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Homework Help: Find the sum of arctan(1/2n^2)

  1. Jun 24, 2005 #1
    [tex]\sum_{i=1}^{n} \arctan(\frac{1}{2n^2})[/tex]

    I have to find sum of first n terms.
    I have no idea what to do with these arctan. Give me please any hint.
     
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  3. Jun 24, 2005 #2

    dextercioby

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    The answer is immediate.

    [tex] \sum_{i=1}^{n} \arctan\left(\frac{1}{2n^{2}}\right) =n\arctan\left(\frac{1}{2n^{2}}\right) [/tex]

    Daniel.
     
  4. Jun 24, 2005 #3
    I got that
    [tex]S_n=\arctan\frac{\tan(S_{n-1})+1/2n^2}{1-\frac{\tan(S_{n-1})}{2n^2}}[/tex]
    [tex]S_n[/tex] is sum of first n terms.
    But i'm not sure if it could help.
    I just started studying series by myself. Could you elaborate your solution, Dexter. I don't understand
     
  5. Jun 24, 2005 #4

    dextercioby

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    I've already given you the answer.Yours is not the answer to the problem written in post #1.

    Daniel.
     
  6. Jun 24, 2005 #5
    Oh, yes. It's my mistake.
    [tex]\sum_{n=1}^{\infty} \arctan(\frac{1}{2n^2})[/tex] is supposed to be. Sorry
     
  7. Jun 24, 2005 #6

    dextercioby

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    I haven't got a clue so far to prove that

    [tex] \sum_{n=1}^{\infty} \arctan\left(\frac{1}{2n^{2}}\right) =\frac{\pi}{4} [/tex]

    Daniel.
     
  8. Jun 24, 2005 #7
    Integration Daniel?
     
  9. Jun 24, 2005 #8

    dextercioby

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    Integrate what...?I don't see a pattern to form a Riemann sum.

    Daniel.
     
  10. Jun 24, 2005 #9
    I tried it out, no light.Where did you get thid problem from?
     
  11. Jun 24, 2005 #10

    Hurkyl

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    Wee, this is a variant of my favorite problem! The way I know how to do it involves the application of a trig identity...
     
    Last edited: Jun 24, 2005
  12. Jun 24, 2005 #11
    Hm. Unfortunately it's too subtle hint for me.
     
  13. Jun 24, 2005 #12
    [tex]\frac{1}{2n^2} = \frac{2}{4n^2}[/tex] u can use this
    to get
    [tex]arctan(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}[/tex]
    You can use the identity for inverse
    and try to figure out rest
     
  14. Jun 24, 2005 #13
    ok just saw ur post::
    arctana+arctanb = arctan(a+b)/(1-ab)

    so u will have arctan{1/22}= arctan(2n+1)-arctan(2n-1)

    now u can substitute for n=1,2,3,......k u will finally have

    arctan{1/n2}= arctan(2k+1)-arctan1
    now as k=> inf u will have pi/2-pi/4 = pi/4


    "what is here may be elsewhere, what is not here is nowhere."
    Prayers are empowered by a clear mind and heart, and purity of intention. With clarity, our consciousness aligns with higher consciousness, where the forces of nature and laws of creation operate according to divine order
     
    Last edited: Jun 24, 2005
  15. Jun 24, 2005 #14

    Hurkyl

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    Well, I was trying not to actually do his homework for him. :tongue2:
     
  16. Jun 24, 2005 #15
    At any rate, it is a very neat problem. I had not seen it before. Thanks to the OP for posting it. I wonder if there are other ways to do it ...?
     
  17. Jun 25, 2005 #16
    [tex]\arctan{1/2n^2}= \arctan(2n+1)-\arctan(2n-1)[/tex]
    Yes, it's very tricky!
    I also got that [tex]S_n=\arctan(\frac{n}{n+1})[/tex] (This was my question)
    This also leads to pi/4 when n->infinity.
    Thank you very much to all.
     
    Last edited: Jun 25, 2005
  18. Jun 25, 2005 #17

    saltydog

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    Thanks guys. Very clever. This is a summary which I take no credit for:

    Using Himanshu's technique above:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})[/tex]

    Let:

    [tex]\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}[/tex]

    Now,

    [tex]Arctan(a)-Arctan(b)=Arctan\left[\frac{a-b}{1+ab}\right][/tex]

    Letting:

    a=(2n+1)
    b=(2n-1)

    We have:

    [tex]Arctan[\frac{1}{2n^2}]=Arctan\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]=Arctan(2n+1)-Arctan(2n-1)[/tex]

    Thus we have:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right][/tex]

    So that using a(x) and a(y) to save space we have:

    [tex]
    \sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]=[/tex]

    [tex](a3-a1)+(a5-a3)+(a7-a5)+(a9-a7)+ . . .+[a(2n+1)-a(2n-1)]+ . . .
    [/tex]

    Noting that all but -a1 and the a(2n+1) terms cancel, we take the limit of this sum and find we are left with:

    [tex]\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]\to a(\infty)-a(1)[/tex]

    Since:

    [tex] Arctan(1)=\frac{\pi}{4}[/tex]

    and:

    [tex]\mathop\lim\limits_{n\to\infty}Arctan(n)=\frac{\pi}{2}[/tex]

    we finally obtain:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}[/tex]

    That is beautiful!
     
