a nice functional equation

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[BrollyyLSSJ]

The check at the end is incorrect - "a" term should be outside the cubing brackets on the LHS, you should pull it out first as a shared factor, then apply sum of cubes formula and distribute it in the second bracket to get the conclusion.

[LouisEmery]

The second term on the RHS (f(x^2)+f(y^2) - f(xy)) was suggestive right off the bat. Looked like the factor of x^3+y^3 to begin with. Functional equations are found in math contest not in math courses (that I took). They are kind of fun since they are mostly procedure with some looking ahead.

[spiderjerusalem]

Just by looking at the form, it reminisces one of the factorization of x³+y³

whence, by letting f(x)=xg(x) for some function g, it suffices to prove that g is a constant function

Upon arriving at f(x³)=xf(x²) (implying g is an even function), move the terms of the original eq

yf(x²)+xf(y²)=(x+y)f(xy)
substituting f(x)=xg(x), then subsequently divide both sides by xy
xg(x²)+yg(y²)=(x+y)g(xy)
(but now i'm frankly confused whether or not it is allowed to substitute values for x, y such that xy=0. The eq is true if x=y=0, but case in which precisely one of them is 0 is uncertain)
subs y=0,
xg(x²)=xg(0)
g(x²)=g(0)
g(x)=g(0), for x≥0
since g is an even function
g(–x)=g(x)=g(0), for x≥0
=> g(x)=g(0) for any x
g≡1

[danielbranscombe]

I used
f(x)=x^(1/3)f(x^(2/3))
if you keep applying it n times you end up with

taking the limit as n goes to infinity we end up with
f(x)=xf(1)
thus if f(1)=c
f(x)=cx
and we can check that this works for any value of c

[julienbernu]

I solved that in a completely different way:
- start like you did to establish, f(x^3) = x f(x^2), and thus f is odd.
- then plug y -> -y in the main equation and divide by x-y on both sides, using f odd and taking the limit as y approaches x, find that f is differentiable on |R* with d/dx f(x^3) = 3 f(x^2)
- now differentiate f(x^3) = x f(x^2) to also find d/dx f(x^3) = f(x^2) + 2x^2 f '(x^2)
- equating the 2 expressions for d/dx f(x^3), find that 2 f(x^2) = 2 x^2 f '(x^2), or f(x) = x f '(x) on |R*+
- above differential equation is easily solved as dx/x = df/f => ln(x) + C = ln(f) => f(x) = C x on |R*+, and with f odd, f(x) = C x on |R.

[whiteboar]

Would have it been wrong to say that, since f(x^2) is even, then x*f(x^2) is odd?

[gaufqwi]

Are there any functional equation problems where the solution turns out to be something more interesting than a constant or linear function?

[adamwho]

By inspection, the original problem is the formula for the sum of cubes. This immediately leads to f(x) = x

[ShaunakDesaiPiano]

A way to prove that it’s an odd function without first proving f(0) = 0: substitute y = -x.

[zyklos]

Wonder, if you take a step back, that not only f, but also x and y can be considered searchable objects. So for any fixated f one could find the x and y that fulfil the equation. Or for given x, y find f.
So "the" solution is not only "f regardless of x and y", but is sort of extendable to a triple (f, set for x, set for y) -> true or false 🤔

[mirokajevskivelevski]

Can you solve this problem one from the 2023 MMO (macedonian math olympiad)? Find f (R to R)

[kishanpaswan]

Everytime you upload any video made me realise you are really genius and I really wait everyday specially for your video

[charleyhoward]

how does he get the result at 4:11 ??? x=x^(1/3) doesn't yield -f(x) when u subs, in f(x^3)=xf(x^2)

[yeech]

Do we need to prove this is the only solution?

[JordHaj]

Such a good, methodical procedure, and such a big flop at the end. f(x^3) = ax^3, not (ax)^3 so the only values of a that satisfy the equation are 0 and ±1.
Edit: a similar flop on my part resulted in my methodical procedure being incorrect, as I also had (ax + ay) as the first factor.

[charleyhoward]

totally confusing; not math as I know it