f(x+y) = f(x)f(y)

In this *improvised* video, I show that if is a function such that f(x+y) = f(x)f(y) and f'(0) exists, then f must either be e^(cx) or the zero function. It's amazing how we can derive all that just from such small assumptions!

This is a must-see for calculus and analysis and algebra aficionados! Enjoy!

[drpeyam]

f(x) = 1 corresponds to the case C = 0
f(x) = 2^x corresponds to C = ln(2)

[Tomaplen]

Dr. Peyam semi-nervous: Lets ASSUME f is differentiable at 0


me as a physicist: pfft, wasn't that obvious because why not?

[blackpenredpen]

“Today I want to do something cool, as

Dr. P.





No, I mean. Chen Lu.

[EpicMathTime]

A huge amount of functions satisfy these types of relations if we don't require continuity (or some other property that prevents pathological functions). However, their construction in many cases requires the axiom of choice and so they cannot be explicitly given, and in any case, there's no reason to believe that they could be expressed in terms of elementary functions.

The point I'm trying to make is that we need to remember that a "function" is a _very_ general object. We work so much with functions represented by equations that we begin to think that the function _is_ the equation, but it's not. By restricting to continuous (or to differentiable) functions, the question is more interesting. Without doing that, it's just an unfathomably huge class of functions that probably can't be described in any kind of nice way.

[mitchkovacs]

This is exactly what I've been studying recently! Specifically I looked into non-logarithm solutions of f(xy) = f(x) + f(y). Let D be the set of all positive rationals raised to a rational exponent. Every element of this set can be uniquely written as ∏ (p_k)^(e_k) for p_k is the kth prime and e_k is a rational exponent.
For q in D, define f:D -> Q by f(q) = Σ (e_k). This function is everywhere discontinuous, but it satisfies f(xy) = f(x) + f(y)

[cookieshade]

You can actually start with much weaker assumptions and still land in the same result.
We can safely assume that f(x) is strictly positive, since f(x)=(f(x/2))^2 is non-negative, and f(x) = 0 anywhere gives the boring zero solution as Peyam showed.
First, by induction we get that f(n*x) = f(x)^n for integers n.
By a little bit of algebra it follows that f(q) = f(1)^q for rational q. To extend this to the reals:
Assume f is continuous *at a single point* c.
Now take any real a and the limit as x->a:
f(x) = f(c+x-a)*f(a-c) --> f(c)*f(a-c) = f(a), i.e. f is continuous everywhere.
Any real number is the limit of a sequence of rational numbers, so for any real x we let x_n be a rational sequence with limit x.
By continuity, f(x) = lim(n->infty) f(x_n) = lim(n->infty) f(1)^x_n = f(1)^x.

Since all the non-zero solutions are positive, we can take the logarithm of the original equation and get equivalently
log f(x+y) = log f(x) + log f(y).
This is a well-studied equation, and under many different weak assumptions we can prove log f(x) = k*x
Interestingly, there are known to be solutions to this which are not continuous anywhere, but they are really complicated to construct:

[MrCigarro]

This is something I did in class when providing a proof of the form of the normal distribution. But I never saw a correct and elegant proof like the one you do here. This is so important to me. Thank you very very much.

[OonHan]

Wow, my guess was right immediately! It's very recognisable in my opinion

[MrRoy]

Hahaha! I knew the answer right at the beginning because in Abstract Algebra the exponential function is the homomorphism that transforms addition into multiplication :)

[rogerkearns]

It seems intractable to begin with, but obvious once you know.

[willnewman]

Assuming f is continuous, these are the only functions you can get. But if f does not need to be continuous, there are many more functions

[buxeessingh]

This is a homework problem in baby Rudin.

[josetomasneumann]

Differentiable at 0 is a very strong condition. A weaker condition is to prove the same with the consideration that f is continuous at one point (non zero of course)

[omgopet]

Physicist here.


I just guessed the functions, checked by plugging each back in the equation, checked that some "kinda exponential" function (like sin(x)) doesn't work and called it a day.

[aryammlg]

great job!
but without any assumption about continuity and differentiability, we can prove that f(x)=a^x on rational numbers and for extending the idea for real numbers, it is sufficient for f to be continuous at merely one point. if f is continuous at one point, we can prove that it is continuous over R and differentiability is not even a necessary condition. And if f is not continuous at any point, then it is equal to a^x for rational numbers and for irrational numbers, some very strange functions could satisfy the equations other than a^x. thank you for posting this video😍😍

[LwLiPp]

You can reduce the assumptions even more. If f is real-valued, then you only need continuity in 0.
If f is cont. in 0, then for any real x and any sequence x_n = x + h_n with x_n -> x, i.e. h_n -> 0, we have lim_n f(x_n) = lim_n f(x)*f(h_n) = f(x) by continuity in 0.
This means that f is continuous everywhere in this case. Now, if f(0)=0 we still deduce that f is the constant 0 function. In the case that f(0)=1.
we can show that f is positive: If f(x) < 0 for some x =|=0, then f(x/2) would have to be a complex number. However, f is real-valued by assumption.
Now consider a = f(1). then f(1/k) = a^(1/k) for all natural numbers k. In particular, we deduce that, since f(0)=1, f(p/q) = a^(m/k) for all integers m and natural numbers k.
By continuity of f and density of the rationals in the reals, we finally find that f(x) = a^x for all real x.

For complex valued f I don't know whether this works though.

[GreenyNeko]

*video starts*
Him: "Thanks for watching!"
Me: "Ok, bye."
*lea- "wait.."

[mokouf]

If I do it, I will not pull out f(x) for differentiating f(x) with fundamental theorem of calculus, in order to cover both f(0)=0 and f(0)=1 cases.
For f'(x) = lim(h→0) f(x)[(f(h)-1)/h] to be valid, it is very obvious that this limit has to be in 0/0 form, as the bottom is always 0.
f(0)=0: f(0)-1 = -1, so f(x) has to be 0 for all x.
f(0)=1: f(0)-1 = 0, f(x) is not yet determined. Using L'H, f'(x) = f(x)*f'(0)=Cf(x) as you did.
Then we can play with differential equation to find f(x) for f(0)=1 case. This is a very simple separable differential equation.

[islandfireballkill]

At 4:22 You had a seperable equation, so why not just write f'(x)/f(x) = C?

Anyway this is pretty cool, I wonder if there are any discontinuous at zero solutions 🤔.

[MathIguess]

"I want to do something really cool as usual" xD nice!