A functional equation from my favorite book.

[@marsgal42]

That the result involves logarithms is clear by inspection. Filling in the details is the hard part.

[@xwtek3505]

True. He even forgot considering negative numbers

[@heavysaur149]

@@xwtek3505 f is defined as a continuous function: f : (0, +inf) -> R So we do not have to consider negative numbers

[@robijnbruinsma4489]

This has an important application in statistical physics. The entropy S(W) of a system of particles depends on the number of accessible configurations W of the system. Any thermodynamical potential must be additive so the total entropy S12 of two uncorrelated systems 1 and 2 with W1, respectively W2, configurations must be the sum S1+S2. Since the number of configurations of the combined system equal W1*W2 it follows that S(W1*W2)=S(W1)+S(W). From thus it follows that S(W) = k ln W with k a constant, which is the Boltzmann constant.

[@Nolys-bk4kd]

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

[@stephenhamer8192]

I did this: First we have f(x) = f(x.1) = f(x) + f(1); whence f(1) = 0 Then for x=/=1, we have 0 = f(1) = f(x.1/x) = f(x) + f(1/x); whence f(1/x) = -f(x) Call this [*] f: x |-> 0 is an obvious solution. Suppose, then, that f(x) =/= 0 for x=x*. By [*], we may assume f(x*) > 0 By a simple induction argument f(x*^n) = n.f(x*); whence 1 < f(x*^n) = n.f(x*) for some n [Archimedian prop of the Reals] We also have f(1/x*) < 0 r(n) st lim r(n) = y) and use the continuity of f. Thus: 0 = f(1) = f(b^0) = f(b^(lim y - r(n)) = lim f[b^(y - r(n))] = lim f[b^y.b^(-r(n))] = lim { f(b^y) + f[b^(-r(n))] } = lim{ f(b^y) + f[1/b^r(n)] } = lim{ f(b^y) - f[(b^r(n)) } = f(b^y) - lim f[(b^r(n)) = f(b^y) - lim r(n) = f(b^y) - y, which is what we need

[@stephenhamer8192]

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

[@ealejandrochavez]

It is easy to show that f(x) = A(ln(x)) where A: R -> R is an additive function. That is, A satisfies A(x+y) = A(x) + A(y). It can be shown that there are lots of Lebesgue non-measurable additive functions whose graph {(x, A(x)) for x in R} is dense in R^2. but as soon as you put any mild condition such that the function is measurable in some range or continuous at a single point then A is linear.

[@heavysaur149]

If A is monotone then A is linear too ! Only adding a little detail can make it linear.

[@byronwatkins2565]

e^x > 1 for x > 0 so the domain of g should be all reals to include (0, 1) in the domain of f. This later fixes itself, in this case, when we compose e^ln(x) for x>0.

[@HagenvonEitzen]

2:58 You rather mean g: R -> R, i.e., the domain of g should be all of R, not just the positives.

[@psioniC_MS]

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.

[@JACabLab]

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

[@psioniC_MS]

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

[@JM-us3fr]

I think you need the domain of g to include negative in order to guarantee the range includes numbers in the interval (0,1)

[@MF-qp9yr]

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

[@TwoGrids]

Let's go! Was recently craving some functional equations :D

[@abrahammekonnen]

Great video, and no I don't think it's too hard, but it does take time to break down and understand everything. And really interesting result. This seems like an extremely useful and interesting tool to have.

[@tsawy6]

One fun little technique is to note that we can use the defining property to show that f(x^2)=2f(x), and repeating this process we get values for all natural powers. Then we can show, using the fact that f(xyz)=f(x)+f(yz)=f(x)+f(y)+f(z) that the same is true for all 1/n powers, and hence for all rationals. Then, continuity gives us all the reals for free. In essence we then have that f(1)=0, and all other values are determined by any other given value (because e^x is surjective on positives). In essence, all possible functions are scalings of one example, e.g the logarithm that we can trivially show works!

