 In such a definition, the first clause (known as the base case) determines the value of \(h\) at 0, while the second clause determines how its value at \(x+1\) depends on its value at \(x\). In this case it is easy to see that the value of \(x\) determines how many times the function \(g\) is iterated (i.e., applied to itself) in determining the value of \(h\). For instance, if \(n = 3\) and \(g(x) = mult(x,x)\), then \(h(x) = 3^{x+1}\)—i.e., the \(x+1\)st iterate of the map \(x \mapsto 3 \times x\).