Mixing
Prove $|f^{(n)}(z_0)|leq Mn!/R^n$, where $|f(z)| = M$
$begingroup$
Given a circunference $Gamma$ with radius $R$ and center $z_0$. Prove that if $f(z)$ is analytic in $Gamma$ and their interior, then
begin{equation*}
|f^{(n)}(z_0)|leq frac{Mn!}{R^n},
end{equation*}
where $M$ is the maximum value of $|f(z)|$ in $Gamma$.
Ok, here my attempt:
Since $f$ is analytic, then by Taylor theorem
begin{equation*}
f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^n,
end{equation*}
taking absolute value we obtain
begin{equation*}
|f(z)| = left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!},
end{equation*}
but, by definition, $Mgeq |f(z)|$ and expanding the summation
begin{equation*}
Mgeq |f^0(z_0)|frac{|z-z_0|^0}{0!}+|f^1(z_0)|frac{|z-z_0|^1}{1!}+cdots+|f^n(z_0)|frac{|z-z_0|^n}{n!}+cdots,
end{equation*}
all terms in the summation are potisives, then if we take only the n-term we can assure that
begin{equation*}
Mgeq |f^{(n)}(z_0)|frac{|z-z_0|^n}{n!}.
end{equation*}
Ok here is my problem, if I say $|z-z_0|^ngeq R^n$, I'm finished, but $|z-z_0|geq R$ is false. So, what did I do wrong?.
complex-analysis proof-verification
asked Jan 13 at 23:45
PatchouliPatchouli
31
$endgroup$
add a comment |
$begingroup$
Given a circunference $Gamma$ with radius $R$ and center $z_0$. Prove that if $f(z)$ is analytic in $Gamma$ and their interior, then
begin{equation*}
|f^{(n)}(z_0)|leq frac{Mn!}{R^n},
end{equation*}
where $M$ is the maximum value of $|f(z)|$ in $Gamma$.
Ok, here my attempt:
Since $f$ is analytic, then by Taylor theorem
begin{equation*}
f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^n,
end{equation*}
taking absolute value we obtain
begin{equation*}
|f(z)| = left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!},
end{equation*}
but, by definition, $Mgeq |f(z)|$ and expanding the summation
begin{equation*}
Mgeq |f^0(z_0)|frac{|z-z_0|^0}{0!}+|f^1(z_0)|frac{|z-z_0|^1}{1!}+cdots+|f^n(z_0)|frac{|z-z_0|^n}{n!}+cdots,
end{equation*}
all terms in the summation are potisives, then if we take only the n-term we can assure that
begin{equation*}
Mgeq |f^{(n)}(z_0)|frac{|z-z_0|^n}{n!}.
end{equation*}
Ok here is my problem, if I say $|z-z_0|^ngeq R^n$, I'm finished, but $|z-z_0|geq R$ is false. So, what did I do wrong?.
complex-analysis proof-verification
asked Jan 13 at 23:45
PatchouliPatchouli
31
$endgroup$
$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16
add a comment |
$begingroup$
Given a circunference $Gamma$ with radius $R$ and center $z_0$. Prove that if $f(z)$ is analytic in $Gamma$ and their interior, then
begin{equation*}
|f^{(n)}(z_0)|leq frac{Mn!}{R^n},
end{equation*}
where $M$ is the maximum value of $|f(z)|$ in $Gamma$.
Ok, here my attempt:
Since $f$ is analytic, then by Taylor theorem
begin{equation*}
f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^n,
end{equation*}
taking absolute value we obtain
begin{equation*}
|f(z)| = left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!},
end{equation*}
but, by definition, $Mgeq |f(z)|$ and expanding the summation
begin{equation*}
Mgeq |f^0(z_0)|frac{|z-z_0|^0}{0!}+|f^1(z_0)|frac{|z-z_0|^1}{1!}+cdots+|f^n(z_0)|frac{|z-z_0|^n}{n!}+cdots,
end{equation*}
all terms in the summation are potisives, then if we take only the n-term we can assure that
begin{equation*}
Mgeq |f^{(n)}(z_0)|frac{|z-z_0|^n}{n!}.
end{equation*}
Ok here is my problem, if I say $|z-z_0|^ngeq R^n$, I'm finished, but $|z-z_0|geq R$ is false. So, what did I do wrong?.
complex-analysis proof-verification
asked Jan 13 at 23:45
PatchouliPatchouli
31
$endgroup$
Given a circunference $Gamma$ with radius $R$ and center $z_0$. Prove that if $f(z)$ is analytic in $Gamma$ and their interior, then
begin{equation*}
|f^{(n)}(z_0)|leq frac{Mn!}{R^n},
end{equation*}
where $M$ is the maximum value of $|f(z)|$ in $Gamma$.
