Results 1 to 5 of 5

Thread: Taylor series question

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Taylor series question

    The question
    For the following function,
    i) Find the largest open disc about the given point $\displaystyle z_0$ in which f is analytic.
    ii) find the first 3 non-zero terms of the Taylor series of f about $\displaystyle z_0$
    iii) Find the coefficient $\displaystyle a_n$ of $\displaystyle (z - z_0)^n$ in the Taylor series of f about $\displaystyle z_0$ either as an explicit function of n, or give a recursive formula for $\displaystyle a_n$ as appropriate.

    f(z) = $\displaystyle \frac{1}{z^2 - z - 6}$, $\displaystyle z_o = 1$

    My attempt
    i) I expanded the function as follows:
    $\displaystyle \frac{1}{(z - 3)(z + 2)}$
    Clearly there's singularities at z = 3, -2
    I draw a quick graph and noticed that the radius of convergence is |z - 1| < 2
    This part I got correct. However, I get stuck for parts ii) and iii)

    ii) I used partial fractions to get:
    $\displaystyle \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2}$
    I then manipulated each fraction so I could write them in terms of known Macluarin series $\displaystyle \frac{1}{z \pm 1}$:
    $\displaystyle \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n)$ and
    $\displaystyle \frac{-1}{15}(\sum_{n = 0}^{\infty}{(-1)^n(\frac{z - 1}{3})^n)$

    Thus I got this for the series:
    $\displaystyle \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n) \frac{-1}{15}(\sum_{n = 0}^{\infinity}{(-1)^n(\frac{z - 1}{3})^n)$

    Now, upon substituting n = 0, 1, 2 to get the first three terms, I get the following:
    $\displaystyle -\frac{1}{5}$, $\displaystyle \frac{13(z - 1)}{180}$, $\displaystyle \frac{19(z - 1)^2}{1080}$

    These are wrong, and I'm not sure why. :/

    Any assistance would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Taylor series question

    $\displaystyle \displaystyle \begin{align*} \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2} &= -\frac{1}{15}\left(\frac{1}{1 - \frac{z}{3}}\right) - \frac{1}{10}\left[\frac{1}{1 - \left(-\frac{z}{2}\right)}\right] \\ &= -\frac{1}{15}\sum_{r = 0}^{\infty}{\left(\frac{z}{3}\right)^r} - \frac{1}{10}\sum_{s = 0}^{\infty}{\left(-\frac{z}{2}\right)^s}\end{align*}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Re: Taylor series question

    It needs to be around $\displaystyle z_0 = 1$. Does yours work for that Prove It?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Re: Taylor series question

    OK I worked out what I did wrong with part ii). I thought the series was $\displaystyle \frac{1}{z \pm 1}$, but it's $\displaystyle \frac{1}{1 \pm z}$. So my solution is correct if I just add a negative to the start of my series.

    However, I'm still not sure what I'm supposed to do for part iii). Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Taylor series question

    Quote Originally Posted by Glitch View Post
    It needs to be around $\displaystyle z_0 = 1$. Does yours work for that Prove It?
    I should have read the question properly :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor Series Question
    Posted in the Calculus Forum
    Replies: 13
    Last Post: Sep 13th 2011, 01:53 PM
  2. Don't get this taylor series question?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 2nd 2010, 02:31 AM
  3. question regarding taylor series
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 16th 2010, 11:17 AM
  4. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  5. Taylor Series Question..
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 8th 2009, 11:38 AM

Search Tags


/mathhelpforum @mathhelpforum