consider the highest coefficient of the polynomial:$ -n! < 0 $
Consider the lowest coefficient of the polynomial:$ (2.3...n)^2+(1.3...n)^2+...(1.2...(n-1))^2 - n! = (1.3...n)^2+...(1.2...(n-1))^2 > 0$
Let $x_1,x_2,x_3....$ be the solutions of$ P(x) $
Assume that equation (1) has no positive solution.
According to Viette's theorem: $x_1 x_2 x_3....x_n = (-1)^n \frac{a_0}{a_n} $
$\frac {a_0}{a_n} < 0 \Rightarrow $if $n$ is an odd number , then $x_1 x_2 x_3....x_n < 0$ , but
$(-1)^n \frac{a_0}{a_n} >0 \Rightarrow $ (nonsense) . Same with $n $being an even number.
So the polynomial must have at least one integer root.
Suppose the polynomial has at least 2 positive roots $x$ and $y$ .
$\frac {1}{y+1} + \frac{1}{2(y+2)} + \frac{1}{3(y+3)} + .... + \frac {1} {n(y+n)} =\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} .$
$\Rightarrow (y-x)( \frac {1}{(x+1)(y+1)}+ \frac {1}{2(x+2)(y+2)} +...+\frac {1}{n(x+n)(y+n)} ) = 0 $
$\Rightarrow x=y \Rightarrow$ (no sense)
So for every $n \geq 2$ , equation $( 1 )$ always has a unique positive solution $x_n$
That's all I did , and I have no idea what to do next , I want to establish a relationship between $x_n$ but for $n \geq 3$ the polynomial's solution is a " bad " solution . " so it's very difficult to build. Hope to get help from everyone. Thanks very much .
| For you first question, a picture may help clarify what is happening. This one comes from Wolfram Alpha when $n=7$
In effect your function, let's call it $f_n(x)$, is the sum of $n$ hyperbolae with $n$ vertical asymptotes at $-1,-2,\ldots, -n$, a locally decreasing function of $x$ where it exists, and with $f_n(x) \to 0^-$ as $x \to - \infty$ and $f_n(x) \to 0^+$ as $x \to + \infty$. So $f_n(x)=1$ gives $n$ real solutions, of which $n-1$ are clearly negative and, since $f_n(0)\ge 1$, one must be non-negative and when $n>1$ is positive.
As others have said the limit as $n\to \infty$ comes when $x=1$ and you get $\sum\limits_{k=1}^\infty \frac{1}{k(k+1)}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Ambiguity in Integration done by substitution. I was solving this integral $$I= \int x\sqrt{1-x^2} \mathrm dx$$ My solution is:
Let $x=\sin y$, then $|\cos y|=\sqrt{1-x^2}$ I now use method of substitution for the cases $\cos y\geq 0$ and $\cos y<0$. The correct answer is $-\frac{(1-x^2)^{3/2}}{3} +C$. I am able to obtain this answer for $\cos y\geq 0$ but in the case where $\cos y <0$, we get it to be $\frac{(1-x^2)^{3/2}}{3} +C$. Where have I gone wrong?
| Suppose $~~I=\int x(\sqrt{1-x^2}) dx $
Let $(1-x^2)=u ~~\Longrightarrow -2xdx=du$
$\Longrightarrow dx=\frac{-1}{2x}du $
Now we have,
$$I=\int x(\sqrt{1-x^2}) dx = \int \frac{-1}{2x}x \sqrt{u}du = \frac{-1}{2} \int\sqrt{u} du $$
$$= \frac{-1}{2} \left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right) +c= \frac{-u^\frac{3}{2}}{3} +c = \frac{-(1-x^2)^\frac{3}{2}}{3}+c $$
where c is integration constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequality on $a,b,c$ :-
$$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$$
$$\rightarrow \frac{a^2+b^2+c^2}{3} \geq \frac{(a+b+c)^2}{9}$$
$$\rightarrow \frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2}$$
I got no progress for the second part. I also have a clue, that $a,b,c$ are sides of a triangle, which I have not used yet. So maybe that should be used somehow, but I am not getting it.
Can anyone help me? Thank You.
| We want to show :
$$2(ab+bc+ca)>a^2+b^2+c^2$$
With :
$$a+b-c\geq0$$
$$a+c-b\geq 0$$
$$c+b-a\geq 0$$
As the inequalities are homogenous we need to show $0<x<1$ and $0<y<1$:
$$(x^2+y^2+1)<2(xy+x+y)$$
With :
$$x+y-1\geq0$$
$$x+1-y\geq 0$$
$$y+1-x\geq 0$$
Or :
$$(x^2+y^2+x+y)<2(xy+x+y)$$
Or :
$$2(x+y)<2(xy+x+y)$$
Done !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{2}{3}\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx$ Proof the following
$$\frac{2}{3}\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx$$
I've tried to use x=tan(u) substitution, but don't know how to proceed with this
$$\frac{2}{3}\int_0^{π/4}\ln(\tan(u))du=\int_0^{π/12}\ln(\tan(u))du$$
Please help
| Utilize the integral
$$\int_0^1 \frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dy =\ln(\tan t)
$$
to prove
\begin{align}
&\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx \\= & \int_0^\frac\pi{12}\ln(\tan t)~dt
= \int_0^1 \int_0^{\frac\pi{12} }
\frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dt \ dy\\= &\ -\frac12\int_0^1\frac1y \tan^{-1}\frac y{1-y^2} dy
= -\frac12\int_0^1\frac{\tan^{-1}y^3}y \overset{y^3\to y}{dy}
-\frac12\int_0^1\frac{\tan^{-1}y}y dy\\=&- \frac23\int_0^1\frac{\tan^{-1}y}ydy
\overset{ibp}
=\frac23\int_0^1\frac{\ln y}{1+y^2}dy
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $\sin^2(\pi x)$ as $x \to \infty$.
The following is my best attempt at formalizing this idea:
Let $f(x) = \sin^2(\pi(x+\frac{1}{x}))$ and $\xi_n = \frac{n+\sqrt{n^2-4}}{2}$, where $n \geq 2$. Notice that $f(\xi_n)=0$.
Now, consider the limit of $\xi_{n+1} - \xi_{n}$ as $n \to \infty$:
$$\begin{align} \lim_{n\to\infty} \xi_{n+1}-\xi_n &= \lim_{n\to\infty}\frac{(n+1)+\sqrt{(n+1)^2-4}}{2}-\frac{n+\sqrt{n^2-4}}{2} \\ &= \frac{1}{2}\lim_{n\to\infty}1+\sqrt{(n+1)^2-4}-\sqrt{n^2-4} \\ &= \frac{1}{2}\lim_{n\to\infty}\frac{2+\frac{2}{n}+2\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}}{\frac{1}{n}+\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}+\sqrt{1-\frac{4}{n^2}}} \\ &= 1 \end{align}$$ What the result above tells us is that as x
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $n \to \infty$.
We can also compute the value of $\int^{1}_{0} \sin^2(\pi x) dx$:
$$\begin{align} \int^{1}_{0} \sin^2(\pi x) dx &= \frac{1}{2}\int^{1}_{0}1-\cos(2\pi x)dx \\ &=\frac{1}{2}\{(1-\frac{1}{2\pi}\sin(2\pi))-(0-\frac{1}{2\pi}\sin(0))\} \\ &= \frac{1}{2}\end{align}$$ Thus, we can state that:
$$\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$$
Notice that we can write $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ as... $$\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx - \int^{\xi_{n}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$$ If we consider $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ as a sequence, then $\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$
tells us that $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is
not a Cauchy sequence, and hence $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent as $n \to \infty$.
Thus, we can finally claim that $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent.
Now, the part of my argument which I find to be weak is:
What the result above tells us is that as $x$
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $x \to \infty$.
How can I make this part more rigorous/stronger? I don't really know how to achieve any more rigor with my current level of mathematics.
| What's wrong with this less formal approach?
Replace the original integrand with sin^2 x , since the factor of pi goes away with a simple change of variable, and since the x term dominates the 1/x term over the interval. Then every occurrence of the new integrand as x increases is non-negative and identical in magnitude, so the integral must diverge.
| {
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"url": "https://math.stackexchange.com/questions/4211726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Necessary condition on $|x-1|$ to assure $|x^2 -1| < \frac{1}{2}$. Is my solution correct? How to get the tightest answer? Question: Determine a condition on $|x-1|$ that will assure that:-
$$|x^2 -1| < \frac{1}{2}$$
My solution:-
Let $f(x) = x^2$
$$\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} x^2 = 1$$
To find $\delta$ so that
$\: \: |f(x) - 1| < \epsilon = \frac{1}{2}\: \:$ whenever $\: \: 0 < |x-1| < \delta$
$$|f(x) - 1| < \epsilon = |x^2-1| = |x+1||x-1|$$
$$|f(x)-1| = |x+1||x-1|$$
$$|f(x)-1| < \delta \: |x+1| \tag {1}$$
$$$$
$$0 < |x-1| < \delta$$
$$|x-1| < \delta$$
$$-\delta < x-1 < \delta$$
$$-\delta +2 < x+1 < \delta + 2$$
$$-\delta-2 < x+1 < \delta + 2$$
$$|x+1| < \delta + 2 \tag{2}$$
Subsitutuing equation (2) in (1)
$$|f(x) -1| < \delta \: |x+1|$$
$$| f(x) - 1| < \delta \: (\delta + 2)$$
Solving $\delta \: (\delta + 2) = \frac{1}{2}$ for $\delta$ we get:
$$\delta = \frac{-2\pm \sqrt{6}}{2}$$
Since $\delta$ can only be positive, the necessary condition is
$$0 < |x-1| < \frac{-2+\sqrt{6}}{2} \approx 0.2247$$
| I like how you are trying to use the concept of continuity here. But the problem is much simpler. One can determine the necessary and sufficient condition on $|x-1|$
by simply solving the inequality directly.
First, an algebraic solution:
$$|x^2-1|<\frac{1}{2}\iff -\frac{1}{2}<x^2-1<\frac{1}{2}
\iff \frac{1}{2}<x^2<\frac{3}{2}\iff \frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}}
$$
And
$$\frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}}\iff x\in
]-\frac{\sqrt{3}}{\sqrt{2}},\frac{\sqrt{3}}{\sqrt{2}}[\, \cap \left( ]\frac{1}{\sqrt{2}},\infty[\,\cap\, ]-\infty,-\frac{1}{\sqrt{2}}[ \right)
$$
$$\iff
x\in
]-\frac{\sqrt{3}}{\sqrt{2}},-\frac{1}{\sqrt{2}}[\,\cup\, ]\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{\sqrt{2}}[.$$
Second, a geometric solution:
You know that $x\mapsto x^2-1$ is a parabola whose axis of symmetry is the $y-axis$,
vertex at $(-1,0)$, and open upwards. Find the points of intersection of the parabola with the two horizontal lines $y=-1/2$ and $y=1/2$. You immediately discover the two intervals where the curve $x^2-1$ lies over the line $y=-1/2$
and below the line $y=-1/2$.
Your original argument is not wrong. It is actually creative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Hard and non-trivial inequality with three variable reals For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\frac{a^4+b^4}{2}}.$$
I tried to use Holder inequality:
$$\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)(a^2+b^2+c^2)(a^2+b^2+c^2)\Big(b^4+c^4+c^4+a^4+a^4+b^4\Big) \geq \text{RHS}^4$$
or
$$(a^2+b^2+c^2)^4 \geq (1+1+1)(a+b+c)^2 \Big(\sum_\text{cyc}{\frac{a^2(b^4+c^4)}{2}} \Big) \geq \text{RHS}^4$$
But got stuck after that (the $\geq$ is reversed).
| Your second attempt gives another solution!
Indeed, for non-negative variables we need to prove that:
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\sum_{cyc}(a^4b^2+a^4c^2)$$ or
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\left(\sum_{cyc}a^2\sum_{cyc}a^2b^2-3a^2b^2c^2\right).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$2(9u^2-6v^2)^4\geq27u^2((9u^2-6v^2)(9v^4-6uw^3)-3w^6)$$ or $f(w^3)\geq0,$ where
$$f(w^3)=u^2w^6+6u^3(3u^2-2v^2)w^3+2(3u^2-2v^2)^4-9u^2v^4(3u^2-2v^2).$$
But it's obvious that $f$ increases, which says that it's enough to prove our inequality for the minimal value of $w^3$, which by $uvw$(about $uvw$ see here: https://artofproblemsolving.com/community/c6h278791 )
happens in the following cases:
*
*$w^3=0.$
Let $c=0$ and $b=1$.
We need to prove: $$2(a^2+1)^4\geq3(a+1)^2(a^4+a^2),$$ which is true by C-S and AM-GM:
$$2(a^2+1)^4=2(a^2+1)(a^2+1)^2(a^2+1)\geq(a+1)^2\cdot4a^2(a^2+1)\geq3(a+1)^2(a^4+a^2);$$
2. Two variables are equal.
Let $b=c=1$.
We need to prove that:
$$2(a^2+2)^4-3(a+2)^2(2a^4+2a^2+2)\geq0$$ or $$(a-1)^2(a^6+2a^5+8a^4+2a^3+5a^2-4a+4)\geq0$$ and we are done.
| {
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"url": "https://math.stackexchange.com/questions/4214439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it just a coincidence that the solution to $y=x$, $y=mx+b$ ($m<1$) is the sum of an infinite geometric series with first term $b$ and ratio $m$? The solution for the system of linear equations of $y = x$ and $y = mx + b$ is
$$x=y= \frac{b}{1-m}.$$
I noticed that this is also the sum of an infinite geometric series, where the first term is $b$ and common ratio is $m$ (granted $m$ is less than $1$):
$$ \frac{b}{1-m} = b + bm + bm^2 + \cdots bm^n + \cdots$$
Is this all a big coincidence or is there some deeper meaning to this relationship?
| It follows by induction.
As $x = mx+b$ we can replace $x$ with $mx+b$ indefinitely.
$y = x \\= mx + b=m(mx+b) + b \\= m^2x + mb + b\\=m^2(mx+b)+mb+b \\= m^3x + m^2b + m b + b\\\dots\\b + mb + m^2b + m^3b + m^4b + ......$
This shouldn't be a surprise. If $|m| < 1$ then $b + mb + m^2b + m^3b + m^4b + ...... \\= b(1 + m + m^2 + m^2 + m^4 + ....)\\=b\frac 1{1-m}=\frac b{1-m}$
And if $x = mx + b$ we must have $x-mx = b$ so $x(1-m)=b$ so $x = \frac b{1-m}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$
I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
| I don't know if this is faster for you, but here's how I did it.
The first thing you need to notice is that higher powers of roots of the equation can be reduced to linear terms only. If $x_i$ is a root of $x = 5 - x^2$, then:
$$\begin{align*}x_i^2 & = 5 - x_i \\ x_i^3 & = 5x_i - x_i^2 \\ & = 6x_i - 5\end{align*}$$
So we can simplify:
$$\begin{eqnarray*}(x_i + 1)^3 & = & x_i^2 + 3 x_i^2 + 3 x_i + 1 \\
& = & (6x_i - 5) + 3 ( 5 - x_i ) + 3 x_i + 1 \\
& = & 6x_i + 11
\end{eqnarray*}$$
And therefore:
$$\begin{eqnarray*}& & \left((x_1 + 1)^3 x - 1\right) \left((x_2 + 1)^3 x - 1\right) \\
& = & \left( (6x_1 + 11) x - 1 \right) \left( (6x_2 + 11) x - 1 \right) \\
& = & \left(36 x_1 x_2 + 66(x_1 + x_2) + 121\right) x^2 - \left(6 (x_1 + x_2) + 22\right) x + 1 \\
& = & -125 x^2 - 16 x + 1\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Challenging geometry problem from AMC 2013 Question 29 from the senior division of the 2013 Australian mathematics competition.
I got every question except this one and have no clue how to solve it.
| Applying Menelaus Theorem a few times:
$$\frac{RY}{RQ}\times \frac{QP}{PT}\times\frac{TX}{XY}=1$$
$$\frac{3}{6}\times \frac{QP}{PT}\times\frac{TX}{XY}=1$$
$$\frac{TX}{XY}=\frac{2PT}{QP}...[1]$$
$$\frac{SQ}{ST}\times \frac{TX}{XY}\times\frac{YZ}{ZQ}=1$$
$$\frac{11}{11+QT}\times \frac{TX}{XY}\times\frac{2}{1}=1$$
$$\frac{TX}{XY}=\frac{11+QT}{22}...[2]$$
$$\frac{TX}{XY}=\frac{2PT}{QP}=\frac{11+QT}{22}...[1]\&[2]...[3]$$
$$\frac{PT}{QT}\times \frac{QY}{YR}\times\frac{RX}{XP}=1$$
$$\frac{PT}{QT}\times \frac{3}{3}\times\frac{5}{4}=1$$
$$\frac{PT}{QT}=\frac{PT}{QP+PT}=\frac{4}{5}$$
$$PT=4QP...[4]$$
$$\frac{2PT}{QP}=\frac{11+QT}{22}...[3]\&[4]$$
$$\frac{2\times 4QP}{QP}=\frac{11+QT}{22}$$
$$QT=165$$
$$ST=176$$
| {
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"url": "https://math.stackexchange.com/questions/4220239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Near-Pythagorean triplets: What are the general solutions to $a^2+b^2=c^2-1$? Obtaining the most general solution to a quadratic Diophantine equation in three variables is often easier if the equation is homogeneous. For example, by focusing on "primitive" solutions, it is easy to show that all Pythagorean triplets can be written as $$a=m(r^2-s^2), b=2mrs, c=m(r^2+s^2)$$ and that if you restrict to $(r,s)=1, r\neq s \mod 2$ no triplet is repeated.
Inhomogeneous equations can be tougher. In two variables, Pell's equation $x^2-ny^2$ is well studied, but I wanted to look at three variables. An example is
$$
a^2+b^2=c^2+1
$$
which is solved by taking an arbitrary integer $r>1$ and an arbitrary integer factor $u | r(r-1)$, with $$u>\sqrt{2r^2-2r-1}-r+\frac12$$ and $u^2<r(r-1)$. (These restrictions ensure uniqueness and non-negativity). The solution is then
$$
a= \frac{r(r-1)}{u}-u \\ b=2r-1 \\ c=\frac{r(r-1)}{u}+u
$$
Note that the solutions are generated requiring no more difficult operations than factoring $r(r-1)$.
The next obvious case to try was $$a^2+b^2=c^2-1$$
We can see that $a$ and $b$ must both be even, and transform this using $a=2x,b=2y,c=2z-1$, to $$x^2+y^2=z^2-z$$ Here, $x$ and $y$ must be of the same parity.
