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$f(z_0)=w_0$ Then $exists epsilon, delta$ such that $f(z)-alpha$ has exactly $m$ simple roots in $B(z_0,delta)$ for $|alpha-w_0|<epsilon$
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Let $f$ be analytic at $z_0 in mathbbC$ and $f(z_0)=w_0$. Suppose that $f(z)-w_0$ has a zero of finite order in $m geq 1$ at $z=z_0$.
We need to show that there exist $epsilon >0$ and $delta >0$ such that for all $alpha in mathbbC$ with $|alpha -w_0|<epsilon$, $f(z)-alpha$ has exactly $m$ simple roots in $B(z_0,delta)$.
Looking for some Hints. I wish to use to use $$frac12pi iint_mathcalCfracf'(z)f(z)dz = textnumber of zeros of f text in Int(mathcalC)$$.
complex-analysis roots holomorphic-functions
edited Aug 3 at 14:11
Batominovski
22.6k22776
asked Aug 3 at 9:25
Arindam
193113
add a comment |Â
up vote
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Let $f$ be analytic at $z_0 in mathbbC$ and $f(z_0)=w_0$. Suppose that $f(z)-w_0$ has a zero of finite order in $m geq 1$ at $z=z_0$.
We need to show that there exist $epsilon >0$ and $delta >0$ such that for all $alpha in mathbbC$ with $|alpha -w_0|<epsilon$, $f(z)-alpha$ has exactly $m$ simple roots in $B(z_0,delta)$.
Looking for some Hints. I wish to use to use $$frac12pi iint_mathcalCfracf'(z)f(z)dz = textnumber of zeros of f text in Int(mathcalC)$$.
complex-analysis roots holomorphic-functions
edited Aug 3 at 14:11
Batominovski
22.6k22776
asked Aug 3 at 9:25
Arindam
193113
$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
was my answer below helpful?
– mathworker21
2 days ago
Yes, thanks. It was helpful.
– Arindam
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f$ be analytic at $z_0 in mathbbC$ and $f(z_0)=w_0$. Suppose that $f(z)-w_0$ has a zero of finite order in $m geq 1$ at $z=z_0$.
We need to show that there exist $epsilon >0$ and $delta >0$ such that for all $alpha in mathbbC$ with $|alpha -w_0|<epsilon$, $f(z)-alpha$ has exactly $m$ simple roots in $B(z_0,delta)$.
Looking for some Hints. I wish to use to use $$frac12pi iint_mathcalCfracf'(z)f(z)dz = textnumber of zeros of f text in Int(mathcalC)$$.
complex-analysis roots holomorphic-functions
edited Aug 3 at 14:11
Batominovski
22.6k22776
asked Aug 3 at 9:25
Arindam
193113
Let $f$ be analytic at $z_0 in mathbbC$ and $f(z_0)=w_0$. Suppose that $f(z)-w_0$ has a zero of finite order in $m geq 1$ at $z=z_0$.
We need to show that there exist $epsilon >0$ and $delta >0$ such that for all $alpha in mathbbC$ with $|alpha -w_0|<epsilon$, $f(z)-alpha$ has exactly $m$ simple roots in $B(z_0,delta)$.
Looking for some Hints. I wish to use to use $$frac12pi iint_mathcalCfracf'(z)f(z)dz = textnumber of zeros of f text in Int(mathcalC)$$.
complex-analysis roots holomorphic-functions
edited Aug 3 at 14:11
Batominovski
22.6k22776
asked Aug 3 at 9:25
Arindam
193113
edited Aug 3 at 14:11
Batominovski
22.6k22776
edited Aug 3 at 14:11
Batominovski
22.6k22776
edited Aug 3 at 14:11
Batominovski
22.6k22776
22.6k22776
asked Aug 3 at 9:25
Arindam
193113
asked Aug 3 at 9:25
Arindam
193113
asked Aug 3 at 9:25
Arindam
193113
193113
$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
was my answer below helpful?
– mathworker21
2 days ago
Yes, thanks. It was helpful.
– Arindam
yesterday
add a comment |Â
$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
was my answer below helpful?
– mathworker21
2 days ago
Yes, thanks. It was helpful.
– Arindam
yesterday
$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
was my answer below helpful?
– mathworker21
2 days ago
was my answer below helpful?
