derivative of the binomial coefficient

[ritamdutta]

who else was expecting gamma function?

[pablomartinezazcona]

As a physicist, if x and k were sufficiently big I would have applied Stirling's approximation on the factorial, in fact, in Statistical Mechanics it is done many times this way. The other option I was thinking about was using the derivative of the gamma function or something like that. Really cool video!

[gergodenes]

Okay, now prove that this is equivalent to the result you get by differentiating the expression using the gamma function definition of the factorial.

[blackpenredpen]

Please see my community post to participate for the “integrals for you” t shirt giftaway!

[jayapandey]

When you see the thumbnail of the video and your mind starts saying plz there be some Gamma and Pi.

[giovanni]

The derivative is actually x choose' k
Also check by integration xD

[paulhaso]

The energy you put into your videos is so amazing

[agrawalnidhu]

If we consider x choose k, 0<k<x; k, x € N U {0}, with x as the variable, to be a function, then it has discrete outputs in the form of whole no.s and hence, it is a discontinous function. If the derivative of a function is the slope of the tangent to its curve/graph at any point of it, then what will the slope of tangent be in the case of *discrete points*. (like in the case when we are differentiating x choose k) Isn't differentiation supposed to be done for continuous functions ?
@blackpenredpen pls reply

[robertcotton]

Final answer not on screen long enough to be seen
I am on a phone so can't go frame by frame to rewind or fast forward.
:(

[davidwright]

I saw at once that 'x choose k' is just a function of x and was plausibly differentiable. the rest I did not think of! thanks. Now I can bamboozle the innocent. If I can find any.

[petelok]

Wow I wouldn't even have imagined a derivative for n choose k. Clever stuff 😊

[ThePharphis]

I think you're overthinking the part at 2:20. It's simply one higher than "x-k" so you get "x-k+1".

[envjvru]

Product rule would be appropriate here, derive it for product of multiple functions: and you see a pattern where the derivative is a sum of the derivative of one function times the rest and derivative of the next function times the rest and so on. And the derivative of each (x-n) is just 1 so you quickly get the answer

[benjaminbrady]

Well, I haven't watched the video yet but the definition is just factorials and multiplication so for the derivative you could just use the digamma functions multiplied by the gamma function

[Nothing_serious]

How does that work? So a factorial function has a slope?

[JohnSmith-chsm]

I'm sure Wikipedia said this, but the concept doesn't make sense. You have to use the Gamma function, because the idea of "x choose k" is that both x and k are positive integers. A derivative of an integer requires a constant difference, not an infinitely small dt. You can only do that if you were to change the values to their gamma functions thus allowing for any x and any k.

[turtlellamacow]

It's important to note that the most common definition of the binomial coefficients only allows for integer x; it's discontinuous everywhere and thus not differentiable anywhere. This only holds true (as explained in the wikipedia article) if you first *define* the binomial coefficients as a polynomial and give meaning to noninteger x. So the first equality BPRP has written (the common definition) is not really true in this context.

[andresxj]

If differentiating is about small changes, limits or so, how can you differentiate a function treating it like it only could receive integer values?

[orenrosenman]

I don't understand why this would be true...
By writing x!=x(x-1)(x-2)(x-3)... you're assuming x is only an integer, making the only valid input of the function an integer - so you can't really evaluate the function's derivative, no?

[Guillaume_Paczek]

hey, I dont understand how can you do these calculs based on an expression only valid for integers? (I mean I know the gamma function and all exists, but how can you write x!= x(x-1)(x-2).. ? )