I have pursued the odd-parity case by transforming to $x=2r-1,y=2s-1,z=8u+2$, which comes down to finding solutions to
$$
r(r-1)+s(s-1)=2u(8u+3)
$$
and with this, for small-ish values of $u$, you can generate
$$ (18,6,19), (30,18,35), (50,10,51), (38,34,51) $$
and so forth. But this does not resolve the question since $$r(r-1)+s(s-1)=2u(8u+3)$$ is just another inhomogeneous quadratic Diophantine equation, and I have not been able to find a generic solution to that either.
So my question is:
What is the general solution to $$\mathbf{a^2+b^2=c^2-1}$$
| This is only a partial answer. I managed to find several families of solutions:
Family 1
$$
\begin{align}
a &= 2\left[(4k^2+1)n \pm k\right],\\
b &= 2\left[(4k^2+1)n^2 \pm 2kn - k^2\right],\\
c &= 2\left[(4k^2+1)n^2 \pm 2kn - k^2\right] + 4k^2 + 1.
\end{align}
$$
Family 2
$$
\begin{align}
a &= 2\left[\left(2k(k+1)+1 \vphantom{1^1} \right)n \pm k^2\right],\\
b &= 2\left[\left(2k(k+1)+1 \vphantom{1^1} \right)n^2 \pm 2k^2n - k\right],\\
c &= 2\left[\left(2k(k+1)+1 \vphantom{1^1} \right)n^2 \pm 2k^2n - k\right] + 2k(k+1)+1.
\end{align}
$$
Here, $k$ and $n$ are integers such that $a,b,c$ are positive integers.
I also found solutions for which I have not yet found a generalization:
$$
2(29n\pm 6),\quad 2(29n^2 \pm12n -6),\quad 2(29n^2 \pm12n -6) +29,\\
2(53n\pm15),\quad 2(53n^2\pm30n -9),\quad 2(53n^2\pm30n -9) + 53,\\
2(73n\pm23),\quad 2(73n^2\pm46n - 11),\quad 2(73n^2\pm46n - 11) + 73.
$$
EDIT
I found a recursion: if $(a, b, c)$ is a solution to $a^2 + b^2 = c^2 - 1$, then the three triples
$$
\begin{align}
a' &= 2(a+c-b) - a\\
b' &= 2(a+c-b) + b\\
c' &= 2(a+c-b) + c,
\end{align}
$$
$$
\begin{align}
a' &= 2(b+c-a) - b\\
b' &= 2(b+c-a) + a\\
c' &= 2(b+c-a) + c,\\
\end{align}
$$
$$
\begin{align}
a' &= 2(a+b+c) - b\\
b' &= 2(a+b+c) - a\\
c' &= 2(a+b+c) + c,\\
\end{align}
$$
are all solutions as well. Starting with the trivial solution $(0,0,1)$, I'm reasonably confident that this will generate all solutions. Unfortunately, I haven't found a closed-form expression.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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How to find characteristic polynomial of a matrix $A$ whose elements are of the form $a_{ij} = a_ia_j$ for all $1\leq i,j \leq n$ For $n > 1$,
$$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix}$$
The characteristic polynomial of matrix $A$ is $x^{n-1}(x-tr(A))$
My question is that how to find characteristic polynomial of matrix $A$
My Attempt:
I noticed that elements of $A$ are of the form $a_{ij} = a_ia_j$ for all $1\leq i,j \leq n$ So $$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}$$
| $$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}$$
Let $A = B×C$, where $$B = C = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}$$
We know that $Rank(BC) \leq \min (Rank (B),Rank(C))$
But here we see that $Rank(B) = Rank(C) =1$. Hence $Rank (A) = 1$. So $n-1$ times eigen values are $0$ and one eigen value is trace($A$). Therefore $Ch_A(x) = x^{n-1}(x-trace(A))$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebra in Geometric Proof of Quadratic Reciprocity I am trying to understand the proof of quadratic reciprocity form the George Andrews textbook on number theory (this proof follows geometry and Eisenstein's thinking). I understand what is happening conceptually, but the finer points of inequality algebra are not obvious to me. We are given sets $\mu_1=\{q,2q,\cdots,\frac{1}{2}(p-1)q\}$ and $\mu_2=\{p,2p,\cdots,\frac{1}{2}(q-1)p\}$, with $\mu_1$ representing the negative least residues mod $p$ and $\mu_2$ representing the negative least residues mod $q$. Ultimately the proof shows $\mu_1+\mu_2$ is odd if and only if $p\equiv q\equiv3\pmod{4}$.
The author illustrates this by considering a hexagon $H$ with vertices $ABCDEF$ that lies within a rectangle $AGDJ$ (drawn in Quadrant $I$) that is bounded by $x=p/2$ and $y=q/2$. The attached picture provides additional information about the components of the rectangle.
For a point $(x,y)$ to lie in $H$, it must satisfy, $0<x<p/2$, $0<y<q/2$, $y<\frac{q}{p}x+\frac{1}{2}$, and $y>\frac{q}{p}x-\frac{q}{2p}$. The next step remarks that if $(m,n)$ is some lattice point in $H$, then so is $(\frac{p+1}{2}-m,\frac{q+1}{2}-n)$, where these two points are $equal$. This is verified by substituting "these coordinates" into the four inequalities above.
I'm not sure what is meant by "these coordinates." I assumed it meant $(m,n)$, so this is what I tried.
$\frac{q}{p}x-\frac{q}{2p}<y<\frac{q}{p}x+\frac{1}{2}\Rightarrow \frac{q}{p}m-\frac{q}{2p}<n<\frac{q}{p}m+\frac{1}{2} \Rightarrow qm-\frac{q}{2}<py<qm+\frac{p}{2}$
$\Rightarrow 2qm-q<2py<2qx+p$.
Any assistance would be appreciated.
| After lots of questions and lots of attempts, here is the definitive algebra justifying the claim from this proof:
Consider the algebraic justification for $(m,n)=\left(\frac{p+1}{2}-m,\frac{q+1}{2}-n\right)$. Let $(m,n)$ be some lattice point within $H$, and let $(u,v)=\left(\frac{p+1}{2}-m,\frac{q+1}{2}-n\right)$ be some coordinate pair within $H$. To show the equality between forms, we must show $0<u<\frac{p}{2}, 0<v<\frac{q}{2}, v<\frac{q}{p}u+\frac{1}{2},$ and $v>\frac{q}{p}u-\frac{q}{2p}$. Call these Cases $1-4$, respectively.
*
*$0<u<\frac{p}{2}$
Assume $0<m<\frac{p}{2}\Rightarrow -\frac{p}{2}<-m<0$. Clearly $\frac{p+1}{2}-\frac{p}{2}=\frac{1}{2}<\frac{p+1}{2}-m<\frac{p+1}{2}$. Now,$\frac{p+1}{2}-m=\frac{p}{2}\Rightarrow m=\frac{1}{2}$, but $m$ is an integer, so $0<\frac{p+1}{2}-m<\frac{p}{2}$, i.e., $0<u<\frac{p}{2}$.
*$0<s<\frac{q}{2}$
Assume $0<n<\frac{q}{2}\Rightarrow -\frac{q}{2}<-n<0$. Clearly $\frac{q+1}{2}-\frac{q}{2}=\frac{1}{2}<\frac{q+1}{2}-n<\frac{q+1}{2}$. Now,$\frac{q+1}{2}-n=\frac{q}{2}\Rightarrow n=\frac{1}{2}$, but $n$ is an integer, so $0<\frac{q+1}{2}-n<\frac{q}{2}$, i.e., $0<v<\frac{q}{2}$.
For the remaining cases, let $n<\frac{q}{p}m+\frac{1}{2}$ and let $n>\frac{q}{p}m-\frac{q}{2p}$.
*$v<\frac{q}{p}u+\frac{1}{2}$
Consider $\frac{q+1}{2}-n<\frac{q}{p}\left(\frac{p+1}{2}-m\right)+\frac{1}{2}$. It is clear $\frac{q}{p}\left(\frac{p+1}{2}-m\right)+\frac{1}{2}$ $=\frac{q(p+1)}{2p}-\frac{q}{p}m+\frac{1}{2}=\frac{q}{2}+\frac{q}{2p}-\frac{q}{p}m+\frac{1}{2}=\frac{q+1}{2}-\left(\frac{q}{p}m-\frac{q}{2p}\right)$
$>v=\frac{q+1}{2}-n\left(\text{this follows from } n>\frac{q}{p}m-\frac{q}{2p}\right)$.
*$v>\frac{q}{p}u-\frac{q}{2p}$
Consider $\frac{q+1}{2}-n>\frac{q}{p}\left(\frac{p+1}{2}-m\right)-\frac{q}{2p}$. It is clear $\frac{q}{p}\left(\frac{p+1}{2}-m\right)-\frac{q}{2p}$ $=\frac{q(p+1)}{2p}-\frac{q}{p}m-\frac{q}{2p}=\frac{q}{2}+\frac{q}{2p}-\frac{q}{p}m-\frac{q}{2p}=\frac{q}{2}-\frac{q}{p}m=\frac{q+1}{2}-\left(\frac{q}{p}m+\frac{1}{2}\right)<v=\frac{q+1}{2}-n$ $\left(\text{this follows from } n<\frac{q}{p}m+\frac{1}{2}\right)$.
| {
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"source": "stackexchange",
"question_score": "3",
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$3\times 2^m + 4 = n^2$ I have been trying to solve this problem but I couldn't find any way to find the answers. Find all natural numbers m and n such that:
$3\times\ 2^m + 4 = n^2$
This is a question from the Moroccan Maths Olympiad (2019.)
I tried bringing $4$ to the other side and factorising to get: $3\times 2^m=(n+2)(n−2),$ and I tried solving for that $n+2$ is a multiple of three then $n-2$ is a multiple of three but I couldn't find anything.
Can anyone help me please?
| $$3\cdot2^{m} + 4 = n^2$$
We take the three cases $m=3a, m=3a+1,$ and $m=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet m=3a$
Let $X = 3\cdot2^a, Y=3n$, then we get $Y^2 =X^3 + 36.$
According to LMFDB, this elliptic curve has $(X,Y)=(-3,\pm 3)$, $(0,\pm 6)$, $(4,\pm 10)$, $(12,\pm 42).$
Hence $(X,Y)= (12, 42) \implies (m,n)=(6,14).$
$\bullet m=3a+1$
Let $X = 6\cdot2^a, Y=6n$, then we get $Y^2 =X^3 + 144.$
This elliptic curve has $(X,Y)=(0,12)$, then there is no natural solution $(m,n).$
$\bullet m=3a+2$
Let $X = 12\cdot2^a, Y=12n$, then we get $Y^2 =X^3 + 576.$
This elliptic curve has $(X,Y)=(-8 , 8 ), (0 , 24 ), (12 , 48 ), (24 , 120 ), (160 , 2024 ).$
Hence $(X,Y)= (12 , 48) \implies (m,n)=(2,4).$
$(X,Y)= (24 , 120) \implies (m,n)=(5,10).$
Thus, there are only natual solutions $(m,n)=(2,4),(5,10),(6,14).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?
My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{x^2} - \frac{1}{x^2} \right) = 0$. I used $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$.
But $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\frac{x^2-\sin^2x}{x^2\sin^2x}$ and if I apply L'Hospital's Rule four times at the second expression, I get $\lim\limits_{x\to 0}\,\left(\frac{8\cos 2x}{24\cos 2x - 32\sin 2x - 8x^2\cos 2x}\right) = \frac {1}{3}$.
What am I missing?
| $$\lim_{x\to0}\left(\csc^2(x)-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\right)$$
Both numerator and denominator tend to zero, and they are continuous and differentiable, so L'Hopital's rule applies. The derivative of the numerator is: $2x-2\sin(x)\cos(x)=2x-\sin(2x)$. The derivative of the denominator is: $2x\sin^2(x)+2x^2\sin(x)\cos(x)=2x\sin^2(x)+x^2\sin(2x)$. Well, these both will still tend to zero in the limit, so we can simply reapply L'Hopital.
The second derivative of the numerator is: $2-2\cos(2x)$. The second derivative of the denominator is:
$$2\sin^2(x)+4x\sin(x)\cos(x)+2x\sin(2x)+2x^2\cos(2x)\\=2\sin^2(x)+4x\sin(2x)+2x^2\cos(2x)$$ Which still tends to zero.
The third derivative of the numerator is $4\sin(2x)$. The third derivative of the denominator is:
$$4\sin(x)\cos(x)+8x\cos(2x)+4\sin(2x)+4x\cos(2x)-4x^2\sin(2x)\\=6\sin(2x)+12x\cos(2x)-4x^2\sin(2x)$$
Well, it is our misfortune to have to apply this rule four times. The fourth derivative of the numerator is: $8\cos(2x)\to8,x\to0$. The fourth derivative of the denominator is: $12\cos(2x)+12\cos(2x)-24x\sin(2x)-8x\sin(2x)-8x^2\cos(2x)\to24,x\to0$. At last!
The original limit is the ratio of the (fourth) derivatives of the numerator and denominator, which amounts to $\frac{8}{24}=\frac{1}{3}$.
A quick visual check with Desmos will confirm this, if the visuals help.
| {
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If $x^3+y^2$ and $x^2+y^3$ are integers, show that $x$, $y$ are integers
Given rational numbers $x$, $y$ such that $x^3+y^2$ and $x^2+y^3$ are integers, prove that $x$, $y$ are integers.
For this problem I don't even know how to start tackling it. I tried various ways:
*
*Letting $x = \frac{a}{b}$, $y = \frac{c}{d}$, which only makes it more complicated
*Doing operations on them: $x^3+y^2+x^2+y^3=x^2(x+1)+y^2(y+1)$, which I have no idea how to continue;
$(x^3+y^2)(x^2+y^3)=x^5+y^5+x^3y^3+x^2y^2$, and it's also probably too complex to break down
$x^3+y^2-x^2-y^3=(x-y)(x^2+xy+y^2)+(y-x)(x+y)$ seems more plausible, but I also can't do anything with this.
I would really appreciate any way of tackling this problem, because I have spent a while thinking about this problem.
EDIT: I have found a solution to this problem (this was edited 2 days after this was posted, so there were also answers before this):
Suppose $x = \frac{a}{b}$, $y = \frac{c}{d}$ in which $a$, $b$, $c$, $d$
are integers and $\gcd(a,b) = \gcd(c,d) = 1$. Then we have
$x^2+y^3=\frac{a^2}{b^2}+\frac{c^3}{d^3}=\frac{a^2d^3+b^2c^3}{b^2d^3}$,
so:
$b^2|a^2d^3+b^2c^3$, which means $b^2|a^2d^3$, and $\gcd(a,b)=1$ so
$b^2|d^3$, and
$d^3|a^2d^3+b^2c^3$, which means $d^3|b^2$, so $b^2=d^3$.
Doing the same to $x^3+y^2$, we get $b^3=d^2$. From that we get
$b^5=d^5$, so $b=d$. Substituting to $b^2=d^3$, we get $b^2=b^3$,
hence $b=d=1$, which implies that $x$ and $y$ are integers.
| Suppose $x=a/d$ and $y=b/d$ with $d\gt0$ and $\gcd(a,b,d)=1$ (i.e., write $x$ and $y$ with the smallest possible common denominator). Then $x^2+y^3=m$ implies $b^3=(md^2-a^2)d$, which implies $d\mid b^3$. Likewise, $x^3+y^2=n$ implies $d\mid a^3$. Thus
$$0\lt d=\gcd(a^3,b^3,d)\le\gcd(a^3,b^3,d^3)=(\gcd(a,b,d))^3=1^3=1$$
hence $d=1$, and so $x$ and $y$ are integers.
| {
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Artist needing to determine geometric angle for sculpture based on platonic solid Dear Mathematicians I need your help for a new sculpture!
I will attach images but first imagine 2 hexagons - where one is rotated 30 deg. They are separated by 12 equilateral triangles. I need to confirm the angle between the 2 faces as show in the image. My computer render calculation makes it 145.222 deg but I need to confirm this is absolutely correct before proceeding to fabrication!
Your help would be greatly appreciated, thank you! Pete
This shows the angle I need to confirm:
This is the computer design angle that I need to confirm:
This is something like what the final sculpture will look like!
This shows how I made it:
| You are requesting a dihedral angle of a (regular) hexagonal antiprism. This is calculated as follows. Consider a hexagon with vertices $v_0, \ldots, v_5$, where $$v_k = \left(\cos \frac{\pi}{3} k, \sin \frac{\pi}{3} k, d \right),$$ and another hexagon with vertices $w_0, \ldots, w_5$ where
$$w_k = \left(\cos \frac{\pi}{6} (2k+1), \sin \frac{\pi}{6} (2k+1), -d\right).$$
The distance these hexagons are separated by is $2d$. In order for the lateral triangles to be equilateral, the distance between, say, $v_0$ and $w_0$, must be $1$. This implies $$1 = \left(1 - \cos \frac{\pi}{6}\right)^2 + \left(0 - \sin \frac{\pi}{6}\right)^2 + (2d)^2,$$ hence $$d = \frac{\sqrt{\sqrt{3} - 1}}{2}.$$ The dihedral angle $\theta$ between triangular faces is therefore given by the equation $$\cos \theta = \frac{\vec a \cdot \vec b}{|\vec a| |\vec b|},$$ where $$\vec a = w_5 - \frac{v_0 + w_0}{2} = \left(\frac{\sqrt{3} - 2}{4}, -\frac{3}{4}, -d\right), \\ \vec b = v_1 - \frac{v_0 + w_0}{2} = \left(-\frac{\sqrt{3}}{4}, \frac{2\sqrt{3} - 1}{4}, d \right).$$
This gives
$$\vec a \cdot \vec b = - \frac{\sqrt{3}}{4} - d^2, \\
|\vec a| |\vec b| = 1 - \frac{\sqrt{3}}{4} + d^2,$$ hence $$\theta = \arccos \frac{1 - 2 \sqrt{3}}{3} \approx 145.2218913319^\circ. \approx 145^\circ 13' 18.80879''.$$ The equivalent radian measure is approximately $2.534600149715126$ radians.
| {
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How can I prove that all quadratic equations are not injective? I was trying to prove that any quadratic formula ($ax^2 + bx + c$) will not be injective, but I have a little problem.
I started by assuming $f(x) = f(y)$.
We can put x and y into the general form of quadratic function, and we get
$ax^2 + bx + c = ay^2 + by + c$
Subtract c from both sides and organise a little bit and you get
$a(x^2-y^2) + b(x-y)=0$
$a(x+y)(x-y) + b(x-y)=0$
Here we can assume that $x\ne y$, hence $x-y\ne 0$, so divide both sides by x-y, and we get
$a(x+y)+b=0$
However, from here I couldn't find any contradictions, which is a problem because there must be a contradiction as quadratic functions are not injective. Can anyone tell me what the contradiction is here?
| This is good stuff so far. You now know that you get $x = y$ if you have $a(x + y) + b \neq 0$, in which case, there is no contradiction. So, let's focus on when $a(x + y) + b = 0$. Can you find two values of $x$ and $y$, that are not equal to each other, but for which $a(x + y) + b = 0$? Or, equivalently, $x + y = -\frac{b}{a}$?
