A functional equation from my favorite book. 

True. He even forgot considering negative numbers

Absolutely! I was amazed to learn that Boltzmann's definition of the entropy was based in such an intuitive and mathematically pleasing way

In the third line, I meant x =/= 0

Does this have any effect on the proof, though? If I'm not missing something, the proof works even if we use g: (0, ∞) -> R.

@@psioniC_MS in 9:11 he uses that g is defined in ln(x), so if x

@@JACabLab Ah, now I see it. Thank you!

Yes, g is perfectly well defined on whole IR, so we can view g as a function of type IR -> IR, no problem.

@@angelmendez-rivera351 yes but this is another problem, and in my opinion it is not interesting because we can not construct such functions without this axiom

​@@angelmendez-rivera351 Unlike assuming that f is differentiable, assuming that f is continuous is perfectly fine since, in the first place, the question is asking about continuous functions that solve that particular functional equation.

@@angelmendez-rivera351 Okay I see your point now, that was not clear for me with your first message. Just to clarify, I didn't say that the axiom of choice is false, I use it too sometimes. What I wanted to say is that this axiom implies the existence of such functions, or more precisely it gives a "method" to construct these functions, but it doesn't provide any other information. In particular we can not evaluate these functions at a random point and I think we don't even know if this construction gives all such functions, in other word if there exists a function satisfying those properties which can not be constructed with this method... For a mathematician this is clearly unsatisfying.

@@angelmendez-rivera351 what do you mean ?

How does F being differentiable imply f being differentiable?

The solutions are not unique: There is a free parameter, which is equivalently the coefficient of the natural logarithm or (for a≠0) the base of the logarithm. For a≠0, a*ln(x) is the logarithm of x to the base exp(1/a). (Taking the limit as a→0+ shows that in a sense, the zero function is like a base-infinity logarithm.)

It's nice, but you'd have to prove that the function has a derivative first.

which is actually not that difficult. just look at (f(x + dx) - f(x))/dx. This can be transformed to f((1 + dx/x)^dx) and the inner limit as dx goes to 0 can be substituted with N = 1/dx to get lim(N->infinity) (1 + (1/x)/N)^N = exp(1/x). Therefore we get f'(x) = f(exp(1/x)).

Best solution. It is much better than taking a guess that g(x) = a*x is the only solution.

@@decare696 how can you transformate (f(x+dx) - f(x)) /dx into f((1+dx/x)^dx)?

@@gennaroponsiglione1098 He actually meant f((1+dx/x)^(1/dx)), here's how to prove it (not that simple tho): f(x+dx) - f(x) = f(1+ dx/x) because f(xy) = f(x)+f(y) Then we need to prove for all a in R, af(x) = f(x^a) To do so you do like Micheal Penn, start proving with a in N, then a in Z then a in Q so you get for all a in Q, af(x) = f(x^a) Then by density of Q, for all a in R, af(x) = f(x^a) Then f(1+ dx/x)/dx = f((1+ dx/x)^1/dx) So this method is not that simpler than Micheal Penn's method either

yes I am quite sure everyone saw that. The proof that ONLY log functions satisfy this is the point of MP's vid

btw, "Spivak" means "Singer" in Ukrainian.

It's because f(x+y)=f(x)+f(y) is a famous functional equation you should know! It's called Cauchy's functional equation, so basically he knew the solution of f(x+y)=f(x)+f(y) before he started the problem, he saw that they were very similiar ant tried to connect them (successfully). Knowing tons of diffent problems is a powerful tool in math.

The domain is positive reals so it can't be an even function

Yeah, this Olympic useless "math" that depends on 95% knowledge, and similar problems solved is not that fun as I thought

You can't apply L'Hopital's rule

@@yannld9524 why's that? as h-->0 you have f(1) which is 0 over 0

How did you prove that the derivative of f exist?

@@whitecape21 this is trivial

@@damosdamianos8731 this is not trivial at all

The goal was to determine all *continuous* functions, so f is not necessarily differentiable.

@@TheEternalVortex42 Please show me your way of solving this equation.

