Gary
GaryVasco
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Post by Gary on Apr 2, 2017 21:13:48 GMT
Vasco, I see you have jumped ahead and are gradually returning to the unfinished Exercise 4. I have been working on it, but I have hit a rough spot in (ii). Explain why the particular shape is consistent with a critical point of order 1.From p. 205, a critical point of order 1 would be a critical point for a function with a power of 2. Near the critical point, angles would be doubled. Since the tangent of a circle makes an angle of Pi/2 with a ray, this would be doubled to Pi, which would make it appear that the f(z) curve reversed its direction at the critical point. That is consistent with the shape that Needham presented. It is not entirely consistent with the shapes that we see in Figure 2 (<- dropbox link). In 2(a) for example, we see a crimp of ~ Pi/2 and in 2(b) and 2(c) we see loops. The first nonzero derivative of f(z) is $f^{(1)}$, so it does not appear that $a$ is a critical point. $f^{(1)}(z) = 3 z^2-(2 a+2 b+2 c) z+(a b+a c+b c)$ $f^{(1)}(a) = a^2-a b-a c+b c$ Since a, b, and c produce analogous first derivatives of f, the function appears to have no critical points. From p. 205, we learn that a critical point has order m-1. The example makes it appear that m is the order of the polynomial, so the order of a critical point of f(z) would be 3-1 = 2. I don't understand why Needham has asked about a critical point of order 1 at this point. For each of three p-points $\nu() = 1$ because f is analytic, but that irrelevant to the question of whether a, b, or c is a critical point. Do you have an interpretation of this yet? Gary |
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Post by Admin on Apr 3, 2017 6:26:52 GMT
Gary
I'm working on this one at the moment.
Here are some hints:
The points $a,b,c$ are the zeros of $f$, so that $f(a)=0$, $f(b)=0$, $f(c)=0$.
The critical points are the points where $f'(z)=0$. So you have to solve this equation $(f'(z)=0)$ to find the critical points.
Critical points are not typically the same points as the roots.
The zero $a$ is also a critical point only if $f=(z-a)^n\Omega(z)$ for $n>1$.
I hope these will help you to make progress.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 3, 2017 15:36:40 GMT
Vasco,
That looks like the information I was seeking. I started to solve for a, but I had the impression that if a was a critical point, all values of a were critical points, so I dropped that approach. Judging from the question, I am guessing that the value of a depends on b and c (I think I calculated (b/c-1)) and will probably fall on the point of the crimp.
Gary
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Post by Admin on Apr 3, 2017 16:08:52 GMT
Vasco, That looks like the information I was seeking. I started to solve for a, but I had the impression that if a was a critical point, all values of a were critical points, so I dropped that approach. Judging from the question, I am guessing that the value of a depends on b and c (I think I calculated (b/c-1)) and will probably fall on the point of the crimp. Gary Gary $a,b,c$ are not dependent on each other. Since $f'$ is a quadratic we have two roots. If we set $f'=0$ and solve, then the roots of the quadratic are the critical points of $f$, the cubic. Are you using $a$ to mean two different things. I am a bit confused by your post. Just realised that there is potential for confusion because Needham (on page 346) uses $a$ to signify the $p$-point of a general analytic function $f(z)$ and in the exercise $a$ is a $p$-point of the cubic as a preimage of $0$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 3, 2017 19:49:38 GMT
Vasco, That looks like the information I was seeking. I started to solve for a, but I had the impression that if a was a critical point, all values of a were critical points, so I dropped that approach. Judging from the question, I am guessing that the value of a depends on b and c (I think I calculated (b/c-1)) and will probably fall on the point of the crimp. Gary Gary $a,b,c$ are not dependent on each other. Since $f'$ is a quadratic we have two roots. If we set $f'=0$ and solve, then the roots of the quadratic are the critical points of $f$, the cubic. Are you using $a$ to mean two different things. I am a bit confused by your post. Just realised that there is potential for confusion because Needham (on page 346) uses $a$ to signify the $p$-point of a general analytic function $f(z)$ and in the exercise $a$ is a $p$-point of the cubic as a preimage of $0$. Vasco Vasco, Thanks, I think I am oriented now. Gary |
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Gary
GaryVasco
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Post by Gary on Apr 8, 2017 22:52:17 GMT
Vasco, I could get no further than (iii) with Exercise 4. Here is what I have: nh.ch7.ex4.pdf (309.29 KB) Gary |
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Post by telemeter on Mar 7, 2020 19:39:14 GMT
Hi Vasco, Gary
I apologise if I'm telling you what you've already surmised!
