Solution to Exercise 1.10:

`> (A 1 10)`

1024

> (A 2 4)

65536

> (A 3 3)

65536

`(f n)`

computes 2*n*.

`(g n)`

computes 2^{n} for *n*>0. For *n*=0, it is 0.

`(h n)`

computes *h*(*n*) such that if *n*=0, *h*(0)=0; otherwise *h*(*n*)=2^{h(n-1)}.

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