关于泰勒公式的专题讨论

$f命题:$$f(Landau不等式)$设$f(x)$在$left( { - infty , + infty } ight)$上二次可微,且${M_k} = mathop {Sup}limits_{x in left( { - infty , + infty } ight)} left| {{f^{left( k ight)}}left( x ight)} ight| <  + infty ,k = 0,2$

证明:${M_1} = mathop {Sup}limits_{x in left( { - infty , + infty } ight)} left| {f'left( x ight)} ight| <  + infty $,且${M_1} le sqrt {2{M_0}{M_2}} $

1

$f命题:$ 设$fleft( x ight) in {C^2}left[ {a,b} ight]$,且$left| {int_a^b {fleft( x ight)dx} } ight| < int_a^b {left| {fleft( x ight)} ight|dx} $,${M_1} = mathop {Sup}limits_{x in left[ {a,b} ight]} left| {f'left( x ight)} ight|,{M_2} = mathop {Sup}limits_{x in left[ {a,b} ight]} left| {f''left( x ight)} ight|$,则[left| {int_a^b {f'left( x ight)dx} } ight| le frac{{{M_1}}}{2}{left( {b - a} ight)^2} + frac{{{M_2}}}{6}{left( {b - a} ight)^3}]

1

$f命题:$设$f(x)$在$(-1,1)$上二阶可导,且$fleft( 0 ight) = f'left( 0 ight) = 0$,若$$left| {f''left( x ight)} ight| le left| {fleft( x ight)} ight| + left| {f'left( x ight)} ight|$$在$(-1,1)$上总成立,则存在$x=0$的某个邻域内$fleft( x ight) equiv 0$

1

$f命题:$设$f(x)$在$[0,1]$上二阶可导,且$fleft( 0 ight) = 0,fleft( 1 ight) = 3,min limits_{x in left[ {0,1} ight]} fleft( x ight) =  - 1$,则存在$cin (0,1)$,使得$f''left( c ight) ge 18$

1

$f命题:$设$f(x)$在$[-1,1]$上二阶可导,$f(0)=0$,则存在$xi in [-1,1]$,使得$f''left( xi   ight) = 3int_{ - 1}^1 {fleft( x ight)dx} $

1

$f命题:$设$f(x)$在$[a,b]$上二次可微,且$f'left( {frac{{a + b}}{2}} ight) = 0$,证明:

(1)存在$xi in(a,b)$,使得[left| {f''left( xi   ight)} ight| geqslant frac{4}{{{{left( {b - a} ight)}^2}}}left| {fleft( b ight) - fleft( a ight)} ight|]

(2)若$f(x)$不恒为常数,则存在$eta in(a,b)$,使得[left| {f''left( eta   ight)} ight| > frac{4}{{{{left( {b - a} ight)}^2}}}left| {fleft( b ight) - fleft( a ight)} ight|]

1

$f(02中南七)$设$f(x)$在$left( { - infty , + infty } ight)$上存在三阶导数${f^{left( 3 ight)}}left( x ight)$,且$fleft( { - 1} ight) = 0,fleft( 1 ight) = 1,{f^{left( 1 ight)}}left( 0 ight) = 0$,证明:$mathop {Sup}limits_{x in left( { - 1,1} ight)} {f^{left( 3 ight)}}left( x ight) geqslant 3$

 

原文地址:https://www.cnblogs.com/ly142857/p/3676713.html