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Maclaurin series

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data
    find the maclaurin series of e^(3x) + e(-3x)

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure about finding taylor and maclaurin series, I understand perfectly how to find the terms of the series...But how do I put it into a general term do I just have to recognize the pattern by looking at it? anyway...

    to attempt this problem I was thinking you use the known maclaurin series for e^x which has a general term of x^n / n! it will be (3x)^n / n! for the first term and (-3x)^n / n! for the 2nd term I'm unsure of combining these The answer is 2*3^2n*x^2n/(2n)! What I do not understand is how they got (2n)! I thought it should be 2(n!) and also I thought there should be a negative in the numerator...can anyone explain to me these parts I do not understand...my textbook has one small paragraph on these series and it does not explain much...
  2. jcsd
  3. Apr 22, 2008 #2

    what happens to the terms in your two sums with n even, what to the terms with n odd?
  4. Apr 22, 2008 #3


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    the terms with odd 'n' cancel each other, so the sum is only over terms n=0,2,4,6,... etc

    then you can rewrite that as a sum over m=0,1,2,3,... etc
    then rename m->n

    [edit: whoops sorry about stealing your thunder, Pere. You posted too quick for me.]
  5. Apr 22, 2008 #4
    ahh I see thank you, I'm wondering though this series looks a little strange if there was one like this that I didn't have a known series for I dont think I would have ever found the general term by just looking at it...and also why does the denominator become (2n)! i thought it would be 2(n!)
  6. Apr 22, 2008 #5
    I don't see how you could get a 2(n!) in the denominator. How did you get it?

    As olgranpappy said, all the odd-n terms cancel. So the n=1 term of the first sum cancels with the n=1 term of the second sum and so on. What happens to the terms with n even..? Show some more work. :smile:
  7. Apr 22, 2008 #6
    well actually I don't think its 2(n!) anymore either I got the general term by adding the two modified E^x general terms together which would still have just a single n! in the denominator..the even terms get multiplied by 2 but that is why I thought it would be 2*(n!) instead of (2n)!
    Last edited: Apr 22, 2008
  8. Apr 22, 2008 #7
    True, you get the n.th general term by adding the n.th general terms of the modified E^x sums, as you put it.

    The result of this "adding" is however zero for odd n.. do you agree?

    So in your final sum, all terms for n odd are zero ...so practically you're only summing over even integers n ...so you can just as well replace n by 2m and some over all integers m ... think about it, write down the first 20 terms or so, it's not hard :smile:
  9. Apr 22, 2008 #8
    I agree that the odd terms are zero because they cancel; I think I am thinking of this (2n)! in the wrong way...so if it was say the 2nd term it would be (2!) correct? why wouldn't it be (4!) because (2*2)! = 4! ? thats where I'm confused...I understand that only the even terms have value so we're basically summing the even terms

    EDIT: I think I am finally getting it because the odd terms drop out then what was once the 3rd term is basically now the 2nd term so then the bottom is (2*1)! which is why it works like this thanks a lot
    Last edited: Apr 22, 2008
  10. Apr 22, 2008 #9
    I think I finally get it because the odd terms drop out what was once the 3rd term is now basically the 2nd term since so now the 2nd term will have (2*1)! and the third term will have (2*2)! because the third term will be the term where n=4
  11. Apr 22, 2008 #10


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    Gold Member

    I don't know if this clarifies anything for you but (2*n)! simple means the number is even and likewise (2n+1)! means its odd.
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