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Thread Subject:
fractional polynomial

Subject: fractional polynomial

From: Frank

Date: 3 Dec, 2010 23:51:06

Message: 1 of 8

Hi,

I have a question about fractional polynomial and I cannot solve it. Can anyone give some suggestions? The problem is as follows:

I have two fractional polynomial, f(z) and g(z), which are in the form of

f(z) = sum_{i=1}^K a_i z^(p_i) + 1

and

g(z) = sum_{i=1}^K a_i z^(q_i) + 1.

They have the same coefficients, a_i. Furthermore, p_i and q_i are real numbers, they can be positive or negative.

If I have z_1, z_2, ..., z_K satisfy both f(z) = 0 and g(z) = 0, can I say that f(z) = q(z), that is, p_1 = q_1, p_2 = q_2, ..., p_K = q_K?

Thanks a lot.

Frank

Subject: fractional polynomial

From: Walter Roberson

Date: 4 Dec, 2010 02:02:09

Message: 2 of 8

On 03/12/10 5:51 PM, Frank wrote:

> I have a question about fractional polynomial and I cannot solve it. Can
> anyone give some suggestions? The problem is as follows:
>
> I have two fractional polynomial, f(z) and g(z), which are in the form of
>
> f(z) = sum_{i=1}^K a_i z^(p_i) + 1
>
> and
>
> g(z) = sum_{i=1}^K a_i z^(q_i) + 1.
>
> They have the same coefficients, a_i. Furthermore, p_i and q_i are real
> numbers, they can be positive or negative.
>
> If I have z_1, z_2, ..., z_K satisfy both f(z) = 0 and g(z) = 0, can I
> say that f(z) = q(z), that is, p_1 = q_1, p_2 = q_2, ..., p_K = q_K?

If z_{i} are particular fixed values that are the same between the two,
then the answer is NO.

Suppose z{i} are all 1, and the coefficients are such that f(z) = 0.
Now for each p{i} let q{i} = -p{i}. As 1^1 is the same as 1^(-1) the
value of the expression will clearly not change so g(z) would also be 0,
and yet you cannot conclude on the basis of that one point that f(z) =
g(z) .

I do not follow your notation "sum_{i=1}^K"


If what you have is sum(a{i}*z{i}^p{i}, i=1..K) + 1, and
sum(a{i}*z{i}^q{i}, i=1..K), then for consider particular terms and
whether they will come out the same:

a*z^p = a*z^q
ln(a) + ln(z) * p = ln(a) + ln(z) * q

If ln(a) is infinite (positive or negative) and ln(z) is not infinite,
the terms would be the same. The simplest case of this is ln(a) =
-infinity, which is to say a == 0.

If ln(a) is not infinite, then it can be subtracted from both sides, leaving

ln(z) * p = ln(z) * q

If ln(z) is infinite, then this is satisfied if p and q are non-zero and
the same sign.

If ln(z) is finite, there is still another trivial solution: if ln(z) is
0, then the terms become equal provided that neither p nor q are
infinite. ln(z) is 0 if z is 1.

You can thus see that there are multiple circumstances under which
individual terms come out exactly the same with different p and q -- and
that's not even taking in to account that there may be particular
combinations of p and q that happen to make the overall expressions the
same.


A trickier question comes if instead of allowing us to choose arbitrary
z that make the equation 0, you say that the system is nilpotent -- that
is, that f(z) = 0 and g(z) = 0 for *all* z. *If* that is the case, then
you are going to have to specify the exact constraints so we know what
we are working with, and you are going to have to be clear as to whether
z is a single numeric value or represents a vector of values equal in
length to the number of p and q.

Subject: fractional polynomial

From: Frank

Date: 4 Dec, 2010 03:33:04

Message: 3 of 8

Thanks, Walter.

First, sum_{i=1}^K means that summation from i equal 1 to K where i is the running index. For example,

f(z) = sum_{i=1}^K a_i z^(p_i) + 1
      = a_1 * z^(p_1) + a_2 * z^(p_2) + ... + a_K * z^(p_K) + 1

g(z) = sum_{i=1}^K a_i z^(q_i) + 1
      = a_1 * z^(q_1) + a_2 * z^(q_2) + ... + a_K * z^(q_K) + 1

where i in a_i is the subscript.

