Is there a theorem in Real analysis similar to Cauchy's theorem in Complex analysis?The distinction between infinitely differentiable function and real analytic functionCauchy's Integral TheoremApplications of Residue Theorem in complex analysis?Complex analysis without Cauchy's theoremUnderstanding Integration in Complex analysisCauchy's integral formula and Green's theorem. Scalar or gradient?Integrating a real function using complex analysisIntegral with two different answers using real and complex analysis
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Is there a theorem in Real analysis similar to Cauchy's theorem in Complex analysis?
The distinction between infinitely differentiable function and real analytic functionCauchy's Integral TheoremApplications of Residue Theorem in complex analysis?Complex analysis without Cauchy's theoremUnderstanding Integration in Complex analysisCauchy's integral formula and Green's theorem. Scalar or gradient?Integrating a real function using complex analysisIntegral with two different answers using real and complex analysis
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$begingroup$
Is there a theorem in Real Analysis similar to Cauchy's theorem/Cauchy's Integral Formula from Complex analysis?
If not, then why? What is it about the complex space that makes Cauchy's theorem true?
real-analysis calculus complex-analysis complex-numbers cauchy-integral-formula
$endgroup$
add a comment
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$begingroup$
Is there a theorem in Real Analysis similar to Cauchy's theorem/Cauchy's Integral Formula from Complex analysis?
If not, then why? What is it about the complex space that makes Cauchy's theorem true?
real-analysis calculus complex-analysis complex-numbers cauchy-integral-formula
$endgroup$
$begingroup$
I have edited my answer as per your request
$endgroup$
– Learnmore
7 hours ago
add a comment
|
$begingroup$
Is there a theorem in Real Analysis similar to Cauchy's theorem/Cauchy's Integral Formula from Complex analysis?
If not, then why? What is it about the complex space that makes Cauchy's theorem true?
real-analysis calculus complex-analysis complex-numbers cauchy-integral-formula
$endgroup$
Is there a theorem in Real Analysis similar to Cauchy's theorem/Cauchy's Integral Formula from Complex analysis?
If not, then why? What is it about the complex space that makes Cauchy's theorem true?
real-analysis calculus complex-analysis complex-numbers cauchy-integral-formula
real-analysis calculus complex-analysis complex-numbers cauchy-integral-formula
asked 8 hours ago
krtgdlkrtgdl
19313 bronze badges
19313 bronze badges
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I have edited my answer as per your request
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– Learnmore
7 hours ago
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I have edited my answer as per your request
$endgroup$
– Learnmore
7 hours ago
$begingroup$
I have edited my answer as per your request
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– Learnmore
7 hours ago
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I have edited my answer as per your request
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– Learnmore
7 hours ago
add a comment
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4 Answers
4
active
oldest
votes
$begingroup$
I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.
Harmonic functions are characterised by the Mean Value Property, i.e., if $Usubseteq mathbbR^n$ is open, $xin U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)subseteq U,$ we have
$$
f(x)=frac1dS(partial B(x,r))int_partial B(x,r) f(y)textrmdS(y),
$$
where $dS$ denotes the surface measure on $partial B(x,r)$. In case $n=2,$ this will be a regular cruve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.
Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.
$endgroup$
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The (possible) reasons are:
(1) If $f$ is analytic(differentiable) in $Bbb C$ then so is its derivative $f^'$ which is not true in $Bbb R$ , example consider the function $f(x)=x^2sin (frac1x)$.
(2) If you integrate an analytic function over a closed domain in $Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $Bbb R$ , example consider $x^2 $ over $[-1,1]$
NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $Bbb R$
$endgroup$
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.
See The distinction between infinitely differentiable function and real analytic function
$endgroup$
add a comment
|
$begingroup$
Several theorems are named after Augustin-Louis Cauchy. Cauchy's theorem may mean:
Cauchy's integral theorem in complex analysis, also Cauchy's integral formula
Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem.
Cauchy's criterion for uniform convergence.
Which Cauchy's theorem are you referring too?
$endgroup$
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
add a comment
|
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.
Harmonic functions are characterised by the Mean Value Property, i.e., if $Usubseteq mathbbR^n$ is open, $xin U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)subseteq U,$ we have
$$
f(x)=frac1dS(partial B(x,r))int_partial B(x,r) f(y)textrmdS(y),
$$
where $dS$ denotes the surface measure on $partial B(x,r)$. In case $n=2,$ this will be a regular cruve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.
Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.
$endgroup$
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.
Harmonic functions are characterised by the Mean Value Property, i.e., if $Usubseteq mathbbR^n$ is open, $xin U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)subseteq U,$ we have
$$
f(x)=frac1dS(partial B(x,r))int_partial B(x,r) f(y)textrmdS(y),
$$
where $dS$ denotes the surface measure on $partial B(x,r)$. In case $n=2,$ this will be a regular cruve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.
Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.
$endgroup$
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.
Harmonic functions are characterised by the Mean Value Property, i.e., if $Usubseteq mathbbR^n$ is open, $xin U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)subseteq U,$ we have
$$
f(x)=frac1dS(partial B(x,r))int_partial B(x,r) f(y)textrmdS(y),
$$
where $dS$ denotes the surface measure on $partial B(x,r)$. In case $n=2,$ this will be a regular cruve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.
Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.
$endgroup$
I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.