There are many possible answers to this, but here's one: we can let $x = -\frac{b}{2a} + 1$ and $y = -\frac{b}{2a} - 1$. These values of $x$ and $y$ could well contradict the definition of injectivity. We have
\begin{align*}
&a\left(-\frac{b}{2a} + 1\right)^2 + b\left(-\frac{b}{2a} + 1\right) + c \\
= \, &a\left(\frac{b^2}{4a} - 2\frac{b}{2a} + 1\right) - \frac{b^2}{2a} + b + c \\
= \, &\frac{ab^2}{4a} - \frac{2ab}{2a} + a - \frac{b^2}{2a} + b + c \\
= \, &\frac{b^2}{4} + a - \frac{b^2}{2a} + c.
\end{align*}
Try substituting in $-\frac{b}{2a} - 1$ instead, and show that it comes to the same value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Sufficient condition for infinite product being $>0$? I'm trying to solve the following problem:
When is $\prod_{k=1}^{\infty}\frac{\left(\frac{1}{2}-\frac{1}{2}k^{-\alpha}\right)^{1/2}+\left(\frac{1}{2}+\frac{1}{2}k^{-\alpha}\right)^{1/2}}{2^{1/2}}>0$, for $\alpha >0$?
Since $\frac{1}{2}-\frac{1}{2}k^{-\alpha}$ converges to $\frac{1}{2}$ as $k\to\infty$, the fraction $\frac{\left(\frac{1}{2}-\frac{1}{2}k^{-\alpha}\right)^{1/2}+\left(\frac{1}{2}+\frac{1}{2}k^{-\alpha}\right)^{1/2}}{2^{1/2}}$ converges to $1$ but I'm having serious trouble with the infinite product. How exactly do I use this observation to find sufficient conditions for $\alpha>0$?
| Taking the logarithm, you see that the product is $>0$ if and only if
$$ \sum_{k=1}^{+\infty}\log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) $$
converges.
As $k\rightarrow +\infty$,
$$ \begin{aligned} \log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) &= \log\left(1+\frac{1}{2k^{2\alpha}}+o\left(\frac{1}{k^{2\alpha}}\right)\right)\sim\frac{1}{2k^{2\alpha}}
\end{aligned} $$
Therefore the sum converges if and only if $\alpha>\frac{1}{2}$, which means that the product is $>0$ if and only if $\alpha>\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text { } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}
\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }
\end{equation}
| If $a + b > 4$, then $a > 2$ or $b > 2$. If, for instance, $a > 2$, then $a^2 +b^2 \geq a^2 > 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Proving Fibonacci sum with induction Let the Fibonacci sequence be defined as $f_1 = f_2 = 1$ and $f_n = f_{n-1} + f_{n-2}$
Prove that $f_1f_2+f_2f_3+f_3f_4+...+f_{2n-1}f_{2n}+f_{2n}f_{2n+1} = f^2_{2n+1} - 1$
I was able to verify that the claim holds for $n=1$, since $(1)(1)+(1)(2) = 2^2 - 1$
Now, assuming that the claim is true for every $n=k$, it should be sufficient to prove by checking if $n=k+1$ also is true.
$f_1f_2+f_2f_3+...+f_{2k-1}f_{2k}+f_{2k}f_{2k+1}+f_{2(k+1)}f_{2(k+1)+1}$
$ f^2_{2k+1} - 1 +f_{2(k+1)}f_{2(k+1)+1}$
I'm not sure how to proceed from here.
| Hint.
If the statement holds for $n$, then
$f_1f_2 + \dots + f_{2n}f_{2n+1}+ f_{2n+1}f_{2n+2}+f_{2n+2}f_{2n+3}= (\boxed{f_{2n+1}^2}-1) +\boxed{f_{2n+1}f_{2n+2}}+f_{2n+2}f_{2n+3}= f_{2n+1}\underbrace{(f_{2n+1} + f_{2n+2})}_{=\text{what is it?}} + f_{2n+2}f_{2n+3} -1 = \dots \text{(similar step)} = f_{2n+3}^2-1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $(1+i)^n-(1-i)^n$ I'm trying to simplify this expression because there is a simpler way of writing this. But the furthest I've got is $$(1+i)^n-(1-i)^n=\sum_{k=0}^{n} \binom nk i^{k}-\sum_{k=0}^{n} \binom nk (-1)^{k}i^{k}.$$
| In the case of you being not used to complex numbers, here is an answer with basic algebra.
Let $1+i = \alpha$, $1-i = \beta$, then $\alpha-\beta = 2i$, $\alpha\beta=2$, $\alpha/\beta = i$ and $\beta^2 = -2i$.
\begin{align*}
z_n:= \alpha^n - \beta^n &= (\alpha- \beta)(\alpha^{n-1} + \alpha^{n-2}\beta + \dots + \alpha \beta^{n-2} + \beta^{n-1}) \\
& = 2i \beta^{n-1}\left(1 + (\alpha/\beta) + \dots + (\alpha/\beta)^{n-1}\right) = 2i \beta^{n-1}(1 + i + \dots + i^{n-1})
\end{align*}
We separate cases of $n$ mod 4 since we have $1+i+i^2 + \dots + i^{n-1}$ term, which is periodic of period 4.
(1) $n$ is multiple of $4$: $1+i+i^2 + \dots + i^{n-1} =0$, so $z_n=0$.
(2) $n = 4k+1$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1$, so $z_n = 2i \beta^{4k} = 2i (-2i)^{2k} = (-1)^k\cdot 2^{2k+1}i$.
(3) $n= 4k+2$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1+ i = \alpha$, so $z_n = 2i \beta^{4k+1}\cdot \alpha = 4i \beta^{4k} = (-1)^k\cdot 4^{k+1}i $.
(4) $n= 4k+3$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1+ i -1= i$, so $z_n = 2i \beta^{4k+2}\cdot i =
-2 (-2i)^{2k+1} = (-1)^k \cdot 4^{k+1}i
$
So
$$ z_n = \begin{cases} 0 & n = 4k \\
(-1)^k \cdot 2^{2k+1} i & n = 4k+1 \\
(-1)^k \cdot 4^{k+1}i & n = 4k+2 \text{ or } 4k+3
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Determining the transformation matrix for Jordan normal form. Let $$A=
\begin{pmatrix}
2 & 0 & -4 & -4 \\
0 &4&2&3\\
2&0&8&4\\
-1&0&-2&2\\
\end{pmatrix}$$
I want to find Jordan normal form of $A$ and the transformation matrix $P$. $(P^{-1}AP=J).$
I could find the Jordan form of $A$ is $$\begin{pmatrix}
4&0&0&0\\
0&4&1&0\\
0&0&4&1\\
0&0&0&4\\
\end{pmatrix}$$
But I have difficulty finding the transformation matrix.
I know the procedure for finding the transformation matrix.
Let $P=(p_1 \quad p_2 \quad p_3 \quad p_4)\ (p_i\in \mathbb R^4)$ and $AP=PJ$.
$$AP=(Ap_1 \quad Ap_2 \quad Ap_3 \quad Ap_4)$$
and
\begin{align}
PJ&=
(p_1 \quad p_2 \quad p_3 \quad p_4)
\begin{pmatrix}
4&0&0&0\\
0&4&1&0\\
0&0&4&1\\
0&0&0&4\\
\end{pmatrix}\\
&=(4p_1 \quad 4p_2 \quad p_2+4p_3 \quad p_3+4p_4)
\end{align}
Thus,
$$\begin{cases}
Ap_1=4p_1 \\
Ap_2=4p_2 \\
Ap_3=p_2+4p_3 \\
Ap_4=p_3+4p_4 \\
\end{cases}$$
$p_1$ and $p_2$ are the solutions of $Ax=4x.$ Solving $Ax=4x,$ I get
$$x=
s
\begin{pmatrix}
2 \\
0 \\
-3 \\
2 \\
\end{pmatrix}
+
t
\begin{pmatrix}
0 \\
1 \\
0\\
0\\
\end{pmatrix}.$$
I have to determine $p_1$ and $p_2$ so that they can be linearly independent, thus I decide $p_1=\begin{pmatrix}
2 \\
0 \\
-3 \\
2 \\
\end{pmatrix}$, $p_2=\begin{pmatrix}
0 \\
1 \\
0\\
0\\
\end{pmatrix}$.
But I don't know how I determine $p_3$.
$p_3$ is the solution of $Ay=p_2+4y$, i.e., $(A-4E)y=\begin{pmatrix}
0 \\
1 \\
0\\
0\\
\end{pmatrix}$
I found the solution of this is $y=\begin{pmatrix}
u-1\\
v \\
\frac{1}{2}-\frac{3}{2}u\\
u
\end{pmatrix}$
I have to decide $u,v$ so that $p_1, p_2, p_3$ can be linely independent and
the forth equation $Az=p_3+4z$ can have solutions for $p_4$.
For example, if I decide $u=v=0,$ then I get $p_3=\begin{pmatrix}
-1 \\
0 \\
\frac{1}{2}\\
0
\end{pmatrix}$ and thus $p_1, p_2, p_3$ are linely independent but $Az=p_3+4z$ doesn't have solutions so I cannot find $p_4.$
How should I determine $u,v$ ?
| You need to be careful with your choice of $p_2$; not just any vector will do. Note that
$$p_2 = (A - 4I)p_3 = (A - 4I)^2 p_4,$$
which means we need $p_2$ to be in the columnspace (or range, if you like) of $(A - 4I)^2$. It also needs to be an eigenvector, i.e. an element of $\operatorname{null} (A - 4I)$ so we ought to compute the intersection $$\operatorname{null}(A - 4I) \cap \operatorname{colspace}(A - 4I)^2.$$
First, we compute:
$$(A - 4I)^2=
\begin{pmatrix}
-2 & 0 & -4 & -4 \\
0 &0&2&3\\
2&0&4&4\\
-1&0&-2&-2
\end{pmatrix}^2 =
\begin{pmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 2 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$$
The columnspace is one-dimensional:
$$\operatorname{span} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix},$$
and as you no doubt determined, this is also an eigevector, i.e. it belongs to $\operatorname{null} (A - 4I)$. So, choosing
$$p_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix}$$
will work.
Note that you had this right, but this was something of a fluke. You could have mixed up the order of $p_1$ or $p_2$, or worse, chosen a basis not containing a multiple of $(0, 1, 0, 0)^\top$, in which case neither choice would work. That is the problem you are running into when finding $p_4$ from $p_3$.
Next, to get $p_3$, note that, by similar reasoning, we need it to be both in $\operatorname{null}(A - 4I)^2$ and $\operatorname{colspace}(A - 4I)$. Indeed, choosing $p_3$ such that $(A - 4I)p_3 = p_2$ will guarantee the former. We just need to choose a solution satisfying the latter.
As you showed, possible solutions to the equation $(A - 4I)p_3 = p_2$ are:
$$p_3=\begin{pmatrix}
u-1\\
v \\
\frac{1}{2}-\frac{3}{2}u\\
u
\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1/2 \\ 0 \end{pmatrix} + u\begin{pmatrix} 1 \\ 0 \\ -3/2 \\ 1 \end{pmatrix} + v\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}.$$
Which ones of these belong to the columnspace of $A - 4I$? Note that:
$$\operatorname{colspace}(A - 4I) = \operatorname{span} \left\{\begin{pmatrix} -2 \\ 0 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}\right\}.$$
Clearly, choosing $u = -1$, and $v$ being whatever we want (e.g. $v = 0$) would work, giving us a good choice of $p_3$:
$$p_3 = \begin{pmatrix} -2 \\ 0 \\ 2 \\ -1 \end{pmatrix}.$$
Finally, we just need to solve $(A - 4I)p_3 = p_4$. This will guarantee that $p_4 \in \operatorname{null}(A - 4I)^3$, and this is the only requirement (the requirement that $p_4 \in \operatorname{colspace}(A - 4I)^0 = \operatorname{colspace} I = \Bbb{R}^4$) is trivial!). This time, we should have a solution, if our construction has been correct so far. We then get the augmented matrix
$$\left(\begin{array}{cccc|c}
-2 & 0 & -4 & -4 & -2 \\
0 & 0 & 2 & 3 & 0 \\
2 & 0 & 4 & 4 & 2\\
-1 & 0 & -2 & -2 & -1
\end{array}\right).$$
This has a clear solution right off the bat:
$$p_4 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
So, our constructed Jordan basis should be
$$\left(\begin{pmatrix} 2 \\ 0 \\ -3 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ 0 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}\right).$$
You should verify that this basis does indeed produce the JNF given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
| We have
$$
\begin{array}{l}
S = \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{5 \cdot 6 \cdot 7}} + \frac{1}{{9 \cdot 10 \cdot 11}} + \cdots = \\
= \sum\limits_{0 \le k} {\frac{1}{{\left( {4k + 1} \right)^{\,\overline {\;3\,} } }}} = \sum\limits_{0 \le k} {\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}}}
= \sum\limits_{0 \le k} {t_{\,k} } \\
\end{array}
$$
where $ x^{\,\overline {\,k\,} } $ represents the Rising Factorial
Then we have
$$
\begin{array}{l}
t_{\,0} = \frac{1}{{1 \cdot 2 \cdot 3}} = \frac{1}{6} \\
\frac{{t_{\,k + 1} }}{{t_{\,k} }} = \frac{{\Gamma \left( {4k + 5} \right)}}
{{\Gamma \left( {4k + 8} \right)}}\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}}
= \frac{{\left( {4k + 1} \right)^{\,\overline {\;4\,} } }}{{\left( {4k + 4} \right)^{\,\overline {\;4\,} } }}
= \prod\limits_{j = 0}^3 {\frac{{\left( {4k + 1 + j} \right)}}{{\left( {4k + 4 + j} \right)}}} = \\
= \prod\limits_{j = 0}^3 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 1 + j/4} \right)}}}
= \prod\limits_{j = 0}^2 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 5/4 + j/4} \right)}}} \\
\end{array}
$$
that is the ratio of the consecutive addenda is a rational function of $k$.
So we can represent the sum as the Generalized Hypergeometric function
$$
S = \frac{1}{6}{}_4F_{\,3} \left( {\left. {\begin{array}{*{20}c}
{1,\frac{1}{4},\;\frac{2}{4},\;\frac{3}{4}\;} \\
{1 + \frac{1}{4},\;1 + \frac{2}{4},\;1 + \frac{3}{4}} \\
\end{array}\;} \right|\;1} \right)
$$
which gives $S=0.1723286 \ldots$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 3
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Expected value of sequence of rv I'm stuck with this problem.
Let $X_n$ be a sequence of random variables with $X_1$ the random number chosen at $(0,1)$. $X_2$ be the random number chosen at $(1-X_1,1)$ and so on. This is, $X_n$ is the random number chosen at $(1 - X_{n-1}, 1)$. What is $E[X_n]$?
I have used conditional expectation in my first approach and order statistics in the second, but I'm not sure about my interpretation of the problem. Any hint?
| First of all compute the conditional expectation:
$$\begin{aligned}E[X_n|X_{n-1}]&=\frac{1}{1-(1-X_{n-1})}\int_{(X_{n-1}-1,1)}xdx=\\
&=\frac{1}{X_{n-1}}\frac{1}{2}(1-(X_{n-1}-1)^2)=\\
&=\frac{1-X_{n-1}^2-1+2X_{n-1}}{2X_{n-1}}=\\
&=\frac{2-X_{n-1}}{2}
\end{aligned}$$
We can recast the problem as a sequential problem:
$$E[X_n]=\frac{2-E[X_{n-1}]}{2}\implies (a_{n})_{n \in \mathbb{N}}:a_{n+1}=\frac{2-a_{n}}{2}=1-\frac{a_n}{2}$$
The pattern becomes clear (recall $a_1=0.5$)
$$\begin{align}a_1&=a_1 \\
a_2&=1-\frac{a_1}{2} \\
a_3&=1-\frac{1}{2}+\frac{a_1}{4} \\
a_4&=1-\frac{1}{2}+\frac{1}{4}-\frac{a_1}{8} \\
a_5&=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{a_1}{16}\\
&\mathrel{\vdots} \\
a_n&=\sum_{k=0}^{n-2}\frac{(-1)^{k}}{2^{k}}+\frac{a_1(-1)^{n-1}}{2^{n-1}}, \quad n\geq 2\\&=\frac{1-\bigl(-\frac{1}{2}\bigr)^{n-1}}{\frac{3}{2}}+\frac{1}{2}\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}+\Bigl(\frac{1}{2}-\frac{2}{3}\Bigr)\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}-\frac{1}{6}\Bigl(-\frac{1}{2}\Bigr)^{n-1}\\
&=\frac{2}{3}+\frac{1}{3}\Bigl(-\frac{1}{2}\Bigr)^{n}\end{align}$$
In the limit $E[X_n]\to 2/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Identity $\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$ In the book, Control Systems Engineering - frequency design, the author used the equality $$\phi_{max}=\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$$
Is this some famous identity?
Am I seriously missing out since I've never seen this formula before.
Edit:
Using the formula on the comments:
$$\sin{\arctan{\theta}}=\frac{x}{\sqrt{1+x^2}}$$
Let $\theta= \frac{1-\beta}{2\sqrt{\beta}}$
$$\arctan{\theta}=\arcsin{\frac{\theta}{\sqrt{1+\theta^2}}}
$$
$$\arctan{\theta}=\arcsin{\frac{1-\beta^2}{\beta^2+2\beta+1}}
$$
Thus
$$\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}
$$
| Another way to show two functions are equal.
Let:
$$f(x)=\arctan{\frac{1-x}{2\sqrt{x}}}$$
$$g(x)=\arcsin{\frac{1-x}{1+x}}$$
Both are defined and $C^1$ for $x>0$.
