Consider the following “function” f de?ned recursivelyas follows: • f (1) = 1 • If n is even, f (n) = f (n/2) • If n is odd, then f (n) = f (3n + 1). Show that the value of the “function” de?ned as aboveis equal to 1 for all positive integers n less than or equal to 10. (After the exam, those who are interested cancome see me about an unsolved problem in computer science/mathematics related to this: It isstill unknown whether this “function” is well-de?nedafter more than 70 years of study by some very smart folks.) Solution f(9) = f(28) = f(14) = f(7) =f(22) = f(11) = f(34) = f(17) = f(52) = f(26) = f(13) = f(40) =f(20) = f(10) = f(5) = f(16) =f(8) = f(4) =f(2) = f(1) = 1 f(6) = f(3) = f(10) = 1 (as shownabove).