Feasible way to find interpolating complex polynomial based on absolute value
$begingroup$
Consider a complex degree-$(n-1)$ polynomial $p(z) = sumlimits_{i=0}^{n-1} a_i z^i$.
- Given a number $0 < m < 2n$ of positions in the complex plane with absolute value requirements, i.e. $|p(z_j)| overset{!}{=} b_j$ (with $b_j geq 0$), is there a practical algorithm to find the coefficients $a_i$ such that $p(z)$ satisfies those requirements? In other words, is there a way to solve a complex polynomial interpolation problem based only on given absolute values, leaving the argument (angle) of the polynomial completely arbitrary at any point?
- How big can $m$ be for such an algorithm? In other words, how many degrees of freedom are gained by only specifying the absolute value instead of a "full" complex number consisting of an absolute value and an argument (angle)?
polynomials optimization numerical-methods interpolation nonlinear-system
asked Dec 9 '18 at 5:30
LasseLasse
12
$endgroup$
add a comment |
$begingroup$
Consider a complex degree-$(n-1)$ polynomial $p(z) = sumlimits_{i=0}^{n-1} a_i z^i$.
- Given a number $0 < m < 2n$ of positions in the complex plane with absolute value requirements, i.e. $|p(z_j)| overset{!}{=} b_j$ (with $b_j geq 0$), is there a practical algorithm to find the coefficients $a_i$ such that $p(z)$ satisfies those requirements? In other words, is there a way to solve a complex polynomial interpolation problem based only on given absolute values, leaving the argument (angle) of the polynomial completely arbitrary at any point?
- How big can $m$ be for such an algorithm? In other words, how many degrees of freedom are gained by only specifying the absolute value instead of a "full" complex number consisting of an absolute value and an argument (angle)?
polynomials optimization numerical-methods interpolation nonlinear-system
asked Dec 9 '18 at 5:30
LasseLasse
12
$endgroup$
$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a smalln. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. Mynis usually something like 1024.
$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48
add a comment |
$begingroup$
Consider a complex degree-$(n-1)$ polynomial $p(z) = sumlimits_{i=0}^{n-1} a_i z^i$.
- Given a number $0 < m < 2n$ of positions in the complex plane with absolute value requirements, i.e. $|p(z_j)| overset{!}{=} b_j$ (with $b_j geq 0$), is there a practical algorithm to find the coefficients $a_i$ such that $p(z)$ satisfies those requirements? In other words, is there a way to solve a complex polynomial interpolation problem based only on given absolute values, leaving the argument (angle) of the polynomial completely arbitrary at any point?
- How big can $m$ be for such an algorithm? In other words, how many degrees of freedom are gained by only specifying the absolute value instead of a "full" complex number consisting of an absolute value and an argument (angle)?
polynomials optimization numerical-methods interpolation nonlinear-system
asked Dec 9 '18 at 5:30
LasseLasse
12
$endgroup$
Consider a complex degree-$(n-1)$ polynomial $p(z) = sumlimits_{i=0}^{n-1} a_i z^i$.
- Given a number $0 < m < 2n$ of positions in the complex plane with absolute value requirements, i.e. $|p(z_j)| overset{!}{=} b_j$ (with $b_j geq 0$), is there a practical algorithm to find the coefficients $a_i$ such that $p(z)$ satisfies those requirements? In other words, is there a way to solve a complex polynomial interpolation problem based only on given absolute values, leaving the argument (angle) of the polynomial completely arbitrary at any point?
- How big can $m$ be for such an algorithm? In other words, how many degrees of freedom are gained by only specifying the absolute value instead of a "full" complex number consisting of an absolute value and an argument (angle)?
polynomials optimization numerical-methods interpolation nonlinear-system
polynomials optimization numerical-methods interpolation nonlinear-system
asked Dec 9 '18 at 5:30
LasseLasse
12
asked Dec 9 '18 at 5:30
LasseLasse
12
edited Dec 22 '18 at 15:44
Lasse
asked Dec 9 '18 at 5:30
LasseLasse
12
asked Dec 9 '18 at 5:30
LasseLasse
12
asked Dec 9 '18 at 5:30
LasseLasse
12
12
$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a smalln. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. Mynis usually something like 1024.
$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48
add a comment |
$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a smalln. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. Mynis usually something like 1024.
$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48
$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a small
n. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. My n is usually something like 1024.$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a small
n. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. My n is usually something like 1024.$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48
add a comment |
1 Answer
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oldest
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$begingroup$
Considering the simple case of $n = 2$, it turns out to be rather trivial to find $m = 3$ conditions $|p(z_j)| overset{!}{=} b_j$ that no polynomial of degree $1$ can satisfy. Hence, $m$ can not be greater than $n$ in the general case. There are however some (not so rare) examples of conditions which can be satisfied.
