# Span of a nonlinear function

Fix vectors $$x,y\in\mathbb{R}^d$$ and a smooth function $$\phi:\mathbb{R}\rightarrow \mathbb{R}$$. Define $$\phi^d: \mathbb{R}^d \rightarrow \mathbb{R}^d$$ as applying $$\phi$$ entrywise (i.e. $$\phi^d(x_1, x_2, \dots) = (\phi(x_1),\phi(x_2), \dots)$$). Under what conditions do we have the equality

$$\mathbb{R}^d=\text{span}(\{\phi^d(\alpha x+y)~\big |~\alpha\in \mathbb{R}\})$$

Clearly if $$\phi$$ is the identity operator $$\phi(x)=x$$ or if any two entries of $$x$$ are equal, this equality does not hold. Main claim is equality holds for generic (i.e. almost all) $$x,\phi$$ choices.

• Suppose the span is not the full $\mathbb{R}^d$, then there exists a normal vector to the span which we can call $\nu$. This means that $\sum \nu_i \phi(\alpha x_i) = 0$ for all $\alpha$. Take $k$ derivatives with respect to $\alpha$ you still get 0. Evaluate at $\alpha = 0$ you get that $\phi^{(k)}(0) \sum \nu_i (x_i)^k = 0$. So provided not too many derivatives of $\phi$ vanish at zero (generically true), such $\nu$ cannot exist for generic $x_i$. – Willie Wong Jan 17 at 22:22
• For the final claim, see en.wikipedia.org/wiki/Vandermonde_matrix – Willie Wong Jan 17 at 22:28
• There could be some typo around, because the question does not make any sense, as it is. – Pietro Majer Jan 18 at 0:10
• @Pietro Majer I think that $\phi (x)=(\phi (x_1),\ldots, \phi (x_d))$. – Jochen Wengenroth Jan 18 at 8:44
• So maybe it is a map $\phi:\mathbb{R}^d\to\mathbb{R}^d$? – Pietro Majer Jan 18 at 8:55