Excellent! I have learnt a lot from this video, not just about the gamma function. The integration techniques you use near the end (square/square root, then making 2 dimensional, and converting to polar coords) is new to me. I can see that it could be very useful in general.
The evaluation of the integral of exp(-u squared) from minus infinity to infinity is easy, because it can be easily transformed to the integral of the PDF of the normal distribution with parameters 0 and 1/sqrt(2).
yeah that is a good observation. However this is pretty much the exact same proof as proving that the normal distribution adds up to 1 over the reals. So using that fact would kind of be begging the question. But it is good you now know the relationship between the normal distribution and gamma function.
Your observation of calculus and prob and stat is very good, but to prove this integral of normal distribution is another more difficult story of calculus, so if someone really want to understand why but not just omit sth already true, video explanation give the most simple explanation I believe.
@Per Persson Here's the answer to your question. See this video: th-cam.com/video/L4Trz6pFut4/w-d-xo.html Apparently the standard definition is due to Bernoulli, but Gauss defined it the way you suggested which seems some how better. Gauss's definition is called the PI function, with a capital Greek PI.
By the very definition of the factorial function; it’s the product of all natural numbers up to n, that is n! = n * (n - 1) * (n - 2) * .... * 2 * 1. If you multiply the right side by (n + 1), then you’ve created the product of all natural numbers up to (n + 1), which is (n + 1)!
Isn't there an issue at 2:20? Even though we know it tends to zero, when we evaluate the result -x^a(e^(-x)) from 0 to infinity, I don't think it is appropriate to just say that since e^(-x) approaches zero as x approaches infinity, then (-x^a)(e^(-x)) also approaches zero (actually, I'm not sure if that was what you meant as "this part", since you were pointing to e^(-x)). Since the front half of the expression tends to negative infinity, it results in an indeterminate form and we can't do arithmetics with infinity. I think we need to represent it as a fraction (-x^a)/e^x and at least show, using L'Hopital for the indeterminate form of limits, that the nth derivative of -x^a is (a)(a-1)(a-2)...(a-n-1)(-1) which is a constant depending on a, whereas the nth derivative of e^(x) is always e^x, so after differentiating the expression n times by L'Hopital's Rule, we can show that the denominator tends to infinity while the numerator is a constant, so the expression tends to 0 when x tends to infinity.
which one reaches its destination faster than the other? the power term or exponential term? the value of the faster term shall govern at some point and dictate the final value. obviously the exponential term is faster then the power term and hence exponential term will approach zero before the power term get huge.
@@mathphschjhb7749 yup! I totally agree, both with your reply and this video’s outcome, but what I’m not particularly satisfied with is how the video explained this using limits, because applying limits to infinity, we get 0 times infinity at around 2:20, which is not defined. Your way of showing that exponential will be greater than power terms as the terms get larger is definitely true, so I think it would be more accurate to describe this either using L’Hopital to decompose the terms or some graphical analysis rather than just taking limits to infinity directly
It is one of the most common limits and assumes the viewer should know it Remember this video is 2 years old and back then he had a more "experienced" core audience as in: Only a handful of his viewers are new to university level maths while the rest already finished at least a semester So I understand why he skipped the explanation of the limit
But how does Gamma(1/2) = (-1/2)! , when you only said Gamma(n+1) = n! for natural numbers n? Could you explain anything about why we can say (-1/2)! = sqrt(pi) when we haven't defined the factorial in terms of the gamma function for non-integer n?
The gamma function is just a function which has the recursive properties like the 'normal' factorial for natural numbers, and for natural numbers it's value matches the factorials as we know them from school, but all in all they are just special cases of the general, the gamma, function.
@@CoderboyPB Yes, but in the video he implies (-1/2)! = Gamma(1/2) = sqrt(pi), which is wrong because n! is only defined for integer n. Of course, writing Gamma(1/2) = sqrt(pi) is fine, because the gamma function IS defined for non-integer inputs.
