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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.. )

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I think most calculators would calc $a^b = 2^{b\log_2 a}$, since they have fast and reliable procedures for calculating $2^{.}$ and $\log_2$. – Myself Jan 22 '11 at 12:15
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On a sidenote, if you're working with really huge numbers and you're still using the standard multiplication, you might want to check out this site: en.wikipedia.org/wiki/Multiplication_algorithm E.g. check the section on karatsuba – Myself Jan 22 '11 at 12:27
Regarding the a^b=2^blog2(a) method, upto what precision will you express log2(a) [assuming I want a^b upto m decimals]?? because this is again an irrational term in the power... – Guanidene Jan 22 '11 at 18:20
This may help you approximate the error term for $\epsilon$ small: $2^{a+\epsilon} = 2^\epsilon \cdot 2^a \approx (1 + \epsilon\ln 2/2) \cdot 2^a \approx 2^a + 0.69 \epsilon 2^a$. Actually I think physicists are good at this kind of stuff, because it can be treated as some sort of measurement uncertainty. Maybe you'll find some good reference in that direction. – Myself Jan 22 '11 at 20:16
@Guanidene, Some irrational numbers are not computable. So you cannot compute this in general. If $a$ and $b$ are computable (whether irrational or not) you can compute $a^b$. – quanta May 12 '11 at 21:11
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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.

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Please elaborate your point – Guanidene May 19 '11 at 17:13
For example, one way to calculate the digits of pi is to choose a series that converges to pi with a know error term. By examining the error term, you can tell how many digits won't change adding additional terms. I had the thought that if you had series that converge to your irrationals from above and from below, you could do the same thing with your expression, but it would be complicated, and different for any two irrational numbers. A practical question is how do you know what your irrational numbers are? – phv3773 May 23 '11 at 20:33
That is what my question is .. when you don't know exactly what your (irrational) numbers are, how to find irr^irr. – Guanidene May 25 '11 at 6:48
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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 l1, l2, and upper approximations u1, u2 to irrationals 1 and 2, we have [l1, u1]*[l2, u2]=[min{l1*l2, l1*u2, u1*l2, u1*u2}, max{l1*l2, l1*u2, u1*l2, u1*u2}] where "*" on the left of the "=" denotes interval number multiplication, and * on the right of the "=" denotes real-number multiplication. We have max{l1*l2, l1*u2, u1*l2, u1*u2} - min{l1*l2, l1*u2, u1*l2, u1*u2} less than or equal to some nonnegative n. Now 1 decimal place corresponds to n ending up less than or equal to .1, 2 decimal places to n ending up less than or equal to .01, and so on. So, for k decimal place precision we just need to find l1, l2, u1, u2 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.

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Needs more LaTeX. (I tried editing it, but I don't really know how it works.) – TRiG Aug 20 '12 at 21:30

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