This is a good question. I am not really sure what a good reference is. Most of what I learned was extrapolated from some notes I found online from a student taking notes in a course. I quickly tried and find these again and failed...

@@MichaelPennMath I really like this sort of math, ranging from Diophantine (especally pythagoras in rationals) and Continued Fractions. So your Discrete Calculus discussion itches the same itch as the other examples.

@@MichaelPennMath How about Concrete Mathematics by Graham, Knuth, and Patashnik

Chapter 1 of 'A Primer of Analytic Number Theory' by Stopple covers discrete calculus.

@@mycrust Section 2.6 I see them doing finite calculus.

g(x+y)=g(x)+g(y)

he got to g(x+y)=g(x)=g(y), so g(2/n)=g(1/n)+g(1/n) = 2g(1/n) and so on. Does that help?

There is two steps to this First one is that f(x+y)=f(x)+f(y) functional equation is well known called Cauchys equation and it would be good if we somehow go to that point Second point is f(x+y)=f(x)+f(y) looks just like our normal log identity. So we have to prove that only log is the possible function. So it makes sense that we define a function like that, and plus while that is exponantial we get f(x+y)=f(x)+f(y) which we want

“g(u) is therefore a linear function in the form of “g(u)=au+b”. Well yes but you have to prove this although it seems obvious. Maybe there are some weird functions which are linear but do not satisfy g(u)=au+b.

yep you saw a typo!!

I also noticed this solution was missing.

@@ifomichev It's there. (a=0)

f(x) = 0 is some sort of _everything,_ in a trivial sense, because you can get to it by multiplying any other function by 0.

I don't see it either. And stating that a function that maps to the Reals only means that it doesn't have complex values... that they it is required to map to all the reals.

He assumed we had f that satisfies the hypotheses and then showed f(x) = a*lnx. The contrapositive: if f is not a*lnx, f does not satisfy the hypotheses.

This showed existence of a family of functions satisfying the hypothesis. I was waiting for the bit of work to prove uniqueness. Still waiting Also, jumping straight to the answer without showing that log is a fanatic candidate and the real question is: is log the only class of functions for which this is true.

@@hybmnzz2658 "if f is not a*lnx, f does not satisfy the hypotheses" where does it happen? can you show me the time codes? because I think there is no such proof

@@user-qo3qm7ud1d A => B is equivalent to not B => not A A result in logic

It's not enough to know that g is continuous and g(x) = ax for x in the naturals, since the natural numbers are not dense in R. For instance, the function h(x) = ax * cos(x/(2 pi)) is continuous and h(x) = ax on the naturals, but h is certainly not linear if a is nonzero.

@@williamtaylor8919 Thank you!

NICE

I thought it was obvious that the proof showed that "if property then logarithm (or zero)" meaning that it does give all functions that satisfy the condition; the only addendum to make is that all logarithms (and the zero function) actually do satisfy the condition, and it's obvious that they do, without needing that mathematical detour: First, if f satisfies the property, then the composition of f with the natural exponential is a continuous additive function. Next, by induction, for any integer n and additive function g, g(n)=n*g(1). Then by simple manipulations, for any rational number n/m, g(n/m)=(n/m)*g(1). If g is additionally continuous (as f of exp is), then for any real number r, g(r)=lim(g(x),x,r)=lim(x*g(1),x,r)=r*g(1), where the limit is taken through a sequence of rationals (which can always be done, because the rationals are dense in the reals). In other words, if g is continuous and additive on the reals, then it is of the form g(x)=a*x, where a=g(1); the converse is obvious, so the continuous additive real functions are precisely the real functions defined by multiplication by a specific real number. Because f(exp(x)) is a continuous additive function on the reals, it has the form a*x, so f(x)=f(exp(ln(x)))=a*ln(x), for that same value of a; this means that all continuous homomorphisms from the multiplicative positive reals to the additive group of the reals have the form f(x)=a*ln(x), where a=f(e). More than that, either a=0 or a has a multiplicative inverse, in which case the change-of-base formula says that f(x) is the logarithm to the base exp(1/a). This means that all continuous homomorphisms from the multiplicative positive reals to the additive group of the reals are either real-based logarithms or the zero function; the converse is obvious, so the set of those homomorphisms is precisely the set of real-based logarithms, together with the zero function. (Also, because the limit of exp(1/a) as a→0+ is +∞, the zero function is, in the limit, a base-infinity logarithm.)

