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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5












$begingroup$


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?










share|cite|improve this question









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  • $begingroup$
    I have edited my answer as per your request
    $endgroup$
    – Learnmore
    8 hours ago

















5












$begingroup$


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?










share|cite|improve this question









$endgroup$













  • $begingroup$
    I have edited my answer as per your request
    $endgroup$
    – Learnmore
    8 hours ago













5












5








5





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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share|cite|improve this question










asked 9 hours ago









krtgdlkrtgdl

19313 bronze badges




19313 bronze badges














  • $begingroup$
    I have edited my answer as per your request
    $endgroup$
    – Learnmore
    8 hours ago
















  • $begingroup$
    I have edited my answer as per your request
    $endgroup$
    – Learnmore
    8 hours ago















$begingroup$
I have edited my answer as per your request
$endgroup$
– Learnmore
8 hours ago




$begingroup$
I have edited my answer as per your request
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– Learnmore
8 hours ago










4 Answers
4






active

oldest

votes


















2














$begingroup$

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.






share|cite|improve this answer









$endgroup$














  • $begingroup$
    Thank you! I like this way of thinking about Cauchy's Integral formula!
    $endgroup$
    – krtgdl
    8 hours ago



















4














$begingroup$

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 $Bbb R$






share|cite|improve this answer











$endgroup$














  • $begingroup$
    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
    9 hours ago







  • 1




    $begingroup$
    @krtgdl; It is because every analytic function has an antiderivative
    $endgroup$
    – Learnmore
    8 hours ago










  • $begingroup$
    Yes, I see! Thank you!
    $endgroup$
    – krtgdl
    8 hours ago


















2














$begingroup$

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






share|cite|improve this answer









$endgroup$






















    -1














    $begingroup$

    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?






    share|cite|improve this answer









    $endgroup$










    • 4




      $begingroup$
      This should really be a comment...
      $endgroup$
      – Pixel
      9 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
      9 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
      9 hours ago










    • $begingroup$
      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
      $endgroup$
      – Lukas Kofler
      8 mins ago













    Your Answer








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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$














    • $begingroup$
      Thank you! I like this way of thinking about Cauchy's Integral formula!
      $endgroup$
      – krtgdl
      8 hours ago
















    2














    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$














    • $begingroup$
      Thank you! I like this way of thinking about Cauchy's Integral formula!
      $endgroup$
      – krtgdl
      8 hours ago














    2














    2










    2







    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 9 hours ago









    WoolierThanThouWoolierThanThou

    1,4511 silver badge8 bronze badges




    1,4511 silver badge8 bronze badges














    • $begingroup$
      Thank you! I like this way of thinking about Cauchy's Integral formula!
      $endgroup$
      – krtgdl
      8 hours ago

















    • $begingroup$
      Thank you! I like this way of thinking about Cauchy's Integral formula!
      $endgroup$
      – krtgdl
      8 hours ago
















    $begingroup$
    Thank you! I like this way of thinking about Cauchy's Integral formula!
    $endgroup$
    – krtgdl
    8 hours ago





    $begingroup$
    Thank you! I like this way of thinking about Cauchy's Integral formula!
    $endgroup$
    – krtgdl
    8 hours ago














    4














    $begingroup$

    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 $Bbb R$






    share|cite|improve this answer











    $endgroup$














    • $begingroup$
      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
      9 hours ago







    • 1




      $begingroup$
      @krtgdl; It is because every analytic function has an antiderivative
      $endgroup$
      – Learnmore
      8 hours ago










    • $begingroup$
      Yes, I see! Thank you!
      $endgroup$
      – krtgdl
      8 hours ago















    4














    $begingroup$

    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 $Bbb R$






    share|cite|improve this answer











    $endgroup$














    • $begingroup$
      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
      9 hours ago







    • 1




      $begingroup$
      @krtgdl; It is because every analytic function has an antiderivative
      $endgroup$
      – Learnmore
      8 hours ago










    • $begingroup$
      Yes, I see! Thank you!
      $endgroup$
      – krtgdl
      8 hours ago













    4














    4










    4







    $begingroup$

    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 $Bbb R$






    share|cite|improve this answer











    $endgroup$



    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 $Bbb R$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered 9 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
      9 hours ago







    • 1




      $begingroup$
      @krtgdl; It is because every analytic function has an antiderivative
      $endgroup$
      – Learnmore
      8 hours ago










    • $begingroup$
      Yes, I see! Thank you!
      $endgroup$
      – 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
      9 hours ago







    • 1




      $begingroup$
      @krtgdl; It is because every analytic function has an antiderivative
      $endgroup$
      – Learnmore
      8 hours ago










    • $begingroup$
      Yes, I see! Thank you!
      $endgroup$
      – 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
    9 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
    9 hours ago





    1




    1




    $begingroup$
    @krtgdl; It is because every analytic function has an antiderivative
    $endgroup$
    – Learnmore
    8 hours ago




    $begingroup$
    @krtgdl; It is because every analytic function has an antiderivative
    $endgroup$
    – Learnmore
    8 hours ago












    $begingroup$
    Yes, I see! Thank you!
    $endgroup$
    – krtgdl
    8 hours ago




    $begingroup$
    Yes, I see! Thank you!
    $endgroup$
    – krtgdl
    8 hours ago











    2














    $begingroup$

    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






    share|cite|improve this answer









    $endgroup$



















      2














      $begingroup$

      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






      share|cite|improve this answer









      $endgroup$

















        2














        2










        2







        $begingroup$

        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






        share|cite|improve this answer









        $endgroup$



        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        Ethan BolkerEthan Bolker

        55.8k5 gold badges63 silver badges134 bronze badges




        55.8k5 gold badges63 silver badges134 bronze badges
























            -1














            $begingroup$

            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?






            share|cite|improve this answer









            $endgroup$










            • 4




              $begingroup$
              This should really be a comment...
              $endgroup$
              – Pixel
              9 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
              9 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
              9 hours ago










            • $begingroup$
              This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
              $endgroup$
              – Lukas Kofler
              8 mins ago















            -1














            $begingroup$

            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?






            share|cite|improve this answer









            $endgroup$










            • 4




              $begingroup$
              This should really be a comment...
              $endgroup$
              – Pixel
              9 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
              9 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
              9 hours ago










            • $begingroup$
              This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
              $endgroup$
              – Lukas Kofler
              8 mins ago













            -1














            -1










            -1







            $begingroup$

            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?






            share|cite|improve this answer









            $endgroup$



            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?







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            answered 9 hours ago









            Phinda HlawePhinda Hlawe

            152 bronze badges




            152 bronze badges










            • 4




              $begingroup$
              This should really be a comment...
              $endgroup$
              – Pixel
              9 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
              9 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
              9 hours ago










            • $begingroup$
              This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
              $endgroup$
              – Lukas Kofler
              8 mins ago












            • 4




              $begingroup$
              This should really be a comment...
              $endgroup$
              – Pixel
              9 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
              9 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
              9 hours ago










            • $begingroup$
              This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
              $endgroup$
              – Lukas Kofler
              8 mins ago







            4




            4




            $begingroup$
            This should really be a comment...
            $endgroup$
            – Pixel
            9 hours ago




            $begingroup$
            This should really be a comment...
            $endgroup$
            – Pixel
            9 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
            9 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
            9 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
            9 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
            9 hours ago












            $begingroup$
            This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
            $endgroup$
            – Lukas Kofler
            8 mins ago




            $begingroup$
            This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
            $endgroup$
            – Lukas Kofler
            8 mins ago


















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