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Jan
21
awarded  Supporter
Jan
21
accepted Is this Perlin Noise?
Jan
21
asked Is this Perlin Noise?
Oct
2
awarded  Scholar
Oct
2
accepted Recurrence relation for a function with an integral of the function?
Sep
20
comment Recurrence relation for a function with an integral of the function?
this looks like the right direction I think the miraculous cancellations are a feature, not a bug haha
Sep
20
comment Recurrence relation for a function with an integral of the function?
well, the more I looked at the "simplified" example I gave the more I realized that it just wouldn't do. The itnegral from 0 to 0 would be 0 and then g(1) would = 0 instead of the given starting condition of 1. The answer is most definitely not constant. I suspect it is a differential equation of the form above.
Sep
20
comment Recurrence relation for a function with an integral of the function?
where a, b and c are all functions of f
Sep
20
comment Recurrence relation for a function with an integral of the function?
A friend suggested that the correct approach is differentiating the function thrice. However, I gave that an elementary stab using the Leibniz rule definitions I found online and I couldn't get rid of the integral itself. The solution should most likely be a differential equation in the form g'''(f) = ag''(f) + bg'(f) + c*g(f), which I should hopefully then be able to solve into g(f)
Sep
20
awarded  Student
Sep
20
revised Recurrence relation for a function with an integral of the function?
added 2 characters in body
Sep
20
comment Recurrence relation for a function with an integral of the function?
the fact that there is an initial condition suggested this is the approach, but I just cannot make it work
Sep
20
awarded  Editor
Sep
20
revised Recurrence relation for a function with an integral of the function?
added 134 characters in body
Sep
20
comment Recurrence relation for a function with an integral of the function?
that is perfect. thank you!
Sep
20
asked Recurrence relation for a function with an integral of the function?
Sep
19
comment Continuous distributions on a vanishing length with an off-limits region
I apologize for any confusion in my terminology. I did mean uniformly, and the fraction is always on the right hand end. You have the same process as I, but my expected remaining lengths take it as a given that the slice actually happened, which why the probability that a slice what taken off in the previous round is actually 100%. So <pre>1-1/2*3/4=0.625</pre>
Sep
19
asked Continuous distributions on a vanishing length with an off-limits region
Sep
19
awarded  Autobiographer