Lmao same

Lol..I'm lucky that I've just learnt factorials before watching this video

Ok Feynman

1!

Pro trick: watch series on Flixzone. Me and my gf have been using it for watching loads of movies lately.

lol

When my friend was teaching me calc, I drew a line through the 2 ds and they just walked away.

or just do: dx!/dx = dx(x-1)!/dx = (x-1)! :^)

No--just write out the factorial in expanded form , multiply all factors and differentiate the resultant polynomial !!!! OOPS ! It's an infinite polynomial . I guess that method is out ? But maybe you could generalize the infinite polynomial somehow ?

This can technically be right, if you let a variable called ! be the derivative of the function x!, or more precisely the limit as h approaches 0 of ((x+h)!-(x)!)/h.

What a weird way the german accent spells 'penis'.

infinity boi

*eats it against its own will* oughh óughh *suspicious cough*

Why did you have to say that? Aaaaaagh!

Euler Mascheroni

Dic Lus is my favorite

It was a reference to Blackpenredpen, I'm pretty sure! Another maths youtuber.

@@tiagobeaulieu1745 no, i'm pretty sure that the first who used this term is Dr.Peyam.

@@issam5201 yes Steve ( black pen red pen ) copied from dr peyam

Is it possible?

​@@lordlix6483 Well, we can differentiate it now, so we can at least try integration by parts.

x! Is just a polynomial in its compressed form..integrate it just the way you integrate any polynomial by expanding it

@@shashank30122000 x! Is absolutely NOT a polynomial, at all. Sure, it may look like x(x-1)(x-2)... but how are you supposed to integrate this term by term? You can't expand it out, you don't know how many terms you actually have. Also, it only holds for positive integer values of x, and integration has to be continuous, or have finitely many points of discontinuity. There are no closed forms for the factorial integral as Wolfram alpha will tell you. You could integrate the gamma function if you wanted to though, but this also has no actual closed form.

@@hOREP245 so that means that we can't really integrate x! ?

any MOXXERS?

lol

@CogitoErgoCogitoSum uhh...

Nah man, use product rule, (d/dx)*x! = ((d/dx)*x)*! + x*((d/dx)*!) = ! + x*((d/dx)*!). Now all that's left to do is find out what (d/dx)*! is.

Bruhhhhhh

Me right now

yeah buddy... also me

did someone just called me

Yeah... KZfaq algorithm is kinda weird.

Möbius? Is that you?

@@PapaFlammy69 how so?

He did not really "break up" the sum, it was just easier to show how you can put the 1/x into the sum that way. Actually splitting the sum into two sums and continuing to calculate would not be correct

@@PapaFlammy69 no it doesn't, you've wrote an equality between two infinite sums at some point in your demunstration! Though I understand why it works in your case, it's poorly justified in the video, and it gives a bad impression.

I agree, I almost died inside watching that. Of course "it works", since you can write it as the limit of a finite sum, do the simplification on the finite sum, mention that the last term of the finite sum goes to zero, hand-wave, ..., done.

Well if in your class you were dealing with functions f: R --> R than she is right in saying that the square root of a negative number doesn't exist. She's implicitly saying "it doesn't exist in this domain we're considering, which is not the Complex domain"

@@someoneuppingdudetechnical6320 It is very unlikely that she specified what the domain or codomain of the function were. Functional notation is not even taught correctly in schools, let alone the unique identification of functions by indication of domain and codomain. If you tell your teacher that f : R -> R with f = x |-> log(x) and g : C -> C with g = x |-> log(x) are different functions by definition, I guarantee that there is more than a 50% that the teacher will say this is wrong, and then try to silence you and berate you. It is more likely that host's teacher just actually does not understand complex numbers as well as she should. At the end of the day, mathematics teachers are teachers first, administrators second, and mathematicians third, so it is not uncommon for math teachers to be bad at mathematics that are not in their own expertise, especially when their own education was bad. This is why I have a gripe with mathematics education within institutions in general. It makes me angry, because mathematics teachers teach way too many incorrect things, and this is universal too, although North American schools often take an extra step at this.

