|
|
Post by mondo on Oct 26, 2022 23:23:22 GMT
Vasco, I believe my description about figure [21a] is valid. The $\sqrt{2}$ can be justified as we need a radius of inversion so that when $[Na]$ is equal to $1$, $[NS]$ must be equal $2$ and only $\sqrt{2}$ can do this -> $I_K(z) = \frac{R^2}{p} = \frac{R^2}{1} = 2 -> R = \sqrt{2} $ Additionally I proved my solution for exercise 5. Please consider this figure  It shows a vertical cross section of a sphare $C$ and a real axis. Some point $Z$ from a real axis lies outside the sphare and was stereographically projected to a point $\hat{Z}$ on the sphare. Next we also projected that point from $\hat{Z}$ to $\tilde{Z}$ on a real axis, according to stereographic projection through the south pole. This is the second part of the exercise. Now, let's assume point $\tilde{Z}$ lies on a real axis at the distance $0.4$ from the origin ($z = 0.4$). In order to show the relation between $Z$ and $\tilde{Z}$ lets find the equation of a line that connects $S$ and $\tilde{Z}$. Its easy to find that this equation is $y = \frac{10}{4}x -1 $ Next let's find the coordinates of a point $\hat{Z}$ on the sphare $C$ -> For that we have: $x^2 + (\frac{10}{4}x -1)^2 = 1$ from which we get coordinates $x = 0.689, y = 0.724$. Now that we have this point we can find equation of a line $N\hat{Z}$ -> $y = -0.4x + 1$ we got negative slope which is aligned with what we see on the figure. Now we are in position to find the $X$ coordinate of $Z$ -> $0 = -0.4x + 1 -> x = \frac{10}{4}$ From this we can see that the point $Z$, in the process of stereographic projection with north pole following by another projection with south pole, got mapped from $\frac{10}{4}$ to $\frac{4}{10}$. Which is an inversion. What kind of inversion is it? It must be $\frac{1}{\bar{z}}$ and not $\frac{1}{z}$ because the later also changes angle. So $Z = 0.4e^{i\theta}$ would be moved to $2.5e^{-i\theta}$. In our example in wouldn't make any difference because the angle is $0$ but is easy to imagine what would happen if it isn't- the image would be rotated by $90$ degree. So this proves what I was intuitively predicting before, this stereographic projection sends points from inside the unit circle to the outside according to $\frac{1}{\bar{z}}$ ; Also, in the book author often time doesn't care about proving his statement. Let's take figure [21b] as an example. Author draw it so it graphicly (more or less) looks like an inversion and states it as a fact in (17) page 143. However there is no proof at all. |
|
|
|
Post by Admin on Oct 27, 2022 8:43:29 GMT
Mondo
This is just not true at all. Needham always provides a proof unless it's an exercise. The proof you are looking for is on page 143 immediately following (17).
Vasco
|
|
|
|
Post by mondo on Oct 27, 2022 17:39:51 GMT
Mondo This is just not true at all. Needham always provides a proof unless it's an exercise. The proof you are looking for is on page 143 immediately following (17). Vasco You are right, sorry about that. I don't know how I missed it.. it's not fully clear for me though. For example, "..pairs of points $z$ and $\tilde{z}$ are symmetric with respect to $C$" - can we really say they are symmetric here? They have different lengths of each side of $C$ (unit circle, a point [1,0]). So unless a symmetry here is defined by $\frac{1}{\bar{z}}$ I would not call them symmetric. Points $0$ and $2$ can be called symmetric with respect to $C$ as both are equally distant from $C$. |
|
|
|
Post by Admin on Oct 27, 2022 20:22:16 GMT
Mondo This is just not true at all. Needham always provides a proof unless it's an exercise. The proof you are looking for is on page 143 immediately following (17). Vasco You are right, sorry about that. I don't know how I missed it.. it's not fully clear for me though. For example, "..pairs of points $z$ and $\tilde{z}$ are symmetric with respect to $C$" - can we really say they are symmetric here? They have different lengths of each side of $C$ (unit circle, a point [1,0]). So unless a symmetry here is defined by $\frac{1}{\bar{z}}$ I would not call them symmetric. Points $0$ and $2$ can be called symmetric with respect to $C$ as both are equally distant from $C$. Mondo Yes symmetry is defined by "... the pair of points $z$ and $\widetilde{z}$ are said to be symmetric with respect to $K$" (page 125). See the basic ideas of Inversion on pages 124-126 and especially the last paragraph on page 125. Vasco |
|
|
|
Post by Admin on Oct 29, 2022 14:38:27 GMT
Mondo Here is a link to a document which I hope will help you understand how Needham shows the result at the bottom of page 142: Stereographic projection preserves circles. link to my documentVasco |
|
|
|
Post by mondo on Oct 30, 2022 3:04:09 GMT
Yes symmetry is defined by "... the pair of points $z$ and $\widetilde{z}$ are said to be symmetric with respect to $K$" (page 125). Ok, clear. Mondo Here is a link to a document which I hope will help you understand how Needham shows the result at the bottom of page 142: Stereographic projection preserves circles. link to my documentVasco Thank you for this Vasco. I like how you pointed out that stereographic projection is a special case of figure [12]. However, your comment on why the radius of $K$ is $\sqrt{2}$ is very laconic. Do you agree to what I wrote at the beginning of my reply #15 about this radius? "So we can find two intersecting spheres that define our circle.." and these two spheres are stereographic projection of our circle that lies on $C$ right? Also, did you have a chance too look at my update too prove for Exercise 5(post #15)? Do you agree with the proposed solution? |
|
|
|
Post by Admin on Oct 30, 2022 15:18:48 GMT
Yes symmetry is defined by "... the pair of points $z$ and $\widetilde{z}$ are said to be symmetric with respect to $K$" (page 125). Ok, clear. Mondo Here is a link to a document which I hope will help you understand how Needham shows the result at the bottom of page 142: Stereographic projection preserves circles. link to my documentVasco Thank you for this Vasco. I like how you pointed out that stereographic projection is a special case of figure [12]. However, your comment on why the radius of $K$ is $\sqrt{2}$ is very laconic. Do you agree to what I wrote at the beginning of my reply #15 about this radius? I don't think $[Na]$ is 1 and I can't see what $p$ is supposed to be. I can't see it on your diagram or in figure 21."So we can find two intersecting spheres that define our circle.." and these two spheres are stereographic projection of our circle that lies on $C$ right? No, they are inverted in $K$ and again intersect in a circle, which shows that our circle $C$ is mapped to a circle on $\Sigma$.Mondo Also, did you have a chance too look at my update too prove for Exercise 5(post #15)? Do you agree with the proposed solution? No because you say that stereographic projection sends points from inside the unit circle to outside the unit circle. Stereographic projection sends points inside the unit circle to points on $\Sigma$ in the southern hemisphere.Vasco |
|
|
|
Post by Admin on Oct 30, 2022 20:17:48 GMT
Mondo
I forgot to say that I have revised my document to explain the radius $\sqrt{2}$.
Vasco
|
|
|
|
Post by mondo on Oct 30, 2022 21:00:54 GMT
Mondo I forgot to say that I have revised my document to explain the radius $\sqrt{2}$. Vasco Thank you. I don't think [Na] is 1 and I can't see what p is supposed to be. I can't see it on your diagram or in figure 21. You are right, [Na] is definitely not 1, I did a shortcut here without explaining. What I actually meant to show is - imagine point a from [21a] is moved to the origin of $C$ then [Na] is $1$ and since we need [Ns] to be $2$ the radius of $K$ must be $\sqrt{2}$ since, as I said in reply #15 - $I_K(z) = \frac{R^2}{p} = \frac{R^2}{1} = 2 -> R = \sqrt{2} $. $p$ is of course a distance from the origin of inversion to the point $z$, (as shown on figure 1 p.124). Here $p$ is a distance [Na] (after a is moved to the origin of $C$) In your updated document you seem to rely on the fact that when $K$ intersects $\Sigma$ at $+/- 1$ we can make a right triangle with sides $1,1, R$ and hence we have: $R = \sqrt{1^2 + 1^2} = \sqrt{2}$ . No because you say that stereographic projection sends points from inside the unit circle to outside the unit circle. Stereographic projection sends points inside the unit circle to points on Σ in the southern hemisphere. I meant, this specific projection from exercise 5 not stereographic projection in general. Beside this controversial sentence, do you think it is enough of a prove? |
|
|
|
Post by Admin on Oct 31, 2022 8:43:49 GMT
|
|
|
|
Post by Admin on Oct 31, 2022 18:46:51 GMT
Mondo I forgot to say that I have revised my document to explain the radius $\sqrt{2}$. Vasco Thank you. I don't think [Na] is 1 and I can't see what p is supposed to be. I can't see it on your diagram or in figure 21. You are right, [Na] is definitely not 1, I did a shortcut here without explaining. What I actually meant to show is - imagine point a from [21a] is moved to the origin of $C$ then [Na] is $1$ and since we need [Ns] to be $2$ the radius of $K$ must be $\sqrt{2}$ since, as I said in reply #15 - $I_K(z) = \frac{R^2}{p} = \frac{R^2}{1} = 2 -> R = \sqrt{2} $. $p$ is of course a distance from the origin of inversion to the point $z$, (as shown on figure 1 p.124). Here $p$ is a distance [Na] (after a is moved to the origin of $C$) In your updated document you seem to rely on the fact that when $K$ intersects $\Sigma$ at $+/- 1$ we can make a right triangle with sides $1,1, R$ and hence we have: $R = \sqrt{1^2 + 1^2} = \sqrt{2}$ . No because you say that stereographic projection sends points from inside the unit circle to outside the unit circle. Stereographic projection sends points inside the unit circle to points on Σ in the southern hemisphere. I meant, this specific projection from exercise 5 not stereographic projection in general. Beside this controversial sentence, do you think it is enough of a prove? Mondo If you tidy it up a bit, it could be OK. Vasco |
|
|
|
Post by mondo on Oct 31, 2022 18:59:48 GMT
Thank you Vasco, I will review later today. If you tidy it up a bit, it could be OK. I will, and for that I have a question regarding to drawing tools, what software do you use for drawing diagrams like figure 1 in your exercise 5 solution? |
|
|
|
Post by mondo on Oct 31, 2022 19:24:09 GMT
Thank you Vasco, I will review later today. I couldn't resist and reviewed it right away. Yes, it is nice and clean. In fact, this is a method I started with but moved to more algebraic one as I fixated on showing one length is reciprocal of the other. |
|
|
|
Post by Admin on Nov 1, 2022 7:39:40 GMT
Mondo
I use Texnic center and Miktex to produce PDF documents with mathematical symbols using Latex and then you can add PStricks to do the diagrams.
All the software is free and you can find details on the internet about downloading it and setting it up. Set up Texnic center etc first and when that is working you can add PStricks.
Vasco
|
|
