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Maclaurin Series

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data
    find the Maclaurin Series:

    f(x)=(e^x - cos(x))/x

    2. Relevant equations

    3. The attempt at a solution

    I'm not really sure what to do I do know the Maclaurin series of cos(x) I was thinking that using this known series then doing e^x - the known series then dividinig the entire thing by x. I'm just not sure if this is allowed. Do you have to do the known series of e^x - the cos(x) series ? Someone please let me know.
  2. jcsd
  3. Apr 17, 2008 #2
    you know i didn't expect that to work but it does. [series(exp(x))-series(cos(x))]/x is the right answer
  4. Apr 17, 2008 #3

    Gib Z

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    Either your question wanted you to derive the series for Taylors theorem straight, or more likely not, show that each of those functions are analytic, as you can only do that trick for analytic functions. Most functions studied in analysis, however, are analytic =]
  5. Apr 17, 2008 #4


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    Are you saying you don't know the Maclaurin series for ex? Surely not! The crucial point here is that the first (constant) term of the Maclaurin series for both ex and cos(x) is 1- subtracting them cancels the constant term so that you can then divide by x and still have a Maclaurin series left!
  6. Apr 17, 2008 #5
    I do know the maclaurin series for e^x and for cos(x) :) oh yes you are right; but I do not know if it is acceptable to leave my answer in a form with just like the terms of the series or if it is standard to put the answer into a summation. I can't really figure out the pattern:( because the x^4 terms cancel out and then I am left with 0+ 1 + x + x^2/3! + x^5/^6! so what I did and I do not know if this is correct is subtracted the general terms and then divided by x to get: (x^(n-1))/n!)-((-1)^n*(x^2n-1))/(2n)!
  7. Apr 17, 2008 #6
    1/x isn't analytic
  8. Apr 17, 2008 #7
    well theres an example in the book with (1-cos(x^2))/x and they did it the same way. I don't really know what an "analytic function is" but I am pretty sure you can do it by subtracting the two series I just don't know if I subtract the general terms or should subtract like the first 5 terms from each series? Because when I subtracted the first 5 terms from each I was left with something kind of odd that I couldn't see a pattern in to put into a summation form.

    the terms I end up getting are:

    1 + x + (x^2 / 3!) + (x^5 / 6!) - (x^7/ 8!)
    Last edited: Apr 17, 2008
  9. Apr 17, 2008 #8


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    No, it isn't. However, since all of the functions in this problem are I don't see how that is relevant!
  10. Apr 17, 2008 #9


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    You know that the general term of the Maclaurin series for ex is 1/n! xn and that the general term of the Maclaurin series for cos(x) is (-1)n/(2n)!x2n or: "0 if n is odd (-1)n/2/n! xn if n is even". The general term of the Maclaurin series for ex- cos(x) is "(1/n!)xn if n is odd, [1/n!- (-1)n/2/(2n+1)!]xn if n is even". Notice that when n is 0 (even), that is 1- 1= 0. Dividing each term by x gives "(1/n!)xn-1 if n is odd, [1/n!- (-1)n/2/(2n+1)!]xn-1 if n is even"
  11. Apr 17, 2008 #10
    ah yes I understand what you are saying here would it be acceptable to write it like that though ? I do not understnad how you got (2n +1) however how did it go from (2n)! to (2n+1)! Since the cos part will be 0 for odd terms can you just say the series is the series you wrote for the even terms since when its odd the cos part will drop out anyway and will give you basically the odd series?
    Last edited: Apr 17, 2008
  12. Apr 17, 2008 #11
    how is the function in question analytic if it is (expx - cosx)*1/x? are you saying simply because it is (expx - cosx ) divided by x f(x) is analytic?
  13. Apr 17, 2008 #12


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    What I believe Halls is saying is that if each of the individual functions are analytic, then one can find a series representation of the entire function, whether the entire function is analytic or not.
  14. Apr 17, 2008 #13
    well this is definitely the way to do it we went over it today in class I asked for help as I was still a little confused but thanks guys I sorta get it now? lol i know that you need two summations one for the even and one for the odd getting to them is a little weird
  15. Apr 17, 2008 #14
    no i understand that but my question still stands.
  16. Apr 17, 2008 #15


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    I'm not sure what the question is, but (exp(x)-cos(x))/x only has a removable singularity at x=0. Once you plug that whole the resulting function is analytic.
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