# Tangent Lines as Approximation

## Tangent Lines 101

There are certainly two things that you should know about tangent lines:

• They're graphs of linear functions, and
• They do the best job of approximating a function at the point of tangency of any other linear function.

Tangent lines are generalized by Taylor Series as Approximations.

## The Math

If we want the tangent line at the point (a,f(a)), it's the graph of

$L(x)=f(a)+f^\prime(a)(x-a)$

This makes perfect sense, showing that L(x) matches f(x) in terms of both the function values and derivatives at x = a:

$\begin{cases} L(a)=f(a)+f^\prime(a)(a-a)=f(a) \\ L^\prime(a) = \left( f(a)+f^\prime(a)(x-a) \right)^\prime |_{x=a} = f^\prime(a) \end{cases}$

## Application: Newton's Method

One of the most important applications of the linearization $\left.L(x)\right.$ is in root-finding: that is, finding zeros of a non-linear function.

Given function $\left.f(x)\right.$, and a guess for a root $\left.x_0\right.$. We would first check to see if $\left.f(x_0)=0\right.$. If it is, we're done; otherwise, we might try to improve the guess. How so?

Use the linearization $L(x)=f(x_0)+f^\prime(x_0)(x-x_0)$, and find a zero of it:

$L(x)=f(x_0)+f^\prime(x_0)(x-x_0)=0 \implies x=x_0-\frac{f(x_0)}{f^\prime(x_0)}$

This is Newton's method, an iterative scheme for improving the approximation of a root: we need a good approximation for the root, $\left.x=x_0\right.$, and then

$x_1=x_0-\frac{f(x_0)}{f^\prime(x_0)}$

Then we compute

$x_2=x_1-\frac{f(x_1)}{f^\prime(x_1)}$

and so on, computing a sequence which we hope will converge on the true root $\left.x=r\right.$.

### For example:

• Let $f(x)=\sqrt{x}-x$; then $f^\prime(x)=\frac{1}{2}x^{-1/2}-1$
• Guess: $\left.x_0=4\right.$ (pretty bad guess!)
• Improved guess: $x_1=4-\frac{2-4}{\frac{1}{4}-1}=4-8/3=4/3$
• Now do it again! After another go, $x_2 \approx 1.0182772543872314$
• Once more: $x_3 \approx 1.0000812810399125$
• Once more: $x_4 \approx 1.0000000016514505$
• Once more: $x_5 \approx 1.0$.

We check, and find that $\left.f(1)=0\right.$ -- egads, we've found one! We've found a root: $\left.r=1\right.$

 Starting guess: x0 = 4 Improved: x1 = 4 / 3 Even Better: x2 = 1.00008 Wish I'd started here: x3 = 1.00000000165. I would have saved myself three steps!

It's like a miracle... but it's just mathematics!