Although addressed indirectly in the other answers here, it seems relevant to point out that, strictly speaking, the resulting series expansion in all cases (so far) is -not- a Taylor series. The ...

Based on common conventions for set builder notation, I believe the "for all" or "any" is implied. For example, to refer to the set $A$ of all real numbers from 0 to 5, it is ...

Recall that $\vec{a} \times \vec{b} = - \, \vec{b} \times \vec{a}$. Now consider each, using your definitions (I have included the unit vector $\hat{e}_i$ explicitly, simply for additional clarity; ...

While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well. Does the “field” over which a ...

A summary of the conclusion drawn from the comments (thanks to @saulspatz) Given that we write $k \odot \vec{v} = k\vec{v}$, in the case of the scalar $-1$, this would be either $$-1 \odot \vec{v} = -... View answer 0 votes The power series$$\sum_{n=0}^{\infty} 0 \, x^n = 0 + 0x + 0x^2 + 0x^3 + \dots clearly converges to zero for any value of $x$. That is, if we call this power series $g(x)$, then we can say $g(x) = 0$...

Taking your comment about “any remainder” in mind, BCLC’s comment is effectively the answer you seem to be looking for, i.e. the remainder $R_n(x)$ is simply the difference between the $n^{th}$ order ...