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MathGroup Archive 2004

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Re: Help on a recursive function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53057] Re: Help on a recursive function
  • From: Smoll Est <smollest at supereva.it>
  • Date: Wed, 22 Dec 2004 04:52:47 -0500 (EST)
  • References: <200412160840.DAA27279@smc.vnet.net> <cpue6s$gb6$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

DrBob wrote:
> If I've copied the definition right, here are a few terms for H = 1, 2, and 3:
> 
> Clear@p
> p[H_][i_Integer] /; 0 <= i < H := p[H][i] = a(1 + b)^i
> p[H_][i_Integer?Positive] := p[H][i] =
>    Simplify[(1 + b)p[H][i - 1] - b p[H][i - H]]
> 
> p[1] /@ Range@5
> {a, a, a, a, a}

Hello Bobby,
thank you for your attention first of all; I made a *mistake* in the 
previous formula; instead of:

>>P(i) = (1+b)P(i-1) - b P(i-H)    i>= H

the true formula is:

P(i) = (1+b)P(i-1) - b P(i-H-1)    i>= H

that is "P(i-H-1)" instead of "P(i-H)"

and the limit where:

a / (1-b)

so, the complete formal is:

P(i) = a(1+b)^i     i=0, ..., H-1

P(i) = (1+b)P(i-1) - b P(i-H-1)    i>= H

any idea for its expression in a closed form?

smoll-est


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