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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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