Mixing
Factoring a complex function
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How can I write the function $f(z)=1-0.7z-0.3z^-1$, $z$ complex, as $f^+(z) f^-(z)$, where
- $f^+(z)$ is free of zeros and singularities outside and on the unit circle
- $f^-(z)$ is free of zeros and singularities inside and on the unit circle.
As can be verified, $f$ has a single pole at $z=0$ and two zeros at $z=1$ and $z=3/7$, so I cannot simply take $f^+(z)=f(z)$ and $f^-(z)=1$.
complex-analysis complex-integration
asked Aug 3 at 14:31
ToniAz
877
add a comment |Â
up vote
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How can I write the function $f(z)=1-0.7z-0.3z^-1$, $z$ complex, as $f^+(z) f^-(z)$, where
- $f^+(z)$ is free of zeros and singularities outside and on the unit circle
- $f^-(z)$ is free of zeros and singularities inside and on the unit circle.
As can be verified, $f$ has a single pole at $z=0$ and two zeros at $z=1$ and $z=3/7$, so I cannot simply take $f^+(z)=f(z)$ and $f^-(z)=1$.
complex-analysis complex-integration
asked Aug 3 at 14:31
ToniAz
877
Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
1
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can I write the function $f(z)=1-0.7z-0.3z^-1$, $z$ complex, as $f^+(z) f^-(z)$, where
- $f^+(z)$ is free of zeros and singularities outside and on the unit circle
- $f^-(z)$ is free of zeros and singularities inside and on the unit circle.
As can be verified, $f$ has a single pole at $z=0$ and two zeros at $z=1$ and $z=3/7$, so I cannot simply take $f^+(z)=f(z)$ and $f^-(z)=1$.
complex-analysis complex-integration
asked Aug 3 at 14:31
ToniAz
877
How can I write the function $f(z)=1-0.7z-0.3z^-1$, $z$ complex, as $f^+(z) f^-(z)$, where
- $f^+(z)$ is free of zeros and singularities outside and on the unit circle
- $f^-(z)$ is free of zeros and singularities inside and on the unit circle.
As can be verified, $f$ has a single pole at $z=0$ and two zeros at $z=1$ and $z=3/7$, so I cannot simply take $f^+(z)=f(z)$ and $f^-(z)=1$.
complex-analysis complex-integration
asked Aug 3 at 14:31
ToniAz
877
asked Aug 3 at 14:31
ToniAz
877
asked Aug 3 at 14:31
ToniAz
877
asked Aug 3 at 14:31
ToniAz
877
877
Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
1
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58
add a comment |Â
Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
1
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58
Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
1
1
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.
answered Aug 3 at 15:41
Alex M.
27.7k103057
add a comment |Â
up vote
0
down vote
$$f(z)=1-0.7z-0.3z^-1=dfrac-7z^2+10z-310z=dfrac-7(z-1)(z-frac37)10z$$
$$f(z)=dfrac-710cdot(z-frac37)left(1-dfrac1zright)$$
then
$$f^+=dfrac-710~~~,~~~f^-=(z-frac37)left(1-dfrac1zright)$$
answered Aug 3 at 14:51
user 108128
18.6k41544
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.
answered Aug 3 at 15:41
Alex M.
27.7k103057
add a comment |Â
up vote
2
down vote
accepted
Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.
answered Aug 3 at 15:41
Alex M.
27.7k103057
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.
answered Aug 3 at 15:41
Alex M.
27.7k103057
Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.
answered Aug 3 at 15:41
Alex M.
27.7k103057
answered Aug 3 at 15:41
Alex M.
27.7k103057
answered Aug 3 at 15:41
Alex M.
27.7k103057
answered Aug 3 at 15:41
Alex M.
27.7k103057
27.7k103057
add a comment |Â
add a comment |Â
up vote
0
down vote
$$f(z)=1-0.7z-0.3z^-1=dfrac-7z^2+10z-310z=dfrac-7(z-1)(z-frac37)10z$$
$$f(z)=dfrac-710cdot(z-frac37)left(1-dfrac1zright)$$
then
$$f^+=dfrac-710~~~,~~~f^-=(z-frac37)left(1-dfrac1zright)$$
answered Aug 3 at 14:51
user 108128
18.6k41544
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
add a comment |Â
up vote
0
down vote
$$f(z)=1-0.7z-0.3z^-1=dfrac-7z^2+10z-310z=dfrac-7(z-1)(z-frac37)10z$$
$$f(z)=dfrac-710cdot(z-frac37)left(1-dfrac1zright)$$
then
$$f^+=dfrac-710~~~,~~~f^-=(z-frac37)left(1-dfrac1zright)$$
answered Aug 3 at 14:51
user 108128
18.6k41544
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$f(z)=1-0.7z-0.3z^-1=dfrac-7z^2+10z-310z=dfrac-7(z-1)(z-frac37)10z$$
$$f(z)=dfrac-710cdot(z-frac37)left(1-dfrac1zright)$$
then
$$f^+=dfrac-710~~~,~~~f^-=(z-frac37)left(1-dfrac1zright)$$
answered Aug 3 at 14:51
user 108128
18.6k41544
$$f(z)=1-0.7z-0.3z^-1=dfrac-7z^2+10z-310z=dfrac-7(z-1)(z-frac37)10z$$
$$f(z)=dfrac-710cdot(z-frac37)left(1-dfrac1zright)$$
then
$$f^+=dfrac-710~~~,~~~f^-=(z-frac37)left(1-dfrac1zright)$$
answered Aug 3 at 14:51
user 108128
18.6k41544
edited Aug 3 at 15:52
answered Aug 3 at 14:51
user 108128
18.6k41544
answered Aug 3 at 14:51
user 108128
18.6k41544
answered Aug 3 at 14:51
user 108128
18.6k41544
18.6k41544
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
add a comment |Â
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ has a zero at $z=1$. Doesn't work :(
– ToniAz
Aug 3 at 15:05
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
$f^-$ still has a zero at $z=1$. My requirements are: no zeros at unit circle.
– ToniAz
Aug 3 at 15:57
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
@user108128, $f^-$ must not have its zeroes on the circle, by requirement. I believe that it would be better to delete this answer.
– Alex M.
Aug 4 at 10:08
add a comment |Â
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Crossposted on MO: mathoverflow.net/q/307483.
– Alex M.
Aug 3 at 15:31
1
Notice that the Wiener-Hopf decomposition can also be about decompositions as sums, not only as products. And the domains of definition of $f^+$ and $f^-$ may be the upper and, respectively, lower half-plane, not only the unit disk and its exterior. SO maybe you'll have to rethink the decomposition itself and the domains of definition.
– Alex M.
Aug 3 at 16:04
@AlexM. Thanks for the comment. The main problem is that I cannot find too much theory on the Discrete-time Wiener-Hopf problem. The original decomposition have to do with half planes, but in discrete-time, they have to do with inside and outside the unit circle. That's my understanding.
– ToniAz
Aug 3 at 16:58