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Re: Re: Re: Re: A bug?......In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}]......Out[1]= 1/2

  • To: mathgroup at smc.vnet.net
  • Subject: [mg41898] Re: [mg41870] Re: [mg41828] Re: [mg41793] Re: A bug?......In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}]......Out[1]= 1/2
  • From: Bobby Treat <drmajorbob-MathGroup3528 at mailblocks.com>
  • Date: Sun, 8 Jun 2003 06:46:06 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

How about 1/(1+x) at x=1 or x/(2 + x) at 2?  In both cases (and 
infinitely many others), the function is well-behaved there but the 
series (the very SAME series) is divergent.

I gather that Cesaro convergence is indicative of that, and that's good 
to know.  (I guess.)

However, as I suspected, the series doesn't represent any function at 
all.  At best, it represents a huge family of functions.

Bobby

-----Original Message-----
From: Michael Williams <williams at vt.edu>
To: mathgroup at smc.vnet.net
Subject: [mg41898] Re: [mg41870] Re: [mg41828] Re: [mg41793] Re: A 
bug?......In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}]......Out[1]= 1/2

1/(1-z) of course. The point is its power series representation (at 
z=0) diverges for ALL z on the unit circle (in the complex plane). If 
we examine the series, using Cesaro sums, we get values for all z on 
the unit circle except z=1 (where the real trouble is!), AND those 
values agree with 1/(1-z). The values &formally& obtained 
remain faithful to the given function (defined on the whole plane).


Michael


On Saturday, June 7, 2003, at 08:54 PM, Bobby Treat wrote:


 What function does that sum represent, then?

 Bobby

 -----Original Message-----
From: Michael Williams &williams at vt.edu
To: mathgroup at smc.vnet.net
Subject: [mg41898] [mg41870] Re: [mg41828] Re: [mg41793] Re: A bug?......In[1]:=  Sum[Cos[x], {x, 0, Infinity, Pi}]......Out[1]= 1/2
 1/(1-z)=Sum[z^n,{n,0,Infinity}] |z|1 lhs at z=-1 = 1/2 rhs 
at z=-1  = 1-1+1-1+1-... The Cesaro sum (e.g.) of a series, 
u1+u2+u3+... with  partial sums, s1,s2,s3, is defined to be the 
limit as n-Infinity  of (s1+s2+s3+...+sn)/n . When a series 
converges, the Cesaro value is  the same as the series sum. It is 
easy to see that the Cesaro sum of  the above series is 1/2 and is 
the correct value for the function that  the series represents. 
Indeed, this is true for all |z|=1, z!=1. The  generalized sum 
(formal sum) provides useful (i.e.  correct) 
information about the function the series represents, even  when 
the series does not converge in the traditional sense. Michael  
Williams Blacksburg,Va,USA On Friday, June 6, 2003, at 09:51 AM, Bobby 
 Treat wrote:  Sum[Cos[x],{x,0,Infinity,Pi}] doesn't 
converge in  any sense that's  useful to most of us, and 
I'm curious what kind  of analysis would  benefit from 
assuming that it does converge  somehow.   Dana's 
computations show how easy it is to formally  
prove that it  converges, however, if we 
misapply a  method that often works.   Bobby 
  -----Original  Message-----  From: Dana 
To: mathgroup at smc.vnet.net
DeLouis delouis at bellsouth.net To:  
mathgroup at smc.vnet.net  To: mathgroup at smc.vnet.net  
Subject: [mg41898]  [mg41870] [mg41828] [mg41793] Re: A bug?......In[1]:= 
Sum[Cos[x], {x,   0, Infinity,  Pi}]......Out[1]= 
1/2   Hello. I am not  an expert, but I came across 
a chapter recently in my  studies of  Fourier Analysis. 
Basically, your series sums the following   terms. (the 
first 10 terms...) Table[Cos[x], {x, 0, 10*Pi, Pi}] {1,  -1, 
 1, -1, 1, -1, 1, -1, 1, -1, 1} You are summing a series of 
 alternating  +1 and -1's. Your series can also be written 
like  this... Plus @@  Table[(-1)^j*r^j, {j, 0, 10}] 1 - r 
+ r^2 - r^3 +  r^4 - r^5 + r^6 - r^7  + r^8 - r^9 + r^10 
With r equal to 1 For  example, if r is 1, then the  first 
10 terms are...  Table[(-1)^j*r^j, {j, 0, 10}] /. r - 1 
{1,  -1, 1, -1, 1,  -1, 1, -1, 1, -1, 1} If you sum this as 
j goes to  infinity, you  get the following. 
Sum[(-1)^j*r^j, {j, 0, Infinity}] 1/(1  + r)  Apparently, 
this is correct and has something to do with Abel's   
method. I still do not understand this topic too well yet though. 
  Anyway, if you set r = 1, then 1/(1+r) reduces to 1/2. 
Although it   doesn't look like it, I believe Mathematica 
is correct -- Dana  DeLouis  Windows XP Mathematica 
$VersionNumber - 4.2  delouis at bellsouth.net =  =
 = = = = = = = = = = = = = = =  Mark 
 nanoburst at yahoo.com  wrote in 
message  news:bb1ua4$9do$1 at smc.vnet.net...  I  
think that the sum does not  converge. Does  the 
following  (from Mathematica for Students,   v. 
4.0.1) reveal a bug?  If so, do you have  any insight 
into this  bug?     
In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}]   
  Out[1]= 1/2   
     ********** 
 1366294709   




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