question about an integral

Originally posted by twhitehead
Given the large number of variables, it would either have to be a weird multidimensional plot, or he would need to fix some of the variables.

Having said that, the Beta function I believe is a special case of it which is graphed on Wikipedia:
https://en.wikipedia.org/wiki/Beta_function

Wow, we get a bent sheet🙂

Originally posted by twhitehead
Given the large number of variables, it would either have to be a weird multidimensional plot, or he would need to fix some of the variables.

Having said that, the Beta function I believe is a special case of it which is graphed on Wikipedia:
https://en.wikipedia.org/wiki/Beta_function

The integral is the indefinite Beta function B(p, q; x), but with integer values for p = S+1 and q = F - 1. At x = 1 it is equal to the full beta function B(p, q) = B(p, q; 1). The graph of the beta function you referenced has p and q as continuous variables, for Humy's function we only have integer variables as abscissae so it's not really the right graph. I'd want to see two axes with multiple curves on, each curve labelled with its p and q values.

Originally posted by DeepThought
I took a quick look at Wikipedia, it gives both notations and says they are both defined by the International standard ISO 31-11, so I wouldn't bother changing anything. I think in Physics the round bracket notation is more common and so I queried the reversed square bracket notation because I'd never seen it before.

So far as I can tell from Wikiped ...[text shortened]... fter 1935. The problem is that it is sometimes used to mean the complement of an open interval.

I think I have made up my mind now and I will change it to the other notation with round brackets. It would take me perhaps something like ~2 hour to iterate through my book to do that but that's not a big deal esp as I will be iterating through my book anyway to make several other changes and improvements.

Originally posted by sonhouse
Can you graph that result? I wondered what it would look like graphed. Really want to see the finished book!

I will show a graph of that result, along with every other result I make in my book.
In this case, I can graph it by treating x and S (for success outcome) and F (for failure outcome) as constants and choose examples of particular examples of values for them and plot a curve for each set of such examples.
But S and F is for when it is for applied maths to a particular type of probability distribution I have discovered, called a Bernoulli poly-paracav (there is also a Bernoulli zeroth-paracav and a Bernoulli mono-paracav and a Bernoulli poly-racav and many more!), which has a cumulative function of;

( (F + S + 1)! / S! ) ∑[r=0, F] (-1)^r h^(S + 1 + r) / ( (S + 1 + r) r! (F – r)! )

I will state the general pure-maths equation it in part comes from, which at least for now I think I will call the natural bets equation which is a specialized and modified form of the incomplete beta function, as:

x ∈ [0, 1], x ∈ ℝ
a, b ∈ ℕ{0}
∫[0, x] t^a (1 – t)^b dt = b! ∑[r=0, b] (-1)^r x^(a + 1 + r) / ( (a + 1 + r) r! (b – r)! )

My thinking behind naming it the natural beta equation is that, unlike with the conventional incomplete beta function where variables a and b are generally real numbers, the variables a and b must be natural numbers here (else the above equation wouldn't work)

Can anyone tell me if that 'natural beta equation' is probably original or has someone probably beat me to it with deriving a similar-enough equation?

Originally posted by humy
[b]I
...
natural bets equation /b]

misprint;
that should be;
"natural beta equation"

Originally posted by humy
misprint;
that should be;
"natural beta equation"

Well then, all bets are off🙂
Hope you are a success!

just noticed that if x=1 in;

x ∈ [0, 1], x ∈ ℝ
a, b ∈ ℕ{0}
∫[0, x] t^a (1 – t)^b dt = b! ∑[r=0, b] (-1)^r x^(a + 1 + r) / ( (a + 1 + r) r! (b – r)! )

then that very neatly simplifies to just;

∫[0, 1] t^a (1 – t)^b dt = a! b! / (a + b + 1)!

-yet another useful equation for my research.
And I just used it to find the mean average of my new probability distribution which I think is;

mean = (S + 1) / (S + F + 2)

Haven't yet checked that with a computer program though.

Originally posted by humy
just noticed that if x=1 in;

x ∈ [0, 1], x ∈ ℝ
a, b ∈ ℕ{0}
∫[0, x] t^a (1 – t)^b dt = b! ∑[r=0, b] (-1)^r x^(a + 1 + r) / ( (a + 1 + r) r! (b – r)! )

then that very neatly simplifies to just;

∫[0, 1] t^a (1 – t)^b dt = a! b! / (a + b + 1)!

-yet another useful equation for my research.
And I just used it to find the mean average of my new probabi ...[text shortened]... s;

mean = (S + 1) / (S + F + 2)

Haven't yet checked that with a computer program though.

The full beta function can be written in terms of the gamma function, the formula is:

B(p, q) = gamma(p)gamma(q)/gamma(p + q)

Where:

gamma(x) = ∫[0, infinity] t^(x - 1) exp(-t) dt

and for integer values:

gamma(n) = (n - 1)!

for integer values of p and q the beta function is then:

B(p, q) = (p - 1)!(q - 1)! / (p + q - 1)!

Your a = p - 1, and b = q - 1

So:

B(a + 1, b + 1) = a! b! / (a + b + 1)!

Which is the expression you wrote down. Note though that you have a difference in definition between your expression (I'll call your function A(a, b)) and the standard beta function:

A(a, b) = ∫[0, 1] t^a (1 – t)^b dt

B(p, q) = ∫[0, 1] t^(p - 1) (1 – t)^(q - 1) dt

So A(a, b) = B(a + 1, b + 1). The binomial distribution is: C(N, m) = N! / [(N - m)! m!], this gives us:

C(N, m) = 1/ (N - 1) A(N - m, m)

and

C(N, m) = 1 / (N + 1)B(N - m + 1, m + 1)