Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 5

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 apply the definition of Taylor series for the given function f(x) = 1/x centered at c=1 , we list f^n(x) using the Power rule for differentiation: d/(dx) x^n= n *x^(n-1) and basic differentiation property: d/(dx) c* f(x)= c * d/(dx) f(x) .
f(x) =1/x
f'(x) = d/(dx) 1/x
= d/(dx) x^(-1)
=-1 *x^(-1-1)
=-x^(-2) or -1/x^2
f^2(x)= d/(dx) -x^(-2)
=-1 *d/(dx) x^(-2)
=-1 *(-2x^(-2-1))
=2x^(-3) or 2/x^3
f^3(x)= d/(dx) 2x^(-3)
=2 *d/(dx) x^(-3)
=2 *(-3x^(-3-1))
=-6x^(-4) or -6/x^4
f^4(x)= d/(dx) -6x^(-4)
=-6 *d/(dx) x^(-4)
=-6 *(-4x^(-4-1))
=24x^(-5) or 24/x^5
Plug-in x=1 , we get:
f(1)=1/1 =1
f'(1)=-1/1^2 = -1
f^2(1)=2/1^3 =2
f^3(1)=-6/1^4 = -6
f^4(1)=24/1^5 = 24
Plug-in the values on the formula for Taylor series, we get:
1/x =sum_(n=0)^oo (f^n(1))/(n!) (x-1)^n
=f(1)+f'(1)(x-1) +(f^2(1))/(2!)(x-1)^2 +(f^3(1))/(3!)(x-1)^3 +(f^4(1))/(4!)(x-1)^4 +...
=1+(-1)*(x-1) +2/(2!)(x-1)^2 +(-6)/(3!)(x-1)^3 +24/(4!)(x-1)^4 +...
=1-(x-1) +2/2(x-1)^2 -6/6(x-1)^3 +24/24(x-1)^4 +...
= 1-(x-1)+ (x-1)^2 -(x-1)^3 + (x-1)^4 +...
The Taylor series for the given function f(x)=1/x centered at c=1 will be:
1/x=1-(x-1)+ (x-1)^2 -(x-1)^3 + (x-1)^4 +...
or
1/x =sum_(n=0)^oo(-1)^n(x-1)^n