closed form for regular superfunction expressed as a periodic function

[Gottfried]

Hi Sheldon -

I recognize your coefficients....

Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into pari-gp to verify it.

Sheldon -

here some Pari/GP-statements to arrive at the coefficients. I just took your last substitution rules in your earlier letter. The "cryptic" function-calls are in my collection of procedures - "VE" means truncation of vectors/matrices, dFac(1,6) creates a 6x6-diagonalmatrix containing the factorials r!^1

Code:

\\ extract coefficients at x into column-vector "pc"
\ps 16    \\ use only series-precision 16 (coefficients) - if more then the product of the exp-taylorseries needs long time...
pc = polcoeffs( exp('a1*x) * exp('a2*x^2) * exp('a3*x^3) * exp('a4*x^4) * exp('a5*x^5)) ~ ;


\\ %box are meta-commands to advise PariTTY to send the output into a matrix-display-window
%box >pc dFac(1,5)*Mat(VE(pc,5) )    \\ show the coefficients scaled by factorials in window "pc"
\\ result:
\\                                         1
\\                                        a1
\\                                 a1^2+2*a2
\\                         a1^3+6*a2*a1+6*a3
\\  a1^4+12*a2*a1^2+24*a3*a1+(12*a2^2+24*a4)

\\ now begin to substitute
a1 = 1
pc1=subst(VE(pc,6),'a1,a1)

%box >pc dFac(1,6)*Mat(pc1 )    \\ show the coefficients scaled by factorials
\\                                                   1
\\                                                   1
\\                                              2*a2+1
\\                                       6*a2+(6*a3+1)
\\                     12*a2^2+12*a2+(24*a3+(24*a4+1))

a2 = 1/2/(L-1)
pc2=subst(VE(pc1,6),'a2,a2)
%box >pc dFac(1,4)*Mat(VE(pc2,4) ) \\ show only 4 rows
                                                    1
                                                    1
                                              L/(L-1)
                         ((6*a3+1)*L+(-6*a3+2))/(L-1)



a3 = (1/6 + a2)/(L^2-1)
pc3=subst(VE(pc2,6),'a3,a3)
%box >pc dFac(1,4)*Mat(VE(pc3,4) )

a4=(1/24 + a3 + a2/2 + a2^2/2)/(L^3-1)
pc4=subst(VE(pc3,6),'a4, a4)
%box >pc dFac(1,5)*Mat(VE(pc4,5) )

a5=(1/5! + a4+a3/2+a3*a2+a2^2/2+a2/3!)/(L^4-1)
pc5=subst(VE(pc4,6),'a5, a5)
%box >pc dFac(1,6)*Mat(VE(pc5,6) )
                                                                             1
\\                                                                             1
\\                                                                       L/(L-1)
\\                                                     (L^3+2*L^2)/(L^3-L^2-L+1)
\\                                 (L^6+5*L^5+6*L^4+6*L^3)/(L^6-L^5-L^4+L^2+L-1)
\\  (L^10+9*L^9+24*L^8+40*L^7+46*L^6+36*L^5+24*L^4)/(L^10-L^9-L^8+2*L^5-L^2-L+1)

\\ see the coefficients at powers of L in the numerators equal that matrix-entries, for instance last row(revert the order and begin at L^4: [24  36  46  40  24  9  1]
\\ denominators can be factored by (L-1),(L^2-1) and so on

Thanks also for the other explanations. It surely helps to have an entry for my own understanding of the procedure once I've got the whole idea behind...

So far -

Gottfried