Mixing
An application of Borel-Caratheodory theorem
Suppose that $f(z)$ is analytic function in the disk $|z|le R$ and $A(r)=max_{|z|=r}Re(z)$ then prove that $r<R$
$$max_{|z|=r}left|frac{f^n(z)}{n!}right|le frac{2^{n+2}R}{(R-r)^{n+1}}{A(r)+|f(0)|}$$
MY attempt
Using Cauchy generalized integral formula
$f^{(n)}(z_0)=frac{n!}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}$
then $max_{|z|=r}left|frac{f^n(z)}{n!}right|=left|max_{|z|=r} frac{1}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}right|$
and i think we can solve by using Borel caratheodory theorem
Let $f(z)$ be holomorphic function on $|z|le R$ and let $M(r)=max_{|z|=r}|f(z)|$ and $A(r)=max_{|z|=r}Re f(z)$ then for $0<r<R$
$$ M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
from here how to processed ?
complex-analysis
asked Nov 20 '18 at 9:19
Inverse Problem
923918
add a comment |
Suppose that $f(z)$ is analytic function in the disk $|z|le R$ and $A(r)=max_{|z|=r}Re(z)$ then prove that $r<R$
$$max_{|z|=r}left|frac{f^n(z)}{n!}right|le frac{2^{n+2}R}{(R-r)^{n+1}}{A(r)+|f(0)|}$$
MY attempt
Using Cauchy generalized integral formula
$f^{(n)}(z_0)=frac{n!}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}$
then $max_{|z|=r}left|frac{f^n(z)}{n!}right|=left|max_{|z|=r} frac{1}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}right|$
and i think we can solve by using Borel caratheodory theorem
Let $f(z)$ be holomorphic function on $|z|le R$ and let $M(r)=max_{|z|=r}|f(z)|$ and $A(r)=max_{|z|=r}Re f(z)$ then for $0<r<R$
$$ M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
from here how to processed ?
complex-analysis
asked Nov 20 '18 at 9:19
Inverse Problem
923918
add a comment |
Suppose that $f(z)$ is analytic function in the disk $|z|le R$ and $A(r)=max_{|z|=r}Re(z)$ then prove that $r<R$
$$max_{|z|=r}left|frac{f^n(z)}{n!}right|le frac{2^{n+2}R}{(R-r)^{n+1}}{A(r)+|f(0)|}$$
MY attempt
Using Cauchy generalized integral formula
$f^{(n)}(z_0)=frac{n!}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}$
then $max_{|z|=r}left|frac{f^n(z)}{n!}right|=left|max_{|z|=r} frac{1}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}right|$
and i think we can solve by using Borel caratheodory theorem
Let $f(z)$ be holomorphic function on $|z|le R$ and let $M(r)=max_{|z|=r}|f(z)|$ and $A(r)=max_{|z|=r}Re f(z)$ then for $0<r<R$
$$ M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
from here how to processed ?
complex-analysis
asked Nov 20 '18 at 9:19
Inverse Problem
923918
Suppose that $f(z)$ is analytic function in the disk $|z|le R$ and $A(r)=max_{|z|=r}Re(z)$ then prove that $r<R$
$$max_{|z|=r}left|frac{f^n(z)}{n!}right|le frac{2^{n+2}R}{(R-r)^{n+1}}{A(r)+|f(0)|}$$
MY attempt
Using Cauchy generalized integral formula
$f^{(n)}(z_0)=frac{n!}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}$
then $max_{|z|=r}left|frac{f^n(z)}{n!}right|=left|max_{|z|=r} frac{1}{2pi i}int _cfrac{f(z)dz}{(z-z_0)^{n+1}}right|$
and i think we can solve by using Borel caratheodory theorem
Let $f(z)$ be holomorphic function on $|z|le R$ and let $M(r)=max_{|z|=r}|f(z)|$ and $A(r)=max_{|z|=r}Re f(z)$ then for $0<r<R$
$$ M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
from here how to processed ?
complex-analysis
complex-analysis
asked Nov 20 '18 at 9:19
Inverse Problem
923918
asked Nov 20 '18 at 9:19
Inverse Problem
923918
edited Nov 20 '18 at 10:33
asked Nov 20 '18 at 9:19
Inverse Problem
923918
asked Nov 20 '18 at 9:19
Inverse Problem
923918
asked Nov 20 '18 at 9:19
Inverse Problem
923918
923918
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
First, observe that, $r<R$ and so for any $ninmathbb{N}$ we have,
$$left(frac{2r}{R-r}right)^n>0text{ and }left(1+frac{r}{R}right)<2$$
$$implies left(1+frac{r}{R}right)<2<2left(frac{2r}{R-r}right)^n$$
$$implies left(1+frac{r}{R}right)<2left(frac{2r}{R-r}right)^n.....(1)$$
Now by Cauchy Inequality, $$|a_n|le frac{M(r)}{r^n}.....(2)$$ where $displaystyle M(r)=max_{|z|=r}{|f(z)|}$ and $a_n=frac{f^n(z)}{n!}$ is the coefficient of power series of $f$ in the given domain.
Now finally using Borel-Caratheodory theorem we have,
$$M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
$$implies |a_n|=|frac{f^n(z)}{n!}|le frac{1}{r^n}frac{R+r}{R-r}{A(R)+|f(0)|}$$
$$implies |frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
This inequality follows directly from (1) by some simple algebraic manipulation. Now since the RHS is independent of $z$ so taking maximum on the boundary $|z|=r$ we obtain, $$max_{|z|=r}|frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
Hope this works.
