[VA] SRC #012b - Then and Now: Root

[J-F Garnier]

(11-09-2022 11:03 PM)Fernando del Rey Wrote:  I’ve written a short program for the HP-71B that will solve the problem for polynomials of degree up to a few hundred, using brute force with the Math ROM’s PROOT function.

What I found out is that, for polynomials of degree 149 and beyond, the roots with the minimum absolute value have always an absolute value of:

0.80651359926

I did the same yesterday evening with 200 terms, and found the same minimum root as you and Werner already found:
Zmin = (-0.645758096347,0.483177676217) for an abs value of 0.806513599261

My understanding is that the PROOT method, with 200 terms, proves that the root Zmin is indeed the smallest in abs value for the complete 10000th degree polynomial:
- PROOT finds all the roots of the given polynomial (here the 200th degree one), there is no missing one,
- we can be sure that the root Zmin above is also a root of the 10000th degree polynomial, in the limits of the numerical accuracy, because the terms Pn.Zmin^n for n>200 will have a modulus less than about 2E-16 and will not contribute (again in the limits of the numerical accuracy) to the value of the complete polynomial,
- the complete polynomial can not have a smaller root Zx, because it would also be a root (in the numerical accuracy limits...) of the 200th degree polynomial, for the same reason that the Zx^n terms for n>200 would be negligible.

Here is my HP-71B program using both the MATH and JPC ROMs, for reference, I didn't check but it should run in a few hours on a physical HP-71B. The HP48 and later series that have equivalent PROOT commands may find the solution in less time due to the faster CPU.

10 ! SRC12B2
20 OPTION BASE 0
30 !
40 N=200
50 DIM P(N)
60 COMPLEX Z(N)
70 ! build the polynomial
75 P(N)=2
80 FOR I=N-1 TO 0 STEP -1
90   P(I)=FPRIM(P(I+1)+1,20000)
100 NEXT I
110 !
120 MAT Z=PROOT(P)
125 A=MAXREAL
130 FOR I=0 TO N-1
140   A=MIN(A,ABS(Z(I)))
150 NEXT I
160 DISP A
170 END

J-F