    Last edited: Jun 25, 2005
  19. Jun 25, 2005 #18

    saltydog

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    Oh yea, I almost forgot. What about these:

    [tex]\sum_{n=1}^{\infty}ArcSin[\frac{1}{2n^2}][/tex]

    [tex]\sum_{n=1}^{\infty}ArcCos[\frac{1}{2n^2}][/tex]

    [tex]\sum_{n=1}^{\infty}ArcCot[\frac{1}{2n^2}][/tex]

    I assume there must be an equivalent relation for the ArcCot. I'll look into it.
    Wait a minute, isn't the cot=1/tan so can I propose that last one is just [itex]\frac{4}{\pi}[/itex]?
     
  20. Jun 25, 2005 #19

    Hurkyl

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    A sum of reciprocals is generally not the reciprocal of the sum.

    One of the advantages of the tangent function is that its sum formula involves only tangents, so the corresponding formula for arctangents is nicer.

    You could try constructing similar things for the others, I suppose. Start with a telescoping series, then apply an addition formula and see what you get.
     
  21. Jun 25, 2005 #20

    saltydog

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    Hello Hurkyl. You got to it before I could make amends. Yea, ArcCot and ArcCos will not converge. However, I suspect the ArcSin will. :smile:

    You know, it's interesting just looking at the functions. Note that ArcTan and ArcSin are 0 at 0 where the other 2 are >0. Thus intutitively one would suspect taking an infinite sum of a function closer and closer to x=0 for functions which are >0 would not converge whereas if the functions are 0 at x=0, the sum I suspect could converge. Suppose that's all embodied in the various tests for convergence. :redface:
     
    Last edited: Jun 25, 2005
  22. Jun 25, 2005 #21

    saltydog

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    I just seem to dig myself a hole in here sometimes. Suppose it's because my purpose for being here is to learn math as opposed to trying to show off. Anyway, concerning ArcSin. Quickest way for me to determine convergence is the integral test and yes I'm lazy and just used Mathematica but only because I've already done a bunch by hand. Anyway, the integral converges. So the limit of the sum exists. Now as far as what it is . . .
     
  23. Jun 26, 2005 #22
    I was wondering if it is possible to use complex variables to solve the original problem with arctan. I've seen some trick where you convert an infinite sum into a contour integral, but I don't remember exactly how it goes. I think it's called a Sommerfeld-Watson transformation, and you can use it to get the result [itex]\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}[/itex]. Would that also work here?
     
  24. Jun 26, 2005 #23

    Hurkyl

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    I have, in fact, seen someone solve a problem very similar to this one using complex analysis... so I'd imagine this one can be done as well. I don't remember the methodology, though. :frown:
     
  25. Jun 27, 2005 #24
    Hurkyl:

    I googled Sommerfeld-Watson transform to find out more about it. Here's an example. If f(z) goes to zero faster than 1/z as z --> infinity, then the sum [itex]\sum_{n=-\infty}^\infty f(n)[/itex] can be found by computing the integral

    [tex]\int_{\Gamma} f(z) \pi \cot \pi z \ dz = 2\pi i (\sum_{n=-N}^N f(n) \ + \ \sum_k \pi \cot \pi z_k * (\mbox{Res of f(z) at } z_k))[/tex]

    where, the integral is taken over the square of side 2N+1, centered at the origin. The integral goes to zero as N goes to infinity, and that allows you to compute the infinite series by summing a few residues.

    Now it seems to me that arctan(1/2z^2) goes to zero like 1/2z^2 as z goes to infinity, so this method might work. However, doesn't arctan have a logarithmic branch point at zero? That might be annoying. And also, does arctan have any poles in the complex plane? I have forgotten how to determine this. Need to do some serious review.

    Of course, I realize that the right way to do the problem is with the trig identity. I'm just curious to see if this way works too.

    edit: as written, the above formula only makes sense if f(z) has simple poles.
     
    Last edited: Jun 27, 2005
  26. Jun 27, 2005 #25

    saltydog

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    "There were two paths in the forest.
    I took the one less traveled and it has made all the difference" :smile:

    Who said that?
     
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