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Find all functions $$f:\Re \rightarrow \Re$$ such that

$$f({ x }^{ 2 }+yf(z))=xf(x)+zf(y)$$.

Please do post some suggestions on how to solve functional equations.

Note by Priyanshu Mishra
2 years ago

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What have you tried? With functional equations, a lot of times it boils down to testing a bunch of values, and then being smart about choosing the variables.

What solutions have you found so far? That could guide you in thinking about the possible approaches.

Check out Functional Equations

Staff - 2 years ago

Let $$P(x,y,z)$$ be the statement $$f(x^2+yf(z)) = xf(x)+zf(y)$$.

First, $$f(x) = 0$$ for all $$x$$ is clearly a solution. For the rest of this solution, assume there exists $$c$$ such that $$f(c) \neq 0$$.

$$f$$ is injective. To see this, suppose $$f(a) = f(b)$$. Then:

$$P(0,c,a) \implies f(cf(a)) = af(c)$$

$$P(0,c,b) \implies f(cf(b)) = bf(c)$$

But $$f(a) = f(b)$$, so $$af(c) = f(cf(a)) = f(cf(b)) = bf(c)$$, or $$af(c) = bf(c)$$. Since $$f(c) \neq 0$$, we have $$a = b$$.

$$P(0,0,0) \implies f(0) = 0$$

$$P(0,x,x) \implies f(xf(x)) = xf(x)$$

$$P(x,0,0) \implies f(x^2) = xf(x)$$

Thus $$f(x^2) = xf(x) = f(xf(x))$$, or $$f(x^2) = f(xf(x))$$. Since $$f$$ is injective, we have $$x^2 = xf(x)$$. For nonzero $$x$$ this gives $$f(x) = x$$. Conveniently, when $$x = 0$$ the same also applies, so we get the only other solution $$f(x) = x$$ for all $$x$$.

As you can see above, some functional equations are easily solved by simply substituting the right values. Occasionally, finding whether the function is injective/surjective also helps, especially when you have nested functions (function application inside function application). Sometimes you can also try ruling out the obvious answers so you can use a particular property (for example, say that all zero function works/doesn't work, so that for the rest of the problem you can assume there exists a nonzero value).

- 2 years ago

Thanks a lot.

- 2 years ago

i think f(x)=x is the solution because first of all the it has to be an odd function ....that can be proved by putting -x in the place of x and subtracting both the eqns.....then....... replace x^2 +y f(z) with -ve of that ....on LHS rewrite -z f(y) as z f(-y).....and on RHS write -x^2 as x f(-x) and now compare...

- 1 year, 12 months ago

You have found 1 solution. How do you know that there are no other solutions?

For example, $$f(x) = 0$$ is another solution. How was that missed in your explanation?

Staff - 1 year, 12 months ago

Ohh...sorry i have cancelled that while solving....yea ...its the other solution... thanks for it btw... and u get that also when u replace x with -x and subtract..

- 1 year, 12 months ago

So, how do you know that there are no other solutions that you "cancelled" out?

Staff - 1 year, 12 months ago