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Finding a suitable transformation function for the picture
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I haven't taken complex analysis yet so there may be many words to sift through and not much concrete mathematical notation. I'll try my best though.
Given a local Euclidean unit square grid how would I go about finding a suitable conformal mapping function that transforms the real and imaginary grid lines to match those as pictured in the right-most square?
Do I need two separate transformation functions, one to transform the real gridlines and another to transform the imaginary gridlines? Would this approach work?
Algebraic-topologically speaking there is an equivalence relation imposed upon the left and right side of the square, and similarly there is an equivalence relation imposed upon the top and bottom of the square, such that all gridpoints incident upon the left side of the square are all "mapped" or "equated" with the top left corner of the square. The same thing is going on for the imaginary gridlines. There are multiple ways to do this, but the transformation seems to rotate each gridline and stretch them too.
So it seems like the transformation function would be a "multi-step operation" such that first, the equation of points incident upon the left side of the square would be made, the equation of points incident upon the right side of the square would be made, and then the whole structure would be rotated 45 degrees clockwise, and then stretched to the corner points opposite via the diagonal of the square. The same sort of description would apply for the imaginary curves. It's a bit too intricate of a transformation for me to figure out the exact function. I'm stuck. If anyone has ideas on how to proceed from here I welcome them.
complex-analysis transformation conformal-geometry quasiconformal-maps
asked Jul 23 at 23:07
George Thomas
259416
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up vote
0
down vote
favorite
I haven't taken complex analysis yet so there may be many words to sift through and not much concrete mathematical notation. I'll try my best though.
Given a local Euclidean unit square grid how would I go about finding a suitable conformal mapping function that transforms the real and imaginary grid lines to match those as pictured in the right-most square?
Do I need two separate transformation functions, one to transform the real gridlines and another to transform the imaginary gridlines? Would this approach work?
Algebraic-topologically speaking there is an equivalence relation imposed upon the left and right side of the square, and similarly there is an equivalence relation imposed upon the top and bottom of the square, such that all gridpoints incident upon the left side of the square are all "mapped" or "equated" with the top left corner of the square. The same thing is going on for the imaginary gridlines. There are multiple ways to do this, but the transformation seems to rotate each gridline and stretch them too.
So it seems like the transformation function would be a "multi-step operation" such that first, the equation of points incident upon the left side of the square would be made, the equation of points incident upon the right side of the square would be made, and then the whole structure would be rotated 45 degrees clockwise, and then stretched to the corner points opposite via the diagonal of the square. The same sort of description would apply for the imaginary curves. It's a bit too intricate of a transformation for me to figure out the exact function. I'm stuck. If anyone has ideas on how to proceed from here I welcome them.
complex-analysis transformation conformal-geometry quasiconformal-maps
asked Jul 23 at 23:07
George Thomas
259416
It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I haven't taken complex analysis yet so there may be many words to sift through and not much concrete mathematical notation. I'll try my best though.
Given a local Euclidean unit square grid how would I go about finding a suitable conformal mapping function that transforms the real and imaginary grid lines to match those as pictured in the right-most square?
Do I need two separate transformation functions, one to transform the real gridlines and another to transform the imaginary gridlines? Would this approach work?
Algebraic-topologically speaking there is an equivalence relation imposed upon the left and right side of the square, and similarly there is an equivalence relation imposed upon the top and bottom of the square, such that all gridpoints incident upon the left side of the square are all "mapped" or "equated" with the top left corner of the square. The same thing is going on for the imaginary gridlines. There are multiple ways to do this, but the transformation seems to rotate each gridline and stretch them too.
So it seems like the transformation function would be a "multi-step operation" such that first, the equation of points incident upon the left side of the square would be made, the equation of points incident upon the right side of the square would be made, and then the whole structure would be rotated 45 degrees clockwise, and then stretched to the corner points opposite via the diagonal of the square. The same sort of description would apply for the imaginary curves. It's a bit too intricate of a transformation for me to figure out the exact function. I'm stuck. If anyone has ideas on how to proceed from here I welcome them.
complex-analysis transformation conformal-geometry quasiconformal-maps
asked Jul 23 at 23:07
George Thomas
259416
I haven't taken complex analysis yet so there may be many words to sift through and not much concrete mathematical notation. I'll try my best though.
Given a local Euclidean unit square grid how would I go about finding a suitable conformal mapping function that transforms the real and imaginary grid lines to match those as pictured in the right-most square?
Do I need two separate transformation functions, one to transform the real gridlines and another to transform the imaginary gridlines? Would this approach work?
