3:00 I mean, this isn't actually a problem, because the gamma function also has that neat little quirk of G(z)=z*G(z-1) that factorials have, and it extends it to all _(most**)_ real numbers. So bringing in that formula is perfectly fine
Not illegal. It's one of the definitional properties of the Gamma function. If anything, it's at least a theorem. The only times "it doesn't work" is with negative integers, because there is a pole of multiplicity one at each one of them, but it's otherwise good and dandy.
It does work. It is actually a property of the gamma/pi function: pi(x) = x pi(x-1) or gamma(x+1) = x gamma(x). And yes, it is the same property as for the factorial.
@@tommoes8447wow, that is actually really surprising. I would like to know, is there any actual reason people say it's illegal within the current parameters of mathematics? Because if it works for all values, I don't see why it shouldn't be allowed.
It is not "illegal", it is actually the *definition* of the gamma function. You cannot evaluate the integral for z = −3/2 (since it diverges), so you simply use the n! = n·(n−1)! property of the factorial to define the values of the gamma function for z < −1. All we really need is to find the values of the function in the (0; 1] interval, and then we can safely use this property.
@@allozovsky It is illegal by definition of factorial, and he has yet to (in his videos) prove that the gamma and pi function has the same property. It takes some (not much) work to prove that property 'sticks' with all values of the gamma/pi functions (except for when we have the /0 case somewhere in the function) and not just the natural numbers.
I think, even by just expanding the (1/2)!=(1/2).(1/2-1)=(1/2).(-1/2)!. Then cancel (-1/2)!, remaining with 1/1/2=2. But thanks for the much needed explanations😊😊
Yeah, that would be more correct to put it as Γ(1−1/2)/Γ(1+1/2) = 2 or Π(−1/2)/Π(1/2) = 2 and use the property (or, rather, definition) of the corresponding gamma/pi functions to do the cancellation. We assume that for these functions the property holds for all real (complex) numbers - we do not have to prove it for negative real numbers, since it is a *definition* of these functions.
You could have done integration by parts by integrating the t^(-1/2) and taking the derivative of e^(-t). You would have ended up with 2* the integral that defines (1/2)! which you calculated in the previous video. Never stop learning.
For Gamma it's a fundamental property that Gamma(z+1) = z·Gamma(z) for (almost) all complex z, so the "illegal" trick is legal here and the answer 2 just falls out. I would not be surprised if those crying foul at "illegal" tricks are also unhappy at something other than a non-negative integer next to the factorial sign. Is using an illegal trick on an illegal factorial twice as illegal, or is it a case of two negatives making a positive?🤔
Would it be so bad for all the parts of mathematics if FACTORIAL method, for instance, would be applied to natural numbers only? What confuses me is the practice of "convenient assumptions" that became bases for further convenient assumptions...and...and then we wonder why we end up with a theory that two "things" can be in multiple places at the same time.
Hi, could you do this question? I think you might enjoy it If g(x) is a polynomial function that satisfies the following relation: g(x) * g(y) = g(x) + g(y) + g(xy) - 2, for all x,y in the real numbers. So, the value of g(3) when g(2)=5 is: a) 8 b) 10 c) 12 d) 15 e) 18
I don't get those Mathematicians. It doesn't matter, if the Funktion isn't defined for that Range (/set). Only if it remains internally consistent. And if it doesn't, if we can find and define a subset for which it is. The interpretation comes later. That is how negative Numbers work. Then complex numbers and less known ones like the NaN's with infinite digits on the left of the decimal point. It really is just that simple.
half cancel out, answer = -! 😂
!! Aslo cancel out
-
- also cancels out with ____
Ok this is funny 😂
if everything else cancels out, then it’s just -/ 😂
"never stop learning those who stop learning they stop living " ❤❤❤❤❤❤❤❤
Never stop teaching, cause those who stop teaching stop developing the world
The entire purpose of extending the fsctorial using the gamma function is to preserve the property that people call "illegal". Like lol.
I agree!
The Bob Ross of math
3:00 I mean, this isn't actually a problem, because the gamma function also has that neat little quirk of G(z)=z*G(z-1) that factorials have, and it extends it to all _(most**)_ real numbers. So bringing in that formula is perfectly fine
Yeah true this formula fails for 0
Thanks for keeping my brain alert!
please do for i! (imaginary factorial) :3 very interesting video sir!
