Laurent series/isolated singularity












0












$begingroup$


I want to classify the singularities of
$$ f(z)=frac{sin(2z)}{(z-1)^3}$$



The Taylor series is: $sin(2z)=sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}$



So:



$ frac{sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}}{(z-1)^3}$.



How can I go on from there?






taylor-expansion residue-calculus laurent-series singularity







share|cite|improve this question











$endgroup$












  • $begingroup$
    The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
    $endgroup$
    – xbh
    Jan 23 at 16:41










  • $begingroup$
    Okay clearly I have guessed wrong.
    $endgroup$
    – xbh
    Jan 23 at 16:55
















0












$begingroup$


I want to classify the singularities of
$$ f(z)=frac{sin(2z)}{(z-1)^3}$$



The Taylor series is: $sin(2z)=sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}$



So:



$ frac{sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}}{(z-1)^3}$.



How can I go on from there?






taylor-expansion residue-calculus laurent-series singularity







share|cite|improve this question











$endgroup$












  • $begingroup$
    The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
    $endgroup$
    – xbh
    Jan 23 at 16:41










  • $begingroup$
    Okay clearly I have guessed wrong.
    $endgroup$
    – xbh
    Jan 23 at 16:55














0












0








0





$begingroup$


I want to classify the singularities of
$$ f(z)=frac{sin(2z)}{(z-1)^3}$$



The Taylor series is: $sin(2z)=sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}$



So:



$ frac{sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}}{(z-1)^3}$.



How can I go on from there?






taylor-expansion residue-calculus laurent-series singularity







share|cite|improve this question











$endgroup$




I want to classify the singularities of
$$ f(z)=frac{sin(2z)}{(z-1)^3}$$



The Taylor series is: $sin(2z)=sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}$



So:



$ frac{sum_{k=0}^{infty}frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}}{(z-1)^3}$.



How can I go on from there?






taylor-expansion residue-calculus laurent-series singularity




taylor-expansion residue-calculus laurent-series singularity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 17:51









José Carlos Santos

165k22132235




165k22132235










asked Jan 23 at 16:27









Steven33Steven33

316




316












  • $begingroup$
    The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
    $endgroup$
    – xbh
    Jan 23 at 16:41










  • $begingroup$
    Okay clearly I have guessed wrong.
    $endgroup$
    – xbh
    Jan 23 at 16:55


















  • $begingroup$
    The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
    $endgroup$
    – xbh
    Jan 23 at 16:41










  • $begingroup$
    Okay clearly I have guessed wrong.
    $endgroup$
    – xbh
    Jan 23 at 16:55
















$begingroup$
The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
$endgroup$
– xbh
Jan 23 at 16:41




$begingroup$
The undefined points, i.e. the possible singularities are $z=1$, so I think you should expand $sin (2z)$ around $z = 1$, not $z = 0$. I guess it is a pole of order $3$.
$endgroup$
– xbh
Jan 23 at 16:41












$begingroup$
Okay clearly I have guessed wrong.
$endgroup$
– xbh
Jan 23 at 16:55




$begingroup$
Okay clearly I have guessed wrong.
$endgroup$
– xbh
Jan 23 at 16:55










1 Answer
1






active

oldest

votes


















1












$begingroup$

You go nowhere from that. Insted, use the fact thatbegin{align}sin(2z)&=sinbigl(2(z-1)+2bigr)\&=sinbigl(2(z-1)bigr)cos(2)+cosbigl(2(z-1)bigr)sin(2)end{align}and expand in power series centered at $1$ the functions $sinbigl(2(z-1)bigr)$ and $cosbigl(2(z-1)bigr)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
    $endgroup$
    – Steven33
    Jan 23 at 16:56












  • $begingroup$
    The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
    $endgroup$
    – Steven33
    Jan 23 at 17:02












  • $begingroup$
    No. The residue is $-2sin(2)$.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 17:45










  • $begingroup$
    The pole of order 3 is right?
    $endgroup$
    – Steven33
    Jan 23 at 18:33












  • $begingroup$
    Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 18:35











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1 Answer
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1 Answer
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1












$begingroup$

You go nowhere from that. Insted, use the fact thatbegin{align}sin(2z)&=sinbigl(2(z-1)+2bigr)\&=sinbigl(2(z-1)bigr)cos(2)+cosbigl(2(z-1)bigr)sin(2)end{align}and expand in power series centered at $1$ the functions $sinbigl(2(z-1)bigr)$ and $cosbigl(2(z-1)bigr)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
    $endgroup$
    – Steven33
    Jan 23 at 16:56












