Solving f(x/y)=f(x)/f(y), A Nice Functional Equation

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FunctionalEquations

PLAYLISTS 🎵 :

[leickrobinson]

3:26 Potentially very confusing for students to re-use “y” to represent “f(x)” when it just meant something different on the previous line! Please use a different letter next time! :-D

[WahranRai]

The hypothesis of any problem must be displayed (domain of variables, function derivable, continuous, integrable ...)

[sebby]

At 2:00 if you differentiate with respect to x, you assume that y is a constant, but later you give y=1/x which is not constant

[darwinvironomy]

sustitute from (x, y) to (xy, y). we get:

f(xy/y) = f(xy)/f(y)
f(x) = f(xy)/f(y)

f(x)f(y) = f(xy)
which the solutions to this is power function. x^c

[jclzjbe]

Cooler solution using (function) conjugation:
x and y are positive, so you can write them as e^a and e^b. Plug that in and take log of both sides, and you get g(a-b) = g(a) - g(b), where g(a) = ln(f(a)) (g is a conjugate of f by exp).
It's easy to see that g is differentiable and linear, so g(a) = ca.
Now you just conjugate back to get f(x) = exp(g(ln(x)) = exp(cln(x)) = x^c.

[emanuellandeholm]

Very nice problem! I used another method to get d/dx f(x/y) = f'(x)/f(y) by handwaving a bit, but I think it should give the same solution.

[menachembloom]

This also proves (a/b)^c = a^c/b^c which is just hilarious

[yanlonghao]

Well. Set a=x/y. It reduce to a very standard function equation. As long as f is continuous, f will be a power function. However, there are also other noncontinuous solutions. I mean, we do not need diffferentiable here.

[chaosredefined]

f(y/x) = f(y)/f(x).
Consider y = e^a, x = e^b
f(e^a / e^b) = f(e^a) / f(e^b)
f(e^(a-b)) = f(e^a)/f(e^b)
Let g(x) = f(e^x)
g(a-b) = g(a)/g(b)
g(a-b)g(b) = g(a)
Set b = 1, and define K = g(1)
Kg(a-1) = g(a).
This is a geometric series with ratio K. We can also see that, if a=b, we get g(0) = g(a)/g(a) = 1, so it's g(x) = K^x
Therefore, f(e^x) = K^x.
Define C = ln(K). Hence K = e^C, and K^x = e^(Cx) = (e^x)^C
Therefore f(e^x) = (e^x)^C.
Let y = e^x
f(y) = y^C
Hence, f(x) = x^C

No derivatives, but I did use the discrete derivative of g(x)

[Eknoma]

It is very simple to see that f must be multiplicative

x=y -> f(1) = 1
x=1 -> f(y^-1) = f(y)^-1
y=z^-1 -> f(xz) = f(x)f(z)

Differentiable => continuous
And it is obvious that the multiplicative functions on R+ which are continuous must be just like the multiplicative functions on Q+

So our candidates are f(x) = x^c for any C

And they are clearly all differentiable

[kostasl]

I did it the same way you did (though I gotta admit it took me some time) with the difference being after minute 3:22. I think there is a more elegant way to warp it up since c/x= f'(x)/f(x) --> (clnx)'=(lnf(x))' --> (lnx^c)'=(lnf(x))'(intergal) --> f(x)=x^c + a, with a=0 since f(1)=1+a=1, so f(x)=x^c, with c delonging to R.

P.S.: I really could never do derivatives with the df(x)/dx fromat, it is really confusing for me :)

[yoav]

Very nice solution!!, it can be shorter because it is well known that the integral of f'/f is ln(f)+C

[klopkerna]

An antiderivative of 1/X is ln |X| + k ( a constant)

[antonyqueen]

I have to correct. At 2:25 after differentiating over x, we set x=y and not y=x. This won’t change the result, but it is the mathematical correct wording because we differentiated based on y being a constant (or parameter >0) independent of x, and x is the variable. Thus x can take the value of y but y is independent of x .

[persiff]

Why you supposed that y=constant although you put each step y=x and we know x it’s variable and we can’t forget that f(1)=1 relied by x=y in each time the x change what do u think 🤔!?

[satyapalsingh]

Very good approach, dear professor ! Keep it up!

[klopkerna]

After solving the differential equation you have to put your solution into your functional equation as you reasoned by implications

[ahmetatar]

Hocam buna benzer her x, y için f(x.y)=f(x)+f(y) eşitliğini saglayan logaritma fonksiyonu var. f(x)=logn(x). onun da ispatlı videosunu yapabilirsiniz çok iyi olur.

[abdulalhazred]

a confusing mess with the ambiguous use of the symbol y

[anestismoutafidis]

log(2)->[16÷4] = log(2)->16÷ log(2)->4 = 2 <=> f=log(2) ; x=16; y=4; f(x;y)=2