Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor expansion given by :
$$T^n_{0} = 9 – \frac{1}{2}(81)^{-1/2}(x) + \frac{1}{4}(81)^{-3/2}\frac{x^2}{2}-\frac{3}{8}(81)^{-5/2}\frac{x^3}{6}\ + \dotsm $$
By the Lagrange remainder, $\exists$ for each $n \in \mathbb{N}$, $c_n \in [0,1]$ such that :
$$R^n(1) = f^{n+1}(c) \frac{1-0^{n+1}}{(n+1)!} = f^{n+1}(c) \frac{1}{(n+1)!} \leq 9.\frac{1}{(n+1)!} \lt 10^{-3}$$
$=> n(+1)! \gt \frac{9}{10^{-3}} = 9000$
So we can take $n = 8$
The approximation seems a little bit tricky to calculate especially without a calculator. I'm wondering if everything above is correct ?
Best Answer
Use $\sqrt{80}=9\sqrt{1-\frac{1}{81}}$. Now the Taylor series converges much faster: you only need $$\sqrt{1-x}=1-\frac12x+O(x^2)$$
We get $$9\left(1-\frac12\cdot\frac{1}{81}\right)=9-\frac{1}{18}=8.94444\ldots$$ with an error of $0.00017\ldots$