This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

# Linear approximations

### From Mathematics Is A Science

We know that the tangent line "approximates" the graph of $y=f(x)$ around $x=a$. There amny straight lines that look OK, the point is that, in fact, this is the best one among all other lines!

Assume $f$ is differentiable at $x=a$. In that case, when you zoom in on the point, the tangent line will (but any other line won't) merge with the graph. This is the geometric meaning of *best approximation*. What about algebra?

What is the equation of the tangent line? $$ \text{slope } = f^{\prime}(a), \text{ passes through }(a,f(a))$$

Point-slope form: $$ y - y_{0} = m(x - x_{0}) $$ or $$ y - f(a) = f^{\prime}(a)(x - a) $$ This is in fact a function. We solve for $y$ to make explicit: $$ y = f(a) + f^{\prime}(a)(x - a) $$ What kind of function is this? Linear (because the power of $x$ is 1).

The *best linear approximation* at $x = a$ (or around $a$) of the (differentiable) function $y = f(x)$ is
$$ f(x) = f(a) + f^{\prime}(a)(x - a) $$

How good is it?

$$ \text{error } = | f(x) - \ell(x) | = \text{ lengths of these segments.} $$ It's 0 at $x = a$, and may grow as you move away from $a$.

**Example:** Approximate $\sqrt{4.1}$. We can't compute $\sqrt{x}$ by hand. So $f(x) = \sqrt{x}$ will have to be approximated at $x = a = 4$.

Start with this: $$ \sqrt{4} = 2, \text{ exactly, or}$$ $$f(4) = 2. $$

Question: What else do we know about 4?

Answer: the derivative. $$\begin{aligned} f^{\prime}(x) &= \frac{1}{2\sqrt{x}} \\ \Rightarrow f^{\prime}(4) &= \frac{1}{2\sqrt{4}} = \frac{1}{4} \end{aligned}$$ So to compute $\sqrt{4.1}$, use the approximation: $$\begin{aligned} \ell(x) &= f(a) + f^{\prime}(a) (x - a) \\ & = 2 + \frac{1}{4} (x - 4) \end{aligned} $$ Good enough.

$$\begin{aligned} \ell(4.1) &= 2 + \frac{1}{4}(4.1 - 4) \\ & = 2 + \frac{1}{4} \cdot 1 \\ & = 2 + 0.025 \\ & = 2.025 \end{aligned} $$ This is a linear approximation (aka 1st degree).

Compare to the "dumb" approximation (aka 0 degree): 4.1 is close to 4, then $\sqrt{4.1}$ is close to $\sqrt{4}$. So $$ \sqrt{4.1} \approx 2. $$

Compare now to the true value: $$ \sqrt{4.1} \approx 2.0248 $$ So, the linear approximation is better. (Taylor polynomials provide even higher degree approximations, see Calc 2).