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Topic: Analytic problem (Read 10106 times) |
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Felix.R
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Analytic problem
« on: Jun 12th, 2007, 2:32pm » |
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Suppose g is analytic on abs(z)<=1 and let abs(g(z)) be maximized for abs(z)<=1 at z0 with abs(z0)=1. Prove that g'(z0) is not zero unless g is constant.
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Michael Dagg
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Re: Analytic problem
« Reply #1 on: Jun 13th, 2007, 9:52am » |
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You should be able to solve this using Schwarz. Notice that
|g(z)| < 1 in |z| < 1 by the maximum principle.
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Michael Dagg
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Felix.R
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Re: Analytic problem
« Reply #2 on: Jun 15th, 2007, 10:51pm » |
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Is it because the function g is not constant that the maximum principle gives the statement you made? Actually this is not that easy to see for one thing. I think. I am not sure what to do with Schwarz's lemma because it does not appear to lead me to a function composed with g that I can work with. So I don't understand how I can solve it using Schwarz. Explain if you will.
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Michael Dagg
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Re: Analytic problem
« Reply #3 on: Jun 17th, 2007, 8:57pm » |
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Yes. You are at liberty to make some assumptions here.
For example, you may assume that g(z0) = z0, z0 = 1.
Noting that |g(z)| < 1 in |z| < 1, and so if we take
g(0) = 0 then by Schwarz we have |g(z)| <= |z|, |z| < 1.
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| « Last Edit: Jun 17th, 2007, 9:00pm by Michael Dagg » |
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Michael Dagg
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Felix.R
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Re: Analytic problem
« Reply #4 on: Jun 18th, 2007, 6:36pm » |
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I do not follow all this. I understand assuming g(1) = 1 and g(0) = 0 but doesn't this require that we also assume that g(0) not zero? Also, I can't use the bar character for absolute value I guess because my language pack seems to display character codes on this web site for that symbol. I have no idea why.
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| « Last Edit: Jun 18th, 2007, 6:37pm by Felix.R » |
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Michael Dagg
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Re: Analytic problem
« Reply #5 on: Jun 18th, 2007, 7:07pm » |
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I just lost a fair amount of typed text here. I am
certainly annoyed by expired pages messages.
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Michael Dagg
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Michael Dagg
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Re: Analytic problem
« Reply #6 on: Jul 10th, 2007, 9:58am » |
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Did you solve this problem?
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Michael Dagg
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Felix.R
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Re: Analytic problem
« Reply #7 on: Jul 11th, 2007, 9:32am » |
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Alas, only partially to a degree with some missing pieces using a Taylor series. We appreciate help!
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Michael Dagg
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Re: Analytic problem
« Reply #8 on: Jul 30th, 2007, 6:56pm » |
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That leads me to guess what you mean but you might be
on to something. Is that series that of g(z) - g(z0) ?
See what you can do with this:
Show that |g'(1)| >= 1 and take g(0) = a, a <> 0. Now
consider
h(z) = (g(z) - a)/(1 - a bar g(z)).
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Michael Dagg
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Michael Dagg
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Re: Analytic problem
« Reply #9 on: Sep 15th, 2007, 8:34am » |
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Did you try to work with this?
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Michael Dagg
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Michael Dagg
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Re: Analytic problem
« Reply #10 on: Sep 29th, 2007, 7:16pm » |
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What happened to you on this problem?
This is a nice problem by the way -- not too hard
but interesting, but maybe you solved it and
forgot about us?
I would like to see what you came up with.
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| « Last Edit: Sep 29th, 2007, 7:30pm by Michael Dagg » |
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Michael Dagg
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Sameer
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Pie = pi * e
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Re: Analytic problem
« Reply #11 on: Sep 29th, 2007, 7:21pm » |
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Yea seems like he is MIA..
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Ghost Sniper
Senior Riddler
Do not hide. It is useless.
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Re: Analytic problem
« Reply #12 on: Dec 13th, 2007, 10:51am » |
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[redacted] im right
// excuse me? --SMQ
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| « Last Edit: Dec 13th, 2007, 11:06am by SMQ » |
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*sob* I miss my mommy... *blows nose* huh, I'm on? oh right.
(thinks to self) Time for my speech to these college kids.
"Reason is more important than all emotions..."
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Michael Dagg
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Re: Analytic problem
« Reply #13 on: Jan 23rd, 2008, 11:35pm » |
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>> [redacted] im right
>> excuse me? --SMQ
Being curious, I had wondered what this exchanged
was referring to but apparently I had miss it at some
point -- perhaps Ghost had proposed solution but
took it back (?).
