We can actually get to value of (1/2)! directly without the gamma function by using the Wallis product: The Wallis product is pi/2 = (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)... which is equivalent to lim m-> ∞ [(2m)!!]^2/[(2m-1)!!][(2m+1)!!] Multiply top and bottom by (2m +1) to get pi/2 = lim m->∞ [(2m)!!/(2m+1)!!]^2 * (2m+1) Take the equation shown at 11:45, (1/2)! = lim m->∞ 2^m * sqrt(m+1) * m! / (2m + 1)!! Note that (2m)!! = 2^m * m! because you can factor a 2 out of each of the even terms in the double factorial. Thus, (1/2)! = lim m->∞ sqrt(m+1) * (2m)!! / (2m + 1)!! Note that we have that (2m)!! / (2m + 1)!! in the formula for the Wallis Product, so let's make our equation look more like that. Square both sides and multiply top and bottom of the RHS by (2m+1) to get: [(1/2)!]^2 = lim m->∞ (m+1)/(2m + 1) * [(2m)!! / (2m + 1)!!]^2 * (2m + 1) The rightmost portion is the Wallis product! We can now substitute: [(1/2)!]^2 = lim m->∞ (m+1)/(2m + 1) * pi/2 This limit is very easy to evaluate and so [(1/2)!]^2 = pi/4 or equivalently, (1/2)! = sqrt(pi)/2
Exactly what I came here to say! I approached the problem differently by using factorial and double factorial relations and then used the Wallis product as a convergence and equivalency argument. 😊
Fun fact: sqrt(pi)/2 is the side length of a square with the same area as a quarter unit circle, so to solve (1/2)! geometrically, we only have to square the circle ;-)
12:38 Using Stirling’s Formula(my favorite math result🤩), and a lot of algebraic manipulation, this limit is the limit of e*sqrt(2π)/2*(1+1/(2m))^(-2m)(1+1/(2m))^(-1)sqrt((m+1)/(2m+1)) which equals e*sqrt(2π)/2*e^(-1)*1*1/sqrt(2) = sqrt(π)/2
This formula can be obtained as a corollary of the Bohr-Mollerup theorem, which states that gamma function is the only log convex extension of the factorials. For more info, see Baby Rudin, the section on Gamma function
As someone who is not Euler (or anywhere close), it's frustrating to look at some of these alternate representations and wonder how the hell they came up with them.
I think the proof makes sense as a way to find this one. You want something missing an n! from the denominator, and you want something that approximates a simple number as m gets much larger than n. If you can arrange for the error to come from adding n instead of some constant to m in each term in a product, that's going to work, and you just have to work backwards to a formula that is tidy enough to give as a definition. Of course, you'd have to be Euler to think of a complete set of actual steps, but if you're really good at all the math of 1729, I don't think it would have been too hard to come up with a plan to find this identity.
I actually derived this one myself and it is not that difficult. I used a kind of standard strategy for extending things like the factorial or the Harmonic numbers and used the fact that if you take the logarithm of the factorial it locally approaches a straight line for big inputs. Assuming that the extendet factorial is "nice" and locally approaches this straight line on non-integers you can go far out (represented by the m!), go a step on the interpolated line that turned into an exponential after removing the logarithm ((m+1)^n) and then go back using the recursion (x+1)! = (x+1) x! which is achieved by everything in the denominator. Hope this helps.
True my friend - very true and! While we have the internet and all its wonders who could Euler turn to in order to discuss? Letters and replies would probably take days or weeks I suppose Euler just hunkered down, reflected on being paid to do what he loved doing, picked up a quill, container of ink then a piece of parchment scratched his head and sorta worked it all out?
I haven't even watched the video, let alone Euler's written account of this formula, but I know that Euler did tend to give some explanation and what he had tried, what did not work and what worked. So I would not be surprised if you could find some hints to how the hell he came up with this in his papers. Gauss worked very differently and compared himself to an architect who does not leave the scaffolding in place once the building is finished.
I remember visiting a friend in hospital. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the Ramanujan-Hardy number."
3:10 to 9:10 Wouldn't it be much easier and shorter to first cancel the m! in the fraction, so that you have (m+1)^n in the numerator and (m+1) (m+2) ... (m+n) in the denominator, then write that as the product of (m+1)/(m+1) times (m+1)/(m+2) times ... (m+1)/(m+n) and then argue that for m to infinity, each of these n fractions obviously goes to 1 ?!?
@@davidg6324 YES THANK YOU--that's what i did.but its stll an approximationSINCE its only as m approaches infinty --because we don't know how big n is remember?
