This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Differentiation Written by MenGen Tsai email: [email protected] 1. Let f be defined for all real x , and suppose that  f ( x ) f ( y )  ≤ ( x y ) 2 for all real x and y . Prove that f is constant. Proof:  f ( x ) f ( y )  ≤ ( x y ) 2 for all real x and y . Fix y ,  f ( x ) f ( y ) x y  ≤  x y  . Let x → y , therefore, ≤ lim x → y f ( x ) f ( y ) x y ≤ lim x → y  x y  = 0 It implies that ( f ( x ) f ( y )) / ( x y ) → 0 as x → y . Hence f ( y ) = 0, f = const . 2. Suppose f ( x ) > 0 in ( a, b ). Prove that f is strictly increasing in ( a, b ), and let g be its inverse function. Prove that g is differentible, and that g ( f ( x )) = 1 f ( x ) ( a < x < b ) . Proof: For every pair x > y in ( a, b ), f ( x ) f ( y ) = f ( c )( x y ) where y < c < x by MeanValue Theorem. Note that c ∈ ( a, b ) and f ( x ) > in ( a, b ), hence f ( c ) > 0. f ( x ) f ( y ) > 0, f ( x ) > f ( y ) if x > y , f is strictly increasing in ( a, b ). Let Δ g = g ( x + h ) g ( x ). Note that x = f ( g ( x )), and thus, ( x + h ) x = f ( g ( x + h )) f ( g ( x )) , 1 h = f ( g ( x ) + Δ g ) f ( g ( x )) = f ( g + Δ g ) f ( g ) . Thus we apply the fundamental lemma of differentiation, h = [ f ( g ) + η (Δ g )]Δ g, 1 f ( g ) + η (Δ g ) = Δ g h Note that f ( g ( x )) > 0 for all x ∈ ( a, b ) and η (Δ g ) → 0 as h → 0, thus, lim h → Δ g/h = lim h → 1 f ( g ) + η (Δ g ) = 1 f ( g ( x )) . Thus g ( x ) = 1 f ( g ( x )) , g ( f ( x )) = 1 f ( x ) . 3. Suppose g is a real function on R 1 , with bounded derivative (say  g  ≤ M ). Fix > 0, and define f ( x ) = x + g ( x ). Prove that f is onetoone if is small enough. (A set of admissible values of can be determined which depends only on M .) Proof: For every x < y , and x, y ∈ R , we will show that f ( x ) 6 = f ( y ). By using MeanValue Theorem: g ( x ) g ( y ) = g ( c )( x y ) where x < c < y, ( x y ) + (( x ) g ( y )) = ( g ( c ) + 1)( x y ) , that is, f ( x ) f ( y ) = ( g ( c ) + 1)( x y ) . ( * ) Since  g ( x )  ≤ M , M ≤ g ( x ) ≤ M for all x ∈ R . Thus 1 M ≤ g ( c ) + 1 ≤ 1 + M , where x < c < y . Take c = 1 2 M , and g ( c ) + 1 > where x < c < y for all x, y . Take into equation (*), and f ( x ) f ( y ) < since x y < 0, that is, f ( x ) 6 = f ( y ), that is, f is onetoone (injective). 2 4. If C + C 1 2 + ... + C n 1 n + C n n + 1 = 0 , where C , ..., C n are real constants, prove that the equation C + C 1 x + ... + C n 1 x n 1 + C n x n = 0 has at least one real root between 0 and 1. Proof: Let f ( x ) = C x + ... + C n n +1 x n +1 . f is differentiable in R 1 and f (0) = f (1) = 0. Thus, f (1) f (0) = f ( c ) where c ∈ (0 , 1) by MeanValue Theorem. Note that f ( x ) = C + C 1 x + ... + C n 1 x n 1 + C n x n ....
View
Full Document
 Spring '09
 ddd
 Calculus, Continuous function, Tn, Xn, lim Dn

Click to edit the document details