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

Jul 12, 2016

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

#### Explanation:

For any function $f \left(x\right)$ the taylor expansion of that function about a variable $a$ 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:
$f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$
$f ' \left(x\right) = {x}^{- \frac{1}{2}}$
$f ' ' \left(x\right) = {x}^{- \frac{3}{2}}$
$f ' ' ' \left(x\right) = {x}^{- \frac{5}{2}}$

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

Thus, we can replace all ${f}^{n} \left(a\right)$ 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!)