       Re: Simplification

• To: mathgroup at smc.vnet.net
• Subject: [mg75440] Re: [mg75431] Simplification
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 1 May 2007 03:18:52 -0400 (EDT)
• References: <200704300740.DAA22559@smc.vnet.net>

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= -1. Mathematica can only do the simplest
case:

FullSimplify[f]
-1

but fails on the next one:

FullSimplify[f]

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, 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:=
> 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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