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f=Ax^2+Bx+C , |f(x)| = < 1 ?
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Alex. Lupas
science forum Guru Wannabe


Joined: 06 May 2005
Posts: 245

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

Instead of
Quote:
(1.1) K[G]=[4*G((a+b)*0.5)-(G(a)+G(b))-2*sqrt(G(a)G(b))]/(b-a)^2 .
Then K>= 0 and on the other hand holds the identity

G(x)=(mx+n)^2+K(x-a)(b-a)
please read

[...]Then K>= 0 and on the other hand holds the identity

G(x)=(mx+n)^2+K(x-a)(b-x)

/Sorry,proposer
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Alex. Lupas
science forum Guru Wannabe


Joined: 06 May 2005
Posts: 245

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

[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((1-f(a))*(1-f(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)) )/(b-a)
n=( b*sqrt(G(a))-a*sqrt(G(b)) )/(b-a)

(1.1) K[G]=[4*G((a+b)*0.5)-(G(a)+G(b))-2*sqrt(G(a)G(b))]/(b-a)^2 .

Then K>= 0 and on the other hand holds the identity

G(x)=(mx+n)^2+K(x-a)(b-a)

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(x-a)(b-x) 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

1-2f((a+b)*0.5)+ (f(a)+f(b))/2 >= sqrt((1-f(a))*(1-f(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

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

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)) )/(b-a)

n=( b*sqrt(G(a))-a*sqrt(G(b)) )/(b-a)

(1.1) K=[G(a)+G(b)-4*G((a+b)*0.5)-2*sqrt(G(a)G(b))]/(b-a)^2

Then K>= 0 and on the other hand holds the identity

G(x)=(mx+n)^2+K(x-a)(b-a)

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(x-a)(b-x) 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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PostPosted: Fri Jul 22, 2005 7:02 am    Post subject: Re: f=Ax^2+Bx+C , |f(x)| = < 1 ? Reply with quote

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((1-f(a))*(1-f(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

PostPosted: Thu Jul 14, 2005 6:26 am    Post subject: f=Ax^2+Bx+C , |f(x)| = < 1 ? Reply with 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((1-f(a))*(1-f(b))) .
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