This is a beta course, so its structure, chapters, and examples may continue to change.
Mean Value Theorems for Derivatives
Mean value theorems connect an average rate of change on an interval with an instantaneous rate of change at some interior point. They underpin many proofs and estimates in calculus.
ξ \xi ξ (Xi):希腊字母,读作“克西”,常用来表示中值定理中存在的那个点。
Rolle’s Theorem
数学定理
定理是数学中经过严格证明的命题,是数学推理的基础。定理通常包含条件和结论,是数学知识体系的重要组成部分。
定理 1
If f f f is continuous on [ a , b ] [a,b] [ a , b ] , differentiable on ( a , b ) (a,b) ( a , b ) , and f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) , then there exists ξ ∈ ( a , b ) \xi \in (a,b) ξ ∈ ( a , b ) with f ′ ( ξ ) = 0 f'(\xi) = 0 f ′ ( ξ ) = 0 .
几何解释If a smooth curve starts and ends at the same height, some interior point must have a horizontal tangent.
证明
By the Extreme Value Theorem, f f f attains a max/min on [ a , b ] [a,b] [ a , b ] .
If an extreme point lies inside ( a , b ) (a,b) ( a , b ) , the derivative there is 0 0 0 .
If both extrema are at endpoints, f f f is constant and any interior point works.
Example
Show that x 3 − 3 x + 1 = 0 x^3 - 3x + 1 = 0 x 3 − 3 x + 1 = 0 has at most one real root in ( 0 , 1 ) (0,1) ( 0 , 1 ) .
Assume two roots x 1 ≠ x 2 x_1 \ne x_2 x 1 = x 2 in ( 0 , 1 ) (0,1) ( 0 , 1 ) .
Then f ( x 1 ) = f ( x 2 ) = 0 f(x_1) = f(x_2) = 0 f ( x 1 ) = f ( x 2 ) = 0 . Rolle gives ξ \xi ξ with f ′ ( ξ ) = 0 f'(\xi) = 0 f ′ ( ξ ) = 0 .
But f ′ ( x ) = 3 ( x 2 − 1 ) < 0 f'(x) = 3(x^2 - 1) < 0 f ′ ( x ) = 3 ( x 2 − 1 ) < 0 on ( 0 , 1 ) (0,1) ( 0 , 1 ) , contradiction.
Lagrange Mean Value Theorem (MVT)
数学定理
定理是数学中经过严格证明的命题,是数学推理的基础。定理通常包含条件和结论,是数学知识体系的重要组成部分。
定理 2
If f f f is continuous on [ a , b ] [a,b] [ a , b ] and differentiable on ( a , b ) (a,b) ( a , b ) , there exists ξ ∈ ( a , b ) \xi \in (a,b) ξ ∈ ( a , b ) such that
f ′ ( ξ ) = f ( b ) − f ( a ) b − a . f'(\xi) = \frac{f(b) - f(a)}{b - a}. f ′ ( ξ ) = b − a f ( b ) − f ( a ) .
几何解释At some point, the tangent slope equals the slope of the secant line joining ( a , f ( a ) ) (a,f(a)) ( a , f ( a )) and ( b , f ( b ) ) (b,f(b)) ( b , f ( b )) .
证明
Define g ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) b − a ( x − a ) g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x-a) g ( x ) = f ( x ) − f ( a ) − b − a f ( b ) − f ( a ) ( x − a ) .
Then g ( a ) = g ( b ) = 0 g(a) = g(b) = 0 g ( a ) = g ( b ) = 0 .
By Rolle, ∃ ξ \exists \xi ∃ ξ with g ′ ( ξ ) = 0 g'(\xi) = 0 g ′ ( ξ ) = 0 .
Hence f ′ ( ξ ) = f ( b ) − f ( a ) b − a f'(\xi) = \dfrac{f(b) - f(a)}{b-a} f ′ ( ξ ) = b − a f ( b ) − f ( a ) .
Examples
Prove ∣ sin x − sin y ∣ ≤ ∣ x − y ∣ |\sin x - \sin y| \le |x - y| ∣ sin x − sin y ∣ ≤ ∣ x − y ∣ .
