Re: Simplification
- To: mathgroup at smc.vnet.net
- Subject: [mg75470] Re: Simplification
- From: dimitris <dimmechan at yahoo.com>
- Date: Wed, 2 May 2007 03:49:38 -0400 (EDT)
- References: <200704300740.DAA22559@smc.vnet.net><f16pnf$72b$1@smc.vnet.net>
I must admit I don't use the other CAS so often; so I lack any serious
knowledge.
But...since you ask
for f[10] we get
Product(cos(Pi*2^j/1023), j= 0..9)/ Product(cos(Pi*2^j/1025),
j=0..10):
p:=value(%);
convert(p, sin);
simplify(%);
Pi 2 Pi 4 Pi 8 Pi 16 Pi 32 Pi
p := cos(----) cos(----) cos(----) cos(----) cos(-----) cos(-----)
1023 1023 1023 1023 1023 1023
64 Pi 128 Pi 256 Pi 511 Pi / /
cos(-----) cos(------) cos(------) cos(------) / |
1023 1023 1023 1023 / \
Pi 2 2 Pi 4 Pi 8 Pi 16 Pi
cos(----) cos(----) cos(----) cos(----) cos(-----)
1025 1025 1025 1025 1025
32 Pi 64 Pi 128 Pi 256 Pi 512 Pi \
cos(-----) cos(-----) cos(------) cos(------) cos(------)|
1025 1025 1025 1025 1025 /
512 Pi 1022 Pi Pi 2
2 sin(------) sin(-------) sin(----)
1023 1023 1025
--------------------------------------------
Pi 511 Pi 2 Pi 1024 Pi
sin(----) sin(------) sin(----) sin(-------)
1023 1023 1025 1025
Pi
2 sin(----)
1025
-----------
2 Pi
sin(----)
1025
which is correct
evalf(%);
1.000004697
for f[17] we get
Product(cos(Pi*2^j/1023), j= 0..9)/ Product(cos(Pi*2^j/1025),
j=0..17):
> p:=value(%);
> convert(p, sin);
> simplify(%);
Pi 2 Pi 4 Pi 8 Pi 16 Pi 32 Pi
p := cos(----) cos(----) cos(----) cos(----) cos(-----) cos(-----)
1023 1023 1023 1023 1023 1023
64 Pi 128 Pi 256 Pi 511 Pi / /
cos(-----) cos(------) cos(------) cos(------) / |
1023 1023 1023 1023 / \
Pi 2 2 Pi 2 4 Pi 2 8 Pi 2 16 Pi 2
cos(----) cos(----) cos(----) cos(----) cos(-----)
1025 1025 1025 1025 1025
32 Pi 2 64 Pi 2 128 Pi 2 256 Pi 512 Pi \
cos(-----) cos(-----) cos(------) cos(------) cos(------)|
1025 1025 1025 1025 1025 /
512 Pi 1022 Pi Pi 2
256 sin(------) sin(-------) sin(----)
1023 1023 1025
----------------------------------------------
Pi 511 Pi 256 Pi 1024 Pi
sin(----) sin(------) sin(------) sin(-------)
1023 1023 1025 1025
Pi
256 sin(----)
1025
-------------
256 Pi
sin(------)
1025
which is incorrect
> evalf(%);
1.110486595
So the performance is not so good as initially I thought!
Dimitris
=CF/=C7 Andrzej Kozlowski =DD=E3=F1=E1=F8=E5:
> 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
> >
> >