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Poles defining a function - do two equivalent expressions share the same poles?
The Next CEO of Stack OverflowWhen is a function satisfying the Cauchy-Riemann equations holomorphic?closed form of $sum frac1z^3 - n^3$The function $p(z)$ has finitely many poles and sum of residues zeroCan it be proved that a Meromorphic function only has a countable number of poles?Finding a Laurent Series involving two polesWhy must a meromorphic function, bounded near infinity, have the same number of poles and zeros?Elliptic Functions, Residue Computation, Same zeros and poles of same ordersPoles of hypergeometric function $_2F_1$Show that the poles of analytic function $f$ cannot have a limit point in $G$.Change of the hyperplane equation in $mathbbC^3$.
$begingroup$
Take a term $f$, with poles at $z=1, z=2$, i.e
$f=dfracg(z-1)(z-2)$.
Now assume that $f$ can be split into two (or more) terms, as
$f equiv h_1 + h_2$
Is it true that $h_1$ and $h_2$ must have the same poles as $f$? Is it true that if not, then their overall form must have the total same poles, i.e if they don't individally have the same poles as $f$, then the terms must have a form like
$h_1 = dfraca(z-2) quad h_2 = dfracb(z-1)$
complex-analysis
asked Mar 18 at 20:29
BradBrad
103
$endgroup$
add a comment |
$begingroup$
Take a term $f$, with poles at $z=1, z=2$, i.e
$f=dfracg(z-1)(z-2)$.
Now assume that $f$ can be split into two (or more) terms, as
$f equiv h_1 + h_2$
Is it true that $h_1$ and $h_2$ must have the same poles as $f$? Is it true that if not, then their overall form must have the total same poles, i.e if they don't individally have the same poles as $f$, then the terms must have a form like
$h_1 = dfraca(z-2) quad h_2 = dfracb(z-1)$
complex-analysis
asked Mar 18 at 20:29
BradBrad
103
$endgroup$
$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00
add a comment |
$begingroup$
Take a term $f$, with poles at $z=1, z=2$, i.e
$f=dfracg(z-1)(z-2)$.
Now assume that $f$ can be split into two (or more) terms, as
$f equiv h_1 + h_2$
Is it true that $h_1$ and $h_2$ must have the same poles as $f$? Is it true that if not, then their overall form must have the total same poles, i.e if they don't individally have the same poles as $f$, then the terms must have a form like
$h_1 = dfraca(z-2) quad h_2 = dfracb(z-1)$
complex-analysis
asked Mar 18 at 20:29
BradBrad
103
$endgroup$
Take a term $f$, with poles at $z=1, z=2$, i.e
$f=dfracg(z-1)(z-2)$.
Now assume that $f$ can be split into two (or more) terms, as
$f equiv h_1 + h_2$
Is it true that $h_1$ and $h_2$ must have the same poles as $f$? Is it true that if not, then their overall form must have the total same poles, i.e if they don't individally have the same poles as $f$, then the terms must have a form like
$h_1 = dfraca(z-2) quad h_2 = dfracb(z-1)$
complex-analysis
complex-analysis
asked Mar 18 at 20:29
BradBrad
103
asked Mar 18 at 20:29
BradBrad
103
asked Mar 18 at 20:29
BradBrad
103
asked Mar 18 at 20:29
BradBrad
103
asked Mar 18 at 20:29
BradBrad
103
103
$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00
add a comment |
$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00
$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00
$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To give a simplified version of the situation, suppose you have a single variable $,z,$ and define the simple poles $,p_0 := 1/z,, p_1 := 1/(z-1),, p_2 := 1/(z-2),,$ and so on. Suppose you have finite set of simple poles $,P := p_i,p_j,dots,p_k,$ and a finite sum of them
