[seqfan] Re: Two roots of trigonometric polynomials
allouche at math.jussieu.fr
allouche at math.jussieu.fr
Wed Oct 2 21:52:54 CEST 2013
This can also be done "by hand". Putting T = 2 cos(2x),
we easily have that:
cos(6x) = (cos(4x))^2 gives
T^4 - 2 T^3 - 4 T^2 + 6 T + 4 = 0
and that
(cos x)^2 + (sin 3x)^2 = 1/2 gives
T^3 - 4 T - 2 = 0
but
T^4 - 2 T^3 - 4 T^2 + 6 T + 4 = (T - 2) (T^3 - 4 T - 2)
jp
Charles Greathouse <charles.greathouse at case.edu> a écrit :
> Zero[aa_] :=
> With[{a = Exp[2 I*TrigToExp[aa]]},
> FullSimplify[Numerator[a] - Denominator[a]] == 0]
>
> Zero[(x /.
> Solve[Cos[6 x] == Cos[4 x]^2 && 0 < x < 1, x,
> Reals][[1]][[1]]) - (x /.
> Solve[Cos[x]^2 + Sin[3 x]^2 == 1/2 && 0 < x < 1, x,
> Reals][[1]][[1]])]
>
> True
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
>
> On Wed, Oct 2, 2013 at 2:44 PM, Robert G. Wilson v <rgwv at rgwv.com> wrote:
>
>> I also "Played" with Mathematica. I first looked that their graphs from 0
>> to
>> Pi. Bob.
>>
>> -----Original Message-----
>> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Charles
>> Greathouse
>> Sent: Wednesday, October 02, 2013 2:21 PM
>> To: Sequence Fanatics Discussion list
>> Subject: [seqfan] Re: Two roots of trigonometric polynomials
>>
>> I was able to prove that the constants are the same by the time-honored
>> method of "playing with Mathematica". I don't yet have anything nice, but
>> you should be able to recycle A197757 while I look for something more
>> presentable.
>>
>> Charles Greathouse
>> Analyst/Programmer
>> Case Western Reserve University
>>
>>
>> On Wed, Oct 2, 2013 at 1:48 PM, Richard J. Mathar
>> <mathar at mpia-hd.mpg.de>wrote:
>>
>> > Is there a way to reduce the trigonometric expressions in
>> > http://oeis.org/A197757
>> > and http://oeis.org/A197488 to demonstrate that the two constants are
>> > the same?
>> >
>> > RJM
>> >
>> > _______________________________________________
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