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Post by mondo on Jan 14, 2023 20:22:27 GMT
Vasco, I am also confused by the sentence in the middle of page 237 "The amplitwist that carries the disc at $p$ to the disc at $f'(p)$ is f''(p)" - why the second derivative and not the fist one? According to the quoted sentence, in order to map a point $p$ (and its neighborhood) to $p'$ we need to calculate a derivative of a derivative - $f''(p)$ which means calculate amplitwist twice. I don't get why twice and not just once.
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Post by Admin on Jan 14, 2023 20:29:57 GMT
Mondo
Look at figure 18 and you will se that there are two mappings illustrated: one using $f$ and one using $f'$
Vasco
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Post by mondo on Jan 14, 2023 20:43:57 GMT
Mondo Look at figure 18 and you will se that there are two mappings illustrated: one using $f$ and one using $f'$ Vasco So this suggests the same amplitwist is used to get the image of the mapping as well to get the derivative of the mapping right? In other words to get the value of $\chi$ from $\xi$ we do amplitwist $\xi$ to get its image under $f(z)$ mapping, namely $\tilde{\xi}$, then we amplitwist $\tilde{\xi}$ again by the same amplification and twist as before to get $\chi$ right? If that is the case I only question if we can use the same amplitwist for two different things? |
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Post by Admin on Jan 14, 2023 20:53:19 GMT
Mondo
No both mappings are applied to $\xi$.
Vasco
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Post by Admin on Jan 14, 2023 20:56:09 GMT
Mondo
You need to study 18 carefully.
Vasco
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Post by mondo on Jan 14, 2023 21:08:11 GMT
Mondo No both mappings are applied to $\xi$. Vasco Mondo You need to study 18 carefully. Vasco There is one more fundamental thing that doesn't make sense to me - on figure [18] - why the $f'$ mapping is applied on $\xi$ and not on its image $\tilde{\xi}$? It's like calculating derivative of the argument say $x'$ instead of f'(x). In the former case it will always be $0$ as every $x$ is a constant so it only really makes sense to calculate a derivative of $f(x)$. |
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Post by Admin on Jan 14, 2023 21:21:13 GMT
Mondo
If you look at (23) you will see that everything is evaluated at $p$.
Vasco
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Post by mondo on Jan 14, 2023 21:43:00 GMT
Mondo If you look at (23) you will see that everything is evaluated at $p$. Vasco yes I see that. I really don't get why $\chi = f''(p)\xi$. From [18] we can see that $\chi$ is a distance between $f'(p)$ and $f'(q)$ so if single amplitwist was needed to get the endpoints of $\chi$ why two amplitwists are needed to get its length? |
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Post by Admin on Jan 14, 2023 21:53:47 GMT
Mondo
it only needs one. the mapping using $f$ allows us to calculate $\widetilde{\xi}$ and the mapping using $f'$ allows us to calculate $\chi$. One amplitwist for each one.
Vasco
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Post by Admin on Jan 14, 2023 22:15:20 GMT
Mondo
The mapping $f'$ gives us the endpoints of $\chi$ and then the amplitwist $f''$ gives us $\chi$.
Vasco
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Post by mondo on Jan 15, 2023 3:55:53 GMT
Mondo The mapping $f'$ gives us the endpoints of $\chi$ and then the amplitwist $f''$ gives us $\chi$. Vasco Yes this is how I understand it too but it is confusing as it looks like author uses amplitwist concept for both "function mapping" and derivatives. For example $f(z) = z^2$ can be understood as an amplitwist with an amplification $2r$ and twist $\theta$, so in light of what is said on page 237 we can also say that $f'(z) = z^2$ hence we have $f(z) = f'(z) = z^2$ this is super confusing.. |
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Post by Admin on Jan 15, 2023 14:37:26 GMT
Mondo
Yes we can define a mapping like $z\mapsto z^2$ as a function which maps one complex plane whose points are $z$ to another complex plane whose points are $w=z^2$, as the first few chapters of the book explain.
However we can also think of a function/mapping as being defined by its amplitwist, because the amplitwist tells us what the mapping does to any infinitesimal vector in the plane.
You will remember when you studied chapter 10 which was about vector fields and these vector fields were introduced as a new way of thinking about functions, using just one complex plane. If we apply a mapping to a vector field and we think of it as being like figures 1 and 2 in chapter 10 but with infinitesimal vectors, then the amplitwist will amplify and twist the vectors in the vector field to give us another vector field which is the same as that which results from the original mapping.
So you can see that the mapping $f$ as well as being defined by the complex function $f=f(z)$, is also defined by its amplitwist.
However what you have written in reply 25 is not correct. We cannot say that $f(z)=f'(z)=z^2$.
