Not to debate this much further, but we were discussing whether the medium is dispersive. That means the speed of sound vs frequency. Your calculations are showing speed of sound with temperature - no debate about that
A temperature gradient will render the medium more dispersive in frequency. For large wavelengths, the waves see an average air density, whereas shorter wavelengths will travel through the gradient and locally see a varying speed of sound:
If the wavelength λ is larger than the length L of the gradient, the effective speed of sound in the medium will be
c1=sqrt(K/<ρ>), where K is the elasticity modulus and <ρ> the average density.
For shorter wavelengths this is a bit more interesting. The travel time is,
T= int_0^L 1/c(x) dx,
where int_0^L is the integral from 0 to L and c(x) the local speed of sound along the gradient. Replacing c(x) with sqrt(K/ρ(x)) we get,
T= 1/sqrt(K) int_0^L sqrt(ρ(x)) dx = L <sqrt(ρ)>/sqrt(K)
And the effective speed of sound is
c2=sqrt(K)/<sqrt(ρ)>
Note that in the first case we have sqrt(<ρ>) and in the second we have <sqrt(ρ)>, which is the square root of the average vs. the average of the square roots. The Jansen inequality implies that c2 > c1. Thus, different wavelengths have strictly different speed of sound in a temperature gradient. Anyway, I guess the difference is ridiculously small in a room.
I hope this clarifies everything and that we can move forward. BUT, if I made a calculation error, please correct me.