Now,
$$\arctan'(x)=\frac{1}{1+x^2}$$
$$\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$$
Therefore
$$f'(x)=\left(\frac{1-x}{2\sqrt{x}}\right)'\frac{1}{1+\left(\dfrac{1-x}{2\sqrt{x}}\right)^2}=\frac{-2\sqrt{x}-(1-x)\frac{1}{\sqrt{x}}}{4x}\cdot\frac{4x}{4x+1-2x+x^2}\\
=\frac{1}{\sqrt x}\cdot\frac{-x-1}{(1+x)^2}=-\frac{1}{(1+x)\sqrt x}$$
$$g'(x)=\left(\frac{1-x}{1+x}\right)'\frac{1}{\sqrt{1-\left(\frac{1-x}{1+x}\right)^2}}=\frac{-(1+x)-(1-x)}{(1+x)^2}\cdot\frac{1+x}{\sqrt{(1+x)^2-(1-x)^2}}\\
=-\frac{2}{1+x}\cdot\frac{1}{\sqrt{4x}}=-\frac{1}{(1+x)\sqrt x}$$
Hence $f'(x)=g'(x)$ for $x>0$. Since $f(1)=g(1)=0$, we have $f(x)=g(x)$ for all $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Refinement about :$\left(\left(1-x\right)^{-\left(2x\right)}-1\right)\left(\left(x\right)^{-\left(2\left(1-x\right)\right)}-1\right)\geq 1$ Claim :
Let $0.5\leq x<1$ then it seems we have :
$$\left(\left(1-x\right)^{-2x}-\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\right)\left(x^{-2\left(1-x\right)}-1\right)\geq 1$$
Background :
It's a refinement of :
$$\left(\left(1-x\right)^{-\left(2x\right)}-1\right)\left(\left(x\right)^{-\left(2\left(1-x\right)\right)}-1\right)\geq 1\quad (I)$$
With the same constraint as above wich is an inequality due to Vasile Cirtoaje .
My refinement is based on one single and simple fact :
Let $0.5\leq x<1$ then we have :
$$\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\geq 1$$
The proof of this fact is not hard taking logarithm and derivative .
Also Vasile Cirtoaje proved the inequality $(I)$ with some tools wich are not sufficient to show the refinement above .Generalising this simple fact it seems work with Prove that if $a+b =1$, then $\forall n \in \mathbb{N}, a^{(2b)^{n}} + b^{(2a)^{n}} \leq 1$. .
Edit : We have the precious inequality wich simplify the rest on $x\in [0.5,0.75]$ :
$$\left(\left(1-x\right)^{-2x}-\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\right)\left(x^{-2\left(1-x\right)}-1\right)\geq\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}} \geq 1$$
It works also on a larger interval but like this we can use Bernoulli's inequality next.
Edit 2 :
The claim is also :
Let $0.5\leq x \leq 0.75$ then we have :
$$1\geq x^{2\left(1-x\right)}\left(\frac{\left(1-x\right)^{4x}}{2^{4}\left(x\left(1-x\right)\right)^{3}}+1\right)$$
We have also :
Let $0.5\leq x \leq 0.75$ then we have :
$$(1-x)^{4x}\leq \left(2^{2\left(1-x\right)}x\left(1-x\right)^{2}\cdot2\right)^{2}$$
And using Gerber's theorem we have $x\in[0.5,0.75]$:
$$f\left(x\right)=0.5^{2\left(1-x\right)}+2\cdot0.5^{2\left(1-x\right)}\cdot2\left(1-x\right)\left(x-0.5\right)+2\cdot0.5^{2\left(1-x\right)}\cdot\left(2\left(1-x\right)-1\right)\cdot2\left(1-x\right)\cdot\left(x-0.5\right)^{2}+\frac{4}{3}\cdot0.5^{2\left(1-x\right)}\cdot\left(2\left(1-x\right)-1\right)\cdot2\left(1-x\right)\cdot\left(2\left(1-x\right)-2\right)\cdot\left(x-0.5\right)^{3}\geq x^{2(1-x)}$$
Last edit :
We the following inequalities $x\in[0.5,0.55]$
$$h(x)=\left(\frac{2^{-2\cdot\left(1-x\right)}x}{1-2^{\left(0.95-1\right)}\left(\left(1-x\right)2x\right)^{0.95}}\right)\geq x^{2(1-x)}$$
And :
$$\left(\frac{\left(2^{2\left(1-x\right)}x\left(1-x\right)^{2}\cdot2\right)^{2}}{2^{4}\left(x\left(1-x\right)\right)^{3}}+1\right)h(x)\leq 1$$
I think it's not hard using derivatives .
Question :
How to show the claim ?
Thanks.
Reference :
VASILE CIRTOAJE, PROOFS OF THREE OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS, Journal of Nonlinear Sciences and Applications, 4 (2011), no. 2, 130-137
| Sketch of a proof:
Denote
$$A = (1 - x)^{2(1 - x)}, \quad B = (1 - x)^{2x - 1},
\quad C = x^{2x}.$$
The desired inequality is written as
$$\left(\frac{A}{(1 - x)^2} - \frac{B}{2^4x(1 - x)^2C}\right)(x^{-2}C - 1) \ge 1.$$
Denote $a = \ln 2$. Let
\begin{align*}
A_1 &= \frac{p_1x^2 + p_2x + p_3}{(8a^3 - 24a^2
+ 48a - 24) x - 4a^3 - 12}, \\
B_1 &= \frac{(2a -2)x^2 + (-3a + 2)x + a}{(4a - 2)x^2 + (-4a + 2)x + a}, \\
C_1 &= \frac{(4a - 2)x^2 + (-4a + 2)x + a}{(2a - 2)x - a + 2},
\end{align*}
where
\begin{align*}
p_1 &= -4a^4 + 16a^3 -24a^2 + 24a - 24, \\
p_2 &= 4a^4 - 24a^3 + 48a^2 - 48a + 36, \\
p_3 &= - a^4 + 8a^3 - 24a^2 + 30a - 24.
\end{align*}
Fact 1: $A \ge A_1 > 0$ for all $x \in [1/2, 1)$.
Fact 2: $B \le B_1$ for all $x \in [1/2, 1)$.
Fact 3: $C \ge C_1 > 0$ for all $x \in [1/2, 1)$.
Fact 4: $x^{-2}C_1 - 1 > 0$ for all $x \in [1/2, 1)$.
By Facts 1-4, it suffices to prove that
$$\left(\frac{A_1}{(1 - x)^2} - \frac{B_1}{2^4x(1 - x)^2C_1}\right)(x^{-2}C_1 - 1) \ge 1$$
which is true (simply a polynomial inequality).
We are done.
| {
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"url": "https://math.stackexchange.com/questions/4238079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area enclosed between $2$ curves
Question: The area enclosed by the curves $$ y^2 = x \ \text{and} \ y^2 = \ 3x - 1, \ \ \ \ \ \text{where} \ \ 0\leq \ x\leq \frac{1}{2} \ $$
My work:
To begin with I re-arranged the curves in terms of $y$ to the power $1$
$$y_{1} = \sqrt{x}$$
and
$$y_{2} = \sqrt{3x-1}$$
and then proceeded to take the integral for both over the interval $0\leq \ x\leq \frac{1}{2}$
$$f(x_{1}) = \frac{1}{3\sqrt{2}}$$
and
$$f(x_{2}) = \frac{5^{\frac{3}{2}}}{9\sqrt{2}} - \frac{2}{9}$$
And therefore the area between the curves is $\displaystyle\int f(x_{1}) - f(x_{2})$ right?
Is this the correct method? Because the answer that was provided to me was $\frac{2\sqrt{2}}{9}$.
| Say $A_1$ is area between parabola $y^2 = x$ and line $x = \frac{1}{2}$, which you write as $f(x_1)$ and $A_2$ is the area between parabola $y^2 = 3x-1$ and line $x = \frac{1}{2}$
You need to multiple $A_1$ by $2$ as the answer you have is only upper / lower half of the area.
Coming to $A_2$, your mistake is that you are integrating between $x = 0$ and $x = \frac{1}{2}$. The vertex of the second parabola is $(\frac{1}{3}, 0)$. So, you should integrate the second one between $x = \frac{1}{3}$ and $\frac{1}{2}$.
So the second integral should be,
$ \displaystyle 2 \int_{1/3}^{1/2} \int_{0}^{ \sqrt {3x-1}} dy ~dx$
That gives you $A = A_1 - A_2 = \frac{2}{3 \sqrt2} - \frac{2}{9\sqrt2} = \frac{2 \sqrt2}{9}$
Also note that you can directly integrate if you go with respect to $dx$ first. Keep the equations in the form $y^2 = x$ and $y^2 = 3x - 1 \implies x = \frac{y^2+1}{3} ~$.
At intersection $y^2 = x = 3x - 1 \implies x = \frac{1}{2}, y = \pm \frac{1}{\sqrt2}$
So the integral to find area between the curves is,
$\displaystyle \int_{- 1/\sqrt2}^{1/\sqrt2} \int_{y^2}^{ (y^2+1) / 3} ~ dx ~dy$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions:
*
*$f(-x)=-f(x)$
*$f(x+1)=f(x)+1$
*$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
I assume $f(x)=x$ satisfies the conditions. I could prove $f(0)=0 $ and $f(n)=n$ for all $n\in{\mathbb{N}}$, then I didn't know how to go any further.
|
$f:\Bbb{R} \to \Bbb{R}(\text{odd}), f(x+1)=f(x)+1, f(\frac{1}{x})=\frac{f(x)}{x^2}.$
\begin{align}
& f\left(1+\dfrac{1}{x}\right)=f\left(\dfrac{1}{x}\right)+1 = \dfrac{f(x)}{x^2}+1. \tag{1}\label{1} \\
& f\left(\dfrac{x+1}{x}\right)=\dfrac{f\left(\dfrac{x}{x+1}\right)}{\left(\dfrac{x}{x+1}\right)^2}\\
& = \dfrac{(x+1)^2}{x^2}f\left( 1-\dfrac{1}{x+1} \right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-f\left(\dfrac{1}{x+1}\right)\right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-\dfrac{f(x+1)}{(x+1)^2}\right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-\dfrac{f(x)+1}{(x+1)^2}\right). \tag{2}\label{2} \\
\ \\
\therefore \; & (\ref{1}) = (\ref{2}) \Rightarrow \dfrac{f(x)}{x^2}+1=\dfrac{(x+1)^2}{x^2}\left( 1-\dfrac{f(x)+1}{(x+1)^2} \right). \\
\Rightarrow \; & f(x)+x^2=(x+1)^2-f(x)-1, 2f(x)=2x, \boxed{f(x)=x}.
\end{align}
| {
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Evaluate $\int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}dx$ using elementary methods I found this problem while doing some integration from my problem practice book (unkown name). It said to evaluate it using elementary methods.
Please help me evaluating the following integral using elementary methods
$$\int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1} \, dx.$$
I tried to factorize the denominator but it failed. I can't think of any substitution too. I tried to input this at wolfram alpha, and it showed the answer as a summation of a complex function formed using complex cube roots of unity.
Substitution of $x=\sec\theta$ also failed.
I am continuously thinking of this but I'm not getting how to start it. If I succeed I will post it. It will be great if someone could give a beautiful solution to this problem preferably using elementary methods.
I am a high school student in India. I know how to evaluate elementary integrals .
| Presented below is a full solution via elementary integration.
The polynomial $x^4-2x^3-6x-1$ admits two real roots, $a=$ -0.1650 and $b=$ 2.8068, computed with the standard quartic-root formulae below
$$a,b=\frac12\left(1+\sqrt r \pm \sqrt{3-r+{14}{r^{-\frac12}}}\right)\tag1
$$
with $ r= 1+ \left(\frac43\right)^{\frac23}\left( \sqrt[3]{\sqrt{129}+9} - \sqrt[3]{\sqrt{129}-9} \right) $.
Thus, the denominator of the integrand factorizes as
$$x^4-2x^3-6x-1=(x-a)(x-b)[x^2+(a+b-2)x-(ab)^{-1}]
$$
leading to following partial fractionalization
\begin{align}
\frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}
=1+ \frac {p}{x-a} + \frac {q}{x-b}+ \frac{(2-p-q )x-\frac1{ab}(2+\frac {p}a+ \frac {q}b)}{x^2+(a+b-2)x-\frac1{ab}}
\end{align}
with $p=\frac{a^3+a^2+3a-1}{2a^3-3a^2-3}$ and $q=\frac{b^3+b^2+3b-1}{2b^3-3b^2-3}$.
Then, integrate to obtain
\begin{align}
& \int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}dx\\
=& \>x +2p \ln|x-a| +2q \ln|x-b|+(1-p-q)\ln\left(x^2+(a+b-2)x-\frac1{ab}\right)\\
& \>-\frac{(1-p-q)(a+b-2)+\frac2{ab}(1+\frac pa + \frac qb)}{\sqrt{ -\frac{(a+b-2)^2}4-\frac1{ab}}}
\tan^{-1}\frac{x+\frac{a+b-2}2}{\sqrt{ -\frac{(a+b-2)^2}4-\frac1{ab}}}+C
\end{align}
The result, though rather involved in appearance, is expressed in terms of $a$ and $b$, the two real roots given in (1).
| {
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"url": "https://math.stackexchange.com/questions/4240992",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "21",
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Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Then the derivative of the given function is $\frac{1}{\sqrt{1-(\frac{x+\sqrt{1-x^2}}{\sqrt 2}})^2}\cdot \frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Simplifying this gives the final answer $\frac{1}{\sqrt{1-x^2}}$. But the simplication process is quite lengthy and involves some bizarre calculations. Is there tricks/ways to solve these kinds of derivatives that do not involve too much calculations like above?
| If you're confused about why your own method seemingly missed that the derivative is discontinuous, see here. The magic happens in line $4$. We have
$$\begin{split}
f'(x)
&= \frac{1}{\sqrt{1-\left(\frac{x+\sqrt{1-x^2}}{\sqrt 2}\right)^2}}\cdot \frac{1}{\sqrt 2}\left(1-\frac{x}{\sqrt{1-x^2}}\right) \\
&= [\ldots] \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \frac{\sqrt{1-x^2} - x}{\sqrt{\left(\sqrt{1-x^2} - x\right)^2}} \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \frac{\sqrt{1-x^2} - x}{\left\lvert\sqrt{1-x^2} - x\right\rvert} \\
&= \frac{1}{\sqrt{1-x^2}}\cdot \operatorname{sign}\left(\sqrt{1-x^2} - x\right).
\end{split} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequality $$5a^2+5b^2\geq10ab\\ 5b^2+\frac 12c^2\geq\sqrt{10}bc\\ \frac 12c^2+5a^2\geq\sqrt{10}ca$$
Summing them gives $$10a^2+10b^2+c^2\geq\sqrt{10}(\sqrt{10}ab+bc+ca)$$
But this doesn't help.
| Another way.
By C-S
$$10a^2+10b^2+c^2=$$
$$=\sqrt{\left(8a^2+\frac{c^2}{2}+8b^2+\frac{c^2}{2}+2a^2+2b^2\right)\left(\frac{c^2}{2}+8a^2+\frac{c^2}{2}+8b^2+2b^2+2a^2\right)}\geq\sqrt{(2ac+2ac+2bc+2bc+2ab+2ab)^2}=\sqrt{16}=4.$$
The equality occurs, when
$$\left(2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\frac{c}{\sqrt2},\sqrt2a,\sqrt2b\right)||\left(\frac{c}{\sqrt2},2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\sqrt2b,\sqrt2a\right),$$ which gives $a=b=\frac{c}{4}$, which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
Find the equation of the tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
This is in the context of learning implicit differentiation.
First, I apply $\frac{dy}{dx}$ operator to both sides of the equation yielding:
$-\sin^{-2}(x) - \sin^{-1}(y)\frac{dy}{dx} = 0$
Second, I want to solve for $\frac{dy}{dx}$.
$\frac{dy}{dx} = -\sin^{-2}(x)\sin(y)$.
Third, I substitute the point $(0,\frac{1}{2})$ into the above equation to find the slope of the tangent line.
$\frac{dy}{dx}\mid_{(0,\frac{1}{2})} = -\sin^{-2}(0)\sin(\frac{1}{2}) = -0.479$
Finally, I substitute the slope into the point-slope equation of the line to obtain
$y = -0.479x + 0.2395$
Is this correct?
| As
$$
\sin\left(\sin^{-1}x+\sin^{-1}y\right)=\sin\left(\frac{\pi}{6}\right)
$$
gives
$$
\sqrt{1-x^2} y+x \sqrt{1-y^2}=\frac{1}{2}
$$
calling $f(x,y) = \sqrt{1-x^2} y+x \sqrt{1-y^2}$ we have the tangent at $p_0=\left(0,\frac 12\right)$ as
$$
(p-p_0)\cdot\nabla f(p_0)=0
$$
with $p = (x,y),\ \ \nabla f(x,y) = \left\{\sqrt{1-y^2}-\frac{x y}{\sqrt{1-x^2}},\sqrt{1-x^2}-\frac{x y}{\sqrt{1-y^2}}\right\}$ giving
$$
\frac{\sqrt{3} x}{2}+y-\frac{1}{2}=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using logarithmic differentiation or otherwise, differentiate $y = (x −1) (x − 2) (x − 3$)? And show that $y' = 3x^2-12x+11$ So far this is my method:
\begin{align*}
y = (x-1)(x-2)(x-3) & \Longleftrightarrow \ln(y) = \ln(x-1) + \ln(x-2) + \ln(x-3)\\\\
& \Longleftrightarrow \frac{y'}{y} = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\\\\
& \Longleftrightarrow
y' = y\left(\frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\right)
\end{align*}
| The product rule is straightforward.
$$ y= (x-1)(x-2)(x-3)$$
$$ y' = (x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)=$$
$$ x^2-5x+6 + x^2-4x+3+x^2-3x+2 =$$
$$ 3x^2-12x+11$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} \right) + \frac {1}{3^{k+1}}< \frac{1}{2}.$$
Multiplying the $n = k + 1$ step by $3$:
$$ \left( 1 + \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} \right) + \frac {1}{3^{k}}< \frac{3}{2}$$
$$\implies \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} + \frac {1}{3^{k}} < \frac{3}{2} - 1 = \frac {1}{2}.$$
We know this to be true from our inductive hypoethsis. Hence, $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}.$
Is this proof correct?
| $$\sum\limits_{i=1}^n \frac{1}{3^i} = \frac{3n-1}{2 \cdot 3^n}$$
So $$\lim\limits_{n \to \infty} \frac{3n-1}{2 \cdot 3^n} = \frac{1}{2}$$
So for finite $k$:
$$\lim\limits_{n \to k} \sum\limits_{i=1}^n \frac{1}{3^i} < \frac{1}{2}$$
or for all finite $n$:
$$\sum\limits_{i=1}^n \frac{1}{3^i} + \underbrace{\sum\limits_{i=n+1}^\infty \frac{1}{3^i}}_{>0} = \frac{1}{2}$$
For $n \to \infty$, we get the equality.
