ganeshie8
  • ganeshie8
show that \[\large \sum_{n=1}^{\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4}\]
Mathematics
jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
spoiler hint : \(\huge \color{white}{🔭}\)
freckles
  • freckles
\[\tan(\frac{\pi}{4})=1 \\ \tan(\sum \arctan(\frac{1}{2n^2}))=1 \\ \tan(\arctan(\frac{1}{2})+\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+\cdots) \text{ w.t.s this is }1 \] but that might be harder to show not sure yet
freckles
  • freckles
\[\tan(\arctan(\frac{1}{2})+s_1)=\frac{\tan(\arctan(\frac{1}{2}))+\tan(s_1)}{1-\tan(\arctan(\frac{1}{2})) \tan(s_1)}=\frac{\frac{1}{2}+\tan(s_1)}{1-\frac{1}{2} \tan(s_1)}\\ =\frac{1+2 \tan(s_1)}{2-\tan(s_1)}\]

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freckles
  • freckles
where \[s_1=\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+ \cdots \]
freckles
  • freckles
yep seems like a dead end for me
ganeshie8
  • ganeshie8
thats prety clever! i think that might actually telescope... im still trying on paper..
ganeshie8
  • ganeshie8
\[\arctan\frac{1}{2n^2} \\~\\= \arctan\frac{2}{(2n)^2} \\~\\= \arctan\frac{2}{1+(2n)^2-1}\\~\\=\arctan\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\]
ganeshie8
  • ganeshie8
\[\arctan(\dfrac{x-y}{1+xy}) = \arctan(x)-\arctan(y)\] let \(x=(2n+1)\) and \(y=2n-1\)
anonymous
  • anonymous
this is nice too, classic
ganeshie8
  • ganeshie8
im biased to telescoping xD but there is another way to work this using complex numbers which is pretty neat too the basic idea is to interpret the "multiplication of complex numbers" geometrically by "adding up the angles"
ganeshie8
  • ganeshie8
\[\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\]
Astrophysics
  • Astrophysics
That sounds fascinating, have you written it down, can you just post a pic of it, then you don't have to write it out
Astrophysics
  • Astrophysics
on here
ganeshie8
  • ganeshie8
it is easy, just need to see it one time that the angles add up when we multiply a complex number by another complex number : |dw:1435211815854:dw|
ganeshie8
  • ganeshie8
that opens us a door for writing infinite series as an infinite product and many other cool things xD
ganeshie8
  • ganeshie8
Keeping in mind that \(\arctan(\frac{b}{a})\) gives the angle of complex number \(a+bi\), we have : \[N=\prod_{k=1}^\infty (a_k+ib_k)~~\implies~~\arg N=\sum_{k=1}^\infty \arctan\frac{b_k}{a_k}\]
freckles
  • freckles
I do like the telecoping one. That was pretty cute stuff.
anonymous
  • anonymous
@freckles you may try this\[\sum_{n=0}^{\infty} \text{Arccot}(n^2+n+1) \]
freckles
  • freckles
\[y=arccot(n^2+n+1) \\ \cot(y)=n^2+n+1 \\ \tan(y)=\frac{1}{n^2+n+1} \\ y=\arctan(\frac{1}{n^2+n+1}) \] \[ \frac{1}{n^2+n+1}=\frac{4}{4n^2+4n+4}=\frac{4}{4n^2+4n+1+3} \\ =\frac{4}{(2n+1)^2+3}\] \[=\frac{1}{(\frac{2n+1}{2})^2+\frac{3}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{3}{4}-1} \\ =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{-1}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2-\frac{1}{4}} \\ =\frac{1}{1+(\frac{2n+1}{2}-\frac{1}{2})(\frac{2n+1}{2}+\frac{1}{2})} \\ \] \[(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})=\frac{1}{2}+\frac{1}{2}=1 \\ \text{ so we have } \\arccot(n^2+n+1)=\arctan(\frac{(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})}{1+(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}-\frac{1}{2})}) \\ arccot(n^2+n+1)=\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2})\]
freckles
  • freckles
\[\sum_{n=0}^\infty arcot(n^2+n+1)=\sum_{n=0}^{\infty}(\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2})) \]...
freckles
  • freckles
\[\arctan(\frac{1}{2}+\frac{1}{2})-\arctan(\frac{1}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{3}{2}+\frac{1}{2})-\arctan(\frac{3}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{5}{2}+\frac{1}{2})-\arctan(\frac{5}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{7}{2}+\frac{1}{2})-\arctan(\frac{7}{2}-\frac{1}{2}) \\ + \\ +\cdots \] \[=-\arctan(\frac{1}{2}-\frac{1}{2})+\lim_{k \rightarrow \infty}\arctan(\frac{2k+1}{2}+\frac{1}{2})\]
freckles
  • freckles
\[=-\arctan(0)+\lim_{k \rightarrow \infty}\arctan(\frac{2k+2}{2}) \\ =-(0)+\lim_{k \rightarrow \infty}\arctan(k+1) \\ =0+\frac{\pi}{2} \\ =\frac{\pi}{2}\] and I only wrote 0 a bunch of times because I thought it looked really pretty
anonymous
  • anonymous
nice job @freckles :-)
freckles
  • freckles
only mimicked what @ganeshie8 did in the previous problem but thank you kindly @mukushla

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