Expressing coefficients of a complex function using integrals












0












$begingroup$


Let $f(z)$ be the function defined by:



begin{equation}
f(z)=a_{-3}z^{-3}+a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3
end{equation}



How can we express the coefficients $a_i$ using integrals?



This is clearly a rational function $P(z)/Q(z)$ of degree 9. The formula for the residue $a_{-1}$ implies derivatives, not integrals. Maybe a quotient of integrals?






complex-analysis complex-integration







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f(z)$ be the function defined by:



    begin{equation}
    f(z)=a_{-3}z^{-3}+a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3
    end{equation}



    How can we express the coefficients $a_i$ using integrals?



    This is clearly a rational function $P(z)/Q(z)$ of degree 9. The formula for the residue $a_{-1}$ implies derivatives, not integrals. Maybe a quotient of integrals?






    complex-analysis complex-integration







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f(z)$ be the function defined by:



      begin{equation}
      f(z)=a_{-3}z^{-3}+a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3
      end{equation}



      How can we express the coefficients $a_i$ using integrals?



      This is clearly a rational function $P(z)/Q(z)$ of degree 9. The formula for the residue $a_{-1}$ implies derivatives, not integrals. Maybe a quotient of integrals?






      complex-analysis complex-integration







      share|cite|improve this question









      $endgroup$




      Let $f(z)$ be the function defined by:



      begin{equation}
      f(z)=a_{-3}z^{-3}+a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3
      end{equation}



      How can we express the coefficients $a_i$ using integrals?



      This is clearly a rational function $P(z)/Q(z)$ of degree 9. The formula for the residue $a_{-1}$ implies derivatives, not integrals. Maybe a quotient of integrals?






      complex-analysis complex-integration




      complex-analysis complex-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 12 '18 at 19:43









      IchVerlorenIchVerloren

      969




      969






















          2 Answers
          2






          active

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          1












          $begingroup$

          For any $n in mathbb{Z}$, $$a_n=int_0^{1}{f(e^{2ipi t})e^{-2inpi t}dt}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! How does that work? I can't find it in my book
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 19:52












          • $begingroup$
            Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 19:57












          • $begingroup$
            I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:02





















          0












          $begingroup$

          The formula you're looking for is sometimes called the generalized Cauchy integral formula. If $f(z) $ is analytic in a deleted neighbourhood of $z_0$ with Laurent series $sum_n a_n (z - z_0)^n$,



          $$ a_n = frac{1}{2pi i} oint_Gamma frac{f(z); dz}{(z-z_0)^{n+1}}$$
          where $Gamma$ is a simple positively oriented closed contour around $z_0$ such that $f$ is analytic on and inside $Gamma$ except possibly at $z_0$. It should be in any complex variables text.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:33










          • $begingroup$
            The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
            $endgroup$
            – Robert Israel
            Dec 12 '18 at 21:19











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          For any $n in mathbb{Z}$, $$a_n=int_0^{1}{f(e^{2ipi t})e^{-2inpi t}dt}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! How does that work? I can't find it in my book
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 19:52












          • $begingroup$
            Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 19:57












          • $begingroup$
            I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:02


















          1












          $begingroup$

          For any $n in mathbb{Z}$, $$a_n=int_0^{1}{f(e^{2ipi t})e^{-2inpi t}dt}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! How does that work? I can't find it in my book
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 19:52












          • $begingroup$
            Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 19:57












          • $begingroup$
            I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:02
















          1












          1








          1





          $begingroup$

          For any $n in mathbb{Z}$, $$a_n=int_0^{1}{f(e^{2ipi t})e^{-2inpi t}dt}.$$






          share|cite|improve this answer









          $endgroup$



          For any $n in mathbb{Z}$, $$a_n=int_0^{1}{f(e^{2ipi t})e^{-2inpi t}dt}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 19:48









          MindlackMindlack

          3,62517




          3,62517












          • $begingroup$
            Hi! How does that work? I can't find it in my book
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 19:52












          • $begingroup$
            Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 19:57












          • $begingroup$
            I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:02




















          • $begingroup$
            Hi! How does that work? I can't find it in my book
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 19:52












          • $begingroup$
            Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 19:57












          • $begingroup$
            I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:02


















          $begingroup$
          Hi! How does that work? I can't find it in my book
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 19:52






