Laurent series of $frac{z}{sin({frac{pi}{z+1}})}$ in the roots of the denominator












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I'd like to compute the first few terms of the Laurent series of $frac{z}{sin(frac{pi}{z+1})}$ at $z=frac{1}{k}-1, kinmathbb{Z}$.
I assume I know the espansion of $frac{1}{sin{z}}$ in $z=0$, so i'm trying to express $frac{pi}{z+1}$ in terms of $[z-(frac{1}{k}-1)], kinmathbb{Z}$, but I can't get anything useful to be put in the expansion of $frac{1}{sin{z}}$.






complex-analysis laurent-series







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    $begingroup$


    I'd like to compute the first few terms of the Laurent series of $frac{z}{sin(frac{pi}{z+1})}$ at $z=frac{1}{k}-1, kinmathbb{Z}$.
    I assume I know the espansion of $frac{1}{sin{z}}$ in $z=0$, so i'm trying to express $frac{pi}{z+1}$ in terms of $[z-(frac{1}{k}-1)], kinmathbb{Z}$, but I can't get anything useful to be put in the expansion of $frac{1}{sin{z}}$.






    complex-analysis laurent-series







    share|cite|improve this question









    $endgroup$















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      1








      1


      1



      $begingroup$


      I'd like to compute the first few terms of the Laurent series of $frac{z}{sin(frac{pi}{z+1})}$ at $z=frac{1}{k}-1, kinmathbb{Z}$.
      I assume I know the espansion of $frac{1}{sin{z}}$ in $z=0$, so i'm trying to express $frac{pi}{z+1}$ in terms of $[z-(frac{1}{k}-1)], kinmathbb{Z}$, but I can't get anything useful to be put in the expansion of $frac{1}{sin{z}}$.






      complex-analysis laurent-series







      share|cite|improve this question









      $endgroup$




      I'd like to compute the first few terms of the Laurent series of $frac{z}{sin(frac{pi}{z+1})}$ at $z=frac{1}{k}-1, kinmathbb{Z}$.
      I assume I know the espansion of $frac{1}{sin{z}}$ in $z=0$, so i'm trying to express $frac{pi}{z+1}$ in terms of $[z-(frac{1}{k}-1)], kinmathbb{Z}$, but I can't get anything useful to be put in the expansion of $frac{1}{sin{z}}$.






      complex-analysis laurent-series




      complex-analysis laurent-series






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 8 at 12:40









      EugenioDiPaolaEugenioDiPaola

      515




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          1 Answer
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          1












          $begingroup$

          Hint:



          Let $z=w+left(frac1k-1right)$ then
          $$
          f(z)=F(w)=frac{w+left(frac1k-1right)}{sinleft(frac{pi}{w+frac1k}right)}=frac{w+left(frac1k-1right)}{sinleft(frac{kpi}{1+kw}right)}
          $$
          Now recall that the expansion of $kpi(1+kw)^{-1}$ at $0$: for $|w|<frac1k$,
          $$
          frac{kpi}{1+kw}=kpisum_{n=0}^{infty} (-1)^n k^nw^n=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n
          $$
          and, for $v in mathbb{C}$,
          $$
          sin(v)=v-frac{v^3}{3!}+frac{v^5}{5!}-dotstag{1}
          $$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
            $endgroup$
            – EugenioDiPaola
            Jan 9 at 8:37












          • $begingroup$
            @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
            $endgroup$
            – Jevaut
            Jan 9 at 16:16










          • $begingroup$
            $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
            $endgroup$
            – EugenioDiPaola
            Jan 15 at 13:37













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          $begingroup$

          Hint:



          Let $z=w+left(frac1k-1right)$ then
          $$
          f(z)=F(w)=frac{w+left(frac1k-1right)}{sinleft(frac{pi}{w+frac1k}right)}=frac{w+left(frac1k-1right)}{sinleft(frac{kpi}{1+kw}right)}
          $$
          Now recall that the expansion of $kpi(1+kw)^{-1}$ at $0$: for $|w|<frac1k$,
          $$
          frac{kpi}{1+kw}=kpisum_{n=0}^{infty} (-1)^n k^nw^n=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n
          $$
          and, for $v in mathbb{C}$,
          $$
          sin(v)=v-frac{v^3}{3!}+frac{v^5}{5!}-dotstag{1}
          $$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
            $endgroup$
            – EugenioDiPaola
            Jan 9 at 8:37












          • $begingroup$
            @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
            $endgroup$
            – Jevaut
            Jan 9 at 16:16










          • $begingroup$
            $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
            $endgroup$
            – EugenioDiPaola
            Jan 15 at 13:37


















          1












          $begingroup$

          Hint:



