Compilation of representations of holomorphic functions












1














Holomorphic functions are my muse. As my muse, I love drawing them different ways. Allow me to frame this as though an artist talking about his muse.



A holomorphic function $f$ on the unit disk $mathbb{D}$ is completely determined (and not only determined, represented) by ${f^{(j)}(0)}_{j=0}^infty$.



$$f(z) = sum_{j=0}^infty f^{(j)}(0) frac{z^j}{j!}$$



Similarly a holomorphic function $f$ on $mathbb{D}$ is completely determined (and not only determined, represented on the inside of $mathcal{C}$) by its values on any contour $mathcal{C} subset mathbb{D}$.



$$f(z) = int_{mathcal{C}} frac{f(zeta)}{zeta - z},dzeta$$



Adding more constraints, and restricting my muse to certain poses, you can find better and more nuanced art:



If $f$ is entire, and $|f(z)| < C e^{tau |z|^rho}$ everywhere, for arbitrary constants $C,rho,tau in mathbb{R}^+$, then $f$ is completely determined (and not only determined, $nearly$ represented) by its zeroes ${a_j}_{j=1}^infty$.



$$f(z) = e^{p(z)}prod_{j=1}^infty (1 - frac{z}{a_j})e^{-frac{z}{a_j} - frac{z^2}{2a_j^2} -...-frac{z^n}{na_j^n}}$$



where $n$ is the closest greatest integer to $rho$ and $p$ is a polynomial of at most degree $n$.



My muse also has rare representations, that bring out specificity and still beauty. Thanks to Ramanujan's careful deliberations,



A holomorphic function $f$ on $mathbb{C}_{Re(z)>0}$, such that $|f(z)|< Ce^{rho |Re(z)| + tau |Im(z)|}$, for arbitrary constants $C, rho, tau in mathbb{R}^+$ with $tau < pi/2$, then $f(z)$ is completely determined (and not only determined, represented) by $f big{|}_{mathbb{N}}$.



$$f(z)Gamma(1-z) = int_0^infty vartheta(-x)x^{-z},dx$$



where $vartheta(x) = sum_{j=0}^infty f(j+1) frac{x^j}{j!}$, $Gamma$ is the Gamma function, and $0 < Re(z) < 1$. This can be extended to the expression



$$f(z)Gamma(1-z) = sum_{j=0}^infty f(j+1)frac{(-1)^j}{j!(j+1-z)} + int_1^infty vartheta(-x)x^{-z},dx$$



which works for all $mathbb{C}_{Re(z) > 0}$.



What other instances do holomorphic functions (on any domain subject to whatever constraints) admit a unique representation theorem based on a sliver of information about the function. Slightly different than identity theorems, as these determine, but rather examples that also represent.



If need be this can be Community wiki.



Thank you and Happy New Year,



Richard Diagram










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    1














    Holomorphic functions are my muse. As my muse, I love drawing them different ways. Allow me to frame this as though an artist talking about his muse.



    A holomorphic function $f$ on the unit disk $mathbb{D}$ is completely determined (and not only determined, represented) by ${f^{(j)}(0)}_{j=0}^infty$.



    $$f(z) = sum_{j=0}^infty f^{(j)}(0) frac{z^j}{j!}$$



    Similarly a holomorphic function $f$ on $mathbb{D}$ is completely determined (and not only determined, represented on the inside of $mathcal{C}$) by its values on any contour $mathcal{C} subset mathbb{D}$.



    $$f(z) = int_{mathcal{C}} frac{f(zeta)}{zeta - z},dzeta$$



    Adding more constraints, and restricting my muse to certain poses, you can find better and more nuanced art:



    If $f$ is entire, and $|f(z)| < C e^{tau |z|^rho}$ everywhere, for arbitrary constants $C,rho,tau in mathbb{R}^+$, then $f$ is completely determined (and not only determined, $nearly$ represented) by its zeroes ${a_j}_{j=1}^infty$.



    $$f(z) = e^{p(z)}prod_{j=1}^infty (1 - frac{z}{a_j})e^{-frac{z}{a_j} - frac{z^2}{2a_j^2} -...-frac{z^n}{na_j^n}}$$



    where $n$ is the closest greatest integer to $rho$ and $p$ is a polynomial of at most degree $n$.



    My muse also has rare representations, that bring out specificity and still beauty. Thanks to Ramanujan's careful deliberations,



    A holomorphic function $f$ on $mathbb{C}_{Re(z)>0}$, such that $|f(z)|< Ce^{rho |Re(z)| + tau |Im(z)|}$, for arbitrary constants $C, rho, tau in mathbb{R}^+$ with $tau < pi/2$, then $f(z)$ is completely determined (and not only determined, represented) by $f big{|}_{mathbb{N}}$.



    $$f(z)Gamma(1-z) = int_0^infty vartheta(-x)x^{-z},dx$$



    where $vartheta(x) = sum_{j=0}^infty f(j+1) frac{x^j}{j!}$, $Gamma$ is the Gamma function, and $0 < Re(z) < 1$. This can be extended to the expression



    $$f(z)Gamma(1-z) = sum_{j=0}^infty f(j+1)frac{(-1)^j}{j!(j+1-z)} + int_1^infty vartheta(-x)x^{-z},dx$$



    which works for all $mathbb{C}_{Re(z) > 0}$.