@@edmundwoolliams1240 I don't know how to explain it in english, but the gamma is an EXTENSION of the SEQUENCE of the faculty to a CONTINIOUS function, which values at zero and the positiv values are their faculties. Try to see it as an INTERPOLATION :-)
Неуверен в правильности вычисления Гамма функции через интегралы при современном полном непонимании деления на ноль... Что мы знаем о факториалах... Для начала мы знаем что факториал следующего числа равен факториалу предыдущего числа умноженному на это самое следующее число... N!= (N-1)!×N или по другому... факториал предыдущего числа равен факториалу следующего числа деленному на это самое следующее число... N!=(N+1)!/(N+1) есть еще вид (N+1)!= N!×(N+1)... значит (N-1)!=N!/N и N=N!/(N-1)! При N=1 получаем 0!=1!/1 и 1=1!/0! При N=0 получаем (-1)!=0!/0 и 0=0!/(-1)! При N=(-1) получаем (-2)!=(-1)!/(-1) и (-1)=(-1)!/(-2)! При N=(-2) получаем (-3)!=(-2)!/(-2) и (-2)=(-2)!/(-3)! При N=(-3) получаем (-4)!=(-3)!/(-3) и (-3)=(-3)!/(-4)! При N=(-4) получаем (-5)!=(-4)!/(-4) и (-4)=(-4)!/(-5)! Видим что вычисление положительных факториалов по действию очень похоже на действие возведения в степень... только множители различные... Исходя из полученных формул отрицательный факториал берется не только от отрицательного значения но и имеет смысл обратных значений для положительных факториалов N... Во всяком случае вполне возможно N!=(N+1)!/(N+1) 0!=1!/1=1 (-1)!=0!/(0)=1/(0)= 1 неделённая единица (-2)!=(-1)!/(-1)= 1/(-1)= -1 (-3)!=(-2)!/(-2)=(-1)/(-2)= 1/2 (-4)!=(-3)!/(-3)=(1/2)/(-3)= -1/6 (-5)!=(-4)!/(-4)=(-1/6)/(-4)= 1/24 (-6)!=(-5)!/(-5)=(1/24)/(-5)= -1/120... Интересно что получаются обратные значения Гамма функциям от положительных значений когда Г(N+1)=N! Г(N+1)=N×Г(N)=N×(N-1)! Немного неожиданно... Получается что для отрицательных Г(-(N+1))=1/Г(N+1)=1/N! Но есть "проблема" со знаком... Видим что постоянно через один изменяется знак при делении "факториалов" от отрицательных значений... Предположу что нужно брать для отрицательных значений N значение по модулю (а для обобщения и для положительных значений N...) N!=(N+1)!/|N+1| (N-1)!=N!/|N| 0!=1/1=1 (-1)!=0!/0=1/0= 0 (относительный ноль) или безотносительно единица неделённая что более верно... Тогда следует (-2)!= (-1)!/|-1|=1 (-3)!=(-2)!/|-2|=1/2 (-4)!=(-3)!/|-3|=1/6 (-5)!=(-4)!/|-4|=1/24... Как видим получаем обратные величины факториалов для положительных значений N... но еще идет сдвиг на один ход относительно факториалов для положительных значений N... Смею предположить что отрицательные факториалы должны считаться по формуле N!=(N+1)!/|N|... Тогда (-1)!=0!/|-1|=1/1=1 (-2)!=(-1)!/|-2|=1/2 (-3)!=(-2)!/|-3|=1/6 (-4)!=(-3)!/|-4|=1/24 (-5)!=(-4)!/|-5|=1/120... и получается что эти значения численно равны коэффициентам для нахождения "обратного факториала"... Кстати по этой же формуле получается 0!=1!/0=1/0=1 единица неделённая что наверное будет более верно... Если уж быть совсем дерзким и исходить из того что график этих значений должен бы быть хоть немного математически красив то возможно факториалы от отрицательных значений должны бы быть и сами отрицательными... Но я пока не нахожу физического смысла отрицательным значениям факториалов... (самим факториалам от отрицательных чисел смысл проявился очень явно)... к тому же придется признать что тогда при этом 0!=1/0=0 равен относительному нулю... Но это пока мои личные фантазии... и в этом надо сначала разобраться... а перед этим хорошенько подумать... Мне все же ближе "вариант с модулями"...
