Quick question on an approximation from physics

Question

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?

Answer 1

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 $\varepsilon$.

Answer 2

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 $$\left((1+x)^{\alpha}\right)'=\alpha (1+x)^{\alpha -1 }$$

Answer 3

That's a special case of the Binomial approximation formula

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

for $|x|$ small.

Answer 4

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 $$ 1-\epsilon\approx (1-2\epsilon)^\frac12\approx(1+2\epsilon)^{-\frac12}. $$

Answer 5

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 $$\binom{-\frac 12}0 + \binom{-\frac 12}1(2\epsilon)$$ $$= \boxed{1-\epsilon}$$