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Expand $fracz-sin zz^3$ Around $z=0$
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Expand $fracz-sin zz^3$ Around $z=0$
$$sin z=sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!$$
So $$fracz-sin zz^3=fracz-sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!z^3=frac1z^2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!=z^-2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!$$
Can we say that the singularity will be a pole of order $3$ (independent on the point around we expand?)
$R=lim_nto infty|fraca_na_n+1|=lim_nto inftyfrac(2n+3)!(2n+1)!=lim_nto infty4n+5=infty$
So the Laurent series converges for all $z$
Is it correct?
complex-analysis laurent-series
asked Jul 20 at 13:59
newhere
759310
add a comment |Â
up vote
0
down vote
favorite
Expand $fracz-sin zz^3$ Around $z=0$
$$sin z=sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!$$
So $$fracz-sin zz^3=fracz-sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!z^3=frac1z^2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!=z^-2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!$$
Can we say that the singularity will be a pole of order $3$ (independent on the point around we expand?)
$R=lim_nto infty|fraca_na_n+1|=lim_nto inftyfrac(2n+3)!(2n+1)!=lim_nto infty4n+5=infty$
So the Laurent series converges for all $z$
Is it correct?
complex-analysis laurent-series
asked Jul 20 at 13:59
newhere
759310
4
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Expand $fracz-sin zz^3$ Around $z=0$
$$sin z=sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!$$
So $$fracz-sin zz^3=fracz-sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!z^3=frac1z^2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!=z^-2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!$$
Can we say that the singularity will be a pole of order $3$ (independent on the point around we expand?)
$R=lim_nto infty|fraca_na_n+1|=lim_nto inftyfrac(2n+3)!(2n+1)!=lim_nto infty4n+5=infty$
So the Laurent series converges for all $z$
Is it correct?
complex-analysis laurent-series
asked Jul 20 at 13:59
newhere
759310
Expand $fracz-sin zz^3$ Around $z=0$
$$sin z=sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!$$
So $$fracz-sin zz^3=fracz-sum_n=0^inftyfrac(-1)^n z^2n+1(2n+1)!z^3=frac1z^2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!=z^-2-sum_n=0^inftyfrac(-1)^n z^2n-2(2n+1)!$$
Can we say that the singularity will be a pole of order $3$ (independent on the point around we expand?)
$R=lim_nto infty|fraca_na_n+1|=lim_nto inftyfrac(2n+3)!(2n+1)!=lim_nto infty4n+5=infty$
So the Laurent series converges for all $z$
Is it correct?
complex-analysis laurent-series
asked Jul 20 at 13:59
newhere
759310
asked Jul 20 at 13:59
newhere
759310
asked Jul 20 at 13:59
newhere
759310
asked Jul 20 at 13:59
newhere
759310
759310
4
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20
add a comment |Â
4
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20
4
4
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$$fracz-sin zz^3=fracdisplaystyle z-sum_nge 0dfrac(-1)^n z^2n+1(2n+1)!z^3=fracdisplaystylesum_nge1dfrac(-1)^n+1 z^2n+1(2n+1)!z^3=sum_nge1dfrac(-1)^n+1 z^2(n-1)(2n+1)!,$$
which can be simplified to
$$sum_nge0dfrac(-1)^n z^2n(2n+3)!.$$
answered Jul 20 at 14:19
Bernard
110k635103
add a comment |Â
up vote
2
down vote
The sine is an odd function and $sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.
The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.
answered Jul 20 at 14:06
Yves Daoust
111k665204
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$fracz-sin zz^3=fracdisplaystyle z-sum_nge 0dfrac(-1)^n z^2n+1(2n+1)!z^3=fracdisplaystylesum_nge1dfrac(-1)^n+1 z^2n+1(2n+1)!z^3=sum_nge1dfrac(-1)^n+1 z^2(n-1)(2n+1)!,$$
which can be simplified to
$$sum_nge0dfrac(-1)^n z^2n(2n+3)!.$$
answered Jul 20 at 14:19
Bernard
110k635103
add a comment |Â
up vote
2
down vote
accepted
$$fracz-sin zz^3=fracdisplaystyle z-sum_nge 0dfrac(-1)^n z^2n+1(2n+1)!z^3=fracdisplaystylesum_nge1dfrac(-1)^n+1 z^2n+1(2n+1)!z^3=sum_nge1dfrac(-1)^n+1 z^2(n-1)(2n+1)!,$$
which can be simplified to
$$sum_nge0dfrac(-1)^n z^2n(2n+3)!.$$
answered Jul 20 at 14:19
Bernard
110k635103
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$fracz-sin zz^3=fracdisplaystyle z-sum_nge 0dfrac(-1)^n z^2n+1(2n+1)!z^3=fracdisplaystylesum_nge1dfrac(-1)^n+1 z^2n+1(2n+1)!z^3=sum_nge1dfrac(-1)^n+1 z^2(n-1)(2n+1)!,$$
which can be simplified to
$$sum_nge0dfrac(-1)^n z^2n(2n+3)!.$$
answered Jul 20 at 14:19
Bernard
110k635103
$$fracz-sin zz^3=fracdisplaystyle z-sum_nge 0dfrac(-1)^n z^2n+1(2n+1)!z^3=fracdisplaystylesum_nge1dfrac(-1)^n+1 z^2n+1(2n+1)!z^3=sum_nge1dfrac(-1)^n+1 z^2(n-1)(2n+1)!,$$
which can be simplified to
$$sum_nge0dfrac(-1)^n z^2n(2n+3)!.$$
answered Jul 20 at 14:19
Bernard
110k635103
answered Jul 20 at 14:19
Bernard
110k635103
answered Jul 20 at 14:19
Bernard
110k635103
answered Jul 20 at 14:19
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
up vote
2
down vote
The sine is an odd function and $sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.
The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.
answered Jul 20 at 14:06
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
The sine is an odd function and $sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.
The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.
answered Jul 20 at 14:06
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The sine is an odd function and $sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.
The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.
answered Jul 20 at 14:06
Yves Daoust
111k665204
The sine is an odd function and $sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.
The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.
answered Jul 20 at 14:06
Yves Daoust
111k665204
answered Jul 20 at 14:06
Yves Daoust
111k665204
answered Jul 20 at 14:06
Yves Daoust
111k665204
answered Jul 20 at 14:06
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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4
I'd rather say that there is no singularity. Expand the first terms of the summation.
– Yves Daoust
Jul 20 at 14:03
@YvesDaoust in the case of removable singularity we will not be able to spot it by the Laurent series as we can expand a function with a removable singularity to an analytic one?
– newhere
Jul 20 at 15:20