0
$\begingroup$

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 !.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

It could be a special case of

$$\frac{1}{1+u} \approx 1-u$$ when $u$ is small. In your case it would apply when $\frac{v}{c}\cos \theta$ is small.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.