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Online I have found the definition:

Taylor's theorem $\quad$ Multivariate case

$$ \begin{align*} f(\mathbf{x}) = f(\mathbf{a}) + \nabla f(\mathbf{a})^T(\mathbf{x}-\mathbf{a}) + \frac{1}{2}(\mathbf{x}-\mathbf{a})^T \nabla^2 f(\mathbf{a})(\mathbf{x}-\mathbf{a}) + R_2(\mathbf{x}) \end{align*} $$

where $R_2(\mathbf{x}) = o(||\mathbf{x}-\mathbf{a}||^2)$ as $||\mathbf{x}-\mathbf{a}|| \to 0$.

What is the norm $\|\cdot\|$ used here, when defining the remainder error. Is it the euclidean norm?

Dylan Dijk
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    Sure you can use the Euclidean norm, but it doesn’t matter (atleast in finite dimensions) since all norms are equivalent. – peek-a-boo Feb 25 '24 at 17:04
  • Ok thank you, but what do you mean by all norms are equivalent? – Dylan Dijk Feb 25 '24 at 17:11
  • I guess like what is referred to in this answer – Dylan Dijk Feb 25 '24 at 17:16
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    Your guess is right. See also https://en.wikipedia.org/wiki/Equivalence_of_metrics#Relation_with_equivalence_of_norms – Anne Bauval Feb 25 '24 at 17:21
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    Two norms $|\cdot|_1$, $|\cdot|_2$ are said to be equivalent if they give rise to the same topology; this is easily shown to be equivalent to the existence of two constants $a,b>0$ such that $a|\cdot|_1\leq |\cdot|_2\leq b|\cdot|_1$. In finite-dimensions, this is always the case; for example on $\Bbb{R}^n$, you can easily verify this for the three common norms: the $\ell^1$ norm (i.e the sum norm), the $\ell^2$ norm (the Euclidean norm) and $\ell^{\infty}$ norm (i.e the max/sup norm), but the point is that this is true for all pairs of norms. – peek-a-boo Feb 25 '24 at 17:37

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