Hence for all finite $n$ the bound holds.
| {
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If $a_n=b_n = \frac{(-1)^n}{n+1}$, how to deduce that the Cauchy product series converges? If $c_n=\sum_{k=0}^{n}a_kb_{n-k}$, we have to prove that,
$$c_n=(-1)^n\frac{2}{j+2}\left(1+\frac{1}{2}+\cdots+\frac{1}{k+1}\right)$$
But what I get is:
$$c_n=(-1)^n\sum_{k=0}^{n}\frac{1}{(k+1)((n-k)+1)}=(-1)^n\left(\frac{1}{1(n-(-1))}+\frac{1}{2(n-0)}+\frac{1}{3(n-1)}+\cdots+\frac{1}{(n+1)(n-(n-1))}\right)$$
How to proceed further? Thanks in advance.
| We have
$$\tag{*}\sum_{n=0}^\infty c_n = \sum_{n=0}^\infty (-1)^n\underbrace{\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}}_{d_n},$$
where $d_n$ can be expressed in terms of the harmonic sum $H_{n+1} = \sum_{k=1}^{n+1} \frac{1}{k}$, viz.
$$d_n =\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}= \sum_{k=0}^n\frac{1}{n+2} \left(\frac{1}{k+1} + \frac{1}{n+1-k} \right) \\= \frac{1}{n+2} \left(\sum_{k=0}^n \frac{1}{k+1} + \sum_{k=0}^n \frac{1}{n+1-k} \right) = \frac{2}{n+2}\sum_{k=0}^n\frac{1}{k+1} = \frac{2H_{n+1}}{n+2}$$
It is straightforward to show that $d_n$ is decreasing and convergent to $0$ since $H_{n+1} \sim \log (n+1)$ as $n \to \infty$.
Thus, (*) is convergent by the alternating series test.
| {
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Show that for all except perfect powers of 2, it's true.
Let $T_n$ denotes the least natural such that
$$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$
Find all naturals $m$ such that $m\ge T_m$.
I think it’s true for any $m$ except perfect powers of $2.$
For perfect powers of $2.$
Let $m=2^k$. We must have $2^{k+1} \mid T_m(T_m+1)$, hence $2^{k+1} \mid T_m$ or
$2^{k+1} \mid T_m+1$. In both cases, $T_m \geq 2^{k+1}_1 >2^k=m$, therefore all powers of $2$ don't satisfy.
| As Adam Rubinson points out in a comment, you also have to show that $n\ge T_n$ when $n$ is not a power of $2$. So given such a positive integer $n$, we want to find $m\le n$ such that $n\mid T_m=\frac12 m(m+1)$, i.e. $2n\mid m(m+1)$.
So write $n$ as $n=a\cdot2^b$ where $a$ is odd. Then we want $a\cdot 2^{b+1}\mid m(m+1)$. We will construct $m\le n$ such that either $a\mid m+1$ and $2^{b+1}\mid m$; or $a\mid m$ and $2^{b+1}\mid m+1$.
Let $c$ be the inverse of $a\bmod 2^{b+1}$, so that $2^{b+1}\mid ac-1$.
*
*If $c\le 2^b$, put $m=ac-1$. Then $a\mid ac=m+1$ and $2^{b+1}\mid m$; also, $m\le a\cdot 2^b-1<n$.
*If $c>2^b$, put $m=2n-ac$. Then $a\mid n$, so $a\mid m$; and $2^{b+1}$ divides both $2n$ and $ac-1$, so $2^{b+1}\mid 2n-(ac-1)=m+1$. Also, $m=2n-ac<2n-a\cdot 2^b=n$.
For instance, suppose $n=112=7\cdot 2^4$. Then $c=7^{-1} \bmod 32=23$. This is greater than $2^b=16$, so we put $m=2n-ac=224-7\cdot 23=63$. And you can check that $224\mid 63\cdot 64$.
I have deliberately left a logical gap here, to give you something to think about: where does this process go wrong when $n$ is a power of $2$? All the steps look valid, even if $a=1$. So where is the gap?
| {
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Number of terms in product of two monomials with common terms I am trying to find the number of terms in the expression
$$(x+y+z)^{20}(w+x+y+z)^2$$.
I understand that the number of terms in $(x+y+z)^{20} = \binom{22}{2}$ and the number of terms in $(w+x+y+z)^2 = \binom{5}{3}$. If the variables were distinct, then you would simply multiply the terms together. However, in this case it seems that the number of terms would be lower as there may be terms that overlap when you multiply them out.
I am very confused on how to find the number of overlapping terms and any help would be appreciated
| This is a variation continuing OPs approach. OP already observed the number of different terms in
\begin{align*}
(x+y+z)^{20}\quad\text{is}\quad\binom{22}{2}\text{.}
\end{align*}
Since
\begin{align*}
(x+y+z)^{20}&(w+\left(x+y+z\right))^2\\
&=(x+y+z)^{20}(w^2+2w\left(x+y+z\right)+\left(x+y+z\right))\\
&=w^2(x+y+z)^{20}+2w(x+y+z)^{21}+(x+y+z)^{22}
\end{align*}
we obtain
\begin{align*}
\binom{22}{2}+\binom{23}{2}+\binom{24}{2}=231+253+276=760
\end{align*}
| {
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Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.
| You will have to make the numerator as a sum of two functions like this
$\int \frac{x+\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx+\int \frac{\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
The first integral will result in $ln$ and the second in $atan$.
$\frac 1 2 \ln (x^2+x+1)+ \frac 1 {\sqrt{3}} \arctan{\frac {(2x+1)} {\sqrt{3}}}+C$
| {
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Question regarding the largest proper divisor of $x^n+1$ Let $x$ and $n$ be positive integers, with $n$ odd, and let $d$ be the largest proper divisor of $x^n+1$.
QUESTION: Is there a way to characterize the $x$ and $n$ satisfying $d > x^{(n+1)/2}$?
Example: $17^3+1=4914$ has largest [proper] divisor $d=2457 > 17^2 = 289$, but $18^3+1=5833$ has largest [proper] divisor $d=307 \not\gt 18^2=324$.
| Lemma: Let $x,n$ be positive integers with $n\ge 3$. Let $d$ be the largest proper divisor of $x^n+1$. Then $d<x^{(n+1)/2}$ if and only if one of the following is true
*
*$x^n+1$ is prime.
*$x^n+1=pq$ for primes $x^{(n+1)/2}>q>p>x^{(n-1)/2}$
Proof: Assume that $d<x^{(n+1)/2}$. If $x^n+1$ is prime, we are done. Otherwise, let $p$ be the smallest prime factor of $x^n+1$, then
$$p=\frac{x^n+1}{d}>\frac{x^{n}+1}{x^{(n+1)/2}}=x^{(n-1)/2}+\frac{1}{x^{(n+1)/2}},$$
so $p>x^{(n-1)/2}$, since it's an integer coprime to $x$.
Suppose that $p^2\mid x^n+1$, then $(x^n+1)/p^2\le x<p$, which means this quotient must be $1$ and $p^2=x^n+1$. However, this has no solutions by Mihăilescu's theorem. We conclude that $p^2\nmid x^n+1$.
Let $q>p$ be the second largest prime factor of $x^n+1$, then again we find that $(x^n+1)/pq\le x<p$, whence this quotient must be one and $x^n+1=pq$. This completes the proof of one implication, the other one is easy.
Note that for $n$ odd, we have
$$
\frac{x^n+1}{x+1} = \frac{(-x)^n-1}{-x-1}=\sum_{j=0}^{n-1}(-x)^j\in\mathbb{Z},
$$
so $x^n+1$ cannot be prime for $x>1$ and $n\ge 3$ and the first case of the lemma doesn't occur (unless $x=1$ of course)
If we're in the second case, then $x+1\ge p>x^{(n-1)/2}\ge x$, so we must have that $n=3$ and $p=x+1$. It follows that $q=x^2-x+1$. We conclude that:
Theorem: Let $x>1$ and $n\ge 3$ be integers, $n$ odd. Let $d$ be the largest proper divisor of $x^n+1$. Then $d>x^{(n+1)/2}$ if and only if $n=3$, and $x+1$ and $x^2-x+1$ are both prime
| {
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Summation notation $ 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots $ How to write the sum
$$
1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots
$$
in summation (∑) notation.There are 2 series here, one the entire 1 + (1/1+2)... series and the other one in denominator of each term. This summation is a sub-part of a Induction problem.
| $$\sum_{n=1}^{\infty}\frac{2}{n(n+1)}$$ which is equivalent to
$$\sum_{n=1}^{\infty}\frac{1}{T_n}$$
$T_n$ being the $n^{th}$ triangle number.
Clearly, it converges to $2$. This is because the sequence is equivalent to
$$\sum_{n=1}^{\infty}\frac{2}{n}-\frac{2}{n+1}$$ which is telescoping.
| {
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Alternative approach:
From the constraints,
$(a + b)^2 = (c + d)^2 \implies$
[since $a^2 + b^2 = c^2 + d^2$]
$2ab = 2cd \implies ab = cd \implies$
$a(c + d - a) = cd$.
This allows a quadratic equation in $a$ to be formed, that may be solved in terms of $c,d$.
$a^2 - (c+d)a + cd = 0.$
Therefore,
$\displaystyle a = \frac{1}{2}\left[
(c + d) \pm \sqrt{(c + d)^2 - 4cd}\right].$
This equals
$$\frac{1}{2}\left[
(c + d) \pm \sqrt{(c - d)^2}\right].\tag 1$$
Due to the symmetry in the expression in (1) above, you have that without loss of generality, $c \geq d$.
Therefore $\displaystyle a = \frac{1}{2}\left[
(c + d) \pm (c - d)\right].$
The two solutions generated will be $a = c$ or $a = d$.
| {
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How to multiply polynomials in $\mathbb Z[x]/(x^n+1)$? Consider the ring
$$R_q=\mathbb Z_q[X]/(X^n + 1),$$
where $q\equiv 1 \bmod 2n$ and $n$ is a power of $2$.
This is the quotient ring where the cosets are represented by polynomials up to $n-1$ in order.
I'd like to compute $c = a · b \bmod (X^n + 1)$, where $a$ and $b$ are polynomials in this ring, using Wolfram Cloud, or Wolfram Alpha, or anything easy to use.
This is for comparing with some code that I'm writing, that does this using NTT (Fast Fourier Transform for rings).
Is it possible?
For example:
$$(1x^0 + 2x^1 + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 7x^6 + 8x^7)*(2x^0 + 5x^1 + 8x^2 + 11x^3 + 14x^4 + 17x^5 + 20x^6 + 23x^7) \mbox{ mod ($x^8+1$)} = ?$$
(where q = large, here, so all the coefficients fit)
| The command you want in the Wolfram language is PolynomialMod. See documentation here. PolynomialMod[(1x^0+2x^1+3x^2+4x^3+5x^4+6x^5+7x^6+8x^7) * (2x^0+5x^1+8x^2+11x^3+14x^4+17x^5+20x^6+23x^7) , {x^8+1,2}]
will return x^2 + x^4 + x^5 after reducing first mod $x^8+1$, and then reducing coefficients mod $2$.
| {
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Proving $\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$ as $y\to 0$ I want to show that
$$\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$$
as $y\to 0$. Here, $f(y) \approx g(y)$ means that $\lim_{y\to 0} \frac{f(y)}{g(y)}=1$.
By computing the integral and finding asymptotic expression using Mathematica, I think the above is true. (The exact value of the integral is a very complicated function involving hypergeometric functions.) But, how can I prove this asymptotic expression in a clever way?
| With $x = ty$ and $s=-t$, we find
\begin{align*}
&\int_{ - 1/y}^{1/y} {\frac{1}{{\left| {t - 1} \right|^{2/3} }}\frac{1}{{\left| t \right|^{1/3} }}dt} = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt} + \int_{ - 1/y}^0 {\frac{1}{{(1 - t)^{2/3} }}\frac{1}{{( - t)^{1/3} }}dt} \\ & \quad + \int_0^1 {\frac{1}{{(1 - t)^{2/3} }}\frac{1}{{t^{1/3} }}dt}
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt} + \int_0^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds} + \frac{{2\pi }}{{\sqrt 3 }}
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt} + \int_0^1 {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds} + \int_1^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds} + \mathcal{O}(1)
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt} + \int_1^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds} + \mathcal{O}(1)
\\ & = \int_1^{1/y} {\frac{{dt}}{t}} + \int_1^{1/y} {\frac{{ds}}{s}} + \int_1^{1/y} {\left( {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right)\frac{1}{{t^{1/3} }}dt} \\ & \quad + \int_1^{1/y} {\left( {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right)\frac{1}{{s^{1/3} }}ds} + \mathcal{O}(1)
\\ & = - 2\log y + \int_1^{1/y} {\left( {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right)\frac{1}{{t^{1/3} }}dt} \\ & \quad + \int_1^{1/y} {\left( {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right)\frac{1}{{s^{1/3} }}ds} + \mathcal{O}(1).
\end{align*}
Now by the mean value theorem
$$
\left| {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right| \le \frac{2}{{3(t - 1)^{5/3} }}
$$
and
$$
\left| {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right| \le \frac{2}{{3s^{5/3} }}.
$$
Using these for $t>2$, say, and $s>1$, we see that the two integrals are $\mathcal{O}(1)$. Hence the final result is $-2\log y+ \mathcal{O}(1)$ as $y\to 0+$.
| {
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Prove $\left(x^y + y^x \right)\left( \frac{1}{x} + \frac{1}{y} \right)\ge 4$ For arbitrary $x, y > 0$, prove
$$
\left(x^y + y^x \right) \left( \frac{1}{x} + \frac{1}{y} \right) \ge 4
$$
By plotting, it seems true and somewhat tight, but I cannot find a proof for it.
If $x\ge 1,\ y \ge 1$ or $x\le 1,\ y\le 1$, we have
$$
\left(x^y + y^x \right) \left( \frac{1}{x} + \frac{1}{y} \right)\ge \left(x^1 + y^1 \right) \left( \frac{1}{x} + \frac{1}{y} \right) \ge 4
$$
So it suffices to show the inequality for the case $0 < x < 1,\ y > 1$.
Fixing $y$, I tried
$$f(x) = \left(x^y + y^x\right) (x + y) - 4xy$$
and find that $f''(x) > 0$, so we can obtain a minimal point $x_0$ on $(0,1)$ such that $f'(x_0) = 0$. Then we need to prove $f(x_0)\ge 0$. But I have no idea for further progress.
| Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$(x^y + y^x)(1/y + 1/x)
\ge \left(\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}}\right)^2.$$
It suffices to prove that
$$\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}} \ge 2. \tag{1}$$
This inequality has been proved
in show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$.
Remark 0: I have just noticed that @Erik Satie gave an alternative proof for (1) which is simpler than mine. See: show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$
Remark 1: In the link (Chinese website, provided by @mengdie1982) https://www.zhihu.com/question/489824366,
qiqi1509 gave another proof for (1).
The sketch of qiqi1509's proof:
(1) is written as
$$\frac{x}{x + y}x^{(y - 1)/2} + \frac{y}{x + y} y^{(x - 1)/2}
\ge \frac{2\sqrt{xy}}{x + y}.$$
Since $u\mapsto \mathrm{e}^u$ is convex, using Jensen's inequality, we have
$$\frac{x}{x + y}x^{(y - 1)/2} + \frac{y}{x + y} y^{(x - 1)/2}
\ge \mathrm{e}^{\frac{x}{x + y}\frac{y - 1}{2}\ln x
+ \frac{y}{x + y}\frac{x - 1}{2}\ln y}.$$
It suffices to prove that
$$\frac{x}{x + y}\frac{y - 1}{2}\ln x
+ \frac{y}{x + y}\frac{x - 1}{2}\ln y
\ge \ln \frac{2\sqrt{xy}}{x + y}. \tag{2}$$
$\phantom{2}$
Remark 2: (2) is written as
$$\ln \frac{x + y}{2} \ge \frac{2x + (1 - x)y}{2x + 2y}\ln x
+ \frac{x + (2 - x)y}{2x + 2y}\ln y. \tag{3}$$
Perhaps, there is a simpler (or alternative) proof for (3) (with $0 < x \le 1 \le y$).
$\phantom{2}$
Remark 3: A similar inequality is
$$\ln \frac{x + y}{2} \ge \frac{x + 1}{2x + 2y}\ln x
+ \frac{y + 1}{2x + 2y}\ln y. \tag{4}$$
See: Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$
I think qiqi1509's idea works for it.
| {
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Can trigonometric substitution be used to solve this integral? The integral in question is $$\int \frac{x+1}{9x^2+6x+5}dx.$$ I first completed the square in the denominator giving $(3x+1)^2+4$ and proceeded to perform a $u$-substitution with $u = 3x+1$, $du=3~dx$, and $x=(u-1)/3$. After simplifying, I was left with $$\frac{1}{9}\int\frac{u+2}{u^2+4}du.$$ It is at this point I used trigonometric substitution with $u = 2\tan\theta$ and $du = 2\sec^2\theta~d\theta$ (I'm aware the integral can be written as $\frac 1 9\int\frac{u}{u^2+4}du+\frac 1 9\int\frac{2}{u^2+4}du$ and solved this way). After performing the trig substitution, I was left with $$\frac 1 9\int(\tan\theta + 1)~d\theta = \frac 1 9 \ln|\sec\theta|+\frac 1 9 \theta + C.$$ Rewriting everything in terms of $x$ gave me $$\frac 1 9\ln\left(\frac{(3x+1)^2+4}{2}\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C,$$ which is incorrect. The correct answer is $$\frac{1}{18}\ln\left(9x^2+6x+5\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C.$$ What went wrong with my trig substitution?
| There is a simpler way to determine this integral: rewrite first the numerator:
$$x+1=\frac 1{18}(18x+6)+\frac 23,$$
split the integral in two:
\begin{align}
\int \frac{x+1}{9x^2+6x+5}\,\mathrm dx&=\frac 1{18}\int \frac{18x+6}{9x^2+6x+5}\,\mathrm dx+\frac23\int\frac{\mathrm dx}{(3x+1)^2+4}\\
&=\frac 1{18}\ln(9x^2+6x+5)+\frac23\int\frac{\mathrm dx}{(3x+1)^2+4}\\
&=\frac 1{18}\ln(9x^2+6x+5)+\frac29\int\frac{\mathrm d(3x+1)}{(3x+1)^2+4},
\end{align}
and use the standard formula
$$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\,\arctan \Bigl(\frac xa\Bigr).$$
| {
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How to calculate sum, if the recurrence is given? Let $a_n=4a_{n-1}-a_{n-2}$, for $n\geq 3$. Prove that $\sum\limits_{n=1}^\infty \cot^{-1}{a_n^2}=\frac{\pi}{12}$, if $a_1=2$, $a_2=8$.
This recurrent formula is easy to solve. Characteristic equation $x^2-4x+1=0$ has two roots, $2\pm \sqrt{3}$. Therefore general solution is $a_n=a(2+\sqrt{3})^n+b(2-\sqrt{3})^n$, where a and b can be obtained from $a_1=2$ and $a_2=8$. But, I have no idea what to do next. How to calculate sum $\sum\limits_{n=1}^\infty \cot^{-1}{a_n^2}$?
