# Quick question on an approximation from physics

I've seen written in a physics problem sheet that for a small $$\epsilon>0$$, the following approximation is considered $$(1+2\epsilon)^{-\frac12}\approx 1-\epsilon.$$ Any reason why this is the case? Is this any special approximation in physics?

This is an instance of Taylor's expansion $$(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\dots,$$ with $$x=2\varepsilon$$ and $$n=-1/2$$. If you truncate the expansion, the error is an infinitesimal of order higher than the last term left. In your case, the error is of order $$\varepsilon^2$$, which is small compared to $$\varepsilon$$.

That's a special case of the Binomial approximation formula

$$(1 + x)^\alpha \approx 1 + \alpha x$$

for $$|x|$$ small.

This comes from the definition of derivative at $$x=0$$ of $$f(x)=(1+x)^\alpha$$: $$(1+x)^{\alpha}=1+\alpha x + o(x)$$ Because $$\left((1+x)^{\alpha}\right)'=\alpha (1+x)^{\alpha -1 }$$

Start with squaring: $$(1-ϵ)^2 = 1-2ϵ-ϵ^2 ≈ 1-2ϵ$$

If $$ϵ$$ is small, then $$ϵ^2$$ is $$very$$ small, too small to care about. (At least physicists think so)

Take the square root on both sides and you get $$1-ϵ ≈ (1-2ϵ)^{1\over 2}$$

This is close to what you got, but not quite.

On the other hand, you have $$(1-ϵ)(1+ϵ)=1-ϵ^2 ≈ 1$$

which means $$(1-ϵ)≈(1+ϵ)^{-1}$$

Applying it to $$(1-2\epsilon)^\frac12$$ you end up with $$1-\epsilon\approx (1-2\epsilon)^\frac12\approx(1+2\epsilon)^{-\frac12}.$$

This and many other cases are covered by Newton's Generalized Binomial Theorem. Here, we have $$(1+2\epsilon)^{-\frac12} = \displaystyle\sum_{i=0}^\infty \binom {-\frac 12}i (2\epsilon)^i$$

And of course, $$\epsilon^i~\forall~i>0$$ is very small in comparison to the rest of the sum so we omit those terms. So, our sum approximates to $$\binom{-\frac 12}0 + \binom{-\frac 12}1(2\epsilon)$$ $$= \boxed{1-\epsilon}$$