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Post by mondo on Jan 17, 2023 8:26:16 GMT
Chapter 3 seems to be even more complicated, author does a lot of shortcuts in his formulas derivation, I follow your supplement document and it is helpful but still not all is clear:
1. Right above the second formula on page 238, "the curvature of the image of a straight line.." but that should be $0$ by definition right? We will end up with a division $\frac{0}{|\xi|}$
2.Right below second formula on the same page, "Now consider the fate of all curves that pass through $p$ in the direction $\hat{\xi}$" - but there can be only one curve that passes thru $p$ in the direction of a tangent unit vector at $p$?
Thank you.
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Post by Admin on Jan 17, 2023 20:29:16 GMT
Mondo
This is my reply to your post #29
First, I would like to point out that in figure 18 none of the vectors or angles are infinitesimals. This applies also to all the other figures in this book. To understand this please read the first part of subsection 4 on pages 20-21 up to the paragraph before the one that starts with the words "We now return...". This explains the meaning of the word "ultimately" in this context.
I'm not sure what your expression for $\widetilde{q}$ means because of the mixture of complex values and the modulus's of complex numbers. I would write:
$\widetilde{q}=f(q)$ as a starting point.
Comments about 1.:
I hope when you read the above it will answer your questions.
Comments about 2.:
We are not looking for the angle of $v$. $\sigma$ is not the angle of $v$.
I will post again regarding your second post number #30.
Vasco
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Post by Admin on Jan 17, 2023 21:12:59 GMT
Mondo Reply to your post number #30 Chapter 3 seems to be even more complicated, author does a lot of shortcuts in his formulas derivation, I follow your supplement document and it is helpful but still not all is clear: 1. Right above the second formula on page 238, "the curvature of the image of a straight line.." but that should be $0$ by definition right? We will end up with a division $\frac{0}{|\xi|}$ The image of a straight line could be any thing, it doesn't have to be a straight line. Not true! look at this link www.desmos.com/calculator/8xfpcgboiwmany circle passing through the same point with the same tangent. Vasco |
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Post by mondo on Jan 18, 2023 5:11:52 GMT
Mondo This is my reply to your post #29 First, I would like to point out that in figure 18 none of the vectors or angles are infinitesimals. This applies also to all the other figures in this book. To understand this please read the first part of subsection 4 on pages 20-21 up to the paragraph before the one that starts with the words "We now return...". This explains the meaning of the word "ultimately" in this context. Huh, none of the vector or angles in the book are infinitesimals? I would argue otherwise, most of figures in this book have at least one infinitesimal value (no surprise as it is fundamental concept in all calculus). Figure [18] is not an exception here - last paragraph of page 245 - "As ilustrated $\xi$ is a snall (ultimately infinitesimal)" complex number...". That's one thing, second it must be a very tiny vector otherwise formula (26) won't work. I see now that "ultimately" is another form of saying that something will equal to some value in a limit. I'm not sure what your expression for $\widetilde{q}$ means because of the mixture of complex values and the modulus's of complex numbers. I would write: $\widetilde{q}=f(q)$ as a starting point. Well this is a side, quick exercise I wanted to do, to see if this content is understood. $\widetilde{q}=f(q)$ is true but what if we want to calculate it only knowing a value of a neighbor point ($p$) plus a derivative ($f'()$) and a distance between that neighbor point and the point of interest ($\xi$). I think the solution then is what I wrote in post #29 - $\tilde{q} = f(p) + f'(p) \cdot |\xi|$. Do you agree with that? It is same as in real calculus, only difference is here we use vectors. Comments about 1.:I hope when you read the above it will answer your questions. Sort of, I mean the use of word "ultimately" suggest, as I wrote above that $\sigma$ is under some limiting behavior. While after we did calculate $f'()$ it is what it is between $f'(p)$ and $f'(q)$ hence the word ultimately seems to be not needed here or I miss something. Comments about 2.:We are not looking for the angle of $v$. $\sigma$ is not the angle of $v$. Right, $\sigma$ is just imaginary part of $v$, we can say so only because of the trick that we have $1 \cdot \sigma$ for the arc length which is $Im[v]$. Clear. Answers to post $30 are all clear. Thank you Vasco! |
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Post by Admin on Jan 18, 2023 18:26:27 GMT
Mondo
The figures themselves do not illustrate infinitesimals but just small-valued quantities. They are only ultimately infinitesimal.
The result (26) is only ultimately correct.
