I Found Out How to Differentiate Factorials!
ฝัง
- เผยแพร่เมื่อ 8 ก.ย. 2024
- Have you ever wondered how to find the derivative of a factorial? In this video I'll show you how to differentiate factorial functions! It's time to find out how to differentiate the factorial of x. \
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#math #brithemathguy #factorial
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
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answer is still no. You are taking the derivative of the gamma not factorial. I also do not think that the gamma function is valid as it breaks the rules of factorial and ends up with imaginary numbers in the negative, which actually do exist for all negative values in factorial. It is just the negative side does not go towards anything nice. Derivatives by definition must be on continuous functions. The gamma function is not continuous.
@@kennethgee2004nah
"That's a good place to start" I undesrtood the reference hahahaha
Me too 😂
I know you...
Michael Penn?
Qué haces aquí :v
@@jimjim3979 yes!
Take the derivative of floor function.
Take derivative of ceiling function.
Take derivative of modulo function.
Take integral of error function.
Take Fourier Transform of Step function (Heaviside).
Take integral of e^e^x.
For the floor and ceil fonction, derivitives areF:R\Z -> R
x |---> 0
1. 0 for x is not an integer
2. 0 for x isn't an integer
3. Modulo has two variables, you need to specify which one is to be derived
4. x*erf(x)+e^(-x^2)+C by integration by parts It's even in the wiki en.wikipedia.org/wiki/Error_function
7. is defined with help of the exponential integral. Ei(e^x)
The first 2 are impossible, they are both defined by not being continuous. If you took the derivative of small sections like between 0 and 1 or something, the derivative would be 0 as they have no "velocity"
@@hassanshaikh3451 it is possible if you use the distributional derivative, but strictly speaking you wouldn't get a function in the normal sense but a distribution which is basically a linear functional on the hilbert space of functions if i remember correctly
I have uncontrollable urge to click on maths videos that intrigue me, and this definitely one of them, very great!
Wow, thanks!
Derivative turned into integration!!! Unexpectedly awesome.
Right?!
@@BriTheMathGuythis should be the case, because each component of e^x is based on a reciprocal of a factorial.
The Gamma function is simply amazing.
I know right?!
No
@@David-km2ie based
I love it too much.
θ
What else can you "not" take the derivative of?
How about differentiating super cube root of x?
Can this be integrated?
(Φ^(x))-(-Φ)^(-x))/√5
Where Φ is the golden number.
Me
x/0
HMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
What's the inverse of the factorial?
Is there a function: f(120) = 5 --> Or generally: f(x) = y where y! = x
And how do you invert the gamma-function (in positive 'R')?
gamma doesnt have inverse
but the gamma function at 1 to infinity?
You cannot invert the factorial function, since it is not injective, nor is it surjective.
@@angelmendez-rivera351 Well, it does have a right inverse (on the domain of non-zero reals because fuck zero). But not a left inverse, which is the one you're likely referring to.
@@andrasfogarasi5014 The word "inverse" in mathematics typically refers to the two-sided inverse, unless otherwise specified. If you ignore 0, which is arbitrary, then it is surjective, but not injective, so it would have a right-inverse, but not a left-inverse, and therefore, no inverse. For it to have an inverse, or alternatively, to be invertible, it must have both a left-inverse and a right-inverse.
@@angelmendez-rivera351 it is bijective in its own domain and range, we want to define it only where x is actually a factorial of something
I am happy to see that your channel is getting bigger. Soon gonna be 100k, thanks for all videos, greetings.
Thank you very much!! Really appreciate the support!
Little precisation that none will care about: swapping the integral with the derivative requires some hyphothesis. For example the integrand must be continous with continous derivative(and here this is easily verified). Otherwise the step may be incorrect.
Good video!
you must find a function g such that for all x and t, |f(x,t)| < g(t) with g being integrable over the same domain as f with respect to t.
Thanks for pointing this out! I tend not to be totally rigorous on TH-cam :)
@@BriTheMathGuy As someone else pointed out in another comment, the purpose of your video is not to focus on such tedious details.
@@clementboutaric3952 I know that it's not the purpouse of the video, I just thought it would be interesting to provide insights on why this actually works. I think that tediousness is a relative concept, for me they are not tedious details.
@@francesco1777 Well once you know that one can not swap however one wants, you can't go back. These details become everything but details.
