Estimate $\sqrt{e}$ with the error smaller than $10^{-5}$ 
Estimate $\sqrt{e}$ with the error smaller than $10^{-5}$ 

My idea to solve this is to use the Maclaurin series around $x=0$ to get this: 
$$e^{\frac 1 2} = 1 +  \frac 1 2+ \frac{1}{2! \cdot 4}+ \frac{1}{3! \cdot 8} + \dots + \frac{1}{n! \cdot 2^n} $$ 
The problem is that I don't know when to stop adding things - namely, when my approximation hits the allowed error. Could you help me to estimate the remainder of this function? Is it possible to do this without integrals?
 A: The Maclaurin series comes with the Lagrange remainder, which for $e^x$  is given by $\frac{e^\xi x^{n+1}}{(n+1)!}$ where $\xi$ is between $0$ and $x$. If you can come up with an upper bound bound for $|e^\xi|$ then you can finish your problem.
A: A very interesting approach to the problem is that you can approximate $\sqrt{e}$ with great accuracy using Newton's method on the function $f(x)=x^2-e=0$, which uses the iterative process
$$a_{n+1}=a_n-\begin{bmatrix}\textbf{D}f(a_{n})\end{bmatrix}^{-1}f(a_n)$$
which applied to our $f(x)$ is equal to 
$$a_{n+1}=a_n-\frac{1}{2a_n}(a_n^2-e)=\frac{1}{2}\begin{pmatrix}a_n+\frac{e}{a_n}\end{pmatrix}$$
Choose an $a_0$ that is less than $e$, suppose 2.  Then, we know 
that this sequence is bounded below, implying that it converges, implying that ${a_n}$ will converge to $\sqrt{e}$:
$$a=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{1}{2}\begin{pmatrix}a_n+\frac{e}{a_n}\end{pmatrix}=\frac{1}{2}\begin{pmatrix}a+\frac{e}{a}\end{pmatrix},\;\;\;\;\text{i.e.,}\;\;\;a=\sqrt{e}$$
You can keep iterating this sequence, which will end up being very few iterations, until you get a decimal $1.64872(...error...)$ where the $10^{-5}$ place of the decimal does not change after another iteration.
Newton's method is very neat to approximate the $n$th root of any number with significant accuracy relatively quickly. 
Hope you find this useful.
