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

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)+(f'(a))/(1!) (x-a)+(f''(a))/(2!)(x-a)^2+(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)+(g'(7))/(1!) (x-7)+(g''(7))/(2!)(x-7)^2+...=`


`=4+ (-1)/(1!) (x-7)+(1)/(2!)(x-7)^2+...=`


`=4-(x-7)+(1)/(2)(x-7)^2+...=`.

Answer 2

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