Re: Re: Vieta infinite product formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg84165] Re: Re: Vieta infinite product formula*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 12 Dec 2007 01:20:15 -0500 (EST)*References*: <20071201233246.712$Zx@newsreader.com> <04798D06-095D-4AA6-8004-57C0E811F1E4@mimuw.edu.pl> <fittmj$8jk$1@smc.vnet.net> <200712112031.PAA26250@smc.vnet.net>

>> 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}]. I do not wish to belobor this point, which I have made many times in the past, but the issue is not so much whether a CAS can provide "the right answer" to a particular problem but whether it does so by using some general algorithm, which can deal with a number of realted problems or whether it uses some heuristic which solves this particular problem and fails on another, just as "simple" one. However, in connection with this particular case note that Mathematica 6.0 gives: Limit[Product[Cos[x]/2^k, {k, 1, n}], n -> Infinity] 0 I hope you are now less disappointed. (Note that this means that in general infinite products are not computed by taking limits but by applying certain general simplifications, which obviously do not include any rules for dealing with such cases.) Andrzej Kozlowski On 12 Dec 2007, at 05:31, David W.Cantrell wrote: > 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 >

**References**:**Re: Vieta infinite product formula***From:*"David W.Cantrell" <DWCantrell@sigmaxi.net>

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