Approximate $\sqrt{17}$ with an error smaller than $10^{-4}$ using the series expansion of $\sqrt{1+x}$ with initial point $x_0 = 15$.
I am able to to the following:
$$\sqrt{17} = \sqrt{16+1} = \sqrt{16}\sqrt{1+1/16} = \sqrt{16}\sqrt{1+x} = \sqrt{16}\sum_{n=0}^\inf {1/2 \choose n}x^n \quad x = 1/16$$
Also:
$$\sqrt{17} = (1/17)^{-1/2} = (1 - 16/17)^{-1/2} = (1 + x)^{-1/2} = \sum_{n=0}^\inf {-1/2 \choose n}x^n \quad x = -16/17$$
Both are correct since $|x| < 1$.
But to answer the question I think I need to derive $\sum_{n=0}^\inf f(n,x) (x-15)^n$ because the initial point is $x_0 = 15$.
Taylor's Theorem goes $$ f(x)=f(a)+(x-a)f'(a)+{(x-a)^2\over2}f''(a)+\dotsb $$ Now let $f(x)=\sqrt{1+x}$, $x=16$, $a=15$.