– mathworker21
2 days ago
Yes, thanks. It was helpful.
– Arindam
yesterday
Yes, thanks. It was helpful.
– Arindam
yesterday
add a comment |Â
2 Answers
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1
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Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-alpha = (f(z)-w_0)-(w_0-alpha)$. You can take $delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2delta)$. Then take $epsilon = frac12sup_z in partial B(z_0,delta) |f(z)-w_0|$, so that $|w_0-alpha| le epsilon < |f(z)-w_0|$ for every $z in partial B(z_0,delta)$.
answered Aug 3 at 9:30
mathworker21
6,4231727
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up vote
1
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I assume that you have the condition $0<|alpha -w_0|<epsilon$; otherwise, as mathworker21 noted, $alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.
From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write
$$f(z)=w_0+(z-z_0)^m,g(z),,$$
where $g$ is a holomorphic function and $g(z_0)neq 0$. Since $g(z_0)neq0$, there exists a positive real number $delta_0$ such that, on the neighborhood $B_2delta_0(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $partial B_delta_0(z_0)$. Write $K$ for the maximum value of $|g'|$ on $partial B_delta_0(z_0)$. Set $$delta:=minleftdelta_0,fracMm,(K+1)right>0,.$$
We first claim that, on $B_delta(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that
$$beginalign
f'(z)&=(z-z_0)^m,g'(z)+m,(z-z_0)^m-1,g(z)
\&=(z-z_0)^m-1,big(m,(z-z_0),g'(z)+g(z)big),.
endalign$$
If $zin B_delta(z_0)setminusz_0$, then
$$beginalignfracbig^m-1&=big|m,(z-z_0),g'(z)+g(z)big|geq big|g(z)big|-m,big|z-z_0big|,big|g'(z)big|
\
&geq M-m,delta, Kgeq M-left(fracMK+1right),K=fracMK+1>0,.
endalign$$
The claim is now established.
Set $epsilon:=delta^m M>0$. For $alpha in B_delta(z_0)$, we see that
$$f(z)-alpha=(z-z_0)^m,g(z)-(alpha-w_0),.$$
When $alphain B_epsilon(w_0)$ and $zinpartial B_delta(z_0)$, we have
$$big|(z-z_0)^m,g(z)big|geq delta^m M=epsilon> |alpha-w_0|,.$$
Using Rouché's Theorem, the number of roots of $f(z)-alpha$ in $B_delta(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_delta(z_0)$, which is $m$.
We now prove that, if $0<|alpha-w_0|<epsilon$, then $f(z)-alpha$ has no repeated roots inside $B_delta(z_0)$. First, by the choice of $delta$, the only roots of $f'(z)$ inside $B_delta(z_0)$ is $z_0$. Second, if $z=zetain B_delta(z_0)$ is a double root of $f(z)-alpha$, then both $f(zeta)-alpha$ and $left(dfractextdtextdz,big(f(z)-alphabig)right)Bigg|_z=zeta=f'(zeta)$ must be both $0$. However, $f'(zeta)=0$ implies $zeta=z_0$, but then $f(z_0)-alpha=w_0-alphaneq 0$. This is a contradiction, whence such $zeta in B_delta(z_0)$ does not exist.
answered Aug 3 at 14:06
Batominovski
22.6k22776
add a comment |Â
2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-alpha = (f(z)-w_0)-(w_0-alpha)$. You can take $delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2delta)$. Then take $epsilon = frac12sup_z in partial B(z_0,delta) |f(z)-w_0|$, so that $|w_0-alpha| le epsilon < |f(z)-w_0|$ for every $z in partial B(z_0,delta)$.
answered Aug 3 at 9:30
mathworker21
6,4231727
add a comment |Â
up vote
1
down vote
Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-alpha = (f(z)-w_0)-(w_0-alpha)$. You can take $delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2delta)$. Then take $epsilon = frac12sup_z in partial B(z_0,delta) |f(z)-w_0|$, so that $|w_0-alpha| le epsilon < |f(z)-w_0|$ for every $z in partial B(z_0,delta)$.
answered Aug 3 at 9:30
mathworker21
6,4231727
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-alpha = (f(z)-w_0)-(w_0-alpha)$. You can take $delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2delta)$. Then take $epsilon = frac12sup_z in partial B(z_0,delta) |f(z)-w_0|$, so that $|w_0-alpha| le epsilon < |f(z)-w_0|$ for every $z in partial B(z_0,delta)$.