It is interesting that this is also a representation of the gamma function. It motivates the gamma function as being the most natural extension of factorials. A video exploring its uniqueness would be appreciated, as it is the only logarithmically convex extension.
@@Mystery_Biscuits The gamma function is the only function that extends the factorial whose logarithm is convex. That is the second derivative of the logarithm of the gamma function is always greater than or equal to zero.
@@Mystery_Biscuits You could connect the dots between points of the factorial many ways. Requiring that extension function to be convex prevents weird zig zag paths and severely limits the options. Log convex is an even more strict condition which leaves only one option. Polynomials like x^3 are convex (concave up), but not log convex because 3*logx is concave (concave down). So the gamma function must 'increase' so fast that that there is only one possible path, deviating in the slightest would make it fail the log convex condition.
No need for logarithms. Performing division by n! and simplifying by m! leaves us with (m+1)^n in the numerator and (m+1)...(m+n) in the denominator, which can be viewed as a product of *finitely many* terms of form (m+1)/(m+k), each of which converges to 1, thus their product does as well. Btw I came here to see a proof for the identity refered to at 11:55, a followup video on that would be cool.
Instead of taking a log, cancel m! from numerator and denominator. Then you have a product of x terms on top and x terms on bottom, each of the form (m+finite number). Multiply them out and divide numerator and denominator by m^x. Now the numerator and denominator are each of the form 1+[a finite number of terms with m in the denominator]. So as m->infinity all the other terms have limit zero and the fraction is 1/1.
Yeap. But then again, most formulas are. We only learn the useful ones, but we forget that some of those were discovered in the same way as with the other useless ones
If you were going to take the ln in the end and cancel out terms, why not just do it right from the start instead of dividing things over unnecessarily?
At 12:20, this is NOT the gamma function; why should its results be equal to the gamma function? A priori, I see no reason why their results should be equal except when x is a non-negative integer. I do not say that their results are different; I only say that these might extend the factorial function differently and that we have not demonstrated that they are the same.
After expanding this double factorial i have following result limit(4^m*sqrt(m+1)/((2*m+1)*binomial(2*m,m)),m=infinity) but can it be simplified even further ?
@@petersievert6830 I tried gnu Calculator on linux variant and it gave (all truncated here to 4 dp) 0.5! ~ 0.8862 1.5! ~ 1.9817 1.99! ~ 1.9817 1.9999! ~ 1.998 3.999! ~ 23.9964 Ah well - for a cell phone calculator it could be ... a bit better ?
As for the exercise, maybe it helps to know en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 because sqrt(pi)/2 all to the square actually does equal pi/4 So if we square that expression and can convert it to 1 - 1/3 + 1/5 - 1/7 +- ... we are good.
We can actually get to value of (1/2)! directly without the gamma function by using the Wallis product:
The Wallis product is pi/2 = (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)... which is equivalent to lim m-> ∞ [(2m)!!]^2/[(2m-1)!!][(2m+1)!!]
Multiply top and bottom by (2m +1) to get pi/2 = lim m->∞ [(2m)!!/(2m+1)!!]^2 * (2m+1)
Take the equation shown at 11:45, (1/2)! = lim m->∞ 2^m * sqrt(m+1) * m! / (2m + 1)!!
Note that (2m)!! = 2^m * m! because you can factor a 2 out of each of the even terms in the double factorial.
Thus, (1/2)! = lim m->∞ sqrt(m+1) * (2m)!! / (2m + 1)!!
Note that we have that (2m)!! / (2m + 1)!! in the formula for the Wallis Product, so let's make our equation look more like that. Square both sides and multiply top and bottom of the RHS by (2m+1) to get:
[(1/2)!]^2 = lim m->∞ (m+1)/(2m + 1) * [(2m)!! / (2m + 1)!!]^2 * (2m + 1)
The rightmost portion is the Wallis product! We can now substitute:
[(1/2)!]^2 = lim m->∞ (m+1)/(2m + 1) * pi/2
This limit is very easy to evaluate and so [(1/2)!]^2 = pi/4 or equivalently, (1/2)! = sqrt(pi)/2
Was on the right track, but didn't know of the Wallis Product, thanks for pointing that out
Wallis product was my first thought.
@@journeymantraveller3338 but i don't think that makes sense
Exactly what I came here to say! I approached the problem differently by using factorial and double factorial relations and then used the Wallis product as a convergence and equivalency argument. 😊
I like this extension as it highlights clearly that the factorial of negative integers is undefined.
