Exercise 1.41.  Define a procedure double that takes a procedure of one 
argument as argument and returns a procedure that applies the original 
procedure twice. For example, if inc is a procedure that adds 1 to its 
argument, then (double inc) should be a procedure that adds 2. What value 
is returned by

(((double (double double)) inc) 5)

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(define (double f) (lambda (x) (f (f x))))

The returned value will be 21.

In pseudo-code (lambdas elided), evaluation proceeds as:

(((double (double double)) inc) 5)
(((double (double (double f))) inc) 5)
(((double (double (double (double f)))) inc) 5)
(((double (double (double (inc (inc x))))) 5)
...