Gary
GaryVasco
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Post by Gary on Mar 26, 2016 21:06:07 GMT
Vasco,
Thank you. I believe I can now verify your derivation of $\mathcal{E}$, but don't have time to write it up at the moment.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 27, 2016 1:44:33 GMT
There were two typos in my document in the last paragraph: $\mathcal{R}$ and $\mathcal{E}$ were in the wrong expressions. I have corrected this. Please download the document again to get an updated version. Vasco Vasco, I didn't notice they were reversed and read them as they should be. Gary |
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Gary
GaryVasco
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Post by Gary on Mar 27, 2016 5:38:15 GMT
Vasco,
I have the following for $\alpha$ (the extra amplification):
$\hspace{1em}\alpha = Re(\nu) = Re(\frac{\mathcal{X}}{f’}) = Re(\frac{f’’\xi}{f’})$
$\hspace{2em}= |f'||\xi| Re(\frac{f’’\hat{\xi}}{f’|f'|}) = |f'||\xi|Re(i\overline{\mathcal{K}}\hat{\xi})$
$\hspace{2em}= |f'||\xi| Re(i (\mathcal{K}\cdot \hat{\xi} + i\mathcal{K} \times \hat{\xi} ) = |f'||\xi| Re(-\mathcal{K} \times \hat{\xi} + i\mathcal{K}\cdot \hat{\xi}) = - |f'||\xi|\mathcal{K} \times \hat{\xi}$
$\hspace{2em}= |f'||\xi|(\hat{\xi} \times \mathcal{K}) = |f'||\xi|\mathcal{E}$
Gary
Gary
But when $\mathcal{E}=0$ which is when $\mathcal{K}$ and $\widehat{\xi}$ are in the same direction, the extra amplification has to be 1, which is why in my derivation I have your result for $\alpha$, plus 1$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 27, 2016 14:25:51 GMT
Vasco, I have the following for $\alpha$ (the extra amplification): $\hspace{1em}\alpha = Re(\nu) = Re(\frac{\mathcal{X}}{f’}) = Re(\frac{f’’\xi}{f’})$ $\hspace{2em}= |f'||\xi| Re(\frac{f’’\hat{\xi}}{f’|f'|}) = |f'||\xi|Re(i\overline{\mathcal{K}}\hat{\xi})$ $\hspace{2em}= |f'||\xi| Re(i (\mathcal{K}\cdot \hat{\xi} + i\mathcal{K} \times \hat{\xi} ) = |f'||\xi| Re(-\mathcal{K} \times \hat{\xi} + i\mathcal{K}\cdot \hat{\xi}) = - |f'||\xi|\mathcal{K} \times \hat{\xi}$ $\hspace{2em}= |f'||\xi|(\hat{\xi} \times \mathcal{K}) = |f'||\xi|\mathcal{E}$ Gary Gary But when $\mathcal{E}=0$ which is when $\mathcal{K}$ and $\widehat{\xi}$ are in the same direction, the extra amplification has to be 1, which is why in my derivation I have your result for $\alpha$, plus 1$. Vasco
Vasco, Yes, I realized on rising this morning that to use it as a multiplier you have to add the 1. I'm attaching a document that explains my thoughts about how to apply the information to a plot. I was defeated by Latex. extra_amplitwist_forum_note.pdf (47.81 KB) Gary
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Post by Admin on Mar 27, 2016 16:04:56 GMT
Gary
Looks good to me.
Vasco
PS I've managed to do the plot for $e^z$ and $\log z$. I'm now looking at $f=z^m$ and then I'll move on to the $f=$Mobius. I found that I had to be careful in the choosing of some values when doing the calculations for the plot associated with $f=\log z$, so that the amplification $1+a \times \sin \beta$ where $a$ is real and $>0$, and $\beta$ is some angle, doesn't become less than zero, because if $\sin\beta$ becomes equal to -1 then if $a>1$ then the amplification becomes $<0$. Disaster!
