Difference between arbitrary constant and free variable

Question

Take the following mathematical sentence $f(x) = ax^2 + bx + c$, before I learnt about first-order-logic I would say, $x$ is a variable and $a,b,c$ are all arbitrary constants. Now after learning about logic I would write down something like $\forall x \in \mathbb{R}: f(x) = ax^2 + bx + c$, where $x$ is a bound variable and $a,b,c$ are free variables. Do these two interpretations agree with each other, or is one wrong?

Answer 1

The comments you have gotten so far are inaccurate. In your first sentence "arbitrary constant" does not have a proper meaning, and is ambiguous or incorrect depending on exactly how you interpreted it. In the first place, it is not even really a "mathematical sentence" as it is missing its context. In the typical contexts, $a,b,c$ are the same for every $x$, and any proper mathematical writing would make that clear such as in:

1. [In a context where $f : ℝ→ℝ$ has already been defined, we may make deductions until..] There are $a,b,c∈ℝ$ such that $f(x) = a·x^2+b·x+c$ for every $x∈ℝ$. Symbolically, $∃a,b,c{∈}ℝ\ ∀x{∈}ℝ\ ( \ f(x) = a·x^2+b·x+c \ )$.
2. [In a context where $a,b,c∈ℝ$ have already been defined..] Let $f : ℝ→ℝ$ such that $f(x) = a·x^2+b·x+c$ for each $x∈ℝ$. Symbolically, [we pick a fresh variable $f$ and then can write down..] $∀x{∈}ℝ\ ( \ f(x) = a·x^2+b·x+c \ )$.

Note that in both cases there is nothing arbitrary about $a,b,c$. In the first case, we prove that there is some specific combination of $a,b,c$ such that $∀x{∈}ℝ\ ( \ f(x) = a·x^2+b·x+c \ )$, and it may be that there is only one. In the second case, $a,b,c$ have been defined earlier, and may be specific too.