Speed of the binomial series for calculating $\sqrt{3}$ Calculating $\sqrt{3}$ by the binomial series for $(4-1)^{\frac{1}{2}}$ seems to converge very much more slowly than the Newton Raphson method for improving an initial approximation of  $\sqrt{3}\approx 2$. Is that right?
This is not a sophisticated question. I ask because I'm trying to understand an algebra textbook from 1826 (Bridge The Elements of Algebra).  So I have done some hand calculations with the binomial series and they seem much slower than the obvious Newton Raphson calculations although Bridge says the binomial series is very fast.  So I want to check with people who know this method.  Should I expect the binomial theorem to be slower than Newton Raphson for this problem?
 A: The binomial series, or more general, Taylor's formula in general converges at an exponential rate, here like $(1/4)^n$ (related to singulaties in the complex plane). Newton's method is in general quadratic or super-exponential fast (faster than any exponential) when you are close enough to the 'true' value.
A: Binomial Series
$$
\begin{align}
\sqrt3
&=2\sqrt{1-\frac14}\\
&=2\left(1+\frac{\frac12}{1!}\left(-\frac14\right)+\frac{\frac12\left(-\frac12\right)}{2!}\left(-\frac14\right)^2+\frac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{3!}\left(-\frac14\right)^3+\dots\right)\\
&=-2\sum_{k=0}^\infty\frac{(2k-3)!!}{8^kk!}
\end{align}
$$
where $n!!$ is the Double Factorial. Note that $(-3)!!=-1$ and $(-1)!!=1$.
Each term in the series is about $\frac14$ the size of the previous term.

Newton Raphson
Newton's Method for $\sqrt3$ gives
$$
x_{n+1}=\frac{x_n^2+3}{2x_n}
$$
For $x_n\gt0$, we have $x_{n+1}\ge\sqrt3$.
Thus,
$$
\begin{align}
x_{n+1}-\sqrt3
&=\frac{\left(x_n-\sqrt3\right)^2}{2x_n}\\
&\le\frac{\left(x_n-\sqrt3\right)^2}{2\sqrt3}
\end{align}
$$
Each term in the sequence is proportional to the square of the previous term.
