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May 2

Analytic function

Can any one please help me with the following question:

Does there exist an analytic function f on C such that:

f(1/2n) = f ( 1/(2n+1) ) = 1/2n for all n>=1.

Much appreciated--Shahab (talk) 09:21, 2 May 2008 (UTC)[reply]

I'm not sure if I should just give the answer, but the fact that the zeros of an analytic function are isolated may be useful. Algebraist 10:32, 2 May 2008 (UTC)[reply]

This is my attempt:Suppose an analytic function satisfying the above conditions exist. Then f is continuous. As

{\frac {1}{2n}}\rightarrow 0

so

f({\frac {1}{2n}})={\frac {1}{2n}}\rightarrow f(0)

. But then

f(0)=0

. Also,

f({\frac {1}{2n+1}})={\frac {1}{2n}}\rightarrow f(0)=0

. If

\epsilon >0

choose n so that

{\frac {1}{2n+1}}<\epsilon <{\frac {1}{2n}}

. I now want to show that

f(\epsilon )=0

. ~~If f was real valued it is obvious but as f is complex valued so~~I am not sure how to proceed.--Shahab (talk) 11:15, 2 May 2008 (UTC)[reply]

The point is that an analytic function is determined by its values on any set with an accumulation point (this is essentially what the isolated zeros principle says). Thus f is determined by its values on the set {1/2n|n in N}. Have you been taught (a form of) the isolated zeros principle? If not, I would call this an unfair question. Algebraist 12:17, 2 May 2008 (UTC)[reply]

I'm confused. I can see this is true if the definition of f holds for all

n\in \mathbb {R}

, but you imply it is also true if the definition applies only for all

n\in \mathbb {N}

. In that case the domain we are talking about is then a set of isolated points, is it not? Wouldn't that mean there are no accumulation points and such a proof cannot hold? --Prestidigitator (talk) 21:36, 2 May 2008 (UTC)[reply]

The domain is the set of complex numbers. f has only been defined on a restricted set of 1/2n and 1/2n+1. The question is that can there be an analytic f defined on whole of C so that whatever its values on other complex numbers are, they are 1/2n on the given set.--Shahab (talk) 06:09, 3 May 2008 (UTC)[reply]

Perhaps we should mention that the identity theorem is useful for such questions. –King Bee ^{(τ • γ)} 15:01, 2 May 2008 (UTC)[reply]

Thank you both. The identity theorem artice told me (if I understood the article properly), says that if there are two analytic functions and the set where they agree possesses an accumulation point in the domain then they are identical. That's fine but the original question was does there exist an analytic function or not. It didn't ask whether the function is unique or not?--Shahab (talk) 16:15, 2 May 2008 (UTC)[reply]

Assume there is a function f such that f(1/2n) = 1/2n for all n >=1. The function g(z) = z has this property as well. What does the identity theorem tell you about this? –King Bee ^{(τ • γ)} 18:11, 2 May 2008 (UTC)[reply]

That f is the identity which cannot be since f ( 1/(2n+1) ) = 1/2n. So there exists no such function. Thank you.--Shahab (talk) 19:12, 2 May 2008 (UTC)[reply]

Thanks for the identity theorem link; I knew we had an article on the result I know as the isolated zeros principle, but I'd completely forgotten where it was hiding. Algebraist 20:57, 2 May 2008 (UTC)[reply]

GMAT score as percentile

Many years ago I did a GMAT test, and I do remember my score was at the 86th. percentile. What would my actual score have been please? Although I did go to business school in the UK and got a MBA, I'm just curious if I had a good enough score to have gone to the best known business schools mentioned in the GMAT article. When I did the test, years before the internet was invented, I had no idea of how good my score was compared with any entrance requirements.

On a side note, I am British and during the test I was rather startled by having to answer questions about the grammar of American-English. So I tried to guess what a typical (stereotypical movie) American would say. Subsequently, I have found out that the grammar of British-English and American-English are either the same or very similar. If I had known that at the time I believe I would have got a higher score. So my second question is, how much difference would getting one (or two, or three) questions right or wrong make to the percentile score? I imagine that towards the tails of ability, the wrong answer to one question makes much more difference to the percentile score than it does nearer the average. Thanks. —Preceding unsigned comment added by 80.0.108.213 (talk) 09:43, 2 May 2008 (UTC)[reply]

Whatever your score was, it was better then 86% of the rest of the candidates.--Shahab (talk) 10:09, 2 May 2008 (UTC)[reply]

Also, answering one or two questions more might or might not affect your percentile score. It depends on the rest of the candidates.--Shahab (talk) 10:11, 2 May 2008 (UTC)[reply]

86th percentile is a z-score around 0.99 and from our GMAT article the mean for 2005/2006 is 533 with a standard deviation of around 100. So your score would be around 632.--droptone (talk) 11:38, 2 May 2008 (UTC)[reply]

Correction, you said 86th percentile and for some reason I looked up the 84th percentile. Your score in the 86th percentile (z-score ~1.08) is around 641. Sorry for the error.--droptone (talk) 11:41, 2 May 2008 (UTC)[reply]

Thanks, how do you get from the percentile score to the z score please? 80.2.197.130 (talk) 21:46, 2 May 2008 (UTC)[reply]

Since these sorts of tests are assumed to be normally distributed: you look at the percentile score (86th percentile), check your handy unit normal table for the z-score which is associated with 100 minus the Percentile (or above Z in the link provided), and this figure will be used to find the proportion in the tail (or above Z). So for this example, it is the 86th percentile so we need to look for 14% (or .14) in the tail (or above the z) which corresponds roughly to a z-score of 1.08.--droptone (talk) 22:12, 2 May 2008 (UTC)--[reply]

Derivation of

I asked this earlier, but I think I didn't put the question in the right form to catch the interest of people who normally watch the math reference desk.