Apply MVT to f ( t ) = sin t f(t) = \sin t f ( t ) = sin t on [ x , y ] [x,y] [ x , y ] , so cos ξ = sin y − sin x y − x \cos \xi = \dfrac{\sin y - \sin x}{y - x} cos ξ = y − x sin y − sin x with ∣ cos ξ ∣ ≤ 1 |\cos \xi| \le 1 ∣ cos ξ ∣ ≤ 1 .
For f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 on [ 1 , 2 ] [1,2] [ 1 , 2 ] , MVT gives 3 ξ 2 = 7 3\xi^2 = 7 3 ξ 2 = 7 , so ξ = 7 3 \xi = \sqrt{\tfrac{7}{3}} ξ = 3 7 .
Cauchy Mean Value Theorem
数学定理
定理是数学中经过严格证明的命题,是数学推理的基础。定理通常包含条件和结论,是数学知识体系的重要组成部分。
定理 3
If f , g f,g f , g are continuous on [ a , b ] [a,b] [ a , b ] , differentiable on ( a , b ) (a,b) ( a , b ) , and g ′ ( x ) ≠ 0 g'(x) \ne 0 g ′ ( x ) = 0 , then ∃ ξ ∈ ( a , b ) \exists \xi \in (a,b) ∃ ξ ∈ ( a , b ) such that
f ′ ( ξ ) g ′ ( ξ ) = f ( b ) − f ( a ) g ( b ) − g ( a ) . \frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}. g ′ ( ξ ) f ′ ( ξ ) = g ( b ) − g ( a ) f ( b ) − f ( a ) .
几何解释For two smooth curves sharing parameter x x x , some pair of points has tangent-slope ratio equal to the secant-slope ratio.
证明
Let h ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) g ( b ) − g ( a ) ( g ( x ) − g ( a ) ) h(x) = f(x) - f(a) - \dfrac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a)) h ( x ) = f ( x ) − f ( a ) − g ( b ) − g ( a ) f ( b ) − f ( a ) ( g ( x ) − g ( a )) .
h ( a ) = h ( b ) = 0 h(a) = h(b) = 0 h ( a ) = h ( b ) = 0 .
By Rolle, h ′ ( ξ ) = 0 h'(\xi) = 0 h ′ ( ξ ) = 0 .
Rearranging yields the conclusion.
Example
Take f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 , g ( x ) = x 3 g(x) = x^3 g ( x ) = x 3 on [ 1 , 2 ] [1,2] [ 1 , 2 ] . Cauchy MVT gives 2 ξ 3 ξ 2 = 3 7 \dfrac{2\xi}{3\xi^2} = \dfrac{3}{7} 3 ξ 2 2 ξ = 7 3 , so ξ = 14 9 \xi = \dfrac{14}{9} ξ = 9 14 .
Relations among the theorems
Rolle is the base case.
Lagrange MVT generalizes Rolle by dropping f ( a ) = f ( b ) f(a)=f(b) f ( a ) = f ( b ) .
Cauchy MVT generalizes Lagrange to two functions; letting g ( x ) = x g(x)=x g ( x ) = x recovers Lagrange.
Applications
Inequalities : e x > 1 + x e^x > 1 + x e x > 1 + x for x > 0 x>0 x > 0 (apply MVT to e x e^x e x on [ 0 , x ] [0,x] [ 0 , x ] ).
Root existence : sign change plus continuity ensures a root (Intermediate Value), and MVT refines counts.
Monotonicity : if f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 on an interval, f f f is strictly increasing there.
Convexity : f ′ ′ ( x ) > 0 f''(x)>0 f ′′ ( x ) > 0 implies f f f is convex (and secants lie above the graph).
Common pitfalls
Forgetting to check continuity on the closed interval.
Forgetting differentiability on the open interval.
Applying Cauchy MVT without ensuring g ′ ( x ) ≠ 0 g'(x)\ne 0 g ′ ( x ) = 0 .
Using an interval where f ( a ) = f ( b ) f(a)=f(b) f ( a ) = f ( b ) fails for Rolle.
