Question

Suppose g is a function which has continuous derivatives, and that g(7)=4,g′(7)=−1, g″(7)=1, g‴(7)=1. (a) what is the taylor polynomial of degree 2 for g near 7? p2(x)=

Answer 1

Taylor series of a function g(x) that can be differentiated indefinitely at "a" (a=complex or real number) is given by:

pn(a) = g(a)+g'(a)(x-a)/1! +g''(a)(x-a)^2/2! + g'''(a)(x-a)^3/3! + g''''(a)(x-a)^4/4! + ...
 Where n= 0,1,2,3,4, ... respectively = degrees of the polynomial series

In the current task,
n=2, a=7

Substituting;

p2(x) = g(7)+g'(7)(x-7)+g''(7)(x-7)^2/2! = 4+(-1)(x-7)+(1)(x-7)^2/2! 
= 4-(x-7)+1/2(x-7)^2

Answer 2

A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function f(x) about a point x=a is given by

$f(x0=f(a)+\dfrac{f'(a)}{1!} (x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+...$.

In your case a=7 and g(7)=4, g′(7)=−1, g″(7)=1, g‴(7)=1.

The Taylor polynomial of degree 2 for g near 7 is:

$g_T(x)=g(7)+\dfrac{g'(7)}{1!} (x-7)+\dfrac{g''(7)}{2!}(x-7)^2+...=$

$=4+ \dfrac{-1}{1!} (x-7)+\dfrac{1}{2!}(x-7)^2+...=$

$=4-(x-7)+\dfrac{1}{2}(x-7)^2+...=$.