Thank For Sharing 😀😊
Meaningful...
Not illegal. It's one of the definitional properties of the Gamma function. If anything, it's at least a theorem.
The only times "it doesn't work" is with negative integers, because there is a pole of multiplicity one at each one of them, but it's otherwise good and dandy.
Does the illegal method work for any value where they exist? Like π and π+1 or -3/2 and -1/2?
It does work. It is actually a property of the gamma/pi function: pi(x) = x pi(x-1) or gamma(x+1) = x gamma(x). And yes, it is the same property as for the factorial.
@@tommoes8447wow, that is actually really surprising. I would like to know, is there any actual reason people say it's illegal within the current parameters of mathematics? Because if it works for all values, I don't see why it shouldn't be allowed.
It is not "illegal", it is actually the *definition* of the gamma function. You cannot evaluate the integral for z = −3/2 (since it diverges), so you simply use the n! = n·(n−1)! property of the factorial to define the values of the gamma function for z < −1. All we really need is to find the values of the function in the (0; 1] interval, and then we can safely use this property.
@@allozovsky neat explanation. Thanks!
@@allozovsky It is illegal by definition of factorial, and he has yet to (in his videos) prove that the gamma and pi function has the same property. It takes some (not much) work to prove that property 'sticks' with all values of the gamma/pi functions (except for when we have the /0 case somewhere in the function) and not just the natural numbers.
I think, even by just expanding the (1/2)!=(1/2).(1/2-1)=(1/2).(-1/2)!. Then cancel (-1/2)!, remaining with 1/1/2=2. But thanks for the much needed explanations😊😊
This is wrong cause this definition can only be used when x is a natural number
Yeah, that would be more correct to put it as Γ(1−1/2)/Γ(1+1/2) = 2 or Π(−1/2)/Π(1/2) = 2 and use the property (or, rather, definition) of the corresponding gamma/pi functions to do the cancellation. We assume that for these functions the property holds for all real (complex) numbers - we do not have to prove it for negative real numbers, since it is a *definition* of these functions.
@@allozovsky exactly
@@Harrykesh630 i checked with wolfram alpha. ((1/n)-1)!/n = (1/n)!
@@allozovsky, thank you for the much needed information 💁♀️ 🙏
Thanks Sir
Welcome
You could have done integration by parts by integrating the t^(-1/2) and taking the derivative of e^(-t). You would have ended up with 2* the integral that defines (1/2)! which you calculated in the previous video. Never stop learning.
For Gamma it's a fundamental property that Gamma(z+1) = z·Gamma(z) for (almost) all complex z, so the "illegal" trick is legal here and the answer 2 just falls out. I would not be surprised if those crying foul at "illegal" tricks are also unhappy at something other than a non-negative integer next to the factorial sign. Is using an illegal trick on an illegal factorial twice as illegal, or is it a case of two negatives making a positive?🤔
Gamma Function!
Can you make a video telling us how the epsilon bet came about?
What about the full cover on differential equations video you promised?
Would it be so bad for all the parts of mathematics if FACTORIAL method, for instance, would be applied to natural numbers only? What confuses me is the practice of "convenient assumptions" that became bases for further convenient assumptions...and...and then we wonder why we end up with a theory that two "things" can be in multiple places at the same time.
Hi, could you do this question? I think you might enjoy it
If g(x) is a polynomial function that satisfies the following relation: g(x) * g(y) = g(x) + g(y) + g(xy) - 2, for all x,y in the real numbers. So, the value of g(3) when g(2)=5 is:
a) 8 b) 10 c) 12 d) 15 e) 18
I don't get those Mathematicians.
It doesn't matter, if the Funktion isn't defined for that Range (/set). Only if it remains internally consistent. And if it doesn't, if we can find and define a subset for which it is.
The interpretation comes later.
That is how negative Numbers work. Then complex numbers and less known ones like the NaN's with infinite digits on the left of the decimal point.
It really is just that simple.
can you solve any number from the Unified State Exam(Russia)?
In the Riemann Paradox and Sphere Geometry System Incorporated...
π = 2
It can solved for 2 = π
there is a nother way to solve this : (-1/2)!/(1/2)! = (-1/2)!/((1/2)*(-1/2)!) = 1/(1/2) = 2
Give the proof of Gaussian integral.
Go watch the video