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Let the function g : R $\rightarrow$ R satisfy $g(xy) = x \cdot g(y) +y \cdot g(x)$ for all real numbers x and y.

Prove $g(u^n) = nu^{n-1}g(u)$, for all positive integers $n$ and all real numbers $u$.

the Inductive step for me is a bit tricky, the base case is straightforward:

$g(1^{1})$ = $1\cdot 1^{0}\cdot g(1)$

$g(1) = g(1)$

Now, I believe I have a fundamental misunderstanding of what induction looks like. Of course in this case it means to try and compute $u+1$ and $n+1$, but plugging it in and doing all the algebra isn't working for me.

Any hints, tips, suggestions?


Going off of what voldemort said, $u$ is fixed and we're only trying to prove the induction step for n+1. I get the following:

$g(u^{n+1}) = (n+1)\cdot u^n\cdot g(u)$

However, I'm unsure of how to algebraically deal with exponents in side functions.

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Tom: you do not need to prove it for $u+1$. Fix $u$ and prove this for all $n$ by using induction. This will show that the result holds for all real numbers $u$. – voldemort Feb 11 '14 at 19:04
It's not clear to me what you mean by "perform induction." Induction would require you to be trying to either prove or define something. You haven't really given us a statement you are trying to prove. – Thomas Andrews Feb 11 '14 at 19:04
Thomas, I'm trying to prove the equation I listed. – Benji_Bombadill Feb 11 '14 at 19:07
Voldy, I'll trying to redo the math. – Benji_Bombadill Feb 11 '14 at 19:08
It seems that there is really only one function involved. Replace every $g$ by $f$ in your question maybe? – Christoph Feb 11 '14 at 19:24
up vote 1 down vote accepted

Let $f:\mathbb R\to\mathbb R$ be a function satisfying $f(xy)=x\cdot f(y)+f(x)\cdot y$ for all $x,y\in\mathbb R$.

Claim: For all $n\in\mathbb N$ and $u\in\mathbb R$ we have $f(u^n) = nu^{n-1} f(u)$.

Proof. Let $u\in\mathbb R$ be any real number. Proceed by induction on $n$.

For $n=1$ the claim is $$ f(u^1) = 1\cdot u^0 f(u), $$ which is true, since $u^0=1$ and $u^1=u$.

Now let the claim hold for a fixed $n\in\mathbb N$, then $$ f(u^{n+1}) = f(u^n u) = f(u^n) u + u^n f(u) = nu^{n-1} f(u) u + u^n f(u) = (n+1)u^n f(u) $$ as desired. $\square$

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when write $f(u^n u) = f(u^n) u + u^n f(u)$, is this just a known function operation? – Benji_Bombadill Feb 11 '14 at 20:00
No, this is the known property of $f$, namely $f(xy) = f(x)\cdot y + x\cdot f(y)$, applied for $x=u^n$, $y=u$. – Christoph Feb 11 '14 at 20:02
Aww. Right in front of my face. – Benji_Bombadill Feb 11 '14 at 20:06

You cannot prove a statement about all $u\in\mathbb R$ by induction (well, some nitpickers might find ways, but ignore that). Proof by induction is for $\mathbb N$, not $\mathbb R$, because induction is the very essence of the Peano axioms - for the natural numbers. So (as already has been said in the comments), let your statement $P(n)$ be:

$P(n)$: For all $u\in\mathbb R$ we have $g(u^n)=nu^{n-1}f(u)$

Then show $P(1)$, i.e.:

For all $u\in\mathbb R$ we have $g(u)=f(u)$.

and show $P(n)\implies P(n+1)$. (However, I don't know how as you do not mention any specific properties of $g$ and $f$ as given)

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Alright, let me add to the problem. – Benji_Bombadill Feb 11 '14 at 19:19
+1 for you cannot prove a statement about the reals using induction. However, I would not back off and say it is possible because it is not. – kleineg Feb 11 '14 at 19:21
Voldemort pointed out that u (the reals) is fixed. – Benji_Bombadill Feb 11 '14 at 19:24
@kleineg Some weird theorem might be provable for $u\in[0,1]$ and then with induction steps $u\to u\pm 1$. – Hagen von Eitzen Feb 11 '14 at 21:01

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