Unfortunately not; letting x or y be 0 shows g(0) = 0, and using induction shows that g(n) = ng(1) for all natural numbers n. g(1+ -1) = 0 = g(1) + g(-1), so g(-1) = -g(1). I won't go on forever, but you can see how you can prove g(x) = g(1)x similarly to how Mike proved his claim, but including negative real numbers this time. I think it's a neat result honestly.

i think it’s interesting that logs are the *only* (continuous) functions that satisfy these properties, and there aren’t any other weird functions that also happen to satisfy them. i think that’s the point of this exercise :p

@@snillie the constant function f(x) =0 also satisfies this constraint.

I think you're missing the point. How do we know that there are not arcane loglike functions buried deep in the mathematical landscape that satisfy the same functional equation? MP proves that this is not the case.

@@theVtuberCh haha, fair. maybe that helps make the point even clearer, actually. there ARE non-logarithmic continuous functions that satisfy the given functional equation. however, we know that all solutions have to look like f(x) = aln(x) for some a, and the only value of a for which this isn't logarithmic is a = 0 (unless we consider the case of f(x) = 0 to be some kind of degenerate logarithm function?) in any case, it still proves that there aren't other strange functions also satisfying the equation

He says that in that video

I picked up that book when he did the derivative of the cosine. I did a first reading of the book since then. Will circle around for a second reading. This book does have a lot of cute 'alternative' methods of proving or linking things which helps build a better foundation of understanding in my opinion.

@@Alex_Deam Must have missed it.

Then you would get lots of pathological, wild and highly discontinuous solutions. Look for Cauchy functional equation for references.

For the linear functions g(x), you can do the exact same arguments for n*y in place of n*1, and y/n in place of 1/n, and m*y/n in place of m*1/n for any number y, and in particular any irrational number y, to get g(x*y)=g(y)x for rational x and arbitrary y. What this means, is that if we take the real numbers as a vector space over the rationals and pick a basis b_i, then we can write g as a diagonal matrix with real entries. This is weird - first because the basis is uncountable which is always trippy - and then even though we're "considering" it as a vector space over the rationals to get the basis, the matrix itself could have real entries instead of just rational entries. Continuity forces all of the diagonal entries of that matrix to be equal, and then there's no need to decompose our number into a basis over the rationals, and we can instead describe g as a one-dimensional matrix times x. I haven't thought about the log functions f(x) specifically, but I imagine they'd be similar with some quirk. If I get the chance, I'll describe them later.

The weird thing about this is it's impossible to explicitly construct such a function, although infinitely many exist! (of course, assuming Axiom of Choice).

@@TheEternalVortex42 even if we can't find such a function, would it be correct to say the graph would have self-similarity?

@@TheEternalVortex42 this is why I find the constructive reals fun. The seams in these crazy functions are too fine to "see", so don't exist in real numbers without strict equality.

My guess is that the book introduces logarithms as the integral of 1/x and uses integration properties to develop logarithm properties and then does some abstractions like in this problem to develop theorems about any functions that have similar properties.

edit ... At first glance a log function is evidently PART OF THE SOLUTION SET

Agreed, but I think that's the only adjustment needed

@@MichaelGrantPhD yeah, you'll have to add a part where you deal with negative numbers too

@@romajimamulo I don't think the rest of the proof needs to be updated. Just say g(x) = f(e^x) is defined on all of R instead of just (0,inf) and the rest of his steps don't change

@@MichaelGrantPhD the way he proved g(x)=ax didn't account for x being 0 or negative though

@@romajimamulo ha ha, got ya. I skipped that part of the proof because I felt it was artificially complex :-)

He did not assume that f was differentiable. All we know is f is continuous. That's why the proof is circonvoluted.