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Math Help - Taylor Polynomial

  1. #1
    Member akhayoon's Avatar
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    Taylor Polynomial

    Question was

    evaluate the 4th taylor polynomial of \sqrt{3+x^{2}} around a=-1


    my answer was

    f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}

    the marker notes however that there was something missing in the 3rd term

    that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...

    but after looking at my derivatives I can't figure out why this is???
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by akhayoon View Post
    Question was

    evaluate the 4th taylor polynomial of \sqrt{3+x^{2}} around a=-1


    my answer was

    f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}

    the marker notes however that there was something missing in the 3rd term

    that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...

    but after looking at my derivatives I can't figure out why this is???
    f'(x) = x(3+x^2)^{-1/2},\ \ f'(-1) = {\textstyle-\frac12}.

    f''(x) = (3+x^2)^{-1/2} - x^2(3+x^2)^{-3/2},\ \ f''(-1) = {\textstyle\frac12 - \frac18 =\frac38}.

    f'''(x) = -3x(3+x^2)^{-1/2} + 3x^3(3+x^2)^{-5/2},\ \ f'''(-1) = {\textstyle\frac38 - \frac3{32} =\frac9{32}}.

    That gives f_3(x) = 2 - \frac{(x+1)}{2} + \frac{3(x+1)^{2}}{(8)2!} + \frac{9(x+1)^{3}}{(32)3!}. I'll leave you to sort out the (x+1)^4 term.
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