Re: [math-fun] Diffie-Hellman vs. exponentiation
What a nice sequence of operations! It would be great if a natural continuous interpolation of operations could be found. --Dan Mike Stay wrote: << The Diffie-Hellman operation a # b = exp(ln(a) ln(b)). This is clearly commutative, unlike exponentiation, but it still distributes over multiplication. This can be extended to an infinite series of operations in both directions: . . . a $ b = ln(exp(a) @ exp(b)) a @ b = ln(exp(a) + exp(b)) = exp(ln(a) $ ln(b)) a + b = ln(exp(a) * exp(b)) = exp(ln(a) @ ln(b)) a * b = ln(exp(a) # exp(b)) = exp(ln(a) + ln(b)) a # b = exp(ln(a) * ln(b)) . . . The operation @ is roughly max (Maslov dequantization can deform it into max). Each operation distributes over the operation above it. What is this sequence called?
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Dan Asimov