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Multiplication of Maclaurin Series

  1. Aug 16, 2008 #1
    I have the following problem:

    find the first 3 non-zero terms in the Maclaurin series for the function:

    e-x2 + Cos[x]

    I know in this case, the series behave like polynomials and I have done the following. The left expression is the first 3 terms of the e portion of the problem, and the second expression is the first 3 terms of Cosx.

    (1 - x2+ x4/2)(1 - x2/2 + x4/24)

    this =

    1 - x2/2 + x4/24 - x2 - x4/2 - x6/24 + x4 - x6/4 + x8/48

    How do I know which terms are the "first 3 non-zero terms" of this series?

    Thanks - the answer is attached, I just don't understand how the polynomial, after multiplied out is consolidated at the end.

    Jeff power series multiplication answer.jpg
     
  2. jcsd
  3. Aug 16, 2008 #2

    tiny-tim

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    Hi Jeff! :smile:

    (You meant e-x2 *Cos[x] :wink:)

    The "first three terms" would be 1 + 0.x - x2/2 .

    "non-zero" simply means that you skip over "0.x" and "0.x3" :smile:
    You can always change the order of the terms of a series (except if you're using an infinite number of terms, in which case there are rules to follow :wink:).
     
  4. Aug 16, 2008 #3
    Tiny-Tim,

    Thanks, and yes - not sure how "+" found its way in there :-)

    2 Questions -

    What is "0.x" and since the 2 original series are infinite, isn't the product of the series infinite as well?

    So, just to confirm: If I take the first 3 terms of each Taylor polynomial and multiply through (line 4 in the image), I can use any 3 non zero terms of that product? What convention compelled them to use:

    1 - 1.5x2 +25/24x4 as the answer to the question?

    Many Thanks,

    Jeff
     
  5. Aug 16, 2008 #4

    tiny-tim

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    I meant 0 times x.

    Yes, it is infinite, but you're only using a few terms at the beginning.
    Nooo … you must use all the terms of the three lowest powers (x0 x2 and x4).

    Those are the "first 3 non-zero terms". :smile:
     
  6. Aug 16, 2008 #5
    Great - that clarifies it perfectly. Thanks for your help, Tim.

    Jeff
     
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