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Math Help - power series representation!

  1. #1
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    power series representation!

    I'm studying for my final exam in calculus and I'm not exactly sure how to turn this particular natural log function into a power series representation..

    f(x) = (x+1)ln(1−x)

    I know what the answer is because it's just a practice question but I have no idea how to get it! help! thank you!
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  2. #2
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    Quote Originally Posted by khood View Post
    I'm studying for my final exam in calculus and I'm not exactly sure how to turn this particular natural log function into a power series representation..

    f(x) = (x+1)ln(1−x)

    I know what the answer is because it's just a practice question but I have no idea how to get it! help! thank you!
    Assume that the logarithm can be expressed as a polynomial. I.e.

    \ln{(1 - x)} = a + bx + cx^2 + dx^3 + \dots

    If x = 0 then

    \ln{1} = a \implies a = 0


    Take the derivative, you should find

    -(1 - x)^{-1} = b + 2cx + 3dx^2 + 4ex^3 + \dots

    If x = 0 then

    -1 = b.


    Take the derivative again, you should find

    -(1 - x)^{-2} = 2c + 6dx + 12ex^2 + 20fx^3 + \dots

    If x = 0 then

    -1 = 2c \implies c = -\frac{1}{2}.


    Take the derivative again, you should find

    -2(1 - x)^{-3} = 6d + 24ex + 60fx^2 + \dots

    If x = 0 then

    -2 = 6d \implies d = -\frac{1}{3}


    Follow this procedure and you should find

    \ln{(1 - x)} = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots.


    So (x + 1)\ln{(1 - x)} = (x + 1)\left(-x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots\right).

    I'll let you do the expanding.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    Find the power series for ln(1-x) Multiply the result by (x+1) collect like terms
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  4. #4
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    i did that but I don't understand how to put the (x+1) back into the series representation for ln(1-x).
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  5. #5
    MHF Contributor chisigma's Avatar
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    The series expansion of ln (1-x) is well known...

    \ln (1-x) = - x - \frac {x^{2}}{2} - \frac{x^{3}}{3} - \dots = - \sum_{n=1}^{\infty} \frac {x^{n}}{n}

    Using simple multiplication rule you obtain...

    (1+x)\cdot \ln (1-x)= - x - (1+\frac{1}{2})\cdot x^{2} - (\frac{1}{2} + \frac{1}{3})\cdot x^{3} - \dots = -x - \sum_{n=2}^{\infty} \frac{2n-1}{n\cdot (n-1)}\cdot x^{n}

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Like what chisigma did --- if you are interested I took a slighlty different approach based on what I used to use to find power series representations to solutions of differential equation, playing with the indices

    See attachment
    Attached Files Attached Files
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