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Math Help - Multiplication of two Taylor series

  1. #1
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    Multiplication of two Taylor series

    I'm trying to multiply two Taylor series together to find the quartic Taylor polynomial about 0 of a function

    <br />
(1+\frac{1}{2}x^2+\frac{1}{6}x^3+....)(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)<br />

    <br />
= 1(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)+\frac{1}{2}x^2(x-\frac{1}{2}x^2+...)+\frac{1}{6}x^3(x-...)<br />

    <br />
= x+(-\frac{1}{2})x^2+(\frac{1}{3}+\frac{1}{2})x^3+(-\frac{1}{4}+\frac{1}{6})x^4<br />

    <br />
= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{12}x^4<br />

    The problem is when I enter the function into my computer maths program(mathcad) I get.

    <br />
= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{3}x^4<br />

    where am I going wrong

    Oh the function by the way is

    <br />
f(x)= (e^x-x)(\ln(1+x))<br />
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  2. #2
    Eater of Worlds
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    Quote Originally Posted by macca101
    I'm trying to multiply two Taylor series together to find the quartic Taylor polynomial about 0 of a function

    <br />
(1+\frac{1}{2}x^2+\frac{1}{6}x^3+....)(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)<br />

    <br />
= 1(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)+\frac{1}{2}x^2(x-\frac{1}{2}x^2+...)+\frac{1}{6}x^3(x-...)<br />

    <br />
= x+(-\frac{1}{2})x^2+(\frac{1}{3}+\frac{1}{2})x^3+(-\frac{1}{4}+\frac{1}{6})x^4<br />

    <br />
= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{12}x^4<br />

    The problem is when I enter the function into my computer maths program(mathcad) I get.

    <br />
= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{3}x^4<br />

    where am I going wrong

    Oh the function by the way is

    <br />
f(x)= (e^x-x)(\ln(1+x))<br />
    You're missing a -1/4 term

    -1/4-1/4+1/6=-1/3


    x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{3}}{2}-\frac{x^{4}}{4}+\frac{x^{4}}{6}=x-\frac{x^{2}}{2}+\frac{5x^{3}}{6}-\frac{x^{4}}{3}
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  3. #3
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    Quote Originally Posted by galactus
    [b]You're missing a -1/4 term
    Of course I am

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