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Math Help - Maclaurin series question

  1. #1
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    Maclaurin series question

    Hi

    I need help on the following question:
    1) Obtain the Maclaurin Series for e^x and ln(1+x). Hence obtain by multiplication the Maclaurin series for e^{x}ln(1+x) giving the first three terms.

    I have worked out the series for the  e^{x} and ln(1+x) by themselves.
    I then tried to multiple the values together, but was unsuccessful.

    e^{x} series is:

    1 + x + \frac{x^2}{2} + \frac{x^3}{6}

    ln(1+x) series is:

    ln(1) + x - \frac{x^2}{2} + \frac{2x^3}{6}

    P.S
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  2. #2
    MHF Contributor matheagle's Avatar
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    It's called the Cauchy product and it's not that clever.
    Just start multiplying the two infinite sums.
    Ln(1)=0, by the way.

     \biggl(a_1+a_2+\cdots \biggr) \biggl(b_1+b_2+\cdots \biggr)=a_1b_1+a_1b_2+a_2b_1+a_2b_2+ \cdots
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  3. #3
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    So if you want only the first three terms, itís a simple case of multiplying multinomials at this point. Evaluate the multiplication:

    \left(1 + x + \frac{x^2}{2}\right)\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)

    and take the first three terms.

    If, on the other hand, you want an expression for the Maclaurin series of e^{x}\ln(1+x), you could multiply together the individual series:

    \left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)\left(  \sum_{j=1}^\infty\frac{(-1)^j}{j}x^j\right)=\sum_{j=1}^\infty\sum_{k=0}^\in  fty\frac{(-1)^jx^{j+k}}{jk!}
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  4. #4
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    ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

    x - \frac{x^2}{2} + \frac{x^3}{3} ,

    x^2 - \frac{x^3}{2} + \frac{x^4}{3} ,

    \frac{x^3}{2} - \frac{x^4}{4} + \frac{x^6}{6}
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Paymemoney View Post
    ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

    x - \frac{x^2}{2} + \frac{x^3}{3} ,

    x^2 - \frac{x^3}{2} + \frac{x^4}{3} ,

    \frac{x^3}{2} - \frac{x^4}{4} + \frac{x^6}{6}

    add these now, it's only one series in the end
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  6. #6
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    Quote Originally Posted by matheagle View Post
    add these now, it's only one series in the end
    This is what i have:

    x - \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^6}{6}

    book's answer is:

    1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...... ,
    x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + .......,
    x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ......,
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    book's answer is:
    1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...... ,
    x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +  .......,
    x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} +  ......,
    This is for all three answers:
    e^x=1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots,
    \ln(1+x)=x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +\ldots, and
    e^x\ln(1+x)=x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} +\ldots

    Quote Originally Posted by Paymemoney View Post
    ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

    x - \frac{x^2}{2} + \frac{x^3}{3} ,

    x^2 - \frac{x^3}{2} + \frac{x^4}{3} ,

    \frac{x^3}{2} - \frac{x^4}{4} + \frac{x^6}{6}
    I believe you have the last one wrong.
    \left(1 + x + \frac{x^2}{2}\right)\left(x - \frac{x^2}{2} +   \frac{x^3}{3}\right)=\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)+\ \left(x^2-\frac{x^3}{2}+\frac{x^4}{3}\right)+\left(\frac{x^3  }{2}-\frac{x^4}{4}+\frac{x^5}{6}\right) =x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{12}+\fra  c{x^5}{6}

    The first three terms here are of course all thatís correct.
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