Is it possible to observe the residue?
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine Galantin
635212
add a comment |
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine Galantin
635212
add a comment |
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine Galantin
635212
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
complex-analysis
asked Nov 28 '18 at 18:47
Marine Galantin
635212
asked Nov 28 '18 at 18:47
Marine Galantin
635212
asked Nov 28 '18 at 18:47
Marine Galantin
635212
asked Nov 28 '18 at 18:47
Marine Galantin
635212
asked Nov 28 '18 at 18:47
Marine Galantin
635212
635212
add a comment |
add a comment |
1 Answer
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No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
answered Nov 28 '18 at 18:59
José Carlos Santos
151k22123224
151k22123224
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
So is there a way to interpret the residue?
– Marine Galantin
Nov 28 '18 at 19:00
2
2
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
– José Carlos Santos
Nov 28 '18 at 19:02
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
– Marine Galantin
Nov 28 '18 at 19:16
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
– José Carlos Santos
Nov 28 '18 at 19:21
1
1
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
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