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 nonzero 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.
