I Found Out How to Differentiate Factorials! 

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

Me too 😂

I know you...

Michael Penn?

Qué haces aquí :v

@@jimjim3979 yes!

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

Wow, thanks!

Right?!

​@@BriTheMathGuythis should be the case, because each component of e^x is based on a reciprocal of a factorial.

I know right?!

No

@@David-km2ie based

I love it too much.

θ

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

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

Thank you very much!! Really appreciate the support!

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 RU-vid :)

@@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.

How u get pi from half factorial squared?

@@paolomorseletto3030 pi=4*(1/2)!^2

Not that I understand it but sounds cool 😎

pure clickbait

🤯 You're very welcome!

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You didn't look very well...

Thanks for sharing and thanks for watching!

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!

On what bounds? It would be a double integral for X and t.

There is not, not really

There are probably different representations but as far as I know that's as simply as it gets 😬

Hope it was worth the wait!

Yes it was.

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.

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).

For example, for f(x)=x!, f'(3) = 3! * (-0.5772 + 1 + 1/2 + 1/3) = 7.5367...

Glad you liked it!

No. The goal is to prove that the gamma function satisfies the same relation as factorial

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

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.

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

Thanks! Have a great day!

Glad you think so!

@@BriTheMathGuy i am a jee advanced aspirant for 2021 💪🤟 the problems there are really unexpected , anythung can come , your videos help in develovping that perspective 🤟

@@mastery4667 Or gamma(x) + n*sin(m*x*pi), where n and m are any integers!

I don't know, but there's a thing called the pi function with the only difference being that the shift isn't present.

I don't know, but there's a thing called the pi function with the only difference being that the shift isn't present.

Are you sure