1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
4
$\begingroup$

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 $c_n=c_{n-1}+c_{n-2}$.

$\endgroup$

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.