Find the Taylor Series expansion up to degree three of....

Question

sqrt(2+x) about a=2

Answer 1

Since we are interested in a Taylor series expansion up to degree three we are looking for...

P(x) = a₀ + a₁(x-2) + a₂(x-2)² + a₃(x-2)³

a_k (k = 0..3) are the successive derivatives of f(x) evaluated at the center point of a = 2 divided by k!

f(a=2) = 2

f'(x) = 1/2 * 1/sqrt(2+x)

f'(a=2) = 1/2 * 1/2 = 1/4

f''(x) = 1/2 * -1/2 * 1/[sqrt(2+x)]³ = -1/4 * 1/[sqrt(2+x)]³

f''(a=2) = -1/32 = -1/4 * 1/8

f'''(x) = 1/2 * -1/2 * -3/2 * 1/[sqrt(2+x)]⁵ = 3/8 * 1/[sqrt(2+x)]⁵

f'''(a=2) = 3/8 * 1/32

P(x) = f(a=2)/0! + f'(a=2)/1! * (x-2) + f''(a=2)/2! * (x-2)² + f'''(a=2)/3! * (x-2)³

P(x) = 2 + 1/4*(x-2) - 1/32*(x-2)² + 3/256*(x-2)³

If you want to simplify further, you can multiply out all the pieces and rearrange collecting like components for x, x², x³ to get another representation of the polynomial.