Still, a pretty cool problem, and in spite of my remarks it has
gone unsolved -- on this forum nevertheless.
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Michael Dagg
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Eigenray
wu::riddles Moderator
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Re: Analytic problem
« Reply #14 on: Mar 14th, 2008, 6:43pm » |
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I don't think you need anything but the power series expansion. We can assume z0=1, and g(1)=1 (if g is not identically 0). Then for some n  1,
g(1+w) - 1 = wn (C + O(w)),
where C = g(n)(1)/n!  0. Writing w=reit, then for t fixed,
Arg [ g(1+w) - 1 ]  nt + Arg[C]
as r 0. If n 2, then for some integer k, we can take t = (2 k-Arg[C])/n  (/2, 3/2). Then Arg[g(1+w)-1] 0, and therefore for r sufficiently small, |g(1+w)| > 1, with 1+w in the unit disk.
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Michael Dagg
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Re: Analytic problem
« Reply #15 on: Mar 21st, 2008, 1:59pm » |
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Geometrically nice!
I think your proof could be made to show that the
derivative at 1 is actually positive (not just non-zero)!
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Michael Dagg
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Eigenray
wu::riddles Moderator
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Re: Analytic problem
« Reply #16 on: Mar 21st, 2008, 2:50pm » |
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Actually, we can think of it like a Lagrange multiplier problem:
 |f|2(z0) =  z0 for some 0,
since otherwise we could increase |f| by moving along the unit circle (or in the -z0 direction, if < 0). Since
|f|2 = 2f bar[f'],
we get that
f'(z0) = t z0/bar[f(z0)]
for some t 0, but we need analycity (analyticality?) of f to get t 0.
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TJMann
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Re: Analytic problem
« Reply #17 on: Mar 22nd, 2008, 11:49pm » |
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Pardon, but how is |f|2(z0) read? I have never seen del used with complex functions before. Also, I thought analyticity is pre-supposed if you write f'(z)?
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Eigenray
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Re: Analytic problem
« Reply #18 on: Mar 23rd, 2008, 12:40am » |
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I'm thinking of h(x,y) = |f(x+iy)|2 = u2 + v2 just as a real-valued function of two real variables. Then h is a vector in  2, which I'm thinking of as the complex number
h =  h/x + ih/y
= (2uux+2vvx) + i(2uuy+2vvy)
= 2(u+iv)(ux+iuy)
= 2f bar[f'],
using ux=vy, uy=-vx. We could also compute
hz = (f bar f)z
= fz (bar f) + f (bar f)z
= f' * bar f,
since bar f is antiholomorphic, and use the fact that h = 2 bar[hz]. I don't know if this is standard notation, though.
My point was just that from the point of view of multivariable calculus, it's clear that h must be perpendicular to the unit circle, but it's not clear that it can't be the zero vector. That is where we need to think of f as a complex analytic function, not just as a function from 2 2.
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| « Last Edit: Mar 23rd, 2008, 12:40am by Eigenray » |
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TJMann
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Re: Analytic problem
« Reply #19 on: Mar 24th, 2008, 9:59am » |
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cool. I never thought of it like that and saves lots of space. In your last paragraph, I don't see why the gradient of h is perpendicular to the unit circle though, I think to the level lines of h it is, {(x,y), h(x,y) = const}
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Eigenray
wu::riddles Moderator
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Re: Analytic problem
« Reply #20 on: Mar 24th, 2008, 1:56pm » |
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This is just because h points in the direction of increasing h. Let v be a unit vector tangent to the unit circle at z0. Then the directional derivative of h in the direction v is just Dv h = <h, v>. If this dot product were positive, then h increases as I move along the unit circle in the v direction, contradicting the fact that h is maximized at z0. Similarly, if it were negative, h would increase in the opposite direction. So we must have <h, v> = 0, i.e., h is perpendicular to the unit circle at z0.
This is a special case of the Lagrange multiplier method. If we want to maximize f given the constraint g=0, then f must be perpendicular to the level curve g=0, so it must be a scalar times g, i.e., f = g.
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TJMann
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Re: Analytic problem
« Reply #21 on: Mar 25th, 2008, 3:55pm » |
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Thx. I see. I thought you were just speaking in general but see that you were referring to the particulars of the given problem.
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Nourhan
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Re: Analytic problem
« Reply #22 on: May 21st, 2014, 5:50pm » |
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You have to use Schwarz Equation !
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