# Taylor expansion for equation simplification

What taylor approximation would be necessary to approximate the equation

$${1 \over 1 + \left(v/c\right)\cos\left(\,\theta\,\right)} \approx 1 - {v \over c}\,\cos\left(\,\theta\,\right)$$

I saw Walter Lewin write down an equation like this, but I don't understand how he came to this conclusion. It comes from the equation$\ldots$

$\displaystyle f' = f\left[1 + {v \over c}\cos\left(\,\theta\,\right)\right]\quad \text{where}\quad f = {c \over \lambda}$.

$\displaystyle λ'\left[1 + {v \over c}\,\cos\left(\,\theta\,\right)\right] = λ$.

This is then somehow changed to $λ' = λ\left[1 - \left(v/c\right)\cos\left(\,\theta\,\right)\right]$. Any explanation ?. Thanks !.

I hope he didn't write down $$\frac1{1+\cos\theta}\approx1-\cos\theta.$$ That's bad.
But in effect he wrote down $$\frac1{1+(v/c)\cos\theta}\approx1-(v/c)\cos\theta.$$ If $v/c$ is small, this is OK. It is $$\frac1{1+x}\approx1-x$$ for small $x$ which is fine: $1-x$ is the beginning of the geometric series for $1/(1+x)$. The error is of the order of $x^2$, so that is fine if you can afford to neglect errors of that size.
$$\frac{1}{1+u} \approx 1-u$$ when $u$ is small. In your case it would apply when $\frac{v}{c}\cos \theta$ is small.