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Coefficient for a Term in a Taylor Expansion for Cosine

  1. Jan 15, 2017 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    The coefficient of the term (x−π)2 in the Taylor expansion for f(x)=cos(x) about x=π is:

    2. Relevant equations
    ##cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}...##

    3. The attempt at a solution

    Unless my taylor series for cosine is incorrect, I'm not seeing any (x-π)2 term. What am I missing here? I see that the 2nd term is a squared term and that the coefficient is -1. Is that the answer?
     
  2. jcsd
  3. Jan 15, 2017 #2

    Ray Vickson

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    You need to apply the correct definition of a Taylor series. You are using the Maclaurin expansion, not the Taylor expansion.
     
  4. Jan 15, 2017 #3

    Drakkith

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    I see. Do you have a good reference that has the correct series? Everything I've found so far has the taylor series for cosine exactly as I have it. Very frustrating. :H
     
  5. Jan 15, 2017 #4

    fresh_42

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  6. Jan 15, 2017 #5

    Drakkith

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    Thanks Fresh. I usually avoid wikipedia when doing homework as I rarely find it helpful, but it appears this time I was incorrect.
     
  7. Jan 15, 2017 #6

    fresh_42

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    They are usually quite correct in the definition section and fast to look up. I mostly look it up in two languages, as it's not a 1:1 translation.
     
  8. Jan 15, 2017 #7

    Drakkith

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    Okay, wiki's definition with a = π:
    ##F(π) + \frac{F'(π)}{1!}(x-π) + \frac{F''(π)}{2!}(x-π)^2 + \frac{F'''(π)}{3!}(x-π)^3 + ...##

    The first few derivatives of cos(x) are:
    ##F(x) = cos(x)##
    ##F'(x) = -sin(x)##
    ##F''(x) = -cos(x)##
    ##F'''(x) = sin(x)##

    So that would be:
    ##cos(π) - sin(π)(x-π) - \frac{cos(π)}{2}(x-π)^2 + \frac{sin(π)}{6}(x-π)^3 + ...##

    or

    ##-1 - 0 - \frac{1}{2}(x-π)^2 + 0 + ...##

    So the coefficient would be -1/2.
     
  9. Jan 15, 2017 #8

    fresh_42

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    I haven't done it, but something like this, yes. Maybe you'll have to take a few terms more, as they all include lower powers of ##x##.
     
  10. Jan 15, 2017 #9

    Mark44

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    As already mentioned what you have is the Maclaurin series for the cosine function. A Maclaurin series is a series in powers of x (or powers of (x - 0)). A Taylor's series is one that is in powers of x - a. A Maclaurin series is a special case of a Taylor's series.

    Your textbook should have definitions of both types of series.
     
  11. Jan 15, 2017 #10

    Drakkith

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    Unfortunately I don't have my old textbook with me, so I can't look it up, and I didn't look at what the homework would be over before I left my house. I've had a really bad start to the semester thanks to some sleep issues and I'm still trying to get everything together and planned out.

    Thanks guys.
     
  12. Jan 16, 2017 #11

    Ray Vickson

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    Since you have already solved the problem I guess it is OK to now point out that we can write ##\cos(x) = -\cos(y)##, where ##y = \pi - x##; that is a standard trigonometric identity (and is obvious geometrically). When ##x## is near ##\pi##, ##y## is near 0, expanding in powers of ##y## makes sense.
     
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