I'm just getting into multivariate and I'm just confused on some notation.

Question

If $T: R^2 \to R^2$ is defined by $T(s,t)=\big(\cos(t^2s),\log{(\sqrt{1+s^2})}\big)$, is $T(s,t)$ also the same as $\cos(t^2s)+\log{(\sqrt{1+s^2})}$? If not, then how to I make $T(s,t)=\big(\cos(t^2s),\log{(\sqrt{1+s^2})}\big)$ into something that excludes that comma?

Answer 1

No, they're different.

To exclude the comma, you could write $T(s,t)=\cos s^2t \bf{i}+\log\sqrt{1+s^2}\bf j$, but now here come the commas again, ${\bf i}{=(1,0)}$ and ${\bf j}{=(0,1)}$.

Or you could write $T(s,t)=\begin{pmatrix}\cos s^2t\\\log\sqrt{1+s^2}\end{pmatrix}$.

So the comma is pretty essential.

Answer 2

Clarifying, $F:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is a function that takes an n-dim. vector as an input and outputs a m-dim. vector. In your example, $T$ maps 2D vectors to 2D vectors.

The other function you're describing can be thought of as $F:\mathbb{R}^2 \rightarrow \mathbb{R}$ defined as $F(s,t)=\cos(t^2 s) + \log (\sqrt{1+s^2})$. This function maps 2D vectors to 1D vectors, and 1D vectors are simply scalars or numbers. This is why there are no commas, since your output vector is 1-dimensional.

Does that make sense?