When thing I just realized: We proof Gamma(alpha +1) by assuming alpha > 0. Then we use alpha = 0 (or rather n = 0) in the base case of the induction?!
![[안방1열 풀캠4K] 베이비몬스터 'DRIP' (BABYMONSTER FullCam)│@SBS Inkigayo 241110](http://i.ytimg.com/vi/AQ4RIghEvUs/mqdefault.jpg)
Excellent! I have learnt a lot from this video, not just about the gamma function. The integration techniques you use near the end (square/square root, then making 2 dimensional, and converting to polar coords) is new to me. I can see that it could be very useful in general.
Good for the Gaussian error function / normal distribution too.
The evaluation of the integral of exp(-u squared) from minus infinity to infinity is easy, because it can be easily transformed to the integral of the PDF of the normal distribution with parameters 0 and 1/sqrt(2).
yeah that is a good observation. However this is pretty much the exact same proof as proving that the normal distribution adds up to 1 over the reals. So using that fact would kind of be begging the question. But it is good you now know the relationship between the normal distribution and gamma function.
Your observation of calculus and prob and stat is very good, but to prove this integral of normal distribution is another more difficult story of calculus, so if someone really want to understand why but not just omit sth already true, video explanation give the most simple explanation I believe.
Hi, I have a comment to make here. Can you explain how is defined $Gamma(-n)$ with $n$ an integer. And what happens with $Gamma(-1-2n)$ for instance?
I think it diverges for negative integers
At what point does the gamma function equal pi?
Hi sir I have a question, can you show that the recursive property for the gamma function also works for negative arguments?, thanks.
He had already shown that, IBP works forza generic real value of alpha😊
Thank you michael! Very clear and to the point.
Does anyone know why the gamma function was defined so that Γ(n+1) = n! and not so that Γ(n) = n! ?
here you go: www.cantorsparadise.com/the-beautiful-gamma-function-and-the-genius-who-discovered-it-8778437565dc
How many times have I asked myself that question. 😁
@@Jack_Callcott_AU He said Obviously.. but for me.. it's obviously not obvious
@Per Persson Here's the answer to your question. See this video:
th-cam.com/video/L4Trz6pFut4/w-d-xo.html
Apparently the standard definition is due to Bernoulli, but Gauss defined it the way you suggested which seems some how better. Gauss's definition is called the PI function, with a capital Greek PI.
@@Jack_Callcott_AU I cannot hear him saying anything of *why* the gamma function is defined with x-1 in the exponent instead of just x.
6:45 how can you switch to 2 different variables? is it allowed tho?
Why not? Substitution is always valid.
@@CliffSedge-nu5fvwouldnt that imply that x is not equal to y? but u^2=u u, where u=u
how (n+1)n!=(n+1)! ? can somebody tell me that equality how it comes in general form.
By the very definition of the factorial function; it’s the product of all natural numbers up to n, that is n! = n * (n - 1) * (n - 2) * .... * 2 * 1. If you multiply the right side by (n + 1), then you’ve created the product of all natural numbers up to (n + 1), which is (n + 1)!
I think a better way to remember the gamma function is to define it as the (n-1)th moment of the exponential distribution with lambda=1.
thanks! pls more easy "shorts" like that.
thank you Sir, very clear explaniation
I am lost
Which part?
Isn't there an issue at 2:20? Even though we know it tends to zero, when we evaluate the result -x^a(e^(-x)) from 0 to infinity, I don't think it is appropriate to just say that since e^(-x) approaches zero as x approaches infinity, then (-x^a)(e^(-x)) also approaches zero (actually, I'm not sure if that was what you meant as "this part", since you were pointing to e^(-x)). Since the front half of the expression tends to negative infinity, it results in an indeterminate form and we can't do arithmetics with infinity.
