2
$\begingroup$

I am trying to find a general expression from a recursion.

Here it goes: $(x+i)P_i = (i+1)P_{i+1} + \frac{x}{2} P_{i-1}$

$i$ goes from $0$ to $S$.

How can I calculate a generic $P_i$ in terms of $P_0$? So, if $P_i = C \times P_0$, what is $P_0$?

Is it possible to solve this in Mathematica? What is the function I'm looking for?

You can assume $P_0 = 1$.

Thanks.

[UPDATE]

I am not quite sure if this would help. I started from $P_0$ and have the following equations: $P_{1}=P_{0}x$

$P_{2}=P_{0}(\frac{1}{2!})(x^2+\frac{x}{2})$

$P_{3}=P_{0}(\frac{1}{3!})(x^3+\frac{3x^2}{2}+x)$

$P_{4}=P_{0}(\frac{1}{4!})(x^4+3x^3+\frac{19x^2}{4}+3x)$

$P_{5}=P_{0}(\frac{1}{5!})(x^5+5x^4+\frac{55x^3}{4}+20x^2+12x)$

I tried to find polynomials that will represent above, but I could not find one. These are basically balance equations for a two dimensional markov chain. I solved it numerically, but I want to have a analytical solution as well.

  • 0
    Isn't the issue here that you have bidirectional recursion so you have to know the value of both $P_0$ and $P_L$?2012-01-12
  • 0
    @Verbeia: Or just any two $P_i$s...2012-01-12
  • 0
    @Simon - yes, and I meant $S$ not $L$ anyway. Not sure where I got $L$ from. :-/2012-01-12
  • 0
    I just updated the question. Thanks for your time.2012-01-12
  • 0
    Ok, so there isn't any dependence on $P_{i+1}$ after all. Have a look at the [`RSolve`](http://reference.wolfram.com/mathematica/ref/RSolve.html) function.2012-01-12
  • 0
    Your new edit is a very special choice of $P_1$. It's equivalent to choosing the initial condition $P_{-1} = 0$ (and $P_0$ unfixed)...2012-01-12
  • 0
    Let me name the polynomial sequences(excluding $1/x!$ )as $C_1, C_2 C_3, C_4, C_5$ $C_2/C_1$ = $x+0.5$ $C_3/C_2$ = $x+1$ remainder: $0.5x$ $C_4 / C_3$ = $x+1.5$ remainder: $1.5x^2+1.5x$ $C_5 / C_4$ = $x+2$ remainder: $3x^3 + 7.5x^2 +6x$ Is this a hyper-geometric series? I am novice in this.2012-01-12
  • 0
    A [hypergeometric function](http://en.wikipedia.org/wiki/Generalized_hypergeometric_function#Notation) can be defined as a exponential generating function of a hypergeometric series. This does not really correspond to what you have above... Although, maybe you mean something different. (Also, these types of comments aren't really appropriate as answers in this site)2012-01-13
  • 0
    @sosruko: I've merged your old and new accounts. If you wish, you can now put the above comment as an edit to your question. Cheers.2012-01-16

1 Answers 1