close
  • Showing 74 of 88
  • Reviews of the site
  • Join StumbleUpon or login to add a review! default avatar
  • Reviewed by k2laz on Sep 09, 3:17am

    For those who are not familiar with math it seems like magic and the magician refuses to show his trick, but the relationship comes from the definition of exp(x+iy)= exp(x)[sin(y)+cos(i*y)], letting x=0 means that exp(0)=1 and where y=pi. However if you plot the original diagram for various value of y it create a semi-circle above the x axis and where y is the distance along the semi-circle starting at (1,0) you will see that when y=pi the curve will have reached (-1,0). The engineer can use the original equation in a Taylor expansion to see this relationship. What is also interesting is that these 3 little numbers are also involved in the statistical normal distribution.
  • Rated by 33Arsenic on Jul 29, 4:15pm

    @People complaining about the "proof is easy." Yes the proof is easy to understand (but I'd love to see you derive it yourself and yes that means you'll have to derive Euler's Formula), but that is not the subject of the quote. The subject is the meaning of such an identity.The consequence of this identity (and more formally Euler's Formula) is that exponential and trigonometric functions are related. Now why that is is another question entirely.
  • Rated by mellocello2003 on Jun 09, 5:44am

    Yes it's cool looking but when you think about it (if you know anything about math), it's really obvious.
  • Reviewed by S0meoneElse on Jun 01, 11:49am

    Grrr... i sqrt(-1) i sqrt(-1) Roots of negative numbers are equal to sets, not single numbers. (-1)^(1/2) = sqrt(-1) = {i, -i} If you forget about that you may get strange contradictions like: (-1) = i * i = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1 Because of that, one shouldn't assume that i = sqrt(-1). The proper definition is i^2 = -1. Not all numbers can be squared as freely as positive real numbers.
  • Reviewed by civver on May 11 2009, 7:33pm

    Unifies almost all of the significant constants in mathematics in one identity.
  • Rated by Kirro on Apr 04 2009, 6:42pm

    This is just pathetic. The proof is easy. To completely understand the proof, you'll probably need to have taken Calculus I and possibly Calculus II. But without that, you can still use the identity e^(ix) = cos(x) + i sin(x). So e^(pi*i) = cos(pi) + i sin(pi) = -1 + i*0 = -1. The longer explanation requires that you prove the initial identity, but that's beyond high school math, and you see it explained here: http://www.math.toronto.edu/mathnet/questionCorner/epii.html
  • Rated by mikegreaves1 on Jan 26 2009, 9:51pm

    I don't have any clue regarding this but seems very pure
  • Rated by themoertel on Jan 25 2009, 12:01am

    Balls. Just...balls.