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Taylor Polynomial question

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Here's a screenshot of the problem: http://puu.sh/2Bta5 [Broken]


    2. Relevant equations



    3. The attempt at a solution
    As can be seen by the screenshot, the answer's already given, but I'm not sure how to go about getting it. This one has me stumped since e-4x and sin(5x) are both Maclaurin series that I can easily find. But I'm not sure where to go from there, or if I'm even supposed to use both of the Maclaurin series.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 16, 2013 #2

    Mark44

    Staff: Mentor

    It looks like you are supposed to write a few terms of the Maclaurin series for e-4x in one area and a few terms for the series for sin(5x) in the other area.
     
    Last edited by a moderator: May 6, 2017
  4. Apr 16, 2013 #3

    Dick

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    Using both the Maclaurin series is probably the easiest way to do it. Just multiply them together. Ignore products that will give you powers greater than 3.
     
    Last edited by a moderator: May 6, 2017
  5. Apr 16, 2013 #4
    Ok, that's what I thought I should do but wasn't sure if that was allowed or not.
    Is this correct so far? :

    http://puu.sh/2BySD [Broken]

    I'll try this too, thanks.
     
    Last edited by a moderator: May 6, 2017
  6. Apr 16, 2013 #5

    Dick

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    No, not correct. You don't multiply series term by term. You multiply them like polynomials. Write out each series up to power x^3 and then foil them. Like polynomials. Drop powers bigger than 3.
     
    Last edited by a moderator: May 6, 2017
  7. Apr 16, 2013 #6
    Ah, ok. I think I understand now.
     
  8. Apr 16, 2013 #7
    Never mind, still not getting it. Looking at these series for e^-4x and sin(5x), wouldn't that mean I need 4 terms (n=0 to n=3) for e^-4x to make that 3rd degree, and 2 terms (n=0 to n=1) for sin(5x) to make that 3rd degree?

    So for e^-4x I have:
    1 - 4x + 8x - (64/6)x3

    For sin(5x):
    5x + (125/6)x3

    FOILing those out and ignoring powers greater than 3, I get:

    5x + (125/6)x3 - 20x2 + 40x3

    Is there something else I'm doing wrong?
     
  9. Apr 16, 2013 #8

    Dick

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    For sin(5x):

    5x - (125/6)x3

    Not plus.
     
  10. Apr 16, 2013 #9
    Ok, never mind, found my error. That + (125/6)x3 should actually be a - (125/6)x3
     
  11. Apr 16, 2013 #10
    Just caught it. Thanks for your help, much appreciated.
     
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