Ok, here my attempt:
Since $f$ is analytic, then by Taylor theorem
begin{equation*}
f(z) = sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^n,
end{equation*}
taking absolute value we obtain
begin{equation*}
|f(z)| = left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!},
end{equation*}
but, by definition, $Mgeq |f(z)|$ and expanding the summation
begin{equation*}
Mgeq |f^0(z_0)|frac{|z-z_0|^0}{0!}+|f^1(z_0)|frac{|z-z_0|^1}{1!}+cdots+|f^n(z_0)|frac{|z-z_0|^n}{n!}+cdots,
end{equation*}
all terms in the summation are potisives, then if we take only the n-term we can assure that
begin{equation*}
Mgeq |f^{(n)}(z_0)|frac{|z-z_0|^n}{n!}.
end{equation*}
Ok here is my problem, if I say $|z-z_0|^ngeq R^n$, I'm finished, but $|z-z_0|geq R$ is false. So, what did I do wrong?.
complex-analysis proof-verification
complex-analysis proof-verification
asked Jan 13 at 23:45
PatchouliPatchouli
31
asked Jan 13 at 23:45
PatchouliPatchouli
31
asked Jan 13 at 23:45
PatchouliPatchouli
31
asked Jan 13 at 23:45
PatchouliPatchouli
31
asked Jan 13 at 23:45
PatchouliPatchouli
31
31
$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16
add a comment |
$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16
$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint By The Cauchy Integral Formula
$$begin{equation*}
|f^{(n)}(z_0)|= left| frac{n!}{2 pi i}int_{Gamma} frac{f(z)}{|z-z_0|^{n+1}} dz right| leq frac{n!}{2pi}int_{Gamma} frac{left|f(z)right|}{|z-z_0|^{n+1}} dz
end{equation*}$$
Now use $|f(z)| leq M$ on $Gamma$. Also, on $Gamma$ what is $|z-z_0|$?
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
$endgroup$
add a comment |
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$begingroup$
Hint By The Cauchy Integral Formula
$$begin{equation*}
|f^{(n)}(z_0)|= left| frac{n!}{2 pi i}int_{Gamma} frac{f(z)}{|z-z_0|^{n+1}} dz right| leq frac{n!}{2pi}int_{Gamma} frac{left|f(z)right|}{|z-z_0|^{n+1}} dz
end{equation*}$$
Now use $|f(z)| leq M$ on $Gamma$. Also, on $Gamma$ what is $|z-z_0|$?
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
$endgroup$
add a comment |
$begingroup$
Hint By The Cauchy Integral Formula
$$begin{equation*}
|f^{(n)}(z_0)|= left| frac{n!}{2 pi i}int_{Gamma} frac{f(z)}{|z-z_0|^{n+1}} dz right| leq frac{n!}{2pi}int_{Gamma} frac{left|f(z)right|}{|z-z_0|^{n+1}} dz
end{equation*}$$
Now use $|f(z)| leq M$ on $Gamma$. Also, on $Gamma$ what is $|z-z_0|$?
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
$endgroup$
add a comment |
$begingroup$
Hint By The Cauchy Integral Formula
$$begin{equation*}
|f^{(n)}(z_0)|= left| frac{n!}{2 pi i}int_{Gamma} frac{f(z)}{|z-z_0|^{n+1}} dz right| leq frac{n!}{2pi}int_{Gamma} frac{left|f(z)right|}{|z-z_0|^{n+1}} dz
end{equation*}$$
Now use $|f(z)| leq M$ on $Gamma$. Also, on $Gamma$ what is $|z-z_0|$?
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
$endgroup$
Hint By The Cauchy Integral Formula
$$begin{equation*}
|f^{(n)}(z_0)|= left| frac{n!}{2 pi i}int_{Gamma} frac{f(z)}{|z-z_0|^{n+1}} dz right| leq frac{n!}{2pi}int_{Gamma} frac{left|f(z)right|}{|z-z_0|^{n+1}} dz
end{equation*}$$
Now use $|f(z)| leq M$ on $Gamma$. Also, on $Gamma$ what is $|z-z_0|$?
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
answered Jan 13 at 23:54
N. S.N. S.
104k7112208
104k7112208
add a comment |
add a comment |
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$begingroup$
The expression $left|sum _{n = 0}^{infty}frac{f^{(n)}(z_0)}{n!} (z-z_0)^nright| = sum _{n = 0}^{infty}|f^{(n)}(z_0)|frac{|(z-z_0)^n|}{n!}$ is not always true. This should not be an equality, but an inequality
$endgroup$
– Denis28
Jan 13 at 23:51
$begingroup$
Read 'Cauchy's estimates' in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 0:08
$begingroup$
@Denis28 Hahaha thanks, I forgot triangle inequality.
$endgroup$
– Patchouli
Jan 14 at 0:16