  • $begingroup$
    The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
    $endgroup$
    – Steven33
    Jan 23 at 17:02












  • $begingroup$
    No. The residue is $-2sin(2)$.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 17:45










  • $begingroup$
    The pole of order 3 is right?
    $endgroup$
    – Steven33
    Jan 23 at 18:33












  • $begingroup$
    Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 18:35
















1












$begingroup$

You go nowhere from that. Insted, use the fact thatbegin{align}sin(2z)&=sinbigl(2(z-1)+2bigr)\&=sinbigl(2(z-1)bigr)cos(2)+cosbigl(2(z-1)bigr)sin(2)end{align}and expand in power series centered at $1$ the functions $sinbigl(2(z-1)bigr)$ and $cosbigl(2(z-1)bigr)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
    $endgroup$
    – Steven33
    Jan 23 at 16:56












  • $begingroup$
    The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
    $endgroup$
    – Steven33
    Jan 23 at 17:02












  • $begingroup$
    No. The residue is $-2sin(2)$.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 17:45










  • $begingroup$
    The pole of order 3 is right?
    $endgroup$
    – Steven33
    Jan 23 at 18:33












  • $begingroup$
    Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 18:35














1












1








1





$begingroup$

You go nowhere from that. Insted, use the fact thatbegin{align}sin(2z)&=sinbigl(2(z-1)+2bigr)\&=sinbigl(2(z-1)bigr)cos(2)+cosbigl(2(z-1)bigr)sin(2)end{align}and expand in power series centered at $1$ the functions $sinbigl(2(z-1)bigr)$ and $cosbigl(2(z-1)bigr)$.






share|cite|improve this answer









$endgroup$



You go nowhere from that. Insted, use the fact thatbegin{align}sin(2z)&=sinbigl(2(z-1)+2bigr)\&=sinbigl(2(z-1)bigr)cos(2)+cosbigl(2(z-1)bigr)sin(2)end{align}and expand in power series centered at $1$ the functions $sinbigl(2(z-1)bigr)$ and $cosbigl(2(z-1)bigr)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 at 16:43









José Carlos SantosJosé Carlos Santos

165k22132235




165k22132235












  • $begingroup$
    I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
    $endgroup$
    – Steven33
    Jan 23 at 16:56












  • $begingroup$
    The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
    $endgroup$
    – Steven33
    Jan 23 at 17:02












  • $begingroup$
    No. The residue is $-2sin(2)$.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 17:45










  • $begingroup$
    The pole of order 3 is right?
    $endgroup$
    – Steven33
    Jan 23 at 18:33












  • $begingroup$
    Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 18:35


















  • $begingroup$
    I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
    $endgroup$
    – Steven33
    Jan 23 at 16:56












  • $begingroup$
    The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
    $endgroup$
    – Steven33
    Jan 23 at 17:02












  • $begingroup$
    No. The residue is $-2sin(2)$.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 17:45










  • $begingroup$
    The pole of order 3 is right?
    $endgroup$
    – Steven33
    Jan 23 at 18:33












  • $begingroup$
    Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    Jan 23 at 18:35
















$begingroup$
I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
$endgroup$
– Steven33
Jan 23 at 16:56






$begingroup$
I see:) Then I can see that I have a pole of order 3. The residue would be $frac{-4}{3}$. Did I calculate that right?
$endgroup$
– Steven33
Jan 23 at 16:56














$begingroup$
The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
$endgroup$
– Steven33
Jan 23 at 17:02






$begingroup$
The residue is $frac{-4}{3} cos2$. The power series of cos has no $(z-1)^{-1}$, so the residue can be only determined by the series of sin(2(z-1))
$endgroup$
– Steven33
Jan 23 at 17:02














$begingroup$
No. The residue is $-2sin(2)$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:45




$begingroup$
No. The residue is $-2sin(2)$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:45












$begingroup$
The pole of order 3 is right?
$endgroup$
– Steven33
Jan 23 at 18:33






$begingroup$
The pole of order 3 is right?
$endgroup$
– Steven33
Jan 23 at 18:33














$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 23 at 18:35




$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 23 at 18:35


















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