Singularity of the function $f(z)=sinleft(cosleft(frac{1}{z}right)right)$ at the point $z=0$ is …












0












$begingroup$



For the function $f(z)=sin(cos(frac{1}{z}))$, the point $z=0$ is



(a) a Removable singularity



(b) a pole



(c) an essential singularity



(d) Non-isolated singularity




I have written Laurent's series expansion $f(z)=sin(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)= (1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)-frac{(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)^3}{3!}+...$
all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.






complex-analysis trigonometry proof-verification singularity







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    For the function $f(z)=sin(cos(frac{1}{z}))$, the point $z=0$ is



    (a) a Removable singularity



    (b) a pole



    (c) an essential singularity



    (d) Non-isolated singularity




    I have written Laurent's series expansion $f(z)=sin(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)= (1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)-frac{(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)^3}{3!}+...$
    all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.






    complex-analysis trigonometry proof-verification singularity







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$



      For the function $f(z)=sin(cos(frac{1}{z}))$, the point $z=0$ is



      (a) a Removable singularity



      (b) a pole



      (c) an essential singularity



      (d) Non-isolated singularity




      I have written Laurent's series expansion $f(z)=sin(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)= (1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)-frac{(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)^3}{3!}+...$
      all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.






      complex-analysis trigonometry proof-verification singularity







      share|cite|improve this question











      $endgroup$





      For the function $f(z)=sin(cos(frac{1}{z}))$, the point $z=0$ is



      (a) a Removable singularity



      (b) a pole



      (c) an essential singularity



      (d) Non-isolated singularity




      I have written Laurent's series expansion $f(z)=sin(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)= (1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)-frac{(1-frac{1}{2!z^2}+frac{1}{4!z^4}-frac{1}{6!z^6}+...)^3}{3!}+...$
      all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.






      complex-analysis trigonometry proof-verification singularity




      complex-analysis trigonometry proof-verification singularity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 22 at 19:32









      José Carlos Santos

      164k22131234




      164k22131234










      asked Nov 5 '17 at 9:35









      Unknown xUnknown x

      2,54011026




      2,54011026






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          No, it is not correct. For two reasons:




          1. the Laurent series of $cosleft(frac1zright)$ at $0$ is$$1-frac1{2!z^2}+frac1{4!z^4}-cdots;$$

          2. you did not prove that the negative powers do not cancel each other after a certain point.


          You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $cosleft(frac1zright)$, if $V$ is a neighborhood of $0$, then the set $W=left{cosleft(frac1zright),middle|,zin Vsetminus{0}right}$ is a dense subset of $mathbb C$ and, since $sin$ is non-constant entire function, $sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for pointing mistake.
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:04










          • $begingroup$
            Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
            $endgroup$
            – StammeringMathematician
            Jan 22 at 16:30








          • 1




            $begingroup$
            What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
            $endgroup$
            – José Carlos Santos
            Jan 22 at 16:44










          • $begingroup$
            ...it is a isolated singularity. Am I right?
            $endgroup$
            – StammeringMathematician
            Jan 22 at 17:21






          • 1




            $begingroup$
            That should be obvious. No: it is an essential singularity.
            $endgroup$
            – José Carlos Santos
            Jan 22 at 17:24



















          1












          $begingroup$

          Notice that the limit $lim_{z to 0} cos(1/z)$ does not exist. Therefore the singularity is of essential type.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:09












          • $begingroup$
            @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
            $endgroup$
            – Rohan
            Nov 5 '17 at 10:37












          • $begingroup$
            Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:42












          • $begingroup$
            math.stackexchange.com/questions/1806582/…
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:46










          • $begingroup$
            this one is essential right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:47











          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          No, it is not correct. For two reasons:




          1. the Laurent series of $cosleft(frac1zright)$ at $0$ is$$1-frac1{2!z^2}+frac1{4!z^4}-cdots;$$

          2. you did not prove that the negative powers do not cancel each other after a certain point.


