Taylor approximation of the Doppler Law for slow-moving emitters

[IntegrateMe]

The relationship linking the emitted frequency F_e and the received frequency F_r is the Doppler Law:

$$F_r = \sqrt \frac{1-\frac{v}{c}}{1-\frac{v}{c}} F_e$$

The Taylor series for the function $$\sqrt\frac{1+x}{1-x}$$ near x = 0 is $$1+x+\frac{x^2}{2}+\frac{x^3}{3}+...$$

On Earth, most objects travel with v much smaller than c. That is, the ratio v/c is very small. Use this fact to obtain the approximation to the Doppler Law for most objects on Earth:

$$F_r ≈ (1-\frac{v}{c})F_e$$

I'm not really sure where to go with this. I know it will involve using the Taylor series 1 + x + x²/2 + ... probably only using the 1 + x part, but I'm not sure what to do from here.

 

[lippyka]

Can someone please explain? Since v/c is very small on Earth, we can assume that the ratio v/c is approximately equal to 0. Therefore, using the Taylor series expansion for the Doppler Law, the approximation for most objects on Earth is: F_r ≈ (1+0) F_e = F_e Thus, the received frequency is approximately equal to the emitted frequency.