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ganeshie8

  • one year ago

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

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  1. ganeshie8
    • one year ago
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    spoiler hint : \(\huge \color{white}{🔭}\)

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

  3. freckles
    • one year ago
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    \[\tan(\arctan(\frac{1}{2})+s_1)=\frac{\tan(\arctan(\frac{1}{2}))+\tan(s_1)}{1-\tan(\arctan(\frac{1}{2})) \tan(s_1)}=\frac{\frac{1}{2}+\tan(s_1)}{1-\frac{1}{2} \tan(s_1)}\\ =\frac{1+2 \tan(s_1)}{2-\tan(s_1)}\]

  4. freckles
    • one year ago
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    where \[s_1=\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+ \cdots \]

  5. freckles
    • one year ago
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    yep seems like a dead end for me

  6. ganeshie8
    • one year ago
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    thats prety clever! i think that might actually telescope... im still trying on paper..

  7. ganeshie8
    • one year ago
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    \[\arctan\frac{1}{2n^2} \\~\\= \arctan\frac{2}{(2n)^2} \\~\\= \arctan\frac{2}{1+(2n)^2-1}\\~\\=\arctan\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\]

  8. ganeshie8
    • one year ago
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    \[\arctan(\dfrac{x-y}{1+xy}) = \arctan(x)-\arctan(y)\] let \(x=(2n+1)\) and \(y=2n-1\)

  9. anonymous
    • one year ago
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    this is nice too, classic

  10. ganeshie8
    • one year ago
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    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"

  11. ganeshie8
    • one year ago
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    \[\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\]

  12. Astrophysics
    • one year ago
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    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

  13. Astrophysics
    • one year ago
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    on here

  14. ganeshie8
    • one year ago
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    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|

  15. ganeshie8
    • one year ago
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    that opens us a door for writing infinite series as an infinite product and many other cool things xD

  16. ganeshie8
    • one year ago
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    Keeping in mind that \(\arctan(\frac{b}{a})\) gives the angle of complex number \(a+bi\), we have : \[N=\prod_{k=1}^\infty (a_k+ib_k)~~\implies~~\arg N=\sum_{k=1}^\infty \arctan\frac{b_k}{a_k}\]

  17. freckles
    • one year ago
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    I do like the telecoping one. That was pretty cute stuff.

  18. anonymous
    • one year ago
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    @freckles you may try this\[\sum_{n=0}^{\infty} \text{Arccot}(n^2+n+1) \]

  19. freckles
    • one year ago
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    \[y=arccot(n^2+n+1) \\ \cot(y)=n^2+n+1 \\ \tan(y)=\frac{1}{n^2+n+1} \\ y=\arctan(\frac{1}{n^2+n+1}) \] \[ \frac{1}{n^2+n+1}=\frac{4}{4n^2+4n+4}=\frac{4}{4n^2+4n+1+3} \\ =\frac{4}{(2n+1)^2+3}\] \[=\frac{1}{(\frac{2n+1}{2})^2+\frac{3}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{3}{4}-1} \\ =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{-1}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2-\frac{1}{4}} \\ =\frac{1}{1+(\frac{2n+1}{2}-\frac{1}{2})(\frac{2n+1}{2}+\frac{1}{2})} \\ \] \[(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})=\frac{1}{2}+\frac{1}{2}=1 \\ \text{ so we have } \\arccot(n^2+n+1)=\arctan(\frac{(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})}{1+(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}-\frac{1}{2})}) \\ arccot(n^2+n+1)=\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2})\]

  20. freckles
    • one year ago
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    \[\sum_{n=0}^\infty arcot(n^2+n+1)=\sum_{n=0}^{\infty}(\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2})) \]...

  21. freckles
    • one year ago
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    \[\arctan(\frac{1}{2}+\frac{1}{2})-\arctan(\frac{1}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{3}{2}+\frac{1}{2})-\arctan(\frac{3}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{5}{2}+\frac{1}{2})-\arctan(\frac{5}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{7}{2}+\frac{1}{2})-\arctan(\frac{7}{2}-\frac{1}{2}) \\ + \\ +\cdots \] \[=-\arctan(\frac{1}{2}-\frac{1}{2})+\lim_{k \rightarrow \infty}\arctan(\frac{2k+1}{2}+\frac{1}{2})\]

  22. freckles
    • one year ago
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    \[=-\arctan(0)+\lim_{k \rightarrow \infty}\arctan(\frac{2k+2}{2}) \\ =-(0)+\lim_{k \rightarrow \infty}\arctan(k+1) \\ =0+\frac{\pi}{2} \\ =\frac{\pi}{2}\] and I only wrote 0 a bunch of times because I thought it looked really pretty

  23. anonymous
    • one year ago
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    nice job @freckles :-)

  24. freckles
    • one year ago
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    only mimicked what @ganeshie8 did in the previous problem but thank you kindly @mukushla

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