0
$\begingroup$

Find the generating function of the sequence with the property $\sum_{i=0}^n a_{i}a_{n-i} = 1$. I'm not sure where to start in this problem.

$\endgroup$
2
$\begingroup$

Hint: If $f(x) = \sum_{n=0}^\infty a_n x^n$ is a formal power series then $$ f(x)^2 = \sum_{n=0}^\infty \left(\sum_{i=0}^n a_{i}a_{n-i} \right) x^n $$ (See Cauchy product.) In your case, the right-hand side is a well-known infinite series.

$\endgroup$
1
$\begingroup$

Assume that: $$ f(x) = \sum_{n\geq 0}a_n x^n.\tag{1} $$ Cauchy's convolution hence gives: $$ f(x)^2 = \sum_{n\geq 0}\left(\sum_{i=0}^{n}a_i a_{n-i}\right) x^n = \sum_{n\geq 0} x^n = \frac{1}{1-x}\tag{2} $$ and we have $f(x)=\frac{1}{\sqrt{1-x}}$, from which: $$ a_n = \binom{-\frac{1}{2}}{n}(-1)^n=\frac{(2n-1)!!}{2^n\,n!}=\frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}.\tag{3} $$

$\endgroup$

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.