answered Aug 3 at 9:30
mathworker21
6,4231727
Well, you can use Rouché's theorem, which follows from the equation you wrote. $f(z)-alpha = (f(z)-w_0)-(w_0-alpha)$. You can take $delta > 0$ so that $z_0$ is the only root of $f$ in $B(z_0,2delta)$. Then take $epsilon = frac12sup_z in partial B(z_0,delta) |f(z)-w_0|$, so that $|w_0-alpha| le epsilon < |f(z)-w_0|$ for every $z in partial B(z_0,delta)$.
answered Aug 3 at 9:30
mathworker21
6,4231727
answered Aug 3 at 9:30
mathworker21
6,4231727
answered Aug 3 at 9:30
mathworker21
6,4231727
answered Aug 3 at 9:30
mathworker21
6,4231727
6,4231727
add a comment |Â
add a comment |Â
up vote
1
down vote
I assume that you have the condition $0<|alpha -w_0|<epsilon$; otherwise, as mathworker21 noted, $alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.
From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write
$$f(z)=w_0+(z-z_0)^m,g(z),,$$
where $g$ is a holomorphic function and $g(z_0)neq 0$. Since $g(z_0)neq0$, there exists a positive real number $delta_0$ such that, on the neighborhood $B_2delta_0(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $partial B_delta_0(z_0)$. Write $K$ for the maximum value of $|g'|$ on $partial B_delta_0(z_0)$. Set $$delta:=minleftdelta_0,fracMm,(K+1)right>0,.$$
We first claim that, on $B_delta(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that
$$beginalign
f'(z)&=(z-z_0)^m,g'(z)+m,(z-z_0)^m-1,g(z)
\&=(z-z_0)^m-1,big(m,(z-z_0),g'(z)+g(z)big),.
endalign$$
If $zin B_delta(z_0)setminusz_0$, then
$$beginalignfracbig^m-1&=big|m,(z-z_0),g'(z)+g(z)big|geq big|g(z)big|-m,big|z-z_0big|,big|g'(z)big|
\
&geq M-m,delta, Kgeq M-left(fracMK+1right),K=fracMK+1>0,.
endalign$$
The claim is now established.
Set $epsilon:=delta^m M>0$. For $alpha in B_delta(z_0)$, we see that
$$f(z)-alpha=(z-z_0)^m,g(z)-(alpha-w_0),.$$
When $alphain B_epsilon(w_0)$ and $zinpartial B_delta(z_0)$, we have
$$big|(z-z_0)^m,g(z)big|geq delta^m M=epsilon> |alpha-w_0|,.$$
Using Rouché's Theorem, the number of roots of $f(z)-alpha$ in $B_delta(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_delta(z_0)$, which is $m$.
We now prove that, if $0<|alpha-w_0|<epsilon$, then $f(z)-alpha$ has no repeated roots inside $B_delta(z_0)$. First, by the choice of $delta$, the only roots of $f'(z)$ inside $B_delta(z_0)$ is $z_0$. Second, if $z=zetain B_delta(z_0)$ is a double root of $f(z)-alpha$, then both $f(zeta)-alpha$ and $left(dfractextdtextdz,big(f(z)-alphabig)right)Bigg|_z=zeta=f'(zeta)$ must be both $0$. However, $f'(zeta)=0$ implies $zeta=z_0$, but then $f(z_0)-alpha=w_0-alphaneq 0$. This is a contradiction, whence such $zeta in B_delta(z_0)$ does not exist.
answered Aug 3 at 14:06
Batominovski
22.6k22776
add a comment |Â
up vote
1
down vote
I assume that you have the condition $0<|alpha -w_0|<epsilon$; otherwise, as mathworker21 noted, $alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.
