Question

Find all functions f:ℝ→ℝ that satisfy
f(x^3+y^3)=(x^2)f(x)+yf(y^2)
for all x,y∈ℝ

Sorry, I don't like littering these forums with a question every few minutes, but they're haaaaarrrd.

Answer 1

Where are these problems coming from? Some competition paper?

Answer 2

Possibly. These questions were compiled for me by a friend, with solutions attached. :P Math club is a cool place.

Answer 3

It is fair to say that
f(x^3)=xf(x^2)-->f(x^2)=xf(x), right?

Answer 4

Zero function is a solution, f(t) = 0

Other constant functions aren't a solution, f(t) = c, c not zero

And straight lines through the origin, f(t) = mt, are solutions because
\[ f(x^3 + y^3) = m(x^3 + y^3) = x^2(mx) + y(my^2) = x^2f(x) + yf(y^2) \]

The next step is harder.

Answer 5

since \(x^2f(x)\) involves terms just involving the variable \(x\), and \(yf(y^2)\) involves terms just involving the variable \(y\), then is it valid to assume:\[f(x^3+y^3)=f(x^3)+f(y^3)\]and:\[f(x^3)=x^2f(x)\]\[f(y^3)=yf(y^2)\]

Answer 6

And from this we can deduce that \( f(x^2) = xf(x) \).

Answer 7

f(y^3)=yf(y^2)-->yf(y^2)=y^2f(y)?

Answer 8

This is pretty much where I got stuck, when I tried solving it earlier. AND I'M NOT LOOKING THROUGH THE SOLUTIONS.

Answer 9

therefore:\[f(x)=mx\]is the only solution - isn't it?

Answer 10

It is the correct answer, but I... don't see the methodical step between f(x^2)=xf(x) and that conclusion. It seems intuitive, but my intuition is pretty bad.

Answer 11

I mean, I see it's correct working backwards, but how can I prove that's the only possible scenario?

Answer 12

if:\[f(x^3)=x^2f(x)\]then f(x) can only contain terms in x such that multiplying it by \(x^2\) terms them into terms involving \(x^3\) - that was my thought process.

Answer 13

*turns them into

Answer 14

There's no non-simple function out there that could possibly have this be true? I get the feeling that there's some f(x)!=ax for which f(x^n)=nf(x^(n-1)) is true for only a few values of n.

Answer 15

I see you believe in conspiracy theories badreferences :)

Answer 16

Haha.

Answer 17

First notice that from asnaseer's identities above, if you set x equal to y then
\[ x^2f(x) = f(x^3) = xf(x^2) \]
Hence for all non-zero \( x \), we have that \[ f(x^2) = xf(x) \] Now, for the moment, I share BR's anxiety there may be other functions out there. Proving their non-existence is not entirely trivial.

Answer 18

I just tried wolfram for fun and got this: http://www.wolframalpha.com/input/?i=solve+f%28x%5E2%29%3Dx*f%28x%29

I wonder why it presents the solution as:\[f(x)=\frac{cx}{e^2}\]

Answer 19

Man, I don't know how to prove nonexistence. >.> I haven't even finished abstract algebra yet, and only just got through introductory proofs.

Answer 20

Ah ... there other functions. For instance:
Define \[ f\colon\mathbb{R}\to\mathbb{R} \] by
$$
f(x) \;=\; \begin{cases} 2x & \text{if }x\text{ is algebraic,} \\ 3x & \text{if }x\text{ is transcendental.}\end{cases}
$$
Both algebraic and transcendental numbers are closed under the operation of squaring, and therefore \( f(x^2) = x\,f(x) \) for all \( x \).

In fact, whenever we can partition the reals into algebraically closed sets, we can define f as different linear functions on each of those sets.

Answer 21

Wow James - you are a true genius!

Answer 22

@JamesJ, that's a lovely fact:
"In fact, whenever we can partition the reals into algebraically closed sets, we can define f as different linear functions on each of those sets."

I'm keeping this in mind. It explains so many special functions now that I think about it.

Answer 23

\(x=y=0\) yields \(f(0)=0\)

setting \(x=0\) gives \(f(y^3)=yf(y^2)\) similarly \(y=0\) gives \(f(x^3)=x^2f(x)\)

so\[f(x^3+y^3)=f(x^3)+f(y^3)\]setting \(u=x^3\) and \(v=y^3\) gives\[f(u+v)=f(u)+f(v)\]which is Cauchy-equation with 2 solutions :

\(f(x)=0\) and \(f(x)=cx\).