Taylor Series?

Question

Given f(x)=x^(1/2) where x is from 1/4 to 1

Find P(x)=ax+b using Taylor expansion at the midpoint of the interval.

Answer 1

Taylor expansion in the point a has the form

f(a) + f'(a) * (x -a) + f''(a) * (x - a)^2 /2 + f'''(a) * (x - a)^3/6 + ...

In your case, a = 5/8, f(a) = sqrt(5/8), f'(a) = 1/[2 sqrt(5/8)], so you get

sqrt(5/8) + (x - 5/8) / (2 sqrt(5/8)) + ...
=
1/2 sqrt(5/8) + x / sqrt(5/8) + ...

So you would find a = 1/sqrt(5/8) = 2/5 sqrt(10) and b = 1/2 sqrt(5/8) = 1/8 sqrt(10).

Answer 2

middle point is (1+1/4)/2=5/8

a=f'(5/8)
b=f(5/8)

f'(x)=1/2x^{-1/2}
f'(5/8)=1/2 (5/8)^{-1/2}

while
f(5/8)=(5/8)^{1/2}