          You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $cosleft(frac1zright)$, if $V$ is a neighborhood of $0$, then the set $W=left{cosleft(frac1zright),middle|,zin Vsetminus{0}right}$ is a dense subset of $mathbb C$ and, since $sin$ is non-constant entire function, $sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for pointing mistake.
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:04










          • $begingroup$
            Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
            $endgroup$
            – StammeringMathematician
            Jan 22 at 16:30








          • 1




            $begingroup$
            What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
            $endgroup$
            – José Carlos Santos
            Jan 22 at 16:44










          • $begingroup$
            ...it is a isolated singularity. Am I right?
            $endgroup$
            – StammeringMathematician
            Jan 22 at 17:21






          • 1




            $begingroup$
            That should be obvious. No: it is an essential singularity.
            $endgroup$
            – José Carlos Santos
            Jan 22 at 17:24
















          1












          $begingroup$

          No, it is not correct. For two reasons:




          1. the Laurent series of $cosleft(frac1zright)$ at $0$ is$$1-frac1{2!z^2}+frac1{4!z^4}-cdots;$$

          2. you did not prove that the negative powers do not cancel each other after a certain point.


          You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $cosleft(frac1zright)$, if $V$ is a neighborhood of $0$, then the set $W=left{cosleft(frac1zright),middle|,zin Vsetminus{0}right}$ is a dense subset of $mathbb C$ and, since $sin$ is non-constant entire function, $sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for pointing mistake.
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:04










          • $begingroup$
            Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
            $endgroup$
            – StammeringMathematician
            Jan 22 at 16:30








          • 1




            $begingroup$
            What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
            $endgroup$
            – José Carlos Santos
            Jan 22 at 16:44










          • $begingroup$
            ...it is a isolated singularity. Am I right?
            $endgroup$
            – StammeringMathematician
            Jan 22 at 17:21






          • 1




            $begingroup$
            That should be obvious. No: it is an essential singularity.
            $endgroup$
            – José Carlos Santos
            Jan 22 at 17:24














          1












          1








          1





          $begingroup$

          No, it is not correct. For two reasons:




          1. the Laurent series of $cosleft(frac1zright)$ at $0$ is$$1-frac1{2!z^2}+frac1{4!z^4}-cdots;$$

          2. you did not prove that the negative powers do not cancel each other after a certain point.


          You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $cosleft(frac1zright)$, if $V$ is a neighborhood of $0$, then the set $W=left{cosleft(frac1zright),middle|,zin Vsetminus{0}right}$ is a dense subset of $mathbb C$ and, since $sin$ is non-constant entire function, $sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.






          share|cite|improve this answer











          $endgroup$



          No, it is not correct. For two reasons:




          1. the Laurent series of $cosleft(frac1zright)$ at $0$ is$$1-frac1{2!z^2}+frac1{4!z^4}-cdots;$$

          2. you did not prove that the negative powers do not cancel each other after a certain point.


          You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $cosleft(frac1zright)$, if $V$ is a neighborhood of $0$, then the set $W=left{cosleft(frac1zright),middle|,zin Vsetminus{0}right}$ is a dense subset of $mathbb C$ and, since $sin$ is non-constant entire function, $sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 5 '17 at 9:51









          Rohan

          27.8k42444




          27.8k42444










          answered Nov 5 '17 at 9:51









          José Carlos SantosJosé Carlos Santos

          164k22131234




          164k22131234












          • $begingroup$
            Thank you for pointing mistake.
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:04










          • $begingroup$
            Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
            $endgroup$
            – StammeringMathematician
            Jan 22 at 16:30








          • 1




            $begingroup$
            What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
            $endgroup$
            – José Carlos Santos
            Jan 22 at 16:44










          • $begingroup$
            ...it is a isolated singularity. Am I right?
            $endgroup$
            – StammeringMathematician
            Jan 22 at 17:21






          • 1




            $begingroup$
            That should be obvious. No: it is an essential singularity.
            $endgroup$
            – José Carlos Santos
            Jan 22 at 17:24


















          • $begingroup$
            Thank you for pointing mistake.
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:04










          • $begingroup$
            Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
            $endgroup$
            – StammeringMathematician
            Jan 22 at 16:30








          • 1




            $begingroup$
            What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
            $endgroup$
            – José Carlos Santos
            Jan 22 at 16:44










          • $begingroup$
            ...it is a isolated singularity. Am I right?
            $endgroup$
            – StammeringMathematician
            Jan 22 at 17:21