I don't like the term "undefined" i think saying it has pole at negative integers has more meaning but whatever i get what you are saying...
@@aweebthatlovesmath4220 yeah that's a more descriptive description (ha) for sure
@@Mystery_BiscuitsI couldn’t describe a better description
@@Mystery_Biscuits The Mittag-Leffler expansion of gamma explicitly shows the poles (there's an explanation on stackexchange, btw).
Fun fact: sqrt(pi)/2 is the side length of a square with the same area as a quarter unit circle, so to solve (1/2)! geometrically, we only have to square the circle ;-)
12:38 Using Stirling’s Formula(my favorite math result🤩), and a lot of algebraic manipulation, this limit is the limit of
e*sqrt(2π)/2*(1+1/(2m))^(-2m)(1+1/(2m))^(-1)sqrt((m+1)/(2m+1))
which equals
e*sqrt(2π)/2*e^(-1)*1*1/sqrt(2)
= sqrt(π)/2
This formula can be obtained as a corollary of the Bohr-Mollerup theorem, which states that gamma function is the only log convex extension of the factorials. For more info, see Baby Rudin, the section on Gamma function
As someone who is not Euler (or anywhere close), it's frustrating to look at some of these alternate representations and wonder how the hell they came up with them.
A lot of brilliant people walked this earth
I think the proof makes sense as a way to find this one. You want something missing an n! from the denominator, and you want something that approximates a simple number as m gets much larger than n. If you can arrange for the error to come from adding n instead of some constant to m in each term in a product, that's going to work, and you just have to work backwards to a formula that is tidy enough to give as a definition. Of course, you'd have to be Euler to think of a complete set of actual steps, but if you're really good at all the math of 1729, I don't think it would have been too hard to come up with a plan to find this identity.
I actually derived this one myself and it is not that difficult. I used a kind of standard strategy for extending things like the factorial or the Harmonic numbers and used the fact that if you take the logarithm of the factorial it locally approaches a straight line for big inputs. Assuming that the extendet factorial is "nice" and locally approaches this straight line on non-integers you can go far out (represented by the m!), go a step on the interpolated line that turned into an exponential after removing the logarithm ((m+1)^n) and then go back using the recursion (x+1)! = (x+1) x! which is achieved by everything in the denominator. Hope this helps.
True my friend - very true and! While we have the internet and all its wonders who could Euler turn to in order to discuss?
Letters and replies would probably take days or weeks
I suppose Euler just hunkered down, reflected on being paid to do what he loved doing, picked up a quill, container of ink then a piece of parchment scratched his head and sorta worked it all out?
I haven't even watched the video, let alone Euler's written account of this formula, but I know that Euler did tend to give some explanation and what he had tried, what did not work and what worked. So I would not be surprised if you could find some hints to how the hell he came up with this in his papers.
Gauss worked very differently and compared himself to an architect who does not leave the scaffolding in place once the building is finished.
From 1729, huh? Isn't that the Hardy Ramanujan number? Weird coincidence.
This was my first thought as well
I remember visiting a friend in hospital. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the Ramanujan-Hardy number."
It’s actually the year Michael Penn was born
@@ronald3836😂
What has the Taxi number to do with factorials? -- interesting. Dumb me!
It looks like Wallis product - sort of
Lines that connect did a great derivation of this formula
The short historical background of the problem at the beginning of the video is nice for putting it into context.
3:10 to 9:10 Wouldn't it be much easier and shorter to first cancel the m! in the fraction, so that you have (m+1)^n in the numerator and (m+1) (m+2) ... (m+n) in the denominator, then write that as the product of (m+1)/(m+1) times (m+1)/(m+2) times ... (m+1)/(m+n) and then argue that for m to infinity, each of these n fractions obviously goes to 1 ?!?
Yes, this seems to be much faster and intuitive than what he did
@@davidg6324 YES THANK YOU--that's what i did.but its stll an approximationSINCE its only as m approaches infinty --because we don't know how big n is remember?
12:54
It is interesting that this is also a representation of the gamma function. It motivates the gamma function as being the most natural extension of factorials. A video exploring its uniqueness would be appreciated, as it is the only logarithmically convex extension.
Could you define what you mean by a "logarithmically convex extension"
@@Mystery_Biscuits The gamma function is the only function that extends the factorial whose logarithm is convex. That is the second derivative of the logarithm of the gamma function is always greater than or equal to zero.
@@spaghetti1383 Interesting. Why is that a desirable property?
