Linear approximation

Near its point of tangency a tangent can stand in for the curve¹. We approximate $f(x) = \sqrt{1 + x}$ around $x = 0$ .

Finding the tangent

From $f(0) = 1$ and $f'(x) = \dfrac{1}{2\sqrt{1 + x}}$ we get $f'(0) = \dfrac{1}{2}$ , so the tangent is the following line.

y = 1 + \frac{x}{2}

In general one approximates $f(x) \approx f(a) + f'(a)(x - a)$ around $x = a$ . That is a linear approximation.

Accuracy

$x$	$\sqrt{1+x}$	Approximation $1 + \dfrac{x}{2}$	Error
$0.01$	$1.0049876$	$1.005$	$0.0000124$
$0.1$	$1.048809$	$1.05$	$0.0012$
$1$	$1.41421$	$1.5$	$0.086$

Shrinking $x$ by a factor of ten shrinks the error by a factor of a hundred, because the error is of the order of $(x - a)^2$ . It breaks down at a distance, so the approximation serves only near the point of tangency.

Using it to compute

From $\sqrt{101} = 10\sqrt{1.01} \approx 10 \times 1.005 = 10.05$ , a number very close to the true value $10.049876$ comes out by mental arithmetic.

How physics uses it

The pendulum approximation $\sin\theta \approx \theta$
The relativistic $\dfrac{1}{\sqrt{1 - v^2/c^2}} \approx 1 + \dfrac{v^2}{2c^2}$
The thin-lens approximation in optics

However unwieldy the original expression, over a small range it can be replaced by a line or by a quadratic.

Improving the accuracy

Adding a term improves it.

\sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}

At $x = 1$ the error shrinks from $0.086$ to $0.039$ . Continuing in this direction gives the Taylor expansion, and the coefficient $-\dfrac{1}{8}$ of the quadratic term comes from $\dfrac{f''(0)}{2}$ ².

The curve on the graph is $y = \sqrt{1 + x}$ , the line is the tangent $y = 1 + \dfrac{x}{2}$ , and the large dot is the point of tangency $(0, 1)$ .

Linear approximation, Wikipedia
Taylor series, Wikipedia