• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Computations with the double-exp series mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 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? « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Single-exp series computation code mike3 0 3,096 04/20/2010, 08:59 PM Last Post: mike3

Users browsing this thread: 1 Guest(s)