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Math Help - Taylor series, please check

  1. #1
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    Taylor series, please check

    given

    f(x) = 36/(6+x)^2 and g(x) = 36/(1+x)^2

    By writing f(x) = 1/(1+1/6x)^2


    I found the Taylor series about 0 for f up to term in x^3,

    = 1 - 1/3x + 1/12x^2-1/54X^3 .........(a)

    Valid for -6<x<6

    Then g(x) can be written as 1/(1+(x-5)/6)^2

    Replace x in solution 1 by x-5 ???????

    Solution

    1 - 1/3(x-5) + 1/12(x-5)^2-1/54(X-5)^3 ......(b)...?????

    Valid for -1<x<11

    Not sure how to check the first four terms in the Taylor seris found in part b by finding the first, second and third derivatives of g and using these to find the cubic Taylor polynomial about 5 for g.

    Thanx a lot.
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  2. #2
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    Quote Originally Posted by fair_lady0072002
    given

    f(x) = 36/(6+x)^2 and g(x) = 36/(1+x)^2

    By writing f(x) = 1/(1+1/6x)^2


    I found the Taylor series about 0 for f up to term in x^3,

    = 1 - 1/3x + 1/12x^2-1/54X^3 .........(a)

    Valid for -6<x<6

    Then g(x) can be written as 1/(1+(x-5)/6)^2

    Replace x in solution 1 by x-5 ???????

    Solution

    1 - 1/3(x-5) + 1/12(x-5)^2-1/54(X-5)^3 ......(b)...?????

    Valid for -1<x<11

    Not sure how to check the first four terms in the Taylor seris found in part b by finding the first, second and third derivatives of g and using these to find the cubic Taylor polynomial about 5 for g.

    Thanx a lot.
    It looks like a right substitution to me. Though the second series is a Taylor's series for 'g' centered at x=5. Do you want to "uncenter" it?
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  3. #3
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    Is the range of validity in the first post correct for the second series. Should it not be -11>x>1? Sorry if thats what you are hinting at in the previous post.
    Just keen to learn!
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