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Math Help - Find Taylor Series

  1. #1
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    Find Taylor Series

    Hi, I could use some help on part (b) of the following question from my Complex Variables, book. I didn't know if this belonged in the Calculus forum or not.

    Question 3.2.5
    (a) Use the identity
    \frac1z = \frac1{(z+1)-1}

    to establish
    \frac1z = - \sum_{n=0}^\infty (z+1)^n \hspace{15 mm} \left|z+1 \right| < 1

    (b) Use the above identity to also establish
    \frac1{z^2} = \sum_{n=0}^\infty (n+1)(z+1)^n \hspace{15 mm} \left|z+1 \right| < 1

    Check the answer by differentiating the series from (a)

    Part (a) was easy:
    \frac1z = \frac{1}{(z+1)-1} = \frac{-1}{1-(z-1)} = -\sum_{n=0}^\infty(z+1)^n

    The check in part (b) is easy, too. But I can't do part (b) using the identity given. I tried squaring the identity, but this is as far as I've gotten (essentially, I just eliminated the negative sign):
    \frac1{z^2} = \frac1{z*z} = \frac1{(z+1)-1}* \frac1{(z+1)-1} = \left(- \sum_{n=0}^\infty (z+1)^n \right)* \left(- \sum_{n=0}^\infty (z+1)^n \right) = \left( \sum_{n=0}^\infty (z+1)^n \right)^2

    I've also got this (but I don't know if it's worth anything, or if it is even correct):

    \ \sum_{n=0}^\infty (z+1)^n \right) * \left( \sum_{n=0}^\infty (z+1)^n

    = \left[1 + (z+1) + (z+1)^2 + ... \right] * \sum_{n=0}^\infty (z+1)^n

    = 1 * \sum_{n=0}^\infty (z+1)^n + (z+1) * \sum_{n=0}^\infty (z+1)^n + (z+1)^2 * \sum_{n=0}^\infty (z+1)^n + ...

    = \sum_{n=0}^\infty (z+1)^n + \sum_{n=0}^\infty (z+1)^{n+1} + \sum_{n=0}^\infty (z+1)^{n+2} + ...

    = \sum_{n=0}^\infty \sum_{m=0}^\infty (z+1)^{m+n}


    I have answered part (a) and the check in part (b) of the following question. But I cannot figure out part (b)
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  2. #2
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    Quote Originally Posted by MSUMathStdnt View Post
    Hi, I could use some help on part (b) of the following question from my Complex Variables, book. I didn't know if this belonged in the Calculus forum or not.

    Question 3.2.5
    (a) Use the identity
    \frac1z = \frac1{(z+1)-1}

    to establish
    \frac1z = - \sum_{n=0}^\infty (z+1)^n \hspace{15 mm} \left|z+1 \right| < 1

    (b) Use the above identity to also establish
    \frac1{z^2} = \sum_{n=0}^\infty (n+1)(z+1)^n \hspace{15 mm} \left|z+1 \right| < 1

    Check the answer by differentiating the series from (a)
    \displaystyle \dfrac{1}{z^2}=\left[ \sum_{k=0}^{\infty}(z+1)^k\right]\times \left[ \sum_{k=0}^{\infty}(z+1)^k\right]

    Now consider how we obtain the coefficient of the term in (z+1)^n when multiplying out the right hand side. We get $$1 from (z+1)^0\times (z+1)^n, and another $$1 from (z+1)\times (z+1)^{n-1} ... and $$1 form (z+1)^n\times (z+1)^0. Which total to (n+1). Hence the coefficient of (z+1)^n in the expansion is (n+1) which is what you need.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    \displaystyle \dfrac{1}{z^2}=\left[ \sum_{k=0}^{\infty}(z+1)^k\right]\times \left[ \sum_{k=0}^{\infty}(z+1)^k\right]

    Now consider how we obtain the coefficient of the term in (z+1)^n when multiplying out the right hand side. We get $$1 from (z+1)^0\times (z+1)^n, and another $$1 from (z+1)\times (z+1)^{n-1} ... and $$1 form (z+1)^n\times (z+1)^0. Which total to (n+1). Hence the coefficient of (z+1)^n in the expansion is (n+1) which is what you need.

    CB
    Thanks, CB. Your reply was a little difficult for me to understand. But I think you're trying to say something similar to what I found, anyway:

    1/(z^2)=\left(-\sum(z+1)^n \right) \cdot \left(-\sum(z+1)^n \right)=\sum(z+1)^n \cdot \sum(z+1)^n=

    Expand the left summation (but not the right one):
    \left[1+(z+1)+(z+1)^2+(z+1)^3+... \right] \cdot \sum(z+1)^n=

    Distribute the right summation:
    1 \cdot \sum(z+1)^n+(z+1) \cdot \sum(z+1)^n+(z+1)^2 \cdot \sum(z+1)^n+(z+1)^3 \cdot \sum(z+1)^n+... =

    \sum(z+1)^n +\sum(z+1)^{n+1}+\sum(z+1)^{n+2}+\sum(z+1)^{n+3}+.  .. =

    Expand each summation on a row (lining them up nice would make the following step easier to see, too):
    \sum(z+1)^{n} \hspace{4mm} = 1+(z+1)+(z+1)^2+(z+1)^3+(z+1)^4...
    \sum(z+1)^{n+1} = \hspace{7mm} (z+1)+(z+1)^2+(z+1)^3+(z+1)^4+...
    \sum(z+1)^{n+2} =  \hspace{24mm} (z+1)^2+(z+1)^3+(z+1)^4+...
    \sum(z+1)^{n+3} =  \hspace{43mm} (z+1)^3+(z+1)^4+...

    Now, draw a line underneath and add like terms vertically (you know, like you did when you first learned addition back in the first grade) and you get:
    1+2(z+1)+3(z+1)^2+4(z+1)^3 +...= \sum_{n=0}^\infty (n+1)(z+1)^n

    So the coefficient of each term is one more than the exponent, giving the summation on the right. Which is what you said.
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