# Weird Induction proof

• Feb 25th 2010, 10:28 PM
morbius27
Weird Induction proof
Suppose f:R-->R satisfies f(xy) = xf(y)+yf(x) for all real numbers x,y. Prove that f(1)=0 and that f(u^n)=n*u^(n-1)*f(u) for all natural numbers n and all real numbers u.

I really have no clue where to start on this proof, except that the hint in the problem says that in using induction on n, consider the case u = 0 separately.

Any help would be much appreciated.
• Feb 25th 2010, 11:56 PM
Jose27
Quote:

Originally Posted by morbius27
Suppose f:R-->R satisfies f(xy) = xf(y)+yf(x) for all real numbers x,y. Prove that f(1)=0 and that f(u^n)=n*u^(n-1)*f(u) for all natural numbers n and all real numbers u.

I really have no clue where to start on this proof, except that the hint in the problem says that in using induction on n, consider the case u = 0 separately.

Any help would be much appreciated.

For the first, let \$\displaystyle y=1\$ then \$\displaystyle f(x)=xf(1)+f(x)\$ and so \$\displaystyle xf(1)=0\$ and since \$\displaystyle x\$ was arbitrary, it follows that \$\displaystyle f(1)=0\$. For the second suppose you know it for \$\displaystyle n\$ then \$\displaystyle f(u^{n+1})=f(u^nu)=u^nf(u)+uf(u^n)=u^nf(u)+unu^{n-1}f(u)=(n+1)u^nf(u)\$