the first argument of B, must be positive
the second argument of B, must be positive
the fraction of integration completion from above, 0 $(LE) x $(LE) 1
It returns I$(SUP C)x(a,b), an element of [0,1].
assert(betaIncompleteCompl(.1, .2, 0) == betaIncomplete(.2, .1, 1));
Regularized incomplete beta function complement I$(SUP C)x(a,b)
Mathematically, if a > 0, b > 0, and 0 $(LE) x $(LE) 1, then I$(SUP C)x(a,b) = $(INTEGRATE x, 1)ta-1(1-t)b-1dt/B(a,b) where B is the beta function. It is also the complement of the cumulative distribution function of the beta distribution. It can be shown that I$(SUP C)x(a,b) = I1-x(b,a).
betaIncompleteCompl(a, b, x) evaluates I$(SUP C)x(a,b).