Schwarz-Christoffel transformation help

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e^x
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I'm struggling to find the 2 angles from the image to compute the integral.

Does any1 have a method for this?
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Atiq629
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Blazy
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(Original post by e^x)
I'm struggling to find the 2 angles from the image to compute the integral.

Does any1 have a method for this?
Travelling anti-clockwise, define the interior angle of the polygon to be  \alpha_{j} \pi and the exterior angle to be  \beta_{j} \pi  = \pi - \alpha_{j} \pi . This means that at a corner when we turn left,  \beta_{j} will be positive, and  \beta_{j} will be negative when we turn right at a corner.

In your question, we want to map (from the z plane to the w plane), z = -1 to w = -1 and z = 1 to w = 0. So the Schwarz Christoffel mapping will be of the form
 f(z) = A + C\int_{}^{z} (t+1)^{-\beta_{1}}(t-1)^{-\beta_{2}} \mathrm d t ,
where I've left the betas, A and C for you to determine. [Hint: A and C fixes your points, i.e. f(-1) = -1 etc].

Technically speaking, we need a third point (by the Riemann Mapping Theorem, we can fix the pre-images of 3 boundary points, i.e. 3 of the xj, and therefore a third angle (in order to close the polygon, imagine going all the way back round at (0, -infty) and (-1, infty)). But we can choose  x_{3} = \infty and "ignore" this point in the formula.
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Apachai Hopachai
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(Original post by Blazy)
Travelling anti-clockwise, define the interior angle of the polygon to be  \alpha_{j} \pi and the exterior angle to be  \beta_{j} \pi  = \pi - \alpha_{j} \pi . This means that at a corner when we turn left,  \beta_{j} will be positive, and  \beta_{j} will be negative when we turn right at a corner.

In your question, we want to map (from the z plane to the w plane), z = -1 to w = -1 and z = 1 to w = 0. So the Schwarz Christoffel mapping will be of the form
 f(z) = A + C\int_{}^{z} (t+1)^{-\beta_{1}}(t-1)^{-\beta_{2}} \mathrm d x ,
where I've left the betas, A and C for you to determine. [Hint: A and C fixes your points, i.e. f(-1) = -1 etc].

Technically speaking, we need a third point (by the Riemann Mapping Theorem, we can fix the pre-images of 3 boundary points, i.e. 3 of the xj, and therefore a third angle (in order to close the polygon, imagine going all the way back round at (0, -infty) and (-1, infty)). But we can choose  x_{3} = \infty and "ignore" this point in the formula.
Is this Complex Analysis?

It looks so interesting... what is the pre requisites
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Blazy
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(Original post by Apachai Hopachai)
Is this Complex Analysis?

It looks so interesting... what is the pre requisites
It is - the Schwarz Christoffel formula is an example of a constructive mapping from the UHP to a polygon. You can use it to solve problems in applied maths, so it's something you'd see in an applied course rather than a pure course. Depending on how in-depth you want to learn this kind of stuff, basic ideas from complex analysis and conformal mappings is probably enough.
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e^x
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(Original post by Blazy)
Travelling anti-clockwise, define the interior angle of the polygon to be  \alpha_{j} \pi and the exterior angle to be  \beta_{j} \pi  = \pi - \alpha_{j} \pi . This means that at a corner when we turn left,  \beta_{j} will be positive, and  \beta_{j} will be negative when we turn right at a corner.

In your question, we want to map (from the z plane to the w plane), z = -1 to w = -1 and z = 1 to w = 0. So the Schwarz Christoffel mapping will be of the form
 f(z) = A + C\int_{}^{z} (t+1)^{-\beta_{1}}(t-1)^{-\beta_{2}} \mathrm d t ,
where I've left the betas, A and C for you to determine. [Hint: A and C fixes your points, i.e. f(-1) = -1 etc].

Technically speaking, we need a third point (by the Riemann Mapping Theorem, we can fix the pre-images of 3 boundary points, i.e. 3 of the xj, and therefore a third angle (in order to close the polygon, imagine going all the way back round at (0, -infty) and (-1, infty)). But we can choose  x_{3} = \infty and "ignore" this point in the formula.
I think i understand what you're trying to say.

If i start from the top of the diagram and as i approach u=-1 i can go clockwise which is -pi/2 the interior angle or i can go anti-clockwise which is pi/2 and the exterior angle, is this correct?

