Originally posted by DeepThoughtThat looks good to me 🙂
There's a big difficulty with the last step, this doesn't solve the problem. But I don't think integration by parts really helps (at first I thought I'd got it and then spotted the problem):
Using my notation from earlier (a, b) rather than (c, e) we have:
F(X; a, b) = ∫x^a (1 - x)^b dx
With the limits of the integral being [0, X]. We can inte ...[text shortened]... have to do the iterations by hand, or find a way of summing the polynomial in my previous post.
I had just independently come to what I think is very similar kind of results although expressed completely differently.
I will now work on developing a numerical approach to check our results and then came back to you.
Basically you can only solve specific cases.
Yes, and I independently came to that same conclusion and that there is simply no way around that. The method always requires first stating the values of variables a and b so to in-effect turn them into constants and then tediously work out by hand what the formula is for that pair of values -simply no sensible way out of it without resorting to a numerical approach. Actually, I now come round to thinking that a numerical approach would generally be better in most cases except a few 'special' cases which I will explain later.
I have completed my mathematical analysis of this problem and finally come to the following conclusions:
for f(x) = (x^v)( (1 – x) ^ (c – v) ) (I have now replace 'e' for 'v' to avoid confusion )
where 0 ≤ x ≤ 1, 0 ≤ v ≤ c, c > 0, and all numbers non-negative.
let M be the x value that cuts the area under the curve for f(x) into exactly two equal areas (with a vertical line on the f(x) graph ) between x = 0 and x = 1.
My conclusion is that, in most cases, generally better to use an iterative numerical approach to find M.
However, there is 3 noteworthy special cases where an algebraic solution is better:
The 3 special cases:
for v = 0, M = 1 – 2^(-1/(c + 1))
for v = c/2, M = ½
for v = c, M = 2^(-1/(c + 1))
I thank everyone for their input here 🙂