Recent content by ADDA

Technically, again I am wrong. I looked back in the book. A decimation in time fast Fourier transform starts with many radix-2 transforms then many radix-4 transforms, then many radix-8, continually until frame size / 2 is reached. So I'm attempting a base 3 or base 9 or base 5 transform, I've...

"To the extent that a fool knows his foolishness, He may be deemed wise. A fool who considers himself wise is indeed a fool." Thanks again Mark44, modens ponens. One other question, say I perform a radix 9 transform, in the transform data, would a peak magnitude index represent 1.0 / 9.0 of a...

Thanks Mark44, I've been experimenting with trying to change the radix of a fast Fourier transform. My solution for x is this: Since squaring both sides could equal: eix2 = eix2 equating the exponents of e would yield ix2 = i * x * 2; -x2 = 2 * i * x; so x = -2 * i. Now if I wanted to do...

Take item (3) and square both sides: e2ix = -1 * (-sin(x) + icos(x))2 expanding the right side: sin2(x) - i2sin(x)cos(x) - cos2(x) apply the double angle identity - isin(2x) - (cos2(x) - sin2(x)) apply the double angle identity -cos(2x) - isin(2x) Did I miss the sign above?

https://en.wikipedia.org/wiki/Euler's_formula (1) eix = cos(x) + isin(x) (2) eixidx = (-sin(x) + icos(x))dx (3) eix = (-sin(x) + icos(x)) / i (4) eix = cos(x) + isin(x) Just lost in circles. Just for fun.. post a solution for x.

newjersyrunner, thanks for the reply, seems like you are as lost as I am as to how to accomplish this.

Yeah, I guess I need to explain my thoughts better. The background was the inner product of the the camera's direction with the computed normal of the surface. The normal of the surface is computed by the cross product of the partial derivative of with respect to u and the partial derivative...

Would there be a way to take a uv map and compute the 2 dimensional position of a viewing plane or eye or camera, without the need to expand the parametric equation into three dimensions? In the above video the uv map in the background yellow is calculated first. Then only the bright portions...

A more efficient and simpler to comprehend is Expectation Maximization: public class EM { public class CONSTANTS { public static final int NFRAMES_COUNT = 3; public static final int NFRAMES_SETS = 2; public static final float THRESH = 1.0f/ 6.0f; }...

cleaner example: Started to get copyright issues on youtube, so I decided to delete everything. I must be reproducing the signal correctly!

Or find a woman to court and spark:

And Rock and Roll:

At times it is best to simplify and get back to the basics:

Thanks, I was sort of misunderstanding you, perhaps: Were you laughing at the video, or my inability to properly do this?

Another example: