Re: Unknown Sum of Series
- To: mathgroup at smc.vnet.net
- Subject: [mg63497] Re: Unknown Sum of Series
- From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
- Date: Tue, 3 Jan 2006 01:25:03 -0500 (EST)
- Organization: The Open University, Milton Keynes, U.K.
- References: <dpb3vu$9tg$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Klaus G." <Karl_boehme_9 at msn.com> a écrit dans le message de news:
dpb3vu$9tg$1 at smc.vnet.net...
| Mathematica 5.0 is not able to compute the symbolic sum:
|
| Sum[(-1)^(1 + n)*(E - ( 1 + (1/n))^n ), {n, 1, Infinity}]
|
| However, Nsum[...] results in 0.4456224031968407..
|
| I tried http://oldweb.cecm.sfu.ca/projects/ISC/ to find hidden
| constants in that number like Pi or E, but without success.
|
| Any idea?
|
| Klaus G.
|
Hi Klaus,
If I have understood correctly your message, you are wondering why *Sum* is
not able to provide an answer whereas *NSum* is. What you expect from *Sum*
is a kind of symbolic result -- an exact result indeed -- that must match
the numeric answer returned by *NSum*. For example, say that *NSum* yields
3.13159 for a given sum, we could expect *Sum* to return Pi for the same
sum.
That would be the case if both functions were using the same set of
algorithms or if NSum[expr] were strictly equivalent to N[Sum[expr]].
Symbolic and numeric algorithms are usually quite different in their
implementation and approach to a given problem.
Compare *Solve* and *FindRoot* for instance. The latter -- thanks to
Newton's Method -- can easily find a numerical approximation to the solution
of a transcendental equation such as x = sin(x), while the former is unable
to find any symbolic solution since an exact solution does not exist.
Coming back to your initial problem, we can conclude that there exits no
closed form for the given sum. However, we can find/compute an approximate
solution to it.
Finally, note that even if the numeric value of the symbolic solution of a
sum seems to be visually in agreement with the numeric solution returned by
*NSum*, that does not mean that both solution are equal at any degree of
precision as you can see in the following example:
In[1]:=
symsum = Sum[1/n^2, {n, 1, Infinity}]
Out[1]=
Pi^2/6
In[2]:=
numsum = NSum[1/n^2, {n, 1, Infinity}]
Out[2]=
1.64493
In[3]:=
N[symsum]
Out[3]=
1.64493
In[4]:=
symsum == numsum
Out[4]=
False
In[5]:=
{N[symsum, 16], InputForm[numsum]}
Out[5]=
{1.644934066848226,1.6449340667600105}
In[6]:=
Precision /@ {symsum, numsum}
Out[6]=
{Infinity, MachinePrecision}
Best regards,
/J.M.