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Re: difficulty using FindRoot

  • To: mathgroup at smc.vnet.net
  • Subject: [mg110290] Re: difficulty using FindRoot
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Sat, 12 Jun 2010 05:31:00 -0400 (EDT)

Daniel Lichtblau wrote:
> Roger Bagula wrote:
>> The question of a q-form infinite exponential series
>> solving to give Pi came up.
>> I had absolutely no luck with infinite sums on this!
>> I tried a plot of the function to narrow it down:
>> Clear[f, x, n, i]
>> f[x_] := 1 + Sum[1/Product[1 - x^i, {i, 1, n}], {n, 1, 100}]
>> Plot[f[x], {x, 1.021831198825114750405873564886860549451,
>>         1.02183648425181683450091441045515239239}, PlotRange -> All]
>>
>> The find root that seemed to work was:
>> q /. FindRoot[1 + Sum[1/Product[1 -
>>         q^i, {i, 1,
>>             n}], {n, 1, 150}] - Pi == 0, {q,
>>               1.0218701842518167}, WorkingPrecision -> 800,
>> AccuracyGoal ->
>>       795]
>> gives:
>> 1.0218311988251147504058736
>>
>> with error messages:
>> \!\(Divide::"infy" \(\(:\)\(\ \)\) "Infinite
>>       expression \!\(3.14159265346825122833252`25.0094071873645\/0\) \
>> encountered."\)
>>
>> \!\(\*
>>   RowBox[{\(FindRoot::"jsing"\), \(\(:\)\(\ \)\), "\<\"Encountered a
>> singular
>>       Jacobian at
>>       the point \\!\\({q}\\) = \
>> \\!\\({1.0218311988251147504058736`25.0094071873645}\\). Try
>> perturbing the \
>> initial point(s). \\!\\(\\*ButtonBox[\\\"More=85\\\", \
>> ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \
>> ButtonData:>\\\"FindRoot::jsing\\\"]\\)\"\>"}]\)
>>
>> 1 + Sum[1/Product[1 - (1.02183119882511475040587356488686054945)^i,
>> {i, 1, n}], {n, 1, 150}]
>> gives
>>  0*10^(-19)
>>
>> It appears there is no real q such that the sum?
>> 1 + Sum[1/Product[1 - q^i, {i, 1, n}], {n, 1, Infinity}]==Pi
>>
>> Respectfully, Roger L. Bagula
>> 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
>> http://www.google.com/profiles/Roger.Bagula
>> alternative email: roger.bagula at gmail.com
>>
> 
> Correct, it cannot be done with real q. Just work with the sum (so the 
> target is Pi-1).
> 
> For -1<=q<=1 the sum does not converge because terms grow in size 
> (slightly different behavior at the endpoints, but same conclusion: 
> divergence).
> 
> For q>1 the sum is alternating and terms strictly decrease in magnitude. 
> So it converges. But the first term is negative, so the result must be 
> negative.
> 
> For q<-1 again it is alternating with terms strictly decreasing in 
> magnitude, hence convergent. This time the first term is between 0 and 
> 1/2, so the result of the sum is between 0 and 1/2.
> 
> Conclusion: for real valued q, the sum cannot be Pi-1.
> 
> Daniel Lichtblau
> Wolfram Research

Okay, I got that wrong. First, for negative q the terms actually 
alternate in pairs after the first. But that's not the real issue. The 
claim that they strictly decrease in magnitude is correct for 
|q|>=sqrt(2). For 1<|q|<sqrt(2) it is a different ballgame. Eventually 
they decrease and so you have convergence. But they can get arbitrarily 
large before that happens, hence it is not trivial to bound the values.

I believe one can show the sum approaches -1 as q approaches 1 from the 
right, and that it decreases in magnitude as q grows. And I think 
something similar happens for q<-1. And these together would indicate 
that you cannot attain a value of pi-1 with real q. But I do not have 
proofs and as I mention above, the case where 1<|q|<sqrt(2) seems tricky.

If I am seeing correctly, the unit circle is a natural boundary of 
convergence for the function f(1/q). That is to say, it is analytic 
inside the circle and does not have an analytic continuation past any 
point on the circle.

Daniel Lichtblau
Wolfram Research




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