Re: Vieta infinite product formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg84150] Re: Vieta infinite product formula*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Tue, 11 Dec 2007 15:31:24 -0500 (EST)*References*: <20071201233246.712$Zx@newsreader.com> <04798D06-095D-4AA6-8004-57C0E811F1E4@mimuw.edu.pl> <fittmj$8jk$1@smc.vnet.net>

Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > On 2 Dec 2007, at 15:02, Andrzej Kozlowski wrote: > > On 2 Dec 2007, at 13:32, David W. Cantrell wrote: > >> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > >>> I just discovered, to my disappointment, that Mathematica does not > >>> know the classic Vieta infinite product formula: > >>> > >>> Sin[x]/x == Product[Cos[x]/2^k, {k, 1, Infinity}] > >>> > >>> Shouldn't something be done about that? [snip] > >> Second, there is a typo. You intended to ask for > >> > >> Product[Cos[x/2^k], {k, 1, Infinity}] > >> > >> instead. But Mathematica (at least version 5.2) leaves that > >> unevaluated. I would have hoped that version 6 would have given > >> Sinc[x] as the result. > > > > Yes, indeed it does that. Having just now installed version 6 myself, I am happy to say that I can now see that for myself. > >> Third, and surely more surprising, Mathematica 5.2 leaves > >> > >> Product[Cos[x]/2^k, {k, 1, Infinity}] -- that is, _with_ the typo > >> > >> unevaluated! I had expected Mathematica to factor out the cosine And the reason that I had expected that was that -- silly me -- I was thinking about an infinite _series_, rather than an infinite product. OTOH, it is true, as you eventually came to agree, that the product should be 0, regardless of x. > >> and thus to treat the product as though it were > >> > >> In[14]:= Cos[x] Product[1/2^k, {k, 1, Infinity}] > >> > >> Out[14]= 0 > >> > >> That Mathematica failed to do that I consider to be more disappointing > >> than its failure to recognize Euler's product as being Sinc[x]. > > > > Not quite. Gentle understatement? > > You get a product of infinitely many Cos[x], so assuming > > that x is real the result is indeed 0. But I suspect Mathemaitca is > > not making this assumption. If you use any real number in place of > > x, e.g 27, you will indeed get: > > > > Product[Cos[27]/2^k, {k, 1, Infinity}] > > 0 > > Of course even if x is complex the answer is still 0, since the > numbers Abs[Cos[x]/2^k] eventually become all less than 1. But > probably this already needs slightly too sophisticated mathematical > reasoning for present day CAS. Hmm. I know of one present-day CAS which quickly gives 0 for the infinite product of cos(x)/2^k. So of course I tend to think that Mathematica too could give 0 for Product[Cos[x]/2^k, {k, 1, Infinity}]. Two other related things: 1) While I'm happy that version 6 simplifies Euler's infinite product to Sinc[x], I'm slightly puzzled by In[5]:= Sum[(-x^2)^i/(2*i + 1)!, {i, 0, Infinity}] Out[5]= Sin[x]/x not also giving Sinc[x]. I'll just assume that Sinc is new enough to Mathematica that it hasn't been fully integrated into the system yet and that, in some future version, the sum above will give Sinc[x]. 2) Thinking of a related sum due to Euler, I'm sorry to see that In[7]:= Sum[Tan[x/2^i]/2^i, {i, 1, Infinity}] Out[7]= Sum[Tan[x/2^i]/2^i, {i, 1, Infinity}] does not give 1/x - Cot[x] in Mathematica yet. David W. Cantrell

**Follow-Ups**:**Re: Re: Vieta infinite product formula***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>