I've come across this compact way to write the Taylor expansion formula:
$$ f(x+ \varepsilon) = e^{\varepsilon \frac{d}{dx}}f(x) $$
With $\varepsilon$ a generic increment. So naturally I wondered if, opportunely modified, this holds in n-dimensions.
I thinked to the formula:
$$ f(\bar{x}+ \bar{\varepsilon}) = e^{\bar{\varepsilon} \cdot \bar{\nabla}}f(\bar{x}) $$
That upon expansion gives:
$$ f(\bar{x}+ \bar{\varepsilon}) = f(\bar{x}) + \bar{\varepsilon} \cdot \bar{\nabla}f(\bar{x}) + \frac{1}{2} (\bar{\varepsilon} \cdot \bar{\nabla})^2 f(\bar{x}) + \dots $$
(I found this same reasoning in some Physics textbook, namely Solid State Physics of Ashcroft-Mermin.)
My problem is that I found it wrong, when confronted with the real deal:
$$ f(\bar{x}+ \bar{\varepsilon}) = f(\bar{x}) + \bar{\varepsilon}^T\bar{\nabla}f(\bar{x}) + \frac{1}{2}\bar{\varepsilon}^T H[f(\bar{x})] \bar{\varepsilon} + \dots $$
With $H[f(\bar{x})]$ the Hessian evaluated at $f(\bar{x})$. Specifically I'm convinced that:
$$ \bar{\varepsilon}^T H[f(\bar{x})] \bar{\varepsilon} \neq (\bar{\varepsilon} \cdot \bar{\nabla})^2 $$
The question is: I'm I right or wrong ? Thanks for your attention
Moreover, is it possible, if the formula doesn't hold, to find an extremely compact way to express it, as the 1 dimensional case ?
We have $$\varepsilon^TH[f]\varepsilon = \sum\limits_{j,k = 1}^n \varepsilon_jH[f]_{j,k}\varepsilon_k = \sum\limits_{j,k = 1}^n \varepsilon_j(\partial_j\partial_kf)\varepsilon_k = \sum\limits_{j = 1}^n \varepsilon_j\partial_j\sum\limits_{k = 1}^n \varepsilon_k\partial_kf = (\varepsilon \cdot \nabla)(\varepsilon \cdot \nabla)f,$$ so they are really the same.