Re: Recursion problem in SymbolicSum
- To: mathgroup at smc.vnet.net
- Subject: [mg60705] Re: [mg60693] Recursion problem in SymbolicSum
- From: Pratik Desai <pdesai1 at umbc.edu>
- Date: Sun, 25 Sep 2005 02:36:14 -0400 (EDT)
- References: <200509240655.CAA13410@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
D.J. Wischik wrote: >I was surprised to get problems (recursion limit exceeded) when executing >a symbolic sum. The terms in the sum depend on a parameter mu. When I >leave mu unspecified and calculate the sum and then substitute a numerical >value for mu, I get the right answer. When I specify mu in the sum, the >symbolic sum fails. (The sum definitely exists and is finite.) I would be >grateful if anyone could explain this behaviour. > >PoissonProb[mu_, k_] = Exp[-mu] mu^k / k!; > >Sum[PoissonProb[mu, k] (k + 1 - 9)/(k + 1), {k, 9, Infinity}] /. > {mu -> 1.05} > >[returns the answer 1.82353 * 10^(-7) as expected] > >Sum[PoissonProb[1.05, k](k + 1 - 9)/(k + 1), {k, 9, Infinity}] > >[ $RecursionLimit::reclim: Recursion depth of 256 exceeded. >$IterationLimit::itlim: Iteration limit of 4096 exceeded. >and then it returns the following. ] > >\!\(0.34993774911115527`\ \((4.298654386611213`*^-6 - > 7.999999999999789`\ \ >Hold[If[MatchQ[Numerator[SymbolicSum`InfiniteDump`expr1$214], >SymbolicSum`a$_ \ >+ SymbolicSum`b$_ /; \(! > FreeQ[SymbolicSum`a$, > K$94]\) && \(! FreeQ[SymbolicSum`b$, K$94]\)], \ >\((SymbolicSum`InfiniteDump`infinitesum[#1, K$94, 0] &)\) /@ > Expand[SymbolicSum`InfiniteDump`expr1$214], > SymbolicSum`InfiniteDump`HypergeometricSeries[ > 1, SymbolicSum`InfiniteDump`expr1$214, \ >SymbolicSum`InfiniteDump`expr2$214, K$94, 0, SymbolicSum`eps$214]]])\)\) > >Damon. > > > This is indeed strange, it seems that your expression gives the Recursion Limit message only for the value 1.05. Infact if I give mu=105/100 then it will evaluate. It will evaluate for 1.06 too! Here is the code I have been using Clear[mu, k] Remove["Global`*"] f[k_] = ((k + 1) - 9)/(k + 1) PoissonProb[mu_, k_] := PoissonProb[mu, k] = (Exp[-mu]* mu^k)/k!*f[k]; sol[mu_, k_] := sol[mu, k] = Sum[PoissonProb[mu, k], {k, 9, Infinity}] sol[1.03, k] // N PS: It even evaluates for mu=1.04999->1.8235339756733727*^-7.............Most peculiar indeed :) Hope this helps Pratik Desai -- Pratik Desai Graduate Student UMBC Department of Mechanical Engineering Phone: 410 455 8134
- References:
- Recursion problem in SymbolicSum
- From: djw1005@cus.cam.ac.uk (D.J. Wischik)
- Recursion problem in SymbolicSum