For $m leq n$, the polynomial can easily be found by choosing a random argument (angle) for each $b_j$ and then solving the resulting system of linear equations. Since a polynomial only exists in some special cases when $m > n$, an algorithm to find that polynomial for arbitrary conditions cannot exist. There still might be a way to determine the space of conditions that yield a polynomial and a corresponding algorithm, though.
answered Dec 22 '18 at 15:59
LasseLasse
12
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Considering the simple case of $n = 2$, it turns out to be rather trivial to find $m = 3$ conditions $|p(z_j)| overset{!}{=} b_j$ that no polynomial of degree $1$ can satisfy. Hence, $m$ can not be greater than $n$ in the general case. There are however some (not so rare) examples of conditions which can be satisfied.
For $m leq n$, the polynomial can easily be found by choosing a random argument (angle) for each $b_j$ and then solving the resulting system of linear equations. Since a polynomial only exists in some special cases when $m > n$, an algorithm to find that polynomial for arbitrary conditions cannot exist. There still might be a way to determine the space of conditions that yield a polynomial and a corresponding algorithm, though.
answered Dec 22 '18 at 15:59
LasseLasse
12
$endgroup$
add a comment |
$begingroup$
Considering the simple case of $n = 2$, it turns out to be rather trivial to find $m = 3$ conditions $|p(z_j)| overset{!}{=} b_j$ that no polynomial of degree $1$ can satisfy. Hence, $m$ can not be greater than $n$ in the general case. There are however some (not so rare) examples of conditions which can be satisfied.
For $m leq n$, the polynomial can easily be found by choosing a random argument (angle) for each $b_j$ and then solving the resulting system of linear equations. Since a polynomial only exists in some special cases when $m > n$, an algorithm to find that polynomial for arbitrary conditions cannot exist. There still might be a way to determine the space of conditions that yield a polynomial and a corresponding algorithm, though.
answered Dec 22 '18 at 15:59
LasseLasse
12
$endgroup$
add a comment |
$begingroup$
Considering the simple case of $n = 2$, it turns out to be rather trivial to find $m = 3$ conditions $|p(z_j)| overset{!}{=} b_j$ that no polynomial of degree $1$ can satisfy. Hence, $m$ can not be greater than $n$ in the general case. There are however some (not so rare) examples of conditions which can be satisfied.
For $m leq n$, the polynomial can easily be found by choosing a random argument (angle) for each $b_j$ and then solving the resulting system of linear equations. Since a polynomial only exists in some special cases when $m > n$, an algorithm to find that polynomial for arbitrary conditions cannot exist. There still might be a way to determine the space of conditions that yield a polynomial and a corresponding algorithm, though.
answered Dec 22 '18 at 15:59
LasseLasse
12
$endgroup$
Considering the simple case of $n = 2$, it turns out to be rather trivial to find $m = 3$ conditions $|p(z_j)| overset{!}{=} b_j$ that no polynomial of degree $1$ can satisfy. Hence, $m$ can not be greater than $n$ in the general case. There are however some (not so rare) examples of conditions which can be satisfied.
For $m leq n$, the polynomial can easily be found by choosing a random argument (angle) for each $b_j$ and then solving the resulting system of linear equations. Since a polynomial only exists in some special cases when $m > n$, an algorithm to find that polynomial for arbitrary conditions cannot exist. There still might be a way to determine the space of conditions that yield a polynomial and a corresponding algorithm, though.
answered Dec 22 '18 at 15:59
LasseLasse
12
answered Dec 22 '18 at 15:59
LasseLasse
12
answered Dec 22 '18 at 15:59
LasseLasse
12
answered Dec 22 '18 at 15:59
LasseLasse
12
12
add a comment |
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$begingroup$
I have never seen an exclamation point placed above an equality sign. What does this combination of symbols mean? In any case, have you tried writing your problem as a real system of non-linear equations for a small value of $n$?
$endgroup$
– Carl Christian
Dec 10 '18 at 9:28
$begingroup$
@CarlChristian My bad. This is indeed pretty nonstandard notation. It's supposed to mean "has to be equal to". It's basically still just an equality. I have tried writing the equation system out for a small
n. The equations are not at all straightforward to solve, which is why I asked the question here. The best algorithms I have found so far that can actually solve this in a practical way are optimization algorithms (differential evolution etc.). Since these generally have exponential complexity, they do not scale well. Mynis usually something like 1024.$endgroup$
– Lasse
Dec 11 '18 at 12:53
$begingroup$
In truth, I do not see an alternative to doing a non-linear solve. I suspect that it will be important not to use square roots needlessly and solve $p_j bar{p}_j = b_j^2$. It might be useful to know what your specific application is.
$endgroup$
– Carl Christian
Dec 11 '18 at 15:48