Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on May 9, 2017 23:49:47 GMT
Vasco, Exercise 18 now available for comment: nh.ch7.ex18.pdf (143.12 KB) Gary |
|
|
|
Post by Admin on May 11, 2017 5:03:48 GMT
Gary I have not produced an answer for this exercise yet, so I will not look at yours yet. Vasco |
|
|
|
Post by Admin on May 29, 2017 11:18:25 GMT
|
|
|
|
Post by Admin on May 29, 2017 13:06:46 GMT
Gary
I don't think we can put the origin inside $\Gamma$ for this exercise because we need to have a positive minimum on $\Gamma$. (see section 2 on page 357).
Also, is there a reason why you have chosen to make $\Gamma$ and $f(\Gamma)$ circles?
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 30, 2017 5:44:22 GMT
Gary I don't think we can put the origin inside $\Gamma$ for this exercise because we need to have a positive minimum on $\Gamma$. (see section 2 on page 357). Also, is there a reason why you have chosen to make $\Gamma$ and $f(\Gamma)$ circles? Vasco Vasco, I made them circles, because I have difficulty drawing loops such as those in [14], and how would one know that the $f$ that created them was analytic? So I made $\Gamma$ a circle because it was easy to plot, and I made $f(\Gamma)$ a circle because $f = z sin(z)$ was a convenient analytic function, which happens to produce a circle. I wanted the analytic function, not the circle, but I got both. I don't see why the question requires a positive minimum on $\Gamma$, but I don't understand much of section 2 on p. 357. For example, sentence 1, paragraph 2, is a complete mystery. Why do you think we need a positive minimum on $\Gamma$? Gary |
|
|
|
Post by Admin on May 30, 2017 6:13:16 GMT
Gary I don't think we can put the origin inside $\Gamma$ for this exercise because we need to have a positive minimum on $\Gamma$. (see section 2 on page 357). Also, is there a reason why you have chosen to make $\Gamma$ and $f(\Gamma)$ circles? Vasco Vasco, I made them circles, because I have difficulty drawing loops such as those in [14], and how would one know that the $f$ that created them was analytic? So I made $\Gamma$ a circle because it was easy to plot, and I made $f(\Gamma)$ a circle because $f = z sin(z)$ was a convenient analytic function, which happens to produce a circle. I wanted the analytic function, not the circle, but I got both. I don't see why the question requires a positive minimum on $\Gamma$, but I don't understand much of section 2 on p. 357. For example, sentence 1, paragraph 2, is a complete mystery. Why do you think we need a positive minimum on $\Gamma$? Gary Gary I see what you mean about the circles etc. Sorry, I meant to write "...put the origin inside $f(\Gamma)$..". The exercise is in two parts and the second asks us to find the result for a positive minimum on $f(\Gamma)$. I see the mystery sentence as meaning project $\Gamma$ upwards onto the modular surface and almost literally cut it out the surface from the rest of the surface along the contour produced by projecting $\Gamma$ upwards. Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 30, 2017 7:49:33 GMT
Vasco, I made them circles, because I have difficulty drawing loops such as those in [14], and how would one know that the $f$ that created them was analytic? So I made $\Gamma$ a circle because it was easy to plot, and I made $f(\Gamma)$ a circle because $f = z sin(z)$ was a convenient analytic function, which happens to produce a circle. I wanted the analytic function, not the circle, but I got both. I don't see why the question requires a positive minimum on $\Gamma$, but I don't understand much of section 2 on p. 357. For example, sentence 1, paragraph 2, is a complete mystery. Why do you think we need a positive minimum on $\Gamma$? Gary Gary I see what you mean about the circles etc. Sorry, I meant to write "...put the origin inside $f(\Gamma)$..". The exercise is in two parts and the second asks us to find the result for a positive minimum on $f(\Gamma)$. I see the mystery sentence as meaning project $\Gamma$ upwards onto the modular surface and almost literally cut it out the surface from the rest of the surface along the contour produced by projecting $\Gamma$ upwards. Vasco Vasco, I still don't see the problem. Apparently, you read exercise 18 to require that all $f(z)$ on or within $\Gamma$ must have positive $|f|$. I don't see why the question or the MMT requires that. Oh! You must be referring to the " positive minimum" in the second part, so no point of $f(z)$ can be zero. I wonder why we can't just say "apart from roots". What analytic function moves all of $f(z)$ on $\Gamma$ beyond the origin? Would adding a simple translation to a typical analytical function suffice? But looking at it again, I think my reading of it confines $f$ to $\Gamma$. Thus, there can be maximums and positive minimums on $f(\Gamma)$ even though there may be zero points in the interior. But since I appealed to the MMT in my answer, it seems I am trying to have it both ways. Where are those nice analytic functions with positive minimums when you need them? I read the mystery sentence the same way, but I don't understand the implications of it. Gary |
|
|
|
Post by Admin on May 30, 2017 16:26:25 GMT
Gary
Consider your sentence in red above: There can only be one minimum value and one maximum value, but they could be attained at more than one point on $\Gamma$.
