Mixing
Cauchy's Estimate for $|z|=R$
$begingroup$
Let $f$ be entire and let $|f(z)|leq M$ for $|z|=R$ and R is fixed. I want to prove that $$|f^{(k)}(re^{itheta})|leq frac{k!M}{(R-r)^k}$$
I don't understand how you get R-r in the denominator. I know the Cauchy's estimate formula.
Thank you for the help!
complex-analysis
asked May 5 '13 at 18:17
d13d13
177218
$endgroup$
add a comment |
$begingroup$
Let $f$ be entire and let $|f(z)|leq M$ for $|z|=R$ and R is fixed. I want to prove that $$|f^{(k)}(re^{itheta})|leq frac{k!M}{(R-r)^k}$$
I don't understand how you get R-r in the denominator. I know the Cauchy's estimate formula.
Thank you for the help!
complex-analysis
asked May 5 '13 at 18:17
d13d13
177218
$endgroup$
$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07
add a comment |
$begingroup$
Let $f$ be entire and let $|f(z)|leq M$ for $|z|=R$ and R is fixed. I want to prove that $$|f^{(k)}(re^{itheta})|leq frac{k!M}{(R-r)^k}$$
I don't understand how you get R-r in the denominator. I know the Cauchy's estimate formula.
Thank you for the help!
complex-analysis
asked May 5 '13 at 18:17
d13d13
177218
$endgroup$
Let $f$ be entire and let $|f(z)|leq M$ for $|z|=R$ and R is fixed. I want to prove that $$|f^{(k)}(re^{itheta})|leq frac{k!M}{(R-r)^k}$$
I don't understand how you get R-r in the denominator. I know the Cauchy's estimate formula.
Thank you for the help!
complex-analysis
complex-analysis
asked May 5 '13 at 18:17
d13d13
177218
asked May 5 '13 at 18:17
d13d13
177218
asked May 5 '13 at 18:17
d13d13
177218
asked May 5 '13 at 18:17
d13d13
177218
asked May 5 '13 at 18:17
d13d13
177218
177218
$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07
add a comment |
$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07
$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By Cauchy's Integral Formula:
$$f^{(k)}(t)=frac{k!}{2pi i}intlimits_Cfrac{f(z)}{(z-t)^{k+1}}dz$$
so putting $,z=re^{itheta},$ , we get by Cauchy's Estimates
$$|f^{(k)}(re^{itheta})|le frac{k!}{|2pi i|}max_{|z|=R}left(frac{|f(z)|}{|z-re^{itheta}|^{k+1}}right)2pi Rle frac{k!MR}{(R-r)^{k+1}}$$
since $,zin{ zinBbb C;;;|z|=R};,;;t=re^{itheta}implies,$
$$frac1{|z-t|}lefrac1{|z|-|re^{itheta}|}=frac1{R-r}$$
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
add a comment |
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$begingroup$
By Cauchy's Integral Formula:
$$f^{(k)}(t)=frac{k!}{2pi i}intlimits_Cfrac{f(z)}{(z-t)^{k+1}}dz$$
so putting $,z=re^{itheta},$ , we get by Cauchy's Estimates
$$|f^{(k)}(re^{itheta})|le frac{k!}{|2pi i|}max_{|z|=R}left(frac{|f(z)|}{|z-re^{itheta}|^{k+1}}right)2pi Rle frac{k!MR}{(R-r)^{k+1}}$$
since $,zin{ zinBbb C;;;|z|=R};,;;t=re^{itheta}implies,$
$$frac1{|z-t|}lefrac1{|z|-|re^{itheta}|}=frac1{R-r}$$
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
add a comment |
$begingroup$
By Cauchy's Integral Formula:
$$f^{(k)}(t)=frac{k!}{2pi i}intlimits_Cfrac{f(z)}{(z-t)^{k+1}}dz$$
so putting $,z=re^{itheta},$ , we get by Cauchy's Estimates
$$|f^{(k)}(re^{itheta})|le frac{k!}{|2pi i|}max_{|z|=R}left(frac{|f(z)|}{|z-re^{itheta}|^{k+1}}right)2pi Rle frac{k!MR}{(R-r)^{k+1}}$$
since $,zin{ zinBbb C;;;|z|=R};,;;t=re^{itheta}implies,$
$$frac1{|z-t|}lefrac1{|z|-|re^{itheta}|}=frac1{R-r}$$
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
$endgroup$
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
add a comment |
$begingroup$
By Cauchy's Integral Formula:
$$f^{(k)}(t)=frac{k!}{2pi i}intlimits_Cfrac{f(z)}{(z-t)^{k+1}}dz$$
so putting $,z=re^{itheta},$ , we get by Cauchy's Estimates
$$|f^{(k)}(re^{itheta})|le frac{k!}{|2pi i|}max_{|z|=R}left(frac{|f(z)|}{|z-re^{itheta}|^{k+1}}right)2pi Rle frac{k!MR}{(R-r)^{k+1}}$$
since $,zin{ zinBbb C;;;|z|=R};,;;t=re^{itheta}implies,$
$$frac1{|z-t|}lefrac1{|z|-|re^{itheta}|}=frac1{R-r}$$
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
$endgroup$
By Cauchy's Integral Formula:
$$f^{(k)}(t)=frac{k!}{2pi i}intlimits_Cfrac{f(z)}{(z-t)^{k+1}}dz$$
so putting $,z=re^{itheta},$ , we get by Cauchy's Estimates
$$|f^{(k)}(re^{itheta})|le frac{k!}{|2pi i|}max_{|z|=R}left(frac{|f(z)|}{|z-re^{itheta}|^{k+1}}right)2pi Rle frac{k!MR}{(R-r)^{k+1}}$$
since $,zin{ zinBbb C;;;|z|=R};,;;t=re^{itheta}implies,$
$$frac1{|z-t|}lefrac1{|z|-|re^{itheta}|}=frac1{R-r}$$
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
edited Jan 21 at 9:14
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
answered May 5 '13 at 19:53
DonAntonioDonAntonio
179k1494232
179k1494232
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
add a comment |
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
What happened to the R in the numerator? How does that cancel with the extra $R-r$ in the denominator?
$endgroup$
– Esmath
Jan 21 at 4:37
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
It is there, I just forgot to write down 6 years ago...And you don't get what the OP asked.
$endgroup$
– DonAntonio
Jan 21 at 9:14
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
If I'm understanding correctly, there shouldn't be an $R$ in the numerator at the end... according to the OP's question.
$endgroup$
– Esmath
Jan 24 at 20:35
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
$begingroup$
@Esmath Perhaps he forgot it (his formula is also missing one exponent more in the denominator), perhaps I just didn't reach what he wanted and he's right , though I doubt it...I just can't remember what was this, sorry.
$endgroup$
– DonAntonio
Jan 24 at 21:26
add a comment |
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$begingroup$
What is Cauchy's estimate formula? Do you know Cauchy's formula (not estimate …) for $f^({k})(z)$?
$endgroup$
– Harald Hanche-Olsen
May 5 '13 at 18:23
$begingroup$
yes its $f^{(k)}(a)= frac{k!}{2pi{i}}int_Cfrac {f(z)dz}{(z-a)^k}$ am i right?
$endgroup$
– d13
May 5 '13 at 18:53
$begingroup$
@d13 It is actually $(z-a)^{k+1}$ in the denominator.
$endgroup$
– user39280
May 5 '13 at 19:07