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# Linear approximations

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).