[@professoraltamir]

It has already been shown that there is a function f(x)= a.log x. But it would be necessary to show the Uniqueness of Solutions for the problem... I don't see that! How would we show uniqueness?

[@JamesLewis2]

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.)

[@Alchiiii]

I found it easier to show that the functional equation implies that f(e^r)=rf(e) for all real numbers r. The process of doing so is similar. First proving for naturals, extending to positive rationals, negative rationals. Then for any real there is a sequence of rationals that converges to it. Taking limit of f(e^qn) as n goes to infinity equals f(e^r) since continuity implies you can pull in the limit into the function. And likewise the limit equals rf(e) since the interior of the limit is qn*f(e) by the above rule proven for rationals. That extends the proof of f(e^r)=rf(e) from rationals to reals. Finally argue that for all x>0, f(x)=f(e^ln(x))=ln(x)f(e) which leaves f(e) as a free parameter which may be any real number.

[@igorvereshch944]

I was actually aware of this equation but not this way of solving it. The solution I know is to take the derivative of initial equation by x (result will also work for all possible x and y): y*f'(xy) = f'(x). Left side is dependent on y and right is not, so we can set x=1 and get y*f'(y) = const which gives us f(y) = A*ln(y) + B By putting this into initial eq you get A - any, B=0 and f(y) = A*ln(y) (this method also can be used to solve g(x+y) = g(x) + g(y) and similar eq for exponent f(x+y) = f(x)*f(y) yet this one is tricky this way)

[@momom6197]

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

[@decare696]

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)).

[@fitbit4518]

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

[@gennaroponsiglione1098]

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

[@yuitora8749]

@@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

[@JamesLewis2]

When you defined g, its natural domain is the entire set of reals, not just the positive reals; also, the proper term for a homomorphism between additive groups is an *additive* function, while it is a theorem (that you proved along the way) that all continuous additive functions are linear. (Discontinuous additive real functions can have graphs that are dense in the plane, as shown by Gelbaum & Olmsted.)

[@MF-qp9yr]

Nice video, but you should have mentioned around 3:18 that the equation you got is the famous Cauchy functional equation (and so in the remaining part of the video you're deriving again the fact that the only say continuous solutions to the Cauchy functional equation are the linear functions).

[@yuitora8749]

A lot of solutions in the comment require f being differentiable. Here is a way to prove it Fix x, and integrate the equation with respect to y from 1 to 2, then do a variable change u = xy, and you get f(x) = (F(2x) -F(x))/x + C where F is an antiderivative of f and thus F is differentiable. Thus f is differentiable. Edit: Yes I assumed f continuous because the question asks for continuous functions. And as someone stated in the replies, assuming axiom of choice there exists non-continuous functions satisfying the equation.

[@yannld9524]

@@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

[@whonyx6680]

​@@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.

[@yannld9524]

@@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.

[@yannld9524]

@@angelmendez-rivera351 what do you mean ?

[@LordoftheFleas]

How does F being differentiable imply f being differentiable?

[@abrahammekonnen]

One thing I do get stuck on in this problem is how would you know to try g(x) = f(exp(x)). The best I have is noticing that multiplication of the inputs leads to addition of the outputs. So you treat exponentiation as the 'multiplication' and multiplication as the 'addition'.

[@secmari8645]

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.

[@Vladimir_Pavlov]

f(xy)=f(x)+f(y) (1) Differentiate (1) separately by x and by y using the differentiation rule of a complex function. We obtain two equations (df(u)/du) * y= df(x)/dx and (df(u)/du) * x= df(y)/dy. where u=x*y. From these equations follows x*df(x)/dx=y*df(y)/dy . One part of the equality depends on x, the other on y. Since x and y are independent variables, this can only be if x*df(x)/dx= C , where C is an arbitrary constant. =>df(x)= C*dx/x => f(x)=C*ln|x|+C1. It follows from the original equation for x=y=1 that f(1) =0. So C1=0. Answer: f(x)= C*ln|x|.