          Let $z=w+left(frac1k-1right)$ then
          $$
          f(z)=F(w)=frac{w+left(frac1k-1right)}{sinleft(frac{pi}{w+frac1k}right)}=frac{w+left(frac1k-1right)}{sinleft(frac{kpi}{1+kw}right)}
          $$
          Now recall that the expansion of $kpi(1+kw)^{-1}$ at $0$: for $|w|<frac1k$,
          $$
          frac{kpi}{1+kw}=kpisum_{n=0}^{infty} (-1)^n k^nw^n=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n
          $$
          and, for $v in mathbb{C}$,
          $$
          sin(v)=v-frac{v^3}{3!}+frac{v^5}{5!}-dotstag{1}
          $$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
            $endgroup$
            – EugenioDiPaola
            Jan 9 at 8:37












          • $begingroup$
            @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
            $endgroup$
            – Jevaut
            Jan 9 at 16:16










          • $begingroup$
            $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
            $endgroup$
            – EugenioDiPaola
            Jan 15 at 13:37
















          1












          1








          1





          $begingroup$

          Hint:



          Let $z=w+left(frac1k-1right)$ then
          $$
          f(z)=F(w)=frac{w+left(frac1k-1right)}{sinleft(frac{pi}{w+frac1k}right)}=frac{w+left(frac1k-1right)}{sinleft(frac{kpi}{1+kw}right)}
          $$
          Now recall that the expansion of $kpi(1+kw)^{-1}$ at $0$: for $|w|<frac1k$,
          $$
          frac{kpi}{1+kw}=kpisum_{n=0}^{infty} (-1)^n k^nw^n=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n
          $$
          and, for $v in mathbb{C}$,
          $$
          sin(v)=v-frac{v^3}{3!}+frac{v^5}{5!}-dotstag{1}
          $$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          Hint:



          Let $z=w+left(frac1k-1right)$ then
          $$
          f(z)=F(w)=frac{w+left(frac1k-1right)}{sinleft(frac{pi}{w+frac1k}right)}=frac{w+left(frac1k-1right)}{sinleft(frac{kpi}{1+kw}right)}
          $$
          Now recall that the expansion of $kpi(1+kw)^{-1}$ at $0$: for $|w|<frac1k$,
          $$
          frac{kpi}{1+kw}=kpisum_{n=0}^{infty} (-1)^n k^nw^n=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n
          $$
          and, for $v in mathbb{C}$,
          $$
          sin(v)=v-frac{v^3}{3!}+frac{v^5}{5!}-dotstag{1}
          $$
          Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 9 at 16:14

























          answered Jan 8 at 14:23









          JevautJevaut

          1,166212




          1,166212












          • $begingroup$
            not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
            $endgroup$
            – EugenioDiPaola
            Jan 9 at 8:37












          • $begingroup$
            @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
            $endgroup$
            – Jevaut
            Jan 9 at 16:16










          • $begingroup$
            $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
            $endgroup$
            – EugenioDiPaola
            Jan 15 at 13:37




















          • $begingroup$
            not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
            $endgroup$
            – EugenioDiPaola
            Jan 9 at 8:37












          • $begingroup$
            @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
            $endgroup$
            – Jevaut
            Jan 9 at 16:16










          • $begingroup$
            $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
            $endgroup$
            – EugenioDiPaola
            Jan 15 at 13:37


















          $begingroup$
          not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
          $endgroup$
          – EugenioDiPaola
          Jan 9 at 8:37






          $begingroup$
          not really, either if I use $1/sin{x}=1/x+x/6+7/360x^3+O{x^5}$ and then expand the geometric series or if I expand the the geometric series and the $sin$ at denominator for the first few terms and then perform long division I only seem to get powers of w with positive exponent. Could you please write some more of what you meant to do?
          $endgroup$
          – EugenioDiPaola
          Jan 9 at 8:37














          $begingroup$
          @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
          $endgroup$
          – Jevaut
          Jan 9 at 16:16




          $begingroup$
          @EugenioDiPaola Let $$v:=pisum_{n=0}^{infty} (-1)^n k^{n+1}w^n$$ and then plug it in equation $(1)$ of my answer.
          $endgroup$
          – Jevaut
          Jan 9 at 16:16












          $begingroup$
          $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
          $endgroup$
          – EugenioDiPaola
          Jan 15 at 13:37






          $begingroup$
          $$sin{left(frac{kpi}{1-(-kw)}right)}=sin{left[kpi^2sum_{n=0}^{infty}{(-1)^n k^{n+1}w^n}right]}={kpi^2left[k-k^2 w+O(w^2)right]-frac{k^3pi^6}{6^3}left[k-k^2 w+O(w^2)right]^3+kpi^2 O[(k-k^2 w+O(w^2))^5]} =kpi+k^6 frac{pi^6}{6^3}+O(k^5)$$, which doesn't contain any $w$
          $endgroup$
          – EugenioDiPaola
          Jan 15 at 13:37




















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