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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Thu Jul 14, 2005 6:26 am Post subject:
f=Ax^2+Bx+C , f(x) = < 1 ?



Let f(x)=Ax^2+Bx+C with real coefficients and infty <a<b<infty.
Prove or disprove that f(x)=< 1 , for all x in [a,b] , if and
only if
f(a)=< 1 , f(b)=< 1 and moreover
 sqrt((1+f(a))*(1+f(b))) =<
=< (f(a)+f(b))/2  2f*((a+b)/2) =<
=< sqrt((1f(a))*(1f(b))) . 

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Posted: Fri Jul 22, 2005 7:02 am Post subject:
Re: f=Ax^2+Bx+C , f(x) = < 1 ?



Alex. Lupas wrote:
Quote:  Let f(x)=Ax^2+Bx+C with real coefficients and infty <a<b<infty.
Prove or disprove that f(x)=< 1 , for all x in [a,b] , if and
only if
f(a)=< 1 , f(b)=< 1 and moreover
 sqrt((1+f(a))*(1+f(b))) =
=< (f(a)+f(b))/2  2f*((a+b)/2) =
=< sqrt((1f(a))*(1f(b))) .

Is there some other condition on C ?
A, B, C ,x real, Ax^2+Bx =y
C = int(y) => f(x) <1 

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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Sat Jul 23, 2005 6:51 am Post subject:
Re: f=Ax^2+Bx+C , f(x) = < 1 ?



HINT: Let G(x):=Px^2+Qx+R with real coefficients. Let us prove that
G >= 0 on [a,b], if and only if : G(a)>=0 ,G(b)>=0 and moreover
(G(a)+G(b))/{2}2*G((a+b)*0.5) >= sqrt(G(a)G(b)) .
Proof:
1). Suppose that
G(a)>= 0 , G(b)>= 0 and
(1)
(G(a)+G(b))/{2}2*G((a+b)*0.5) >= sqrt(G(a)G(b)) .
Denote
m= ( sqrt(G(b))sqrt(G(a)) )/(ba)
n=( b*sqrt(G(a))a*sqrt(G(b)) )/(ba)
(1.1) K=[G(a)+G(b)4*G((a+b)*0.5)2*sqrt(G(a)G(b))]/(ba)^2
Then K>= 0 and on the other hand holds the identity
G(x)=(mx+n)^2+K(xa)(ba)
which implies
(2) G(x) >= 0 , forall x in [a,b].
2). Further assume (2) as verified . Then G(a)>= 0,G(b)>=0
and moreover G(x) may be written as (Lukacs's Theorem)
(3) G(x)=(m_1x+n_1)^2+ K_1(xa)(bx) with K>= 0 ,
m_1, n_1 being real constants .
If we select in (3) x=a , x=b and then x=(a+b)*0.5
then we find
(m_1a+n_1)^2 =G(a)
(*)
(m_1b+n_1)^2 =G(b)
and K_1=K , K being as in (1.1) . Note that the system (*) furnish
us
m_1 ,n_1 and we must have K>= 0 , therefore inequalities (1) are
verified.
========
Further apply the equivalence
G(x)>= 0 <====> G(a)>= 0, G(b)>=0 , K= >= 0 ,
to the polynomial functions G_1(x):=1 f(x) and to
G_2(x)=f(x)+1 . 

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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Sat Jul 23, 2005 12:41 pm Post subject:
Re: f=Ax^2+Bx+C , f(x) = < 1 ?



[Correction]
QUESTION: Let f(x)=Ax^2+Bx+C with real coefficients and infty
<a<b<infty.
Prove or disprove that f(x)=< 1 , for all x in [a,b] , if and only
if
f(a)=< 1 , f(b)=< 1 and moreover
1 + sqrt((1f(a))*(1f(b))) =<
=< (f(a)+f(b))/2  2f*((a+b)/2) =<
=< 1 sqrt((1+f(a))*(1+f(b) )
SOLUTION: we need following
Proposition: Let G(x):=Px^2+Qx+R with real coefficients. Then
G >= 0 on [a,b], if and only if : G(a)>=0 ,G(b)>=0 and moreover
2*G((a+b)*0.5) (G(a)+G(b))/2 >= sqrt(G(a)G(b)) .
Proof: 1). Suppose that
G(a)>= 0 , G(b)>= 0 and
(1)
2*G((a+b)*0.5) (G(a)+G(b))/{2} >= sqrt(G(a)G(b)) .
Denote
m= ( sqrt(G(b))sqrt(G(a)) )/(ba)
n=( b*sqrt(G(a))a*sqrt(G(b)) )/(ba)
(1.1) K[G]=[4*G((a+b)*0.5)(G(a)+G(b))2*sqrt(G(a)G(b))]/(ba)^2 .
Then K>= 0 and on the other hand holds the identity
G(x)=(mx+n)^2+K(xa)(ba)
which implies
(2) G(x) >= 0 , forall x in [a,b].
2). Further assume (2) as verified . Then G(a)>= 0,G(b)>=0
and moreover G(x) may be written as (Lukacs's Theorem)
(3) G(x)=(m_1x+n_1)^2+ K_1(xa)(bx) with K>= 0 ,
m_1, n_1 being real constants .
If we select in (3) x in {a ,(a+b)*0.5, b} , then we find
(m_1a+n_1)^2 =G(a)
(*)
(m_1b+n_1)^2 =G(b)
and K_1=K[G], K[G] being as in (1.1) . Note that the system (*)
furnish us
m_1 ,n_1 and we must have K>= 0 , therefore inequalities (1) are
verified.
========
Further apply the equivalence
G(x)>= 0 <====> G(a) >= 0, G(b) >=0 , K[G] >= 0 ,
to the polynomial functions G_1(x):=1 f(x) , G_2(x)=F(x)+1 .
Inequalities G_1 >= 0 and G_2 >=0 on [a,b], are satisfied iff
f(a) =< 1 , f(b)=< 1 and moreover K[G_1] >= 0 and K[G_2]>= 0.
But the last two inequalities are equivalent with
12f((a+b)*0.5)+ (f(a)+f(b))/2 >= sqrt((1f(a))*(1f(b)))
1+2f((a+b)*0.5)(f(a)+f(b))/2 >= sqrt((1+f(a))*(1+f(b)))
which completes the solution. 

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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Sat Jul 23, 2005 1:17 pm Post subject:
Re: f=Ax^2+Bx+C , f(x) = < 1 ?



Instead of
Quote:  (1.1) K[G]=[4*G((a+b)*0.5)(G(a)+G(b))2*sqrt(G(a)G(b))]/(ba)^2 .
Then K>= 0 and on the other hand holds the identity
G(x)=(mx+n)^2+K(xa)(ba)
please read 
[...]Then K>= 0 and on the other hand holds the identity
G(x)=(mx+n)^2+K(xa)(bx)
/Sorry,proposer 

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