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Black body polylog application

Black_body_Integrand u 3^e u -^* 1 e u -^- / u int ≡

x sol x 3^e x -^* 1 e x -^- / x int maple :

x sol x x x 1 x - exp - ln * 32 x - exp polylog * - (* 63 x - exp polylog * - (* 64 x - exp polylog * -

Jonquière x n Ji x k^k n^/ k 140 sum :

x c 11 - x ≤ (x 1 ≤ (& cases :

x Jonquiere x 1 Ji x c * x 2 Ji x c * x 3 Ji x c * x 7 Ji x c * 41 sys :

x Jonquiere

Substitute each polylog term by its "Jonquière" Σseries ... i.e: the integral curve.
Next => by deriving  JonQ(x), we should retrieve the original Black Body integrand. x JonQ x x x 1 x - exp - ln * 3 x - exp 2 Ji * - (* 6 x - exp 3 Ji * - (* 6 x - exp 4 Ji * - :

x JonQ

x f x JonQ :

h 0.01 :

t0 1 time :

Incremental  Derivative x h f deriv 0.5 x h - f x f - x h - (x - / x f x h - f - x x h - (- / + (* (:

slow, incremental rescue project

x h f deriv

fails to derive, but no complain

x f x diff

1 time t0 - min 1.1

Interpretation:

The maths of the Black Body looks confusing. Experimental absorbsion sums, i.e:

resembles the ∫Curve, from which a model must be best approximated [the data

set is not so perfect !]. That the Integrand(u) represents the derivative wrt to

wavelength is valid. Otherwise, one will have to go back to square 1 original works.

The point is about Maple solution, not digestible. JonQ(x) is very close and considered

the solution, unfortunately not supported by the derivative operator.

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u Integrand u 3^e u -^* 1 e u -^- / :

Discrete  ∫ over linear path ∫Curve z 1015 -^12 0.125 range : i 1 z rows range C i el u Integrand u z 1 el z i el int : for z C augment 31 line :

∫Curve