I thought you were going to use Gamma(n+1) as the factorial function and integrate from there, But it was interesting to see this as a polynomial function. That's a great generalization as well
One of your best videos yet...Beautifully demonstrated the symmetry of the function by building on the premise of the practical integration example that would not have been as easy to convey otherwise. Thanks Prime Netwons!
Why does the expansion of x! Into the polynomial work? Isnt the factorial only defined like that for an integer x? Also if x is between 0 and 4 how could it x! Have an x-5 factor? Wouldnt it make it negative?
Expansion of x! works for any real value of x (check wolfram typing "x choose 5" for instance). Also: it is sometimes negative (like for x=1.5) and sometimes it is not (like for x=2.5) - in fact that's why the integral goes to zero.
You can show it algebriacally: let u = 4 - x, separate the integral into 0 to 2 and 2 to 4, then substitute for u in the second one. You get the same expression but with the limits of integration reversed.
Let n = 2m+1 for m a non-negative integer f(x) = n! * xCn = x(x-1)(x-2)...(x-n+2)(x-n+1) = x(x-1)(x-2)...(x-2m+1)(x-2m) f(2m-x) = (2m-x)(2m-x-1)(2m-x-2)...(2m-x-2m+1)(2m-x-2m) = (-1)^(2m+1) * (x-2m)(x-2m+1)(x-2m+2)...(x-1)x = - f(x) Integral[0 to n-1: f(x)] = Integral[0 to 2m: f(x)] = Integral[0 to m: f(x)] + Integral[m to 2m: f(x)] Let u = 2m-x in the second integral: = Integral[0 to m: f(x)] + Integral[m to 0: f(2m-u)(-du)] = Integral[0 to m: f(x)] + Integral[0 to m: - f(u)] = 0 So Integral[0 to n-1: xCn] = Integral[0 to n-1: f(x) / n!] = 0
I would say that it is better (u=x-2) than "seeing the graph" because on the graph we can only "feel" that there is a symetry, but in fact it could be an illusion. On the graph cosh is like (2/3)x^2+1, but it is not.
Ahh... Yes, just looking at the factorization 𝑥(𝑥 − 1)(𝑥 − 2)(𝑥 − 3)(𝑥 − 4) we see that all the zeros are evenly spaced, which definitely means that it's symmetrical about the median of the zeros, and because there's an odd number of zeros it then has to be an odd function.
need detailed explanation: because (0;4) interval is not symmetric.... We know that integral of an odd function in the interval of (-a; a) equals to zero.....
The function is symetric by (2,0) point but yeah... in a strict sence we should "move" the function and integral limits first to make it symetric by (0,0) point. But this is yt example video "how to think", not dissertation.
There is another way to somehow find an integration to the function (X n) which initially defined on the non-negative integers. And it is by using the intuitive fact that the factorial function is a restriction of the gamma function. And then trying to calculate its integral. But Taking in mind that il the integral would be a fraction of the form Γ(X) )/ Γ( x-a) then you will realize that its a mission impossible.
does x need to be greater than 5 in order to be able to create those factorials? otherwise if x < 5 then (x-5)! is a factorial of a negative. Or does it not matter because of the gamma function?
