Use a taylor polynomial to show that , for a large wavelength, Planck's Law gives approximately the same values as the Rayleigh-Jeans Law. see the attached problem #2© BrainMass Inc. brainmass.com December 24, 2021, 4:42 pm ad1c9bdddf
Recall that the Taylor expansion of a function around a point x = a is
f(x) = f(a) + f'(a)(x-a) + 1/2 f''(a)(x-a)^2 + 1/6 f'''(a)(x-a)^3 + ...
I think the approach would be to calculate the Taylor polynomial for the Plancks Law and then show that for large lambda Plancks law is consistent with Rayleigh-Jeans.
f(lamba) = 8 Pi h c lambda^(-5) / (exp(hc/lambda K T) - 1)
Ok. First off, just going at it and trying to calculate the first, second, third derivative and so on is not the right plan of attack. (If you try it, things get complicated fast).
A better way is to build up the Taylor polynomial from the Taylor ...
This solution explains how to use Taylor polynomials to prove Planck's Law gives approximately the same values as the Rayleigh-Jeans Law.