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Math Help - Taylor/Maclaurin polynomials

  1. #1
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    Taylor/Maclaurin polynomials

    Hey guys, I just wanted to make sure that I did this question correctly:

    The fourth-order Taylor polynomial for f(x) = \frac {1}{x} based at x = 1 is  P4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4. Find the fourth-order Maclaurin polynomial for  g(x) = \frac {1}{(1 + x)}

    So all I did was place (1 + x) wherever I see an x up in the original function, which would mean...

     1 - ((1 + x)-1) + ((1 + x)-1)^2 - ((1 + x) - 1)^3 + ((1 + x) -1)^4

    which would then mean the final answer is:
     1 - x + x^2 - x^3 + x^4 right?
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  2. #2
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    Quote Originally Posted by larson View Post
    Hey guys, I just wanted to make sure that I did this question correctly:

    The fourth-order Taylor polynomial for f(x) = \frac {1}{x} based at x = 1 is  P4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4. Find the fourth-order Maclaurin polynomial for  g(x) = \frac {1}{(1 + x)}

    So all I did was place (1 + x) wherever I see an x up in the original function, which would mean...

     1 - ((1 + x)-1) + ((1 + x)-1)^2 - ((1 + x) - 1)^3 + ((1 + x) -1)^4

    which would then mean the final answer is:
     1 - x + x^2 - x^3 + x^4 right?
    You could have also rewritten

    \frac{1}{1+x}=\frac{1}{1-(-x)} This is the def of a geometric series with r =-x so you would get

    \sum_{n=0}^\infty(-x)^n=\sum_{n=0}^\infty(-1)^nx^n

    if you write out the first few terms you get...

    1-x+x^2-x^3+x^4
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