Re: Re: a tricky limit
- To: mathgroup at smc.vnet.net
- Subject: [mg15415] Re: [mg15387] Re: a tricky limit
- From: "Richard Finley" <rfinley at medicine.umsmed.edu>
- Date: Wed, 13 Jan 1999 20:57:29 -0500
- Sender: owner-wri-mathgroup at wolfram.com
I think there may be a more direct way to see the limit of this
function... .if you multiply the product terms out you will see that
they approach 2/n as x->1 (or 2/(n+1) as Paul found which is the same
thing since n->Infinity) ....since the limit (if it exists) cannot
depend on how x approaches 1 from below, let x = (n-1)/n which gives
1/(1-x) = n . Finally, then the whole function limit is 2 n/n as n->
Infinity or limit ->2. That is, the limit is exactly 2 (although I
haven't strictly proven that the limit exists). This is why it
appeared to be near 2!
regards, R Finley
>>> Paul Abbott <paul at physics.uwa.edu.au> 01/12 2:14 AM >>>
Arnold Knopfmacher wrote:
> I wish to obtain a numerical estimate (say 8 decimal digits) of the
> limit as x tends to 1 from below of the function
> h[x]=(Product[(1-fm[x]/(m+1)),{m,2,Infinity}])/(1-x) where
> fm[x]=x^(m-m/d) and d is the smallest divisor of m that is greater
than
> 1. The problem is that when I replace Infinity by say 1000 as the upper
> limit of the product, the function blows up near 1. Visual inspection
> of the graph of h[x] for 0<x<0.9 say, suggests that the limit should
> have a value around 2.1. Can anyone help?
With the definitions,
In[1]:= h[n_][x_] := Product[1 - f[m][x]/(m + 1), {m, 2, n}]/(1 - x)
In[2]:= f[m_][x_] := x^(m - m/Divisors[m][[2]])
series expansion of truncated products
In[3]:= h[10][x] + O[x, 1]
Out[3]=
-2 1153
----------- + ---- + O[-1 + x]
11 (-1 + x) 1155
In[4]:= h[20][x] + O[x, 1]
Out[4]=
-2 40020454
----------- + -------- + O[-1 + x]
21 (-1 + x) 33948915
indicates the leading (problem) term and the low-order approximations to
the limit you are after. You can easily show that the leading
(problem) term, i.e.,
In[5]:= Product[m/(m + 1), {m, 2, n}] Out[5]=
2
-----
1 + n
vanishes as n tends to Infinity.
Approximations to the limit you are after can be found by determining
the coefficient of x in the Product with x replaced by 1-x, e.g.,
In[6]:= Coefficient[Product[1 - f[m][1 - x]/(m + 1), {m, 2, 10}], x]
Out[6]=
1153
----
1155
Alternatively, simple series rearrangement gives
In[7]:= 2/(1 + 10) Sum[1 - 1/Divisors[n][[2]], {n, 2, 10}] Out[7]= 1153
----
1155
Hence the number you are after is the same as
Lim[2/(1 + p) Sum[1 - 1/Divisors[n][[2]], {n, 2, p}],p->Infinity]
I am not aware of a way of computing this expression exactly. However,
you can easily find numerical approximations:
In[8]:= Timing[N[2/(1+10000) Sum[1 - 1/Divisors[n][[2]], {n,2,10000}]]]
Out[8]= {25.9 Second, 1.33959}
In[9]:= Timing[N[2/(1+20000) Sum[1 - 1/Divisors[n][[2]], {n,2,20000}]]]
Out[9]= {65.85 Second, 1.33972}
Cheers,
Paul
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia Nedlands WA 6907
mailto:paul at physics.uwa.edu.au AUSTRALIA
http://www.physics.uwa.edu.au/~paul
God IS a weakly left-handed dice player
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