# How to show that the remainder of this Taylor expansion of this homogeneous function is zero?

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 $$\frac{(P \cdot \nabla)^3 f(\tau P)}{6} = 0$$?

• 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 $t\rightarrow 0$. – 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}