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struggle4progress

(118,234 posts)
9. The relation it comes from
Thu Jan 31, 2013, 06:50 PM
Jan 2013

e^(i*x) = cos x + i sin x

is even better, since it provides a way to compute many trig identities, such as

sin^2 x + cos^2 x = (cos x + i sin x)(cos x - i sin x) = e^(i*x) * e^(-i*x) = 1

or

cos 2x = (e^(2i*x) + e^(-2i*x))/2 = (e^(i*x)^2 + e^(-i*x)^2)/2 = <-2 + (e^(i*x) + e^(-i*x))^2>/2
= -1 + 2cos^2 x = -sin^2 x - cos^2 x + 2 cos^2 x = cos^2 x - sin^2 x

having studied mathematics and statistics for a number of years, I appreciate elegance in DrDan Jan 2013 #1
I find Eulers identity wonderful intaglio Jan 2013 #2
It's even more beautiful when expressed using tau pokerfan Jan 2013 #3
Beauty is in the eye of the beholder, I suppose. I prefer the equation with pi because . . . Geoff R. Casavant Jan 2013 #4
OK, how about this pokerfan Jan 2013 #5
Thanks tama Jan 2013 #8
+1 tama Jan 2013 #7
The relation it comes from struggle4progress Jan 2013 #9
Yeah, we EE's get that beaten into our skulls at an early age pokerfan Feb 2013 #16
My favorites pokerfan Jan 2013 #6
Viewed from a sufficiently advanced PoV the Liebniz formula is by itself a proof struggle4progress Jan 2013 #10
The last "surprising one" is (of course) seen false, if computed well enough struggle4progress Jan 2013 #11
spoilsport pokerfan Jan 2013 #12
oops struggle4progress Jan 2013 #13
This is a fun read... pokerfan Jan 2013 #14
thanks! that is fun! struggle4progress Jan 2013 #15
aka Ramanujan's constant ... to see why ... eppur_se_muova Feb 2013 #17
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