I want to show that he taylor series of $f(z)=\frac{1}{1-z+z^2}$ at zero is: $$\sum_{n}^{\infty}a_nz^n$$

where $a_0=a_1=0$, $a_2=0$ and $a_{n+3}=-a_n$. what is the radius of convergence ?

My attempt;(by the way I would much rather hints or explanations rather than full answers if possible)

First Using Taylor's theorem we have that $a_n=\frac{1}{2\pi i}\int\frac{f(z)}{(z-z_0)^{n+1}}dz$

Then by Cauchy's integral formula for derivatives we can say $a_n=\frac{1}{2\pi i}\int\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{f^n(z_0)}{n!}$ and using this neat little formula we can compute the coefficients




{Personal Question 1: how could one go about proving $a_{n+3}=-a_n$, I thought to do something like $a_{n+3}=\frac{f^{n+3}(0)}{(n+3)!}=\frac{1}{(n+3)!}\frac{d^n(f^3(0)}{dz^n}$ But this comes out as zero.}

Next I calculated the radius of convergence by the ration test $R=\lim_{n \to\infty}|\frac{a_{n+1}}{a_n}|=f'(0)/n+1=0$

And finially in attempting to show that the taylor series representation is given by $\sum_{n}^{\infty}a_nz^n$,I used a Taylor expansion to show $f(z)=f(0)+f'(0)z+\frac{f''(0)z^2}{2}+...+\frac{f^n(z)}{n!}$

Then I used the formula found from combining the Taylor and Cauchy theorems as above where we see that $a_n=\frac{f^n(z_0)}{n!}$. which means $f(z)=\sum a_nz^n$

Personal question number 2: is it cheating to use a Taylor expansion as i did here or do you think this is what the question was looking for ?


$$\frac{1}{1-z+z^2}=\frac{1+z}{1+z^3}=\frac{(1+z)(1-z^3)}{1-z^6} = (1+z-z^3-z^4)\sum_{k\geq 0} z^{6k} $$ is enough and it clearly shows that the radius of convergence is $1$.


\begin{align}\frac1{1-z+z^2}&=\frac{1+z}{1+z^3}\\&=(1+z)(1-z^3+z^6-z^9+\cdots)\text{ (if $|z|<1$)}\\&=1+z-z^3-z^4+z^6+z^7-z^9-z^{10}+\cdots\end{align} Besides, note that if you wish to express this as $\displaystyle\sum_{n=0}^\infty a_nz^n$, then the sequence of coefficients is defined by$$a_n=\begin{cases}1&\text{ if }n=0\vee n=1\\0&\text{ if }n=2\\-a_{n-3}&\text{ otherwise.}\end{cases}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.