Your equation for $\widetilde{q}$ should be
$\widetilde{q}=f(q)=f(p)+(q-p)f'(p)+...$ ultimately $\widetilde{q}=f(q)=f(p)+\xi f'(p)$
This illustrates what I mean above. The value of $q-p$ is only ultimately equal to $\xi$, not in the diagram.
Vasco
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Post by mondo on Jan 18, 2023 19:15:37 GMT
Mondo The figures themselves do not illustrate infinitesimals but just small-valued quantities. They are only ultimately infinitesimal. The result (26) is only ultimately correct. Hmm but infinitesimals as I understand them are just a " small-valued" quantities. For the result (26) - so what must happen for it to be ultimately correct? Or in other words, when it is not correct? Maybe what values could we substitute for (26) variables to make it not true? Your equation for $\widetilde{q}$ should be $\widetilde{q}=f(q)=f(p)+(q-p)f'(p)+...$ ultimately $\widetilde{q}=f(q)=f(p)+\xi f'(p)$ So the only difference is I used $|\xi|$ while you have $\xi$ and I think you are right because we need a vector to move to proper $\tilde{q}$ locaton. Also, formula (26) is for amplification, what about if I would like to write a similar one for the twist, $\tilde{\epsilon} = \epsilon \cdot \text{twist} = \epsilon \cdot Arg(f'(p))$, is that correct? Thank you. |
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Post by Admin on Jan 18, 2023 20:18:34 GMT
Mondo
If we put any actual numeric values into (26) it is always an approximation although it becomes very accurate for very small values.
I disagree that (26) is about amplification. It is an equation for the modulus or length of $\widetilde{\xi}$ in terms of the modulus or length of $\xi$.
So the other equation you are asking for is about arg$(\widetilde{\xi})$ in terms of arg$(\xi)$ not the twist.
So it will be arg$(\widetilde{\xi})=\text{arg}(f'(p))+\text{arg}(\xi)$ or
arg$(\widetilde{\xi})=\text{(twist)}+\text{arg}(\xi)$.
So I don't agree with your equation. Also you should be using $\xi$ not $\epsilon$. Also don't forget that arg is an angle.
Vasco
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Post by mondo on Jan 19, 2023 19:02:03 GMT
Mondo If we put any actual numeric values into (26) it is always an approximation although it becomes very accurate for very small values. Accurate enough so author put $=$ sign instead of $\approx$ in (26) which proves that $|\xi|$ is infinitesimal  I disagree that (26) is about amplification. It is an equation for the modulus or length of $\widetilde{\xi}$ in terms of the modulus or length of $\xi$. Agree, I did a shortcut in thinking because amplification is involved in this equation. So the other equation you are asking for is about arg$(\widetilde{\xi})$ in terms of arg$(\xi)$ not the twist. Actually no, I try to write a formula for $\tilde{\epsilon}$ knowing only $f'$ and $\epsilon$. Making use of [18] it should be $\tilde{\epsilon} = \epsilon + arg(f''(p))$? So it will be arg$(\widetilde{\xi})=\text{arg}(f'(p))+\text{arg}(\xi)$ or arg$(\widetilde{\xi})=\text{(twist)}+\text{arg}(\xi)$. I agree with that. It is a good, supplementing exercise. Also don't forget that arg is an angle. Yes I know, so in example in $arg(f''(p))$ I used above it means take an angle of a second derivative, a twist in this case, which should be $\sigma$. Thank you. Vasco |
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Post by mondo on Jan 21, 2023 17:45:07 GMT
Actually no, I try to write a formula for $\tilde{\epsilon}$ knowing only $f'$ and $\epsilon$. Making use of [18] it should be $\tilde{\epsilon} = \epsilon + arg(f''(p))$? Vasco do you agree with the formula for $\tilde{\epsilon}$? |
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Post by Admin on Jan 23, 2023 11:08:35 GMT
Actually no, I try to write a formula for $\tilde{\epsilon}$ knowing only $f'$ and $\epsilon$. Making use of [18] it should be $\tilde{\epsilon} = \epsilon + arg(f''(p))$? Vasco do you agree with the formula for $\tilde{\epsilon}$? Mondo No, I don't agree: it can't be correct because $\sigma$ is a small (ultimately infinitesimal) quantity and $\arg(f''(p))$ is not this kind of quantity. Near the bottom of page 237 we have $\displaystyle\sigma=\text{Im}\bigg[\frac{f''(p)\xi}{f'(p)}\bigg]$, and so using (27) on page 237 we find that $\displaystyle\widetilde{\epsilon}=\epsilon+\text{Im}\bigg[\frac{f''(p)\xi}{f'(p)}\bigg]$ Vasco |