Extensions
Integral mean value theorem : ∫ a b f ( x ) d x = f ( ξ ) ( b − a ) \int_a^b f(x)\,dx = f(\xi)(b-a) ∫ a b f ( x ) d x = f ( ξ ) ( b − a ) for continuous f f f .
Taylor’s theorem remainder : ∃ ξ \exists \xi ∃ ξ between x x x and x 0 x_0 x 0 such that
f ( x ) = ∑ k = 0 n f ( k ) ( x 0 ) k ! ( x − x 0 ) k + f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 . f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}. f ( x ) = ∑ k = 0 n k ! f ( k ) ( x 0 ) ( x − x 0 ) k + ( n + 1 )! f ( n + 1 ) ( ξ ) ( x − x 0 ) n + 1 .
Exercises
练习 1
Assume f f f is continuous on [ a , b ] [a,b] [ a , b ] , differentiable on ( a , b ) (a,b) ( a , b ) , and f ( a ) = f ( b ) f(a)=f(b) f ( a ) = f ( b ) . Show that ∃ ξ ∈ ( a , b ) \exists \xi \in (a,b) ∃ ξ ∈ ( a , b ) with f ′ ( ξ ) = 0 f'(\xi)=0 f ′ ( ξ ) = 0 .
Reference Answer (1 个标签)
mean value theorems
Rolle’s theorem applies directly because all hypotheses are satisfied, so f ′ ( ξ ) = 0 f'(\xi)=0 f ′ ( ξ ) = 0 for some ξ ∈ ( a , b ) \xi\in(a,b) ξ ∈ ( a , b ) .
练习 2
Prove ∣ cos x − cos y ∣ ≤ ∣ x − y ∣ |\cos x - \cos y| \le |x - y| ∣ cos x − cos y ∣ ≤ ∣ x − y ∣ .
Reference Answer (1 个标签)
mean value theorems
Apply MVT to f ( t ) = cos t f(t)=\cos t f ( t ) = cos t on [ x , y ] [x,y] [ x , y ] : − sin ξ = cos y − cos x y − x -\sin \xi = \dfrac{\cos y - \cos x}{y - x} − sin ξ = y − x cos y − cos x . Since ∣ sin ξ ∣ ≤ 1 |\sin \xi|\le 1 ∣ sin ξ ∣ ≤ 1 , the inequality follows.
练习 3
Verify Cauchy MVT for f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 , g ( x ) = x 3 g(x) = x^3 g ( x ) = x 3 on [ 1 , 2 ] [1,2] [ 1 , 2 ] .
Reference Answer (1 个标签)
mean value theorems
2 ξ 3 ξ 2 = 3 7 \dfrac{2\xi}{3\xi^2} = \dfrac{3}{7} 3 ξ 2 2 ξ = 7 3 ⇒ ξ = 14 9 \xi = \dfrac{14}{9} ξ = 9 14 .
Summary
本文出现的符号
符号 类型 读音/说明 在本文中的含义 ξ \xi ξ 希腊字母 Xi(克西) 中值定理中存在性的那个点
中英对照
中文术语 英文术语 音标 说明 微分中值定理 mean value theorem /miːn ˈvæljuː ˈθɪərəm/ 平均变化率等于某点瞬时变化率 罗尔定理 Rolle’s theorem /rəʊlz ˈθɪərəm/ 端点相等时存在水平切线 拉格朗日中值定理 Lagrange mean value theorem /ləˈɡrɑːndʒ ˌmiːn ˈvæljuː ˈθɪərəm/ 割线斜率等于某点切线斜率 柯西中值定理 Cauchy mean value theorem /ˈkəʊʃi ˌmiːn ˈvæljuː ˈθɪərəm/ 两函数导数比等于函数增量比 割线 secant line /ˈsiːkənt laɪn/ 连接区间端点的直线 单调性 monotonicity /ˌmɒnəʊtəˈnɪsɪti/ 函数增减趋势 凸函数 convex function /ˈkɒnvɛks ˈfʌŋkʃən/ 二阶导数大于零的函数