Simplification of $ \sqrt{(1-x^2)}$ to $(1-\frac{x^2}{2})$

Question

While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:

$$ \sqrt{1-x^2} \approx \left(1-\frac{x^2}{2}\right) $$

I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).

Answer 1

Term to look for: linear approximation

In general, the best linear approximation for a differentiable function near a point $c$ is $$ f(x) \approx f(c) + f'(c)\;(x-c) $$ This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.

Now if $f(x) = \sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-\frac{1}{2}$. So $$ \sqrt{1-x} \approx 1 - \frac{x}{2} $$ To get your case, substitute $x^2$ for $x$.

Answer 2

Use the Taylor series expansion at $x=0$ to get $\sqrt{1-x^2}\approx1-\frac{x^2}2+o(x^4)$.

Answer 3

Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).

$$\sqrt{1-t}\approx 1-\frac t2.$$

Hence for small $x$,

$$\sqrt{1-x^2}\approx 1-\frac{x^2}2.$$

The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)

$$\sqrt{1-t}\approx 1-\frac t2-\frac{t^2}8,$$ and $$\sqrt{1-x^2}\approx 1-\frac{x^2}2-\frac{x^4}8,$$