    What other instances do holomorphic functions (on any domain subject to whatever constraints) admit a unique representation theorem based on a sliver of information about the function. Slightly different than identity theorems, as these determine, but rather examples that also represent.



    If need be this can be Community wiki.



    Thank you and Happy New Year,



    Richard Diagram










    share|cite|improve this question









    New contributor




    Richard Diagram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1












      1








      1







      Holomorphic functions are my muse. As my muse, I love drawing them different ways. Allow me to frame this as though an artist talking about his muse.



      A holomorphic function $f$ on the unit disk $mathbb{D}$ is completely determined (and not only determined, represented) by ${f^{(j)}(0)}_{j=0}^infty$.



      $$f(z) = sum_{j=0}^infty f^{(j)}(0) frac{z^j}{j!}$$



      Similarly a holomorphic function $f$ on $mathbb{D}$ is completely determined (and not only determined, represented on the inside of $mathcal{C}$) by its values on any contour $mathcal{C} subset mathbb{D}$.



      $$f(z) = int_{mathcal{C}} frac{f(zeta)}{zeta - z},dzeta$$



      Adding more constraints, and restricting my muse to certain poses, you can find better and more nuanced art:



      If $f$ is entire, and $|f(z)| < C e^{tau |z|^rho}$ everywhere, for arbitrary constants $C,rho,tau in mathbb{R}^+$, then $f$ is completely determined (and not only determined, $nearly$ represented) by its zeroes ${a_j}_{j=1}^infty$.



      $$f(z) = e^{p(z)}prod_{j=1}^infty (1 - frac{z}{a_j})e^{-frac{z}{a_j} - frac{z^2}{2a_j^2} -...-frac{z^n}{na_j^n}}$$



      where $n$ is the closest greatest integer to $rho$ and $p$ is a polynomial of at most degree $n$.



      My muse also has rare representations, that bring out specificity and still beauty. Thanks to Ramanujan's careful deliberations,



      A holomorphic function $f$ on $mathbb{C}_{Re(z)>0}$, such that $|f(z)|< Ce^{rho |Re(z)| + tau |Im(z)|}$, for arbitrary constants $C, rho, tau in mathbb{R}^+$ with $tau < pi/2$, then $f(z)$ is completely determined (and not only determined, represented) by $f big{|}_{mathbb{N}}$.



      $$f(z)Gamma(1-z) = int_0^infty vartheta(-x)x^{-z},dx$$



      where $vartheta(x) = sum_{j=0}^infty f(j+1) frac{x^j}{j!}$, $Gamma$ is the Gamma function, and $0 < Re(z) < 1$. This can be extended to the expression



      $$f(z)Gamma(1-z) = sum_{j=0}^infty f(j+1)frac{(-1)^j}{j!(j+1-z)} + int_1^infty vartheta(-x)x^{-z},dx$$



      which works for all $mathbb{C}_{Re(z) > 0}$.



      What other instances do holomorphic functions (on any domain subject to whatever constraints) admit a unique representation theorem based on a sliver of information about the function. Slightly different than identity theorems, as these determine, but rather examples that also represent.



      If need be this can be Community wiki.



      Thank you and Happy New Year,



      Richard Diagram










      share|cite|improve this question









      New contributor




      Richard Diagram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Holomorphic functions are my muse. As my muse, I love drawing them different ways. Allow me to frame this as though an artist talking about his muse.



      A holomorphic function $f$ on the unit disk $mathbb{D}$ is completely determined (and not only determined, represented) by ${f^{(j)}(0)}_{j=0}^infty$.



      $$f(z) = sum_{j=0}^infty f^{(j)}(0) frac{z^j}{j!}$$



      Similarly a holomorphic function $f$ on $mathbb{D}$ is completely determined (and not only determined, represented on the inside of $mathcal{C}$) by its values on any contour $mathcal{C} subset mathbb{D}$.



      $$f(z) = int_{mathcal{C}} frac{f(zeta)}{zeta - z},dzeta$$



      Adding more constraints, and restricting my muse to certain poses, you can find better and more nuanced art:



      If $f$ is entire, and $|f(z)| < C e^{tau |z|^rho}$ everywhere, for arbitrary constants $C,rho,tau in mathbb{R}^+$, then $f$ is completely determined (and not only determined, $nearly$ represented) by its zeroes ${a_j}_{j=1}^infty$.



      $$f(z) = e^{p(z)}prod_{j=1}^infty (1 - frac{z}{a_j})e^{-frac{z}{a_j} - frac{z^2}{2a_j^2} -...-frac{z^n}{na_j^n}}$$



      where $n$ is the closest greatest integer to $rho$ and $p$ is a polynomial of at most degree $n$.



