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MathGroup Archive 2007

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Re: Simplification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75441] Re: [mg75431] Simplification
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 1 May 2007 03:19:23 -0400 (EDT)
  • References: <200704300740.DAA22559@smc.vnet.net> <C2F81A55-BA9C-4D0F-B153-07518BAFADA3@mimuw.edu.pl>

Corrections:

Product[Cos[(2^j*Pi)/(2^(k + 1) - 1)], {j, 0, k}] == -2^(-(k+1))
Product[Cos[(2^j*Pi)/(2^(k + 1) + 1)], {j, 0, k}]  ==   2^(-(k+1))

(The only proofs I know use algebraic number theory.)


On 1 May 2007, at 00:55, Andrzej Kozlowski wrote:

> *This message was transferred with a trial version of CommuniGate 
> (tm) Pro*
> Actually, this is a special case of a general fact:
>
> Let
>
> f[k_] := Product[Cos[(2^j*Pi)/(2^(k + 1) - 1)], {j, 0, k}]/Product 
> [Cos[(2^j*Pi)/(2^(k + 1) + 1)], {j, 0, k}]
>
> then for every integer k>0, f[k]= -1. The problem that you have  
> stated is showing that f[9]= -1. Mathematica can only do the  
> simplest case:
>
> FullSimplify[f[1]]
> -1
>
> but fails on the next one:
>
>
> FullSimplify[f[2]]
>
> Cos[Pi/7]*Cos[(2*Pi)/7]*Cos[(4*Pi)/7]*Sec[Pi/9]*Sec[(2*Pi)/9]*Sec 
> [(4*Pi)/9]
>
>
> However, one can prove a more exact theorem:
>
> Product[Cos[(2^j*Pi)/(2^(k + 1) - 1)], {j, 0, k}] == -2^(k+1) and
> Product[Cos[(2^j*Pi)/(2^(k + 1) + 1)], {j, 0, k}]  ==   2^(k+1)
>
> defining
>
> g[k_] := Product[Cos[(2^j*Pi)/(2^(k + 1) + 1)], {j, 0, k}]
>
> and
>
> h[k_] := Product[Cos[(2^j*Pi)/(2^(k + 1) - 1)], {j, 0, k}]
>
> we see that Mathematica actually performs a little better when  
> dealing with g and h:
>
>
> Table[FullSimplify[g[k]], {k, 1, 3}]
>
> {1/4, 1/8, 1/16}
>
> with h, however, this happens:
>
>
> Table[FullSimplify[h[k]], {k, 1, 3}]
>
>
> {-(1/4), Cos[Pi/7]*Cos[(2*Pi)/7]*Cos[(4*Pi)/7], -(1/16)}
>
> That's as much as Mathematica seems to be able to do. It would be  
> interesting to know if the other CAS can go beyond f[9], and if it  
> can, whether it takes it longer to do so.
>
>
> Andrzej Kozlowski
>
>
>
>
> On 30 Apr 2007, at 16:40, dimitris wrote:
>
>> This appeared in another forum.
>>
>> (Converting to Mathematica InputForm.)
>>
>> In[2]:=
>> oo = Product[Cos[(2^j*Pi)/1023], {j, 0, 9}]/Product[Cos[(2^j*Pi)/
>> 1025], {j, 0, 9}];
>>
>> The expression can be simplified to -1.
>>
>> Indeed, adopted by someone's reply, in another CAS, we simply have
>>
>> Product(cos(Pi*2^j/1023), j= 0..9)/ Product(cos(Pi*2^j/1025), j=
>> 0..9):
>>  p:=value(%):
>>  convert(p, sin):
>>  simplify(%);
>>                                                    -1
>>
>> However, no matter what I tried I was not able to succeed in
>> simplifying above expression
>> to -1 with Mathematica, in reasonable time. Futhermore, even the much
>> more simpler of
>> showing oo==-1 didn't work.
>>
>> So I would really appreciate if someone pointing me out:
>> 1) A way to show (in Mathematica!) that oo is simplified to -1
>> 2) That the equality oo==-1 (or oo-1==0 alternatively) can be
>> simplified
>> to True.
>>
>> Any ideas?
>>
>> BTW, I found the function convert of the other CAS, very useful.
>> Has anyone implementated a similar function in Mathematica?
>> (I ain't aware of a Mathematica built-in function, similar to convert
>> from the other CAS.)
>>
>> Dimitris
>>
>>
>



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