Determination of the order of a pole
In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?
I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.
complex-analysis taylor-expansion singularity
edited Nov 27 '18 at 21:51
Did
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
add a comment |
In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?
I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.
complex-analysis taylor-expansion singularity
edited Nov 27 '18 at 21:51
Did
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50
add a comment |
In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?
I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.
complex-analysis taylor-expansion singularity
edited Nov 27 '18 at 21:51
Did
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?
I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.
complex-analysis taylor-expansion singularity
complex-analysis taylor-expansion singularity
edited Nov 27 '18 at 21:51
Did
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
edited Nov 27 '18 at 21:51
Did
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
edited Nov 27 '18 at 21:51
Did
246k23220454
edited Nov 27 '18 at 21:51
Did
246k23220454
edited Nov 27 '18 at 21:51
Did
246k23220454
246k23220454
asked Nov 27 '18 at 21:35
pter26
307111
asked Nov 27 '18 at 21:35
pter26
307111
asked Nov 27 '18 at 21:35
pter26
307111
307111
Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50
add a comment |
Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50
Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50
add a comment |
1 Answer
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We have
$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
add a comment |
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We have
$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
add a comment |
We have
$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
add a comment |
We have
$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
We have
$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$
$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$
and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
edited Nov 27 '18 at 21:54
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
answered Nov 27 '18 at 21:48
DonAntonio
177k1491225
177k1491225
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
add a comment |
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15
add a comment |
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Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40
All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50