Mixing
Laurent Series Of $fracz(z+1)(z+2)$ Around $z=-1$
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Expand $fracz(z+1)(z+2)$ Around $z=-1$
$$fracz(z+1)(z+2)=-frac1z+1+frac2z+2=-(z+1)^-1+frac2z+2=\=-(z+1)^-1+frac21-(-z-1)=-(z+1)^-1+2sum_n=0^infty(-1)^n(z+1)^n$$
Can I write all inside the sum operator? the first terms are:
$frac-1(z+1)+2-2(z+1)+2(z+1)^2+...$
So the function as a simple pole and the radius of convergence is
$lim_nto inftyfrac1sqrt[n](-1)^n=1$
So $|z+1|<1$ and on $|z+1|=1$ we get $sum_n=0^infty(-1)^n$ which diverges?
complex-analysis laurent-series
asked Jul 22 at 21:14
newhere
747310
 |Â
show 1 more comment
up vote
0
down vote
favorite
Expand $fracz(z+1)(z+2)$ Around $z=-1$
$$fracz(z+1)(z+2)=-frac1z+1+frac2z+2=-(z+1)^-1+frac2z+2=\=-(z+1)^-1+frac21-(-z-1)=-(z+1)^-1+2sum_n=0^infty(-1)^n(z+1)^n$$
Can I write all inside the sum operator? the first terms are:
$frac-1(z+1)+2-2(z+1)+2(z+1)^2+...$
So the function as a simple pole and the radius of convergence is
$lim_nto inftyfrac1sqrt[n](-1)^n=1$
So $|z+1|<1$ and on $|z+1|=1$ we get $sum_n=0^infty(-1)^n$ which diverges?
complex-analysis laurent-series
asked Jul 22 at 21:14
newhere
747310
What exactly is your question?
– anomaly
Jul 22 at 21:17
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
1
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Expand $fracz(z+1)(z+2)$ Around $z=-1$
$$fracz(z+1)(z+2)=-frac1z+1+frac2z+2=-(z+1)^-1+frac2z+2=\=-(z+1)^-1+frac21-(-z-1)=-(z+1)^-1+2sum_n=0^infty(-1)^n(z+1)^n$$
Can I write all inside the sum operator? the first terms are:
$frac-1(z+1)+2-2(z+1)+2(z+1)^2+...$
So the function as a simple pole and the radius of convergence is
$lim_nto inftyfrac1sqrt[n](-1)^n=1$
So $|z+1|<1$ and on $|z+1|=1$ we get $sum_n=0^infty(-1)^n$ which diverges?
complex-analysis laurent-series
asked Jul 22 at 21:14
newhere
747310
Expand $fracz(z+1)(z+2)$ Around $z=-1$
$$fracz(z+1)(z+2)=-frac1z+1+frac2z+2=-(z+1)^-1+frac2z+2=\=-(z+1)^-1+frac21-(-z-1)=-(z+1)^-1+2sum_n=0^infty(-1)^n(z+1)^n$$
Can I write all inside the sum operator? the first terms are:
$frac-1(z+1)+2-2(z+1)+2(z+1)^2+...$
So the function as a simple pole and the radius of convergence is
$lim_nto inftyfrac1sqrt[n](-1)^n=1$
So $|z+1|<1$ and on $|z+1|=1$ we get $sum_n=0^infty(-1)^n$ which diverges?
complex-analysis laurent-series
asked Jul 22 at 21:14
newhere
747310
asked Jul 22 at 21:14
newhere
747310
asked Jul 22 at 21:14
newhere
747310
asked Jul 22 at 21:14
newhere
747310
747310
What exactly is your question?
– anomaly
Jul 22 at 21:17
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
1
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08
 |Â
show 1 more comment
What exactly is your question?
– anomaly
Jul 22 at 21:17
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
1
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08
What exactly is your question?
– anomaly
Jul 22 at 21:17
What exactly is your question?
