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Urgend Geometric series question

  1. Feb 22, 2006 #1
    Hi

    I have the following problem:

    show that

    1/(1+x^2)) = 1-x^2 + x^4 + (-1)^n*(x^2n-2) + (-1)^n * (x^2n)/(1+x^2)

    I that know this arctan function can be expanded as a geometric series by using:

    1 + q + q^2 + q^3 + .... + = 1/(1-q)


    Then by putting q = -x^2. I get:


    1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + .... +

    My question is how do I proceed from this to get the desired result???

    Sincerley and Best Regards

    Fred
     
  2. jcsd
  3. Feb 22, 2006 #2

    arildno

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    No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2
     
  4. Feb 22, 2006 #3
    I still get first result, but has it something to do that the series converges? And then I then add a sum to the first result?

    /Fred
     
  5. Feb 22, 2006 #4

    arildno

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    No, you simply haven't done it correctly. Try again.
    Insert (-x^2) on the q-places in the following expression:
    [tex]1+q+q^2+q^3+++[/tex]
    What do you get?
     
  6. Feb 22, 2006 #5
    This is wrong. Try again.
     
  7. Feb 22, 2006 #6
    Hello by inserting q = -x^2 into the above I get:


    1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... +

    but how do I proceed from there ?

    Sincerley

    Fred
     
  8. Feb 22, 2006 #7
    Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.
     
  9. Feb 22, 2006 #8

    arildno

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    Great! One flaw though:
    You've been working with 1/(1-(-x^2)), not 1/(1-x^2)!

    Now, find the n'th term, and scrutinize the infinite part of the series starting with the n'th term.
     
  10. Feb 22, 2006 #9

    Hello I can see that the the function changes in the following way

    (-1)^n * (x^2)^n, then


    1/(1-x^2) = 1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... + (-1)^n * (x^2)^n


    Do I then rewrite the above result to get the result required ???

    Sincerely
    Fred
     
  11. Feb 22, 2006 #10

    arildno

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    You have now found the general term of the series expansion of 1/(1-(-x^2))
    For all N>=n, draw out the common factor (-1)^(n)x^(2n)
    What are you left with then?
     
  12. Feb 22, 2006 #11
    The n-1 term ???

    /Fred


     
  13. Feb 22, 2006 #12

    arildno

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    No.
    Let's see:
    We have:
    [tex](-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)[/tex]
    Try to deduce how the +++++ will look like..:wink:
     
  14. Feb 22, 2006 #13
    The n + 1 term ???

    /Fred

     
  15. Feb 22, 2006 #14

    arildno

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    What are you talking about?
    Try to figure out the rest by yourself
     
  16. Feb 22, 2006 #15
    I'v have been looking at mathworld, and then arrieved at the following idear:
    Then I need to do it for infinity????

    /Fred


    [tex]S_n = \sum_{k=1} ^ {\infty} (-1)^k \frac{1}{1-x^k} [/tex]


     
    Last edited: Feb 22, 2006
  17. Feb 22, 2006 #16

    arildno

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    Try to tackle it this way:
    The infinite series starting with the n'th term is of the form:
    [tex]\sum_{i=n}^{\infty}(-1)^{i}x^{2i}[/tex]
    Now, introduce the new index j=i-n, that is i=j+n
    Then, the series can be written in the form:
    [tex]\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}[/tex]
    What can you do about this expression?
     
  18. Feb 22, 2006 #17
    I rewrite it as a product of a two Sums ??

    /Fred

     
  19. Feb 22, 2006 #18

    arildno

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    No, draw out the common factor!
    [tex]\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}[/tex]
    What is the j-sum equal to?
     
  20. Feb 22, 2006 #19
    ??


    [tex]\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} x^{2n} [/tex]

    /Fred
     
    Last edited: Feb 22, 2006
  21. Feb 22, 2006 #20

    arildno

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    Please write out the first few terms of the series [itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j}[/itex]
     
    Last edited: Feb 22, 2006
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