I don't see a result that my book say it's straightforward. Here's my try:

Prove that the coefficients of the Taylor series of the function $$f(z)=\frac{1}{1-z-z^2}$$around $z=0$ verify $$c_0=1,\\ c_1=1, \\ c_{n+2}=c_{n+1}+c_n, n\geq 0.$$

From here, what I've done is to find first $c_0$ and $c_1$ as follows:

$$c_0=\frac{f^{0)}(0)}{0!}=\frac{1}{1-0-0^2}=1\\c_1=\frac{f^{1)}(0)}{1!}=\frac{-1\cdot(-1-(2\cdot 0))}{(1-0-0^2)^2}=1$$

I can take both results as straightforward, but my book's solution only says: "identifying coefficients, we have the result." That's the only information I have and I don't see how can we prove that $\ c_{n+2}=c_{n+1}+c_n, n\geq 0.$

Thanks for your time.

  • $\begingroup$ Consider forming the product of $f(x)$ with development of $f(x)$ with $1-z-z^2$. $\endgroup$
    – user65203
    May 24 '20 at 12:53

Note that since $f(z)=1+f(z)z+f(z)z^2$, for $n\ge2$, comparing the coefficient of $z^n$, we have that $c_n=c_{n-1}+c_{n-2}$.


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