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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[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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