       Re: plotting groups of polynomial roots

• To: mathgroup at smc.vnet.net
• Subject: [mg51285] Re: plotting groups of polynomial roots
• From: Roger Bagula <tftn at earthlink.net>
• Date: Tue, 12 Oct 2004 01:57:48 -0400 (EDT)
• References: <ckajb5\$m4j\$1@smc.vnet.net> <ckd674\$52u\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```This idea was inspired by the w cummutator of Potter matrix groups:
( in this month's Math Monthly journal)
A.B=wB.A
-->
A^q+B^q=(A+B)^q
When A and B are unitary matrices it suggests
the root group:
x^q+1=(x+1)^q
They are an interesting group of polynomials.
The de Moivre relationship is:
(Cos[t]+I*Sin[t])^n=Cos[n*t]+I*Sin[n*t]=Exp[I*n*t]
So the relationship to the unit circle isn't unexpected,
but it is nice.
The other roots are what makes the result interesting.
If nothing else it is a new way to look at Pascal's triangle.
Narasimham G.L. wrote:

>Roger Bagula <tftn at earthlink.net> wrote in message news:<ckajb5\$m4j\$1 at smc.vnet.net>...
>
>
>>If you take the first and last term away from a binomial polynomial and
>>set the result equal to zero,
>>you get a number of strange roots.
>>This method allows you to plot such roots.
>>I didn't know it would work when I wrote it up,
>>but I plan to use it in the future
>>on some other polynomial root structures.
>>
>>(* root group where x^q+1=(x+1)^q: binomial expansion without x^q and 1*)
>>digits=21
>>s[q_]=Sum[(q!/((q-k)!*k!))*x^(q-k),{k,1,q-1}]
>>ExpandAll[s]
>>ExpandAll[s]
>>a=Flatten[Table[x/. NSolve[s[n]==0,x],{n,2,digits}]];
>>a0=Floor[Abs[a]]
>>Dimensions[a][]
>>
>>
>
>continued...
>
>b = Table[{Re[a[[n]]], Im[a[[n]]]}, {n, 1, Dimensions[a][]}];
>ChopEnds = ListPlot[b, PlotRange -> {{-1, 2}, {-1, 1}}];
>central = ParametricPlot[{Cos[t], Sin[t]}, {t, 0, 2 Pi}];
>displ = ParametricPlot[{Cos[t] - 1, Sin[t]}, {t, 0, 2 Pi}];
>Show[ChopEnds, central, displ];
>
>Hi Roger,
>
>The roots are neatly herded onto unit circles centered on (0,0) and
>(-1,0), (except one point (-0.5, +/-1), as may be expected for complex
>roots of z^(1/n),(z+1)^(1/n) somehow with only negative real parts,
>|x|<1 .
>
>Regards. Nara
>
>
>

--
Respectfully, Roger L. Bagula