Theory

Theory 1

Linear approximation is the technique of approximating a specific value of a function, say $f(x_1)$, at a point $x_1$ that is close to another point $x_0$ where we know the exact value $f(x_0)$. We write $\mathrm{\Delta }x$ for $x_1-x_0$, and $y_0=f(x_0)$, and $y_1=f(x_1)$. Then we write $dy=f^{\prime }(x_0)\cdot \mathrm{\Delta }x$ and use the fact that:

$$y_1\approx y_0+dy=y_0+f^{\prime }(x_0)\cdot \mathrm{\Delta }x$$

Computing a linear approximation

For example, to approximate the value of $\sqrt{4.01}$, set $f(x)=\sqrt{x}$, set $x_0=4$ and $y_0=2$, and set $x_1=4.01$ so $\mathrm{\Delta }x=0.01$.

Then compute: $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ So $f^{\prime }(x_0)=1/4$.

Finally:

$$y_1\approx y_0+f^{\prime }(x_0)\cdot \mathrm{\Delta }x\gg \gg y_1\approx 2+\frac{1}{4}\cdot 0.01=2.0025$$

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Now recall the linearization of a function, which is itself another function:

Given a function $f(x)$, the linearization $L(x)$ at the basepoint $x=c$ is the functional form of the tangent line, the line passing through $(x_0,y_0)=(c,f(x))$ with slope $m=f^{\prime }(c)$:

$$\begin{array}{cccc}& y& =y_0+m(x-x_0)& \text{(point-slope eqn. of line)}\\ & & & \\ & \gg \gg L(x)& =f(c)+f^{\prime }(c)(x-c)& \text{(linearization fcn.)}\end{array}$$

The graph of this linearization $L(x)$ is the tangent line to the curve $y=f(x)$ at the point $(c,f(c))$.

The linearization $L(x)$ may be used as a replacement for $f(x)$ for values of $x$ near $c$. The closer $x$ is to $c$, the more accurate the approximation $L(x)$ is for $f(x)$.

Computing a linearization

We set $f(x)=\sqrt{x}$, and we let $c=4$.

We compute $f(c)=2$, and $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ so $f^{\prime }(c)=\frac{1}{4}$.

Plug everything in to find $L(x)$:

$$L(x)=f(c)+f^{\prime }(c)(x-c)\gg \gg L(x)=2+\frac{1}{4}(x-4)$$

Now approximate $f(4.01)\approx L(4.01)$:

$$L(4.01)=2+\frac{1}{4}(4.01-4)=2.0025$$

Theory 2

Taylor polynomials

The Taylor polynomials $T_N(x)$ of a function $f(x)$ are the partial sums of the Taylor series of $f(x)$:

$$\begin{array}{cc}{T}_N(x)& =\sum _{i=0}^N\frac{f^{(i)}(c)}{i!}{(x-c)}^i\\ & \\ & =f(c)+\frac{f^{\prime }(c)}{1!}(x-c)+\cdots +\frac{f^{(N)}(c)}{N!}{(x-c)}^N\end{array}$$

These polynomials are generalizations of linearization. Specifically, $f(c)=T_0(x)$, and $L(x)=T_1(x)$.

The Taylor series $T_n(x)$ is a better approximation of $f(x)$ than $T_i(x)$ for any $i<n$.

Facts about Taylor series

The series $T_n(x)$ has the same derivatives as $f(x)$ at the point $x=c$. This fact can be verified by visual inspection of the series: apply the power rule and chain rule, then plug in $x=c$ and all factors left with $(x-c)$ will become zero.

The difference $f(x)-T_n(x)$ vanishes to order $n$ at $x=c$:

$$\begin{array}{cc}f(x)-T_n(x)& =\frac{f^{(n)}(c)}{n!}{(x-c)}^n+\frac{f^{(n+1)}(c)}{(n+1)!}{(x-c)}^{n+1}+\cdots \\ & \\ & ={(x-c)}^n(\frac{f^{(n)}(c)}{n!}+\frac{f^{(n+1)}(c)}{(n+1)!}(x-c)+\cdots )\end{array}$$

The factor ${(x-c)}^n$ drives the whole function to zero with order $n$ as $x\to c$.

If we only considered orders up to $n$, we might say that $f(x)$ and $T_n(x)$ are the same near $c$.