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 27, 2016 16:36:12 GMT
Gary Looks good to me. Vasco PS I've managed to do the plot for $e^z$ and $\log z$. I'm now looking at $f=z^m$ and then I'll move on to the $f=$Mobius. I found that I had to be careful in the choosing of some values when doing the calculations for the plot associated with $f=\log z$, so that the amplification $1+a \times \sin \beta$ where $a$ is real and $>0$, and $\beta$ is some angle, doesn't become less than zero, because if $\sin\beta$ becomes equal to -1 then if $a>1$ then the amplification becomes $<0$. Disaster! Vasco, Thanks for the heads up. I have a basic question, maybe a stupid question. Consider the points c, c', c'+$\zeta$, $\tilde{c},\tilde{c'}$, as described in your document, plus $\widetilde{c' + \zeta}$ I am wondering if I should get the same result for the $\tilde{c},\tilde{c'}, \widetilde{c' + \zeta}$ with (1) and (2), and if not, why not? (1) apply f(z) to c, c', c'+$\zeta$ (2) apply f(z) to c, apply amplitwist to $\xi$ and add it to f(c), and apply the amplitwist with the extra amplitwist to $\zeta$ and add the result to the previous step. Gary |
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Post by Admin on Mar 28, 2016 6:47:58 GMT
Gary
When I was considering your post above I realised that my document contained an error. The $\zeta$ in the top right-hand circle should be $\widetilde{\zeta}$. I have made this correction.
Assuming this correction has been made, then if I understand you correctly, we can re-write your question as:
Consider the points $c,c',c'+\zeta,\widetilde{c},\widetilde{c'}$, as described in your document, plus $\widetilde{c'}+\widetilde{\zeta}$.
I am wondering if I should get the same result for the $\widetilde{c},\widetilde{c′}, \widetilde{c'}+\widetilde{\zeta}$with (1) and (2), and if not, why not?
(1) apply f(z) to c, c', c'+ζ
(2) apply f(z) to c, apply amplitwist to ξ and add it to f(c), and apply the amplitwist with the extra amplitwist to ζ and add the result to the previous step.
(1)
$\widetilde{c}=f(c)$
$\widetilde{c'}=f(c')$
$\widetilde{c'}+\widetilde{\zeta}=f(c'+\zeta)$
(2)
$\widetilde{c}=f(c)$
$\widetilde{\xi}+f(c)=\widetilde{\xi}+\widetilde{c}=\widetilde{c'}$
If you now apply the amplitwist with the extra bit to $\zeta$ and add the result to $\widetilde{c'}$ then you will have achieved your goal!.
At first I thought "previous step" meant (1).
Vasco
PS I have made a few more small changes to my document to hopefully make some things easier to follow.
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Post by Admin on Mar 31, 2016 15:14:36 GMT
Gary I have written a draft version of my attempt to answer part(vi) of exercise 17 in chapter 5, which can be downloaded from here. For each of the four cases I have shown the maths, which is based on my initial thoughts. The document which contained my initial ideas is now the first part of this latest document. In the case of $f=z^m$ I have included a diagram for $m=1,2,3$. The right hand part of the diagram for $m=3$ is scaled by a factor of 2 so that the circles and associated levers on the rim of the wheel are not too small to see. As I do not have access to any sophisticated drawing software I have used a spreadsheet to generate the numbers, based on the maths in the document, and then I have used PStricks to generate the diagrams from the numbers - not too laborious. I have checked the results to a certain extent and both the spreadsheet and the diagrams verify figure 21 on page 240 and are in agreement with (31) on page 240. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 31, 2016 15:43:03 GMT
Gary When I was considering your post above I realised that my document contained an error. The $\zeta$ in the top right-hand circle should be $\widetilde{\zeta}$. I have made this correction. Assuming this correction has been made, then if I understand you correctly, we can re-write your question as: Consider the points $c,c',c'+\zeta,\widetilde{c},\widetilde{c'}$, as described in your document, plus $\widetilde{c'}+\widetilde{\zeta}$.