I actually don’t care about derivating the factorial function, but I knew this was going to be about the gamma function and the demonstration of why it’s a good candidate to extending the factorial to |R really was impressive
Very interesting (something like that) :
With sum of inverse factorial you get e
With half factorial squared you can get pi
Thats very very interesting for me!!
How u get pi from half factorial squared?
@@paolomorseletto3030 pi=4*(1/2)!^2
I will advertise this channel as much as possible. It helped motivate me to get through precalc and now discrete because I get to see the cool stuff I get to do later.
Wow thanks so much!
@@BriTheMathGuy so question, if you have time. Have you taken any interest in cryptology?
And we can also write this as a product of the digamma function and the gamma function, since Ψ(z+1)=Γ'(z+1)/Γ(z+1). Thus, d/dx(x!)=x!•Ψ⁰(x+1)
The derivatives of the gamma function become extremely self-referential and nested within each other, it's actually really cool! (Look up the digamma and polygamma functions)
Not that I understand it but sounds cool 😎
There are lots of continuous functions that match the factorial function on the integers. The (shifted) gamma function is certainly the most popular one, and indeed in some circles people do use the ! notation to refer to it, but calling this the derivative of the factorial function seems like a stretch.
pure clickbait
As a followup it might be nice to introduce the digamma function (or polygamma functions in general). One thing I particularly like is the derivative of the gamma function at integer points which involves the ubiquitous Euler-Mascheroni constant.
I still didn't take calculus or higher maths until now, but I still kinda understood. Your explanation is great! I like the idea.
For us statisticians this function is very important. We use this expression thousands of times throughout our careers, but I never thought of the derivative of this function. Very ilustrative, very impressive. Thank you.
Thanks for sharing and thanks for watching!
I swear two days ago i looked for video about derived of factorial and found nothing literally nothing so thank u very much for reading my mind
🤯 You're very welcome!
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You didn't look very well...
For a little intuition for how fast the factorial function is growing, it's Stirling's approximation to the rescue:
ln(x!) ~= x ln(x) - x
d(ln(x!))/dx ~= ln(x)
dx!/dx ~= x! ln(x)
U can substitute y=x!
y=x(x-1)(x-2)(x-3)....
Taking natural logarithm on both sides
lny=lnx(x-1)(x-2)(x-3).... ln(ab)=lna+lnb
lny=lnx+ln(x-1)+ln(x-2)+ln(x-3)......
Differentiating on both sides
(1/y)dy/dx=1/x +1/x-1 +1/x-2 +1/x-3.......
dy/dx=y(1/x +1/x-1 +1/x-2 +1/x-3...)
Substituting y we get
dy/dx=x!(1/x +1/x-1 +1/x-2 +1/x-3...)
I like the graphics but I realized watching this versus the vids where the formulas are written on the board by hand that this one seems like it’s slightly too fast paced, there’s not enough time to read and really think about the equations as they appear on screen. When they were written by hand it created a naturally slower pace to the script where something would be explained since it took a little more time to write things out.
I think the best of both worlds is keep the new graphics, they’re great, but intentionally speak a little more slowly and include some slight pauses to give the viewer time to digest what just appeared.
I totally agree. This was my first time trying a totally animated video, so I'm still getting the pacing down. I really appreciate the detailed and thoughtful critique! Thank you for watching!
03:50 Polynomial (or power function) ("t^x"), natural logarithm ("ln(t)") and expotential function ("e^(-t)") in one place (and all three multiuplied) - amazing!
Btw It is interesting what are the results of these questions:
1) d/dx( x^2 * ln(x) * e^x ) = ?
2) ʃ ( x^2 * ln(x) * e^x ) dx = ?
1') d/dx( x^2 * log.2(x) * 2^x ) = ?
2') ʃ ( x^2 * log.2(x) * 2^x ) dx = ?
Great approach to an interesting problem. The question is how to address the integral? Is there a follow up video? Thanks
I really approciate you explaining everything and not just saying _'trust me bro!'_ . Helps a lot understanding the video at the first watch. Subscribed!
I have shown this in 2017 to my theoretical physics prof and her assistent. They once asked if anyone could do the derivative of x!. I said yes and 1 thing is if x is discreet, we can do the log (f(x)) = log(x!) derivative , but in this case there are only integers llowed, and the other way for all x is working with the derivative!
Im Glad that someone else did the exact same thing like me! Keep it up friend ! You are awesome!
Hi, I've been working on a proof myself, and stumbled on to x!, having to differenciate it. I knew about the gamma function and it's derivatives, but I was wondering If whe restricted the x value to be an integer, would the derivative have a non-integral form? I couldn't found anything about it, so I thought maybe with complex analysys use the fact that Gamma is meromorphyc and then have a power series. But then again i trouble myself with calculating the n'th derivative in a certain point (integer) because i couldn't find a non-integer form of an n'th derivative of the gamma function on an integer
My brain at 3a.m. : let's differentiate factorials XD
Easy, x! Is also (x^1) * !
So using just your standard power rule, bring the power down, subract one, and then you get
1*(x^0)*!
Therefore, d/dx(x!) = !
gamma function can also be represented as infininite multiplacation, which is gamma(x)=(1/x) * (product from n=1 to infinity){(1+1/n)^x / (1+x/n)}.
since x! = gamma(x+1), x! = (1/(x+1)) * (prod. n=1 to inf){(1+1/n)^(x+1) / (1+(x+1)/n)}
Take natural logarithm both side and differentiate to get
(x!)'/x! = -1/(x+1) + (sigma n=1 to inf){ln(1+1/n) - 1/(n+x+1)} = lim(b->inf){ln(b+1) - (sigma n=0 to b){1/(n+x+1)}} = lim(b->inf){ln(b+1) - (sigma n=1 to b+1){1/(n+x)}}
= lim(b->inf){ln(b+1) - (sigma n=1 to b+1){1/n - 1/n + 1/(n+x)} = lim(b->inf){(ln(b+1) - (sigma n=1 to b+1){1/n}) + (sigma n=1 to b+1){1/n - 1/n+x}} = -γ + H_x
,where γ=0.5772... is euler mascheroni constant (also being a derivative of x! at x=0 times -1), and H_x is harmonic numbers (H_n is 1+1/2+...+1/n for natural n, generalized to real numbers)
Thus, (x!)' = x! * (-γ + H_x).
For example, for f(x)=x!, f'(3) = 3! * (-0.5772 + 1 + 1/2 + 1/3) = 7.5367...
3:19 we can slip the derivative inside in this specific case using a theorem but it's not always true unfortunately
This kind of video always overlook the most difficult and tedious part : proving that the function is defined on R+*, proving that you can make the integration by part, proving that you can derivate inside the integral.
Yes, and that is intentional. These videos are not meant to be rigorous. If you want rigor, then you should just watch university channel videos instead.
@@angelmendez-rivera351 well you're right.
1) The integrand is non-negative so the integral exists, and is finite because "the exponential function with minus in the exponent decreases much faster than any polynomial" (some criterion about convergence of integrals around plus infinity)
2) Do people still check assumptions for per-partes? I've done it maybe once in my lifetime.
3) A bit tricky, but the integrand is non-negative, so the absolute value of the integrand is the integrand itself, and the integral itself is finite (see point 1) so you can interchange, basically. There are a few more assumptions (like the derivative of the integrand exists almost surely) but these seem trivial since we deal with polynomials and an exponential.
Nice video! Now we need to integrate the factorial (if possible)
On what bounds? It would be a double integral for X and t.
By taking logs of both sides of y = x! and differentiating we find
y’ = x! (1/x + 1/(x-1) + 1/(x-2) + … )
= x! ψ(x) where ψ(x) is the digamma function, the differential of ln(Γ(x+1))
And someone who’s currently taking ap calc ab right now. I can’t comprehend any of this
its the gamma function Gamma(z+1)=z! and d/dz Gamma(z) = Gamma(z)Digamma(z) so d/dz z! is technically Gamma(z+1)Digamma(z+1) where Digamma(z)=d/dz(ln(Gamma(z)))=-gamma-sum(n=0,inf)(1/(n+1)-1/(x+n)) where gamma is the Euler-Mascheroni constant
I wonder if there are other functions f with this "factorial feature", that f(x) = x*f(x-1), but with other variations. Like you have to shift the Gamma function by -1, that it matches the original factorial values.
There are
@@andrewgrebenisan6141 Do they have a certain name?
Look at the Pi function. It is defined so that it matches the factorial exactly. In other words, Π(x) = Γ(x + 1).
There are infinitely many continuous functions which can fit to the discrete factorial function. The gamma function is one of them, but what this video lacks is why the gamma function is the best choice.
Good point. The answer would be that it's meromorphic. There is only one such meromorphic function.
yes
See the Bohr-Mollerup theorem.