@@angelmendez-rivera351 Wow that's extremely disappointing... math teachers saying a function mapping the reals to the reals and a function mapping complex numbers to complex numbers are the same... and berate student who says otherwise... damn... that's pretty disgusting...f**ked up...

@@PapaFlammy69 huh sad, you should do a vid about that anyways, or talking about the properties of gamma function if you haven't done it yet!

:D

Shashank Shekhar No, it is not. The degree is not fixed.

@@angelmendez-rivera351 so what ? It's still can be considered one. Can you explain your point in more detail?

Shashank Shekhar Polynomials, by definition, are of the form Sum[a(n)x^n] from n = a > -1 to n = b > a, where b is finite. However, if b depends on x, then it no longer is a polynomial, it becomes a super-exponential expression. Again, this is all just definition, so I do not see why you think it is wise to debate this.

Nah, sadly not ^^'

Why are you using the infinity emoji instead of the actual ∞ symbol

PERMANENT PAPER SIGN

It is not only smooth everywhere in R, 1/Γ(x) is what is known as an entire function, which means it is holomorphic everywhere in C, and equivalently, complex analytic everywhere in C. It is quite literally in the nicest class of transcendental functions that it can be in. The only nicer class of functions is the class of polynomial functions, due to their algebraic simplicity, but as far as analytic properties go, they are equally as nice. This puts 1/Γ(x) in the same class of functions as exp(x).

@@angelmendez-rivera351 I did not know that, thanks ! Makes me wonder what kind of analysis one does on the Poincaré sphere/complex projective line, and if there's an equivalent of entire functions in that space where Γ(x) is "entire" as well (since every element on the Poincaré sphere is invertible)

@@mathephilia The Poincaré sphere? Do you mean perhaps the Riemann sphere? If a function is entire in the Riemann sphere, then it is a constant function, and this is a consequence of Louiville's theorem in complex analysis. This is because the theorem states that any bounded entire function is constant.

@@angelmendez-rivera351 Oh yes of course, sorry, Riemann sphere. That's interesting. I wouldn't have expected the only entire functions on it to be constants. Is there some less restrictive notion than "entire function" for which you get classes of functions with some interesting properties on the Riemann sphere ? Or a gradation of such notions ? (I'm not much of a complex analyst yet personally, so sorry if this sounds trivial)

@@mathephilia To be honest, I have not really thought about that carefully.

@@PapaFlammy69 no problem. I can still enjoy it.

No, they are completely unrelated.

WUT... lol... It just slid right past me...

fmakofmako Uh... he mentioned this in his video

@@angelmendez-rivera351 Assuming you misread what I wrote: when I see the thumbnail I assume differentiating the gamma function. So I was surprised he went a different way and thought it was neat. If you did not misread what I wrote then I must have missed his discussion on the infinite series. Not sure what else you could be talking about.

fmakofmako He did differentiate the Gamma function. That infinite product is the Gamma function too. Anyway, I was not referring to that. You definitely missed the part in which he said you can only use linearity to differentiate the infinite sum if it converges uniformly. He also did say prior to this the multiplication rule for logarithms only works if the product converges.

@@angelmendez-rivera351 I will watch again. Just to be clear, he didnt differentiate the integral form of the gamma function.

It is papa's improvised poll box

wowww they should actually also teach the fun,unique and interesting stuff

You could, but the idea is to let x be a real number, not a natural number.

Gamma'(51) ≈ 1.1928375×10^65 Gamma(51)×ln(51-½) ≈ 1.1928326×10^65

Oops... ! I forgotten that Gamma is shifted by one! So ln(x+½) become ln(x-½) and my last example is true for 51 not 50 :v

Did you ever find an answer to this?

I second this Also Fourier series> Taylor series Come at me

nothing, x is not discrete^^