04/20/2010, 07:32 PM
I was thinking of trying to make a code to compute with the double-exp series mentioned here:
http://math.eretrandre.org/tetrationforu...78#pid4678
wondering if it may be better than the current approach, but I have some problems. To do the Ansus formula iteration, we need several things, done in this order:
1. continuum summation from \( 0 \) to \( z-1 \)
2. exponential function to base \( b \)
3. multiply by \( \log(b)^z \)
4. integration from -1 to \( z \)
5. normalize (divide by the value at 0)
The first, of course, is easy. However it produces a non-exponential linear term. This is fine, we can save it and work it implicitly (same as I do with the periodic approximation in single-exp series). The next one is another challenge. We have linear + another double-exp series. Applying the exponential function should yield exponential * base-b exponential of the double-exp series. This latter should have the same "periodicity" characteristics, but how can it be computed efficiently and to good precision? We could then bring the exponential into the exp-series (just add \( \log(b) \) to the exponents of the \( e^{(...)} \) parts of the terms.).
Then we multiply by \( \log(b)^z \). We could add this \( \log(\log(b)) \) into the exponents of the exp-series. Finally, we can do the integral. This requires mixing the multiplying exponential into the exponentials of the series and then integrating. But also leaves a nonexponential linear term, too. Thus the result here is no longer a series of the form given! We've added \( \log(b) + \log(\log(b)) \) to the exponents and even worse, have a non-exponential linear term! What can be done about this?
http://math.eretrandre.org/tetrationforu...78#pid4678
wondering if it may be better than the current approach, but I have some problems. To do the Ansus formula iteration, we need several things, done in this order:
1. continuum summation from \( 0 \) to \( z-1 \)
2. exponential function to base \( b \)
3. multiply by \( \log(b)^z \)
4. integration from -1 to \( z \)
5. normalize (divide by the value at 0)
The first, of course, is easy. However it produces a non-exponential linear term. This is fine, we can save it and work it implicitly (same as I do with the periodic approximation in single-exp series). The next one is another challenge. We have linear + another double-exp series. Applying the exponential function should yield exponential * base-b exponential of the double-exp series. This latter should have the same "periodicity" characteristics, but how can it be computed efficiently and to good precision? We could then bring the exponential into the exp-series (just add \( \log(b) \) to the exponents of the \( e^{(...)} \) parts of the terms.).
Then we multiply by \( \log(b)^z \). We could add this \( \log(\log(b)) \) into the exponents of the exp-series. Finally, we can do the integral. This requires mixing the multiplying exponential into the exponentials of the series and then integrating. But also leaves a nonexponential linear term, too. Thus the result here is no longer a series of the form given! We've added \( \log(b) + \log(\log(b)) \) to the exponents and even worse, have a non-exponential linear term! What can be done about this?