[@TheEternalVortex42]

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

[@Vladimir_Pavlov]

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

[@damosdamianos8731]

Nice! I proved it another way: Set y=1+h/x in the functional equation, then remove the term f(x) from the rhd to the lhs devide by h and take the limit h-->0. Then the left hand side will be just the derivative f'(x) and in the rhs applying L' Hopital's rule to get f'(1)/x.Integrating the resulting differential equation we get f(x)=a ln|x|, a=f'(1).

[@yannld9524]

You can't apply L'Hopital's rule

[@damosdamianos8731]

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

[@whitecape21]

How did you prove that the derivative of f exist?

[@damosdamianos8731]

@@whitecape21 this is trivial

[@yannld9524]

@@damosdamianos8731 this is not trivial at all

[@joaoppdarocha]

I wish I was smart enough to enjoy your content

[@doctor_no_]

I sort of set x*y = a, where a is a constant and using f(x y) = f(x) + f(y) I got f(x) + f(a/x) = f(a). Setting a = 1 yields f(x) + f(1/x) = f(1), but f(1) = 0 as you can prove setting y = 1 in f(x y) = f(x) + f(y). Thus f(x) + f(1/x) = 0 and from there I just noticed that if f(x) = c ln(x), with c a free constant the whole thing works. Would that be a valid solution?

[@flikkie72]

This might be stupid, but i set x=1, y free and got f(y) = f(1) + f(y), of course there are more functions that can satisfy this, but f(x) = ln(x) is the first that comes to mind. Setting x=2, y free, results in f(2y) = f(2) + f(y), narrowing the options further down to a logarithmic function, i guess adding a multiplication factor (a) makes sense because it's a family of functions. Is that a good place to stop?

[@davidholle6569]

Immediately saw it was a log property, interesting to see how we can find all functions

[@carlosayam]

at uni, studying with Spivak was tough, but the depth of its contents is still unparalleled

[@italyball2166]

I don't get the step at 6:57. Isn't Michael supposed to multiply all the 1/n instead of adding them? Saying this because 1/n is the variable of g and g(x) + g(y) = g(xy) by definition

[@tensor131]

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

[@criskity]

Me looking at initial question: "Isn't that just log?"

[@tensor131]

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

[@nileshbharsawade3242]

The continuity get in the role very smoothly!!

[@dmytro_shum]

but how to prove that ONLY a logarithmic function is the answer? or I dont get something?

[@adamwho9801]

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.

[@patrickabney9995]

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.

[@dmytro_shum]

@Joji Joestar "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

[@dmytro_shum]

@Joji Joestar "if f is not a*lnx, f does not satisfy the hypotheses" you just wrote this in more short and universal way. The question I asked - where does this happened? (proof that ONLY a logarithmic function is the answer)

[@laurentpaget5388]

Michael started well with g(u)=f(e^x) and g(v)=f(e^y), then went a complex way, in my humble opinion. Noticing that g(u+v)=f(e^(x+y))=f((e^x)*(e^y))=f(e^x)+f(e^y)=g(u)+g(v) Thus, g(u+v)=g(u)+g(v) g(u) is therefore a linear function in the form of g(u)=au+b, with (a,b) two constants f(x.1)=f(x)+f(1) => f(1)=0 And, g(0)=f(e^0)=f(1)=0 Hence, g(u)=au, with b=0 Finally, f(e^x)=g(u)=g(x)=ax f(e^ln(x))=f(x)=a.ln(x) f(x)=a.ln(x), q.e.d.

[@remysmets]

“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.

[@malcolms1441]

Could someone help me with my solution? I started by taking x=1 -> f(1)=0 then x = y -> f(y^2) = 2f(y) and f(y^1) = 1*f(y) is trivial. Then using induction hypothesis f(y^n) = nf(y) you take x = y^n and thus f(y^(n+1)) = f(y^n) + f(y) = nf(y) + f(y) = (n+1)f(y) and from there it's easy to see that f is the logarithm. And since log b (a) = ln(a) / ln(b) the final answer is (a real number C) * ln(xy). But I don't think that using induction for only integers is enough of an argument to show that the function is indeed c*ln(xy) what am I missing?

