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Post by Admin on Mar 24, 2017 20:57:35 GMT
Gary
To answer your first question, there are 3 tabs at the bottom left, one for each part of the question. There should be an arrow to allow scrolling.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 21:52:35 GMT
Gary To answer your first question, there are 3 tabs at the bottom left, one for each part of the question. There should be an arrow to allow scrolling. Vasco Vasco, Got it. I've never used the spread sheet of LibreOffice before. I've just been looking at your answer to (iii) and noticed that you do not consider negative p-points. These should also have lengths that fit within $|z| \leq 2$. Gary |
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Post by Admin on Mar 24, 2017 22:16:51 GMT
Gary
I've just realised that I made a mistake when typing up part (iii) - I think I must have missed out one of my handwritten pages. Anyway I will rewrite it first thing tomorrow round midnight your time and publish again. Sorry for the inconvenience caused.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 22:36:42 GMT
Gary I've just realised that I made a mistake when typing up part (iii) - I think I must have missed out one of my handwritten pages. Anyway I will rewrite it first thing tomorrow round midnight your time and publish again. Sorry for the inconvenience caused. Vasco Vasco, No problem. There are lots of little things to consider and oversights are inevitable. I would very much like to know how you can read the multiplicity directly off the polynomial form of cos(z). I don't recall anything about this in the text. I reposted with a bit of discussion of this in (iii). Gary |
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Post by Admin on Mar 24, 2017 22:48:57 GMT
Gary
Have a look at subsection 1 of section IV especially page 347.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 24, 2017 23:31:54 GMT
Gary Have a look at subsection 1 of section IV especially page 347. Vasco Vasco, I can see that the material is related, but I don't see the direct application. If we consider cos(z) as a Taylor's series, we find that all the terms but the initial constant vanish at a zero p-point. Gary |
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Post by Admin on Mar 25, 2017 0:02:12 GMT
Gary
If $f=\cos z$ then $f'=-\sin z, f''=-\cos z, f'''=\sin z$ etc so not all the derivatives are zero when $z=0$, just every other derivative.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 25, 2017 1:38:56 GMT
Gary If $f=\cos z$ then $f'=-\sin z, f''=-\cos z, f'''=\sin z$ etc so not all the derivatives are zero when $z=0$, just every other derivative. Vasco Vasco, I was referring to your document, p. 4, four lines up from the second header. "From page 85 we know that the Taylor series for $sin z^4 = z^4 + ...$ and so the multiplicity of the p-point zero is +4." But I see it now. The first term of the fourth derivative is $\Delta^4$ and the remaining terms go to zero, so the algebraic multiplicity of f on 0 is +4. I was erroneously thinking they would always all go to zero. And I think I was confusing (iii) with (ii). $f' = 4z^3\Delta + ...$ $f'' = \frac{12z^2\Delta^2}{2!} + ...$ $f''' = \frac{24z\Delta^3}{3!} + ...$ $f'''' = \Delta^4 + 0 + ...$ Gary |
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Post by Admin on Mar 25, 2017 10:52:41 GMT
Gary
I have corrected the errors in my document (I hope).
Vasco
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Post by Admin on Mar 25, 2017 11:46:15 GMT
Gary If $f=\cos z$ then $f'=-\sin z, f''=-\cos z, f'''=\sin z$ etc so not all the derivatives are zero when $z=0$, just every other derivative. Vasco Vasco, I was referring to your document, p. 4, four lines up from the second header. "From page 85 we know that the Taylor series for $sin z^4 = z^4 + ...$ and so the multiplicity of the p-point zero is +4." But I see it now. The first term of the fourth derivative is $\Delta^4$ and the remaining terms go to zero, so the algebraic multiplicity of f on 0 is +4. I was erroneously thinking they would always all go to zero. And I think I was confusing (iii) with (ii). $f' = 4z^3\Delta + ...$ $f'' = \frac{12z^2\Delta^2}{2!} + ...$ $f''' = \frac{24z\Delta^3}{3!} + ...$ $f'''' = \Delta^4 + 0 + ...$ Gary Gary Applying the Taylor series for $\sin z$ we find that $sin z^4=z^4-z^{12}/3!+z^{20}/5!-z^{28}/7!+...$, so not all the derivatives of $\sin z^4$ are zero at $z=0$. Vasco |
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Gary
GaryVasco
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Post by Gary on Mar 25, 2017 15:25:16 GMT
Vasco, I was referring to your document, p. 4, four lines up from the second header. "From page 85 we know that the Taylor series for $sin z^4 = z^4 + ...$ and so the multiplicity of the p-point zero is +4." But I see it now. The first term of the fourth derivative is $\Delta^4$ and the remaining terms go to zero, so the algebraic multiplicity of f on 0 is +4. I was erroneously thinking they would always all go to zero. And I think I was confusing (iii) with (ii). $f' = 4z^3\Delta + ...$ $f'' = \frac{12z^2\Delta^2}{2!} + ...$ $f''' = \frac{24z\Delta^3}{3!} + ...$ $f'''' = \Delta^4 + 0 + ...$ Gary Gary Applying the Taylor series for $\sin z$ we find that $sin z^4=z^4-z^{12}/3!+z^{20}/5!-z^{28}/7!+...$, so not all the derivatives of $\sin z^4$ are zero at $z=0$. Vasco Vasco, Yes, that was what I was illustrating with $f'''' = \Delta^4 + ..."$ above. Gary |
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Gary
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Post by Gary on Mar 25, 2017 15:31:29 GMT
Vasco, I blame it on an extra glass of wine. I think it is fixed now. Gary Gary I'm envious but it's too early in the morning here. What do you get when you calculate the topological multiplicity from the Taylor series, rather than looking at the diagram? Vasco Vasco, If you are referring to (iii) at critical point (a), I get 4. Gary |
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Gary
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Post by Gary on Mar 25, 2017 15:41:23 GMT
Vasco,
On reading your document, I realized that I failed to consider the repeats generated by the fourth root of the argument of the sine.
Is it possible to have so many winds about p when one can only count 8 or 9 intersections of a ray with the loop?
Gary
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Post by Admin on Mar 25, 2017 16:26:02 GMT
Vasco, On reading your document, I realized that I failed to consider the repeats generated by the fourth root of the argument of the sine. Is it possible to have so many winds about p when one can only count 8 or 9 intersections of a ray with the loop? Gary Gary Don't forget that the loop is periodic with period $\pi/2$ and so the same loop is traversed 4 times and so the number of intersections with a ray in the graph will look like the actual number/4 Vasco |
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Post by Admin on Mar 25, 2017 16:48:06 GMT
Gary
I think that with complex-valued equations that can have multiple solutions and with multifunctions, you need to make sure you don't miss any solutions. Even though it is clear that a complex equation like $\cos z=1$ has some real solutions, you need to make sure you pick up any complex solutions, as in part (iii) where a real number to the power $(1/4)$ has two imaginary roots, as well as 2 real roots.
Vasco
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