From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write
$$f(z)=w_0+(z-z_0)^m,g(z),,$$
where $g$ is a holomorphic function and $g(z_0)neq 0$. Since $g(z_0)neq0$, there exists a positive real number $delta_0$ such that, on the neighborhood $B_2delta_0(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $partial B_delta_0(z_0)$. Write $K$ for the maximum value of $|g'|$ on $partial B_delta_0(z_0)$. Set $$delta:=minleftdelta_0,fracMm,(K+1)right>0,.$$
We first claim that, on $B_delta(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that
$$beginalign
f'(z)&=(z-z_0)^m,g'(z)+m,(z-z_0)^m-1,g(z)
\&=(z-z_0)^m-1,big(m,(z-z_0),g'(z)+g(z)big),.
endalign$$
If $zin B_delta(z_0)setminusz_0$, then
$$beginalignfracbig^m-1&=big|m,(z-z_0),g'(z)+g(z)big|geq big|g(z)big|-m,big|z-z_0big|,big|g'(z)big|
\
&geq M-m,delta, Kgeq M-left(fracMK+1right),K=fracMK+1>0,.
endalign$$
The claim is now established.
Set $epsilon:=delta^m M>0$. For $alpha in B_delta(z_0)$, we see that
$$f(z)-alpha=(z-z_0)^m,g(z)-(alpha-w_0),.$$
When $alphain B_epsilon(w_0)$ and $zinpartial B_delta(z_0)$, we have
$$big|(z-z_0)^m,g(z)big|geq delta^m M=epsilon> |alpha-w_0|,.$$
Using Rouché's Theorem, the number of roots of $f(z)-alpha$ in $B_delta(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_delta(z_0)$, which is $m$.
We now prove that, if $0<|alpha-w_0|<epsilon$, then $f(z)-alpha$ has no repeated roots inside $B_delta(z_0)$. First, by the choice of $delta$, the only roots of $f'(z)$ inside $B_delta(z_0)$ is $z_0$. Second, if $z=zetain B_delta(z_0)$ is a double root of $f(z)-alpha$, then both $f(zeta)-alpha$ and $left(dfractextdtextdz,big(f(z)-alphabig)right)Bigg|_z=zeta=f'(zeta)$ must be both $0$. However, $f'(zeta)=0$ implies $zeta=z_0$, but then $f(z_0)-alpha=w_0-alphaneq 0$. This is a contradiction, whence such $zeta in B_delta(z_0)$ does not exist.
answered Aug 3 at 14:06
Batominovski
22.6k22776
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I assume that you have the condition $0<|alpha -w_0|<epsilon$; otherwise, as mathworker21 noted, $alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.
From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write
$$f(z)=w_0+(z-z_0)^m,g(z),,$$
where $g$ is a holomorphic function and $g(z_0)neq 0$. Since $g(z_0)neq0$, there exists a positive real number $delta_0$ such that, on the neighborhood $B_2delta_0(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $partial B_delta_0(z_0)$. Write $K$ for the maximum value of $|g'|$ on $partial B_delta_0(z_0)$. Set $$delta:=minleftdelta_0,fracMm,(K+1)right>0,.$$
We first claim that, on $B_delta(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that
$$beginalign
f'(z)&=(z-z_0)^m,g'(z)+m,(z-z_0)^m-1,g(z)
\&=(z-z_0)^m-1,big(m,(z-z_0),g'(z)+g(z)big),.
endalign$$
If $zin B_delta(z_0)setminusz_0$, then
$$beginalignfracbig^m-1&=big|m,(z-z_0),g'(z)+g(z)big|geq big|g(z)big|-m,big|z-z_0big|,big|g'(z)big|
\
&geq M-m,delta, Kgeq M-left(fracMK+1right),K=fracMK+1>0,.
endalign$$
The claim is now established.
Set $epsilon:=delta^m M>0$. For $alpha in B_delta(z_0)$, we see that
$$f(z)-alpha=(z-z_0)^m,g(z)-(alpha-w_0),.$$
When $alphain B_epsilon(w_0)$ and $zinpartial B_delta(z_0)$, we have
$$big|(z-z_0)^m,g(z)big|geq delta^m M=epsilon> |alpha-w_0|,.$$
Using Rouché's Theorem, the number of roots of $f(z)-alpha$ in $B_delta(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_delta(z_0)$, which is $m$.
