Re: Help on a recursive function

*To*: mathgroup at smc.vnet.net*Subject*: [mg52929] Re: [mg52895] Help on a recursive function*From*: DrBob <drbob at bigfoot.com>*Date*: Fri, 17 Dec 2004 05:18:48 -0500 (EST)*References*: <200412160840.DAA27279@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

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} (So your result for H=1 looks wrong.) p[2] /@ Range@5 {a*(1 + b), a*(1 + b + b^2), a*(1 + b + b^2 + b^3), a*(1 + b + b^2 + b^3 + b^4), a*(1 + b + b^2 + b^3 + b^4 + b^5)} p[3] /@ Range@5 {a*(1 + b), a*(1 + b)^2, a*(-b + (1 + b)^3), a*(1 + 2*b + 4*b^2 + 4*b^3 + b^4), a*(1 + b)^2* (1 + 3*b^2 + b^3)} As for the limit, try H=3, b=0.5, for i = 100 and i = 250: {p[3][100], 1/(1-b)}/.b->0.5 {67.5556 a, 2.} {p[3][250], 1/(1-b)}/.b->0.5 {167.556 a, 2.} The limit CLEARLY depends on a, and 1/(1-b) doesn't look viable even if a=1. Make sure you have the definition right, and THEN you can look for closed forms. Bobby On Thu, 16 Dec 2004 03:40:30 -0500 (EST), <smollest at supereva.it> wrote: > > I'm interested on the behaviour on a H-length circular queue, empty before i=0, equal to P(0)=a and P(i)= b Sum(j=i-1, i-H [P(j)]), that is: > > P(i) = a(1+b)^i i=0, ..., H-1 > > P(i) = (1+b)P(i-1) - b P(i-H) i>= H > > > I got just the following results: > > P(i) = a (b(i+1) - 1)/(b - 1) *if H=1* > > and > > lim(i->Inf [ P(i) ]) = 1 / (1-b) for each H, if | b | <1 > > my question is: is it possible to express this function in closed form for each H? > > thank you in advance, > smoll-est > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Help on a recursive function***From:*smollest@supereva.it