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Math Help - Maclaurin Polynomial

  1. #1
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    Maclaurin Polynomial

    Good morning,

    I am working on Taylor Polynomials this morning, and am not entirely sure I understand what I am supposed to be doing. Would someone mind telling me if the following makes sense?

    Compute the Maclaurin Polynomials of degree one, two, and three for f(x) = e^{-2x}

    As I understand this the fact that this is a Maclaurin Polynomial simply means that the term (x-a)^n becomes simply x^n .

    I get the following:

    f'(x)=-2e^{-2x}


    f''(x)=4e^{-2x}


    f'''(x)=-8e^{-2x}


    f'(0)=-2


    f''(0)=4


    f'''(0)=-8


    T_1(x)=1-2x


    T_2(x)=(2x-1)^2


    T_3(x)=-8x^3+4x^2-2x+1

    Am I way off target here?
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  2. #2
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    Don't forget to divide by the factorial.
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  3. #3
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    Ah, indeed...

    Which instead produces:


    T_1(0)=-2x+1


    T_2(0)=2x^2-2x+1


    T_3(0)=-\frac{4}{3}x^3+2x^2-2x+1


    I had to look up the notation n!. Apparently it means n(n-1)(n-2)...(2)(1)). Is that correct?


    Is that all there is to Taylor Polynomials?
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  4. #4
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    Looking good there, now. That is what the factorial means. By convention, 0!=1. The generalization of the factorial is the gamma function, which is the reason for that convention. It also makes the MacLaurin expansion easier to write:

    f(x)=\displaystyle{\sum_{k=0}^{\infty}\frac{f^{(k)  }(0)}{k!}\,x^{k}},

    rather than separating out the earlier terms.

    Another convention is that the zeroth derivative of a function is simply the function itself.
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  5. #5
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    As much as I hate asking "is this right"...

    Is there an easy to verify a Taylor Polynomial so that I don't need to ask if my solution is correct?
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  6. #6
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    You can always go to WolframAlpha and use the Series command.

    Series[Exp[-2x],{x,0,3}]

    will give you the 3rd order MacLaurin expansion of the function in this thread.
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