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Math Help - multipication of power series

  1. #1
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    multipication of power series

    I was wondering if someone could please check my work because i keep getting this one wrong:

    so the function is:
     e^8x ln\left(1 - \frac{x}{3}\right)

    and so far i have
    e^8x = 1 + 8x + \frac{64x^2}{2!} + ...
     ln\left(1 - \frac{x}{3}\right) = -x - \frac{x^2}{6}<br />
 - \frac{x^3}{27}

    and then when i multiple (i only need first 3 nonzero terms) and collect like terms i get:
     = -x - \frac{49x^2}{6} - \frac{37x^3}{27}
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  2. #2
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    Hello,
    Quote Originally Posted by Tascja View Post
    I was wondering if someone could please check my work because i keep getting this one wrong:

    so the function is:
     e^8x ln\left(1 - \frac{x}{3}\right)

    and so far i have
    e^8x = 1 + 8x + \frac{64x^2}{2!} + ...
     ln\left(1 - \frac{x}{3}\right) = -x - \frac{x^2}{6}<br />
 - \frac{x^3}{27}

    and then when i multiple (i only need first 3 nonzero terms) and collect like terms i get:
     = -x - \frac{49x^2}{6} - \frac{37x^3}{27}
    that's because you got wrong in the logarithm power series :
    \ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots

    so for \ln(1-x/3), it should be -{\color{red}\frac x3}-\frac{({\color{red}x/3})^2}{2}-\frac{({\color{red}x/3})^3}{3}-\dots=-\frac x3-\frac{x^2}{18}-\frac{x^3}{81}-\dots

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  3. #3
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    oh okay thanks that makes a lot more sense!
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