We now prove that, if $0<|alpha-w_0|<epsilon$, then $f(z)-alpha$ has no repeated roots inside $B_delta(z_0)$. First, by the choice of $delta$, the only roots of $f'(z)$ inside $B_delta(z_0)$ is $z_0$. Second, if $z=zetain B_delta(z_0)$ is a double root of $f(z)-alpha$, then both $f(zeta)-alpha$ and $left(dfractextdtextdz,big(f(z)-alphabig)right)Bigg|_z=zeta=f'(zeta)$ must be both $0$. However, $f'(zeta)=0$ implies $zeta=z_0$, but then $f(z_0)-alpha=w_0-alphaneq 0$. This is a contradiction, whence such $zeta in B_delta(z_0)$ does not exist.
answered Aug 3 at 14:06
Batominovski
22.6k22776
I assume that you have the condition $0<|alpha -w_0|<epsilon$; otherwise, as mathworker21 noted, $alpha=w_0$ will cause trouble. The problem also inherently assumes that $f$ is a nonconstant function.
From the condition that $z=z_0$ is a root of $f(z)-w_0$ of order $m$, we can write
$$f(z)=w_0+(z-z_0)^m,g(z),,$$
where $g$ is a holomorphic function and $g(z_0)neq 0$. Since $g(z_0)neq0$, there exists a positive real number $delta_0$ such that, on the neighborhood $B_2delta_0(z_0)$ of $z_0$, $g$ does not vanish. Write $M>0$ for the minimum value of $|g|$ on the boundary $partial B_delta_0(z_0)$. Write $K$ for the maximum value of $|g'|$ on $partial B_delta_0(z_0)$. Set $$delta:=minleftdelta_0,fracMm,(K+1)right>0,.$$
We first claim that, on $B_delta(z_0)$, $f'(z)$ has at most one root $z=z_0$. (This root has multiplicity $m-1$.) To see this, we note that
$$beginalign
f'(z)&=(z-z_0)^m,g'(z)+m,(z-z_0)^m-1,g(z)
\&=(z-z_0)^m-1,big(m,(z-z_0),g'(z)+g(z)big),.
endalign$$
If $zin B_delta(z_0)setminusz_0$, then
$$beginalignfracbig^m-1&=big|m,(z-z_0),g'(z)+g(z)big|geq big|g(z)big|-m,big|z-z_0big|,big|g'(z)big|
\
&geq M-m,delta, Kgeq M-left(fracMK+1right),K=fracMK+1>0,.
endalign$$
The claim is now established.
Set $epsilon:=delta^m M>0$. For $alpha in B_delta(z_0)$, we see that
$$f(z)-alpha=(z-z_0)^m,g(z)-(alpha-w_0),.$$
When $alphain B_epsilon(w_0)$ and $zinpartial B_delta(z_0)$, we have
$$big|(z-z_0)^m,g(z)big|geq delta^m M=epsilon> |alpha-w_0|,.$$
Using Rouché's Theorem, the number of roots of $f(z)-alpha$ in $B_delta(z_0)$ is the same as the number of roots of $(z-z_0)^mg(z)$ in $B_delta(z_0)$, which is $m$.
We now prove that, if $0<|alpha-w_0|<epsilon$, then $f(z)-alpha$ has no repeated roots inside $B_delta(z_0)$. First, by the choice of $delta$, the only roots of $f'(z)$ inside $B_delta(z_0)$ is $z_0$. Second, if $z=zetain B_delta(z_0)$ is a double root of $f(z)-alpha$, then both $f(zeta)-alpha$ and $left(dfractextdtextdz,big(f(z)-alphabig)right)Bigg|_z=zeta=f'(zeta)$ must be both $0$. However, $f'(zeta)=0$ implies $zeta=z_0$, but then $f(z_0)-alpha=w_0-alphaneq 0$. This is a contradiction, whence such $zeta in B_delta(z_0)$ does not exist.
answered Aug 3 at 14:06
Batominovski
22.6k22776
edited yesterday
answered Aug 3 at 14:06
Batominovski
22.6k22776
answered Aug 3 at 14:06
Batominovski
22.6k22776
answered Aug 3 at 14:06
Batominovski
22.6k22776
22.6k22776
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$alpha = w_0$??
– mathworker21
Aug 3 at 9:26
you still have $alpha = w_0$
– mathworker21
Aug 3 at 9:28
just corrected it
– Arindam
Aug 3 at 9:28
was my answer below helpful?
– mathworker21
2 days ago
Yes, thanks. It was helpful.
– Arindam
yesterday