Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

given the general recurrence equation $ a_{n+1}-a_{n}=f(n)a_{n+2}$ (1)

is this possible to find a function $ g(x)$ so $ g(x)= \sum_{n=0}^{\infty}a_{n}x^{n}$ ?? where the $ a_{n}$ are the solutions of the recurrence (1)

in case $ f(n)=const$ i know how to get it but for a non constant function $f(n) $ i have no idea , for $ f(n)$ a Polynomial i guess that $ g(x)$ will satisfy a differential equation

share|improve this question
1  
Do you really want the $f(n)$ to go on the $a_{n+2}$ term? If so, a necessary condition for a unique solution is that $f(n) \neq 0$ for all $n$. –  mrf Apr 21 '12 at 9:33
add comment

1 Answer 1

up vote 3 down vote accepted

Indeed, when $f$ is a polynomial you will get a differential equation. Define

$$g(x)=\sum_{n=0}^\infty a_n x^n.$$

Then

$$h(x):=\frac{g(x)-a_0}{x}-g(x)=\sum_{n=0}^\infty (a_{n+1}-a_n)x^n=\sum_{n=0}^\infty a_{n+2}f(n) x^n. \tag{1}$$

Now the falling factorials form a $\Bbb Q$-basis for the vector space of rational-coefficient polynomials, thus we can write $f(n)=\sum_k c_k (n)_k$ and obtain

$$h(x) = \sum_k c_k\sum_{n=0}^\infty a_{n+2} (n)_k x^n = \left(\sum_k c_k x^k \frac{d^k}{dx^k}\right)\underbrace{\sum_{n=0}^\infty a_{n+2}x^n}_{\ell(x)}. \tag{2}$$

Note that

$$\ell(x)=\frac{g(x)-a_0-a_1x}{x^2}.$$

Combining $(1)$ and $(2)$ gives the desired differential equation. Similar algebra works when $f$ is a combination of powers and exponentials as well. I'm not sure if there's a general solution to the problem, though...

share|improve this answer
    
Nice argument, Anon! –  Giuseppe Tortorella Apr 21 '12 at 9:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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