Introduction to Mathematical Analysis I - Second Edition

162 Solutions and Hints for Selected Exercises Therefore, f ( n + 1 ) ( 0 ) = 0. We have proved that for every n ∈ N , f is n times differentiable and, so, f ∈ C n ( R ) . Here we do not need to prove the continuity of f ( n ) because the differentiability of f ( n ) implies its continuity. In a similar way, we can also show that the function f ( x ) = ( e − 1 x , if x > 0; 0 , if x ≤ 0 is n times differentiable for every n ∈ N . SECTION 4.5 Exercise 4.5.1 . Let f ( x ) = e x . By Taylor’s theorem, for any x > 0, there exists c ∈ ( 0 , x ) such that f ( x ) = e x = m ∑ k = 0 f ( k ) ( 0 ) k ! x k + f ( m + 1 ) ( c ) ( m + 1 ) ! c m + 1 = m ∑ k = 0 x k k ! + e c ( m + 1 ) ! c m + 1 > m ∑ k = 0 x k k ! . Exercise 4.5.5 . (a) Observe that a simpler version of this problem can be stated as follows: If f is differentiable on ( a , b ) and ¯ x ∈ ( a , b ) , then lim h → 0 f ( ¯ x + h ) − f ( ¯ x ) h = f 0 ( ¯ x ) 1! . This conclusion follows directly from the definition of derivative. Similarly, if f is twice differentiable on ( a , b ) and ¯ x ∈ ( a , b ) , then lim h → 0 f ( ¯ x + h ) − f ( ¯ x ) − f 0 ( ¯ x ) h 1! h 2 = f 00 ( ¯ x ) 2! . We can prove this by applying the L’Hospital rule to get lim h → 0 f ( ¯ x + h ) − f ( ¯ x ) − f 0 ( ¯ x ) h 1! h 2 = lim h → 0 f 0 ( ¯ x + h ) − f 0 ( ¯ x ) 2 h = f 00 ( ¯ x ) 2! . It is now clear that we can solve part (a) by using the L’Hospital rule as follows: lim h → 0 f ( ¯ x + h ) − f ( ¯ x ) − f 0 ( ¯ x ) h 1! − f 00 ( ¯ x ) h 2 2! h 3 = lim h → 0 f 0 ( ¯ x + h ) − f 0 ( ¯ x ) − f 00 ( ¯ x ) h 1! 3 h 2 = f 000 ( ¯ x ) 3! . Note that the last equality follows from the previous proof applied to the function f 0 . (b) With the analysis from part (a), we see that if f is n times differentiable on ( a , b ) and ¯ x ∈ ( a , b ) , then lim h → 0 f ( ¯ x + h ) − ∑ n − 1 k = 0 f ( k ) ( ¯ x ) h k k ! h n + 1 = f ( n ) ( ¯ x ) n ! . This conclusion can be proved by induction. This general result can be applied to obtain the Taylor expansion with Peano’s remainder in Exercise 4.5.6 .