[@samuelbosman9572]

I don't get the g(1/n) part. Could someone explain?

[@tensor131]

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?

[@sjswitzer1]

This all makes sense except the very first step. Why choose g(x) = f(e^x) to begin? I mean, it’s somehow based on recognizing at the outset that logarithms are the solution, but what exactly is the thought process here?

[@daburak2559]

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

[@hazalouldi7130]

how do you show that g(1)=a?it's the first a=f(e)

[@pharaohgarmar5611]

I wonder if Intuitionalism started as a result of a mathematician saying to his students that he would leave the extension from the rationals to the teals as a homework exercise.

[@ZiyadAllawi]

I’ve read this problem in that book. It was interesting proof of the natural logarithm…

[@johnalley8397]

I like the yellow arrowed comments added in post. very cool.

[@manucitomx]

Every time you use Spivak I’m brought 30-odd years back to college. Thank you, professor!

[@renyxadarox]

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

[@Seriouslyfunny1]

I'm an engineer. I'd have done it differently. I'd have taken partial derivatives of the equation, one equation of x derivative and the other of y derivative. Dividing one equation by the other would give me xf'(x) = yf'(y) A function at two arbitrary inputs has same value only when it is a constant. This gives me xf'(x) = constant c Integration gives me f(x) = c log(x) + C2. Now, f(1) = 0 means C2 = 0 So, f(x) = c log(x)

[@DeanCalhoun]

unrelated to the video, but seriously when will youtube add some way to properly format math notation. canvas can do it surely youtube can as well

[@sigmainclination9483]

Better to put f(x) as c log |x| as it can be easily shown the functional relation forces function to be even ( by putting y as -1 ) .Nice work Michael ❤️🙏

[@sujals7108]

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

[@nerdatmath]

Does this mean the constant function f(x) = 0 should be considered "some sort of logarithm"?

[@ifomichev]

I also noticed this solution was missing.

[@calvindang7291]

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

[@Tzizenorec]

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.

[@gibbogle]

At 5:04, g(k) = ak, not nk. This creates confusion.

[@tensor131]

yep you saw a typo!!

[@abrahammekonnen]

5:16 I think you mean g(k) = k*a because we are doing induction over n.

[@davorinmestric969]

In order to solve this problem, we will first magically bring in the solution at 1.11. (Or an inverse of the solution, as it happens here)

[@alang.2054]

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

[@VincentEngler]

Question about when we're proving the g(x)=ax for real num a (8:12): we get from proof that a can be any rational to a can be any real b/c g(x) is defined as continuous, but couldn't we have done the same after proving the natural numbers? Like, if I've proven that all naturals are valid for the claim, then can't I just jump straight to all Reals b/c it's continuous?

[@williamtaylor8919]

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.

[@VincentEngler]

@@williamtaylor8919 Thank you!

[@antoinedubedout5963]

Sorry guys (and Michael), the function f(x)=a.ln(x) cannot be a solution because it is not defined for x=0

[@minwithoutintroduction]

البرهان رائع.واصل

[@rachidmsmdi6433]

nice idea to put f(x)=g(ex). thanks

[@ingiford175]

Checked out your book list on amazon. I did not see anything that might relate to your Discrete Calculus video six months ago. What would be a good reference book that you would recommend on that topic?

[@MichaelPennMath]

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...

[@ingiford175]

@@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.

[@mycrust]

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

[@martinepstein9826]

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

[@ingiford175]

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

[@mrmaabs5013]

couldnt we just say a^x rather than e^x?

[@tensor131]

NICE

[@xwtek3505]

You forgot to check for the case when x is negative or zero

[@JalebJay]

Interesting sybermath did a similar video today in the reverse direction

[@goodplacetostop2973]

11:10 Whoever reading this, don’t forget to smile today. Life is too short to be unhappy. ☀️

[@ostavaites_zdorovimi.]

f(x)=lnx

[@kareolaussen819]

2:55

[@Creativemathlearning]

Đây là phương trình hàm cauchy nhé bạn.

