# Proving identity about k times differentiable function

I have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(tx) = t^{k}f(x)$ where $f$ is $k$ times differentiable, for any $t, x \in \mathbb{R}$. I was supposed to prove $f(x) = \frac{f^{(k)}(0) x^{k}}{k!}$. The book told me to differentiate $k$ times with respect to $t$ the identity $f(tx) = t^{k}f(x)$, but I don't understand how or why I can do that. If I was going to differentiate with respect to $t$, what would that look like?

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Are you happy differentiating with respect to $x$? If so, then make believe $f(tx)=x^kf(t)$, differentiate with respect to $x$, and then swap all the $x$s and $t$s in the answer. – Gerry Myerson Feb 7 '12 at 22:51
What book is that, Pedro? – Georges Elencwajg Feb 7 '12 at 23:49

Think of $x$ as a constant $-$ equivalently, look at what happens at a fixed value of $x$ $-$ and use the chain rule: $$\frac{d}{dt}\Big(f(tx)\Big)=f\,'(tx)\cdot\frac{d}{dt}(tx)=xf\,'(tx)\;,\tag{1}$$ and $$\frac{d}{dt}\Big(t^kf(x)\Big)=f(x)\frac{d}{dt}(t^k)=kf(x)t^{k-1}\;.\tag{2}$$ Repeating the process $k$ times gives you $x^kf^{(k)}(tx)$ from $(1)$ and $k!f(x)t^0=k!f(x)$ from $(2)$. Now what happens when $t=0$?

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Fix $x_0\in\mathbb R$, you can consider the function $F:\mathbb R\to \mathbb R$ given by $F(t)=f(tx_0)$. $F$ is clearly $C^k$ as is a composition of $C^k$. What is $F^k(0)$?

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$F^{k}(0) = x^{k}f^{k}(0)$? – Pedro Feb 7 '12 at 22:55
Well, Pedro, there's no $x$ in azarel's question, only an $x_0$, so there can't be an $x$ in the answer, right? – Gerry Myerson Feb 8 '12 at 5:47

$f(t)=f(t\cdot 1)=t^kf(1)=at^k$. The hypothesis of differentiability of $f$ is irrelevant.

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