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# Rudin_5 - Differentiation Written by Men-Gen Tsai email...

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Unformatted text preview: Differentiation Written by Men-Gen 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 Mean-Value 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 one-to-one 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 Mean-Value 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 one-to-one (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 Mean-Value Theorem. Note that f ( x ) = C + C 1 x + ... + C n- 1 x n- 1 + C n x n ....
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    2. If an answer to that mathematical question is provided in the text

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TYPE: <one of the following values>
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- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# Rudin_5 - Differentiation Written by Men-Gen Tsai email...

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Unformatted text preview: Differentiation Written by Men-Gen 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 Mean-Value 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 one-to-one 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 Mean-Value 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 one-to-one (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 Mean-Value Theorem. Note that f ( x ) = C + C 1 x + ... + C n- 1 x n- 1 + C n x n ....
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The provided text is a set of solutions to problems from a mathematics textbook, specifically on the topic of differentiation. It contains multiple questions and their corresponding answers.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS