코시 평균값 정리
매개곡선 ( g ( t ) , f ( t ) ) (g(t), f(t)) ( g ( t ) , f ( t )) f ( t ) = t 2 f(t) = t^2 f ( t ) = t 2 g ( t ) = t 3 g(t) = t^3 g ( t ) = t 3 ξ = 14 / 9 \xi = 14/9 ξ = 14/9
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 ) ξ = 14 9 ≈ 1.556 \xi = \frac{14}{9} \approx 1.556 ξ = 9 14 ≈ 1.556 예제
보통의 평균값 정리가 g ( x ) = x g(x) = x g ( x ) = x
f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ξ ) g ′ ( ξ ) → g ( x ) = x f ′ ( ξ ) = f ( b ) − f ( a ) b − a \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)} \xrightarrow{g(x)=x} f'(\xi) = \frac{f(b)-f(a)}{b-a} g ( b ) − g ( a ) f ( b ) − f ( a ) = g ′ ( ξ ) f ′ ( ξ ) g ( x ) = x f ′ ( ξ ) = b − a f ( b ) − f ( a ) ξ = 14 / 9 ≈ 1.56 \xi = 14/9 \approx 1.56 ξ = 14/9 ≈ 1.56 2 / ( 3 ξ ) = 3 / 7 2/(3\xi) = 3/7 2/ ( 3 ξ ) = 3/7 증명 보기
코시 평균값 정리: f , g f, g f , g [ a , b ] [a, b] [ a , b ] ( a , b ) (a, b) ( a , b ) g ′ ≠ 0 g' \neq 0 g ′ = 0 ξ ∈ ( a , b ) \xi \in (a, b) ξ ∈ ( a , b ) f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ξ ) g ′ ( ξ ) \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)} g ( b ) − g ( a ) f ( b ) − f ( a ) = g ′ ( ξ ) f ′ ( ξ )
g ( x ) = x g(x) = x g ( x ) = x g ( b ) − g ( a ) = b − a g(b) - g(a) = b - a g ( b ) − g ( a ) = b − a g ′ ( ξ ) = 1 g'(\xi) = 1 g ′ ( ξ ) = 1 f ( b ) − f ( a ) b − a = f ′ ( ξ ) 1 = f ′ ( ξ ) \frac{f(b)-f(a)}{b-a} = \frac{f'(\xi)}{1} = f'(\xi) b − a f ( b ) − f ( a ) = 1 f ′ ( ξ ) = f ′ ( ξ ) 이것이 바로 보통의 평균값 정리입니다 — 코시 형태에서 g ( x ) = x g(x) = x g ( x ) = x
코시 평균값 정리로 lim x → 0 sin x x = 1 \lim_{x \to 0}\frac{\sin x}{x} = 1 lim x → 0 x s i n x = 1
sin x − sin 0 x − 0 = cos ξ x 1 , 0 < ξ x < x ⇒ cos ξ x → 1 \frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1}, \quad 0 < \xi_x < x \;\Rightarrow\; \cos \xi_x \to 1 x − 0 sin x − sin 0 = 1 cos ξ x , 0 < ξ x < x ⇒ cos ξ x → 1 ξ = 14 / 9 ≈ 1.56 \xi = 14/9 \approx 1.56 ξ = 14/9 ≈ 1.56 3 7 \frac37 7 3 증명 보기
x > 0 x > 0 x > 0 sin x x = sin x − sin 0 x − 0 \frac{\sin x}{x} = \frac{\sin x - \sin 0}{x - 0} x s i n x = x − 0 s i n x − s i n 0 f ( t ) = sin t f(t) = \sin t f ( t ) = sin t g ( t ) = t g(t) = t g ( t ) = t [ 0 , x ] [0, x] [ 0 , x ] g ′ = 1 ≠ 0 g' = 1 \neq 0 g ′ = 1 = 0 ξ x ∈ ( 0 , x ) \xi_x \in (0, x) ξ x ∈ ( 0 , x ) sin x − sin 0 