MATH SOLVE

5 months ago

Q:
# 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)=

Accepted Solution

A:

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

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