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Post by mondo on Jul 25, 2023 6:15:19 GMT
Mondo I will produce a short document which shows this and post a link to it later today. Vasco PS Where are you situated in the world? I am in the UK. I overlooked your PS, now that I go over this thread again, I noticed it - I am currently in San Francisco. Pitty you are that far as I owe you a good lunch! |
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Post by Admin on Jul 25, 2023 6:21:33 GMT
Mondo
If you take the visual (geometric) approach you are less likely to make those kind of mistakes.
Vasco
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Post by mondo on Jul 25, 2023 6:24:52 GMT
I just needed to see complex derivative in polar form to understand it. Also using this document we can see derivatives in both rectangular and polar form are equal: $e^{-i\theta}[2|z|cos(2\theta) + i2|z|sin(2\theta)] = e^{-i\theta}2|z|e^{2i\theta} = 2|z|e^{i\theta} = 2z$  Also, when I read chapter 4 again I stumbled upon another paragraph that I don't get - second paragraph of subsection two on page 208 Since the transformation of a circle is linear I would expect it to transform that circle to another one right? |
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Post by Admin on Jul 25, 2023 6:27:15 GMT
Mondo I will produce a short document which shows this and post a link to it later today. Vasco PS Where are you situated in the world? I am in the UK. I overlooked your PS, now that I go over this thread again, I noticed it - I am currently in San Francisco. Pitty you are that far as I owe you a good lunch! Mondo That's OK. You make me re-visit some things I did year's ago and look at them in a new way. Cheers. Vasco |
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Post by mondo on Jul 25, 2023 6:28:03 GMT
Mondo If you take the visual (geometric) approach you are less likely to make those kind of mistakes. Vasco But look I started this way - I visualized a unit circle for myself even draw it with a vector $z$ and said to myself "ok it doesn't matter how much I rotate it the vector length is always the same" From this I concluded it is legit to calculate these derivatives separately. Where did I make a mistake in this reasoning? |
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Post by Admin on Jul 25, 2023 15:56:38 GMT
Mondo
Can you explain exactly what you are asking me in these last few posts. I am a bit confused. I understand that if we have a point $z=re^{i\theta}$ in the complex plane and then the mapping $z\mapsto w=z^2$. So we can write $w=z^2=[re^{i\theta}]^2=r^2e^{2i\theta}$.
I also agree that it then follows that $r=(x^2+y^2)^{1/2}$ and $\theta=\tan^{-1}(y/x)$, so $r$ and $\theta$ both depend on $x$ and $y$, which means that $r^2=(x^2+y^2)$ and $2\theta=2\tan^{-1}(y/x)$ also depend on $x$ and $y$.
So what is it then that you are asking about?
Vasco
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Post by mondo on Jul 25, 2023 16:23:56 GMT
Also, when I read chapter 4 again I stumbled upon another paragraph that I don't get - second paragraph of subsection two on page 208 Since the transformation of a circle is linear I would expect it to transform that circle to another one right? Vasco, this is confusing now. |
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Post by Admin on Jul 25, 2023 17:56:09 GMT
Mondo
Why do you think a linear mapping transforms a circle to a circle?
Vasco
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Post by Admin on Jul 26, 2023 10:17:24 GMT
Mondo
If you think about it, you will see that in general a linear transformation maps a circle to an ellipse.
Vasco
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Post by mondo on Jul 27, 2023 6:16:30 GMT
Mondo If you think about it, you will see that in general a linear transformation maps a circle to an ellipse. Vasco I have a problem visualizing it, "The local linear transformation is a stretch in the direction of d, another stretch perpendicular to it, and finally a twist." So if these "stretching" steps are equal we should get a circle again. Especially, the word "uniform" is suggesting that. From the algebraic perspective, if the equation of a circle is $r = \sqrt{x^2 + y^2}$ how can I modify it to model this linear transformation to obtain an ellipse? If I modify one of the variables say $x$ by dividing it by some coefficient say $3$ then I will indeed get an ellipse. This probably corresponds to a different stretching in $x$ direction then was done in $y$ direction. |
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Post by Admin on Jul 27, 2023 10:41:50 GMT
Mondo
In your latest reply above:
Why do you think "uniform" means that the two stretchings are the same? Needham explains what it means in paragraph 1 of subsection 2 on page 208. It does not mean that circles are mapped to circles.
Seems OK
Vasco
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Post by mondo on Jul 27, 2023 17:21:08 GMT
Mondo In your latest reply above: Why do you think "uniform" means that the two stretchings are the same? Needham explains what it means in paragraph 1 of subsection 2 on page 208. It does not mean that circles are mapped to circles. Ok I think I misunderstood this phrase "corresponding to a constant matrix" - this basically means, the transformation coefficients do not change during the mapping, and NOT that they are all the same as I initially thought. And conversely, the "non-uniform" mapping corresponds to a matrix whose coefficients change during the mapping. |
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