Irrational$^\text{Irrational}$

Question

How do I compute $\text{(irrational)}^{\text{(irrational)}}$ up to a required number of decimals say m, in the fastest way ? (one way is of course compute both the irrational numbers to a precision much larger than m and then solve it... but you never know how much excess of m you will need to calculate the irrationals.. )

Answer 1

As to the number of decimals m, I think that it should be possible to create an interval between upper and lower approximations using rational numbers. You can then examine how the interval between upper and lower decreases with more digits in the rational approximations.

Answer 2

This might work something like phv3773's idea, this probably works far too slowly for what you want, but maybe it's a decent idea. First as an example, consider calculating the product of two irrationals. We might use lower and upper approximations here. If we use lower approximations $l_1, l_2$, and upper approximations $u_1, u_2$ to irrationals $1$ and $2$, we have $$[l_1, u_1]*[l_2, u_2]=[\min\{l_1l_2, l_1u_2, u_1l_2, u_1u_2\},\max\{l_1l_2, l_1u_2, u_1l_2, u_1u_2\}]$$ where $*$ denotes interval number multiplication. We have $$\max\{l_1l_2, l_1u_2, u_1l_2, u_1u_2\} - \min\{l_1l_2, l_1u_2, u_1l_2, u_1u_2\}\le n$$ for some nonnegative $n$. Now $1$ decimal place corresponds to $n\le.1$, $2$ decimal places to $n\le.01$, and so on. So, for $k$ decimal place precision we just need to find $l_1, l_2, u_1, u_2$ such that $n$ is less than or equal to $10^{-k}$. You don't need to know what the irrationals are exactly, you just need to know that the irrationals lie between some rational numbers.

Now, interval exponentiation, like other interval arithmetic operations, if we can do such (if it "exists"), will yield an interval of numbers $[a, b]$. So, for $k$ decimal place precision we just need $n=b-a$ less than or equal to $10^{-k}$.

Unfortunately, I don't know how to do interval exponentiation. So, maybe it comes as a better idea to figure out how it works on the reals before trying this approach.