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!
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
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.
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.
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.
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.
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?
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.
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
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!
Thank you
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
Really love your videos and teaching style.
I am writting from Venezuela and I love your style for teaching maths
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.
I'm not racist or anything but you sir can be the first black person I have seen on a math video.
I've seen many others here on RUclips. I'm sure you'll soon see them too. Thanks.
You are a racist person
I went to an HBCU and majored in math. All of my professors were brilliant Black mathematicians. I aced the GRE math subject exam.
No way bro 💀
Naw😭💀
Oh that's really clever! I thought we would have to use the gamma function or something, but this is so much nicer!
Instead of multiplying we can substitute u=x-2
and we probably get odd integrand on interval symmetric around zero
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.
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.
x! can be extended to the reals with the gamma function
What a fantastic perspective! Brilliant work.
very interesting example, love your teaching style!
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.
These videos are so fun to watch, I would love to see even more integrals in the future, perhaps some from the MIT integration bee?
Those are so hard ong
Supercalifragilisticexpialidocious. I love you videos they are not just Educational but also Entertainment 🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰
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.
Is this allowed? In the Choose function, you treat x as an integer, but in the integral as a real number? Does it work for any real value of x?
I think we should have used P&C logic first and then put to integral because 4C5 is always 0 because you cannot arrange 4 things in unique order of 5.
Cool! very nice
Can we say that this is an odd function because the highest power of the integrand is 5?
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?
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.
You do not need to multiply all that out. You could have easily evaluated the integral from [0,4]. O-0=0
I need to know this. Please refer me to a video or text.
Good
I love your emotion and way of teaching mr newtons, truly fantastic! Never stop making videos, and ofcourse never stop learning!
Solve the problems is very time consuming ...... Thanks
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