          • 1




            $begingroup$
            That should be obvious. No: it is an essential singularity.
            $endgroup$
            – José Carlos Santos
            Jan 22 at 17:24
















          $begingroup$
          Thank you for pointing mistake.
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:04




          $begingroup$
          Thank you for pointing mistake.
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:04












          $begingroup$
          Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
          $endgroup$
          – StammeringMathematician
          Jan 22 at 16:30






          $begingroup$
          Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance.
          $endgroup$
          – StammeringMathematician
          Jan 22 at 16:30






          1




          1




          $begingroup$
          What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
          $endgroup$
          – José Carlos Santos
          Jan 22 at 16:44




          $begingroup$
          What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $fbigl(Usetminus{0}bigr)$ would not be dense in $mathbb C$. Since it is…
          $endgroup$
          – José Carlos Santos
          Jan 22 at 16:44












          $begingroup$
          ...it is a isolated singularity. Am I right?
          $endgroup$
          – StammeringMathematician
          Jan 22 at 17:21




          $begingroup$
          ...it is a isolated singularity. Am I right?
          $endgroup$
          – StammeringMathematician
          Jan 22 at 17:21




          1




          1




          $begingroup$
          That should be obvious. No: it is an essential singularity.
          $endgroup$
          – José Carlos Santos
          Jan 22 at 17:24




          $begingroup$
          That should be obvious. No: it is an essential singularity.
          $endgroup$
          – José Carlos Santos
          Jan 22 at 17:24











          1












          $begingroup$

          Notice that the limit $lim_{z to 0} cos(1/z)$ does not exist. Therefore the singularity is of essential type.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:09












          • $begingroup$
            @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
            $endgroup$
            – Rohan
            Nov 5 '17 at 10:37












          • $begingroup$
            Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:42












          • $begingroup$
            math.stackexchange.com/questions/1806582/…
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:46










          • $begingroup$
            this one is essential right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:47
















          1












          $begingroup$

          Notice that the limit $lim_{z to 0} cos(1/z)$ does not exist. Therefore the singularity is of essential type.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:09












          • $begingroup$
            @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
            $endgroup$
            – Rohan
            Nov 5 '17 at 10:37












          • $begingroup$
            Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:42












          • $begingroup$
            math.stackexchange.com/questions/1806582/…
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:46










          • $begingroup$
            this one is essential right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:47














          1












          1








          1





          $begingroup$

          Notice that the limit $lim_{z to 0} cos(1/z)$ does not exist. Therefore the singularity is of essential type.






          share|cite|improve this answer









          $endgroup$



          Notice that the limit $lim_{z to 0} cos(1/z)$ does not exist. Therefore the singularity is of essential type.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 5 '17 at 9:51









          RohanRohan

          27.8k42444




          27.8k42444












          • $begingroup$
            consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:09












          • $begingroup$
            @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
            $endgroup$
            – Rohan
            Nov 5 '17 at 10:37












          • $begingroup$
            Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:42












          • $begingroup$
            math.stackexchange.com/questions/1806582/…
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:46










          • $begingroup$
            this one is essential right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:47


















          • $begingroup$
            consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:09












          • $begingroup$
            @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
            $endgroup$
            – Rohan
            Nov 5 '17 at 10:37












          • $begingroup$
            Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:42












          • $begingroup$
            math.stackexchange.com/questions/1806582/…
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:46










          • $begingroup$
            this one is essential right?
            $endgroup$
            – Unknown x
            Nov 5 '17 at 10:47
















          $begingroup$
          consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:09






          $begingroup$
          consider $frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:09














          $begingroup$
          @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
          $endgroup$
          – Rohan
          Nov 5 '17 at 10:37






          $begingroup$
          @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z) $, then $lim_{z to z_0} f (z) = infty $.
          $endgroup$
          – Rohan
          Nov 5 '17 at 10:37














          $begingroup$
          Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:42






          $begingroup$
          Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:42














          $begingroup$
          math.stackexchange.com/questions/1806582/…
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:46




          $begingroup$
          math.stackexchange.com/questions/1806582/…
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:46












          $begingroup$
          this one is essential right?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:47




          $begingroup$
          this one is essential right?
          $endgroup$
          – Unknown x
          Nov 5 '17 at 10:47


















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