Karlisbad
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let be "A" a big number, my question is, is there an efficient algorith (with only a few steps) to evaluate e^{A} 
The discussion revolves around finding efficient algorithms to evaluate the exponential function e^A, particularly for large values of A. Participants explore various methods, including approximations and computational techniques, without reaching a consensus on the best approach.
Participants express various viewpoints on the efficiency and methods for computing e^A, with no clear consensus on a single best approach. Disagreements exist regarding the applicability of certain techniques and the trade-offs involved in accuracy and computational time.
Participants highlight limitations related to accuracy and the size of A, as well as the dependence on computational methods and definitions of efficiency. The discussion does not resolve these issues.
This discussion may be of interest to those involved in computational mathematics, algorithm design, or programming, particularly in contexts where efficient evaluation of exponential functions is required.
0rthodontist said:I'm certainly no expert on this, but you could get an order-of-magnitude approximation by taking the logarithm. Also, you could reduce the number of multiplications used in an exact answer to only log A by using the same technique used in RSA fast exponentiation.
No, not that technique, the one that splits up the exponent. It would take me some time to remember it but the basic idea is that if you have an exponent like 2n, it computes x^n and then squares it.NateTG said:The RSA thing only works for modular arithmetic.
0rthodontist said:No, not that technique, the one that splits up the exponent. It would take me some time to remember it but the basic idea is that if you have an exponent like 2n, it computes x^n and then squares it.
The goal, of course, is to get that "few minutes" down to "few milliseconds".cells said:* if your accuracy is only to a few sig figures , perhaps upto 10sig fig, then you can write a computer program to give you e^x where x is any number, the program if written well will not take very long to run, probably in the order of a few min max
If your exponent is a C integer, then there's a nice optimization you can do; you can simply precompute0rthodontist said:Yes, that's it. It's called "square and multiply" exponentiation.