Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 27

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'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...
To evaluate the given function f(x) =sqrt(x) , we may express it in terms of fractional exponent. The function becomes:
f(x) = (x)^(1/2) .
Apply the definition of the Taylor series by listing the f^n(x) up to n=3.
We determine each derivative using Power Rule for differentiation: d/(dx) x^n = n*x^(n-1) .
f(x) = (x)^(1/2)
f'(x) = 1/2 * x^(1/2-1)
= 1/2x^(-1/2) or1/(2x^(1/2) )
f^2(x) = d/(dx) (1/2x^(-1/2))
= 1/2 * d/(dx) (x^(-1/2))
= 1/2*(-1/2x^(-1/2-1))
= -1/4 x^(-3/2) or -1/(4x^(3/2))
f^3(x) = d/(dx) (-1/4x^(-3/2))
= -1/4 *d/(dx) (x^(-3/2))
= -1/4*(-3/2x^(-3/2-1))
= 3/8 x^(-5/2) or 3/(8x^(5/2))
Plug-in x=4 , we get:
f(x) = (4)^(1/2)
= 2
f'(4)=1/(2*4^(1/2))
=1/(2*2)
=1/4
f^2(4)=-1/(4*2^(3/2))
= -1/(4*8)
= -1/32
f^3(4)=3/(8*4^(5/2))
= 3/(8*32)
= 3/256
Applying the formula for Taylor series centered at c=4 , we get:
sum_(n=0)^3 (f^n(4))/(n!)(x-4)^n
=f(4) + f'(4) (x-4)+ (f'(4))/(2!) (x-4)^2+ (f'(4))/(3!) (x-4)^3
=2+ (1/4) (x-4)+ (-1/32)/(2!) (x-4)^2+ (3/256)/(3!) (x-4)^3
=2+ (1/4) (x-4)+ (-1/32)/(2!) (x-4)^2+ (3/256)/(3!) (x-4)^3
=2+ 1/4 (x-4)-1/(32*2) (x-4)^2+ 3/(256*6) (x-4)^3
=2+ 1/4 (x-4)-1/64 (x-4)^2+ 3/1536 (x-4)^3
=2+ 1/4 (x-4)-1/64 (x-4)^2+ 1/512 (x-4)^3
The Taylor polynomial of degree n=3 for the given function f(x)=sqrt(x) centered at c=4 will be:
P(x) =2+ 1/4 (x-4)-1/64 (x-4)^2+ 1/512 (x-4)^3