Algebraic-topologically speaking there is an equivalence relation imposed upon the left and right side of the square, and similarly there is an equivalence relation imposed upon the top and bottom of the square, such that all gridpoints incident upon the left side of the square are all "mapped" or "equated" with the top left corner of the square. The same thing is going on for the imaginary gridlines. There are multiple ways to do this, but the transformation seems to rotate each gridline and stretch them too.
So it seems like the transformation function would be a "multi-step operation" such that first, the equation of points incident upon the left side of the square would be made, the equation of points incident upon the right side of the square would be made, and then the whole structure would be rotated 45 degrees clockwise, and then stretched to the corner points opposite via the diagonal of the square. The same sort of description would apply for the imaginary curves. It's a bit too intricate of a transformation for me to figure out the exact function. I'm stuck. If anyone has ideas on how to proceed from here I welcome them.
complex-analysis transformation conformal-geometry quasiconformal-maps
asked Jul 23 at 23:07
George Thomas
259416
edited Jul 24 at 18:12
asked Jul 23 at 23:07
George Thomas
259416
asked Jul 23 at 23:07
George Thomas
259416
asked Jul 23 at 23:07
George Thomas
259416
259416
It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39
add a comment |Â
It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39
It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39
add a comment |Â
1 Answer
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This is an outline of what I would do:
First.- In the left graph, the red vertical lines are defined as
$$x=k$$ with k=[0,1]
Second.- Apply a linear transformation
$$x1=k1$$ with k1=[0,$pi$/2]
Third.- Apply the tangent function
$$x2=tan(x1)$$ with k2=[0,infinite)
Fourth.- Apply $$v=u^(x2)$$ with u=[0,1] and with v=[0,1]
answered Jul 24 at 20:06
cgiovanardi
679411
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is an outline of what I would do:
First.- In the left graph, the red vertical lines are defined as
$$x=k$$ with k=[0,1]
Second.- Apply a linear transformation
$$x1=k1$$ with k1=[0,$pi$/2]
Third.- Apply the tangent function
$$x2=tan(x1)$$ with k2=[0,infinite)
Fourth.- Apply $$v=u^(x2)$$ with u=[0,1] and with v=[0,1]
answered Jul 24 at 20:06
cgiovanardi
679411
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
add a comment |Â
up vote
0
down vote
This is an outline of what I would do:
First.- In the left graph, the red vertical lines are defined as
$$x=k$$ with k=[0,1]
Second.- Apply a linear transformation
$$x1=k1$$ with k1=[0,$pi$/2]
Third.- Apply the tangent function
$$x2=tan(x1)$$ with k2=[0,infinite)
Fourth.- Apply $$v=u^(x2)$$ with u=[0,1] and with v=[0,1]
answered Jul 24 at 20:06
cgiovanardi
679411
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is an outline of what I would do:
First.- In the left graph, the red vertical lines are defined as
$$x=k$$ with k=[0,1]
Second.- Apply a linear transformation
$$x1=k1$$ with k1=[0,$pi$/2]
Third.- Apply the tangent function
$$x2=tan(x1)$$ with k2=[0,infinite)
Fourth.- Apply $$v=u^(x2)$$ with u=[0,1] and with v=[0,1]
answered Jul 24 at 20:06
cgiovanardi
679411
This is an outline of what I would do:
First.- In the left graph, the red vertical lines are defined as
$$x=k$$ with k=[0,1]
Second.- Apply a linear transformation
$$x1=k1$$ with k1=[0,$pi$/2]
Third.- Apply the tangent function
$$x2=tan(x1)$$ with k2=[0,infinite)
Fourth.- Apply $$v=u^(x2)$$ with u=[0,1] and with v=[0,1]
answered Jul 24 at 20:06
cgiovanardi
679411
answered Jul 24 at 20:06
cgiovanardi
679411
answered Jul 24 at 20:06
cgiovanardi
679411
answered Jul 24 at 20:06
cgiovanardi
679411
679411
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
add a comment |Â
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
Thanks this is a good idea
– George Thomas
Jul 24 at 21:29
add a comment |Â
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It should not be all too difficult to set up a map which is $C^r$ in the interior, implements your figures, and possesses all intended symmetries. But I don't think it can be done conformally. If the map were conformal then in the neighborhood of the corners of the RH figure the passing curves would be curved the other way round.
– Christian Blatter
Jul 24 at 18:36
what do you mean by $ C^r $ ?
– George Thomas
Jul 24 at 21:28
It means $r$ times continuously differentiable, in other words: smooth and beautiful.
– Christian Blatter
Jul 25 at 6:39