Given a system of linear differential equations:

${\frac {d\mathbf {x} }{dt}}=\mathbf {Ax} +\mathbf {Bu}$

with initial condition $\mathbf {x} =\mathbf {0}$ , we want to solve the problem

$\min _{\mathbf {u} }\int _{0}^{\tau }\|\mathbf {u} (t)\|_{2}^{2}dt$

with the constraint

$\mathbf {x} (\tau )=\mathbf {x} _{0}$ .

It turns out that this minimum value is:

$\mathbf {x} _{0}^{T}\mathbf {W} _{c}^{-1}\mathbf {x} _{0}$

where

$\mathbf {W} _{c}=\int _{0}^{\infty }e^{\mathbf {A} t}\mathbf {BB} ^{T}e^{\mathbf {A^{T}} t}dt$

and I've used the matrix exponential. but I don't know how to derive it. Any advice?

moink (talk) 12:34, 2 May 2008 (UTC)[reply]

Okay, I tried to attack this with standard calculus of variations and didn't get very far, because I couldn't figure out how to put in the constraints. I kept getting

\mathbf {u} =\mathbf {0}

, which is obviously wrong. I think the answer must have something to do with Pontryagin's minimum principle or the Hamilton-Jacobi-Bellman equation, and I would be very interested in a complete answer to this question. —Keenan Pepper 04:44, 3 May 2008 (UTC)[reply]

This site doesn't answer my question but does at least show that you can use

\mathbf {W} _{c}

(called Q there) to get a value of u to meet the boundary conditions. There's a wikibook on this topic but it states the equations without derivation. This Stanford lecture note may have a derivation starting on 16-20; still reading it to see if it skips any steps. One thing I've noticed is that all of these sites seem to put different limits on the integral in the equation for

\mathbf {W} _{c}

, so the ones I put may not be right. moink (talk) 11:17, 3 May 2008 (UTC)[reply]

I think that lecture note has a derivation that's good enough for me. Also, in this formulation of the problem, I'm pretty sure the equation should actually be:

\mathbf {W} _{c}=\int _{0}^{\tau }e^{\mathbf {A} t}\mathbf {BB} ^{T}e^{\mathbf {A^{T}} t}dt

and that the improper integral is used to study the properties of the system more generally. moink (talk) 11:31, 3 May 2008 (UTC)[reply]

Is the Set theory useful for anything practical?

I'm not "dissing" the theory here. I don't mind if the set theory is useful for purely theoretical purposes. From what I've seen, the set theory is used mainly for defining and proving various other math concepts down to the most basic elements. —Preceding unsigned comment added by 199.76.153.227 (talk) 15:22, 2 May 2008 (UTC)[reply]

Most things in maths are defined in terms of sets. Whether you consider that "practical" or not is a matter of semantics, I guess. For example, a group is defined as a set with an operation satisfying certain axioms - you couldn't have group theory without set theory, and group theory certainly has practical applications. --Tango (talk) 18:02, 2 May 2008 (UTC)[reply]

Well, you could have group theory without set theory -- the fact that groups are standardly defined as certain kinds of sets, today, doesn't mean they have to be. What set theory provides is a common framework, making it easier to apply techniques from one branch of mathematics to problems from a different branch. I think of it as sort of "the operating system of mathematics". Different operating systems are certainly possible, but this one works pretty well, and has intrinsic attractions as well.

Those who study set theory, of course, are rarely thinking about the fact (even though it's arguably true) that their work may give theoretical clarity to some aspect of some field of pure mathematics that may have applications in some other field of pure mathematics that may have applications in very-slightly-applied mathematics that ... eventually makes a minuscule contribution to building some improved "smart flange" in the year 2721. --Trovatore (talk) 18:12, 2 May 2008 (UTC)[reply]

Well, you don't have to explicitly mention sets, but if you have group theory and just strip out all the stuff about operations, you'll find you've suddenly invented set theory. You wouldn't necessarily end up ZF set theory, but you would have something basic theory about sets. --Tango (talk) 19:32, 2 May 2008 (UTC)[reply]

I don't think it's useful. It was really created to formalize mathematics after some paradoxes about sets were discovered and people wondered how to solve them. – b_jonas 18:16, 2 May 2008 (UTC)[reply]

That's false history that a lot of people have unfortunately heard. No, set theory was started by Cantor, before the paradoxes were discovered, and the initial motivations came from real analysis. The paradoxes don't really have anything to do with formalization versus not-formalization; they have to do with getting the right intuitive notion rather than some wrong one. --Trovatore (talk) 18:24, 2 May 2008 (UTC)[reply]

I view math as an upside down pyramid. At the very top there are all the applications – the math that is “practical”. However, that builds on the other math, such as set theory, which is much closer to the base than most things you’ll find “practical”. Without it, much of the “practical” math would become more difficult, if even possible.

On another note, what is practical? I’d include being a research or teaching professor of set theory practical. Thus it is useful for making a living for the few people lucky enough to be a professor of set theory. GromXXVII (talk) 19:21, 2 May 2008 (UTC)[reply]

I guess it's arguable whether it's "practical" itself, but programming language theory uses set theory now and then, among various other mathematical formalisms. Although still a kind of theory, PLs theory does have a fairly close connection to at least some kinds of programming-language design, and therefore practical implications. --Delirium (talk) 13:42, 3 May 2008 (UTC)[reply]

Set theory is also used in program specification and verification, which may be useful in keeping your aeroplane from dropping out of the sky because of an overlooked race condition that did not come out in extensive testing. It is also used in the theory of relational databases. Neither is "deep" use in any sense, it is mainly elementary, "naive" set theory. --Lambiam 22:15, 3 May 2008 (UTC)[reply]