Mixing
Zeros of $f + t g$ goes to the zeros of $g$ as $t to infty$?
$begingroup$
Suppose $f, g in mathbb C[x]$ with $deg(f) > deg(g)$. Without loss of generality, let $deg(f) = n$ and $deg(g) = n-1$. I read a statement without proof: If $h(x) = f(x) + t g(x)$ with $t in [0, infty)$, the zeros of $h$ will start from the zeros of $f(x)$ ($t=0$) and converge to the zeros of $g$ ($t to infty$) and of course, one of them will tend to infinity. Intuitively, I understand this should be true since if $t$ is sufficiently large, the zeros should be the ones of $g(x)$. I am wondering what a rigorous way to argue this point?
complex-analysis polynomials roots
asked Jan 9 at 18:03
user1101010user1101010
8191730
$endgroup$
add a comment |
$begingroup$
Suppose $f, g in mathbb C[x]$ with $deg(f) > deg(g)$. Without loss of generality, let $deg(f) = n$ and $deg(g) = n-1$. I read a statement without proof: If $h(x) = f(x) + t g(x)$ with $t in [0, infty)$, the zeros of $h$ will start from the zeros of $f(x)$ ($t=0$) and converge to the zeros of $g$ ($t to infty$) and of course, one of them will tend to infinity. Intuitively, I understand this should be true since if $t$ is sufficiently large, the zeros should be the ones of $g(x)$. I am wondering what a rigorous way to argue this point?
complex-analysis polynomials roots
asked Jan 9 at 18:03
user1101010user1101010
8191730
$endgroup$
$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40
add a comment |
$begingroup$
Suppose $f, g in mathbb C[x]$ with $deg(f) > deg(g)$. Without loss of generality, let $deg(f) = n$ and $deg(g) = n-1$. I read a statement without proof: If $h(x) = f(x) + t g(x)$ with $t in [0, infty)$, the zeros of $h$ will start from the zeros of $f(x)$ ($t=0$) and converge to the zeros of $g$ ($t to infty$) and of course, one of them will tend to infinity. Intuitively, I understand this should be true since if $t$ is sufficiently large, the zeros should be the ones of $g(x)$. I am wondering what a rigorous way to argue this point?
complex-analysis polynomials roots
asked Jan 9 at 18:03
user1101010user1101010
8191730
$endgroup$
Suppose $f, g in mathbb C[x]$ with $deg(f) > deg(g)$. Without loss of generality, let $deg(f) = n$ and $deg(g) = n-1$. I read a statement without proof: If $h(x) = f(x) + t g(x)$ with $t in [0, infty)$, the zeros of $h$ will start from the zeros of $f(x)$ ($t=0$) and converge to the zeros of $g$ ($t to infty$) and of course, one of them will tend to infinity. Intuitively, I understand this should be true since if $t$ is sufficiently large, the zeros should be the ones of $g(x)$. I am wondering what a rigorous way to argue this point?
complex-analysis polynomials roots
complex-analysis polynomials roots
asked Jan 9 at 18:03
user1101010user1101010
8191730
asked Jan 9 at 18:03
user1101010user1101010
8191730
edited Jan 9 at 18:21
user1101010
asked Jan 9 at 18:03
user1101010user1101010
8191730
asked Jan 9 at 18:03
user1101010user1101010
8191730
asked Jan 9 at 18:03
user1101010user1101010
8191730
8191730
$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40
add a comment |
$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40
$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)ne0$ if $0<|z-a|le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t,g$ have the same number of zeros on ${|z-a|<r}$. Letting $rto0$, this shows that $f-t,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
$endgroup$
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)ne0$ if $0<|z-a|le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t,g$ have the same number of zeros on ${|z-a|<r}$. Letting $rto0$, this shows that $f-t,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
$endgroup$
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
add a comment |
$begingroup$
Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)ne0$ if $0<|z-a|le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t,g$ have the same number of zeros on ${|z-a|<r}$. Letting $rto0$, this shows that $f-t,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
$endgroup$
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
add a comment |
$begingroup$
Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)ne0$ if $0<|z-a|le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t,g$ have the same number of zeros on ${|z-a|<r}$. Letting $rto0$, this shows that $f-t,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
$endgroup$
Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)ne0$ if $0<|z-a|le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t,g$ have the same number of zeros on ${|z-a|<r}$. Letting $rto0$, this shows that $f-t,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
edited Jan 9 at 19:01
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
answered Jan 9 at 18:53
Julián AguirreJulián Aguirre
68.5k24094
68.5k24094
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
add a comment |
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
Thanks. Could you give a reference of "Touche-Frobenius"? I have not met this theorem before and google does not lead me to anywhere.
$endgroup$
– user1101010
Jan 9 at 18:58
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
You mean this theorem, right? I can't find any "Touché-Frobenius", though apparently "Rouché-Frobenius" is sometimes the name of an unrelated theorem connected to Rouché.
$endgroup$
– Henning Makholm
Jan 9 at 18:59
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Yes, it was a typo.
$endgroup$
– Julián Aguirre
Jan 9 at 19:03
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
$begingroup$
Could you give a few more words on the detail you omitted? I think you mean we look at $|(t-1)g+f - g| = |(t-2)g + f| {color{red}<} |(t-1) g + f| + |g|$ on the boundary of the disk. But why do we have the strict inequality? Thanks.
$endgroup$
– user1101010
Jan 9 at 19:09
1
1
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
$begingroup$
Compare $t,g$ and $h=f+t,g$. Then $h-t,g=f$. Since $min_{|z-a|=r}|g(z)|>0$, $t$ can be taken large enough to have $|f|<t,|g|$ on ${|z-a|=r}$.
$endgroup$
– Julián Aguirre
Jan 9 at 19:14
add a comment |
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$begingroup$
$h(x)$ at the start, with $t=1$ won't have the zeros of $f$. You probably meant $tin[0,infty)$
$endgroup$
– Calvin Khor
Jan 9 at 18:09
$begingroup$
Intuitively it feels like it would probably be simpler to consider $lambda f(x)+g(x)$ as $lambdato 0$.
$endgroup$
– Henning Makholm
Jan 9 at 18:19
$begingroup$
@CalvinKhor: Yes. I meant $t=0$. Thanks.
$endgroup$
– user1101010
Jan 9 at 18:20
$begingroup$
@HenningMakholm yeah I noticed (but not soon enough for you to see it :) )
$endgroup$
– Calvin Khor
Jan 9 at 18:21
$begingroup$
I think one can complete a rigorous argument using Rouché's theorem, with $g(x)$ being the "dog walker" as $x$ goes in a small circle around a root of $g$ and $frac1t f(x)$ being the "leash". By increasing $t$ we can make the leash short enough for the theorem to apply.
$endgroup$
– Henning Makholm
Jan 9 at 18:40