$,S := c_i,p_i + c_j,p_j + cdots + c_k,p_k.,$ Then the poles are linearly independent and so any two sums $,S,$ and $,S',$ are equal iff they have the same poles with the same coefficients. You can have higher order poles such as
$,p_i^2,$ and $, p_j^3,$ and so on, but they are all linearly independent. Think of P as a set of independent variables used to construct multi-variable polynomials. Linear algebra rules here. Any rational function of
$,z,$ with poles in $,P,$ can be uniquely split into the sum of a polynomial in the simple poles $,P,$ and a polynomial in $,z.,$
As a simple example:
$$ frac2 + z + 3 z^2 + 4 z^3 + 5 z^4z, (z - 1), (z - 2) =
frac1z + frac-15z-1 + frac64z-2 + 19 + 5,z.$$
P.S. Try this decomposition yourself. Here is some Wolfram Mathematica code:
decompRatFunc[r_, z_] := Module[ poles, poly, poles =
DeleteCases[r // Factor // Apart, _?NumberQ | z^n_. _. /; n > 0];
poly = r - poles // Simplify // Expand; poles, poly];
answered Mar 18 at 21:46
SomosSomos
14.7k11337
$endgroup$
add a comment |
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$begingroup$
To give a simplified version of the situation, suppose you have a single variable $,z,$ and define the simple poles $,p_0 := 1/z,, p_1 := 1/(z-1),, p_2 := 1/(z-2),,$ and so on. Suppose you have finite set of simple poles $,P := p_i,p_j,dots,p_k,$ and a finite sum of them
$,S := c_i,p_i + c_j,p_j + cdots + c_k,p_k.,$ Then the poles are linearly independent and so any two sums $,S,$ and $,S',$ are equal iff they have the same poles with the same coefficients. You can have higher order poles such as
$,p_i^2,$ and $, p_j^3,$ and so on, but they are all linearly independent. Think of P as a set of independent variables used to construct multi-variable polynomials. Linear algebra rules here. Any rational function of
$,z,$ with poles in $,P,$ can be uniquely split into the sum of a polynomial in the simple poles $,P,$ and a polynomial in $,z.,$
As a simple example:
$$ frac2 + z + 3 z^2 + 4 z^3 + 5 z^4z, (z - 1), (z - 2) =
frac1z + frac-15z-1 + frac64z-2 + 19 + 5,z.$$
P.S. Try this decomposition yourself. Here is some Wolfram Mathematica code:
decompRatFunc[r_, z_] := Module[ poles, poly, poles =
DeleteCases[r // Factor // Apart, _?NumberQ | z^n_. _. /; n > 0];
poly = r - poles // Simplify // Expand; poles, poly];
answered Mar 18 at 21:46
SomosSomos
14.7k11337
$endgroup$
add a comment |
$begingroup$
To give a simplified version of the situation, suppose you have a single variable $,z,$ and define the simple poles $,p_0 := 1/z,, p_1 := 1/(z-1),, p_2 := 1/(z-2),,$ and so on. Suppose you have finite set of simple poles $,P := p_i,p_j,dots,p_k,$ and a finite sum of them
$,S := c_i,p_i + c_j,p_j + cdots + c_k,p_k.,$ Then the poles are linearly independent and so any two sums $,S,$ and $,S',$ are equal iff they have the same poles with the same coefficients. You can have higher order poles such as
$,p_i^2,$ and $, p_j^3,$ and so on, but they are all linearly independent. Think of P as a set of independent variables used to construct multi-variable polynomials. Linear algebra rules here. Any rational function of
$,z,$ with poles in $,P,$ can be uniquely split into the sum of a polynomial in the simple poles $,P,$ and a polynomial in $,z.,$
As a simple example:
$$ frac2 + z + 3 z^2 + 4 z^3 + 5 z^4z, (z - 1), (z - 2) =
frac1z + frac-15z-1 + frac64z-2 + 19 + 5,z.$$
P.S. Try this decomposition yourself. Here is some Wolfram Mathematica code:
decompRatFunc[r_, z_] := Module[ poles, poly, poles =
DeleteCases[r // Factor // Apart, _?NumberQ | z^n_. _. /; n > 0];
poly = r - poles // Simplify // Expand; poles, poly];
answered Mar 18 at 21:46
SomosSomos
14.7k11337
$endgroup$
add a comment |
$begingroup$
To give a simplified version of the situation, suppose you have a single variable $,z,$ and define the simple poles $,p_0 := 1/z,, p_1 := 1/(z-1),, p_2 := 1/(z-2),,$ and so on. Suppose you have finite set of simple poles $,P := p_i,p_j,dots,p_k,$ and a finite sum of them
$,S := c_i,p_i + c_j,p_j + cdots + c_k,p_k.,$ Then the poles are linearly independent and so any two sums $,S,$ and $,S',$ are equal iff they have the same poles with the same coefficients. You can have higher order poles such as
$,p_i^2,$ and $, p_j^3,$ and so on, but they are all linearly independent. Think of P as a set of independent variables used to construct multi-variable polynomials. Linear algebra rules here. Any rational function of
$,z,$ with poles in $,P,$ can be uniquely split into the sum of a polynomial in the simple poles $,P,$ and a polynomial in $,z.,$
As a simple example:
$$ frac2 + z + 3 z^2 + 4 z^3 + 5 z^4z, (z - 1), (z - 2) =
frac1z + frac-15z-1 + frac64z-2 + 19 + 5,z.$$
P.S. Try this decomposition yourself. Here is some Wolfram Mathematica code:
decompRatFunc[r_, z_] := Module[ poles, poly, poles =
DeleteCases[r // Factor // Apart, _?NumberQ | z^n_. _. /; n > 0];
poly = r - poles // Simplify // Expand; poles, poly];
answered Mar 18 at 21:46
SomosSomos
14.7k11337
$endgroup$
To give a simplified version of the situation, suppose you have a single variable $,z,$ and define the simple poles $,p_0 := 1/z,, p_1 := 1/(z-1),, p_2 := 1/(z-2),,$ and so on. Suppose you have finite set of simple poles $,P := p_i,p_j,dots,p_k,$ and a finite sum of them
$,S := c_i,p_i + c_j,p_j + cdots + c_k,p_k.,$ Then the poles are linearly independent and so any two sums $,S,$ and $,S',$ are equal iff they have the same poles with the same coefficients. You can have higher order poles such as
$,p_i^2,$ and $, p_j^3,$ and so on, but they are all linearly independent. Think of P as a set of independent variables used to construct multi-variable polynomials. Linear algebra rules here. Any rational function of
$,z,$ with poles in $,P,$ can be uniquely split into the sum of a polynomial in the simple poles $,P,$ and a polynomial in $,z.,$
As a simple example:
$$ frac2 + z + 3 z^2 + 4 z^3 + 5 z^4z, (z - 1), (z - 2) =
frac1z + frac-15z-1 + frac64z-2 + 19 + 5,z.$$
P.S. Try this decomposition yourself. Here is some Wolfram Mathematica code:
decompRatFunc[r_, z_] := Module[ poles, poly, poles =
DeleteCases[r // Factor // Apart, _?NumberQ | z^n_. _. /; n > 0];
poly = r - poles // Simplify // Expand; poles, poly];
answered Mar 18 at 21:46
SomosSomos
14.7k11337
edited Mar 19 at 0:47
answered Mar 18 at 21:46
SomosSomos
14.7k11337
answered Mar 18 at 21:46
SomosSomos
14.7k11337
answered Mar 18 at 21:46
SomosSomos
14.7k11337
14.7k11337
add a comment |
add a comment |
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$begingroup$
You can always replace $(h_1, h_2)$ by $(h_1 + phi, h_2 - phi)$ with an arbitrary function $phi$ (having as much poles as you want).
$endgroup$
– Martin R
Mar 18 at 21:00