Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 26

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)=1/x^2 centered at c=2 , we may apply the definition of Taylor series.
For the list of f^n(x) up to n=4 , we may apply Law of Exponent: 1/x^n = x^-n and Power rule for derivative: d/(dx) x^n= n *x^(n-1) .
f(x) = 1/x^2 or x^(-2)
f'(x) = d/(dx)x^(-2)
=-2 *x^(-2-1)
=-2x^(-3) or -2/x^3
f^2(x)= d/(dx) -2x^(-3)
=-2 *d/(dx) x^(-3)
=-2 *(-3x^(-3-1))
=6x^(-4) or 6/x^4
f^3(x)= d/(dx) 6x^(-4)
=6 *d/(dx) x^(-4)
=6 *(-4x^(-4-1))
=-24x^(-5) or -24/x^5
f^4(x)= d/(dx) -24x^(-5)
=-24 *d/(dx) x^(-5)
=-24 *(-5x^(-5-1))
=120x^(-6) or 120/x^6
Plug-in x=2 , we get:
f(2)=1/2^2 =1/4
f'(2)=-2/2^3 = -1/4
f^2(2)=6/2^4 =3/8
f^3(2)=-24/2^5 = -3/4
f^4(2)=120/2^6= 15/8
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
=1/4+(-1/4)(x-2) +(3/8)/(2!)(x-2)^2 +(-3/4)/(3!)(x-2)^3+(15/8)/(4!)(x-2)^4
=1/4-1/4(x-2) +(3/8)/2(x-2)^2 -(3/4)/6(x-2)^3+(15/8)/24(x-2)^4
=1/4-1/4(x-2) + 3/16(x-2)^2 -1/8(x-2)^3+5/64(x-2)^4
The Taylor polynomial of degree n=4 for the given function f(x)=1/x^2 centered at c=2 will be:
P_4(x)=1/4-1/4(x-2) + 3/16(x-2)^2 -1/8(x-2)^3+5/64(x-2)^4