Creating a continuous evaluation of a phase graph
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I have a function
$$
R(xi) := prod_{i=0}^{n} (e^{-ixi} - z_k)
$$
where $z_{k} in mathbb{C}$. The $z_{k}$ are random, but clustered vaguely around the unit circle, if that is relevant. I need to calculate the phase $Psi$ of $R$, where we define $Psi$ via
$$
R(xi) = exp(-iPsi(xi))|R(xi)|
$$
If we do not care about continuity of $Psi$ in $xi$, then this is easy: We could write (in C++)
auto Psi = [&](Real xi)->Real {
Real phi = 0;
Complex ii = {0,1};
Complex z = exp(-ii*xi);
for (zk in roots)
{
phi -= arg(z-zk);
}
return phi;
};
However, I have an additional constraint that the phase $Psi$ must be continuous on the interval $[0, 2pi)$. The arg function always returns a value in $[-pi, pi]$, and as such the phase has discontinuities at seemingly random locations, depending on the values of the roots ${z_{k}}$.
How can I patch up my code/figure out how to make $Psi$ a continuous function of $xi$ on the interval $[0, 2pi)$?
complex-analysis polynomials complex-numbers winding-number
asked Nov 17 at 21:22
user14717
3,7731020
add a comment |
up vote
0
down vote
favorite
I have a function
$$
R(xi) := prod_{i=0}^{n} (e^{-ixi} - z_k)
$$
where $z_{k} in mathbb{C}$. The $z_{k}$ are random, but clustered vaguely around the unit circle, if that is relevant. I need to calculate the phase $Psi$ of $R$, where we define $Psi$ via
$$
R(xi) = exp(-iPsi(xi))|R(xi)|
$$
If we do not care about continuity of $Psi$ in $xi$, then this is easy: We could write (in C++)
auto Psi = [&](Real xi)->Real {
Real phi = 0;
Complex ii = {0,1};
Complex z = exp(-ii*xi);
for (zk in roots)
{
phi -= arg(z-zk);
}
return phi;
};
However, I have an additional constraint that the phase $Psi$ must be continuous on the interval $[0, 2pi)$. The arg function always returns a value in $[-pi, pi]$, and as such the phase has discontinuities at seemingly random locations, depending on the values of the roots ${z_{k}}$.
How can I patch up my code/figure out how to make $Psi$ a continuous function of $xi$ on the interval $[0, 2pi)$?
complex-analysis polynomials complex-numbers winding-number
asked Nov 17 at 21:22
user14717
3,7731020
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a function
$$
R(xi) := prod_{i=0}^{n} (e^{-ixi} - z_k)
$$
where $z_{k} in mathbb{C}$. The $z_{k}$ are random, but clustered vaguely around the unit circle, if that is relevant. I need to calculate the phase $Psi$ of $R$, where we define $Psi$ via
$$
R(xi) = exp(-iPsi(xi))|R(xi)|
$$
If we do not care about continuity of $Psi$ in $xi$, then this is easy: We could write (in C++)
auto Psi = [&](Real xi)->Real {
Real phi = 0;
Complex ii = {0,1};
Complex z = exp(-ii*xi);
for (zk in roots)
{
phi -= arg(z-zk);
}
return phi;
};
However, I have an additional constraint that the phase $Psi$ must be continuous on the interval $[0, 2pi)$. The arg function always returns a value in $[-pi, pi]$, and as such the phase has discontinuities at seemingly random locations, depending on the values of the roots ${z_{k}}$.
How can I patch up my code/figure out how to make $Psi$ a continuous function of $xi$ on the interval $[0, 2pi)$?
complex-analysis polynomials complex-numbers winding-number
asked Nov 17 at 21:22
user14717
3,7731020
I have a function
$$
R(xi) := prod_{i=0}^{n} (e^{-ixi} - z_k)
$$
where $z_{k} in mathbb{C}$. The $z_{k}$ are random, but clustered vaguely around the unit circle, if that is relevant. I need to calculate the phase $Psi$ of $R$, where we define $Psi$ via
$$
R(xi) = exp(-iPsi(xi))|R(xi)|
$$
If we do not care about continuity of $Psi$ in $xi$, then this is easy: We could write (in C++)
auto Psi = [&](Real xi)->Real {
Real phi = 0;
Complex ii = {0,1};
Complex z = exp(-ii*xi);
for (zk in roots)
{
phi -= arg(z-zk);
}
return phi;
};
However, I have an additional constraint that the phase $Psi$ must be continuous on the interval $[0, 2pi)$. The arg function always returns a value in $[-pi, pi]$, and as such the phase has discontinuities at seemingly random locations, depending on the values of the roots ${z_{k}}$.
How can I patch up my code/figure out how to make $Psi$ a continuous function of $xi$ on the interval $[0, 2pi)$?
complex-analysis polynomials complex-numbers winding-number
complex-analysis polynomials complex-numbers winding-number
asked Nov 17 at 21:22
user14717
3,7731020
asked Nov 17 at 21:22
user14717
3,7731020
edited Nov 18 at 1:36
asked Nov 17 at 21:22
user14717
3,7731020
asked Nov 17 at 21:22
user14717
3,7731020
asked Nov 17 at 21:22
user14717
3,7731020
3,7731020
add a comment |
add a comment |
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