x − 0 = cos ξ x 1 = cos ξ x \frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1} = \cos \xi_x x − 0 s i n x − s i n 0 = 1 c o s ξ x = cos ξ x 0 < ξ x < x 0 < \xi_x < x 0 < ξ x < x x → 0 + x \to 0^+ x → 0 + ξ x → 0 \xi_x \to 0 ξ x → 0 cos \cos cos cos ξ x → cos 0 = 1 \cos \xi_x \to \cos 0 = 1 cos ξ x → cos 0 = 1 x < 0 x < 0 x < 0 [ x , 0 ] [x, 0] [ x , 0 ] ξ x ∈ ( x , 0 ) \xi_x \in (x, 0) ξ x ∈ ( x , 0 ) cos ξ x → 1 \cos \xi_x \to 1 cos ξ x → 1 lim x → 0 sin x x = 1 \lim_{x \to 0}\frac{\sin x}{x} = 1 lim x → 0 x s i n x = 1
lim x → 0 e x − 1 − x x 2 = 1 2 \lim_{x \to 0}\frac{e^x - 1 - x}{x^2} = \frac12 lim x → 0 x 2 e x − 1 − x = 2 1 0 / 0 0/0 0/0
e x − 1 − x x 2 = e ξ − 1 2 ξ = e ξ 2 2 → x → 0 + 1 2 \frac{e^x-1-x}{x^2} = \frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} \xrightarrow{x \to 0^+} \frac12 x 2 e x − 1 − x = 2 ξ e ξ − 1 = 2 e ξ 2 x → 0 + 2 1 ξ = 14 / 9 ≈ 1.56 \xi = 14/9 \approx 1.56 ξ = 14/9 ≈ 1.56 2 / ( 3 t ) 2/(3t) 2/ ( 3 t ) 증명 보기
f ( x ) = e x − 1 − x f(x) = e^x - 1 - x f ( x ) = e x − 1 − x g ( x ) = x 2 g(x) = x^2 g ( x ) = x 2 f ( 0 ) = g ( 0 ) = 0 f(0) = g(0) = 0 f ( 0 ) = g ( 0 ) = 0 [ 0 , x ] [0, x] [ 0 , x ] x > 0 x > 0 x > 0 ( 0 , x ) (0, x) ( 0 , x ) g ′ ( t ) = 2 t ≠ 0 g'(t) = 2t \neq 0 g ′ ( t ) = 2 t = 0 코시 평균값 정리에 의해 어떤 ξ ∈ ( 0 , x ) \xi \in (0, x) ξ ∈ ( 0 , x ) f ( x ) g ( x ) = f ′ ( ξ ) g ′ ( ξ ) = e ξ − 1 2 ξ \frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)} = \frac{e^{\xi}-1}{2\xi} g ( x ) f ( x ) = g ′ ( ξ ) f ′ ( ξ ) = 2 ξ e ξ − 1 f 1 ( t ) = e t − 1 f_1(t) = e^t - 1 f 1 ( t ) = e t − 1 g 1 ( t ) = 2 t g_1(t) = 2t g 1 ( t ) = 2 t [ 0 , ξ ] [0, \xi] [ 0 , ξ ] f 1 ( 0 ) = g 1 ( 0 ) = 0 f_1(0) = g_1(0) = 0 f 1 ( 0 ) = g 1 ( 0 ) = 0 g 1 ′ = 2 ≠ 0 g_1' = 2 \neq 0 g 1 ′ = 2 = 0 ξ 2 ∈ ( 0 , ξ ) \xi_2 \in (0, \xi) ξ 2 ∈ ( 0 , ξ ) e ξ − 1 2 ξ = e ξ 2 2 \frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} 2 ξ e ξ − 1 = 2 e ξ 2
0 < ξ 2 < ξ < x 0 < \xi_2 < \xi < x 0 < ξ 2 < ξ < x x → 0 + x \to 0^+ x → 0 + ξ 2 → 0 \xi_2 \to 0 ξ 2 → 0 e t e^t e t e ξ 2 → 1 e^{\xi_2} \to 1 e ξ 2 → 1 1 2 \frac12 2 1 x < 0 x < 0 x < 0 [ x , 0 ] [x, 0] [ x , 0 ] lim x → 0 e x − 1 − x x 2 = 1 2 \lim_{x \to 0}\frac{e^x-1-x}{x^2} = \frac12 lim x → 0 x 2 e x − 1 − x = 2 1
9. 미분