On part iii) I think you are close to a proof. If you consider the transformation you propose to be to that of an equilateral triangle then the ellipse must become o circle because the ellipse is centred at the centroid from your diagram Figure 3. As there is no other ellipse-like figure centred on the centroid of an equilateral triangle and touching the midpoints, then the ellipse must be unique. ie if you reverse the transformation you can only regain your original ellipse and no other. On reflection, this proves that if there is such an ellipse then it must be centred at the centroid which is helpful for what follows.
On part iv) it seems you have got a large part of the way because your expression for the critical points is (a+b+c)/3 +/- sqrt( )
ie you have shown them to be symmetric about the centre of the ellipse which is the centroid (a+b+c)/3.
If we work backwards and assume critical points are the foci then your sqrt expression is the normal relationship between the semi-major and semi-minor axis lengths (say, q and r)of the ellipse.
Let the foci p1, p2 be at +/-p from the centre then p^2 = q^2 - r^2 = (a^2+b^2+c^2 -(ab+ac+bc))/9. (It is suggestive is it not to see the semi-major axis length as sqrt(a^2 + b^2 + c^2)/3...but we don't need to guess that...)
We also know that the sum of the distances from the mid-points of a,b,c to the foci are constant on the ellipse ie mod(AB-p1)+mod(AB-p2)=mod(BC-p1)+mod(BC-p2)=mod(AC-p1)+mod(AC-p2)=2q. So all that is needed for the proof is to crunch the algebra and show that these relations do indeed work for your critical points.
Just thinking how best to do that.
Telemeter
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Post by telemeter on Mar 8, 2020 10:39:02 GMT
Hi again,
Thinking about this overnight, moving the centroid to the origin via a translation (of - (a+b+c)/3) simplifies the position of the critical points a bit (looking at Vasco's first expression for z' on p3 of his notes) to
z'=+/-sqrt(-(a'b'+a'c'+b'c')/3) (using ' to deonte translated points)
We can remove the - in the sqrt by roatating axes by by pi/2 to give
z'=+/-sqrt((a'b'+a'c'+b'c')/3)
which looks as if it should be tractable. So what we need is mod(c'+ sqrt((a'b'+a'c'+b'c')/3)) + mod(c'-sqrt((a'b'+a'c'+b'c')/3)). Time for another think!
Having moved the centroid we now have a'+b'+c' = 0 which may be useful.
(Meanwhile I note that moving the centroid to the origin removes the squared term in the original cubic as in Ex 1 ii) on p45, and the centroid is a critical point of f'. Although the significance of this escapes me)
Further thoughts....
This is useful because the original equation becomes
f(z')= z'^3 +(a'b'+b'a'+a'c')z'-a'b'c'
Telemeter
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Post by telemeter on Mar 8, 2020 11:07:38 GMT
On part ii) you discuss above (and in Vasco's note) why Needham talks about a critical point of order 1 when f(z) is a cubic.
What I think he is getting at is that when you consider the behaviour of f(z) near critical point a, it is natural to use a Taylor expansion. If you do, the first term in the series (that is not a constant) is the second derivative or higher. That is, to see the behaviour near the critical point we look at f(z-a)-f(a) = f''(a) (z-a)^2/2 +.... . Here, the behaviour will thus be dominated by (z-a)^2 (as (z-a) is small), and the critical point is of order 1.