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Post by mondo on Jan 25, 2023 6:36:29 GMT
No, I don't agree: it can't be correct because $\sigma$ is a small (ultimately infinitesimal) quantity and $\arg(f''(p))$ is not this kind of quantity. Near the bottom of page 237 we have $\displaystyle\sigma=\text{Im}\bigg[\frac{f''(p)\xi}{f'(p)}\bigg]$, and so using (27) on page 237 we find that $\displaystyle\widetilde{\epsilon}=\epsilon+\text{Im}\bigg[\frac{f''(p)\xi}{f'(p)}\bigg]$ Vasco, your formula is exactly what is in the book, and I agree with it. The motivation of my formula was $f''(z)$ represents an amplitwist of an amplitwist. Hence it will have its own amplification and twist. Now from figure [18] $\sigma$ sims to be exactly that "second" twist - it will tell us how the first twist changes when the argument $z$ slightly changes and I was hoping to get that twist with $arg(f''(z))$. This is because $arg(f'(z))$ gives us the "first" twist right? So naturally I would expect $arg(f''(z))$ gives a twist of the mapping amplitwist. So why do you say: In addition I tried to calculate it "manually" but gave up after a first partial derivative since I don't see a hope with that method - here is what I did for the $z^2$ mapping: $z^2 = r^2e^{2i\theta}$; $u = r^2\cos(2\theta)$ hence $\partial_{x} u =2rr'\cos(2\theta) -2r^2\sin(2\theta)\theta' =...= 2r\cos{\theta}$; $\partial_{x}^2u = 2r'cos(\theta) - 2r\sin(\theta) = 2\cos^2(\theta) - 2r\sin(\theta)$ And here I stopped as it doesn't give a simple twist. |
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Post by Admin on Jan 25, 2023 13:43:32 GMT
Mondo
The reason I said this is because you said you wanted to calculate $\sigma$ and $\sigma$ is an infinitesimal quantity, but clearly $f''(p)$ is not an infinitesimal quantity and so cannot represent $\sigma$.
To do the amplitwist of an amplitwist we could apply the function $f'$ to $\widetilde{p}$ and write $\mu=f''(\widetilde{p})\widetilde{\xi}$
Since $\widetilde{\xi}=f'(p)\xi$, we can write this as $\mu=f''(\widetilde{p})f'(p)\xi$.
So the total twist applied to $\xi$ is $\arg{[f''(\widetilde{p})]}+\arg{[f'(p)]}$, where $\widetilde{p}=f(p)$.
However more generally using the second function as $g$ this would be $\mu=g'(\widetilde{p})f'(p)\xi$.
So the total twist applied to $\xi$ is $\arg{[g'(\widetilde{p})]}+\arg{[f'(p)]}$, where $\widetilde{p}=f(p)$
Vasco
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Post by Admin on Jan 25, 2023 18:59:43 GMT
Mondo
I have made a correction to the last line in the above so that "$\widetilde{p}=f(p)"$
Vasco
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Post by mondo on Jan 25, 2023 21:36:01 GMT
Thank you Vasco! The reason I said this is because you said you wanted to calculate $\sigma$ and $\sigma$ is an infinitesimal quantity, but clearly $f''(p)$ is not an infinitesimal quantity and so cannot represent $\sigma$. I agree, also, we can wrote $\sigma$ as $\sigma = arg(f'(q)-f'(p))$ and get the angle between these two vectors. To do the amplitwist of an amplitwist we could apply the function $f'$ to $\widetilde{p}$ and write $\mu=f''(\widetilde{p})\widetilde{\xi}$ I agree with the equation but you rather wanted to say "apply the function $f''$" instead of $f'$ right? Because the later can only be applied to $p$ and represents the mapping amplitwist, so we cannot use it again to $\tilde{p}$ as that would mean $f(f(p))$ right? So the total twist applied to $\xi$ is $\arg{[f''(\widetilde{p})]}+\arg{[f'(p)]}$, where $\widetilde{p}=f(p)$. Okay, so now we corrected my previous claim that $\sigma = arg(f''(p))$ to $\sigma = arg(f''(\tilde{p}))$. Am I right that it shows an important difference between a real analysis and complex one? Because in real analysis if you want to calculate say, an acceleration at a point $p$ you do $f''(p)$ while here it seems to be $f''(f(p)) = f''(\tilde{p})$ |
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Post by mondo on Jan 26, 2023 1:46:37 GMT
My last statement about the difference between real and complex analysis is no a revelation though - we must apply the second derivative to a different point because the first derivative mapped it to a different one. Unlikely in real analysis where each derivative calculated or mapping does not change the argument location. Do you agree Vasco?
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