Difference between revisions of "Talk:Euler's Formula"

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::Mathematically, you can derive this formula by doing standard mathematics (i.e., calculating series) in the field '''R'''[x]/[x²+1]. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 18:08, 16 August 2015 (EDT)
 
::Mathematically, you can derive this formula by doing standard mathematics (i.e., calculating series) in the field '''R'''[x]/[x²+1]. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 18:08, 16 August 2015 (EDT)
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Test:
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*x: <math>x</math>
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*x^2: <math>x^2</math>
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*x^2_1: <math>x^2_1</math>
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*\frac{x}{y}: <math>\frac{x}{y}</math>
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*\hat{x}: <math>\hat{x}</math>

Revision as of 15:43, August 17, 2015

I would disagree that this is Euler's most famous formula. I would plump for either e^(i*pi) = -1 or the F+V -E = 2 for convex polyhedra. Google search brings up these two ... --CatWatcher 17:16, 7 April 2007 (EDT)

e^(i*pi) = -1 is derived from euler's formula by set .Jaques 17:21, 7 April 2007 (EDT)
granted, but I have heard it claimed that this formula is the "e=Mc^2" od mathematics, in that it ties together four of the most infamous numbers in the whole of maths. -1, as for centuries, it was thought that numbers could not be negative, pi, which was always thought to be fractional (hence squaring the circle), i, which was thought to be impossible, but now is termed imaginary, and e, the base of natural logarthims, which underpins calculus.
I would certainly say this was Euler's masterpiece, even though I am a mainly a Graph Theorist, and the other 'Euler's Formula' is the one I am most familiar with.--CatWatcher 18:39, 7 April 2007 (EDT)

Arbitrary definition for using the imaginary square root of -1

Andy, what do you think is "arbitrary" about the square root of -1? Extending a field by the root of an equation is a well understood procedure (and i or -i will lead to the same result....) --AugustO (talk) 14:43, 16 August 2015 (EDT)

The square root of negative one is imaginary; Euler's formula is simply a definition rather than a mathematically derived theorem. It's interesting that the comments above compared this with E=mc2, because in both cases it is simply a redefinition rather than a derivation.--Andy Schlafly (talk) 17:57, 16 August 2015 (EDT)
Mathematically, you can derive this formula by doing standard mathematics (i.e., calculating series) in the field R[x]/[x²+1]. --AugustO (talk) 18:08, 16 August 2015 (EDT)

Test:

  • x:
  • x^2:
  • x^2_1:
  • \frac{x}{y}:
  • \hat{x}: