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Thread: Multiplying power series - Maclaurin series

  1. #1
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    Multiplying power series - Maclaurin series

    Hi, I've run into some trouble trying to answer this power series question that asks:

    Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the given function.

    $\displaystyle e^{7x}ln(1-x/5)$

    This is my work so far (all sums are n=0 to infinity):
    $\displaystyle e^x$ = $\displaystyle \sum$ $\displaystyle \frac{x^n}{n!}$

    $\displaystyle e^{7x}$ = $\displaystyle \sum$$\displaystyle \frac{7^nx^n}{n!}$

    and
    $\displaystyle
    ln(1+x)$ = $\displaystyle \sum$ $\displaystyle (-1)^{n-1}\frac{x^n}{n}$

    $\displaystyle ln(1-x/5)$ = $\displaystyle \sum$ $\displaystyle (-1)^{n-1}(-1)^n\frac{5^{-n}x^n}{n}$
    $\displaystyle ln(1-x/5)$ = $\displaystyle \sum$ $\displaystyle (-1)^{2n-1}\frac{x^n}{n5^n}$

    To find the first 3 non-zero terms, I think I'm supposed to do something like the following:

    n=3 $\displaystyle \longrightarrow$ $\displaystyle a_0b_3+a_1b_2+a_2b_1+a_3b_0$
    n=2 $\displaystyle \longrightarrow$ $\displaystyle a_0b_2+a_1b_1+a_2b_0$
    n=1 $\displaystyle \longrightarrow$ $\displaystyle a_0b_1+a_1b_0$
    n=0 $\displaystyle \longrightarrow$ $\displaystyle a_0+b_0$

    What I got from doing that was:
    ($\displaystyle (-1)^{2n-1}$ becomes -1 for all n)

    n=3
    $\displaystyle -\frac{(7x)^0}{0!}*\frac{x^3}{3(5^3)}-\frac{(7x)^1}{1!}*\frac{x^2}{2(5^2)}-\frac{(7x)^2}{2!}*\frac{x^1}{1(5^1)}-\frac{(7x)^3}{3!}*\frac{x^0}{0(5^0)} $
    = $\displaystyle -\frac{x^3}{375}-\frac{7x^3}{50}-\frac{49x^3}{10}-\frac{343x^3}{0}$
    = $\displaystyle -\frac{3781x^3}{375}-\frac{343x^3}{0}$

    n=2
    $\displaystyle -\frac{(7x)^0}{0!}*\frac{x^2}{2(5^2)}-\frac{(7x)^1}{1!}*\frac{x^1}{1(5^1)}-\frac{(7x)^2}{2!}*\frac{x^0}{0(5^0)}$
    = $\displaystyle -\frac{x^2}{50}-\frac{7x^2}{5}-\frac{49x^2}{0}$
    = $\displaystyle -\frac{71x^2}{50}-\frac{49x^2}{0}$

    n=1
    $\displaystyle -\frac{(7x)^0}{0!}*\frac{x^1}{1(5^1)}-\frac{(7x)^1}{1!}*\frac{x^0}{0(5^0)}$
    = $\displaystyle -\frac{x}{5}-\frac{7x}{0}$

    n=0
    $\displaystyle -\frac{(7x)^0}{0!}-\frac{x^0}{0(5^0)}$
    $\displaystyle -1-\frac{1}{0}$

    How do I get my 3 terms from the above? I have a lot of fractions where I'm dividing by 0.
    Edit: I just realized that I changed $\displaystyle e^{7x}$ into a power series, would the Maclaurin series for that function be the same?
    Last edited by triggy; Mar 7th 2009 at 06:04 AM. Reason: Fixing grammar
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  2. #2
    Moo
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    Hello,

    I haven't checked your formulae, it looks long


    However, note that the power series of the exponential starts at n=0, and the power series of the logarithm stars at n=1.
    This should solve the problem of dividing by 0.

    The MacLaurin series is a power series. The expansion of a function into a series is unique. So expanding a function as a power series is equivalent to getting the MacLaurin series. (I don't know if I'm clear here ? :s)



    To simplify calculations, you can use the Cauchy product for series. But again, be careful that the series should start at the same point
    Last edited by Moo; Mar 8th 2009 at 12:51 PM. Reason: removing a useless parenthesis
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  3. #3
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    Alright, so if I changed the index for the logarithmic series to start at 1, the following:

    n=3
    n=2
    n=1
    n=0

    ..would change to:

    n=3 $\displaystyle \longrightarrow$ $\displaystyle a_0b_3 + a_1b_2 + a_1b_1$
    n=2 $\displaystyle \longrightarrow$ $\displaystyle a_0b_2 + a_1b_1$
    n=1 $\displaystyle \longrightarrow$ $\displaystyle a_0b_1$

    Would that be correct?
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  4. #4
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    Actually, forget about the last reply.

    I rechecked my work and fixed the power series for the ln function. I'll show my work for it.

    $\displaystyle ln(1+x) = \sum (-1)^{n+1}\frac{x^{n+1}}{n+1}$

    $\displaystyle ln(1-x/5) = \sum (-1)^{n+1}\frac{(-\frac{x}{5})^{n+1}}{(n+1)}$
    $\displaystyle ln(1-x/5) = \sum (-1)^{n+1}(-1)^{n+1}\frac{x^{n+1}}{(n+1)5^{n+1}}$
    $\displaystyle ln(1-x/5) = \sum (-1)^{2n+2}\frac{x^{n+1}}{(n+1)5^{n+1}}$
    $\displaystyle ln(1-x/5) = \sum \frac{x^{n+1}}{(n+1)5^{n+1}}$


    $\displaystyle a_n$ = $\displaystyle \sum \frac{(7x)^n}{n!}$
    $\displaystyle b_n$ = $\displaystyle \sum \frac{x^{n+1}}{(n+1)5^{n+1}}$

    With the n=3 $\displaystyle \longrightarrow a_0b_3 + a_1b_2 + ...$ etc. and same for n=2, n=1, n=0

    I got:
    $\displaystyle n=0 \longrightarrow 1+\frac{x}{5}$
    $\displaystyle n=1 \longrightarrow \frac{71x^2}{50}$
    $\displaystyle n=2 \longrightarrow \frac{3782x^3}{750}$
    $\displaystyle n=3 \longrightarrow \frac{89568x^4}{7500}$

    So are the first 3 non-zero terms of the series: $\displaystyle 1, \frac{x}{5},$ and $\displaystyle \frac{71x^2}{50}$?
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  5. #5
    Moo
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    There is a little problem in the Taylor series for the logarithm !

    The formula is $\displaystyle \ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$
    which is $\displaystyle \sum_{n=0}^\infty (-1)^{\color{red}n} \frac{x^{n+1}}{n+1}$

    so $\displaystyle \ln(1-x/5)={\color{red}-} \sum_{n=0}^\infty \frac{\left(\tfrac x5\right)^{n+1}}{n+1}$


    this last modification should give you the correct answers


    Edit : there is another problem, $\displaystyle c_0=a_0b_0=-\frac x5$. There is no 1
    Last edited by Moo; Mar 8th 2009 at 12:21 AM.
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