a2 - n such that there is a fixed point of f in the...

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York University EECS 3101 September 17, 2015 Homework Assignment #2 Due: September 24, 2015 at 4:00 p.m. 1. In this question we will consider functions f : IR IR. A value x IR is called a fixed point of f if f ( x ) = x . A function f is called strictly decreasing if, for all x, y IR, x < y f ( x ) > f ( y ). In solving this problem, you may use the following result from calculus. Bolzano’s Theorem Let g : IR IR be a continuous function, and let a and b be real numbers with a < b , g ( a ) < 0 and g ( b ) > 0. Then, there exists a real number c such that a < c < b and g ( c ) = 0. For the remainder of this assignment, let f : IR IR be any continuous, strictly decreas- ing function f that has f (0) > 0. (a) Let f : IR IR be a continuous strictly decreasing function and let a and b be real numbers such that a < b , f ( a ) a and f ( b ) < b . Prove that there exists a fixed point c of f such that a c < b . (b) Show that f has a fixed point x such that 0 x < f (0). (c) Provide pseudocode for an algorithm that outputs an integer
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Unformatted text preview: n such that there is a fixed point of f in the interval [ n,n + 1) = { x ∈ IR : n ≤ x < n + 1 } . The number of times your algorithm calls the function f should be O (log | f (0) | ). Your pseudocode should include pre- and post-conditions. If you use a loop, you should also include a loop invariant in the pseudocode that you can use for part (c), below. (d) Prove that the algorithm given in part (b) is correct. Note: you do not have to prove that your algorithm calls f O (log | f (0) | ) times, but it should be true in order for your algorithm to get full marks. Example: The graph below shows the strictly decreasing function f ( x ) = (7-x ) 3 30 . It has a fixed point between 2 and 3, so if your algorithm is run with this function f , it should output 2. 1...
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