In my physics class today, we were looking at the expression $(1-(u/v))^{-1}$. In a single step, the professor showed that this expression equals $1+u/v$. How is that? Is it the binomial theorem?
If $u$ is much less than $v$, as needed to write out the expression with a Taylor series approximation, how can that be? I thought that $v$ was the speed through a material, so shouldn’t it be lower than $u$? And on that note, if you don’t mind, what exactly is $u$?
Assuming your professor had assumed $\frac{u}{v}<1$, let us write $x=\frac{u}{v}$ and let us denote $f(x)=\frac{1}{1-x}$. Now if you know how Taylor expansion works then you can find it easily and approximate f(x) up to linear order in $x$. Let me also clarify you that $f(x)\approx 1+x$ not both are equal and generally for practical purpose linear approximation is enough and higher order terms/corrections on $x$ gives more and more accuracy to $f(x)$. From high school we already know infinite geometric series,So we can write for $x<1$ $$1+x+x^2+\dots +x^n+\dots=\frac{1}{1-x}$$ where $x$ is common ratio of the infinite geometric series. $$\implies f(x)=1+x+o(x^2)$$ Thus $f(x)\approx 1+x$.
I hope this helps.
It is the first order Taylor expansion and is valid for $\frac{u}{v} \ll1$.
$v^2=\frac{GM}{r}$. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$