Introduction to Mathematical Analysis I - Second Edition

159 with a similar strategy. 1. Prove that | cos ( a ) − cos ( b ) | ≤ | a − b | for all a , b ∈ R . 2. Prove that | ln ( 1 + e 2 a ) − ln ( 1 + e 2 b ) | ≤ 2 | a − b | for all a , b ∈ R . Exercise 4.2.4 . Let us define f : [ − π , π ] → R by f ( x ) = x + n ∑ k = 1 ( a k sin kx + b k cos kx ) . We want to find c ∈ ( − π , π ) such that f ( c ) = 0. Now, consider the function g ( x ) = x 2 2 + n ∑ k = 1 − a k cos ( kx ) k + b k sin ( kx ) k . Observe that g ( − π ) = g ( π ) and g 0 = f . The conclusion follows from Rolle’s Theorem. Exercise 4.2.5 . Use the identity 1 g ( b ) − g ( a ) f ( a ) f ( b ) g ( a ) g ( b ) = f ( a ) g ( b ) − f ( b ) g ( a ) g ( b ) − g ( a ) = f ( a ) g ( a ) − f ( b ) g ( b ) 1 g ( a ) − 1 g ( b ) . Then apply the Cauchy mean value theorem for two functions φ ( x ) = f ( x ) g ( x ) and ψ ( x ) = 1 g ( x ) on the interval [ a , b ] . Exercise 4.2.6 . 1. Apply Rolle’s theorem to the function f ( x ) = a 1 x + a 2 x 2 2 + · · · + a n x n n on the interval [ 0 , 1 ] . 2. Apply Rolle’s theorem to the function f ( x ) = n ∑ k = 0 sin ( 2 k + 1 ) x 2 k + 1 on the interval [ 0 , π / 2 ] . Exercise 4.2.8 . (a) Given ε > 0, first find x 0 large enough so that a − ε / 2 < f 0 ( x ) < a + ε / 2 for x > x 0 . Then use the identity f ( x ) x = f ( x ) − f ( x 0 )+ f ( x 0 ) x − x 0 + x 0 = f ( x ) − f ( x 0 ) x − x 0 + f ( x 0 ) x − x 0 1 + x 0 x − x 0 , and the mean value theorem to show that, for x large, a − ε < f ( x ) x < a + ε . (b) Use the method in part (a). (c) Consider f ( x ) = sin ( x ) .