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Thread: 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 $\displaystyle e^x$ and $\displaystyle ln(1+x)$. Hence obtain by multiplication the Maclaurin series for $\displaystyle e^{x}ln(1+x)$ giving the first three terms.

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

    $\displaystyle e^{x}$ series is:

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

    $\displaystyle ln(1+x)$ series is:

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

    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.

    $\displaystyle \biggl(a_1+a_2+\cdots \biggr) \biggl(b_1+b_2+\cdots \biggr)=a_1b_1+a_1b_2+a_2b_1+a_2b_2+ \cdots$
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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:

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

    and take the first three terms.

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

    $\displaystyle \left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)\left( \sum_{j=1}^\infty\frac{(-1)^j}{j}x^j\right)=\sum_{j=1}^\infty\sum_{k=0}^\in fty\frac{(-1)^jx^{j+k}}{jk!}$
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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.

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

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

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

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

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

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

    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:

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

    book's answer is:

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

    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.

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

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

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