For which models was 2^3>8?
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04-04-2017, 10:26 PM
Post: #15
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RE: For which models was 2^3>8?
(04-04-2017 05:51 PM)Dieter Wrote: The explanation is simple and straightforward: yx is calculated as ex · ln y. How about all the way? It's really quite efficient if you use exponentiation by squaring: x^n = (x^(n/2))^2 if n is even, and x*(x^((n-1)/2))^2 if n is odd; apply recursively or iteratively. The algorithm is O(log(n)). This is how Free42 gets 2^3=8 exactly, despite not using extended precision for intermediate results. |
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