Any help is welcome. Thanks in advance.
| Note: This is one part of a solution, the other part is in cineel's answer.
The first elements of the sequence are
$$
a_n = 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, \ldots
$$
Using the identity
$$
\cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1} \left( \frac{xy-1}{x+y}\right)
$$
we compute the first few partial sums:
$$
\begin{align}
\sum_{k=1}^2 \cot^{-1}(a_k^2) &= \cot^{-1}(2^2)+\cot^{-1}(8^2) = \cot^{-1}\left( \frac{4 \cdot 64 -1}{4+64}\right)
= \cot^{-1}\left( \frac{15}{4}\right) \\
\sum_{k=1}^3 \cot^{-1}(a_k^2) &= \cot^{-1}\left( \frac{15}{4}\right)
+ \cot^{-1}(30^2) = \cot^{-1}\left( \frac{56}{15}\right) \\
\sum_{k=1}^4 \cot^{-1}(a_k^2) &= \cot^{-1}\left( \frac{56}{15}\right)
+ \cot^{-1}(112^2) = \cot^{-1}\left( \frac{209}{56}\right) \\
\sum_{k=1}^5 \cot^{-1}(a_k^2) &= \cot^{-1}\left( \frac{209}{56}\right)
+ \cot^{-1}(418^2) = \cot^{-1}\left( \frac{780}{209}\right)
\end{align}
$$
Now the arguments of $\cot^{-1}$ on the right happen to be ratios of consecutive $a_n$:
$15/4 = a_3/a_2$, $56/15 = a_4/a_3$, $209/56 = a_5/a_4$, etc. This leads to the following
Conjecture: $$\sum_{k=1}^n \cot^{-1}(a_k^2)= \cot^{-1}\left( \frac{a_{n+1}}{a_n}\right) $$
If that were true then the sum of the infinite series can be computed as
$$
\sum_{k=1}^\infty \cot^{-1}(a_k^2) = \lim_{n \to \infty} \cot^{-1}\left( \frac{a_{n+1}}{a_n}\right) = \cot^{-1}(2 + \sqrt 3) = \frac{\pi}{12} \, .
$$
and we are done. So it remains to prove the conjecture.
The formula is true for $n=1$. In order to prove it via induction, we have to show that
$$
\frac{\frac{a_{n+1}}{a_n} a_{n+1}^2-1}{\frac{a_{n+1}}{a_n} + a_{n+1}^2}
= \frac{a_{n+2}}{a_{n+1}}
$$
holds for all $n$, or equivalently,
$$ \tag{$*$}
\frac{a_{n+1}^3-a_n}{1+a_n a_{n+1}} = a_{n+2} \, .
$$
I have verified $(*)$ with PARI/GP for all $n \le 30$, so I am fairly sure that it is correct. However, I haven't been able yet to prove it from the recursion formula for the $a_n$.
Addendum: $(*)$ has been proved in cineel's answer.
| {
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"url": "https://math.stackexchange.com/questions/4270022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^k \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}\right)$$
Then calculating a few numbers for $n$ we get:
$$\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} \right) + \left(\frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) + \left(\frac{1}{6} - \frac{1}{4} + \frac{1}{10}\right) . . . \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{n+2}\right)$$
The first two fractions cancel out in the first bracket and we're left with $\frac{1}{6}$, as for the second bracket the first fraction is cancelled out by the second fraction in the third bracket.
I have noticed that the first fractrion so $\frac{1}{2n}$ cancel out by every even term in the denominator for $-\frac{1}{n+1}$ so the equation becomes:
$$\left(-\frac{1}{2n+1}+\frac{1}{n+2}\right) = \left(\frac{n-1}{(2n+1)(n+2)} \right)$$
Have I approached this correctly? I would greatly appreciate some assistance on any improvements!
| Your partial fraction is incomplete. At least you can further decompose $\frac{1}{n(n+2)}$. Anyway the correct decomposition is $\frac{1}{2}(\frac{1}{n} + \frac{1}{n+2}) - \frac{1}{n+1}$.
Now for any $3 \le n \le k$, $\frac{1}{n}$ will appear twice with coefficient $\frac{1}{2}$ and once with coefficient $-1$, therefore they all cancel. What are left? Only $1/1, 1/2, 1/(k+1), 1/(k+2)$ have nonzero coefficients. And they are $1/2, 1/2-1, 1/2-1, 1/2$ separately.
BTW, this type of problems can be usually quickly "solved" by WolframAlpha like this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}$ Edit: A friend of mine used a contour integral to solve this problem and his final answer matches with mine. It seems that Wolphram´s is incorrect
I saw the following integral here and went to proof it.
$$\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}$$
I started using the following identity
$$\frac{1}{e^{2\pi x}+1}=\frac{1}{e^{2\pi x}-1}-\frac{2}{e^{4\pi x}-1} \tag{1}$$
To rewrite the integral as following
$$\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2 \pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{4\pi x}-1}\,dx \tag{2}$$
In my previous post I established the value of the first integral of the R.H.S. of $(2)$
$$\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{2 \pi x}-1} d x=\zeta^{\prime}(-1)+\frac12\ln 2 +\frac12 \ln \pi-\frac{3}{4} \tag{3}$$
So it remains to evaluate
$$\int_0^\infty\frac{x \ln(1+x^2)}{e^{4\pi x}-1}\,dx$$
To this end, I followed the same procedure I used in the previous post. Consider the integral
$$J(z)=\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{4\pi x}-1}\,dx \tag{4}$$
And differentiate it w.r. to $z$ to get
$$J^{\prime}(z)=\int_0^\infty\frac{2zx }{(z^2+x^2)(e^{4\pi x}-1)}\,dx \tag{5}$$
Then comparing it with Binet´s formula
$$\int_0^\infty\frac{x }{(z^2+x^2)(e^{2\pi x}-1)}\,dx=\frac{\log(z)}{2}-\frac{\psi(z)}{2}-\frac{1}{4z} \tag{6}$$
And after a change in parameter and a change of variable we find that
$$\int_0^\infty\frac{2sx }{(s^2+x^2)(e^{4\pi x}-1)}\,dx=s \ln(2s)-s\psi(2s)-\frac{1}{4} \tag{7}$$
Now, integrate $(7)$ with respect to $s$ from $0$ to $z$ and then let $z \to 1$, I got
$$\int_0^\infty\frac{x \ln(1+x^2)}{e^{4\pi x}-1}\,dx=\frac{\ln \pi}{4}-\frac{36}{48}+\frac{35}{48} \ln2+\frac14 \zeta^{\prime}(-1) \tag{7}$$
Since I could not find the answer of this integral nowhere, I went to Wolfram Alpha to see whether it matches with it´s solution, but it doesn´t! Nevertheless, I sticked to it, and together with $(3)$ I plugged it´s values in $(2)$.
To get the final answer
$$\begin{aligned}
\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx&=\zeta^{\prime}(-1)+\frac12\ln 2 +\frac12 \ln \pi-\frac{3}{4}-2\left(\frac{\ln \pi}{4}-\frac{36}{48}+\frac{35}{48}\ln2+\frac14 \zeta^{\prime}(-1) \right)\\
&=\frac12\zeta^{\prime}(-1)-\frac{23}{24}\ln 2+\frac{18}{24}\\
&=\frac12\left(\frac{1}{12}-\ln A \right)-\frac{23}{24}\ln 2+\frac{18}{24}\\
&=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2} \qquad \blacksquare
\end{aligned}
$$
Now, this answer matches with the answer from where I saw the integral, but it does not match Wolfram Alpha´s answer.
The obvious question is: Which one is correct??
| It can be shown by the Abel-Plana formula, hoping for convergence, that:
$$\sum_{x\ge0} f(x)=\frac{f(0)}2 +\int_0^\infty f(x)dx+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}dx$$
Note that when $f(x)=-\frac i2 x \ln(1-x^2)\implies f(ix)-f(-ix)=x\ln(x^2+1)$:
Let’s plug in to the formula:
$$\sum_{x\ge0} -\frac i2 x \ln(1-x^2) =\frac{-\frac i2 0 \ln(1-0^2)}2 +\int_0^\infty -\frac i2 x \ln(1-x^2) dx+i\int_0^\infty\frac{-\frac i2 x \ln(1-x^2)- -\frac i2 x \ln(1-x^2)}{e^{2\pi x}-1}dx\implies \sum_{x\ge0} -\frac i2 x \ln(1-x^2) -0- \int_0^\infty -\frac i2 x \ln(1-x^2) dx =i\int_0^\infty\frac{x\ln(x^2+1)}{e^{2\pi x}-1}dx$$
Let’s collect our results:
$$-i\sum_{x\ge0} -\frac i2 x \ln(1-x^2) - -i\int_0^\infty -\frac i2 x \ln(1-x^2) dx =\int_0^\infty\frac{x\ln(x^2+1)}{e^{2\pi x}-1}dx= \frac12\int_0^\infty x \ln(1-x^2) dx -\frac12 \sum_{x\ge0} x \ln(1-x^2) $$
This was an attempt using the formula. There should be a better method to use. Please correct me and give me feedback!
| {
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"url": "https://math.stackexchange.com/questions/4272846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Error of $\left(1+\frac{1}{n}\right)^n$ For a positive $\varepsilon$, how do you find $n$ that satisfies $e-\left(1+ \frac{1}{n}\right)^n<\varepsilon$?
Below is the context.
Definition
$$e=\lim_{n\rightarrow\infty}\left(1+ \frac{1}{n}\right)^n$$
Attempt
It is proved that $e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$. On the other hand, $\left(1+ \frac{1}{n}\right)^n = 1+1+\left(1-\frac{1}{n}\right)\frac{1}{2!}+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\frac{1}{3!}+\dots+\left(1-\frac{1}{n}\right)\dots\left(1-\frac{n-1}{n}\right)\frac{1}{n!}$.
So,
$$\begin{aligned}
e-\left(1+ \frac{1}{n}\right)^n &= \sum_{k=1}^n \left( 1 - \prod_{i=1}^k \left(1-\frac{i}{n}\right)\right)\frac{1}{(k+1)!} +r(n)\\
&<\sum_{k=1}^n \left( 1 - \left(1-\frac{k}{n}\right)^{k}\right)\frac{1}{(k+1)!} + r(n)
\end{aligned}$$
$r(n)$ can be evaluated using Taylor's theorem. The problem is to find $n$ which satisfies $\sum_{k=1}^n \left( 1 - \left(1-\frac{k}{n}\right)^{k}\right)\frac{1}{(k+1)!}<\varepsilon$.
Motivation and where the question comes from
This is a natural question. It is normal to evaluate the error for all limits.
My background
Undergraduate.
| This is an alternate way to derive an $n$ that works, reasonable close to the optimal one given in Claude Leibovici's answer, without using any special function.
Notice for $x > 0$,
$$\left(\frac1x\right)' = -\frac1{x^2} < 0\quad\text{ and }\quad \left(\frac1x\right)'' = \frac{2}{x^3} > 0$$
$\frac1x$ is strictly decreasing and strictly convex over $(0,\infty)$. This implies for any $n > 0$, its integral over $[n,n+1]$ is sandwiched between its values at lower end and midpoint. i.e
$$
\frac1n > \int_n^{n+1} \frac{dx}{x} = \log\left(1 + \frac1n\right) >\frac{1}{n+\frac12}$$
This is equivalent to
$$\left(1+\frac1n\right)^n < e < \left(1+\frac1n\right)^{n+\frac12}$$
From this, we can deduce
$$0 < e - \left(1+\frac1n\right)^n
< \left(1+\frac1n\right)^n\left(\sqrt{1+\frac1n}-1\right) < \frac{e}{2n}$$
This implies for any integer $n \ge \left\lceil \frac{e}{2\epsilon}\right\rceil$, we will have $0 < e - \left(1+\frac1n\right)^n < \epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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common ratio of a geometric sequence formed from three terms of an arithmetic sequence The problem is to prove that if we have an arithmetic sequence and we choose $3$ numbers from it which form a geometric sequence as well, and also in the new geometric sequence these $3$ numbers are after each other with no gap, the common ratio ($r$) can be calculated like this:
$A(m),A(n)$ & $A(p)$ are the chosen numbers, so the common ratio is $= \dfrac{p-n}{n-m}$
Please help me prove it!
Here is the original question in Persian:
| Let $(a_k)$ be an arithmetic sequence with initial term $a_1$ and common difference $d$. Then its $k$th term is given by $a_k = a_1 + d(k - 1)$. Hence,
\begin{align*}
a_m & = a_1 + d(m - 1)\\
a_n & = a_1 + d(n - 1)\\
a_p & = a_1 + d(p - 1)
\end{align*}
Since $a_m, a_n, a_p$ are successive terms of a geometric sequence with common ratio $r$,
$$r = \frac{a_p}{a_n} = \frac{a_n}{a_m}$$
since $r = \dfrac{a_{k + 1}}{a_k}$ for each positive number $k$. Substituting for $a_m, a_n,$ and $a_p$ yields
$$r = \frac{a_1 + d(p - 1)}{a_1 + d(n - 1)} \tag{1}$$
and
$$r = \frac{a_1 + d(n - 1)}{a_1 + d(m - 1)} \tag{2}$$
If we multiply both sides of equation $1$ by $a_1 + d(n - 1)$, we obtain
\begin{align*}
ra_1 + rd(n - 1) & = a_1 + d(p - 1)\\
ra_1 + nrd - rd & = a_1 + pd - d \tag{3}\\
\end{align*}
If we multiply both sides of equation $2$ by $a_1 + d(m - 1)$, we obtain
\begin{align*}
ra_1 + rd(m - 1) & = a_1 + d(n - 1)\\
ra_1 + mrd - rd & = a_1 + nd - d \tag{4}
\end{align*}
Subtracting equation $4$ from equation $3$ yields
$$nrd - mrd = pd - nd$$
If $d \neq 0$, we obtain
\begin{align*}
nr - mr & = p - n\\
(n - m)r & = p - n\\
r & = \frac{p - n}{n - m}
\end{align*}
which is valid if the arithmetic sequence is not constant.
If $d = 0$, both sequences are constant, and the claim fails to hold. For instance, let $(a_k)$ be the arithmetic sequence with $a_1 = 1$ and $d = 0$. Then $a_k = 1$ for each positive integer $k$, so $a_1 = a_2 = a_6 = 1$ and $r = 1$, but
$$r = 1 \neq \frac{6 - 2}{2 - 1} = \frac{4}{1} = 4$$
| {
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"url": "https://math.stackexchange.com/questions/4274012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How could I use telescoping to find $f(1) + f(2) + f(3) + \cdots + f(100) $? How could I use telescoping to find $f(1) + f(2) + f(3) + \cdots + f(100) $?
Let $f(x)$ be a function defined by $f(x) = x^6 - 3x^5 + 5x^4 - 5x^3 + 3x^2 - x$. Compute the sum of the base-ten digits of the sum $f(1) + f(2) + f(3) + \cdots + f(100)$.
I tried telescoping by creating a new function $g(x)$ such that $f(x) = g(x) - g(x-1) $, which would then lead to the initial sum becoming $g(100) - g(0)$, but I can't figure out what $g(x)$ is.
Any help is appreciated!
| The key observation is that $$7f(x) = 7x^6 - 21x^5 + 35x^4 - 35x^3 + 21x^2 - 7x$$ has coefficients that look suspiciously like terms in Pascal's triangle; specifically, $\binom{7}{k}$ for $k \in \{1, 2, \ldots, 6\}$. So in fact $$7f(x) = x^7 - (x-1)^7 - 1,$$ and we no longer need to define $g$ because $$\sum_{x=1}^{100} f(x) = \frac{1}{7} \sum_{x=1}^{100} \left( x^7 - (x-1)^7 - 1 \right),$$ and the rest is straightforward.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx
&= \int_0^{\frac{\pi}{2}} \frac{x}{1-\cos^2 x}\sin x\ dx \\
&= \sum_{n=0}^{\infty} \int_0^{\frac{\pi}{2}} x\sin x \ \cos^{2n}x \ dx \\
&= \sum_{n=0}^{\infty}\frac{1}{2n+1} \int_0^{\frac{\pi}{2}} \cos^{2n+1}x \ dx \\
&= \sum_{n=0}^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}
\end{align*}
which is close but not quite there.
| $$
\begin{aligned}
I&=\int_{0}^{\pi / 2} \frac{x}{\sin (x)} d x\\&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} d x \\
&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} \frac{1}{\frac{\cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}} d x \\
&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos ^{2}\left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)} d x\\
&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x \sec ^{2}\left(\frac{x}{2}\right)}{\tan \left(\frac{x}{2}\right)} d x\\
&=2 \int_{0}^{1} \frac{\arctan (x)}{x} d x \qquad\left(\tan \left(\frac{x}{2}\right) \to x \right)\\
&=2 \int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)} x^{2 k} d x \\
&=2 \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2}}\\
&=2 G \qquad \blacksquare
\end{aligned}
$$
| {
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"url": "https://math.stackexchange.com/questions/4281469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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Proof recursion is a subset of Lucas numbers I need to prove that the recursion $a_n=\frac{a_{n-1}^2+5}{a_{n-2}}$ for $a_0=2,a_1=3$ are the Lucas numbers with even index. I would like to use induction, but I got a fraction that I'm not sure how to simplify into the clean recursion for the Lucas numbers. Is induction the way to go here, or is there a way to manipulate some formula for the Lucas numbers to show that the recursion works?
| We know that
$$L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n \quad \forall\ n \in \mathbb{N}$$
you can see it here for example. To simplify the calculations, let
$$x=\frac{1+\sqrt{5}}{2}\quad and\quad y=\frac{1-\sqrt{5}}{2}$$
where $x^2+y^2=3$ and $xy=-1$. Then, (with $n\geq 2$)
\begin{align*}
a_{n}\cdot a_{n-2}-a_{n-1}^2&=L_{2n}\cdot L_{2n-2}-L_{2n-1}^2\\
&=(x^{2n}+y^{2n})\cdot (x^{2n-2}+y^{2n-2})-(x^{2n-1}+y^{2n-1})^2\\
&=x^{4n-2}+x^2\underbrace{(xy)^{2n-2}}_{1}+y^2\underbrace{(xy)^{2n-2}}_{1}+y^{4n-2}-x^{4n-2}-2\underbrace{(xy)^{2n-1}}_{-1}-y^{4n-2}\\
&=x^2+y^2+2\\
&=5
\end{align*}
Thus we have $a_n\cdot a_{n-2}-a_{n-1}^2=5$, that is,
$$a_n=\frac{a_{n-1}^2+5}{a_{n-2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Triangular prism in a sphere A sphere of unit radius touches all edges of some triangular prism. What can be the volume of this prism?