Gamma function are the unique function which respects.
f(1) = 1
f(x+1) = x*f(x) , for all x>0
f(x) are log-convex for all x>0, which means that the second derivative of logarithm of f(x) are always positive.
One can use Stirling's Approximation as well. X! ≈ [√(2πX)] • (X/e)^X
The answer is way too long and complicated to write here. You might try it once.
My answer is (it is gigantic and it definitely could be wrong) :
(-√(2π) xˣ√x)/eˣ + (√(2π)xˣ√x(2xlnx + 2x + 1))/2xeˣ
It's terrific, thanx, I'll show it to my sonny for the scientific high school math exam preparation
Hi Bri. You have "formally solved" the problem, but I think quite obviously, you have left the hard part, i.e. the integral, unsolved. To use this "solution" in, say, an engineering model, one would actually have to solve it by some method, albeit perhaps approximation. [This is one of the many frustrating aspects of mathematics education.] The video is quite helpful, but I assert, incomplete.
Is the Gamma function the only function that satisfies Gamma(x+1) = x*Gamma(x) and Gamma(1) = 1? Or is there a whole group of functions that satisfy this?
now time to figure out how to take the partial derivative of single variable function
Well, if you had something like f(x)=2x-5 in the single-variable case, we could make this a multivariable equation, instead, if we wanted to (i.e., f(x,y)=2x-5) (This is now a function in terms of both x and y). Then, taking the first-order partial derivatives of this function, we get that ∂f/∂x=2 (The derivative of 2x (with respect to x) is just 2, and the -5 is a constant, so its derivative with respect to anything is just equal to zero (0)) and ∂f/∂y=0 (Here, since we are differentiating with respect to y, we treat x like a constant, but since there are no y's in this function, the whole 2x-5 is treated like a constant, so the partial derivative of this function (2x-5) with respect to y is just equal to zero (0)). We could even go further and make this a function of 3 (i.e., f(x,y,z)=2x-5, in the case of the function being a function of 3 variables) or more variables, if we wanted to, and the first-order partial derivatives with respect to those other variables would also be equal to zero (0), for the exact same reasoning as when we took the (first-order) partial derivative of the above function with respect to y (So, if we had f(x,y,z)=2x-5, then ∂f/∂z=0, etc.).
@@herbcruz4697
f(x, y) = 2x - 5
f(x) = 2x - 5
f(x, y) = f(x)
wait what? did I do something wrong here? how can a multivariable func be single variable at the same time?
@@mastershooter64 It just reduces down to the single-variable case.
The partial derivative of a single variable function is well-defined. We just call it the ordinary derivative, though.
Derivative of the W-Lambert funktion W(x)
how did you turn a derivative of a factorial to an integral?! This is amazing!
I was expecting some pi(-ish) thing from this derivative but it isn't. But, as always nice video
Glad you liked it!
U could have used the expression of 1 over gamma function where the euler-mascheroni constant shows up and derive from it the expression of the digamma function . Hence derivkng the formula of the derivative of x factorial ( WHAT THE FACTORIAL)
What values of x could you find a value for the indefinite integral at the end? The weird power log exponential product looks like finding an exact answer would be annoying
I've got some news for you. The problem why diescrete funcions like the factorial have no derivative is not about continouity. In fact, every function defined on integers is continous. You can just check the definition. The problem is: the function is not defines on an open set, therefor, there is no good way a derivative can be defined
You can actually find the factorial derivative without defining gamma function, just use the limit definition of derivative and you'll find that if f(x) = x! then f'(x) = f(x)*(f'(0) + H_x)
Now to evaluate that horrible looking integral
so much so fast! I cant learn math like this. I had to pause so often, and even then... sheesh. Looks like I need to brush up on partial derivatives.
Going by the title alone: Seems like it would be fairly easy, since the gamma function (and thus the pi function) is defined as an integral. So just don't integrate it.
Is there a way to then simplify that integral at all?
There is not, not really
There are probably different representations but as far as I know that's as simply as it gets 😬
Here o was thinking he’d use the Stirling Approximation for factorials and then differentiate that
A little playing around with actually trying to calculate that integral would have been desirable. I'm not saying it's easy, but perhaps a little something could be done to get closer to something more known.
He asks at the end was that what u were expecting like yeah sure cuz i worked it out with my best friend and photomath lol
i was thinking about this thing since i was 13 years old.
