How can we get Nth digit of any irrational number in base 2 using Taylor expansion?

Question

For example:

$$\sqrt e = e^{0.5} = \sum^\infty_{n=0}\frac{(0.5^n)}{n!}$$

I know, we can easily get nth term of the series, but how can we get nth binary digit of that irrational number namely infinite sum of that infinite terms?

Or, are there more efficient approaches to get them?

For example, I want to get bits whose locations(orders) are in interval [100000-104096)

namely: 100000th bit, 104095th bit and every bits between them in order.

Answer 1

There is not a general method to compute any irrational number, in fact there is provably no method to compute "most" irrational numbers (most in terms of measure). Also, for a lot of constants there is no known method to directly convert its Taylor series to the terms of its binary or decimal expansion. For most numbers, the only approach is likely to compute the number to a desired accuracy and then round its decimals.

For example, with $\pi=4\arctan 1$, we can express it with the Taylor series, $\arctan x=\sum\limits_{i=0}^\infty\frac{\left(-1\right)^{i}x^{2i+1}}{{2i+1}}$, but so far, there is no known way to convert this to a "spigot algorithm" that returns a particular decimal of $\pi$. Such a result would be publishable but interestingly, the BBP formula is a spigot algorithm for base-16.