help in taylor series

**hurz** · April 28th 2011, 09:46 AM

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?

**Opalg** · April 29th 2011, 12:07 AM

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.

**FernandoRevilla** · April 29th 2011, 01:41 AM

Originally Posted by Opalg

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,

$\displaystyle\sum_{n=0}^{+\infty}F_nz^n=\dfrac{1}{ 1-z-z^2}\qquad (\forall z\in D)$

**TheCoffeeMachine** · April 29th 2011, 06:29 PM

Originally Posted by hurz

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:

$\begin{aligned} \therefore ~ \frac{f^{(n)}(0)}{n!} & = \frac{(-1)^n2^{n+1}}{\sqrt{5}\left(\sqrt{5}+1\right)^{n+1} }+\frac{2^{n+1}}{\sqrt{5}\left(\sqrt{5}-1\right)^{n+1}} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n +1}+\frac{1}{ \sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n+1} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{\sqrt{5}-1}{2}\bigg)^{n+1}+\frac{1}{\sqrt{5}}\bigg(\frac{ \sqrt{5}+1}{2}\bigg)^{n+1} \\& = C_{n}.\end{aligned}$

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