5第五讲微分与不等式.docx
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5第五讲微分与不等式
关于泰勒中值定理的不等式
将f(α)在β处展开
α,β是区间端点a,b或区间中点a+b或极值点(包括最值点)c或任意点x
2
条件结论特殊点
设函数f
(x)在[0,1]上二阶可导,且满足
f'(x)
≤1,f
(x)在区间(0,1)内取得最大值1
4
证明:
f(0)+
f
(1)<1
将区间端点的值在最值点展开
设f(c)=maxf
0(x),c∈(0,1)
⇒f'(c)=0
f(0)=f
(c)
-
cf
'(c)+c
2
2
f'(ξ1)
0<ξ1f
(1)=f
(c)+(1-c)f'(c)+
(1-c)2
2
f'(ξ2)
c<ξ2<1
f(0)=
1+c2
42
f'(ξ1)
≤1+c
42
f'(ξ1)
≤1+c
42
f
(1)=
1(1-c)2
+
42
f'(ξ2)
≤1+
4
(1-c)2
2
f'(ξ2)
≤1+
4
(1-c)2
2
f(0)+
f
(1)
≤1+c
+(1-c)2
=1+
(c+1-c)2
-2c(1-c)<
1+1
222222
f(x)二次可微,f
(0)
=f
(1)
=0,maxf
0≤x≤1
(x)=
2,证明:
minf'(x)≤
0≤x≤1
-16
设f(c)=maxf(x),c∈[0,1]
0≤x≤1
⇒c∈(0,1)⇒f'(c)=0
将区间端点的值在最值点展开
f(0)=
f(c)
-
cf
'(c)
+
c2
2
f'(ξ1)
0<ξ1f
(1)=f(c)+(1-c)f'(c)+
(1-c)2
2
f'(ξ2)
c<ξ2<1
⇒f'(ξ1
)=-4
c2
f'(ξ2
)=-
4
(1-c)2
c与1-c当中必有一个≤1
-4与-
4当中必有一个≤-4
f'(ξ1
)与f'(ξ2
2c2
)当中必有一个≤-16
(1-c)2
ç⎛1⎫⎪
⎝2⎭
设f(x)在[a,b]上二阶可导,f'(a)=f'(b)=0,证明:
∃ξ∈(a,b),使得
f'(ξ)
≥4
(b-a)2
f(a)-f
(b)
在区间端点展开
f(x)=f
(a)+(x-a)f'(a)+
1(x-a)2
2
f'(ξ1)
a<ξ1f(?
)待定xf(x)
f(x)=f
(b)+(x-b)f'(b)+
1(x-b)2
2
f'(ξ2)
x<ξ1
0=f
(a)-f
(b)+
1(x-a)2
2
f'(ξ1
)-1(x-b)2
2
f'(ξ2)
f(a)-f
(b)=
1(x-a)2
2
f'(ξ1
)-1(x-b)2
2
f'(ξ2)
≤1(x-a)2
2
f'(ξ1)
+1(x-b)2
2
f'(ξ2)
(x-a)2
≤
+(x-b)2
2
f'(ξ)
f'(ξ)
=max{
f'(ξ1
),f'(ξ2
)},ξ∈{ξ1,ξ2}
f'(ξ)
≥
(x-a)2
2
+(x-b)2
f(a)-f(b)
取x=
a+b2
f(x)在[0,1]上二阶可导,f(0)=f
(1),f'(x)≤2,证明:
当0f(0)=
f(x)-xf
'(x)+x
2
2
f'(ξ1)
0<ξ1将区间端点的值在任意点展开
f
(1)=f(x)+(1-x)f'(x)+
(1-x)2
(1-x)2
2
x2
f'(ξ2)
x<ξ2<1
0=f'(x)+
2f'(ξ2)-
2f'(ξ1)
f'(x)=
(1-x)2
2
2
f'(ξ2)-2
f'(ξ1)
(1-x)2
≤
2
f'(ξ2)
+
x2
2
f'(ξ1)
≤(1-x)2+x2
=(1-x+x)2-2x(1-x)<1
设函数f
(x)在(+∞,-∞)内二阶可导,并且
f(x)
≤M0
,f'(x)
≤M2
证明:
f'(x)≤
(y-x)2
将任意点的值在任意点展开
f(y)=f(x)+(y-x)f'(x)+
2f'(ξ1)
ξ1介于x、y之间
f(x+h)
=f(x)+hf
'(x)+h
2
2
f'(ξ1)
ξ1介于x、x
+
h之间
f'(x)=
f(x+
h)-
f(x)
h
-
h2
2
f'(ξ1)
f'(x)=
f(x+h)-f(x)-h
2
f'(ξ1)
f(x+h)+
≤
f(x)
+
h2
2
f'(ξ1)
2
2M0+M2
≤2
hhh
h2h2
2M0+
2M2
22M0⋅
M2h2M
≥2=2
hh
M0M2
当2M0=
2M2
时取等号即当h=±2
0时取等号
M
2
设函数f
(x)在(+∞,-∞)内二阶可导,并且
f(x)
≤M0
,f'(x)
≤M2
证明:
f'(x)≤
2
f(x+h)=f(x)+hf'(x)+h
2
f'(ξ1)
ξ1介于x、x+h之间
h>0
f(x-h)
=f(x)-hf
'(x)+h
2
2
f'(ξ2)
ξ2介于x、x
-
h之间
f(x+
h)-
f(x
-h)=
2hf
'(x)+h
2
2
(f'
(ξ1
)-f
'(ξ2))
f'(x)=
f(x+
h)-
f(x
-
h)
-
h2
2
2h
(f'
(ξ1
)-f
'(ξ2))
f'(x)=
f(x+
h)-
f(x
-
h)
-
h2
2
(f'
(ξ1
)-f
'(ξ2))
f(x+h)+
≤
f(x-h)
+h2(
2
f'(ξ1)+
f'(ξ2))
2h
≤2M0
2h
+h2M2h
设函数f
(x)在(+∞,-∞)内二阶可导,并且
f(x)
≤M0
,f'(x)
≤M2
证明:
f'(x)≤
2M+h2M
22M⋅h2M2M
02
2h
≥02=
2h
当2M0
=h2M
2时取等号即当h=
0时取等号
M
2
2M+h2M
f'(x)
≤02
2h
设函数f(x)在(+∞,-∞)内二阶可导,并且
f(x)
≤M0
,f''(x)
≤M3
证明:
f'(x)≤
f(x
+
h)
=f(x)+hf
'(x)+h
2
f'(x)+h
6
f''(
ξ1)
x<ξ1
h>0
f(x-h)
=f(x)-hf
'(x)+h
2
2
f'(x)
-
h3
6
f''(ξ2)
x-h
<ξ2f(x+
h)-
f(x
-h)=
2hf
'(x)+h
3
6
(f''(
ξ1)+f
''(ξ2))
f'(x)=
f(x+
h)-
f(x
-
h)
-
h3
6
2h
(f''(
ξ1)+f
''(ξ2))
f'(x)=
f(x+
h)-
f(x
-
h)
-
h3
6
(f''(
ξ1)+f
''(ξ2))
f(x+h)+
≤
f(x-h)
+h3(
6
f''(ξ1)+
f''(ξ2))
2h2h
2M0+M3
≤3
2h夜雨教你数学竞赛
设函数f(x)在(+∞,-∞)内二阶可导,并且
f(x)
≤M0
,f''(x)
≤M3
证明:
f'(x)≤
2M0
h3
M
33
=M0
h2M
+3
=M0
+M0
h2M
+3
≥33
M0
⋅M0
h2M
⋅3=
2hh6
MMh2M
2h2h6
3M
2h2h6
当0
2h
=0
2h
=3
6
h3
时取等号即当h=3
0时取等号
M3
f'(x)
2M0+
≤
3M3
2h
下凸函数的加权琴生不等式
设f'(x)≥0,λ1
+λ2
+⋅⋅⋅+λn
=1且λk
>0,k
=1,2,⋅⋅⋅,n
则f(λ1x1
+λ2x2
+⋅⋅⋅+λnxn
)≤λ1f
(x1
)+λ2f
(x2
)+⋅⋅⋅+λnf
(xn)
设x0=λ1x1
+λ2x2+⋅⋅⋅+λnxn
(x-x)2
i若xk
≠x0
f(xk)=f
(x0
)+(xk
-
x0
)f'(x0
)+k0f'(ξ)
2k
ξk介于x
k、x
0之间
⇒f(xk
)≥f
(x0
)+(xk
-
x0
)f'(x0)
⇒λkf
(xk
)≥λkf
(x0
)+λk
(xk
-
x0
)f'(x0)
ii若xk
=x0
f(xk
)=f
(x0
)+(xk
-
x0
)f'(x0)
⇒λkf
(xk
)=λkf
(x0
)+λk
(xk
-
x0
)f'(x0)
故总有λkf
(xk
)≥λkf
(x0
)+λk
(xk
-
x0
)f'(x0)
∑λkf
k=1
n
(xk
)≥f
(x0
)∑λkk=1
+f'(x0
)∑λkk=1
n
(xk
-x0)
nn
⇒∑λkf
k=1
(xk
)≥f
(x0)
∑λkk=1
(xk
-
x0
)=∑λkxk
k=1
-∑λkx0k=1
=x0
-x0=0
下凸函数的加权琴生不等式
设f'(x)≥0,λ1
+λ2
+⋅⋅⋅+λn
=1且λk
>0,k
=1,2,⋅⋅⋅,n
则f(λ1x1
+λ2x2
+⋅⋅⋅+λnxn
)≤λ1f
(x1
)+λ2f
(x2
)+⋅⋅⋅+λnf
(xn)
上凸函数的加权琴生不等式
设f'(x)≤0,λ1
+λ2
+⋅⋅⋅+λn
=1且λk
>0,k
=1,2,⋅⋅⋅,n
则f(λ1x1
+λ2x2
+⋅⋅⋅+λnxn
)≥λ1f
(x1
)+λ2f
(x2
)+⋅⋅⋅+λnf
(xn)
下凸函数的琴生不等式
设f'(x)≥0则f(x1
+x2
+⋅⋅⋅+xn
)≤f(x1
)+f(x2
)+⋅⋅⋅+f(xn)
nn
上凸函数的琴生不等式
设f'(x)≤0则f(x1
+x2
+⋅⋅⋅+xn)≥f(x1)+f(x2)+⋅⋅⋅+f(xn)
nn
x+x+⋅⋅⋅+xn
均值不等式:
xk
>0,则
≥12n≥
n
≥
1+1
x1x2
+⋅⋅⋅+1
xn
x+x+⋅⋅⋅+x
x2+x2+⋅⋅⋅+x2
⎛x+x+⋅⋅⋅+x⎫
≥12n
⇔12n≥ç12n⎪
设f(x)=
n
x2,f'(x)=2>0
n⎝n⎭
x1+x2
+⋅⋅⋅+xn≥
n
⇔lnx1
+x2
+⋅⋅⋅+xn
n
≥lnx1
+lnx2
n
+
⋅⋅⋅+lnxn
设f(x)=lnx,f'(x)=-1<0
x2
1+1
+⋅⋅⋅+1
≥
1+1
x1x2
n⇔
+⋅⋅⋅+1
xn
≤x1x2xn
n
x+x+⋅⋅⋅+x
nsinx
⎛sinx⎫n
0<πk
=1,2,⋅⋅⋅,n记x
=12n
,证明:
∏k
≤ç0⎪
k0n
k=1xk
⎝x0⎭
nsinx
⎛sinx⎫n
nsinx
sinx
1nsinx
sinx
∏k
≤ç0⎪
⇔∑lnk
≤nln0⇔∑lnk
≤ln0
k=1xk
⎝x0⎭
k=1xk
x0n
k=1xkx0
设f(x)=lnsinx
x
f'(x)=
xcosx-sinxxsinx
=cotx-1
x
f'(x)=-csc2
x+1
x2
=1-
x2
1<0
sin2x
1111ab
a,b>0,p,q
>1,+
pq
=1,证明:
apbq≤+
pq
11ab11⎛11⎫
apbq
≤+⇔
pq
lna+
p
lnb≤lnçq⎝
⋅a+
p
⋅b⎪
q⎭
设f(x)=lnx,f'(x)=-1<0
x2
pq+r
正数p,q,r满足2p=q+r,证明:
≤1
qqrr
pq+r
qqrr
≤1⇔
pq+r
≤qqrr
⇔⎛ç
⎝
q+r2
q+r
⎪
⎭
≤qqrr
⇔(q+r)ln
q+r2
≤qlnq+rlnr
⇔q+rlnq+r≤qlnq+rlnr
设f(x)=
xlnx,f'(x)=1>0
x
222