How can you find the taylor expansion of sqrt (x) x about x=1?

1 Answer
Jul 12, 2016

sqrt(x) = sum_(n=0)^infty (x-1)^n /(n!)x=n=0(x1)nn!

Explanation:

For any function f(x)f(x) the taylor expansion of that function about a variable aa will be:
f(x) = f(a) + f'(a)(x-a) +f''(a) (x-a)^2 /(2!) +...
f(x) = sum_(n=0)^infty f^n(a) (x-a)^n / (n!)

The function sqrt(x) presents us with an easily exploitable property. All of its derivatives, evaluated at x = 1 are also 1:
f(x) = sqrt(x) = x^(1/2)
f'(x) = x^(-1/2)
f''(x) = x^(-3/2)
f'''(x) = x^(-5/2)

It should be immediately obvious that 1^n = 1 for any n.

Thus, we can replace all f^n(a) terms with 1. From that point forward, writing the solution is easy:
f(x) = 1 + (x-1) + (x-1)^2 / (2!) +...
f(x) = sum_(n=0)^infty (x-1)^n /(n!)