Re: Infinite series problem
- To: mathgroup at smc.vnet.net
- Subject: [mg35958] Re: [mg35917] Infinite series problem
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Fri, 9 Aug 2002 05:18:06 -0400 (EDT)
- References: <200208081006.GAA12939@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Seung Soon Jang wrote: > > Hi, Group, > > I am just beginner to start using Mathematica to solve my problem. > MY goal is to get some asymptotic value of the following series. But it > doesn't give the answer and > I have no clue how to resolve the error message I met. > > Please give me the idea to go through this hard thing. > > Thanks, > > Seung Soon > > In[9]:= > Sum[1/(2^(0.5)*(Pi*k)^2* > > NIntegrate[(x/2)^(-0.5)*(1-(x/2))^(-0.5)*Cos[Pi*k*x],{x,0,1}]),{k,1, > Infinity}] > > NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x} > = \ > {0.5}. > > NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x} > = \ > {0.5}. > > NIntegrate::inum: Integrand 2.3094 Cos[1.5708 k] is not numerical at {x} > = \ > {0.5}. > > General::stop: Further output of NIntegrate::inum will be suppressed > during \ > this calculation. > > -- > ---------------------------------------------------------- > > Seung Soon Jang, Ph.D. > California Institute of Technology > Chemistry 139-74 > Pasadena, CA 91125 > TEL) 1-626-395-8147 > FAX) 1-626-585-0917 > e-mail) jsshys at wag.caltech.edu > home page) http://www.wag.caltech.edu/home/jsshys > > ---------------------------------------------------------- One approach would be to break the sum into a piece from, say, 1 to 100, and another from 101 to Infinity. Compute the first in roughly as you now do. For the second, approximate by only integrating from 0 to 2/k and throw away the slowly changing part of the denominator, (1-x/2)^(1/2). Offhand I do not have an error estimate for how good an approximation this is to your actual integral. But after some experimentation I believe it is reasonable, perhaps three or so decimal places. In[126]:= ff[k_?NumberQ, opts___] := NIntegrate[(x/2)^(-1/2)*(1-(x/2))^(-1/2)*Cos[Pi*k*x], {x,0,1}, opts] In[127]:= firstpart = Sum[1/(2^(N[1/2,40])*(Pi*k)^2*ff[k, WorkingPrecision->40, PrecisionGoal->10]), {k,1,100}] Out[127]= 0.17722511803 In[130]:= rest = N[Sum[1/(2^(1/2)*(Pi*k)^2* Integrate[(x/2)^(-1/2)*Cos[Pi*k*x], {x,0,2/k}, Assumptions->k>0]), {k,101,Infinity}], 40] Out[130]= 0.01463711889688464408318349581501263184547 In[131]:= approx = firstpart + rest Out[131]= 0.19186223693 Daniel Lichtblau Wolfram Research
- References:
- Infinite series problem
- From: Seung Soon Jang <jsshys@wag.caltech.edu>
- Infinite series problem