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.

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TopNewestWhat 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.

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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).

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Thanks a lot.

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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...

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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?

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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..

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