The mystery sentence can be understood by looking at the RHS of figure 14 on page 356 and noting that there is an origin centred circle through $Q$ and another through $T$, so that the max and min lie on the edge of $\Gamma$.
I am changing my solution slightly to allow $f(z)=0$ on $\Gamma$ in the first part of the exercise.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 31, 2017 0:54:10 GMT
Vasco,
Yes, that is what I meant. I forgot that nominalizing minimum or maximum could be construed as either the point (my intent) or the value.
I'm afraid I must be extremely dense. The mystery sentence has to do with cutting out the piece of modular surface lying above $\Gamma$ and its interior, but Figure [14] does not show that the modular surface |f| lies above $\Gamma$. I wonder if it should be changed to read "the piece of the modular surface lying above $f(\Gamma)$". With that modification, it is clear that the minimum and maximum will lie on the edge unless there is an interior 0-point of f. In any case, [14] illustrates why the function with the positive minimum in Exercise 18 can have no interior 0-points. I will look for another function.
Gary
|
|
|
|
Post by Admin on May 31, 2017 4:31:45 GMT
Vasco, Yes, that is what I meant. I forgot that nominalizing minimum or maximum could be construed as either the point (my intent) or the value. I'm afraid I must be extremely dense. The mystery sentence has to do with cutting out the piece of modular surface lying above $\Gamma$ and its interior, but Figure [14] does not show that the modular surface |f| lies above $\Gamma$. I wonder if it should be changed to read "the piece of the modular surface lying above $f(\Gamma)$". With that modification, it is clear that the minimum and maximum will lie on the edge unless there is an interior 0-point of f. In any case, [14] illustrates why the function with the positive minimum in Exercise 18 can have no interior 0-points. I will look for another function. Gary Gary The sentence seems to me to be correct as written. We know that Figure 14 does not represent the modular surface, but the function $f$. Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 31, 2017 5:05:09 GMT
Vasco, Yes, that is what I meant. I forgot that nominalizing minimum or maximum could be construed as either the point (my intent) or the value. I'm afraid I must be extremely dense. The mystery sentence has to do with cutting out the piece of modular surface lying above $\Gamma$ and its interior, but Figure [14] does not show that the modular surface |f| lies above $\Gamma$. I wonder if it should be changed to read "the piece of the modular surface lying above $f(\Gamma)$". With that modification, it is clear that the minimum and maximum will lie on the edge unless there is an interior 0-point of f. In any case, [14] illustrates why the function with the positive minimum in Exercise 18 can have no interior 0-points. I will look for another function. Gary Gary The sentence seems to me to be correct as written. We know that Figure 14 does not represent the modular surface, but the function $f$. Vasco Vasco, The modular surface lies above $f$, and there is no indication that $f$ lies above $\Gamma$ and its interior, so how can one know about a piece of the modular surface lying above $\Gamma$ and its interior? Gary |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 31, 2017 5:11:09 GMT
Vasco,
I have amended my answer to Ex 18 and the plot to provide for a positive minimum. I did it by a simple translation of the original $f$. If the applicability of the Minimum Modulus Theorem depends on a simple translation that puts all of $f(\Gamma)$ and its interior in a position where 0 is exterior, is it really saying something fundamental about the function?
Gary
|
|
|
|
Post by Admin on May 31, 2017 14:05:57 GMT
|
|
|
|
Post by Admin on May 31, 2017 16:01:20 GMT
Gary
I have slightly revised my answer to exercise 18 to make it clearer and to correct a typo.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 31, 2017 16:27:41 GMT
Vasco, That clears up my confusion. I'm not sure when I first made the mistake of thinking of $|f|$ as a quantity plotted above $f$. It even occurred to me that they should all look the same, but I never pursued that thought. I am posting a new plot that I think meets all the criteria, and it's a little more interesting, as it has two maximums. Gary |
|