I am wondering if I should get the same result for the $\widetilde{c},\widetilde{c′}, \widetilde{c'}+\widetilde{\zeta}$with (1) and (2), and if not, why not?
(1) apply f(z) to c, c', c'+ζ
(2) apply f(z) to c, apply amplitwist to ξ and add it to f(c), and apply the amplitwist with the extra amplitwist to ζ and add the result to the previous step.(1) $\widetilde{c}=f(c)$ $\widetilde{c'}=f(c')$ $\widetilde{c'}+\widetilde{\zeta}=f(c'+\zeta)$ (2) $\widetilde{c}=f(c)$ $\widetilde{\xi}+f(c)=\widetilde{\xi}+\widetilde{c}=\widetilde{c'}$ If you now apply the amplitwist with the extra bit to $\zeta$ and add the result to $\widetilde{c'}$ then you will have achieved your goal!. At first I thought "previous step" meant (1). Vasco PS I have made a few more small changes to my document to hopefully make some things easier to follow. Vasco, Thank you. My implementation did not work out even in a very bare-bones version with only one $\xi$, so perhaps I have a programming error. Sometimes sophisticated software just allows one to display one's ignorance more quickly. Gary |
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Post by Admin on Apr 1, 2016 5:46:23 GMT
Gary
I found a couple of typos in my draft solution to exercise 17 part(vi) chapter 5:
In stage 3 of the maths for $f=z^m$ and $f=(az+b)/(cz+d)$ I omitted the factor $|\xi|$ from the distance that $\widetilde{Q}$ moves.
These omissions are not carried through in the subsequent maths and so are not of any great significance. However I have updated my document and the link in my previous post now accesses this corrected version.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 1, 2016 7:04:10 GMT
Vasco,
I noted the change to the document.
On p. 3, you have $\mathcal{E} = |\mathcal{K}|\sin \gamma$, then two lines down you have $|\mathcal{K}| = e^x$, but then four lines down you have $1 + e^x |\xi|e^{-x}\sin \gamma$, where $e^{-x}$ appears to be substituted for $|\mathcal{K}|$. Is the minus sign in $e^{-x}$ a typo?
Gary
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Post by Admin on Apr 1, 2016 7:51:19 GMT
Vasco, I noted the change to the document. On p. 3, you have $\mathcal{E} = |\mathcal{K}|\sin \gamma$, then two lines down you have $|\mathcal{K}| = e^x$, but then four lines down you have $1 + e^x |\xi|e^{-x}\sin \gamma$, where $e^{-x}$ appears to be substituted for $|\mathcal{K}|$. Is the minus sign in $e^{-x}$ a typo? Gary Gary Yes, but the typo is in the equation for $|\mathcal{K}|$. The line should read: "...since we know from part(ii) that $|\mathcal{K}|=e^{-x}$, it follows..." I will make the correction. Thanks. Vasco |
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Post by Admin on Apr 1, 2016 14:08:00 GMT
Gary
I have found a mistake in the part referring to $f=z^m$. I am about to correct it and also the three accompanying diagrams. It will probably take me about an hour.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 1, 2016 14:09:47 GMT
> Yes, but the typo is in the equation for $|\mathcal{K}|$.
I signed on to the forum this morning to correct myself on that point.
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Post by Admin on Apr 2, 2016 7:16:18 GMT
Gary
I have corrected my document for part (vi) of exercise 17 chapter 5. You will need to download it again. The diagrams for the case $f=z^m$, $m=1,2,3$ are all drawn to the same scale. For $m=3$ it gets rather large. Maybe I could do without a diagram when $m=1$ since it is almost trivial and then scale down the other two by the same amount. Sorry it took so long to make this last change.
Vasco
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