      My muse also has rare representations, that bring out specificity and still beauty. Thanks to Ramanujan's careful deliberations,



      A holomorphic function $f$ on $mathbb{C}_{Re(z)>0}$, such that $|f(z)|< Ce^{rho |Re(z)| + tau |Im(z)|}$, for arbitrary constants $C, rho, tau in mathbb{R}^+$ with $tau < pi/2$, then $f(z)$ is completely determined (and not only determined, represented) by $f big{|}_{mathbb{N}}$.



      $$f(z)Gamma(1-z) = int_0^infty vartheta(-x)x^{-z},dx$$



      where $vartheta(x) = sum_{j=0}^infty f(j+1) frac{x^j}{j!}$, $Gamma$ is the Gamma function, and $0 < Re(z) < 1$. This can be extended to the expression



      $$f(z)Gamma(1-z) = sum_{j=0}^infty f(j+1)frac{(-1)^j}{j!(j+1-z)} + int_1^infty vartheta(-x)x^{-z},dx$$



      which works for all $mathbb{C}_{Re(z) > 0}$.



      What other instances do holomorphic functions (on any domain subject to whatever constraints) admit a unique representation theorem based on a sliver of information about the function. Slightly different than identity theorems, as these determine, but rather examples that also represent.



      If need be this can be Community wiki.



      Thank you and Happy New Year,



      Richard Diagram







      cv.complex-variables soft-question






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      edited 1 hour ago





















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      asked 1 hour ago









      Richard Diagram

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          The distinction between "determined" and "represented" is not clear.
          Consider a function $f$ analytic in domain $U$ containing, say, $0$.
          The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent
          the coefficients $a_k$ of the Maclaurin series recursively by limit operations:
          $$ a_k = lim_{n to infty} left(p_n^{-k} f(p_n) - sum_{j=0}^{k-1} a_j p_n^{j-k}right) $$
          and then of course $f(z) = sum_{k=0}^infty a_k z^k$ for $z$ within the radius of convergence.






          share|cite|improve this answer





















          • + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
            – Richard Diagram
            49 mins ago












          • And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
            – Richard Diagram
            45 mins ago











          Your Answer





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          1 Answer
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          The distinction between "determined" and "represented" is not clear.
          Consider a function $f$ analytic in domain $U$ containing, say, $0$.
          The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent
          the coefficients $a_k$ of the Maclaurin series recursively by limit operations:
          $$ a_k = lim_{n to infty} left(p_n^{-k} f(p_n) - sum_{j=0}^{k-1} a_j p_n^{j-k}right) $$
          and then of course $f(z) = sum_{k=0}^infty a_k z^k$ for $z$ within the radius of convergence.






          share|cite|improve this answer





















          • + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
            – Richard Diagram
            49 mins ago












          • And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
            – Richard Diagram
            45 mins ago
















          2














          The distinction between "determined" and "represented" is not clear.
          Consider a function $f$ analytic in domain $U$ containing, say, $0$.
          The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent
          the coefficients $a_k$ of the Maclaurin series recursively by limit operations:
          $$ a_k = lim_{n to infty} left(p_n^{-k} f(p_n) - sum_{j=0}^{k-1} a_j p_n^{j-k}right) $$
          and then of course $f(z) = sum_{k=0}^infty a_k z^k$ for $z$ within the radius of convergence.






          share|cite|improve this answer





















          • + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
            – Richard Diagram
            49 mins ago












          • And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
            – Richard Diagram
            45 mins ago














          2












          2








          2






          The distinction between "determined" and "represented" is not clear.
          Consider a function $f$ analytic in domain $U$ containing, say, $0$.
          The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent
          the coefficients $a_k$ of the Maclaurin series recursively by limit operations:
          $$ a_k = lim_{n to infty} left(p_n^{-k} f(p_n) - sum_{j=0}^{k-1} a_j p_n^{j-k}right) $$
          and then of course $f(z) = sum_{k=0}^infty a_k z^k$ for $z$ within the radius of convergence.






          share|cite|improve this answer












          The distinction between "determined" and "represented" is not clear.
          Consider a function $f$ analytic in domain $U$ containing, say, $0$.
          The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent
          the coefficients $a_k$ of the Maclaurin series recursively by limit operations:
          $$ a_k = lim_{n to infty} left(p_n^{-k} f(p_n) - sum_{j=0}^{k-1} a_j p_n^{j-k}right) $$
          and then of course $f(z) = sum_{k=0}^infty a_k z^k$ for $z$ within the radius of convergence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Robert Israel

          41.2k48117




          41.2k48117












          • + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
            – Richard Diagram
            49 mins ago












          • And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
            – Richard Diagram
            45 mins ago


















          • + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
            – Richard Diagram
            49 mins ago












          • And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
            – Richard Diagram
            45 mins ago
















          + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
          – Richard Diagram
          49 mins ago






          + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could.
          – Richard Diagram
          49 mins ago














          And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
          – Richard Diagram
          45 mins ago




          And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. –
          – Richard Diagram
          45 mins ago










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