[@Omar-hm6pu]

Homomorphism from (R*+ , . ) in (R,+)

[@moonlightcocktail]

What if we allowed f to be noncontinuous?

[@ondrejdarmovzal4976]

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

[@SlipperyTeeth]

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.

[@TheEternalVortex42]

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).

[@moonlightcocktail]

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

[@adamnevraumont4027]

@@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.

[@naffouri]

How are we sure of the uniqueness of the solution g(x)=ax? Couldn't there possibly be abother solution satisfying g(x+y)=g(x)+g(y)?

[@timstigation9620]

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.

[@clementeromano5691]

This is another solution Let f be a solution, let y be a positive real number that's different then 1. f(y^n) = nf(y) easily by induction, this works for all n in Z. Let m,n in Z with n not being = 0, then we have that n * f(y^(m/n) ) = f( (y^(m/n)) ^ n) = f(y ^ m) = m * f(y) Dividing by n we get that f(y^(m/n)) = (m/n) * f(y) So, if p is in Q then f(y^p) = p * f(y) By the continuity of f and the density of Q we easily get that f(y^a) = a * f(y) Now let a = log_y(x), we then get f(x) = log_y(x) * f(y) Because log_y(x) = ln(x) / ln(y) we then get that f(x) = C * ln(x) With C being a real number So if f is a solution of the equation then is of the form C * ln(x), conversely every function of the form C * ln(x) is a solution so those are the only solutions of the equations

[@thatkindcoder7510]

What topics do the textbook not cover lol

[@stephenbeck7222]

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.

[@popodori]

"when a function has the properties of log, then the function is a log". wow , just wow

[@snillie]

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

[@theVtuberCh]

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

[@tensor131]

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.

[@snillie]

@@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

[@romajimamulo]

Hey, G(x) has a range of all real numbers, not just the positive ones. This is because e to a negative number is still positive So you need to make your proof about g a bit wider

[@MichaelGrantPhD]

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

[@romajimamulo]

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

[@MichaelGrantPhD]

@@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

[@romajimamulo]

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

[@MichaelGrantPhD]

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

[@ujwalsmanhas1093]

At first glance it is a log function

[@tensor131]

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

[@foxinio4225]

Jesus how much ads can you put in one video, I've already watched like 6 and to hasn't even been 4 min.

[@theimmux3034]

once again the chad f(x) = 0 makes an appearance

[@whonyx6680]

Fun fact, the proof you showed recently about e being transcendental is the same proof that is used in Spivak's book.

[@Alex_Deam]

He says that in that video

[@ingiford175]

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.

[@whonyx6680]

@@Alex_Deam Must have missed it.

[@adamnevraumont4027]

YP!

[@antonuitert5608]

This is great! I wonder how it is obvious that the solution gives ALL functions that satisfy the condition.

[@JamesLewis2]

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.)

[@RedRad1990]

TL;DR: They are what we thought they were.

[@qdrtytre]

The squiggles on the screen bother me.

[@erfanmohagheghian707]

One of your worst videos to be honest (no doubt in your math knowledge, that being said). First look, log(x) of any base is a solution. I expect a proof for the uniqueness

[@amirb715]

oh man, this is way too long...first easily show that f(x/y)=f(x)-f(y) (because f(1)=0) then let y=x+\delta x and divide both sides with \deltax and let \deltax approach zero. Using a taylor series on the right for f(1+\deltax / x) knowing that f(1)=0, you easily end up with f'(x)=k/x so f(x)=k log(x)

[@melainekerfaou8418]

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

[@Liridiona]

69 views nice

[@mokouf3]

After finding out g(x+y) = g(x)+g(y), I will go this route. Assume y is any real constant, g(y) is also real constant g'(x+y) = g'(x) = C for all x, y integrate back g(x) = C*x + K g(x+y) = C*(x+y) + K put it back to original equation C*(x+y) + K = C*x + K + C*y + K K = 0 g(x) = C*x