THANKS
Fantastic can you please explain. CArdano
Method of solving
Cubic equation
Why not use the pi function?
How about differentiating super cube root of x?
You mean the inverse of x^x^x ? It can be done through using a formula for differentiating inverse functions, after you manage to find a derivative of x^x^x :-)
@@MrMatthewliver yes
Is there an inverse function to gamma(x)?
i know that there technically infinite inverses but i only care about roots for gamma(x) where x>1 so there is only 1 root in this function.
Q: So, what if you differentiate x! ?
A: It becomes worse
0:44 should 0! be one? Why is there a vertical asymptote?
My thought was that you may say (n/e)^n * sqrt(2пn) is equivalent to n! for big ns, so you may differentiate the latter
Derivative of a fractal curve.
Something like the Weierstrass function
Please explain why did you apply by parts integration method to the gamma function? Like why integrate it?
No. The goal is to prove that the gamma function satisfies the same relation as factorial
I'm fascinated by what this is don't understand any of it how it works or what it actually solves. But I am watching and again find it very interesting. Great video.
Just as an experiment, i tried doing integral of x! and apparently it's almost the exact same thing but the ln(t) appears in the denominator
Pls can you take indefinite integral of x! I need it
Looking for some help... Can anyone explain the answer to this question. "The local zoning ordinance requires that any property with impervious coverage that discharges more than 100,000 gallons of water in a 4-inch rainstorm must have a retention pond to handle the water. The proposed subject property will have 260,000 sf. of impervious surface. A cubic foot of water contains 7.48 gallons and retention ponds can hold no more that 500,000 gallons for safety reasons. There is no retention pond an the plan. What will this property discharge in a 4-inch rainstorm. Trying to use an HP C12 calculator... thank you.
I was waiting for this.
Hope it was worth the wait!
Yes it was.
I wish people would use the Pi function instead of the Gamma function for factorial since it needlessly over complicates things and the Pi function is a little tidier than the Gamma function
How about the derivative of x^(1/x^(1/x^(1/x ... up to infinity?
Try differentiating e^(phi*x) with respect to x over and over again. You'll get a pretty cool series
Are you sure
e^x is always a sum of factorial reciprocals
I hate that the gamma function is more popular than the pi function (not the 3.14 pi). The formula of the pi function looks nicer as it doesn't have an x-1 but just an x, and pi(x) = x! and not (x-1)! as us the case for gamma(x).
It's just nicer as an extension of the factorial function.
I understand why the Gamma function is x! (For discrete values) and as a result can be generalised to non-discrete values but how does one go about discovering the Gamma function?
why is the limit for x=r and r going to infinity from r^x * e^-r /x =0 ? for x=0 i understand but that i dont understand
Brilliant thinking🤔
Thanks! Have a great day!
Numeric differentiation was not considered so this is kind of half an answer if we want to consider factorials over integers and not a continuous function. Yes it uses numeric approximation methods, but it would be interesting if taking the approximation as the number of discrete terms considered goes to infinity... how far off would this method be from the continuous version? Honestly a very interesting idea IMO
good solution and i learned gamma function
I feel dead inside while you differentiate under integral sign without any reference of something like the dominated convergence theorem. Personally, it took me three days to show this dude!
good one
“Factorials _!_ “
Very great explanation!!
Thanks!
Needed to pause the video (several times) so my ears could catch their breath; remember, speed kills!!!
The derivative is: d/dx(x!) = Γ(x + 1) polygamma(0, x + 1).
1:46 where does the x come from
me: has an out loud realization about deriving factorials
my 5th grade friend: bruuuuh
But what is the use of this as factorial is for only natural numbers that meanse it is not a cuntinous fucntion in any domain then how come it's differential can be calculated 🤔
It was exactly what I thought it would be, that math degree came in handy
How do you solve that integral that comes from this derivative. I know I can change the variable t into W(u) but I don’t know how to integrate W. Where does the series for W lambert come from?
Wow, that was pretty easy), thanks
i was thinking that u r going to use product rule lol
😂😂😂
Soooo before starting the video I got the idea of using the gamma function... I gave up xD.
I also started thinking that we can use gamma function for this
@@LearnwithTejeshwar I failed to use it though. But ehh, I got halfway xD
for differentiate an integrating with different variable don't you need to use Leibniz integral rule ?
Easier than the expectations
Super interesting, a little too fast for me, I'll be watching over before I try it myself. Thanks for sharing!