[Date Index]
[Thread Index]
[Author Index]
Re: a tricky limit
*To*: mathgroup at smc.vnet.net
*Subject*: [mg15821] Re: a tricky limit
*From*: Daniel Lichtblau <danl>
*Date*: Fri, 12 Feb 1999 18:39:42 -0500 (EST)
*Organization*: Wolfram Research, Inc.
*References*: <199901080915.EAA03988@smc.vnet.net.> <79e810$96b@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Jurgen Tischer wrote:
>
> Arnold,
> I think this is the last time I comment about your tricky limit. This
> nice David Lichtblau read all my code and comments and pointed out to
> me some errors I made and showed me that my upper limit in fact should
> be much better as I thought. In the end there was little I had to
> correct, essentially it amounts to that my corrected lower bound is
> much better so here is the result:
>
> upper bound: 2.292173695248657289096168980403 (as before)
>
> lower bound: 2.292173695221057149738189560492 (much better)
>
> this makes it ten digits and an error less than 3 10^-11. I wonder what
> you need that thing for, but it was lots of fun.
>
> Jurgen
>
> If someone wants the notebook, please email me.
>
> 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?
> >
> > Thanks
> > Arnold Knopfmacher
> > Dept of Computational and Applied Math Witwatersrand University
> > South Africa
I have a few comments.
First let me congratulate Jurgen Tischer on finding a much closer
approximation to the limit than I could find, or had even thought would
be computationally feasible. He did this by judiciously rewriting the
series for the logarithm, collecting terms with equal denominators, and
then carefully formulating limit results for such terms.
What I consider to be a tricky part of the job is to approximate the sum
below.
Sum[c[k]/b[k]*Log[1-1/Prime[k]], {k,1,Infinity}]
The auxiliary functions c[] and b[] are defined, as per JT post, as
b[k_] := Product[Prime[j], {j,1,k}]
c[k_] := Product[Prime[j]-1, {j,1,k-1}]
These were correct as stated by JT, proof or lack thereof
notwithstanding.
He and I had some differences as to how one might bound this. I am not
sure whether we reached an agreement on what he termed the lower bound,
but I think I can provide a simple explanation which will yield an
excellent approximation for the original limit. We will estimate the
sum above by truncating it, and then bound the magnitude of the error.
estimate[n_] := Sum[c[k]/b[k]*Log[1-1/Prime[k]], {k,1,n}] error[n_] :=
Sum[c[k]/b[k]*Log[1-1/Prime[k]], {k,n+1,Infinity}]
As I recall, my estimate[n] corresponds to JT's ub[] function. He did
the calculation for n somewhere larger than 10^9, so I will defer to
his result for that computation and just show the error.
To get this error bounded, recall that Prime[k] is asymptotically
approximated by k*Log[k] and |Log[1-1/m]| is asymptotically
approximated by 1/m. Also note that the products c[k]/b[k] are
conveniently bounded by 1/Prime[k] (use a telescoping argument to see
this). In fact, for k>20000, one can check that they are bounded by
.049/Prime[k] just using a numerical computation.
So Abs[error[10^9]] is bounded by
Sum[(.049/Prime[k])*(1.00001/Prime[k]), {k,10^9,Infinity}]
which in turn is bounded by
Sum[1.0001^2*(.049/(k*Log[k]))*(1.00001/(k*Log[k])), {k,10^9,Infinity}]
obtained by throwing in factors of 1.0001 to acount for error in the
asymptotic approximations. This summand is seen to be less than .05 *
1/(k*Log[k])^2. We now use the fact that the sum is dominated by an
integral from one less than 10^9.
In[33]:= N[1/20 * Integrate[1/(k*Log[k])^2, {k,10^9-1,Infinity}]]
-13
Out[33]= 1.06562 10
I threw in a fair amount of slop factors so one can actually conclude
the error is less than 10^(-13). Since JT computed a value that is near
unity, even after exponentiating he has a result that is correct to
twelve places. This I find to be quite impressive, as I had doubts as
to whether one could even get six.
-----------------------------
I had tried to tangle with this using truncated power series without
rearranging terms by denominators. One can see why this is inferior, as
a lower bound on the error from truncating after n terms is O(1/n). I
will show why this is the case below. Again, the messy part is a sum of
the form (using a meta-notation for the inner sum)
Sum[x^n * (1/n-Sum[1/k+1, {f[k]==n}]]
where f[k_] := k-k/Divisors[k][[2]]
This basically corresponds to a power series for -Log[1-x] added to
Sum[-x^f[k]/(k+1), {k,3,Infinity}]
that is, I am dropping terms after first order contributed from logs in
the original log-of-product. I did this because bounding the
contributions from these later terms is feasible. In fact, JT found an
exact limit for them.
It is easy to see that the convergence, on truncation, cannot be better
than O(1/n). This is because for odd k the coefficients are
(k+1)/(2*k^2+k) which is around 1/(2*k). Actually from this we cannot
even conclude the infinite sum converges, but let us acceept that it
does (as we already know to be the case anyway). SO what more can we
say about the coefficients? I can show that the kth coefficient is
always bounded by log(log(k))/k (logs taken base 2) but can they be
shown to be O(1/k)?
Another curious point is that the signs seem to alternate
positive/negative/positive/negative several times, followed
sporadically by three negatives in a row. Is this always the case? Is
there a pattern as to how many +/- sequences will precede a triple
negative?
Daniel Lichtblau
Wolfram Research
Prev by Date:
**Position Available**
Next by Date:
**Re: BracketingBar for output**
Previous by thread:
**Re: a tricky limit**
Next by thread:
**FindRoot behaving funny?**
| |