I need to compute the addition of four arctangents:
\alpha = \arctan(x_1) + \arctan(x_2) + \arctan(x_3) + \arctan(x_4)
Rather than call four arctangent functions, I was thinking of using the arctangent addition formula:
\arctan(u) + \arctan(v) = \arctan(\frac{u + v}{1 - uv}) + \pi n...
Hello chiro,
Yes, this is for a computer graphics application. I'm tracing a ray through an opaque volume and integrating the lighting effects along the way. I march along the ray in fixed size steps so this boils down to a very long summation where each term in that summation may be quite...
Hello mathman, I've been thinking about your reply but it seems to me that randomly choosing samples from the discrete set without replacement is going to skew the statistics.
Let's say I have a discrete random variable X with probability density f_X(x) that exists in a discrete set χ with N...
Hello all,
I'm aware of the Monte Carlo Summation method in discrete spaces, where you can approximate a very long summation over the entire space by a shorter one with only a few randomly selected terms from the original summation (weighted by the inverse probability density of them being...
Thank you both - I understand it now. I didn't imagine the solution was so simple as to just multiply the inverse square with the absorption (after application of the exponential to the latter).
Hi Sophie, you mean that there is an effective absorption coefficient given by σ(t)/t^2 ? I don't see how that would work because in an optically transparent medium (σ = 0), the inverse square term would vanish together with the σ, which we know is not true.
manuel
Hello all,
I already know that the radiant intensity of a point light source falls off with the inverse square of the distance to the source. This, however, only happens in a vacuum. My question is, what is the more general law for a point source inside an opaque medium with a known...