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：$mathop {Sup}limits_{x in left( { - 1,1} ight)} {f^{left( 3 ight)}}left( x ight) geqslant 3$