The prism is regular because it is inscribed in the sphere. We express the radius by the Pythagorean theorem $1=\sqrt{(0,5 a)^{2}+\left(\frac{a}{\sqrt{3}}\right)^{2}}$. So, $a_{1}=-\frac{2 \sqrt{21}}{7}, a_{2}=\frac{2 \sqrt{21}}{7}$. And we have $\left(\frac{2 \sqrt{21}}{7}\right)^{2} \times 0,5 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{7}$. Then the answer is $\frac{18 \sqrt{7}}{49}$. Am I right?
| If the edge length is $2 a $, then the circumcircle of the the triangular face (which is an equilateral triangle) is
$ R = \sqrt{ 1^2 - a^2 }$
but $ R = \dfrac{2a}{\sqrt{3}} $
Hence, $1 - a^2 = \dfrac{4 a^2}{3} $
making $a^2 = \dfrac{3}{7} $
The volume is $(2 a) \dfrac{\sqrt{3}}{4} (4 a^2) = (6) (\dfrac{1}{\sqrt{7}}) \dfrac{3}{7} = \dfrac{ 18 }{7 \sqrt{7}} = \dfrac{ 18 \sqrt{7}}{49} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\cos^{-1}(-x)=\pi-\cos^{-1}x$ without geometry. Let
$$
\cos^{-1}x=a
\implies x=\cos a
$$
and
$$
\cos^{-1}(-x)=b
\implies -x=\cos b
$$
Hence we have
$$
\cos a+\cos b=0
$$
Using
$$
\cos(A+B)+\cos(A-B)=2\cos A\cos B
$$
with
$$
A=\frac{a+b}{2} \\
\text{ and } \\
B=\frac{a-b}{2}
$$
we get
$$
\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)=0
$$
with which
$$
\cos\left(\frac{a+b}{2}\right)=0 \text{ or } \cos\left(\frac{a-b}{2}\right)=0\\
\implies a=\pi-b \text{ or } a=\pi+b
$$
Now, to choose between the two, I'm making the below argument:
To find the inverse of a function, the function has to be one-to-one. Hence in our case, both $\cos a$ and $\cos b$ have to be one-to-one, which is possible only when
$$
n\pi\leq a,b \leq (n+1)\pi \text{, }n\in\mathbb{Z}
$$
From the above, we get,
$$
-n\pi \leq \pi-b \leq (-n+1)\pi
$$
and
$$
(n+1)\pi \leq \pi+b \leq (n+2)\pi
$$
The ranges can be reconciled for $a$ and $\pi-b$ by taking $n=0$ and we get
$$
0 \leq a \leq \pi \\
0 \leq \pi-b \leq \pi
$$
But no value of $n\in\mathbb{Z}$ can simultaneously reconcile the ranges of $a$ and $\pi+b$.
Therefore, we conclude that
$$
\cos^{-1}(-x)=\pi-\cos^{-1}(x)
$$
From this I also learnt that the inverse function is meaningful when the angle is discussed in the range $[0,\pi]$.
Is this line of argument mathematically fool-proof?
| Let $f(x) = \cos^{-1}(-x) + \cos^{-1}x$. Note that
$$
f^{\prime}(x) = -\dfrac{1}{\sqrt{1 - (-x)^2}}\cdot (-1) - \dfrac{1}{\sqrt{1 - x^2}} = 0 \quad \Rightarrow \quad f(x) = k, \quad k \in \mathbb{R}
$$
For $x = 0$, we have $k = \cos^{-1}(-0) + \cos^{-1}0 = \pi/2 + \pi/2 = \pi$. Thus,
$\cos^{-1}(-x) = \pi - \cos^{-1} x$.
| {
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"url": "https://math.stackexchange.com/questions/4285963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the general solution to $y''+2y'+4y=\sin x \cos x + 3x^2$ Given this equation $y''+2y'+4y=\sin x \cos x + 3x^2$, how do I find the general solution? It is obvious that this is a differential equation, so that's a start. However, this is a second order differential equation. And it doesn't have any elementary solutions. But then I realized I can convert $\sin x \cos x$ to $\frac{\sin (2x)}{2}$. Now, I can solve for $\lambda^2 + 2\lambda + 4 = 0$.
Now I have started on solving this equation and trying to get a general solution. So now I need some help, thank you!
| First, solve the homogeneous equation. Here, you got $\lambda^2+2\lambda+4=0$, which gives $\lambda_1 = -1+ i\sqrt3, \lambda_2 = -1-i\sqrt3,$ so the homogeneous solution is $$y_h(x)=c_1 e^{-x}\cos(\sqrt3 x) + c_2 e^{-x}\sin(\sqrt3 x).$$
Now to get the particular solution, note that we have $f(x)=\frac{1}{2} \sin(2x) + 3x^2$, so using undetermined coefficients we will guess a solution of the form $y_p(x)=a_1 + a_2 x + a_3 x^2 + a_4 \cos(2x) + a_5 \sin(2x).$
Note that then:
$$y'_p(x) =a_2 +2a_3x - 2a_4 \sin(2x) + 2a_5 \cos(2x)$$
and
$$y''_p(x) = 2a_3 - 4a_4 \cos(2x) -4a_5 \sin(2x).$$
Replacing in the original ODE:
$$2a_3 - 4a_4 \cos(2x) -4a_5 \sin(2x) + 2(a_2 +2a_3x - 2a_4 \sin(2x) + 2a_5 \cos(2x)) + 4 (a_1 + a_2 x + a_3 x^2 + a_4 \cos(2x) + a_5 \sin(2x)) = \frac{1}{2} \sin(2x) + 3x^2$$
simplifying:
$$4a_1 + 2a_2 +2a_3 + (4a_2 + 4a_3) x + 4a_3 x^2 + 4a_5 \cos(2x) - 4a_4 \sin(2x) = 3x^2 + \frac{1}{2} \sin(2x)$$
Comparing coefficients:
\begin{align}
4a_1 + 2a_2 +2a_3 &= 0\\
4a_2 + 4a_3 &= 0\\
4a_3 &= 3\\
4a_5 &= 0\\
-4a_4 &= 1/2\\
\end{align}
Solving this system gives $a_1 = a_5 = 0, a_2 = -a_3 = 3/4, a_4=-1/8$
Replacing those values in $y_p(x)$ gives
$$y_p(x) = \frac{3}{4} x^2 - \frac{3}{4}x - \frac{1}{8} \cos(2x)$$
In that case, the general solution will be
$$y(x) = y_h(x) + y_p(x)$$
or
$$y(x) = \frac{3}{4} x^2 - \frac{3}{4}x - \frac{1}{8} \cos(2x) + c_1 e^{-x} \cos(\sqrt3 x) + c_2 e^{-x} \sin(\sqrt3 x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Fourth Degree Pell Equation I got stuck at the following of my research problem:
Prove that only solution to equation $4b^2-3a^4=1$ for odd positive integers $a$, $b$ is $(1,1)$.
I made factorization -->$3a^4 = (2b-1)(2b+1)$
from here since 2b-1 and 2b+1 are coprime,
$$2b-1 = 3x^4, 2b+1 = y^4$$
or
$$2b-1 = x^4, 2b+1 = 3y^4$$
obviously first one has no solution from modulo 3. Second one yields $x^4+2 = 3y^4$ which is more specific than original equation.
Another approach that might be useful is to rewrite the original equation as $(2b)^2 - 3(a^2)^2 = 1$ , we get a recursive equation for values of $a^2$ by pell and it is sufficient to prove that there is no odd perfect squares other than 1 in this recursive sequence (a similar method is used here https://arxiv.org/pdf/1705.03011.pdf )
I am seeking an elementary solution which doesn't include algebraic NT. Thanks.
| Comment: May be this idea helps:
$4b^2-3a^4=1 \Rightarrow 3(b-a^2)(b+a^2)=(1-b)(1+b)$
We consider following cases:
Case 1:
$$\begin{cases} 3b-3a^2=1-b\\b+a^2=1+b\rightarrow
a^2=1\rightarrow a=\pm 1\end {cases}$$
which gives $b= 1$ for $a=+1$
Case 2:
$$\begin{cases} 3b+3a^2=1+b\\b-a^2=1-b\end {cases}$$
which gives no integer solution.
Case 3:
$$\begin{cases}3b-3a^2=1-b\rightarrow 2b-3a^2=1\\b^2+a^2=1-b \end{cases}$$
this system also has no integer solution.
We may consider other cases.
| {
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"source": "stackexchange",
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prove that: $(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\ge16abc$
Let $a,b,c\ge0$, prove that: $$(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\ge16abc$$
I tried AM-GM as below:
$$a^2+1\ge2a; a^2+b^2\ge2ab; a^2+b^2+c^2\ge ab+bc+ca$$ The rest is proving the inequality which is not true for all non-negative real numbers: $a(2ab+1)(ab+bc+ca+1)\ge8abc$
By the way, equality do not hold for a=b=c, so I guess my approach seems lost. Is there any better idea to help me deal with problem? Thank you!
| If you write it as a quadratic in $c$ and find its discriminant, then it suffices to prove:
$$(a^2+1)^2(a^2+b^2+1)^3\geq 64a^2b^2.$$
This can be done with AM-GM since:
$$(a^2+1)^2 = \left(a^2+\frac 13+\frac 13+\frac 13\right)^2\geq (4\cdot3^{- 3/4}a^{ 1/2})^2=\dfrac{16}{3\sqrt{3}}a$$
and so
$$(a^2+b^2+1)^3 = \left(b^2+\dfrac{a^2+1}{2}+\dfrac{a^2+1}{2}\right)^3\geq \dfrac{27}{4}b^2(a^2+1)^2\geq12\sqrt{3}b^2a$$
and when you multiply them together, we obtain the desired. The equality is reached at:
$$b = \sqrt{2}a = \dfrac{\sqrt{2}}{\sqrt{3}}$$
and from this you can deduce when is your original inequality is attained.
EDIT:
If one insists on using purely AM-GM, then:
$$c^2+a^2+b^2+1\geq 2c\sqrt{a^2+b^2+1}$$ and therefore:
$$(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\geq 2c(a^2+1)\sqrt{(a^2+b^2+1)^3}\geq$$
$$\geq 2c(a^2+1)\dfrac{3\sqrt{3}}{2}b(a^2+1)=3\sqrt{3}bc(a^2+1)^2\geq 3\sqrt{3}bc\cdot\dfrac{16}{3\sqrt{3}}a=16abc.$$
| {
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"source": "stackexchange",
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Finding all primes $p$ satisfying $p=a^2+b^2$ and $p$ divides $a^3+b^3-4$ .
Determine all primes $p$ such that there exist integers $a, b$ satisfying $p=a^2+b^2$ and $a^3+b^3-4$ is divisible by $p$.
So, $p | ab(a+b)+4$ after writing $a^3+b^3-4=(a+b)(p-ab)-4$ and $p=2$ is a solution if $p|4$. If $p$ doesn't divide $4$, $ab(a+b)\equiv -4 \pmod p$. Also, $p=4m+1$ for some integer $m$ by Fermat's 2-square theorem.It seems like $5$ and $2$ are the only primes that satisfy the conditions but I can't seem to prove it.
I've tried manipulating the stuff around but I'm not getting anything. Can anyone give a small hint? Thanks.
| Assume $p>2$, and let $s=a+b$. Then
$$ab=\frac{(a+b)^2-(a^2+b^2)}2\equiv \frac{s^2}2\pmod p.$$
Now,
$$a^3+b^3=(a+b)(a^2-ab+b^2)\equiv (a+b)(-ab)\equiv -\frac{s^3}2\pmod p.$$
This implies that $s^3\equiv -8$ modulo $p$. This means that either $p\mid s+2$ or $p\mid s^2-2s+4$.
In the first case,
$$s\geq p-2\implies p=a^2+b^2\geq \frac{(a+b)^2}2\geq \frac{(p-2)^2}2,$$
which implies $p\leq 5$. In the second case,
$$p\mid s^2-2s+4-p=2ab-2(a+b)+4.$$
Since $p>2$, this means $p\mid ab-a-b+2$, and thus $p\leq ab-a-b+2<ab$. However, $a^2+b^2\geq ab$, so this can't occur.
| {
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"source": "stackexchange",
"question_score": "4",
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} |
Find domain and range of $\log_2\left(\sin(2x)+\cos(2x)\right)$ I'm having trouble finding domain and range of this function.
Can somebody give me a HINT please? Thanks.
| \begin{align}
\cos(2x)+\sin(2x)
&=\cos(2x)\,r\cos \phi + \sin(2x)\,r\sin \phi \\
&= r\cos(2x-\phi)
\end{align}
\begin{cases}
r\cos \phi =1\\
r\sin \phi =1
\end{cases}
Finding $r$:
\begin{align}
r^2 (\sin^2\phi +\cos^2 \phi) &= 1^2+1^2\\
r &=\sqrt{2}
\end{align}
Finding $\phi$:
\begin{align}
\frac{r\sin \phi}{r\cos \phi} &= \frac{1}{1}\\
\tan \phi &= 1\\
\phi &= 45^\circ
\end{align}
Return to the main. Argument of logarithm must be positive.
\begin{align}
\cos(2x)+\sin(2x)&>0\\
\sqrt{2}\cos(2x-45^\circ) &>0\\
\cos(2x-45^\circ) &>0
\end{align}
\begin{align}
-90^\circ + 360^\circ k < 2x-45^\circ < 90^\circ + 360^\circ k\\
-45^\circ + 360^\circ k < 2x < 135^\circ + 360^\circ k\\
-22.5^\circ + 180^\circ k < x < 67.5^\circ + 180^\circ k
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the measure of the largest angle of a triangle formed by the reciprocals of the altitudes of a given triangle For reference:
The measures of the sides of a triangle
are $3$, $5$ and $7$. Calculate the measure of the largest angle
of a triangle whose sides are the inverses of the heights of the first triangle (Answer:$120^\circ$)
*I checked in geogebra and the answer is correct.
My progress: I tried but I don't know if that's the way to go.
$p = \frac{3+5+7}{2}=\frac{15}{2}\\p(p-a)p(p-b)(p-c)=\frac{15}{2}(\frac{9.}{2}\frac{5}{2}.\frac{1}{2}) = \frac{675}{16}$
Heron's Formula: $
h_b=\frac{2}{7}\sqrt{\frac{675}{16}}=\frac{2.15\sqrt3} {7.4}=\frac{15\sqrt3}{14}$
Similarly:
$h_a = \frac{2}{5}\cdot\frac{15\sqrt3}{4} = \frac{3\sqrt3}{2}\\h_c = \frac{2}{3}\cdot\frac{15\sqrt3}{4}=\frac{5\sqrt3}{2} \\
\frac{1}{h_a}=\frac{2}{3\sqrt3}\implies (\frac{1}{h_c})^2 = \frac{4}{27}\\
\frac{1}{h_b} = \frac{14}{15\sqrt3}\implies (\frac{1}{h_b})^2=\frac{196}{675}\\
\frac{1}{h_c} = \frac{2}{5\sqrt3} \implies (\frac{1}{h_c})^2=\frac{4}{75}$
By Law of Cosines:
$\frac{196}{675} = \frac{4}{27}+\frac{4}{75} - 2\cdot\frac{2}{3\sqrt3}\frac{2}{5\sqrt3}\cdot\cos B\implies\\
\cos B=-0.5\\ \therefore B = 120^\circ$
| Elaborating @Stinking Bishop's hint, consider triangle $ABC$ with sides $a$, $b$ and $c$ (in standard notation). Let altitudes from each vertex to the opposite side be $h_a$, $h_b$ and $h_c$. Now, if the area of the triangle is $A$, we get, $$\frac1{h_a}=\frac{a}{2A}$$ $$\frac1{h_b}=\frac{b}{2A}$$ $$\frac1{h_c}=\frac{c}{2A}$$
It is clear that the triangle formed by the inverses of the heights is similar to the original triangle with similarity ratio $\frac1{2A}$.
Now we can find the required angle considering the above fact.
The angle opposite to the longest side is the largest angle. Now using law of cosines, $$\cos\theta=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac12$$ $$\implies\theta=120^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$$
(Only use AM-GM, Cauchy-Schwarz inequalities)
My progress:
Till now, I have not made much of a progress besides finding out that $x^2+y^2+z^2\geq1$ and $xyz\leq\frac{\sqrt{3}}{9}$ and turn the LHS into $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
| Another way.
Let $x^2+y^2+z^2=t(xy+xz+yz).$
Thus, $t\geq1$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}\left(\frac{x}{\sqrt{1+x^2}}+\frac{1}{x^2}\right)=\sum_{cyc}\frac{2x}{2\sqrt{(x+y)(x+z)}}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x^2y^2}{x^2y^2z^2}\geq$$
$$\geq\sum_{cyc}\frac{2x}{2x+y+z}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{xyz}=\sum_{cyc}\frac{2x^2}{2x^2+xy+xz}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{xyz}\geq$$
$$\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(2x^2+xy+xz)}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{\sqrt{\left(\frac{\sum\limits_{cyc}xy}{3}\right)^3}}=\frac{t+2}{t+1}+3\sqrt{3(t+2)}=$$
$$=\frac{t+2}{t+1}+3\sqrt{(1+2)(t+2)}\geq\frac{t+2}{t+1}+3(\sqrt{t}+2)\geq$$
$$\geq\frac{t+2}{t+1}+3\left(\frac{2t}{t+1}+2\right)=\frac{5t}{t+1}+8\geq\frac{5t}{t+t}+8=\frac{21}{2}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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What is the sum of arcus cotangents? Say I wish to find the sum of three arcus cotangents. I wanted to start with a smaller example, say
$$\mathrm{arccot}{x} + \mathrm{arccot}{y}.$$
I know that
$$
\mathrm{arccot}{x} =
\begin{cases}
\arctan{\frac{1}{x}} & \text{if $x>0$} \\
\pi + \arctan{\frac{1}{x}} & \text{if $x<0$}
\end{cases}
$$
And that
$$
\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}
$$
Is this the right path to find the expression for $\mathrm{arccot}{x} + \mathrm{arccot}{y}$?
Simply consider every case of $x,y,xy$? How to get to the sum of three arcus cotangents from here?
Was also looking for a "ready" formulation online but I ultimately failed. So out of curiosity - are arcus cotangents that rarely used?
|
I know that $$ \mathrm{arccot}(x) = \begin{cases}
\arctan{\frac{1}{x}} & \text{if $x>0$} \\ \pi + \arctan{\frac{1}{x}} &
\text{if $x<0$} \end{cases} $$
And a small addition: $$=\frac\pi2\quad\text{if }x=0.$$
Based on this definition (there are two common definitions of arccot), we have that on $\mathbb R,$ $$\arctan(x)\equiv\frac {\pi}2-\mathrm{arccot}(x).$$
and that $$\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Is this the right path to find the expression for $\mathrm{arccot }(x) + \mathrm{arccot }(y)$?
