0

Given the following function:

$$ P_N = \sum_{n=1}^N (A_n + \sum_{m=0}^{n-1} P_mB_{nm}) $$

Where: $$ A_{n} = (1-\alpha)^{n+1} $$ $$ B_{nm} = (1-\beta)^{n-m} $$

And $ N \in \aleph$

Given a known $ P_0 $ I want to remove the recursion created by $\sum_{m=0}^n P_mB_{nm}$ of the function $ P_N $.

For a similar single sum function, I was able to reduce the recursion to something like:

$$ \prod_{n=1}^{N-1} 1 + B_n $$

But the double sum made it much more difficult.

Please help.

Thank you.

Joel
  • 1

1 Answers1

0

1.- let $B_k= (1-\beta)^{k}$

2.- Modify into a polynomic form: $$P_N = (A_1 +P_0B_1) + (A_2 + P_0B_2 + P_1B_1) + (A_3 + P_0B_3 + P_1B_2 + P_2B_1) ... + (A_N + P_0B_N + P_1B_{N-1} + P_2B_{N-2} ... + P_{N-1}B_1)$$ $$-> P_N= \sum_{n=1}^N ((\sum_{m=1}^{n}B_m)P_{N-n}+A_n)= \sum_{n=1}^N(C_nP_{N-n} + A_n)$$

3.- Create a "Generating function":

https://en.wikipedia.org/wiki/Generating_function

https://www.youtube.com/watch?v=Pp4PWCPzeQs

$$P_1 = C_1P_0 + A_1, P_2 = C_1(C_1P_0 + A_1)+C_2P_0 + A_2, P_3 = C_1(C_1(C_1P_0 + A_1)+C_2P_0 + A_2) + C_2(C_1P_0 + A_1) + C_3P_0 + A_3 ... $$

Let: $$F(x) = P_0 +P_1x + P_2x^2 + P_3x^3...+P_Nx^N $$ Multiply F(x) by $1-\sum_{n=1}^NC_nx^n$ gives:

$$(1-\sum_{n=1}^NC_nx^n)F(x) = P_0 + \sum_{n=1}^NA_nx^n$$ $$->F(x) = \frac{P_0 + \sum_{n=1}^NA_nx^n}{1-\sum_{n=1}^NC_nx^n}$$

$P_N$ will be the coeficient of $x^N$

Joel
  • 1