# Recurrence - using power series

Could you help me in solving this recursion( a closed form ) using power series

$\mu(n)=\mu(n−1)k_0+(n−1)\mu(n−2) k_1 \tag 1$,

where $k_0,k_1$ are constants $\mu(0)=3,\mu(1)=5$

HINT: We can think about a substitution such that it will be converted to a solvable non varying coefficient case.. Just thinking..

NB :: Avoid any methods where $k_i$s come as denominator in any steps. We are looking for a higher dimension like matrices, compatible solution. So a method based on series will be ideal for upward compatibility(Avoid ODE methods,Issue is upward compatibility with matrices. All matrices are not commutative. So it will be an issue when solving ODE.). Keep the positions of $k_i$ as it is in the question(by seeing the upward compatibility for matrices). A solution which satisfies the constraints in the question for non matrices can be easily transformed to matrices.. Thanks

• do you need a closed-form expression for $\mu(n)$?
– Alex
Aug 8, 2014 at 16:44
• Yes.. A closed form is ideal Aug 8, 2014 at 16:54
• Consider that $\mu(n)=\sum_{n=0}^\infty a_nx^n$ and just replace in the equation and you'll find the $a_n$ that will gives you the solution.
– idm
Aug 8, 2014 at 17:05
• Would you elaborate bit Please..I guess I tried that already Aug 8, 2014 at 17:06
• $\mu$ parameter is bit confusing with summation index.. Please make it correct Aug 8, 2014 at 17:24

Hint: If $g(x) = \sum_{n=0}^\infty \mu(n) x^n$, the recursion corresponds to a differential equation for $g(x)$. Solve that differential equation with initial conditions $g(0) = 3$, $g'(0) = 5$.

EDIT: Actually that won't work in this case, because the differential equation has a nasty singularity at $0$.

Note that with $\mu(n) = k_0^{n} u(n)$ and $k_1 = c k_0^2$ the recursion becomes

$$u(n) = u(n-1) + c (n-1) u(n-2),\ n \ge 2$$

Rather than an ordinary generating function, consider an exponential generating function

$$G(z) = \sum_{n=0}^\infty u(n) z^n/n!$$

Using Maple's gfun package, I find that two linearly independent egf's satifying this recurrence are

$$G_1(z) = \exp((cz+1)^2/(2c)),\ G_2(z) = \text{erf}((cz+1)/\sqrt{2c}) \exp((cz+1)^2/(2c))$$

The appropriate linear combination of these will satisfy your initial conditions.

• I know that solution. That is why specially mentioned only using Power series . Issue is upward compatibility with matrices. All matrices are not commutative. So it will be an issue when solving ODE.. Thanks sir Aug 8, 2014 at 17:13
• If you have a particular matrix recursion in mind, why not ask about that directly rather than trying to adapt the solution to one problem to a different problem? Aug 8, 2014 at 23:29
• It is easy that people may get confused with matrix calculus. I had experience here in the same regard. So I posted a recursion with constraints so that,it can be upgraded easily to matrix. And may invite more thoughts of the solution from people who are not familiar with matrix calculus. Thanks Aug 9, 2014 at 5:40