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Thu Jun 19, 2014, 07:40 AM

ERB: Nye vs. Newton



Given that integral(sec(x), dx) = ln( abs( sec(x) + tan(x) ) ) + C


One finds

integral(sec(x), dx, {0, Pi/6}) = ln( abs( sec(Pi/6) + tan(Pi/6) ) ) - ln( abs( sec(0) + tan(0) ) )

sec(0) = 1
tan(0) = 0

integral(sec(x), dx, {0, Pi/6}) = ln( abs( sec(Pi/6) + tan(Pi/6) ) ) - ln( abs( 1 + 0 ) )

abs( 1 + 0 ) = 1

integral(sec(x), dx, {0, Pi/6}) = ln( abs( sec(Pi/6) + tan(Pi/6) ) ) - ln( 1 )

ln( 1 ) = 0

integral(sec(x), dx, {0, Pi/6}) = ln( abs( sec(Pi/6) + tan(Pi/6) ) ) - 0

sec(Pi/6) = 1/( cos(Pi/6) ) = 2 /((3)^(1/2))
tan(Pi/6) = 1/((3)^(1/2))

integral(sec(x), dx, {0, Pi/6}) = ln( abs( sec(Pi/6) + tan(Pi/6) ) ) = ln( abs( (2/((3)^(1/2))) + (1/((3)^(1/2))) ) )

integral(sec(x), dx, {0, Pi/6}) = ln ( ((3)^(1/2)) ) = ln( (3)^(1/2) ) * 1


Since ln( (3)^(1/2) ) * 1 = ln( (3)^(1/2) ) * ((k)^64) implies 1 = k^64 or k^64 = 1, one must find the sixty-fourth roots of unity.


The sixty-fourth roots of unity are the elements of the set {e^((2*Pi*i*n)/64)} with n being an element of the set {0, 1, 2, 3, ..., 63}.

Euler's formula states that e^(i*m) = cos(m) + i*sin(m). If n = 16, (2*Pi*i*16)/64 = i*Pi/2 and m = Pi/2.

cos(Pi/2) = 0 and sin(Pi/2) = 1, so e^((2*Pi*i*16)/64) = 0 + i*1 = i.

Thus, the solution in the video corresponds to the 17th of the sixty-fourth roots of unity -i.e., the one that has the value n = 16.


Would anyone care to analyze the prism's refraction of light (i.e., the displayed order of colors) relative to the orientation of the prism? It is difficult to tell, but the intended refracted beam looks a lot like it is reflecting off one of the lateral faces or lateral edges of the prism.

This is a really neat video and a cool rap.









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