Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Aug 17, 2016 20:58:35 GMT
Vasco, The question looks easy, but I didn't see any straightforward way derive [M]. My approach was to work backwards. If there is a more straightforward answer, I would like to see it. (Vasco replied to this question and I rewrote and replaced this file. Sept 3, 2016) nh.ch6.ex4.pdf (124.67 KB) Gary |
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Post by Admin on Aug 18, 2016 17:28:01 GMT
Vasco, The question looks easy, but I didn't see any straightforward way derive [M]. My approach was to work backwards. If there is a more straightforward answer, I would like to see it. View AttachmentGary Gary I do have a more straightforward answer which I will publish as soon as I can. Vasco |
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Post by Admin on Aug 19, 2016 16:24:35 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Aug 19, 2016 18:02:58 GMT
Vasco,
It looks interesting. I'll give it some study after the weekend.
Gary
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Post by Admin on Aug 20, 2016 7:59:52 GMT
Gary
I have just made some minor changes in the presentation - particularly near the end.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Aug 22, 2016 21:29:01 GMT
Vasco, I studied your most recent version and rewrote my own. nh.ch6.ex4.pdf (124.67 KB) Gary |
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Post by Admin on Aug 29, 2016 13:43:33 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Aug 30, 2016 4:39:57 GMT
Vasco,
Thank you very much. They are all good suggestions and comments. I think it is all fixed now. I'm afraid I don't follow your suggestion in 3. Regarding 4, I wonder if you approve my interim notation for $[\tilde{M}]$, i.e. $[\frac{\mathfrak{m}\tilde{z}}{1}]$. Re. 5, I take your point that there is no need to normalize the matrix. I hadn't noticed that. Yet, isn't it more than a coincidence that the exponentials in the final result reflect the elements of the normalized matrix, so that deriving it in this way may reflect the thought that led to this particular presentation of the problem?
Gary
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Post by Admin on Aug 30, 2016 13:47:13 GMT
Vasco, Thank you very much. They are all good suggestions and comments. I think it is all fixed now. I'm afraid I don't follow your suggestion in 3. Regarding 4, I wonder if you approve my interim notation for $[\tilde{M}]$, i.e. $[\frac{\mathfrak{m}\tilde{z}}{1}]$. Re. 5, I take your point that there is no need to normalize the matrix. I hadn't noticed that. Yet, isn't it more than a coincidence that the exponentials in the final result reflect the elements of the normalized matrix, so that deriving it in this way may reflect the thought that led to this particular presentation of the problem? Gary Gary I think it's pretty clear what you mean with your interim notation. My suggestion in 3 is that it is neater and shorter to write $F\circ M=\widetilde{M}\circ F$ $M=F^{-1}\circ F\circ M=F^{-1}\circ\widetilde{M}\circ F$. I'm still thinking about the normalisation, as a result of your comment above Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Aug 30, 2016 14:29:32 GMT
Vasco,
Agreed. My way was very round about.
Gary
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