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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: [mg41891] Re: [mg41870] Re: [mg41828] Re: [mg41793] Re: A bug?......In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}]......Out[1]= 1/2
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sun, 8 Jun 2003 06:45:52 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

Actually one can do these things with greater generality.

Let S be a function which takes two natural numbers as arguments and 
such that
Limit[S[n,m],n->Infinity] = 1  for every m.

For example we can take S = a, where

a[n_, m_] := UnitStep[1 - m/n]*(1 - m/n)

(this will give Cesaro sum)

Limit[a[n,m],n->Infinity]

1

or S=b where

b[n_, m_] := (n/(n + 1))^m


Limit[b[n,m],n->Infinity]

1

(there are many more possibilities of this type)

Suppose we are given an infinite sum of terms v[n], then we can define 
it's SSum to be

SSum[v_, a_] := Limit[Sum[Evaluate[a[n, m]
     v[m]], {m, 1, Infinity}], n -> Infinity]

It is  easily to show that if the usual Sum[v[m],{m,1,Infinity}] 
converges then SSum is the same as Sum. For example with


v[n_]=1/2^n

SSum[v,a]

1

SSum[v,b]

1

We can also compute SSum for divergent series. Taking

v[n_] := (-1)^n

SSum[v, b]

-(1/2)

Unfortunately Mathematica can't evaluate SSum[v,a] (Cesaro-sum) :

SSum[v, a]


Limit[Sum[(-1)^m*(1 - m/n)*UnitStep[1 - m/n],
    {m, 1, Infinity}], n -> Infinity]

but loading


<< NumericalMath`NLimit`

and waiting for a while we get:


NLimit[Sum[(-1)^m*(1 - m/n)*UnitStep[1 - m/n],
    {m, 1, Infinity}], n -> Infinity]


-0.4999990591617879

These things become even more interesting for series with random 
coefficients, for example,  J-P Kahane, "Some Random Series", D.C. 
Heath and Co., 1968.

Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/



On Sunday, June 8, 2003, at 12:44  am, Michael Williams wrote:

> 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 DeLouis <delouis at bellsouth.net>
To: mathgroup at smc.vnet.net
>> To: mathgroup at smc.vnet.net
>> Subject: [mg41891] [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 -&gt; 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 -&gt; 4.2 delouis at bellsouth.net 
>> =
>> = = = = = = = = = = = = = = = = &quot;Mark&quot;
>> &lt;nanoburst at yahoo.com&gt; wrote in message
>> news:bb1ua4$9do$1 at smc.vnet.net... &gt; I think that the sum does not
>> converge. Does &gt; the following (from Mathematica for Students, &gt;
>> v. 4.0.1) reveal a bug? If so, do you have &gt; any insight into this
>> bug? &gt; &gt; &gt; In[1]:= Sum[Cos[x], {x, 0, Infinity, Pi}] &gt; 
>> &gt;
>> Out[1]= 1/2 &gt; &gt; &gt; &gt; &gt; &gt; ********** &gt; 1366294709
>> &gt;
>>
>
>
>