@@Mystery_Biscuits You could connect the dots between points of the factorial many ways. Requiring that extension function to be convex prevents weird zig zag paths and severely limits the options. Log convex is an even more strict condition which leaves only one option. Polynomials like x^3 are convex (concave up), but not log convex because 3*logx is concave (concave down). So the gamma function must 'increase' so fast that that there is only one possible path, deviating in the slightest would make it fail the log convex condition.
@@spaghetti1383 I see. Thank you for the explanation!
No need for logarithms. Performing division by n! and simplifying by m! leaves us with (m+1)^n in the numerator and (m+1)...(m+n) in the denominator, which can be viewed as a product of *finitely many* terms of form (m+1)/(m+k), each of which converges to 1, thus their product does as well.
Btw I came here to see a proof for the identity refered to at 11:55, a followup video on that would be cool.
Instead of taking a log, cancel m! from numerator and denominator. Then you have a product of x terms on top and x terms on bottom, each of the form (m+finite number). Multiply them out and divide numerator and denominator by m^x. Now the numerator and denominator are each of the form 1+[a finite number of terms with m in the denominator]. So as m->infinity all the other terms have limit zero and the fraction is 1/1.
Another method can be to first prove for n=1(which is quite simple) and then use induction
yes. But need lim(n+1 form)/lim(n form)=n+1, which is not hard to prove. The value is (n+1)\dfrac{n+m+1-n}{n+m+1} which goes to n+1.
I don't know where Michael Penn grew up but I really like the way he pronounces "towards" and "orange".
impractical ass formula
Yeap. But then again, most formulas are. We only learn the useful ones, but we forget that some of those were discovered in the same way as with the other useless ones
In my book there is extension for this formula which even works for negative non integer
If you were going to take the ln in the end and cancel out terms, why not just do it right from the start instead of dividing things over unnecessarily?
Euler wrote this in 1729? A very prophetic date!
Isn't gamma function also called euler's first integral function or something like that? Isn't it also his doing as well?
At 12:20, this is NOT the gamma function; why should its results be equal to the gamma function? A priori, I see no reason why their results should be equal except when x is a non-negative integer. I do not say that their results are different; I only say that these might extend the factorial function differently and that we have not demonstrated that they are the same.
Euler was 21 when he wrote that... WOW.
Lowercase m and n as a pair of variables in an explanation is a rather unfortunate choice - m and n look similar, and they sound similar.
The limit for 1/2! can be evaluated easily using Striling's formula
* (1/2)!
* Stirling's formula
@@forcelifeforce wtf is wrong with you bro
aesthetically pleasing ❤️
It would be nice to know how Euler found this limit and understand the history of mathematics
Any Euler historians/experts or math history experts out there?
1729 was a very interesting year...
we can use sterling equivalence
After expanding this double factorial i have following result
limit(4^m*sqrt(m+1)/((2*m+1)*binomial(2*m,m)),m=infinity)
but can it be simplified even further ?
Yes! The Central binomial coefficient (2m nCr m) has an asymptotic expantion as 4^m/sqrt(Pi*m)*(1+O(m^-1)). From there it's a short way to sqrt(Pi)/2.
Euler's extension matches the Gamma function on ℝ? Do we know that?
Oh. You kinda answer that question.
Is this limit and the gamma representation always equal?
Offset by one, like the Gamma function usually is. (At least from what I can tell.)
Thank you, as usual, for not saying Yooler.
You mean you'd have a non positive integer there in the denominator. But this is really for yt algo.
Wow dude you really buried the punch line
Isn't that defining factorial 'through' factorial? x!=...m!?
Well ... my Samsung calculator gives (0.5)! as undefined !?*?!
apparently it can only do it for natural numbers... windows calculator does it alright anyway
@@petersievert6830 I tried gnu Calculator on linux variant and it gave (all truncated here to 4 dp)
0.5! ~ 0.8862
1.5! ~ 1.9817
1.99! ~ 1.9817
1.9999! ~ 1.998
3.999! ~ 23.9964
Ah well - for a cell phone calculator it could be ... a bit better ?
@@petersievert6830 as a ps - gnu calculator also gave me results for negative values
-0.5! ~ 0.8862 to 4 dp
Wow! Defining a factorial in terms of a much larger factorial!
The question is:"what's the factorial of a function f(x)"?
Γ(f(x) + 1) where Γ is the gamma function
As for the exercise, maybe it helps to know en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
because sqrt(pi)/2 all to the square actually does equal pi/4
So if we square that expression and can convert it to 1 - 1/3 + 1/5 - 1/7 +- ... we are good.