On Friday 28 April 2006 23:28, Emeric Deutsch wrote:
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Hardly the only definition. You could, e.g., work along these lines: - log x = integral from 1 to x of 1/t dt [x real, >0] - exp x = inverse function of log - find the Taylor series for exp - plug in complex numbers as well as real ones - behold, exp (i pi) + 1 = 0. Or these: - exp is defined to be anything (C -> C) satisfying exp 0 = 1 and exp' = exp - this turns out to be unique - and to satisfy exp (i pi) + 1 = 0. What's really startling isn't so much any single result as the fact that many different approaches all lead to the *same* results. It's not surprising that exp(x+y) = exp(x) exp(y) if you start by defining e and then define exp x as e^x; it's very surprising if you start from the Taylor series. It's not surprising that exp (i pi) = -1 if you start from the definition Emeric wrote above; it's very surprising if you start from the differential equation. It's not surprising that exp' = exp if you start from the Taylor series; it's very surprising if you start with e and define exp(x) = e^x. The real surprise is that all these things fit together so neatly. Which, more generally, is one reason why I find this sort of favourite-theorem question very difficult. Sure, Goedel's theorem is lovely, but an order of magnitude lovelier is the whole constellation of ideas that surrounds it: incompleteness, undecidability, equivalence of formal systems to computing systems, diagonal arguments, ... . Likewise for most of the plausible candidates for "best theorem"; they're gems, but they only really sparkle properly when their brilliance is reflected and re^n-reflected off all the other gems located nearby in theorem-space. So which one do you pick to receive the credit? -- g