So, the behaviour is not determined by the power of the original function but by its first non-zero derivative at the critical point.
When I read p205 the first time around I was not entirely clear what he was getting at and thought that the order of the critical point was to do with the polynomial power, and so when I first looked at this question I too was puzzled...however, the discussion of algebraic multiplicity from the bottom of p346 to the end of the section on p347 I think illuminates the meaning of his p 205 remarks and justifies the above. In the p205 section it was just an unhelpful 'coincidence' that the example he chose, z^2 has a critical point of order 1.
Regards
Telemeter
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 8, 2020 14:56:50 GMT
telemeter,
Thank you for the contributions. I will follow up and I'm fairly certain that Vasco will, but it may take me a couple of days to reorient to chapter 7.
Gary
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Post by Admin on Mar 8, 2020 16:40:09 GMT
Hi again, Thinking about this overnight, moving the centroid to the origin via a translation (of - (a+b+c)/3) simplifies the position of the critical points a bit (looking at Vasco's first expression for z' on p3 of his notes) to Telemeter Which of my notes are you referring to here? Thanks Vasco |
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Post by telemeter on Mar 8, 2020 17:47:31 GMT
Hi The note I refer to is nh.ch7.ex4.pdf which is linked to above in your post of Apr 8, 2017 at 11:52pm. I've found that the ellipse result is standard (Marden's Theorem) and follows immediately from the expressions in my above post, in that if you have a cubic function whose roots are say, a,b,c then when expressed in the form f(z) = z^3 +pz +q , then the foci of the "inellipse" are at +/- sqrt(-p/3). Exactly what your result ties into as per my earlier post. If you are prepared to accept Marden's theorem www.su.se/polopoly_fs/1.229312.1426783194!/menu/standard/file/marden.pdf then I think we have proved what was needed. It took a fair bit of blind kludging around and I couldn't have gone any further than the serendipitous simplifications I got from translating to the origin ...so finding this doc was a godsend. Do I feel a bit like I've "cheated" by relying on Marden for the final bit...yes I guess so. Let me know your view. Anyway a fun day on Q4...onto Q5! Regards Telemeter |
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Post by Admin on Mar 8, 2020 19:08:19 GMT
telemeter
That is not my note. It is Gary's, and so is the post.
Vasco
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Post by telemeter on Mar 9, 2020 17:42:01 GMT
Ah so it is. Apologies to both of you for over and under attribution respectively.
telemeter
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 11, 2020 18:36:28 GMT
Hi The note I refer to is nh.ch7.ex4.pdf which is linked to above in your post of Apr 8, 2017 at 11:52pm. I've found that the ellipse result is standard (Marden's Theorem) and follows immediately from the expressions in my above post, in that if you have a cubic function whose roots are say, a,b,c then when expressed in the form f(z) = z^3 +pz +q , then the foci of the "inellipse" are at +/- sqrt(-p/3). Exactly what your result ties into as per my earlier post. If you are prepared to accept Marden's theorem www.su.se/polopoly_fs/1.229312.1426783194!/menu/standard/file/marden.pdf then I think we have proved what was needed. It took a fair bit of blind kludging around and I couldn't have gone any further than the serendipitous simplifications I got from translating to the origin ...so finding this doc was a godsend. Do I feel a bit like I've "cheated" by relying on Marden for the final bit...yes I guess so. Let me know your view. Anyway a fun day on Q4...onto Q5! Regards Telemeter telemeter,
I have just gotten reoriented to the section, and what you say all sounds reasonable, so I will try to work through it and I will take a close look at your page on Carlson's proof. The first time through I was totally out of my depth, but it looks slightly less daunting at a distance of three years.
What you say about my confusion of the order of $P$ vrs. $P^{(1)}$ also sounds reasonable.
Gary
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