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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First, observe that, $r<R$ and so for any $ninmathbb{N}$ we have,
$$left(frac{2r}{R-r}right)^n>0text{ and }left(1+frac{r}{R}right)<2$$
$$implies left(1+frac{r}{R}right)<2<2left(frac{2r}{R-r}right)^n$$
$$implies left(1+frac{r}{R}right)<2left(frac{2r}{R-r}right)^n.....(1)$$
Now by Cauchy Inequality, $$|a_n|le frac{M(r)}{r^n}.....(2)$$ where $displaystyle M(r)=max_{|z|=r}{|f(z)|}$ and $a_n=frac{f^n(z)}{n!}$ is the coefficient of power series of $f$ in the given domain.
Now finally using Borel-Caratheodory theorem we have,
$$M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
$$implies |a_n|=|frac{f^n(z)}{n!}|le frac{1}{r^n}frac{R+r}{R-r}{A(R)+|f(0)|}$$
$$implies |frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
This inequality follows directly from (1) by some simple algebraic manipulation. Now since the RHS is independent of $z$ so taking maximum on the boundary $|z|=r$ we obtain, $$max_{|z|=r}|frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
Hope this works.
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
add a comment |
First, observe that, $r<R$ and so for any $ninmathbb{N}$ we have,
$$left(frac{2r}{R-r}right)^n>0text{ and }left(1+frac{r}{R}right)<2$$
$$implies left(1+frac{r}{R}right)<2<2left(frac{2r}{R-r}right)^n$$
$$implies left(1+frac{r}{R}right)<2left(frac{2r}{R-r}right)^n.....(1)$$
Now by Cauchy Inequality, $$|a_n|le frac{M(r)}{r^n}.....(2)$$ where $displaystyle M(r)=max_{|z|=r}{|f(z)|}$ and $a_n=frac{f^n(z)}{n!}$ is the coefficient of power series of $f$ in the given domain.
Now finally using Borel-Caratheodory theorem we have,
$$M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
$$implies |a_n|=|frac{f^n(z)}{n!}|le frac{1}{r^n}frac{R+r}{R-r}{A(R)+|f(0)|}$$
$$implies |frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
This inequality follows directly from (1) by some simple algebraic manipulation. Now since the RHS is independent of $z$ so taking maximum on the boundary $|z|=r$ we obtain, $$max_{|z|=r}|frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
Hope this works.
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
add a comment |
First, observe that, $r<R$ and so for any $ninmathbb{N}$ we have,
$$left(frac{2r}{R-r}right)^n>0text{ and }left(1+frac{r}{R}right)<2$$
$$implies left(1+frac{r}{R}right)<2<2left(frac{2r}{R-r}right)^n$$
$$implies left(1+frac{r}{R}right)<2left(frac{2r}{R-r}right)^n.....(1)$$
Now by Cauchy Inequality, $$|a_n|le frac{M(r)}{r^n}.....(2)$$ where $displaystyle M(r)=max_{|z|=r}{|f(z)|}$ and $a_n=frac{f^n(z)}{n!}$ is the coefficient of power series of $f$ in the given domain.
Now finally using Borel-Caratheodory theorem we have,
$$M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
$$implies |a_n|=|frac{f^n(z)}{n!}|le frac{1}{r^n}frac{R+r}{R-r}{A(R)+|f(0)|}$$
$$implies |frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
This inequality follows directly from (1) by some simple algebraic manipulation. Now since the RHS is independent of $z$ so taking maximum on the boundary $|z|=r$ we obtain, $$max_{|z|=r}|frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
Hope this works.
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
First, observe that, $r<R$ and so for any $ninmathbb{N}$ we have,
$$left(frac{2r}{R-r}right)^n>0text{ and }left(1+frac{r}{R}right)<2$$
$$implies left(1+frac{r}{R}right)<2<2left(frac{2r}{R-r}right)^n$$
$$implies left(1+frac{r}{R}right)<2left(frac{2r}{R-r}right)^n.....(1)$$
Now by Cauchy Inequality, $$|a_n|le frac{M(r)}{r^n}.....(2)$$ where $displaystyle M(r)=max_{|z|=r}{|f(z)|}$ and $a_n=frac{f^n(z)}{n!}$ is the coefficient of power series of $f$ in the given domain.
Now finally using Borel-Caratheodory theorem we have,
$$M(r)le frac{2r}{R-r}A(R)+frac{R+r}{R-r}|f(0)|$$
$$implies |a_n|=|frac{f^n(z)}{n!}|le frac{1}{r^n}frac{R+r}{R-r}{A(R)+|f(0)|}$$
$$implies |frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
This inequality follows directly from (1) by some simple algebraic manipulation. Now since the RHS is independent of $z$ so taking maximum on the boundary $|z|=r$ we obtain, $$max_{|z|=r}|frac{f^n(z)}{n!}|le frac{2^{n+1}R}{(R-r)^{n+1}}{A(R)+|f(0)|}$$
Hope this works.
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
answered Nov 20 '18 at 9:55
Sujit Bhattacharyya
945318
945318
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
add a comment |
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
...very nice way of thinking and answer ..thankyou...
– Inverse Problem
Nov 20 '18 at 10:06
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
welcome! basically its just an application of Borel-Caratheodory theorem
– Sujit Bhattacharyya
Nov 20 '18 at 10:28
add a comment |
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