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by AVt
December 28th, 2013, 6:21 pm
Forum: Technical Forum
Topic: Can you solve this in terms of special functions?
Replies: 69
Views: 9051

Can you solve this in terms of special functions?

Dividing by x^p (and renaming) one gets 1+a*x = x^r with 0 < r, as suggestedby EBal, which admits easy coding for Newton (or its accelerated version).PS: I would not use a series (in positive centers), since x=0 is a branch point,limiting any radius of convergence
by AVt
December 27th, 2013, 8:56 pm
Forum: Technical Forum
Topic: Can you solve this in terms of special functions?
Replies: 69
Views: 9051

Can you solve this in terms of special functions?

<t>Yes, there is a need: both MMA and Maple use variations of that, IIRCAnd - may be - one wants to use it beyond those CAS.But if Alan already has a superb initial guess, then all other methodsare likely out of consideration Edited for PS: NB that it is complex differentiation, while the real onewo...
by AVt
December 27th, 2013, 6:44 pm
Forum: Technical Forum
Topic: Can you solve this in terms of special functions?
Replies: 69
Views: 9051

Can you solve this in terms of special functions?

<t>On the original task (in the first posts):Alan, Do you mean p real *and positive* (or allow negatives and restict x to bepositive)? And what is c - a positive Real?You can not expect an explicit solution: x^p*(1+x)-1 has Galois group S(p+1), atleast for p=4,5,6 and thus has no solutions by radica...
by AVt
December 11th, 2013, 8:17 pm
Forum: Student Forum
Topic: Simple Compounding, why?
Replies: 12
Views: 6476

Simple Compounding, why?

<r>I *guess* (like T4A) that it is historical, in German "Zins" and "Zehent"may even have common roots - may be you dig in Etymology and Bible texts:<URL url="https://de.wikipedia.org/wiki/Zinsverbot#Zinsverbot_im_Alten_TestamentYou"><LINK_TEXT text="https://de.wikipedia.org/wiki/Zinsverbo ... stame...
by AVt
December 3rd, 2013, 8:23 pm
Forum: Brainteaser Forum
Topic: Number series
Replies: 28
Views: 8966

Number series

I go with Alan: Given a sequence of integers one always has a polynomial reproducing it.And one can even prove that the 'minimal' one maps integers to integers.The linked OEIS (or similar) tries to find solutions beyond that approach.
by AVt
November 16th, 2013, 7:59 pm
Forum: Programming and Software Forum
Topic: Mathematica software
Replies: 40
Views: 60735

Mathematica software

<r>I doubt that a usual CAS like Mathematica or Maple allows the notion ofmeasures and measurable sets or functions. They even do not do Riemannintegration. They implement algorithms or formulae or approximations. They do not do Math or proofs. Mostly.Scanning <URL url="http://en.wikipedia.org/wiki/...
by AVt
November 9th, 2013, 9:05 pm
Forum: Off Topic
Topic: Watching forwarded emails
Replies: 23
Views: 5966

Watching forwarded emails

for all careless users of Acrobat ... forbid javascript and external sources
by AVt
November 9th, 2013, 8:49 pm
Forum: Off Topic
Topic: Watching forwarded emails
Replies: 23
Views: 5966

Watching forwarded emails

<t>Besides "why you want to track ?" for original question: I doubt one can avoiddistributions using reasonable pdfs and hope for valid & legal tracking. One wayis like Google tries it for looking at books as images - not a pdf.As soon as the URL for the pdf is in the page one can find it. And t...
by AVt
September 16th, 2013, 5:33 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

He posted it 'here' in 2003, An Exact Treatment of Discrete Dividends, I did a crosscheck for the values
by AVt
September 15th, 2013, 5:33 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

<t>Concerning some integrals from finance For the VG you need to able to compute (a variation of) the Gamma integral:M:= (a,b,c) -> Int( N(a/sqrt(u)+b*sqrt(u)) * u^(c-1)*exp(-u),u=0..infinity)/GAMMA(c)is denoted by Psi in their paper (Appendix), N = cumulative normal, and that isdominated by the sig...
by AVt
September 13th, 2013, 6:45 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

<t>Concerning integration tasks of course characteristic functions come to my mind.They are in the oscillating country, not the very hunting ground for Gauss methods.For the Normal Inverse Gauss model NIG there are real valued integrands, IIRC(but they involve Bessel functions) and for the Variance ...
by AVt
September 12th, 2013, 7:08 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

So we had: an ugly boundary (cos stuff), a nasty infinity (Gamma), a hard task (by Trefethen), and some periodics. Is it ok so far for checking for possible limitations?silent side remark: for easy Gamma over (positive) Reals I would use cephes.
by AVt
September 6th, 2013, 9:43 am
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

<t>One can 'understand' it by taking a single exponential transform (instead of DE)and say we have the singularity already in x=0, so take x = exp(-t).Then the algebraic singularity increases towards infinity, but is damped by expuntil it no longer contributes to the numerical integral.For the very ...
by AVt
September 4th, 2013, 7:18 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

It is nasty. Here the classical would work as well (for the task singular in 0):Substract the singular part ( -cos(255/512*Pi)/x^(255/256) ) from the integrandand do that numerically while doing the singular part analytically, then add them.PS: I do not mind having too less salad.
by AVt
September 3rd, 2013, 5:21 pm
Forum: Numerical Methods Forum
Topic: Looking for tough integrals
Replies: 148
Views: 16961

Looking for tough integrals

<t>Using the analytic solution (wileysw, Ultraviolet) for a complied programmthe result will be 'wrong' (as already said), it is like 1.36739704794349070.I can sketch a way to do it (almost) numerical. Edited: the sketchRoughly: the suspected integral over 0 .. Pi/2 is known to zero, due to analytic...
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