Block, SkipCaptcha, Automoderated users, edit, rollback
5,023
edits
Changes
Jump to navigation
Jump to search
Line 20:
Line 20: + + + + + + + + + + + + + +
→Arbitrary definition for using the imaginary square root of -1: @Aschlafly
:I have to think about that further. Raising <math>i</math> to an exponential power seems to require an additional assumption.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 14:36, 17 August 2015 (EDT)
:I have to think about that further. Raising <math>i</math> to an exponential power seems to require an additional assumption.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 14:36, 17 August 2015 (EDT)
::Just plug it in and see what happens - I suppose, that is what Euler has done :-)
::*<math>i^2 =-1, i^3= -i, i^4 = 1, etc.</math>
::*<math>\cos(x) := \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} </math>
::*<math>\sin(x) := \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} </math>
::*<math>\exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}</math>
::Then <math>\exp(i \pi) := \sum_{n=0}^\infty \frac{(i \pi)^n}{n!}</math>
::<math>=\frac{i^0 \pi^0}{1} + \frac{i^1 \pi^1}{1} + \frac{i^2 \pi^2}{2} + \frac{i^3 \pi^3}{6} + \frac{i^4 \pi^4}{24} + \ldots</math>
::<math>=1 + i \pi - \pi^2/2 - i \pi^3/6 + \pi^4/ 24 + \dots </math>
::<math>= \sum_{n=0}^{\infty} (-1)^n \frac{\pi^{2n}}{(2n)!} + i \sum_{n=0}^{\infty} (-1)^n \frac{\pi^{2n+1}}{(2n+1)!}</math>
::<math>= \sin(\pi) + i \cos(\pi)</math>
::<math>= -1 + i0 = -1</math>
::Granted, it took two centuries to get all the details right (convergence, resorting of infinite sums, etc.), but that's the gist. Straightforward and beautiful. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 16:04, 17 August 2015 (EDT)