3.2 Taylor expansion with Lagrange remainder

On the real line, if we have a function $f(x)$, and we wish to know the value of this function at a point $x=b$ given the value of $f\left(x\right)$ and its derivatives at another point say $x=a$, then we write

$$f(b)=f(a)+(b-a)f^{\prime }(a)+\frac{{(b-a)}^2{f}^{\prime \prime }(a)}{2!}+\cdots$$

Now suppose we want to find the value of the function at arbitrary point $x$ given the value of $f(x)$ and its derivative at another point say $x=a$, then we replace $b$ by $x$ above and write

$$f(x)=f(a)+(x-a)f^{\prime }(a)+\frac{{(x-a)}^2{f}^{\prime \prime }(a)}{2!}+\cdots +R_n$$

Where

$$R_n=\frac{{(x-a)}^{n+1}}{(n+1)!}{f}^{(n+1)}\left(\xi \right)$$

Where $\xi$ is some point between $x$ and $a$

If $x-a=h$, we can write the above as

$$\tilde{f}(x)=f(a)+h{f}^{\prime }(a)+\frac{h^2{f}^{\prime \prime }(a)}{2!}+\frac{h^3{f}^{\prime \prime }(a)}{3!}+\cdots +\frac{h^{n+1}}{(n+1)!}{f}^{(n+1)}\left(\xi \right)$$

Figure 3.7:example

note: If the point of expansion is zero, Taylor series is called maclaurin series.

$$\tilde{f}(x)=f(a)+x{f}^{\prime }(0)+\frac{x^2{f}^{\prime \prime }(0)}{2!}+\frac{x^3{f}^{\prime \prime }(0)}{3!}+\cdots +\frac{x^{n+1}}{(n+1)!}{f}^{(n+1)}\left(\xi \right)$$

Why do we use Taylor series for? To express a function as a series. This can allow one to more easily manipulate it. Also, if the function is non-linear, by expressing it in series, and dropping low order non-linear terms (h must be very small to have good approximation), then we have linearized a non-linear function in the vicinity of a point of expansion. Hence around the point of expansion, we can approximate the non-linear function by its linear Taylor series terms for the purpose of doing further linear system analysis (as it is easier to work with linear functions than non-linear ones).

3.2.1 Finding Error in Taylor series approximation

Things to know: How to find how many terms in Taylor series to approximate some given function to some accuracy?

Idea of solution: Express $E_n$, this is the error term, or the remainder. Make $\left|E_n\right|<ϵ$ where $ϵ$ is the accuracy needed. Find smallest $n$ which makes this true

Example: How many terms needed to find $\ln (2)$ to accuracy of $ϵ={10}^{-8}$?

Expand $\ln (x)$ at $x=1$, hence $h=2-1=1$

$$\begin{array}{rl}\ln (x)& =(x-1)-\frac{1}{2}{(x-1)}^2+\frac{1}{3}{(x-1)}^3+\cdots +E_n\\ \ln (2)& =1-\frac{1}{2}+\frac{1}{3}+\cdots +E_n\end{array}$$

We want $\left|E_n\right|<{10}^{-8}$, but $E_n=\frac{1}{n+1}<{10}^{-8}\Rightarrow n\ge {10}^8$, hence at least $100$ million terms would be needed to computer $\ln (2)$ using Taylor series with accuracy of ${10}^{-8}$

3.2.2 Mean value theorem

if $f(x)$ is continues on $[a,b]$, and if $f^{\prime }\left(x\right)$ exist on the open interval $(a,b)$ then there exists a point $\xi$ between $b,a$ s.t.$$f(b)-f(a)=f^{\prime }\left(\xi \right)(b-a)$$

3.2.3 Rolle’s theorem

if $f(x)$ is continuous on $[a,b]$ and if $f^{\prime }\left(x\right)$ exist on $(a,b)$ and if $f(a)=f(b)$ then $f^{\prime }\left(\xi \right)=0$ for some point in $(a,b)$