Furthermore, a_i and z_i are finite. If z_1, z_2, ..., z_K, which are the roots of f(z) and g(z), are in the form of z_1 = e^{j*x_1}, z_2 = e^{j* x_2}, ..., z_K = e^{j* x_K}. Here, e^{j*x_1} is the exponential with power j*x_1 and j = sqrt(-1). Moreover, x_1, x_2,...,x_K are finite and z_1, z_2, ..., z_K are not equal to 1.

If we now know that z_i s have magnitude 1 but not equal 1, then we have the conclusion that p_i = q_i, i=1,2,...,K?

Thanks a lot.

Frank


Walter Roberson <roberson@hushmail.com> wrote in message <CihKo.3503$%83.2715@newsfe16.iad>...
> On 03/12/10 5:51 PM, Frank wrote:
>
> > I have a question about fractional polynomial and I cannot solve it. Can
> > anyone give some suggestions? The problem is as follows:
> >
> > I have two fractional polynomial, f(z) and g(z), which are in the form of
> >
> > f(z) = sum_{i=1}^K a_i z^(p_i) + 1
> >
> > and
> >
> > g(z) = sum_{i=1}^K a_i z^(q_i) + 1.
> >
> > They have the same coefficients, a_i. Furthermore, p_i and q_i are real
> > numbers, they can be positive or negative.
> >
> > If I have z_1, z_2, ..., z_K satisfy both f(z) = 0 and g(z) = 0, can I
> > say that f(z) = q(z), that is, p_1 = q_1, p_2 = q_2, ..., p_K = q_K?
>
> If z_{i} are particular fixed values that are the same between the two,
> then the answer is NO.
>
> Suppose z{i} are all 1, and the coefficients are such that f(z) = 0.
> Now for each p{i} let q{i} = -p{i}. As 1^1 is the same as 1^(-1) the
> value of the expression will clearly not change so g(z) would also be 0,
> and yet you cannot conclude on the basis of that one point that f(z) =
> g(z) .
>
> I do not follow your notation "sum_{i=1}^K"
>
>
> If what you have is sum(a{i}*z{i}^p{i}, i=1..K) + 1, and
> sum(a{i}*z{i}^q{i}, i=1..K), then for consider particular terms and
> whether they will come out the same:
>
> a*z^p = a*z^q
> ln(a) + ln(z) * p = ln(a) + ln(z) * q
>
> If ln(a) is infinite (positive or negative) and ln(z) is not infinite,
> the terms would be the same. The simplest case of this is ln(a) =
> -infinity, which is to say a == 0.
>
> If ln(a) is not infinite, then it can be subtracted from both sides, leaving
>
> ln(z) * p = ln(z) * q
>
> If ln(z) is infinite, then this is satisfied if p and q are non-zero and
> the same sign.
>
> If ln(z) is finite, there is still another trivial solution: if ln(z) is
> 0, then the terms become equal provided that neither p nor q are
> infinite. ln(z) is 0 if z is 1.
>
> You can thus see that there are multiple circumstances under which
> individual terms come out exactly the same with different p and q -- and
> that's not even taking in to account that there may be particular
> combinations of p and q that happen to make the overall expressions the
> same.
>
>
> A trickier question comes if instead of allowing us to choose arbitrary
> z that make the equation 0, you say that the system is nilpotent -- that
> is, that f(z) = 0 and g(z) = 0 for *all* z. *If* that is the case, then
> you are going to have to specify the exact constraints so we know what
> we are working with, and you are going to have to be clear as to whether
> z is a single numeric value or represents a vector of values equal in
> length to the number of p and q.

Subject: fractional polynomial

From: Bruno Luong

Date: 4 Dec, 2010 11:56:05

Message: 4 of 8

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <idbvpa$iok$1@fred.mathworks.com>...
> Hi,
>
> I have a question about fractional polynomial and I cannot solve it. Can anyone give some suggestions? The problem is as follows:
>
> I have two fractional polynomial, f(z) and g(z), which are in the form of
>
> f(z) = sum_{i=1}^K a_i z^(p_i) + 1
>
> and
>
> g(z) = sum_{i=1}^K a_i z^(q_i) + 1.
>
> They have the same coefficients, a_i. Furthermore, p_i and q_i are real numbers, they can be positive or negative.
>
> If I have z_1, z_2, ..., z_K satisfy both f(z) = 0 and g(z) = 0, can I say that f(z) = q(z), that is, p_1 = q_1, p_2 = q_2, ..., p_K = q_K?

