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I came across this problem in class where I had calculate the Taylor expansion of $f(\theta + \theta')$ where $f \in C^3(\mathbb{R}^3)$ using the hessian matrix $H_f$.

I was told that the solution was $$ f(\theta) + Df(\theta)^\top\theta' + \frac{1}{2}{\theta'}^\top H_f \theta' + o\left({\theta'}^3\right) $$ where o(-) denotes the little notation.

I can't figure out how to get to this result. I've tried rewriting the Taylor formula but I haven't been able to rearrange anything so that it matches the given solution. Any help would be appreciated.

EDIT: I'm using this definition:

Let $f:\mathbb{R}^n \to \mathbb{R}$ a $k$-time continuously differentiable function at the point $a \in \mathbb{R}^n$. Then there exists $h_\alpha : \mathbb{R}^n \to \mathbb{R}$ so that

\begin{align} & f(\boldsymbol{x}) = \sum_{|\alpha|\leq k} \frac{D^\alpha f(\boldsymbol{a})}{\alpha!} (\boldsymbol{x}-\boldsymbol{a})^\alpha + \sum_{|\alpha|=k} h_\alpha(\boldsymbol{x})(\boldsymbol{x}-\boldsymbol{a})^\alpha, \\ & \mbox{and}\quad \lim_{\boldsymbol{x}\to \boldsymbol{a}}h_\alpha(\boldsymbol{x})=0. \end{align}

jaregax432-awinceo
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  • This is pretty much what the multivariable Taylor's theorem tells us, so I'm not sure what you're asking to be proven. (Unles you want a proof of Taylor's theorem itself, in which case you should just look that up in a decent multivariable textbook). Edit: I just remembered I actually wrote several answers on Taylor's theorem's statement and proof and a whole bunch of other stuff related to it; so you should probably start there and go through all the similarly linked posts. – peek-a-boo Dec 26 '21 at 17:43
  • @peek-a-boo My textbook gives me the definition I just added as the multivariable Taylor's theorem. I'm not too comfortable with hessian matrices yet so I can't really tell how I link the two together... – jaregax432-awinceo Dec 26 '21 at 17:56
  • In your summation, can you explicitly write out the $|\alpha|=1$ terms? THen, try the same with $|\alpha|=2$. If you want, assume you're working in $\Bbb{R}^2$ to simplify matters slightly. The pattern should become clearer. – peek-a-boo Dec 26 '21 at 17:57

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