Results 1 to 3 of 3

Thread: Find Taylor Series

  1. #1
    Member
    Joined
    Feb 2010
    From
    New Jersey
    Posts
    81

    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
    $\displaystyle \frac1z = \frac1{(z+1)-1}$

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

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

    Check the answer by differentiating the series from (a)

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

    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):
    $\displaystyle \frac1{z^2} = \frac1{z*z} = \frac1{(z+1)-1}* \frac1{(z+1)-1} = \left(- \sum_{n=0}^\infty (z+1)^n \right)* \left(- \sum_{n=0}^\infty (z+1)^n \right) = \left( \sum_{n=0}^\infty (z+1)^n \right)^2$

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

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

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

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

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

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


    I have answered part (a) and the check in part (b) of the following question. But I cannot figure out part (b)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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
    $\displaystyle \frac1z = \frac1{(z+1)-1}$

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

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

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

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

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    From
    New Jersey
    Posts
    81
    Quote Originally Posted by CaptainBlack View Post
    $\displaystyle \displaystyle \dfrac{1}{z^2}=\left[ \sum_{k=0}^{\infty}(z+1)^k\right]\times \left[ \sum_{k=0}^{\infty}(z+1)^k\right]$

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

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

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

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

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

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

    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:
    $\displaystyle 1+2(z+1)+3(z+1)^2+4(z+1)^3 +...= \sum_{n=0}^\infty (n+1)(z+1)^n$

    So the coefficient of each term is one more than the exponent, giving the summation on the right. Which is what you said.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find derivative using Taylor Series
    Posted in the Calculus Forum
    Replies: 16
    Last Post: Feb 19th 2011, 02:44 AM
  2. Find the Taylor series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 7th 2009, 07:30 PM
  3. Find the taylor series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 15th 2009, 02:30 AM
  4. Find convergence by using taylor series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 20th 2009, 02:08 PM
  5. Find limit using taylor series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 9th 2009, 06:49 AM

Search Tags


/mathhelpforum @mathhelpforum