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From: Peter on 20 May 2010 12:42
Hi,


I am working with a rational function where both the numerator and
denominator are quadratic polynomials,
P(x) = q(x)/r(x)
where
q(x) = x^2 + b*x + c, and q(x) > 0 for all x in R
r(x) = 1 + x^2

I can show treat the problem as a simple unconstrained optimization
problem and find that the stationary points of P are the roots to a
quadratic eqn, one root of which is the max and the other is the min.

I am wondering however if there is a simple way one can argue the
existence of a max and min based on the general properties of the
denominator and numerator given.

TIA,

Peter
From: James Waldby on 20 May 2010 14:52
On Thu, 20 May 2010 09:42:39 -0700, Peter wrote:

> Hi,
>
>
> I am working with a rational function where both the numerator and
> denominator are quadratic polynomials, P(x) = q(x)/r(x)
> where
> q(x) = x^2 + b*x + c, and q(x) > 0 for all x in R
> r(x) = 1 + x^2
>
> I can show treat the problem as a simple unconstrained optimization
> problem and find that the stationary points of P are the roots to a
> quadratic eqn, one root of which is the max and the other is the min.
>
> I am wondering however if there is a simple way one can argue the
> existence of a max and min based on the general properties of the
> denominator and numerator given.

I presume the difficulty is showing that max and min are attained;
it is easy to show that the function is bounded, since if |x| > 2,
|x| > |b|, and x^2 > c, we have 0 < (x^2 + b*x + c)/(1 + x^2) <=
(x^2 + |b*x| + c)/(1 + x^2) < (2*x^2 +c)/x^2 = 2 + (c/x^2) < 3.

--
jiw
From: Arturo Magidin on 20 May 2010 15:23
On May 20, 11:42 am, Peter <pwoly...(a)yahoo.com> wrote:
> Hi,
>
> I am working with a rational function where both the numerator and
> denominator are quadratic polynomials,
> P(x) = q(x)/r(x)
> where
> q(x) = x^2 + b*x + c,   and  q(x) > 0 for all x in R
> r(x)  = 1 + x^2
>
> I can show treat the problem as a simple unconstrained optimization
> problem and find that the stationary points of P are the roots to a
> quadratic eqn, one root of which is the max and the other is the min.
>
> I am wondering however if there is a simple way one can argue the
> existence of a max and min based on the general properties of the
> denominator and numerator given.

The function is defined and continuous on all the real numbers, since
r(x) is never equal to 0. Since the limits as x->oo and x->-oo are
both equal to 1, the function is bounded: there exists M>0 such that
for all x with |x|>M, you have |P(x)-1|<1. Thus, on (-oo,-M) and on
(M,oo), the function only takes values on (0,2). Since [-M,M] is
finite and closed, we know the function P(x) achieves a maximum and a
minimum there. So the function P(x) is bounded above and below.

That means that there is a supremum and infimum to the values of P(x),
and they are both real numbers. The only difficulty would be to show
that they are achieved. The only way in which they would not be
achieved would be if the function is always less than 1, or always
more than 1 (so that the maximum and minimum occur "at the limit"). In
particular, at least one of the max and min must be achieved (if the
function is ever less than 1, then the infinimum is less than 1 and
cannot occur "at the limit"; similarly if the function is ever more
than 1). So all you need to do is show that the function takes values
that are more than and less than 1 and that will imply that the
function achieves its supremum and infimum values.

--
Arturo Magidin
From: Gerry Myerson on 20 May 2010 18:44
In article
<e219ed9c-64e9-45e9-a300-a60505c35072(a)y21g2000vba.googlegroups.com>,
Arturo Magidin <magidin(a)member.ams.org> wrote:

> On May 20, 11:42�am, Peter <pwoly...(a)yahoo.com> wrote:
> > Hi,
> >
> > I am working with a rational function where both the numerator and
> > denominator are quadratic polynomials,
> > P(x) = q(x)/r(x)
> > where
> > q(x) = x^2 + b*x + c, � and �q(x) > 0 for all x in R
> > r(x) �= 1 + x^2
> >
> > I can show treat the problem as a simple unconstrained optimization
> > problem and find that the stationary points of P are the roots to a
> > quadratic eqn, one root of which is the max and the other is the min.
> >
> > I am wondering however if there is a simple way one can argue the
> > existence of a max and min based on the general properties of the
> > denominator and numerator given.
>
> The function is defined and continuous on all the real numbers, since
> r(x) is never equal to 0. Since the limits as x->oo and x->-oo are
> both equal to 1, the function is bounded: there exists M>0 such that
> for all x with |x|>M, you have |P(x)-1|<1. Thus, on (-oo,-M) and on
> (M,oo), the function only takes values on (0,2). Since [-M,M] is
> finite and closed, we know the function P(x) achieves a maximum and a
> minimum there. So the function P(x) is bounded above and below.
>
> That means that there is a supremum and infimum to the values of P(x),
> and they are both real numbers. The only difficulty would be to show
> that they are achieved. The only way in which they would not be
> achieved would be if the function is always less than 1, or always
> more than 1 (so that the maximum and minimum occur "at the limit"). In
> particular, at least one of the max and min must be achieved (if the
> function is ever less than 1, then the infinimum is less than 1 and
> cannot occur "at the limit"; similarly if the function is ever more
> than 1). So all you need to do is show that the function takes values
> that are more than and less than 1 and that will imply that the
> function achieves its supremum and infimum values.

Have you considered the case b = 0, c = 2?

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Arturo Magidin on 20 May 2010 21:45
On May 20, 5:44 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>
wrote:
> In article
> <e219ed9c-64e9-45e9-a300-a60505c35...(a)y21g2000vba.googlegroups.com>,
>  Arturo Magidin <magi...(a)member.ams.org> wrote:
>
>
>
> > On May 20, 11:42 am, Peter <pwoly...(a)yahoo.com> wrote:
> > > Hi,
>
> > > I am working with a rational function where both the numerator and
> > > denominator are quadratic polynomials,
> > > P(x) = q(x)/r(x)
> > > where
> > > q(x) = x^2 + b*x + c,   and  q(x) > 0 for all x in R
> > > r(x)  = 1 + x^2
>
> > > I can show treat the problem as a simple unconstrained optimization
> > > problem and find that the stationary points of P are the roots to a
> > > quadratic eqn, one root of which is the max and the other is the min.
>
> > > I am wondering however if there is a simple way one can argue the
> > > existence of a max and min based on the general properties of the
> > > denominator and numerator given.
>
> > The function is defined and continuous on all the real numbers, since
> > r(x) is never equal to 0. Since the limits as x->oo and x->-oo are
> > both equal to 1, the function is bounded: there exists M>0 such that
> > for all x with |x|>M, you have |P(x)-1|<1. Thus, on (-oo,-M) and on
> > (M,oo), the function only takes values on (0,2). Since [-M,M] is
> > finite and closed, we know the function P(x) achieves a maximum and a
> > minimum there. So the function P(x) is bounded above and below.
>
> > That means that there is a supremum and infimum to the values of P(x),
> > and they are both real numbers. The only difficulty would be to show
> > that they are achieved. The only way in which they would not be
> > achieved would be if the function is always less than 1, or always
> > more than 1 (so that the maximum and minimum occur "at the limit"). In
> > particular, at least one of the max and min must be achieved (if the
> > function is ever less than 1, then the infinimum is less than 1 and
> > cannot occur "at the limit"; similarly if the function is ever more
> > than 1). So all you need to do is show that the function takes values
> > that are more than and less than 1 and that will imply that the
> > function achieves its supremum and infimum values.
>
> Have you considered the case b = 0, c = 2?

Or for that matter, any irreducible quadratic in the numerator which
is strictly larger (or strictly smaller) than x^2+1 for all values of
x. Obviously, I hadn't. Certainly, at least one of the max and min
must be achieved (as I noted). But both need not in general.

On the other hand, the OP says he found that there are two local
extremes, one a local max and the other a local min. If you had
(x^2+2)/(x^2+1), the derivative would have numerator 2x(x^2+1) -
2x(x^2+2) = 2x, so the function would have only one stationary point.
The same will hold in the other situations I describe.

If you assume that the function will have two stationary points, both
of which are relative extremes, then you can't have both extremes be
larger than 1 (that would require a third local maximum); similarly
you can't have both be smaller than 1. So that will give you that the
extremes are necessarily achieved.

--
Arturo Magidin
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