Is it possible to map from a parameter to a trajectory?

Question

If I have a polynomial trajectory

$$y(t)=at^2+bt$$

The idea I would like to express is that by fixing $a$ and $b$, I will get a unique trajectory $y(t)$. Can I say that there exists a map $M$ such that the trajectory $y(t)$ is $$y(t) = M(a,b)?$$ If not, how can I express the idea in proper math terms?

Answer 1

Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:

• One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $\mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define $$ M: \mathbb{R}^2 \to \mathbb{R}[t] : (a,b) \mapsto at^2 + bt. $$ Here, $\mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example, $$M(3,-2) = 3t^2 - 2t. $$
• Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write $$ M: \mathbb{R}^2 \to C(\mathbb{R}) : \varphi(a,b) \mapsto y_{a,b}, $$ where $y_{(a,b)}$ is the function defined by the formula $$ y_{a,b}(t) = at^2 + bt. $$ Here, $C(\mathbb{R})$ denotes the set of continuous functions on $\mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write $$ M(a,b) = (t \mapsto at^2 + bt). $$ In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.
• If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions $$ \{ y_{a,b}\}_{a,b\in\mathbb{R}}, $$ where $y_{a,b}$ is the function defined by $$ y_{a,b}(t) = at^2 + bt. $$ On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.

• Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e. $$ M : \mathbb{R}^2 \times [0,\infty) \to \mathbb{R} : (a,b,t) \mapsto at^2 + bt. $$ I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = \text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).

  Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.

Answer 2

I would express it as:

Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.

Answer 3

You can just use $$y_{a,b}(t)=at^2+bt$$

This implies that $y$ is a function taking some $t\in T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.