The Chen Lu Equation

Here I solve an interesting question: For which f does the fake chen lu hold? In other words, for which f do we have (fog)'(x) = f'(x)g'(x) for all x. Watch this video to find out!

[persiantheory]

not sure if you're Iranian but your videos make me ( as a persian student i am) have some hope to become someone like you in the future

[cassandramonge]

I love the way you teach this beautiful branch of science.

[joshuakaufman]

Can you answer the same question, but for finding g(x) rather than f(x)? Obviously, g(x) = x is a solution, but I’m curious if there are any other solutions.

[armando-rpg]

Unrelated: Dr Peyam, do you think you could do a video on Implicit Function Theorem//Inverse Function Theorem?

[orenfivel]

I think you love the Back propagation in Deep Learning & Neural Network, cuz there are a lot of Chen Lu

[bertrandspuzzle]

Wow. I was watching this and thinking "this reminds me of the f(2x)=f(x) video" and welp. Lol

[ScottBlomquist]

It's "chain rule", isn't it?

[curtiswfranks]

1:05

Do not cancel the g'; just move everything to one side and factor. ℝ has no zero-divisors, so either ∃x: g'(x) = 0, or our thing of interest holds. If g' ≠ 0 uniformly (as a function), then we get our result, for we do not care about pointwise exceptions; else, this result holds ∀f because the composition of any function with a constant function is a constant function, so the lhs = 0 just as the rhs does.

[xyzain_]

Nice! The math didn't go the way I thought it would haha

[divyaraval]

I can't pass day without your video (math)
:0

[nuklearboysymbiote]

if f(x) = ln(x), f'(kx) = f'(x) = 1/x

[igxniisan]

fog(x) = ln(x) + x

Guess the functions f(x) and g(x).















Hint: W(g(x)) = x

[txikitofandango]

But (ln(2x))' = (ln(x))' ... ?

[scienceandnature]

a= golden ratio (φ)= (1+√5)/2, b= golden ratio conjugate (Φ)= ( √5 - 1 )/2 ,
My new formula's

We know,
beta [ m, n ] =[ Γ(m)Γ(n) ]/ [ Γ(m+n) ]

1) Beta [ b, 1] = [ Γ(b)Γ(1) ]/ [ Γ(1+b) ] = Γ(b)/Γ(a)
= [ 2 Γ(2b) ]/ [ Γ(a+b) ] = [ a Γ(2b+1) ]/ [ Γ(a+b)] = a

2) Beta [ a, 1] = [ Γ(a)Γ(1) ]/ [ Γ(1+a) ] = Γ(a)/ Γ(a^(2))
= [ -2 Γ(-2b) ]/ [ Γ(-a-b) ] = [ b Γ(1-2a) ]/ [ Γ(-a-b)]
= b

3) (a)! = a * ( (b)! )
4) (b)! = b * ( (a)! )

[ChefSalad]

Have you heard of converting functions? No, you haven't because I invented it (probably not really, but it is my name for the concept). Let's say you have some function like sin(ax). Find a function f(x), such that f(sin(ax))=sin(cx), where f need not be invertible. Then f(x) is a converting function because it converts some kind of function to another. Or more generally, find some f(x) such that f(g(ax+b))=h(cx+d), where g(x) and h(x) are fixed functions and f(x) need not be invertible (although it probably is). There is an answer and it's pretty easy. I'll explain below.


For f(sin(ax))=sin(cx). First, let y=sin(ax), then x=arcsin(y)/a. Make the substitution on both sides: f(y)=sin(c/a*arcsin(y)). For the general case, if f(g(ax+b))=h(cx+d), the f(y)=h(c/a*(g^-1(y)-b)+d). Perhaps even more generally, for f(g(x))=h(x), f(y)=h(g^-1(y)), but that makes it look too obvious what's going on, so it's less fun.