Estimate $\sqrt{10}$ with an error smaller than $0.01$ I´m trying to find this using Lagrange Error bound.
$$|R|\leq \frac{f^{n+1}(t) (x-a)^{n+1}}{(n+1)!}$$
Basically I thought of using the function $f(x) = \sqrt{x+1}$ centered on $a= 8$ (but when I was calculating the derivatives I couldn't find a "pattern")
So, I tried using $f(x) = \sqrt{x+1}$, $a= 0$ but in Langrage Error bound I need to work with the $n+1$ derivate and I couldn't find a pattern or how to work with it (I know that $(x-a)^{n+1}$ in this case is equal to $9^{n+1}$ and I have to try with differents "$n$" to know what order I need to work with). I'd appreciate some help to know how to deal with $f^{n+1}(t)$ (with t between $0$ and $9$ in this case). (I tried to find a general Taylor Polynomial but it has double factorial $!!$ and I get confused a little bit)
 A: Note that $$\sqrt {10}=\sqrt{9+1}=\sqrt{9\left(1+\frac{1}{9}\right)}=3\sqrt{1+\frac{1}{9}}$$
Without using the binomial coefficient, you can express $\sqrt{1+x}$ as follows:
$$\sqrt{1+x}=1+\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\frac{(2k-3)!!}{2^k}x^k $$
A: It is not entirely clear what you are looking for.
If you apply Newton's method to $f(x) = x^2-10$ and start from some $x_0 > \sqrt{10}$, say $x_0 = 4$, then the sequence of iterates $x_{n+1} = {1 \over 2} (x_n+{10 \over x_n})$ decreases to the positive solution $x_n \downarrow \sqrt{10}$. The sequence converges quadratically since $f'(\sqrt{10}) \neq 0$.
Here is a crude formula to bound the error in terms of $f$. Note that the iterates satisfy $x_n \in [3,4]$,
Note that for $x \in [3,4]$, we have $f(x) = (x-\sqrt{10})(x+\sqrt{10}) \ge 6(x-\sqrt{10}) $ and so
$x-\sqrt{10} \le {1 \over 6} (x^2-10)$.
Starting with $x_0=4$, we get $x_1-\sqrt{10}\le 0.9375$, $x_2-\sqrt{10} \le 0.0012452...$, with $x_2 =3.163461...$.
(Not that it matters, but $x_3-\sqrt{10} \le 0.00000023351...$.)
A: The Taylor formula of order $n$ at point $a$, with Lagrange remainder, for the function  $f(x) = x^N$, ($N$ not necessarily a natural number) is
$$x^N = \sum_{\ell=0}^n \binom{N}{\ell} a^{N-\ell} (x-a)^{\ell} + \binom{N}{n+1} \xi^{N-n-1} (x-a)^{n+1}$$
where $\xi \in [a, x]$. So it looks  like the binomial formula ( we have $\binom{N}{\ell} = \frac{N(N-1) \cdots (N-\ell+1)}{\ell!}$, and the remainder term would look like the next one, except that instead of $x$ we have $\xi$ between $x$ and $x+a$).
Now consider $N=\frac{1}{2}$, $x=10$, $a=9$, $x-a=1$, so we estimate $(10)^{\frac{1}{2}}$. An approximation ( in excess)  with error $< |\binom{\frac{1}{2}}{2} 9^{\frac{1}{2}-2}| < 0.01$ is
$$\sqrt{10} \simeq 9^{\frac{1}{2}} + \frac{1}{2} 9^{\frac{1}{2}-1}\cdot 1 =3+\frac{1}{6}$$
