Results 1 to 4 of 4

Math Help - help in taylor series

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    45

    help in taylor series

    In order to find c_n in :

    \frac{1}{1-z-z^2} =  \sum_{n=0}^\infty c_n z^n
    By partial fractions i got
    \frac{1}{1-z-z^2} = \frac{1}{\sqrt{5}}  \[\frac{1}{z-\frac{1+\sqrt{5}}{2}}  -\frac{1}{z+\frac{\sqrt{5}-1}{2}}\]
    inside \[ \] we have 2 geometric series
    \frac{1}{\sqrt{5}} \[ \frac{1}{-\frac{1+\sqrt5}{2}(1-\frac{2z}{1+\sqrt{5}})} + \frac{1}{-\frac{\sqrt{5}-1}{2}(1+\frac{2z}{\sqrt{5}-1})}\]

    the series for \frac{1}{1-\frac{2z}{1+\sqrt{5}}} is \sum_{n=0}^\infty (\frac{2}{1+\sqrt{5}})^n z^n

    and for \frac{1}{1+\frac{2z}{\sqrt{5}-1}} we have \sum_{n=0}^\infty (-1)^n (\frac{2}{\sqrt{5}-1})^n z^n

    so the coeficients of z^n are -\frac{1+\sqrt5}{2}(\frac{2}{1+\sqrt{5}})^n + -\frac{\sqrt{5}-1}{2}(-1)^n (\frac{2}{\sqrt{5}-1})^n times \frac{1}{\sqrt{5}}

    but the unswer is c_n = (\frac{1+\sqrt{5}}{2})^n + (-1)^n (\frac{\sqrt{5}-1}{2})^n times \frac{1}{\sqrt{5}}

    i did something wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    You did not necessarily do anything wrong. Notice that \frac2{\sqrt5+1} = \frac{\sqrt5-1}2, as you can check by multiplying out the fractions.

    Incidentally, the coefficients c_n in that power series are the Fibonacci numbers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    Quote Originally Posted by Opalg View Post
    Incidentally, the coefficients c_n in that power series are the Fibonacci numbers.

    Certainly, if (F_n) is the Fibonacci's sequence, and f(z)=\sum_{n\geq 0}F_nz^n then, in the open disk of convergence D

    f(z)-zf(z)-z^2f(z)=

    \displaystyle\sum_{n=0}^{+\infty}F_nz^n-\displaystyle\sum_{n=0}^{+\infty}F_nz^{n+1}-\displaystyle\sum_{n=0}^{+\infty}F_nz^{n+2}=

    \left(F_0+F_1z+\displaystyle\sum_{n=2}^{+\infty}F_  nz^n\right)-\left(F_0z+\displaystyle\sum_{n=2}^{+\infty}F_{n-1}z^n\right)-\displaystyle\sum_{n=2}^{+\infty}F_{n-2}z^n=

    F_0+F_1z-F_0z+\displaystyle\sum_{n=2}^{+\infty}(F_{n}-F_{n-1}-F_{n-2})z^n=1+z-z+\displaystyle\sum_{n=2}^{+\infty}0z^n=1

    So,

    \displaystyle\sum_{n=0}^{+\infty}F_nz^n=\dfrac{1}{  1-z-z^2}\qquad (\forall z\in D)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by hurz View Post
    In order to find c_n in :

    \frac{1}{1-z-z^2} =  \sum_{n=0}^\infty c_n z^n
    f(z) = \frac{1}{1-z-z^2} = \frac{2}{\sqrt{5}\left(\sqrt{5}+1+2z\right)}+\frac  {2}{\sqrt{5}\left(\sqrt{5}-1-2z\right)}, thus:

    f^{(n)}(z) = \frac{(-1)^n2^{n+1}n!}{\sqrt{5}\left(\sqrt{5}+1+2z\right)^  {n+1}}+\frac{2^{n+1}n!}{\sqrt{5}\left(\sqrt{5}-1-2z\right)^{n+1}}, set z = 0, then:

    \begin{aligned} \therefore ~ \frac{f^{(n)}(0)}{n!} & = \frac{(-1)^n2^{n+1}}{\sqrt{5}\left(\sqrt{5}+1\right)^{n+1}  }+\frac{2^{n+1}}{\sqrt{5}\left(\sqrt{5}-1\right)^{n+1}} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n  +1}+\frac{1}{ \sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n+1} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{\sqrt{5}-1}{2}\bigg)^{n+1}+\frac{1}{\sqrt{5}}\bigg(\frac{ \sqrt{5}+1}{2}\bigg)^{n+1} \\& = C_{n}.\end{aligned}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 08:32 AM
  2. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  3. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  4. Taylor Series / Power Series
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 4th 2009, 01:56 PM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum