推 wheremi :推!! 06/25 09:58
※ 引述《burgerking07 (joy)》之銘言：
: Show:
: If a function f satisfies f'''>0 on the real line,
: then for any t1< t2< t3< t4,
: we have
: 1 1 1 1
: det ( t1 t2 t3 t4 ) >0 . (証行列式值為正)
: t1^2 t2^2 t3^2 t4^2
: f(t1) f(t2) f(t3) f(t4)
: Thanks!
: ====================================================================
: P.S.
: 如果題目改成 f''>0, t1<t2<t3,
: 我可以証
: 1 1 1
: det ( t1 t2 t3 ) >0 ,
: f(t1) f(t2) f(t3)
: 不過上面題目還是証不出來.
Let
Ri=[t1^i t2^i t3^i t4^i] i=0,1,2
ri=[t1^i t2^i t3^i] i=0,1,2
F =[f(t1) f(t2) f(t3) f(t4)]
E =[f(t1) f(t2) f(t3)]
Since
[ 1 1 1 ]
A = det[ t1 t2 t3 ]=(t3-t2)(t3-t1)(t2-t1)>0
[t1^2 t2^2 t3^2]
Hence {r0,r1,r2} linearly independent,
So we can find a,b,c such that E=a*r0+b*r1+c*r2
Let g(t)=f(t)-(a+b*t+c*t^2)
Clearly g(ti)=0 for i=1,2,3
[ R0 ] (Row operation) [ R0 ]
[ R1 ] F--->F-a*R0-b*R1-c*R2 [ R1 ]
[ R2 ] -----------------------> [ R2 ]
[ F ] [0 0 0 g(t4)]
Hence
[ R0 ]
Det[ R1 ]=A*g(t4)
[ R2 ]
[ F ]
Since g(t)=f(t)-(a+b*t+c*t^2)
So g'''(t)=f'''(t)>0
By Mean Value Theorem
We can find zi,i=1,2,3 such that
t1 < z1 < t2 < z2 < t3 < z3 < t4 and
g(t2)-g(t1) g(t3)-g(t2) g(t4)-g(t3)
g'(z1)= ----------- = 0 g'(z2)= ----------- = 0 g'(z3)= -----------
t2-t1 t3-t2 t4-t3
By Mean Value Theorem
We can find yi,i=1,2 such that
z1 < y1 < z2 < y2 < z3 and
g'(z2)-g'(z1) g'(z3)-g'(z2) g'(z3)
g''(y1)= ------------- = 0 g''(y2)= ------------- = ----------
z2-z1 z3-z2 (z3-z2)
By Mean Value Theorem again,
We can find x such that y1 < x < y2 and
g''(y2)-g''(y1) g''(y2) g'(z3)
g'''(x)= --------------- = ----------- = -------------
y2-y1 (y2-y1) (y2-y1)(z3-z2)
g(t4)
= -----------------------
(y2-y1)(z3-z2)(t4-t3)
Since g'''(x)>0
So g(t4)>0
Hence
[ R0 ]
Det[ R1 ]=A*g(t4)>0 Q.E.D.
[ R2 ]
[ F ]
Clearly we can extend this problem to the more general cases.
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◆ From: 140.114.203.189
※ 編輯: cometic 來自: 140.114.203.189 (06/25 04:04)
※ 編輯: cometic 來自: 140.114.203.189 (06/25 04:12)
※ 編輯: cometic 來自: 140.114.203.189 (06/25 04:14)