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Finding closed form for the next series:

  1. Mar 30, 2007 #1

    MathematicalPhysicist

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    i need to find the closed form expressiosn for the next sums:
    [tex]g(x)=\sum_{n=0}^{\infty}\frac {x^{2n}}{(2n)!}[/tex]
    and [tex]f(x)=\sum_{n=0}^{\infty} \frac {x^{4n+1}}{4n+1}[/tex]
    well for the second one i thought differentiating, i.e [tex]f'(x)= \sum_{n=0}^{\infty} x^{4n}=1/(1-x^4)=\frac {1}{(1+x^2)*(1-x^2)}[/tex], now i could integrate it by parts but in my course i havent yet got to integrals, so i cannot use integration by parts, do you have any other way to compute this sum.
    for the first one, it looks like the power series of e^x, i tried to recursively get to the expression but so far to no success.
    any pointers would be helpful.
     
    Last edited: Mar 30, 2007
  2. jcsd
  3. Mar 30, 2007 #2

    HallsofIvy

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    The first one is, in fact,
    [tex]cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex]

    For the second one, your idea of differentiating to get a geometric series and then integrating works nicely (after you learn to integrate, of course!) using partial fractions, not "by parts". But it gives a function involving arctan and logarithms so I doubt you will find any simple way to do that.
     
  4. Mar 30, 2007 #3

    Gib Z

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    Just in case you wanted it, the second ones

    [tex]\frac{1}{4} (2\arctan x -\ln (x^2-1))[/tex] I think.
     
  5. Mar 30, 2007 #4

    Hurkyl

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    The summand for g(x) looks similar to the summand for e^x -- this suggests you might be able to express the power series for g(x) in terms of the power series for e^x.

    The summand for f(x) looks similar to the summands for log(1+x) and for arctan x -- this suggests you might be able to express the power series for g(x) in terms of the power series for log(1+x) and arctan x.
     
  6. Mar 31, 2007 #5

    MathematicalPhysicist

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    after one hour i got it by myself.
    but i appreciate your help.
    btw, iv'e got something like this:
    i need to find a power series for (1+x^2)*arctg(x).
    i know the power series of arctg(x), but when im multiplying this sum with (1+x^2), i get x+sum, i.e where the sum starts from n=1, but i cant put the x into the sum. perhaps it should be this way?
     
  7. Mar 31, 2007 #6

    Gib Z

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    What is arctg(x)????
     
  8. Mar 31, 2007 #7

    MathematicalPhysicist

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    the same as arctan(x).
    it's one of those shorcuts,like for sinh, there's sh, etc.
     
  9. Mar 31, 2007 #8

    Gib Z

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    So are you saying arctg(x) is the same as arctan(x)? Or that its a similar pattern, its the inverse of g(x)? Because thats argcosh(x).
     
  10. Mar 31, 2007 #9

    MathematicalPhysicist

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    yes arctg(x) is the same as arctan(x), now back on topic.
     
  11. Mar 31, 2007 #10

    Gib Z

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    Well then yes, it can be the way you stated, it doesn't matter and isnt wrong.
     
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