Question

17.3.1

Jllow pt (a applieg to -a)(r -b)2, and find c explicitly. 17.3 the maxim the interval [0,.1]? 1 Determine explicitly the Taylor polynomial T2(x) for e-z at 0. What is um error made in using this as an approximation to the function over

Answer 1

The funciton f(x) is

Derivatives of f(x) are

$f^1(x)=(-1)^1e^{-x}=-e^{-x}$

$f^2(x)=(-1)^2e^{-x}=e^{-x}$

$f^n(x)=(-1)^ne^{-x}$

This implies,

$f^n(0)=(-1)^ne^{-0}=(-1)^n$

Therefore,

$T(x)=f(0)+\frac{f^1(0)}{1!}x+\frac{f^2(0)}{2!}x^2+\frac{f^3(0)}{3!}x^3+\frac{f^4(0)}{4!}x^4+...$

Therefore,

$T(x)=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+...$

The polynomial of order 2 is

$T_{2}(x)=1-x+\frac{x^2}{2}$

From Taylors Remainder Theorem, error bound for the T2(x) is

$|error| \leq \frac{Max|f^{n+1}(c)||x|^{n+1}}{(n+1)!} \ where \ 0 < c < x$

For T2 polynomial

$|error| \leq \frac{Max|f^{2+1}(c)||x|^{2+1}}{(2+1)!} => |error| \leq \frac{Max|f^{3}(c)||x|^{3}}{3!}$

Consider the third derivative of f(x)

The function f(x) is a decreasing function in x over the interval [0,0.1] implies the maximum for f^3(x) exists at x=0

$Max|f^3(x)|=e^{-0}=1$

Therefore,

$|error| \leq \frac{(1)(0.1)^{3}}{3!} => error \leq 1.6667 \times 10^{-4}$

The above is maximum error for T2 polynomial