In Calculus of Several Variables, Third Edition, by Serge Lang, this is exercise 1 in chapter 6, section 5:

Let $f$ be a function of two variables. Assume that $f(O) = 0$, and also that $f(t P) = t^2 f(P)$ for all points $P$ in $\mathbb{R}^2$. Show that for all points $P$ we have $f(P) = \frac{(P \cdot \nabla)^2 f(O)}{2!}$.

Section 5 is on Taylor's formula. I'm guessing that it should be assumed that $f$ has continuous partial derivatives up to order 3 so that the Taylor expansion at $O$ up to degree 2 can be calculated. So, $f(P) = f(O) + (P \cdot \nabla)f(O) + \frac{(P \cdot \nabla)^2 f(O)}{2} + \frac{(P \cdot \nabla)^3 f(\tau P)}{6}$ for some $\tau \in [0, 1]$.

Using the chain rule, we get $\frac{d}{dt}f(t P) = \frac{d}{dt}\left[t^2 f(P)\right] \iff \nabla f(t P) \cdot P = 2 f(P)t \implies P \cdot \nabla f(O) = 0 \iff (P \cdot \nabla)f(O) = 0$. Thus, the Taylor expansion simplifies to $\frac{(P \cdot \nabla)^2 f(O)}{2} + \frac{(P \cdot \nabla)^3 f(\tau P)}{6}$. How can it be shown that $\frac{(P \cdot \nabla)^3 f(\tau P)}{6} = 0$?

  • $\begingroup$ Maybe in your implication you can divide both sides by $t$ and obtain an identity $\frac{\nabla f(tP)\cdot P}{t} =2f(P)$ and let $t\rightarrow 0$. $\endgroup$ – TheWildCat Jul 19 at 2:56

Temporarily define $g: \Bbb{R} \to \Bbb{R}$ by $g(t) = f(tP)$. Then, by the standard Taylor's theorem in single variable calculus, using the Lagrange form of the remainder, we have that \begin{align} g(1) &= g(0) + g'(0) (1-0) + \dfrac{g''(0)}{2!}(1-0)^2 + \dfrac{g'''(\tau)}{3!}(1-0)^3 \tag{$*$} \end{align} for some $\tau \in [0,1]$. Now, note that $g(t) := f(tP) = t^2f(P)$. So, we have that for any $t \in \Bbb{R}$, \begin{align} \begin{cases} g'(t) &= 2t \cdot f(P) \\ g''(t) &= 2f(P) \\ g'''(t) &= 0 \end{cases} \end{align} In particular, $g(0) = g'(0) = g'''(\tau) = 0$. Hence, equation $(*)$ reduces to $g(1) = \dfrac{1}{2} g''(0)$. But, $g(1) = f(P)$, so we now have that \begin{align} f(P) = g(1) = \dfrac{1}{2}g''(0). \end{align} Now, once again, recall that $g(t) = f(tP)$; this time compute the second derivative of $g$, but using the chain rule. You'll find that \begin{align} \dfrac{g''(0)}{2} = \dfrac{1}{2} \left( P \cdot \nabla \right)^2f (O). \end{align} Hence \begin{align} f(P) = \dfrac{1}{2}\left( P \cdot \nabla \right)^2f (O). \end{align}

So, more directly, you can verify that \begin{align} g'''(\tau) = \left( P \cdot \nabla \right)^3f (\tau P), \end{align} and by what I showed above, $g''' = 0$. Hence, this term vanishes.

Also, now it should be easy enough to prove the following generalization:

Let $f: \Bbb{R}^n \to \Bbb{R}$ be $C^{k+1}$, and fix a point $P \in \Bbb{R}^n$. Suppose that for all $t \in \Bbb{R}$, $f(tP) = t^k f(P)$. Then, \begin{align} f(P) = \dfrac{1}{k!} \left( P \cdot \nabla \right)^kf (O) \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.