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quizsol08

# quizsol08 - x For any sequence x n in R converging to x f x...

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Math 104, Solution to Quiz 8 Instructor: Guoliang Wu July 21, 2009 1. (15 points) Prove that the following equation has at least one real solution in (0,1): e x 2 = x + 3 2 . Proof. Let f ( x ) = e x 2 - ( x + 3 2 ) . Then f is continuous on [0 , 1] . f (0) = 1 - 3 2 = - 1 2 < 0; f (1) = e - 5 2 > 0 . By the Intermediate Value Theorem, there exists x 0 [0 , 1] such that f ( x 0 ) = 0 . Clearly x 0 = 0 or 1 since f (0) = 0 , f (1) = 0 . Therefore, therefore x 0 (0 , 1) and f ( x 0 ) = 0 e x 2 0 = x 0 + 3 2 . 2. (15 points) Proof by induction that for any n N , f ( x ) = x n is continuous on R . Proof. (i) n = 1 . Let any x 0 R . We prove that f ( x ) = x is continuous at
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Unformatted text preview: x . For any sequence ( x n ) in R converging to x , f ( x n ) = x n converges to x trivially. Hence f ( x ) = x is continu-ous. (ii) Suppose x n is continuous on R , we want to show that x n +1 is also continuous. Note that x n +1 = ( x n ) x. Hence x n +1 is also continuous on R since both x n and x are. By induction, for any n ∈ N , f ( x ) = x n is continuous on R . 1...
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