          $begingroup$
          Hi! How does that work? I can't find it in my book
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 19:52














          $begingroup$
          Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
          $endgroup$
          – Mindlack
          Dec 12 '18 at 19:57






          $begingroup$
          Just replace $f$ by its expression. Pick, say, $n=1$ to understand what happens.
          $endgroup$
          – Mindlack
          Dec 12 '18 at 19:57














          $begingroup$
          I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 20:02






          $begingroup$
          I can see that it works. Could you give me any source on this? Does this formula have a name? It looks like magic. Thank you!
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 20:02













          0












          $begingroup$

          The formula you're looking for is sometimes called the generalized Cauchy integral formula. If $f(z) $ is analytic in a deleted neighbourhood of $z_0$ with Laurent series $sum_n a_n (z - z_0)^n$,



          $$ a_n = frac{1}{2pi i} oint_Gamma frac{f(z); dz}{(z-z_0)^{n+1}}$$
          where $Gamma$ is a simple positively oriented closed contour around $z_0$ such that $f$ is analytic on and inside $Gamma$ except possibly at $z_0$. It should be in any complex variables text.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:33










          • $begingroup$
            The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
            $endgroup$
            – Robert Israel
            Dec 12 '18 at 21:19
















          0












          $begingroup$

          The formula you're looking for is sometimes called the generalized Cauchy integral formula. If $f(z) $ is analytic in a deleted neighbourhood of $z_0$ with Laurent series $sum_n a_n (z - z_0)^n$,



          $$ a_n = frac{1}{2pi i} oint_Gamma frac{f(z); dz}{(z-z_0)^{n+1}}$$
          where $Gamma$ is a simple positively oriented closed contour around $z_0$ such that $f$ is analytic on and inside $Gamma$ except possibly at $z_0$. It should be in any complex variables text.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:33










          • $begingroup$
            The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
            $endgroup$
            – Robert Israel
            Dec 12 '18 at 21:19














          0












          0








          0





          $begingroup$

          The formula you're looking for is sometimes called the generalized Cauchy integral formula. If $f(z) $ is analytic in a deleted neighbourhood of $z_0$ with Laurent series $sum_n a_n (z - z_0)^n$,



          $$ a_n = frac{1}{2pi i} oint_Gamma frac{f(z); dz}{(z-z_0)^{n+1}}$$
          where $Gamma$ is a simple positively oriented closed contour around $z_0$ such that $f$ is analytic on and inside $Gamma$ except possibly at $z_0$. It should be in any complex variables text.






          share|cite|improve this answer











          $endgroup$



          The formula you're looking for is sometimes called the generalized Cauchy integral formula. If $f(z) $ is analytic in a deleted neighbourhood of $z_0$ with Laurent series $sum_n a_n (z - z_0)^n$,



          $$ a_n = frac{1}{2pi i} oint_Gamma frac{f(z); dz}{(z-z_0)^{n+1}}$$
          where $Gamma$ is a simple positively oriented closed contour around $z_0$ such that $f$ is analytic on and inside $Gamma$ except possibly at $z_0$. It should be in any complex variables text.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 20:11

























          answered Dec 12 '18 at 20:05









          Robert IsraelRobert Israel

          322k23212465




          322k23212465












          • $begingroup$
            I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:33










          • $begingroup$
            The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
            $endgroup$
            – Robert Israel
            Dec 12 '18 at 21:19


















          • $begingroup$
            I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
            $endgroup$
            – IchVerloren
            Dec 12 '18 at 20:33










          • $begingroup$
            The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
            $endgroup$
            – Robert Israel
            Dec 12 '18 at 21:19
















          $begingroup$
          I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 20:33




          $begingroup$
          I thought that Cauchy's formula works if $z_0$ is neither a singularity nor a branch point. Cauchy formula works if $z_0$=0 but isn't 0 a singularity of f(z)?
          $endgroup$
          – IchVerloren
          Dec 12 '18 at 20:33












          $begingroup$
          The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
          $endgroup$
          – Robert Israel
          Dec 12 '18 at 21:19




          $begingroup$
          The generalized formula works for an isolated singularity, giving you the coefficients of the Laurent series.
          $endgroup$
          – Robert Israel
          Dec 12 '18 at 21:19


















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