Re: Vieta infinite product formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg83849] Re: Vieta infinite product formula*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 2 Dec 2007 04:12:22 -0500 (EST)*References*: <20071201233246.712$Zx@newsreader.com>

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? > > First, the result you have in mind is due to Euler actually. Well, actually this formula is rather trivial to derive so so I do not think Euler would have minded my misattribution. Besides, one should keep in mind the famous "Arnold principle" (http://pauli.uni-muenster.de/~munsteg/arnold.html ) which states: If a notion bears a personal name, then this name is not the name of the discoverer. So, in calling this Vieta's infinite product formula I simply followed this well established principle. Actually, what is usually called Vieta's infinite product formula is what you get when you substitute x=Pi/2 and express the right hand side in tems of radicals. (I came across this while reading on a train Mark Kac's - Statistical independence in probability theory, analysis and number theory). > > > 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. > > > 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 > 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]. > > David W. Cantrell Not quite. 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 Andrzej Kozlowski

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**Re: Vieta infinite product formula**

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