How does $φ(n,x) = (2x)^n$ constitute a dynamical system?

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I'm trying to make sense of the example in the following passage.

Definition 1.1. On the topological space X, if there exists a map φ : T × X → X satisfying the following conditions, then (X, φ) is called a dynamical system: (where T is, for example, R, Z, N, etc.)

(1) φ(x, 0) = x for any x ∈ X.

(2) φ(t, φ(s, x)) = φ(s + t, x) for any x ∈ X, s, t ∈ T

When T = R, t ∈ T is meant as continuous time. That is, φ(x, t) represents the location in X after a time t starting from the initial point x. On the other hand, T = Z or N is meant as a discrete time or number of trial iterations. For the example f(x) = 2x we saw at the beginning, if we consider φ(n, x) = fn(x), we can confirm that the definition of the dynamical system is satisfied.

I don't understand how this example constitutes a dynamical system; it doesn't seem to meet the first requirement, although it does meet the second.

If x is set to zero (n>0), I get...

φ(n, 0) = fn(0) = (2*0)n = 0n = 0 ≠ n.

If I suppose that the variables are in the wrong order and instead set n to zero (x≠0), I get...

φ(0, x) = f0(x) = (2x)0 = 1 ≠ x.

Am I misunderstanding something? Am I screwing up basic calculations somehow and need to get a full night's sleep? Or is the text wrong?

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Not sure what $f^n(x)$ means, but it seems that it is not $\left(f(x)\right)^n$. Probably, it is the iterative application of the function $f$, e.g., $f^3(x)=f\left(f\left(f(x)\right)\right)$. Then it works for me since $f^0(x)=x$. Anyway, the required $\phi$ should be $\phi(n,x) = 2^nx$ for $n \in N$.