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Let's say I need to approximate the expression $$\frac{1}{2}mv^2\left(\frac{M}{M+m}+1\right)$$ when $m<<M$.

Here is what I would do: $$\frac{1}{2M}mv^2\left(\frac{M}{1+\frac{m}{M}}+M\right) =\frac{1}{2M}mv^2\left[M\left(1-\frac{m}{M}+O(m^2)\right) +M\right]$$ Therefore approximating this, by dropping the higher-order terms in $m$ in the bracket I get: $$\frac{1}{2M}mv^2(M+M) = mv^2$$

Ok, so when doing this are there any rules that we need to follow for example:

If the very small term is on the bottom of a fraction, use the binomial expansion to bring it up. (Is this always true?)

Are their any more, e.g. if we get something like $Mm$ what do we do? What about $Mm^2$ etc? When do we turn $M+m$ into just $M$?

Travis Willse
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1 Answers1

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If I may, I would suggest you first rewrite $$\frac{1}{2}mv^2\left(\frac{M}{M+m}+1\right)=\frac{1}{2}mv^2\left(\frac{1}{1+\frac{m}{M}}+1\right)$$ and that using that, for small values of $x$, $$\frac{1}{1+x} \simeq 1- x+ x^2-x^3 +\cdots$$ replace in the above $x$ by $\frac{m}{M}$ and get finally $$\frac{1}{2}mv^2\left(\frac{M}{M+m}+1\right) =m v^2-\frac{m^2 v^2}{2 M}+\frac{m^3 v^2}{2 M^2}+O\left(m^4\right)$$ that is to say $$\frac{1}{2}mv^2\left(\frac{M}{M+m}+1\right) =m v^2\Big(1-\frac{1}{2}\frac{m}{M}+\frac{1}{2}(\frac{m}{M})^2+\cdots\Big)$$

In the same spirit, $Mm=M^2 \frac{m}{M}$ $Mm^2=M^3 \Big(\frac{m}{M}\Big)^2$

Claude Leibovici
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