No.

k = 1;
p = 2;
q = 6;
 f=@(z) sum(z.^p)+1
 g=@(z) sum(z.^q)+1

 z=1i

>> f(z)

ans =

     0

>> g(z)

ans =

     0

% However p != q

Bruno

Subject: fractional polynomial

From: Frank

Date: 5 Dec, 2010 06:14:05

Message: 5 of 8

Thanks.

In your example, you chose K = 1 and z = 1i.
What if I choose K greater than 1, e.g. K = 3?
Is there any proof of the general case of arbitrary K?

Thanks a lot.


"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <idda8k$cql$1@fred.mathworks.com>...
> "Frank " <allinone_2003@yahoo.com.hk> wrote in message <idbvpa$iok$1@fred.mathworks.com>...
> > Hi,
> >
> > I have a question about fractional polynomial and I cannot solve it. Can anyone give some suggestions? The problem is as follows:
> >
> > I have two fractional polynomial, f(z) and g(z), which are in the form of
> >
> > f(z) = sum_{i=1}^K a_i z^(p_i) + 1
> >
> > and
> >
> > g(z) = sum_{i=1}^K a_i z^(q_i) + 1.
> >
> > They have the same coefficients, a_i. Furthermore, p_i and q_i are real numbers, they can be positive or negative.
> >
> > If I have z_1, z_2, ..., z_K satisfy both f(z) = 0 and g(z) = 0, can I say that f(z) = q(z), that is, p_1 = q_1, p_2 = q_2, ..., p_K = q_K?
>
> No.
>
> k = 1;
> p = 2;
> q = 6;
> f=@(z) sum(z.^p)+1
> g=@(z) sum(z.^q)+1
>
> z=1i
>
> >> f(z)
>
> ans =
>
> 0
>
> >> g(z)
>
> ans =
>
> 0
>
> % However p != q
>
> Bruno

Subject: fractional polynomial

From: Walter Roberson

Date: 5 Dec, 2010 06:36:14

Message: 6 of 8

On 05/12/10 12:14 AM, Frank wrote:

> In your example, you chose K = 1 and z = 1i.
> What if I choose K greater than 1, e.g. K = 3?
> Is there any proof of the general case of arbitrary K?

Following Bruno's example,

let p(i) and q(i) be positive integers >= 2 such that mod(p(i),4) = 2
and mod(p(i),4) = 2 for odd i, and mod(p(i),4) = 0 and mod(q(i),4) = 0
for even i.

Let z = sqrt(-1)

Then z^p(i) and z^q(i) will be -1 for odd i, and will be 1 for odd i

If k is odd, then the leading even terms will contribute alternating -1
and +1 which will add to 0. The final odd term will contribute -1 which
will balance out the constant +1 to get a final 0 for the summation.

Obviously this will be true even when p and q are completely different,
as long as they have that simple mod 4 relationship.


I would have to think more about the case where K is even to see if I
could come up with a 0 at all under the condition that |z| = 1.

Subject: fractional polynomial

From: Walter Roberson

Date: 5 Dec, 2010 06:48:01

Message: 7 of 8

On 05/12/10 12:36 AM, Walter Roberson wrote:

> Then z^p(i) and z^q(i) will be -1 for odd i, and will be 1 for odd i

Sorry... and will be 1 for even i. This is, note, under the assumption
that I slipped in that p and q are each >=2 .

Subject: fractional polynomial

From: Bruno Luong

Date: 5 Dec, 2010 10:15:08

Message: 8 of 8

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <idfajd$6q2$1@fred.mathworks.com>...
> Thanks.
>
> In your example, you chose K = 1 and z = 1i.
> What if I choose K greater than 1, e.g. K = 3?
> Is there any proof of the general case of arbitrary K?
>

If someone wants to disproof your claim then the person only needs to provide *one* counter example, which is the case.

If you want to change your claim to make it more restrictive accordingly to the counter example, then we have to start over again. But I think this is very inefficient way to finding a property of the problem are face of. It does not seem like the claim is based on any solid foundation at first, so suppose I find the second one are you going to restrict again the assumption of your claim? It can go on forever: When do we stop the escalating?

Until you find something more solid argument that can backup the claim, no I'll not try to find another counter example.

PS: Your problem recalls me a lot of aliasing phenomenon in DFT: if you sample the signal too slow, you can't capture signal: in other word there is always a signal that have a frequency content higher and gives exactly the same data.

Bruno

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