I need to transform an equation into another using taylor series

Question

I have an astrophysics class and I am trying to finalize some exercices. My astrophysics teacher didn't teach us any of this saying that we should have learned it from other classes by other teachers (which I did not). I need to transform this equation

$$v+Δv=(v+Δv')/(1+(vΔv')/c^2) \tag{1}$$

into this

$$Δv≈Δv'(1-v^2/c^2) \tag{2}$$

My instructions say that I should first transform the denominator of (1) into a Taylor series. Then I need to eleminate every (Δv’)^2 to obtain (2). I have no clue how I am supposed to do so. (1) is the velocity addition formula

Answer 1

hint

If $ X$ is closer to zero, then

$$\frac{1}{1+X}=1-X+X^2-X^3+...$$ $$\approx 1-X$$

So

$$\frac{1}{1+\frac{v\Delta v'}{c^2}}\approx 1-\frac{v\Delta v'}{c^2}$$

Answer 2

I think I got it, for future references heres what I did after using hamam's hint: $$v+Δv≈(v+Δv')(1−(vΔv′)/c^2) $$ $$v+Δv≈v+Δv′-((vΔv'^2)/c^2)-((v^2Δv′)/c^2) $$ $$Δv≈Δv′-(vΔv'^2)/c^2)-((v^2Δv′)/c^2) $$ $$Δv≈Δv′(1-(vΔv'/c^2)-(v^2/c^2)) $$ Using the fact that
$$v+Δv≈(v+Δv')(1−(vΔv′)/c^2)$$
I find that
$$(v+Δv)/(v+v′)≈ 1-(vΔv'/c^2)$$ $$Δv≈Δv′((v+Δv)/(v+Δv')-(v^2/c^2)) $$ I then know that $$ (v+Δv)/(v+v′)≈ 1$$ Which leaves me with $$Δv≈Δv′(1-(v^2/c^2)) $$ P.S. if this is wrong please tell