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By choosing a suitable value of x, sum the series

  1. Dec 9, 2011 #1
    The question:
    10535uf.jpg


    Here's my thought so far:
    162a9e9.jpg


    This is pretty much what I have come up with. However, I don't know how to find a general solution for the sum of the series. Can someone help please. (the decimal part is where I'm stuck the most)
     
  2. jcsd
  3. Dec 9, 2011 #2

    SammyS

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    [itex]\displaystyle\frac{1}{1\cdot3}-\frac{3}{5\cdot7}+\frac{5}{9\cdot11}-\frac{7}{13\cdot15}+\dots[/itex]

    What do we see?

    The signs of the terms alternate.

    Each successive numerator is the next odd natural number. If n is the index and starts at 0, then each numerator is (2n-1).

    How are the two factors in each denominator related to each numerator ? (Consider what 2 times the numerator is for each.)
     
  4. Dec 9, 2011 #3
    My guess is that those are not decimal points - they are multiplication dots.

    Presumably you have noticed the alternating signs on the terms.

    There is a simple relationship between the numerator and the two terms in the denominator in each term which will probably help.
     
  5. Dec 9, 2011 #4
    Um... to be honest, I can't see how each denominator relates to each numerator, doesn't seem like there's a sequence. :confused:
     
  6. Dec 9, 2011 #5
    Yes I have noticed the alternating signs, but couldn't find any relationship between the numerator and the denominator though :(
     
  7. Dec 9, 2011 #6
    Well, calculate out the first ten terms, then come back and post them all.

    [itex]+\frac{1}{1\times3}[/itex]

    [itex]-\frac{3}{5\times7}[/itex]

    [itex]+\frac{5}{9\times11}[/itex]

    [itex]-\frac{7}{13\times15}[/itex]

    [itex]\ldots[/itex]
     
  8. Dec 9, 2011 #7
    You might also put in the known values for a, b and c (1,1,3) in your formula for [itex]n_{m}[/itex] and simplify.
     
  9. Dec 9, 2011 #8

    SammyS

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    Compare the factors in each denominator to 2 times the numerator !!
     
  10. Dec 10, 2011 #9
    Just a small typo--I'm guess you meant the numerator is 2n+1.

    bubokribuck: What is the first part of this problem (that you've cut off from the image you posted)? The statement of the problem ("By choosing a suitable value of x...") indicates that in previous parts of this problem you've found a series representation for some function, which you are supposed to use to solve this part.
     
  11. Dec 10, 2011 #10
    Hi spamiam, here's the full question, but I don't like it has anything to do with the previous parts though.

    4ihjxd.jpg
     
  12. Dec 10, 2011 #11
    Hi Joffan, here it is.

    bf2o9c.jpg
     
  13. Dec 10, 2011 #12
    If the first term is n=1, then the numerator of the first term would be 2n-1 right? (as 2(1)-1=1)
     
  14. Dec 10, 2011 #13
    Ah, I think I am getting it now!

    Let N denotes the numerator, and D denotes the denominator, then D=(2N-1)(2N+1). Is this right? So for the nth term, it would be something like (-1)n+1[itex]\frac{2n-1}{(4n-3)(4n-1)}[/itex]
     
    Last edited: Dec 10, 2011
  15. Dec 10, 2011 #14
    It definitely has to do with the previous part! What could you be choosing x for if not to plug into the Fourier series you found in an earlier part? So what did you get for the first 2 parts? In particular, what did you get for the Fourier series of the function?

    You formula for the general term of the series looks correct.
     
  16. Dec 10, 2011 #15

    SammyS

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    If the index, n, starts at zero, (like I said), then, yes, that should be (2n+1).
     
  17. Dec 10, 2011 #16
    Hi spamiam, here's what I've got for the Fourier series:
    1p8dix.jpg

    And here's the sigma equation I've come up with for the last part of the question.
    s30ltd.jpg
     
  18. Dec 10, 2011 #17
    Hm, that's not what I get for the Fourier series. Could you show your calculations? Also, your series has a couple of oddities. For one thing, when n=1, you have terms of the form 1/0. Hopefully this doesn't pose a problem since these terms should cancel out. Which brings me to the second observation--you can simplify your sum since terms cancel out when n is odd.
     
  19. Dec 10, 2011 #18
    Hi, the calculations are quite long, so I've attached a document. :)
     

    Attached Files:

  20. Dec 11, 2011 #19
    Hey bubokribuck,

    I (and I'm guessing this is the case with most other members from the 0 views your attachment has gotten) am wary of downloading zip files as a general rule. Could you upload it in another format like .doc, .pdf, or .jpg?
     
  21. Dec 11, 2011 #20
    Hi, the doc exceeds the upload size limit, therefore I can only upload png. Thanks :)
     

    Attached Files:

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