What does it mean for two [multivalued] complex functions to be equal?
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For $f:Xsubseteqmathbb{R}tomathbb{R}$ and $g:Xsubseteqmathbb{R}tomathbb{R}$, we may say that $f$ and $g$ are equivalent if $forall xin{X}.f(x)=g(x)$.
But for many complex functions $f:Zsubseteqmathbb{C}tomathbb{C}$ and $g:Zsubseteqmathbb{C}tomathbb{C}$, there may be more than one $winmathbb{C}$ such that $f(z)=w$ , so the statement $f(z)=g(z)$ may not be true even if $f$ and $g$ are the same thing.
For example, suppose $f(z)=left(z^2+2iright)^frac{1}{2}$ and $g(z)=left(z^2+2iright)^frac{1}{2}$. Depending on how you evaluate $f$ and $g$, you might arrive at the conclusion that $f(z)=-g(z)implies fnotequiv g$. But then, $f$ and $g$ have the same domain, the same codomain, and are defined by the same equation.
How is the "sameness" of $f$ and $g$ expressed given that the equivalence of the values at $z$ is not always guaranteed? Is there a 'proper' way to describe the de-facto of equivalence of complex, multivalued functions that eliminates this sort of confusion?
complex-analysis definition multivalued-functions
asked Nov 21 at 17:02
R. Burton
855
add a comment |
up vote
0
down vote
favorite
For $f:Xsubseteqmathbb{R}tomathbb{R}$ and $g:Xsubseteqmathbb{R}tomathbb{R}$, we may say that $f$ and $g$ are equivalent if $forall xin{X}.f(x)=g(x)$.
But for many complex functions $f:Zsubseteqmathbb{C}tomathbb{C}$ and $g:Zsubseteqmathbb{C}tomathbb{C}$, there may be more than one $winmathbb{C}$ such that $f(z)=w$ , so the statement $f(z)=g(z)$ may not be true even if $f$ and $g$ are the same thing.
For example, suppose $f(z)=left(z^2+2iright)^frac{1}{2}$ and $g(z)=left(z^2+2iright)^frac{1}{2}$. Depending on how you evaluate $f$ and $g$, you might arrive at the conclusion that $f(z)=-g(z)implies fnotequiv g$. But then, $f$ and $g$ have the same domain, the same codomain, and are defined by the same equation.
How is the "sameness" of $f$ and $g$ expressed given that the equivalence of the values at $z$ is not always guaranteed? Is there a 'proper' way to describe the de-facto of equivalence of complex, multivalued functions that eliminates this sort of confusion?
complex-analysis definition multivalued-functions
asked Nov 21 at 17:02
R. Burton
855
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For $f:Xsubseteqmathbb{R}tomathbb{R}$ and $g:Xsubseteqmathbb{R}tomathbb{R}$, we may say that $f$ and $g$ are equivalent if $forall xin{X}.f(x)=g(x)$.
But for many complex functions $f:Zsubseteqmathbb{C}tomathbb{C}$ and $g:Zsubseteqmathbb{C}tomathbb{C}$, there may be more than one $winmathbb{C}$ such that $f(z)=w$ , so the statement $f(z)=g(z)$ may not be true even if $f$ and $g$ are the same thing.
For example, suppose $f(z)=left(z^2+2iright)^frac{1}{2}$ and $g(z)=left(z^2+2iright)^frac{1}{2}$. Depending on how you evaluate $f$ and $g$, you might arrive at the conclusion that $f(z)=-g(z)implies fnotequiv g$. But then, $f$ and $g$ have the same domain, the same codomain, and are defined by the same equation.
How is the "sameness" of $f$ and $g$ expressed given that the equivalence of the values at $z$ is not always guaranteed? Is there a 'proper' way to describe the de-facto of equivalence of complex, multivalued functions that eliminates this sort of confusion?
complex-analysis definition multivalued-functions
asked Nov 21 at 17:02
R. Burton
855
For $f:Xsubseteqmathbb{R}tomathbb{R}$ and $g:Xsubseteqmathbb{R}tomathbb{R}$, we may say that $f$ and $g$ are equivalent if $forall xin{X}.f(x)=g(x)$.
But for many complex functions $f:Zsubseteqmathbb{C}tomathbb{C}$ and $g:Zsubseteqmathbb{C}tomathbb{C}$, there may be more than one $winmathbb{C}$ such that $f(z)=w$ , so the statement $f(z)=g(z)$ may not be true even if $f$ and $g$ are the same thing.
For example, suppose $f(z)=left(z^2+2iright)^frac{1}{2}$ and $g(z)=left(z^2+2iright)^frac{1}{2}$. Depending on how you evaluate $f$ and $g$, you might arrive at the conclusion that $f(z)=-g(z)implies fnotequiv g$. But then, $f$ and $g$ have the same domain, the same codomain, and are defined by the same equation.
How is the "sameness" of $f$ and $g$ expressed given that the equivalence of the values at $z$ is not always guaranteed? Is there a 'proper' way to describe the de-facto of equivalence of complex, multivalued functions that eliminates this sort of confusion?
complex-analysis definition multivalued-functions
complex-analysis definition multivalued-functions
asked Nov 21 at 17:02
R. Burton
855
asked Nov 21 at 17:02
R. Burton
855
asked Nov 21 at 17:02
R. Burton
855
asked Nov 21 at 17:02
R. Burton
855
asked Nov 21 at 17:02
R. Burton
855
855
add a comment |
add a comment |
1 Answer
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0
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Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.
answered Nov 21 at 17:14
Eduardo Elael
1615
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.
answered Nov 21 at 17:14
Eduardo Elael
1615
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
add a comment |
up vote
0
down vote
Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.
answered Nov 21 at 17:14
Eduardo Elael
1615
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
add a comment |
up vote
0
down vote
up vote
0
down vote
Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.
answered Nov 21 at 17:14
Eduardo Elael
1615
Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.
answered Nov 21 at 17:14
Eduardo Elael
1615
edited Nov 21 at 17:23
answered Nov 21 at 17:14
Eduardo Elael
1615
answered Nov 21 at 17:14
Eduardo Elael
1615
answered Nov 21 at 17:14
Eduardo Elael
1615
1615
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
add a comment |
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
Would this mean that $f:mathbb{C}tomathbb{C}mid f(z)=z^frac{1}{2}implies f(4+0i)={2,-2}$?
– R. Burton
Nov 21 at 17:16
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
As a set-valued function, the codomain would be something like $2^mathbb{C}$ (power set) not just $mathbb{C}$. But yes, in that case $f(4) = {2, -2}$.
– Eduardo Elael
Nov 21 at 17:22
add a comment |
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