I think we need to represent it as a fraction (-x^a)/e^x and at least show, using L'Hopital for the indeterminate form of limits, that the nth derivative of -x^a is (a)(a-1)(a-2)...(a-n-1)(-1) which is a constant depending on a, whereas the nth derivative of e^(x) is always e^x, so after differentiating the expression n times by L'Hopital's Rule, we can show that the denominator tends to infinity while the numerator is a constant, so the expression tends to 0 when x tends to infinity.
which one reaches its destination faster than the other?
the power term or exponential term? the value of the faster term shall govern at some point and dictate the final value. obviously the exponential term is faster then the power term and hence exponential term will approach zero before the power term get huge.
@@mathphschjhb7749 yup! I totally agree, both with your reply and this video’s outcome, but what I’m not particularly satisfied with is how the video explained this using limits, because applying limits to infinity, we get 0 times infinity at around 2:20, which is not defined.
Your way of showing that exponential will be greater than power terms as the terms get larger is definitely true, so I think it would be more accurate to describe this either using L’Hopital to decompose the terms or some graphical analysis rather than just taking limits to infinity directly
@@asel8189 very true & totally agree
It is one of the most common limits and assumes the viewer should know it
Remember this video is 2 years old and back then he had a more "experienced" core audience as in: Only a handful of his viewers are new to university level maths while the rest already finished at least a semester
So I understand why he skipped the explanation of the limit
thank you very much sir!
Magic 🎉
Pretty good
But how does Gamma(1/2) = (-1/2)! , when you only said Gamma(n+1) = n! for natural numbers n? Could you explain anything about why we can say (-1/2)! = sqrt(pi) when we haven't defined the factorial in terms of the gamma function for non-integer n?
The gamma function is just a function which has the recursive properties like the 'normal' factorial for natural numbers, and for natural numbers it's value matches the factorials as we know them from school, but all in all they are just special cases of the general, the gamma, function.
@@CoderboyPB Yes, but in the video he implies (-1/2)! = Gamma(1/2) = sqrt(pi), which is wrong because n! is only defined for integer n. Of course, writing Gamma(1/2) = sqrt(pi) is fine, because the gamma function IS defined for non-integer inputs.
@@edmundwoolliams1240 I don't know how to explain it in english, but the gamma is an EXTENSION of the SEQUENCE of the faculty to a CONTINIOUS function, which values at zero and the positiv values are their faculties.
Try to see it as an INTERPOLATION :-)
@@edmundwoolliams1240
That's not what happened, though.
love
Неуверен в правильности вычисления Гамма функции через интегралы при современном полном непонимании деления на ноль...
Что мы знаем о факториалах...
Для начала мы знаем что
факториал следующего числа равен факториалу предыдущего числа умноженному на это самое следующее число...
N!= (N-1)!×N
или по другому... факториал предыдущего числа равен факториалу следующего числа деленному на это самое следующее число...
N!=(N+1)!/(N+1)
есть еще вид (N+1)!= N!×(N+1)...
значит (N-1)!=N!/N и N=N!/(N-1)!
При N=1 получаем 0!=1!/1 и 1=1!/0!
При N=0 получаем (-1)!=0!/0 и 0=0!/(-1)!
При N=(-1) получаем (-2)!=(-1)!/(-1) и (-1)=(-1)!/(-2)!
При N=(-2) получаем (-3)!=(-2)!/(-2) и (-2)=(-2)!/(-3)!
При N=(-3) получаем (-4)!=(-3)!/(-3) и (-3)=(-3)!/(-4)!
При N=(-4) получаем (-5)!=(-4)!/(-4) и (-4)=(-4)!/(-5)!
Видим что вычисление положительных факториалов по действию очень похоже на действие возведения в степень...