– anomaly
Jul 22 at 21:17
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
1
1
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
Perhaps would be simpler using $w=z+1$ then
beginalign
fracz(z+1)(z+2)
&=fracw-1w(w+1)\
&=fracw-1wfrac11+w\
&=frac1-w-wleft(1-w+w^2-w^3+cdotsright)\
&=frac1-wleft(1-2w+2w^2-2w^3+cdotsright)\
&=-frac1w+2-2w+2w^2-2w^3+cdots\
&=-frac1z+1+2-2(z+1)+2(z+1)^2-2(z+1)^3+cdots
endalign
answered Jul 22 at 21:29
Nosrati
19.4k41544
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
add a comment |Â
up vote
2
down vote
Your Laurent Series is correct. $$fracz(z+1)(z+2)=-(z+1)^-1+2sum_n=0^infty (-1)^n(z+1)^n$$
You have to be careful about the region of convergence which is $0<|z+1|<1$
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
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
Perhaps would be simpler using $w=z+1$ then
beginalign
fracz(z+1)(z+2)
&=fracw-1w(w+1)\
&=fracw-1wfrac11+w\
&=frac1-w-wleft(1-w+w^2-w^3+cdotsright)\
&=frac1-wleft(1-2w+2w^2-2w^3+cdotsright)\
&=-frac1w+2-2w+2w^2-2w^3+cdots\
&=-frac1z+1+2-2(z+1)+2(z+1)^2-2(z+1)^3+cdots
endalign
answered Jul 22 at 21:29
Nosrati
19.4k41544
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
add a comment |Â
up vote
2
down vote
Perhaps would be simpler using $w=z+1$ then
beginalign
fracz(z+1)(z+2)
&=fracw-1w(w+1)\
&=fracw-1wfrac11+w\
&=frac1-w-wleft(1-w+w^2-w^3+cdotsright)\
&=frac1-wleft(1-2w+2w^2-2w^3+cdotsright)\
&=-frac1w+2-2w+2w^2-2w^3+cdots\
&=-frac1z+1+2-2(z+1)+2(z+1)^2-2(z+1)^3+cdots
endalign
answered Jul 22 at 21:29
Nosrati
19.4k41544
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Perhaps would be simpler using $w=z+1$ then
beginalign
fracz(z+1)(z+2)
&=fracw-1w(w+1)\
&=fracw-1wfrac11+w\
&=frac1-w-wleft(1-w+w^2-w^3+cdotsright)\
&=frac1-wleft(1-2w+2w^2-2w^3+cdotsright)\
&=-frac1w+2-2w+2w^2-2w^3+cdots\
&=-frac1z+1+2-2(z+1)+2(z+1)^2-2(z+1)^3+cdots
endalign
answered Jul 22 at 21:29
Nosrati
19.4k41544
Perhaps would be simpler using $w=z+1$ then
beginalign
fracz(z+1)(z+2)
&=fracw-1w(w+1)\
&=fracw-1wfrac11+w\
&=frac1-w-wleft(1-w+w^2-w^3+cdotsright)\
&=frac1-wleft(1-2w+2w^2-2w^3+cdotsright)\
&=-frac1w+2-2w+2w^2-2w^3+cdots\
&=-frac1z+1+2-2(z+1)+2(z+1)^2-2(z+1)^3+cdots
endalign
answered Jul 22 at 21:29
Nosrati
19.4k41544
answered Jul 22 at 21:29
Nosrati
19.4k41544
answered Jul 22 at 21:29
Nosrati
19.4k41544
answered Jul 22 at 21:29
Nosrati
19.4k41544
19.4k41544
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
add a comment |Â
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
Always good to have another way, but it still not within one sum (aka $sum a_n(z+1)^n$
– newhere
Jul 22 at 22:02
add a comment |Â
up vote
2
down vote
Your Laurent Series is correct. $$fracz(z+1)(z+2)=-(z+1)^-1+2sum_n=0^infty (-1)^n(z+1)^n$$
You have to be careful about the region of convergence which is $0<|z+1|<1$
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
Your Laurent Series is correct. $$fracz(z+1)(z+2)=-(z+1)^-1+2sum_n=0^infty (-1)^n(z+1)^n$$
You have to be careful about the region of convergence which is $0<|z+1|<1$
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your Laurent Series is correct. $$fracz(z+1)(z+2)=-(z+1)^-1+2sum_n=0^infty (-1)^n(z+1)^n$$
You have to be careful about the region of convergence which is $0<|z+1|<1$
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
Your Laurent Series is correct. $$fracz(z+1)(z+2)=-(z+1)^-1+2sum_n=0^infty (-1)^n(z+1)^n$$
You have to be careful about the region of convergence which is $0<|z+1|<1$
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:48
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
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What exactly is your question?
– anomaly
Jul 22 at 21:17
@anomaly if I can write all as $sum a_n (z+1)^n$ if the radius of convergence is $|z+1|<1?$
– newhere
Jul 22 at 21:20
Sure, why not?$phantom$
– anomaly
Jul 22 at 21:22
@Bernard Are you sure about the partial fractions decomposition? how can I write it within a sum
– newhere
Jul 22 at 22:01
1
@newhere: Sorry, I hadn't checked with pencil and paper. Something looked weird to me, but I was wrong. Sorry for the trouble, I'll delete my comment.
– Bernard
Jul 22 at 22:08