And as i approach u=0 i can again take the interior angle which is -3*pi/2 (
clockwise) or i can take the exterior angle which is pi/2 (anti-clockwise), is this correct?
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e^x
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In this Q should both the angles be pi/2?

I'm starting from the left and approaching W1 when I get to w1 I turn pi/2 anti-clockwise to continue and then when I get to w2 I turn by pi/2 anti-clockwise to continue.
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Blazy
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(Original post by e^x)
I think i understand what you're trying to say.

If i start from the top of the diagram and as i approach u=-1 i can go clockwise which is -pi/2 the interior angle or i can go anti-clockwise which is pi/2 and the exterior angle, is this correct?

And as i approach u=0 i can again take the interior angle which is -3*pi/2 (
clockwise) or i can take the exterior angle which is pi/2 (anti-clockwise), is this correct?
Essentially, you should think about going around the polygon in an anti-clockwise direction. As you approach (u,v)= (-1,0) from the top, you turn left (left in the sense that if you are on the line going around the polygon in an anticlockwise direction) by pi/2 and then you turn right by pi/2 at the origin. In general, turning left gives a positive beta and turning right gives a negative one.

Perhaps this wasn't the wisest way of explaining this. I'll elaborate with the mathematical definition I posted earlier for.

At (u,v) = (-1,0), the interior angle is  \frac{\pi}{2} , so  \beta_{1}\pi = \pi - \frac{\pi}{2} , so  \beta_{1} = \frac{1}{2} . At the origin, the interior angle is  \frac{3\pi}{2} , so  \beta_{2} = -\frac{1}{2} . Note that my exterior angle is positive when I turn left and negative when I turn right.

We then use the SCF which is:
 f(z) = A + C\int_{}^{z} \prod_{j}^{} (t-x_{j})^{-\beta_{j}} \mathrm d t
where x_i are the points on the UHP you are mapping. You plug in x1 = -1, beta 1 = 1/2, etc and solve the integral.

(Original post by e^x)
In this Q should both the angles be pi/2?

I'm starting from the left and approaching W1 when I get to w1 I turn pi/2 anti-clockwise to continue and then when I get to w2 I turn by pi/2 anti-clockwise to continue.
See above - try and identify the exterior angles.
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e^x
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(Original post by Blazy)
Essentially, you should think about going around the polygon in an anti-clockwise direction. As you approach (u,v)= (-1,0) from the top, you turn left (left in the sense that if you are on the line going around the polygon in an anticlockwise direction) by pi/2 and then you turn right by pi/2 at the origin. In general, turning left gives a positive beta and turning right gives a negative one.

Perhaps this wasn't the wisest way of explaining this. I'll elaborate with the mathematical definition I posted earlier for.

At (u,v) = (-1,0), the interior angle is  \frac{\pi}{2} , so  \beta_{1}\pi = \pi - \frac{\pi}{2} , so  \beta_{1} = \frac{1}{2} . At the origin, the interior angle is  \frac{3\pi}{2} , so  \beta_{2} = -\frac{1}{2} . Note that my exterior angle is positive when I turn left and negative when I turn right.

We then use the SCF which is:
 f(z) = A + C\int_{}^{z} \prod_{j}^{} (t-x_{j})^{-\beta_{j}} \mathrm d t
where x_i are the points on the UHP you are mapping. You plug in x1 = -1, beta 1 = 1/2, etc and solve the integral.



See above - try and identify the exterior angles.
Okay i understand what you're saying. I imagine that in walking on the line starting from the top and if I turn left my angle will be positive and if I turn right my angle will be negative.

So for the first case the exterior angle is pi/2 and because I'm turning left this will be positive.

And when I approach the origin the origin the exterior angle is again pi/2 but because I'm turnin right this angle will be negative.

Is that correct?

Also if I decide to start at the other end of the graph (not starting at the top). Then the sign of my angle will change, is this correct?
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Blazy
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(Original post by e^x)
Okay i understand what you're saying. I imagine that in walking on the line starting from the top and if I turn left my angle will be positive and if I turn right my angle will be negative.

So for the first case the exterior angle is pi/2 and because I'm turning left this will be positive.

And when I approach the origin the origin the exterior angle is again pi/2 but because I'm turnin right this angle will be negative. (

Is that correct?
Correct.

Also if I decide to start at the other end of the graph (not starting at the top). Then the sign of my angle will change, is this correct?
I don't mean to be unhelpful, but I don't quite understand what you're asking here...it doesn't matter where you start, the importance is identifying the exterior angles whilst going anti-clockwise around the polygon.
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