только множители различные...
Исходя из полученных формул отрицательный факториал берется не только от отрицательного значения но и имеет смысл обратных значений для положительных факториалов N...
Во всяком случае вполне возможно
N!=(N+1)!/(N+1)
0!=1!/1=1
(-1)!=0!/(0)=1/(0)= 1 неделённая единица
(-2)!=(-1)!/(-1)= 1/(-1)= -1
(-3)!=(-2)!/(-2)=(-1)/(-2)= 1/2
(-4)!=(-3)!/(-3)=(1/2)/(-3)= -1/6
(-5)!=(-4)!/(-4)=(-1/6)/(-4)= 1/24
(-6)!=(-5)!/(-5)=(1/24)/(-5)= -1/120...
Интересно что получаются обратные значения Гамма функциям от положительных значений когда
Г(N+1)=N!
Г(N+1)=N×Г(N)=N×(N-1)!
Немного неожиданно...
Получается что для отрицательных Г(-(N+1))=1/Г(N+1)=1/N!
Но есть "проблема" со знаком...
Видим что постоянно через один изменяется знак при делении "факториалов" от отрицательных значений...
Предположу что нужно брать для отрицательных значений N значение по модулю (а для обобщения и для положительных значений N...)
N!=(N+1)!/|N+1| (N-1)!=N!/|N|
0!=1/1=1
(-1)!=0!/0=1/0= 0 (относительный ноль)
или безотносительно единица неделённая что более верно...
Тогда следует (-2)!= (-1)!/|-1|=1
(-3)!=(-2)!/|-2|=1/2
(-4)!=(-3)!/|-3|=1/6
(-5)!=(-4)!/|-4|=1/24...
Как видим получаем обратные величины факториалов для положительных значений N...
но еще идет сдвиг на один ход относительно факториалов для положительных значений N...
Смею предположить что отрицательные факториалы должны считаться по формуле
N!=(N+1)!/|N|...
Тогда
(-1)!=0!/|-1|=1/1=1
(-2)!=(-1)!/|-2|=1/2
(-3)!=(-2)!/|-3|=1/6
(-4)!=(-3)!/|-4|=1/24
(-5)!=(-4)!/|-5|=1/120...
и получается что эти значения численно равны коэффициентам для нахождения "обратного факториала"...
Кстати по этой же формуле получается
0!=1!/0=1/0=1 единица неделённая
что наверное будет более верно...
Если уж быть совсем дерзким и исходить из того что график этих значений должен бы быть хоть немного математически красив то возможно факториалы от отрицательных значений должны бы быть и сами отрицательными...
Но я пока не нахожу физического смысла отрицательным значениям факториалов...
(самим факториалам от отрицательных чисел смысл проявился очень явно)...
к тому же придется признать что тогда при этом 0!=1/0=0 равен относительному нулю...
Но это пока мои личные фантазии...
и в этом надо сначала разобраться...
а перед этим хорошенько подумать...
Мне все же ближе "вариант с модулями"...
Факториал неопределен для отрицательных целых, а гамма функция там расходится, а (n+1)! = (n+1)*n! верно лишь для неотрицательных целых n
@@lukandrate9866 кем не определен... почему до сих пор не определен... и что вообще должен определять факториал от отрицательного значения... 🤔
When thing I just realized: We proof Gamma(alpha +1) by assuming alpha > 0. Then we use alpha = 0 (or rather n = 0) in the base case of the induction?!
The base case is calculated for n=1 using the definition of the gamma function. The property is used in the induction step.
How dA =rdrd0
PRIYA YADAV do the Jacobian and you’ll see how dA becomes rdrd\theta
Isn't it ½rdrdtheta? Because in the limit, it approaches area of triangle.
top
Γ(z)= ∫₀ ᪲ xᶻ⁻¹ e⁻ˣ dx
Ok
Γ(α)= ∫₀ ᪲ xᵅ⁻¹ e⁻ˣ dx