Does the Doppler effect formula derive from classical speed addition?

[asfd]

Doppler effect revisited

Am I right when I say that the doppler effect formula "f=f0((v+vo)/(v+vs))" is derivated from the classical theorem of speed addition and this is why the doppler effect for light and EM waves is different?

 

[Integral]

Because light does not behave as a classical particle, it has a constant speed for all observers. It is frequency and wavelength which undergo a Doppler shift in light.

 

[asfd]

I'm a little confused here...

I don't get how on this page: hyperphysics they get the first equation... I don't see the relativistic doppler effect written that way anywhere and I don't understand how they get ot that equation and how to expand it with maple...Anybody can help?

 

[Doc Al]

That equation is just the relativistic Doppler formula rewritten in a form convenient for deriving the low speed approximation. They took something that usually appears in a form like:
$$\sqrt{\frac{1 + x}{1 - x}}$$
And rewrote it like:
$$\frac{\sqrt{1 - x^2}}{1 - x}$$
These expressions are equivalent.