Substituting in the above identity and noting that $\arctan$ is an odd function: $$\mathrm{arccot }(x)+\mathrm{arccot }(y) = \begin{cases}\pi+\arctan\left(\dfrac{x+y}{xy-1}\right), &xy < 1 \\
\arctan\left(\dfrac{x+y}{xy-1}\right), &x>0,\; y>0,\; xy>1 \\
2\pi + \arctan\left(\dfrac{x+y}{xy-1}\right), &x<0,\; y<0,\; xy > 1.\end{cases}$$
| {
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"source": "stackexchange",
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High dimensional generalizations of $\int_{\sqrt{2} }^{\sqrt{3} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } dy =\frac{5\pi^2}{96}$ Let $Q=n_1+n_2+n_3+1$,$\mathbf{s}=(n_1,n_2,n_3)$.
Define
$$
A(\mathbf{s})
=\int_{D}\prod_{n=1}^{Q-1}\frac{1}{1+x_n^2}\int_{1}^{\infty}\left(Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y
\text{d}x_i.
$$
Where $D=[0,\infty]^{n_1}\times[0,1]^{n_2}
\times[1,\infty]^{n_3}\subset\mathbb{R}^{Q-1}$,
$\mathrm{d}x_i
=\prod_{n=1}^{Q-1} \text{d}x_n$.
*
*Special case.
For $\mathbf{s}=(0,0,n_3)$, we have
$$
A(\mathbf{s})
=\frac{1}{Q}\left ( \frac{\pi}{4} \right )^Q.
$$
*Question 1
Prove
$$
\pi^{-Q}A(\mathbf{s})\in\mathbb{Q}.
$$
My ultimate goal is to evaluate $A(\mathbf{s})$. Numerical calculations suggest that
$$A(1,0,0)=\frac{\pi^2}{12} \quad A(0,1,0)=\frac{5\pi^2}{96}\quad A(0,0,1)=\frac{\pi^2}{32}$$
$$A(2,0,0)=\frac{\pi^3}{32} \quad A(1,1,0)=\frac{\pi^3}{80}$$
$$A(0,3,0)=\frac{93\pi^4}{35840} \qquad A(0,4,0)=\frac{193\pi^5}{322560}$$
Actually, if we explicit calculate the multiple integrals, it yields
$$
\begin{aligned}
&\int_{\sqrt{2} }^{\sqrt{3} }
\frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
=\frac{5\pi^2}{96},\\
&\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} }
\frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
-\int_{2}^{\sqrt{5} }
\frac{\displaystyle{\arctan (y)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} }
\text{d}y=\frac{11\pi^3}{5760},\\
&\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} }
\frac{\arctan\left(y\sqrt{2+y^2} \right)}{(y^2-1)\sqrt{y^2-2} } \text{d}y
-\int_{2}^{\sqrt{5} }
\frac{\displaystyle{\arctan\left(y\sqrt{2+y^2}\right)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} }
\text{d}y=\frac{\pi^3}{420}.
\end{aligned}
$$
Are there any other simple results? For $Q=4$, we may meet some 'troubles', such as this one:
$$
\int_{0}^{1} \int_{1}^{\sqrt{2} }
\frac{u\left(\pi-2\arctan\sqrt{u^4-1}-2\arctan \sqrt{\frac{u^2-1}{u^2+1} }
\right) \arctan \sqrt{4+u^2+v^2} }
{(1+v^2)\sqrt{1+u^2}(2+u^2) \sqrt{4+u^2+v^2} } \text{d}u\text{d}v.
$$
I can hardly convert into a 'simple' form.
*
*Question 2.
Can we evaluate a more general family of this kind of integrals?
$$A(\alpha,\mathbf{s})
=\int_{D}\prod_{n=1}^{Q-1}\frac{1}{\alpha^2+x_n^2}\int_{1}^{\infty}\left(\alpha^2 Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y
\text{d}x_i.$$
| I've found a direct way.
Let me give an example,
$$
I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2} }\text{d}x\text{d}y\text{d}z.
$$
Use the fact that
$$
\frac{1}{\sqrt{x} } =\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-xt^2}\text{d}t
$$,
and rewrite the integral into
$$I = \frac{2}{\sqrt{\pi}}
\int_{0}^{\infty}e^{-4t^2}
\left ( \int_{0}^{1} \frac{e^{-x^2t^2}}{1+x^2} \text{d}x \right )^3
\text{d}t.$$
The inner integral is simply equal to
$$\int_{0}^{1} \frac{e^{-x^2t^2}}{1+x^2} \text{d}x
=\frac{\pi}{4}e^{t^2}\left(1-\operatorname{erf}(t)^2\right).$$
Then,$$I=\frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi}}
\int_{0}^{\infty}e^{-t^2}\left(1-\operatorname{erf}(t)^2\right)^3\text{d}t.$$
Note that the integrand has a polynomial of $\operatorname{erf}(t)$. That is to say, if we set
$$P(x)=a_n x^n+a_{n-1} x^{n-1}+...+a_{1} x+a_0$$,
we get
$$\begin{aligned}
I & = \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} }
\int_{0}^{\infty}e^{-t^2}P(\operatorname{erf}(t))\text{d}t \\
& = \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} }
\int_{0}^{\infty}e^{-t^2}\sum_{k=0}^{n}a_k\operatorname{erf}(t)^k\text{d}t\\
&= \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} }
\sum_{k=0}^{n}a_k \int_{0}^{\infty}e^{-t^2}\operatorname{erf}(t)^k\text{d}t\\
&=\frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} }
\sum_{k=0}^{n}a_k\cdot\frac{\sqrt{\pi} }{2(k+1)} \\
&=\frac{\pi^3}{64} \sum_{k=0}^{n} \frac{a_k}{k+1}
\end{aligned}$$
The finite sum is easy to evaluate, and we conclude that
$$I=\frac{\pi^3}{140} .$$
These four integrals have the similar expression:
$$\begin{aligned}
&\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2+y^2+z^2)^2}\text{d}x\text{d}y\text{d}z\\
&\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(2+x^2+y^2+z^2)^{3/2}}\text{d}x\text{d}y\text{d}z\\
&\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(1+y^2)(3+x^2+y^2+z^2)}\text{d}x\text{d}y\text{d}z\\
&\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2}}\text{d}x\text{d}y\text{d}z.
\end{aligned}$$
Their limits can rewrite as $[1,\infty],[0,\infty]$ and still have similar results, of course.
| {
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"url": "https://math.stackexchange.com/questions/4303148",
"timestamp": "2023-03-29T00:00:00",
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In the figure, determine $x$ as a function of $R$. For reference: In the figure, determine $x$ as a function of $R$.
My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't
$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr-3r^2 = 4Rr\\6R-3r=4R \implies \boxed{r = \frac{2R}{3}}\\\triangle FGD: (R+r)^2 = (x+r)^2+(x+R)^2\\R^2+\frac{4R^2}{3} + \frac{4R^2}{9} = x^2+\frac{4xR}{3}+\frac{4R^2}{9}+x^2+2xR+R^2\\\frac{4R^2}{3} =2x^2+\frac{10xR}{3}\\\frac{4R^2}{3} = \frac{6x^2+10xR}{3}\implies 4R^2 = 6x^2+10xR \implies\\ 3x^2+5xR - 2R^2 = 0\\\therefore \boxed{\color{red}x = \frac{R}{3}}$
| If the radius of the other half circle is $r$, you already showed that $ \small \displaystyle r = \frac{2R}{3}$.
Now applying Descartes' theorem,
$ \small \displaystyle \left(- \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R/3} + \frac{1}{x} \right)^2 = 2 \left(\frac{1}{4R^2} + \frac{1}{R^2} + \frac{1}{(2R/3)^2} + \frac{1}{x^2}\right)$
Simplifying, $ \small \displaystyle \left(\frac{2}{R} + \frac{1}{x} \right)^2 = \frac{7}{R^2} + \frac{2}{x^2}$
Solving, $ \small \displaystyle x = \frac{R}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$?
$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$ ?
My approaches which didn't yield anything
I tried manipulating $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$
Observations
*
*$\frac{x+1}{x-1}=\frac{(x+1)(x^3-x^2+2x-2)}{(x-1)(x^3-x^2+2x-2)}=\frac{x^4+x^2-2}{x^4-2x^3+3x^2-4x+2}=\frac{\frac{11}{5}-2}{\frac{11}{5}-2x^3+2x^2-4x+2}$ which is not giving any assistance to find the required value.
*let a=$\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$
then we observe that
$a^3=2\frac{x^2+1}{x^2-1}+3a$ which again is not useful
how should we proceed then to get the required ?
| Let $a = ...$ "which again is not useful", wrong here, it is.
Let $a = x^2$
$a^2 + a - \frac{11}{5} = 0$
$a = \frac{-1 \pm \sqrt{1 - 4 * 1 * \frac{-11}{5}}}{2}$
$a = \frac{-1 \pm \sqrt{\frac{49}{5}}}{2}$
$a = \frac{-1 \pm \frac{7\sqrt{5}}{5}}{2}$
$a = \frac{-5\pm {7\sqrt{5}}}{10}$
$x^2 = \frac{-5\pm {7\sqrt{5}}}{10}$
$x = \pm\sqrt{\frac{-5+{7\sqrt{5}}}{10}}$
Let $y$ = $^3\sqrt{\frac{x+1}{x-1}}+$ $^3\sqrt{\frac{x-1}{x+1}}$
From your observation, $y^3 = \frac{2x^2+2}{x^2-1} + 3y$
Sub the value found for $x$:
$y^3 - 3y - 14\sqrt{5}-32 = 0$
Use the cubic formula without the $bx^2$ term in there. See origins of cubic formula for more info.
Without the $b$ constant we get a simplified version of the formula.
$y =$ $^3\sqrt{\frac{-d}{2a} + \sqrt{(\frac{-d}{2a})^2 + (\frac{c}{3a})^3}} +$ $^3\sqrt{\frac{-d}{2a} - \sqrt{(\frac{-d}{2a})^2 + (\frac{c}{3a})^3}}$
Note how this resembles the original part of the question we were finding for.
Another way to solve it:
Let $f(y) = y^3 - 3y - 14\sqrt{5}-32$
Cubics at most have $3$ real roots (atleast $1$ real, and the other $2$ complex).
In their factored form, $(x - a)(x - b)(x - c)$, $a*b*c = d$ in $ax^3+bx^2+cx+d$. Therefore, our roots must multiply to $-14\sqrt{5}-32$, or $abc = -14\sqrt{5}-32$. But before that we must define how many roots (1, 2 or 3) are there.
Taking $\frac{dy}{dx} = 0$, there is a maximum at $(-1, 61.305)$ and minimum at $(1, 65.305)$. Notice how the curves are way below the horizontal axis, so the $f(y)$ has only 1 root, for $y > 0$.
This means the $f(y)$ can be factored into a form resembling $(y + a)(y^2 + cy + d)$, where is $-a$ is the root and $(y^2 + cy + d)$ is irreducible.
So, $ay^2 + cy^2 = 0$, as there is no squared term. Therefore, $c = -a$.
$ad = -14\sqrt{5}-32$
$acy + dy = -3y$
If you simultaneously solve, you will get $a = - 2 - \sqrt{5}$. Therefore, $(y - 2 - \sqrt{5}) = 0$ and your root is $2 + \sqrt{5}$.
| {
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"url": "https://math.stackexchange.com/questions/4305689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and $BQ$ is free of $\alpha$.
This question appeared in my exam today. The way I did it is first constructed the equation of two perpendicular line to $2x \cos \alpha - 3y \sin \alpha = 6$ which goes through the points $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$. In this way, for $A$ and $B$, I got the following equations respectively-
$$3x \sin \alpha + 2y \cos \alpha -3 \sqrt{5}\sin \alpha=0 \qquad(1)$$$$3x \sin \alpha + 2y \cos \alpha +3 \sqrt{5}\sin \alpha=0 \qquad(2)$$
Then I found that the line $(1)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$P\left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
and the line $(2)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$Q\left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
Now using distance formula $$AP = \sqrt{ \left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}- \sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$ and $$ BQ= \sqrt{ \left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} +\sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$
Now the multiplication product $AP \cdot BQ$ indeed gives a constant value of $4$ which is free of the arbitrary variable $\alpha$ as you can see here is the simplified version of product of those two quantity. But this is tedious and I do not think this is the only way to do it and an appropriate way to follow in exam with limited time. So, I am looking for an alternative, time saving proof of it. I was wondering if using parametric form would help me, but I think it would get as difficult equally.
I created a visualisation for you on desmos to help my problem understand better.
| Check out my analysis of your question.
Equation is 2xcosα−3ysinα=0. This equation passes through origin and points A and B are on x axis. Assume slope of equation be tan $\phi$.
I have illustrated the diagram here
now product of length of two perpendiculars will be 5 sin$^2$ $\phi$. From equation of line slope of line is $$tan\phi =\frac{2cos\alpha}{3sin\alpha}$$
get value of $sin^2\phi$ from $tan\phi$. You will see it is not free of alpha.You can see that in graph too here (I have updated it). As graph is symmetrical ,length will depend on $\phi$.
Maybe I interpreted the question incorrectly. Correct me if it's wrong
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logarithm laws, we have
$$\lim_{x\rightarrow \infty} \left(\dfrac{1}{\ln\frac{x+1}{x}}-x\right).$$
Simplyfing further, we have:
$$\lim_{x\rightarrow \infty} \dfrac{1-x\ln\frac{x+1}{x}}{\ln\dfrac{x+1}{x}}= \lim_{x\rightarrow \infty} \frac{1-\ln(1+\frac{1}{x})^x}{\ln(1+\frac{1}{x})} = \frac{1-e}{0} \rightarrow -\infty.$$
However, the answer sheet tells me that it's $1/2$, and I don't really see where I did something wrong in the solution. Thanks.
| \begin{gather*}
\lim _{x\rightarrow \infty }\frac{1}{\ln( 1+x) -\ln x} -x\\
=\lim _{x\rightarrow \infty }\frac{1}{\ln\left( 1+\frac{1}{x}\right)} -x\\
=\lim _{x\rightarrow \infty }\frac{1-x\cdotp \ln\left( 1+\frac{1}{x}\right)}{\ln\left( 1+\frac{1}{x}\right)}\\
Let\ us\ make\ a\ substitution\ h=\frac{1}{x}\\
Then,\ h\rightarrow 0\ as\ x\rightarrow \infty \\
=\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \ln( 1+h)}{\ln( 1+h)}\\
We\ know\ that\ \\
\ln( 1+h) =h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...\\
Hence,\ \\
\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \ln( 1+h)}{\ln( 1+h)}\\
=\lim _{h\rightarrow 0}\frac{1-\frac{1}{h} \cdotp \left( h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...\right)}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}\\
=\lim _{h\rightarrow 0}\frac{1-1+\frac{h}{2} -\frac{h^{2}}{3} +\frac{h^{3}}{4} -...}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}\\
=\lim _{h\rightarrow 0}\frac{\frac{h}{2} -\frac{h^{2}}{3} +\frac{h^{3}}{4} -...}{h-\frac{h^{2}}{2} +\frac{h^{3}}{3} -...}
\end{gather*}
Can you take it from here? Good luck!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing the formal identity $(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$
How to show this formal identity (or you can assume $|x|<1$)?
$$(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$$
I can show that the latter is $$=1-x+x^3-x^4+x^6-x^7+\cdots$$ but how to show this is equal to the infinite product. I think it has something to do with residue of the exponent modulo $3$.
| Notice that $(1+x+x^2)(1-x+x^2)=(x^2+1)^2-x^2=1+x^2+x^4$. So: multiply both sides by $1+x+x^2$:
$$\begin{array}{rcl}(1+x+x^2)\cdot(\text{LHS})&\equiv&(1+x+x^2)(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^2+x^4)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^4+x^8)(1-x^4+x^8)\cdots\\&&\vdots\end{array}$$
The second row shows that the formal product on the left has no (nonzero) odd terms. The third row shows that it doesn't even have the terms $x^2, x^6$ etc. i.e. all the (nonzero) terms are of the form $x^{4k}$, then in the next step all the (nonzero) terms will actually be of the form $x^{8k}$ etc. - so by using induction (or otherwise) you can easily see that the formal product does not have any nonzero terms (except for the constant, which is obviously $1$). Thus, this formal product is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308837",
"timestamp": "2023-03-29T00:00:00",
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if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then does $a | 2^m-1$ or $a| 2^m+1$
Determine whether for any integer $a$, if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then $a | 2^m-1$ or $a| 2^m+1$.
Suppose $a\not |2^m-1$ and $a\not| 2^m+1$. Then there exist primes $p_1$ and $p_2$ so that $p_1 | 2^m-1, p_1 \not | 2^m+1, p_2 | 2^m+1, p_2 \not | 2^m-1.$ Since $2^m+1$ and $2^m-1$ are coprime, so are $p_1$ and $p_2$. By Bezout's identity, there exist integers $s,t$ so that $sp_1 + tp_2 = 1$. Since $p_1 | 2^m - 1$ and $p_2 | 2^m + 1$ there exist integers $c,d$ with $cp_1 + 1 = 2^m, dp_2 - 1= 2^m$ and so $dp_2 - cp_1 = 2$. Since $k > 1, p_1^2 | 2^m-1$ and $p_2^2 | 2^m+1$ and similarly there exist integers $c,d$ with $cp_1^2 + 1 = 2^m, dp_2^2 - 1= 2^m$ and so $dp_2^2 - cp_1^2 = 2$. In fact, both $2^m+1$ and $2^m-1$ must have prime factorizations where the exponent of each prime is at least $2$; if one prime factor of either has an exponent of one, then it is not possible that $a^k = (2^m-1)(2^m+1)$ because all prime powers in the prime factorization of $a^k$ have exponent at least $2$.
But I'm not sure how to get a contradiction from this. Maybe there's a counterexample?
I also know the following theorem: If $a$ and $b$ are positive integers, then there exists a number $g$ such that every multiple of $\gcd(a,b)$ greater than $g$ may be written in the form $ra+sb$ where $r$ and $s$ are nonnegative integers, but I don't think it's necessary for this problem.
| Let $a$, $k$ and $m$ be positive integers with $k\geq2$ and
$$a^k=(2^m-1)(2^m+1).$$
The two factors on the right hand side are coprime, because they are both odd and their difference is $2$. Then by unique factorization, both are $k$-th powers. That is, there exist positive integers $b$ and $c$ such that
$$2^m-1=b^k\qquad\text{ and }\qquad 2^m+1=c^k.$$
Subtracting the former from the latter shows that
$$c^k-b^k=2,$$
but of course there are no two positive powers that differ by $2$.
So the answer is vacuously yes, because there are no such integers.
| {
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"url": "https://math.stackexchange.com/questions/4313324",
"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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