Harmonic functions are characterised by the Mean Value Property, i.e., if $Usubseteq mathbbR^n$ is open, $xin U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)subseteq U,$ we have
$$
f(x)=frac1dS(partial B(x,r))int_partial B(x,r) f(y)textrmdS(y),
$$
where $dS$ denotes the surface measure on $partial B(x,r)$. In case $n=2,$ this will be a regular cruve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.
Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.
answered 8 hours ago
WoolierThanThouWoolierThanThou
1,4411 silver badge8 bronze badges
1,4411 silver badge8 bronze badges
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
$begingroup$
Thank you! I like this way of thinking about Cauchy's Integral formula!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The (possible) reasons are:
(1) If $f$ is analytic(differentiable) in $Bbb C$ then so is its derivative $f^'$ which is not true in $Bbb R$ , example consider the function $f(x)=x^2sin (frac1x)$.
(2) If you integrate an analytic function over a closed domain in $Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $Bbb R$ , example consider $x^2 $ over $[-1,1]$
NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $Bbb R$
$endgroup$
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The (possible) reasons are:
(1) If $f$ is analytic(differentiable) in $Bbb C$ then so is its derivative $f^'$ which is not true in $Bbb R$ , example consider the function $f(x)=x^2sin (frac1x)$.
(2) If you integrate an analytic function over a closed domain in $Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $Bbb R$ , example consider $x^2 $ over $[-1,1]$
NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $Bbb R$
$endgroup$
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The (possible) reasons are:
(1) If $f$ is analytic(differentiable) in $Bbb C$ then so is its derivative $f^'$ which is not true in $Bbb R$ , example consider the function $f(x)=x^2sin (frac1x)$.
(2) If you integrate an analytic function over a closed domain in $Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $Bbb R$ , example consider $x^2 $ over $[-1,1]$
NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $Bbb R$
$endgroup$
The (possible) reasons are:
(1) If $f$ is analytic(differentiable) in $Bbb C$ then so is its derivative $f^'$ which is not true in $Bbb R$ , example consider the function $f(x)=x^2sin (frac1x)$.
(2) If you integrate an analytic function over a closed domain in $Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $Bbb R$ , example consider $x^2 $ over $[-1,1]$
NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $Bbb R$
edited 7 hours ago
answered 8 hours ago
LearnmoreLearnmore
18.5k3 gold badges29 silver badges114 bronze badges
18.5k3 gold badges29 silver badges114 bronze badges
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
$begingroup$
Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $mathbbR $ - calculus from $mathbbC$ - Calculus?
$endgroup$
– krtgdl
8 hours ago
1
1
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
@krtgdl; It is because every analytic function has an antiderivative
$endgroup$
– Learnmore
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
$begingroup$
Yes, I see! Thank you!
$endgroup$
– krtgdl
7 hours ago
add a comment
|
$begingroup$
The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.
See The distinction between infinitely differentiable function and real analytic function
$endgroup$
add a comment
|
$begingroup$
The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.
See The distinction between infinitely differentiable function and real analytic function
$endgroup$
add a comment
|
$begingroup$
The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.
See The distinction between infinitely differentiable function and real analytic function
$endgroup$
The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.
See The distinction between infinitely differentiable function and real analytic function
answered 8 hours ago
Ethan BolkerEthan Bolker
55.8k5 gold badges63 silver badges134 bronze badges
55.8k5 gold badges63 silver badges134 bronze badges
add a comment
|
add a comment
|
$begingroup$
Several theorems are named after Augustin-Louis Cauchy. Cauchy's theorem may mean:
Cauchy's integral theorem in complex analysis, also Cauchy's integral formula
Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem.
Cauchy's criterion for uniform convergence.
Which Cauchy's theorem are you referring too?
$endgroup$
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
add a comment
|
$begingroup$
Several theorems are named after Augustin-Louis Cauchy. Cauchy's theorem may mean:
Cauchy's integral theorem in complex analysis, also Cauchy's integral formula
Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem.
Cauchy's criterion for uniform convergence.
Which Cauchy's theorem are you referring too?
$endgroup$
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
add a comment
|
$begingroup$
Several theorems are named after Augustin-Louis Cauchy. Cauchy's theorem may mean:
Cauchy's integral theorem in complex analysis, also Cauchy's integral formula
Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem.
Cauchy's criterion for uniform convergence.
Which Cauchy's theorem are you referring too?
$endgroup$
Several theorems are named after Augustin-Louis Cauchy. Cauchy's theorem may mean:
Cauchy's integral theorem in complex analysis, also Cauchy's integral formula
Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem.
Cauchy's criterion for uniform convergence.
Which Cauchy's theorem are you referring too?
answered 8 hours ago
Phinda HlawePhinda Hlawe
172 bronze badges
172 bronze badges
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
add a comment
|
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
2
2
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
This should really be a comment...
$endgroup$
– Pixel
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
$begingroup$
I apologise for the confusion. I thought I had made it clear that I am refering to Cauchy's Integral formula/Cauchy's theorem from Complex analysis.
$endgroup$
– krtgdl
8 hours ago
1
1
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
$begingroup$
I am not allowed to make comments on posts I did not write I'm still new to this.
$endgroup$
– Phinda Hlawe
8 hours ago
add a comment
|
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$begingroup$
I have edited my answer as per your request
$endgroup$
– Learnmore
7 hours ago