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ganeshie8
 one year ago
show that \[\large \sum_{n=1}^{\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4}\]
ganeshie8
 one year ago
show that \[\large \sum_{n=1}^{\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4}\]

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3spoiler hint : \(\huge \color{white}{🔭}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\tan(\frac{\pi}{4})=1 \\ \tan(\sum \arctan(\frac{1}{2n^2}))=1 \\ \tan(\arctan(\frac{1}{2})+\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+\cdots) \text{ w.t.s this is }1 \] but that might be harder to show not sure yet

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\tan(\arctan(\frac{1}{2})+s_1)=\frac{\tan(\arctan(\frac{1}{2}))+\tan(s_1)}{1\tan(\arctan(\frac{1}{2})) \tan(s_1)}=\frac{\frac{1}{2}+\tan(s_1)}{1\frac{1}{2} \tan(s_1)}\\ =\frac{1+2 \tan(s_1)}{2\tan(s_1)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1where \[s_1=\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+ \cdots \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yep seems like a dead end for me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3thats prety clever! i think that might actually telescope... im still trying on paper..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\arctan\frac{1}{2n^2} \\~\\= \arctan\frac{2}{(2n)^2} \\~\\= \arctan\frac{2}{1+(2n)^21}\\~\\=\arctan\dfrac{(2n+1)(2n1)}{1+(2n+1)(2n1)}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\arctan(\dfrac{xy}{1+xy}) = \arctan(x)\arctan(y)\] let \(x=(2n+1)\) and \(y=2n1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is nice too, classic

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3im 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That sounds fascinating, have you written it down, can you just post a pic of it, then you don't have to write it out

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it 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
 one year ago
Best ResponseYou've already chosen the best response.3that opens us a door for writing infinite series as an infinite product and many other cool things xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Keeping 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
 one year ago
Best ResponseYou've already chosen the best response.1I do like the telecoping one. That was pretty cute stuff.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles you may try this\[\sum_{n=0}^{\infty} \text{Arccot}(n^2+n+1) \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[y=arccot(n^2+n+1) \\ \cot(y)=n^2+n+1 \\ \tan(y)=\frac{1}{n^2+n+1} \\ y=\arctan(\frac{1}{n^2+n+1}) \] \[ \frac{1}{n^2+n+1}=\frac{4}{4n^2+4n+4}=\frac{4}{4n^2+4n+1+3} \\ =\frac{4}{(2n+1)^2+3}\] \[=\frac{1}{(\frac{2n+1}{2})^2+\frac{3}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{3}{4}1} \\ =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{1}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2\frac{1}{4}} \\ =\frac{1}{1+(\frac{2n+1}{2}\frac{1}{2})(\frac{2n+1}{2}+\frac{1}{2})} \\ \] \[(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}\frac{1}{2})=\frac{1}{2}+\frac{1}{2}=1 \\ \text{ so we have } \\arccot(n^2+n+1)=\arctan(\frac{(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}\frac{1}{2})}{1+(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}\frac{1}{2})}) \\ arccot(n^2+n+1)=\arctan(\frac{2n+1}{2}+\frac{1}{2})\arctan(\frac{2n+1}{2}\frac{1}{2})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{n=0}^\infty arcot(n^2+n+1)=\sum_{n=0}^{\infty}(\arctan(\frac{2n+1}{2}+\frac{1}{2})\arctan(\frac{2n+1}{2}\frac{1}{2})) \]...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\arctan(\frac{1}{2}+\frac{1}{2})\arctan(\frac{1}{2}\frac{1}{2}) \\ + \\ \arctan(\frac{3}{2}+\frac{1}{2})\arctan(\frac{3}{2}\frac{1}{2}) \\ + \\ \arctan(\frac{5}{2}+\frac{1}{2})\arctan(\frac{5}{2}\frac{1}{2}) \\ + \\ \arctan(\frac{7}{2}+\frac{1}{2})\arctan(\frac{7}{2}\frac{1}{2}) \\ + \\ +\cdots \] \[=\arctan(\frac{1}{2}\frac{1}{2})+\lim_{k \rightarrow \infty}\arctan(\frac{2k+1}{2}+\frac{1}{2})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[=\arctan(0)+\lim_{k \rightarrow \infty}\arctan(\frac{2k+2}{2}) \\ =(0)+\lim_{k \rightarrow \infty}\arctan(k+1) \\ =0+\frac{\pi}{2} \\ =\frac{\pi}{2}\] and I only wrote 0 a bunch of times because I thought it looked really pretty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice job @freckles :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1only mimicked what @ganeshie8 did in the previous problem but thank you kindly @mukushla
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