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Post by Admin on Dec 6, 2016 10:50:40 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Feb 9, 2017 21:30:25 GMT
Vasco, Here is my attempt, with no previous consultation. I'll take a look at yours now. My confidence level on this one is not high. nh.ch6.ex23.pdf (178.42 KB) Gary OK, I have checked now. Your more succinct solutions look good to me. Part (ii) has been fixed. |
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Post by Admin on Feb 10, 2017 15:21:13 GMT
Gary
While your solution to part (i) seems absolutely fine to me, part (ii) looks wrong.
On page 1 your first sentence "in both parts...p. 147." cannot be, because in part (ii) the sphere plays no part in the solution, we just have the PD and formula (46) is applicable to any two points on the disc.
There is no need for any angles such as $\phi$, as there is no stereographic projection involved in this, just the PD, and the PD is obtained from the PUHP using the Möbius transformation $D$ on page 317. There are no spheres involved in any of this.
In your last paragraph you ask the question "Why should this arc be equal to the h-distance $\mathcal{H}\{a,z\}$?". But it couldn't be if you think about it because if you look at (45) on page 318 the h-distance of a point $z$ from the origin becomes extremely large as $z$ approaches $1$ and so could never be the same length as an arc of the unit circle/sphere.
Also the mathematical manipulations on page 3 seem to contain errors. I would suggest a complete re-think of this part keeping well clear of the sphere and complex plane and concentrating on the PD.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 10, 2017 16:03:36 GMT
Gary While your solution to part (i) seems absolutely fine to me, part (ii) looks wrong. On page 1 your first sentence "in both parts...p. 147." cannot be, because in part (ii) the sphere plays no part in the solution, we just have the PD and formula (46) is applicable to any two points on the disc. There is no need for any angles such as $\phi$, as there is no stereographic projection involved in this, just the PD, and the PD is obtained from the PUHP using the Möbius transformation $D$ on page 317. There are no spheres involved in any of this. In your last paragraph you ask the question "Why should this arc be equal to the h-distance $\mathcal{H}\{a,z\}$?". But it couldn't be if you think about it because if you look at (45) on page 318 the h-distance of a point $z$ from the origin becomes extremely large as $z$ approaches $1$ and so could never be the same length as an arc of the unit circle/sphere. Also the mathematical manipulations on page 3 seem to contain errors. I would suggest a complete re-think of this part keeping well clear of the sphere and complex plane and concentrating on the PD. Vasco Vasco, Everything you say sounds right to me. I'll rethink it, but of course I'll have the benefit of having looked at your solution. Gary |
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