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Post by Admin on Jan 8, 2023 20:22:59 GMT
Vasco, I understand it from Jacobian point of view but can you explain what geometric relations are needed to show the length of the infinitesimal vector $\epsilon$ was increased to $2r\epsilon$? Mondo You are right that the Jacobean analysis enables us to draw [5], because it tells us that this mapping multiplies all infinitesimal vectors by $2r$ and rotates them through an angle $\theta$, so yes you are right. Figure [5] illustrates that in the case of this mapping $z\mapsto z^2$ all infinitesimal vectors emanating from a given point have the same amplitwist. But this is true for all analytic functions. Vasco |
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Post by Admin on Jan 9, 2023 10:29:09 GMT
Mondo
Another way of looking at this is to remember that we know from subsection 2 about the Jacobean, that all infinitesimal vectors are expanded by the same factor and rotated through the same angle.
This means that when we draw [5] we can choose an infinitesimal vector at $z(r,\theta)$, that points in the same direction as $z(r,\theta)$ and then it is easy to see that the vector of length $r+\epsilon$ and angle $\theta$ will be mapped to the vector of length $(r+\epsilon)^2$ with angle $2\theta$.
So the length of the transformed infinitesimal vector will be $(r+\epsilon)^2-r^2=2r\epsilon+\epsilon^2\rightarrow 2r\epsilon$ as $\epsilon\rightarrow 0$, which shows geometrically that any infinitesimal vector emanating from $z$ is multiplied by $2r$ regardless of its direction.
Vasco
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Post by mondo on Jan 12, 2023 5:38:34 GMT
Mondo Another way of looking at this is to remember that we know from subsection 2 about the Jacobean, that all infinitesimal vectors are expanded by the same factor and rotated through the same angle. This means that when we draw [5] we can choose an infinitesimal vector at $z(r,\theta)$, that points in the same direction as $z(r,\theta)$ and then it is easy to see that the vector of length $r+\epsilon$ and angle $\theta$ will be mapped to the vector of length $(r+\epsilon)^2$ with angle $2\theta$. So the length of the transformed infinitesimal vector will be $(r+\epsilon)^2-r^2=2r\epsilon+\epsilon^2\rightarrow 2r\epsilon$ as $\epsilon\rightarrow 0$, which shows geometrically that any infinitesimal vector emanating from $z$ is multiplied by $2r$ regardless of its direction. Vasco Vasco, I like the $(r+\epsilon)^2-r^2=2r\epsilon+\epsilon^2\rightarrow 2r\epsilon$ example but there is one problem, $2r\epsilon$ also tends to $0$ as $\epsilon$ tends to $0$. Even though the $\epsilon$ term tends faster (a square) I am not sure if we can use this as a "proof". |
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Post by Admin on Jan 13, 2023 10:02:53 GMT
Mondo
Yes, $2r\epsilon$ tends to $0$ but is never equal to zero. When we say $\epsilon\rightarrow 0$ we expressly exclude $\epsilon=0$.
Vasco
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Post by mondo on Jan 13, 2023 23:33:27 GMT
Makes sense. Thank you Vasco for the help in this thread, I think all is clear now  PS: I think author should have started with Figure[12] page 230 in place of figure [5]. The later doesn't explain the amplitwist at all just visualizes Jacobian effect. |
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Post by Admin on Jan 14, 2023 9:36:20 GMT
Mondo Glad to be of assistance to you. I don't agree with your PS above because [5] explains what the amplitwist is by using a simple function $z^2$ that has an amplitwist, and then goes on to show how it can be generalised for analytic functions. Non-analytic functions don't have an amplitwist . Vasco |
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Post by mondo on Jan 14, 2023 18:31:09 GMT
Mondo Glad to be of assistance to you. I don't agree with your PS above because [5] explains what the amplitwist is by using a simple function $z^2$ that has an amplitwist, and then goes on to show how it can be generalised for analytic functions. Non-analytic functions don't have an amplitwist . Vasco Vasco, in my "PS" comment I wanted to say the figure [12] from page 230 actually shows a nice "geometric/visual" method of getting the amplitwist, while figure [5] barely presents the result of Jacobian matrix. Yes, you were right that in figure [12] and accompanying calculations we can substitute $2$ for $a$ in $z^a$ to have a a graphical method of getting the amplitwist but I wish this was shown on page 194 and figure [5] so I haven't spent a long time trying to understand how in the world the amplitwist was taken from figure [5] alone. |
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