On May 10, 4:30=A0pm, "Jon G." <jon8...@[EMAIL PROTECTED]
> wrote:
> 1 + x + x^2 + x^3 + ... + x^N =3D 0
> for N=3D124 to N=3D136 the roots lie on the curve,
Of course they lie on a curve. The curve on which the roots
lie is commonly known as the _unit circle_.
> REAL:
> 27N/50000 +(11/1000)sin(2piN/5 - 5pi/12) +3 - pi
> COMPLEX
> N/1000 + (1/100)(2piN/5 - 7pi/5) - 113/103
For one thing, you do know that Nth-degree polynomials
usually have N roots, right? So why do you list only one root?
Someone in the other thread already gave a complete solution:
REAL:
cos(2pik/(N+1)), k=3D1,2,3,4,5,6,7,...,N
COMPLEX:
sin(2pik/(N+1)), k=3D1,2,3,4,5,6,7,...,N
Your chart only gives one root for each value of N, omitting the
other N-1 roots.
> allowing for the extrapolation of the next 1500 roots,
> =A0 =A0 =A0 n=3D1570 : -1.00001783767897 + -0.482423186522175I
Which is clearly wrong since it doesn't lie on the unit circle,
since its real part has absolute value greater than 1.
> =A0 =A0 =A0 n=3D1575 : -1.00271783767897 + -0.487423186522175I
Which is clearly wrong since it doesn't lie on the unit circle,
since its real part has absolute value greater than 1.
> =A0 =A0 =A0 n=3D1579 : -1.00024368297496 + -0.481912621359222I
> =A0 =A0 =A0 n=3D1580 : -1.00541783767897 + -0.492423186522175I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1584 : -1.00294368297496 + -0.486912621359224I
> =A0 =A0 =A0 n=3D1585 : -1.00811783767897 + -0.497423186522174I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1589 : -1.00564368297496 + -0.491912621359224I
> =A0 =A0 =A0 n=3D1590 : -1.01081783767897 + -0.502423186522174I
> =A0 =A0 =A0 n=3D1591 : -1.00130834910847 + -0.499790473882148I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1594 : -1.00834368297496 + -0.496912621359224I
> =A0 =A0 =A0 n=3D1595 : -1.01351783767897 + -0.507423186522175I
> =A0 =A0 =A0 n=3D1596 : -1.00400834910847 + -0.504790473882148I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1599 : -1.01104368297496 + -0.501912621359224I
> =A0 =A0 =A0 n=3D1600 : -1.01621783767897 + -0.512423186522175I
> =A0 =A0 =A0 n=3D1601 : -1.00670834910847 + -0.509790473882149I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1603 : -1.00029012928825 + -0.496402056196272I
> =A0 =A0 =A0 n=3D1604 : -1.01374368297496 + -0.506912621359221I
> =A0 =A0 =A0 n=3D1605 : -1.01891783767897 + -0.517423186522175I
> =A0 =A0 =A0 n=3D1606 : -1.00940834910846 + -0.514790473882147I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D1608 : -1.00299012928824 + -0.501402056196272I
> =A0 =A0 =A0 n=3D1609 : -1.01644368297496 + -0.511912621359221I
> =A0 =A0 =A0 n=3D1610 : -1.02161783767897 + -0.522423186522175I
> =A0 =A0 =A0 n=3D1611 : -1.01210834910846 + -0.519790473882147I
> =A0 =A0 =A0 n=3D1612 : -1.00180326889832 + -0.5090347688363I
> =A0 =A0 =A0 n=3D1613 : -1.00569012928824 + -0.506402056196272I
> =A0 =A0 =A0 n=3D1614 : -1.01914368297496 + -0.516912621359225I
> =A0 =A0 =A0 n=3D1615 : -1.02431783767897 + -0.527423186522175I
> =A0 =A0 =A0 n=3D1616 : -1.01480834910847 + -0.524790473882147I
> =A0 =A0 =A0 n=3D1617 : -1.00450326889832 + -0.514034768836298I
> =A0 =A0 =A0 n=3D1618 : -1.00839012928824 + -0.511402056196272I
> =A0 =A0 =A0 n=3D1619 : -1.02184368297496 + -0.521912621359223I
> =A0 =A0 =A0 n=3D1620 : -1.02701783767897 + -0.532423186522175I
> =A0 =A0 =A0 n=3D1621 : -1.01750834910847 + -0.529790473882149I
> =A0 =A0 =A0 n=3D1622 : -1.00720326889832 + -0.519034768836299I
> =A0 =A0 =A0 n=3D1623 : -1.01109012928824 + -0.516402056196271I
> =A0 =A0 =A0 n=3D1624 : -1.02454368297496 + -0.526912621359223I
> =A0 =A0 =A0 n=3D1625 : -1.02971783767897 + -0.537423186522175I
> =A0 =A0 =A0 n=3D1626 : -1.02020834910847 + -0.53479047388215I
> =A0 =A0 =A0 n=3D1627 : -1.00990326889832 + -0.524034768836301I
> =A0 =A0 =A0 n=3D1628 : -1.01379012928825 + -0.521402056196272I
> =A0 =A0 =A0 n=3D1629 : -1.02724368297496 + -0.53191262135922I
> =A0 =A0 =A0 n=3D1630 : -1.03241783767897 + -0.542423186522174I
> =A0 =A0 =A0 n=3D1631 : -1.02290834910847 + -0.539790473882152I
> =A0 =A0 =A0 n=3D1632 : -1.01260326889832 + -0.529034768836303I
> =A0 =A0 =A0 n=3D1633 : -1.01649012928825 + -0.526402056196272I
> =A0 =A0 =A0 n=3D1634 : -1.02994368297496 + -0.536912621359218I
> =A0 =A0 =A0 n=3D1635 : -1.03511783767897 + -0.547423186522174I
Which are clearly wrong since they don't lie on the unit circle,
since their real parts have absolute value greater than 1.
> =A0 =A0 =A0 n=3D65530 : -35.538417837679 + -64.4424231865221I
> =A0 =A0 =A0 n=3D65531 : -35.5289083491086 + -64.4397904738822I
> =A0 =A0 =A0 n=3D65532 : -35.5186032688984 + -64.4290347688364I
> =A0 =A0 =A0 n=3D65533 : -35.5224901292882 + -64.4264020561962I
> =A0 =A0 =A0 n=3D65534 : -35.5359436829749 + -64.4369126213591I
> =A0 =A0 =A0 n=3D65535 : -35.541117837679 + -64.4474231865221I
> =A0 =A0 =A0 n=3D65536 : -35.5316083491085 + -64.4447904738822I
Which lie nowhere near the unit circle.
I truly don't understand the purpose of this thread at all!
The OP may be curious to know that another sci.math poster,
Timothy Golden, once invented solutions to these equations,
which he called "polysigned numbers."
For N=3D1, the only solution is -1 (same as standard, of course)
For N=3D2, the solutions are known as -1, +1.
For N=3D3, the solutions are known as -1, +1, *1.
For N=3D4, the solutions are known as -1, +1, *1, #1.
Unfortunately, Golden doesn't post at sci.math anymore, but
if he were still here, I'm sure he'd be more than eager to give
you a quick introduction. Much of his work can still be found
via the following link:
http://www.bandtechnology.com/PolySigned/
Although one may not wish to use the polysigned numbers,
at least his solutions are more consistent than the OP's.
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