Needham is arguing that if $f$ is an analytic function then so are all its derivatives. So this means that $f'$ is an analytic function in its own right, and so $f'$ has an amplitwist as well. So we can apply the mapping $f$ to the complex plane and we can also apply the mapping $F=f'$ to the same complex plane. Each function, $f$ and $F=f'$, has a different amplitwist and produces a different result.
Vasco
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Post by mondo on Jan 15, 2023 19:18:24 GMT
Vasco, thank you for an extended answer. Yes I agree with all you said. We can not say $f(z) = f'(z) = z^2$ because that would mean both $f(z)$ and $f'(z)$ have the same amplitwist and this is only the case for $f(z) = e^z$. So this is only true for this particular function but not in general.
However, going back to figure [18] - when we do $f'$ we get the function that describes how the amplitwist of the "function mapping, f(z)" varies. That in essence gives us the angle $\delta$ that we need but author claims that in order to get this angle $f''$ is needed - I don't see it. To me, $f''$ described how the amplitwist of the amplitwist varies (i.e acceleration). Does it make sense to you? Could you please explain?
In addition to that, two things that seem to be wrong either on figure [18] or in calculations that follow:
1. The direction of $f'$ arrow - this is something I already have mentioned in previous posts, I think $f'$ arrow should not point from the left part of the picture to the right but from the upper part, where $f$ mapping is presented to the right most image. Otherwise it does not make sense. Why? Because derivative $f'$ acts on a function and not on an argument, so we can't calculate $f'$ if there wasn't $f$ first - hence $f'$ should point from the top part to the right hand side of the image.
2. "The amplitwist that carries the disc at $p$ to the disc at $f'(p)$ is $f''(p)$" but figure [18] shows that the there is only one step on a way of getting $f'(p)$ disc from $p$ disc, namely $f'$ arrow.
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Post by Admin on Jan 16, 2023 15:53:08 GMT
Mondo Vasco, thank you for an extended answer. Yes I agree with all you said. We can not say $f(z) = f'(z) = z^2$ because that would mean both $f(z)$ and $f'(z)$ have the same amplitwist and this is only the case for $f(z) = e^z$. So this is only true for this particular function but not in general. We agree. (I will assume that you don't really mean $\delta$, but that you mean $\sigma$.) Yes it makes perfect sense to me. Since $f'(z)$ is the amplitwist of the mapping $f(z)$, it follows that $f''(z)$ is the amplitwist of the mapping $f'(z)$. No, you are completely wrong here. The functions $f(z)$ and $f'(z)$ are both functions of $z$, and so we can map $z$ using $f(z)$ or $f'(z)$. We are not differentiating when we use the mapping $f'$ or the mapping $f''$, we already know what the derivative is and we are using it as a function to map $z$. So there is nothing wrong here. . Yes, but this is perfectly correct. The function $f'$ is the mapping and $f''$ is the corresponding amplitwist. Take the example $f(z)=z^3$ $f'(z)=3z^2$ and $f''(z)=6z$ We can write $g(z)=f'(z)=3z^2$ and $h(z)=6z$. So $f,g,h$ are all functions which can be used to map $z$, and $g$ is the amplitwist for $f$, and $h$ is the amplitwist for $g$. If we did what you suggest we would have things like $f'(f(z))$ rather than $f'(z)$, and this is not what we want. Vasco |
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Post by mondo on Jan 17, 2023 4:30:14 GMT
Yes it makes perfect sense to me. Since $f'(z)$ is the amplitwist of the mapping $f(z)$, it follows that $f''(z)$ is the amplitwist of the mapping $f'(z)$. Ok, I feel like I am starting to get it. Let's see - $|\tilde{\xi}| = |f'(p)| \cdot |\xi|$ this is (26) and if we multiply by $|f''(p)|$ instead, we get $\chi$ which is now the distance between the derivatives. However, what if I would like to calculate the value of $\tilde{q}$ in this way? Would it be - $\tilde{q} = f(p) + f'(p) \cdot |\xi|$, according to the definition of derivative? There are two more questions I have for this page (237): 1. Just before $\sigma$ calculations - "Because $\sigma$ equals the (ultimately) vertical arc though 1..." - my question is about the word "ultimately", we are dealing here with the infinitesimals so this angle should already be there right. Also, it already is vertical even on the bottom graph of [18], so I don't know why author is emphasizing it? 2. $\sigma = arc = Im(v)$ - So the arc part is understood since the arc length will be $1 \cdot \sigma$ so it must be sigma. It also should be clear that it is $Im(v)$ but I have a moment of doubt here, as this will only give us the angle if $v$ is in a polar form. Hence I think it should rather be $arg(v)$, do you agree Vasco? Thank you. |
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