Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 29

Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of f^n(x) centered at x=c . The general formula for Taylor series is:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...
To determine the Taylor polynomial of degree n=4 from the given function f(x)=ln(x) centered at x=2 , we may apply the definition of Taylor series.
We list f^n(x) up to n=4 as:
f(x) = ln(x)
f'(x) = d/(dx)ln(x) =1/x
Apply Power rule for derivative: d/(dx) x^n= n *x^(n-1) .
f^2(x) = d/(dx) 1/x
= d/(dx) x^(-1)
=-1 *x^(-1-1)
=-x^(-2) or -1/x^2
f^3(x)= d/(dx) -x^(-2)
=-1 *d/(dx) x^(-2)
=-1 *(-2x^(-2-1))
=2x^(-3) or 2/x^3
f^4(x)= d/(dx) 2x^(-3)
=2 *d/(dx) x^(-3)
=2 *(-3x^(-3-1))
=-6x^(-4) or -6/x^4
Plug-in x=2 , we get:
f(2) =ln(2)
f'(2)=1/2
f^2(2)=-1/2^2 = -1/4
f^3(2)=2/2^3 =1/4
f^4(2)=-6/2^4 = -3/8
Applying the formula for Taylor series, we get:
sum_(n=0)^4 (f^n(2))/(n!) (x-2)^n
=f(2)+f'(2)(x-2) +(f^2(2))/(2!)(x-2)^2 +(f^3(2))/(3!)(x-2)^3 +(f^4(2))/(4!)(x-2)^4
=ln(2)+1/2(x-2) +(-1/4)/(2!)(x-2)^2 +(1/4)/(3!)(x-2)^3 +(-3/8)/(4!)(x-2)^4
=ln(2)+1/2(x-2) -(1/4)/2(x-2)^2 +(1/4)/6(x-2)^3 -(3/8)/24(x-2)^4
=ln(2)+1/2(x-2) -1/8(x-2)^2 + 1/24(x-2)^3 -1/64(x-2)^4
The Taylor polynomial of degree n=4 for the given function f(x)=ln(x) centered at x=2 will be:
P_4(x)=ln(2)+1/2(x-2) -1/8(x-2)^2 + 1/24(x-2)^3 -1/64(x-2)^4