<p>Actually in hindsight thats fairly rigorous…the end needs a little more explination…but eh</p>
<p>Is it clear that this function is continuous?</p>
<p>…Mmmm…making a small change…where I have f(b)-f(a)/2>x
make that >x for all x in l,m…and eh…I am fairly sure its continuous…but im not exactly sure, I guess I was kind of taking that for granted…and wow fast! Mm…ill write up something…better…in LaTeX tommorow and make a pretty post script file, however right now it is 2:44…and I slept for three hours last night (Number theory problem set…which I thought was due next week was actually due today[EPGY])…soooooo sleep time.</p>
<p>: ) good luck.</p>
<p>Well…<a href=“http://www.wired-designs.net/localMaxima.pdf[/url]”>www.wired-designs.net/localMaxima.pdf</a> is a little better…I suppose its still not rigorously continuous…mmm…once I have some caffeine in my system I will do a little more…its become a bit of a personal vendetta</p>
<p>It is not clear that the set of local maxima is indeed the Cantor set, nor do I understand the sentence “create a continuous function m = stuff, n = stuff and such that f(m) = f(n) = (f(b)−f(a))/2 > x for all x in (m, n)” (i.e. the sentence is nonsense as written).</p>
<p>Hint: it is hard to prove something false :)</p>
<p>Are you sure it isnt true?..Eh I will rewrite it in a bit…I am not wonderful with mathematical writing…well writing in general, I have difficulties putting down what I am thinking of…mm…again basically the idea of that sentence is instead of an ititeration of the devils staircase being flat (constant) we create a function such that each step is a parabola with the endpoints being local maxima…eh I can work on it a bit after I finish school (online school). However I think it is very clear that it is the cantor set, as the cantor set operates on removing the middle thirds of the interval [0,1] and continue ad infinitum, the cantor set consists of all points not removed. In this case we have turned the middle third of an interval (a,b) into a parabola such that the endpoints are local maxima so the endpoints would be the points which would be left after a complete runthrough of the cantor set. <a href=“http://en.wikipedia.org/wiki/Cantor_set[/url]”>http://en.wikipedia.org/wiki/Cantor_set</a> also, it is possible we have a confusion on the definition of the devils staircase…Wikipedia and google both tell me different things for that, so I am guessing that there are multiple definitions.</p>
<p>Maybe a better summary would almost be that…you are constructing the cantor set from the devils staircase by turning the middle third into a parabola such that the end points are strict local maxima and turning the middle third of the two parts left into a parabola again such that the endpoints are local maxima and continuing ad infinitum.</p>
<p>Is every point in the Cantor set the endpoint of some interval removed during the construction? It seems you would have to show at least that, and I don’t think it’s true.</p>
<p>See if you can give the other way a whack – try to prove no such function exists. Recall that a strict local maximum is a point x such there is some interval (a,b) containing x with f(x) > f(y) for all y in (a,b).</p>
<p>Agh! what a horrible oversight on my part, consider x=1/4, part of the cantor set but not of the construction because as you remove thirds…eh oh well…and I definetly wasnt considering that a proof as much as…a heuristic argument I am fairily sure for it to be rigorous I would have to show that but eh…I am not certain it is true, clearly because the cantor set removes open sets, (1/3,2/3) from [0,1] etc etc…so the endpoints are obviously there but again something like 1/4 is part of the cantor set yet not covered in the construction…Oh well it was an interesting attempt. My original thought was actually no there was not…as far as proving that it does not exist…</p>
<p>Statement: Given a finite function of a Real Variable F, the set of points at which it assumes a strict local maxima are at most enumerable.</p>
<p>Proof: Consider the set A for the points at which the the function F a ssumes a strict local maximum, Let a sub n denote for all positive integers n the set of points such that, F(j)<F(x) holds for each points J not equal to X on the interval, ((x-1)/2,(x+1)/2). Clearly the set A sub n is isolated and therefore at most enumerable. However we know by definition that A = Sigma_n A sub N, it follows that the set A is at most enumerable.</p>
<p><a href=“http://www.wired-designs.net/localExtrema.pdf[/url]”>http://www.wired-designs.net/localExtrema.pdf</a> …All latexified so a bit easier to read…mmm I feel rather silly now</p>
<p>I am quite new to calculus, but can’t help posting…</p>
<p>Isn’t f(x) = x sine (1/x) where x belongs to (0,1]
and f(0)=0
such a function?</p>
<p>Definition: thread hijacking :p</p>
<p>Nope Sarang, it does “oscillate” (I use quotes because it doesnt technically oscilate between two values) for a while, but its not uncountable at all you could verify this fairly quickly by looking at the derivative, or just a graph (though having the derivative might help) Actually a day or so after posting <a href=“http://www.wired-designs.net/localExtrema.pdf[/url]”>http://www.wired-designs.net/localExtrema.pdf</a> (2 replies back) I found a similar proof of the same in a book on integration (Actually it was cited as well known and that a proof appeared in said book from a mathematics article I was reading so I found a pdf of the book).</p>
<p>How about just sin(1/x) in the interval (0,1]</p>
<p>only countably many zeroes. easy to place them into bijection with the integers. infinite is not enough. uncountable is needed.</p>
<p>Robb – sorry I didn’t respond right away. It seems like you sort of have the right idea, but your proof seems to assume you can isolate each maximum inside its own open interval, and this need not be possible… The idea is to cover the maxima by intervals (possibly overlapping) with rational endpoints, and there are only countably many of those.</p>
<p>By the time you get to college, make sure to polish up your mathematical writing. It’s no use understanding it if you can’t explain it lucidly to others.</p>
<p>Yeah the mathematical writing is a bit of an issue for me. The number theory course I am taking through EPGY is pretty writing intensive so that is helping a good bit (The problem sets have been ranging from 12-16 pages…not horribly long but its a good bit of mathematical writing. I always find it funny the largest amount of time is not spent on the proof itself, or the idea, but going back and making the flow alot more coherent…eh it is really a bad habit at this point. I have 3 notebooks on my desk full of random mathematical proofs/ideas and it is incredibly disorganized and cluttered. Previously it had been something that I just do for myself so when attempting to mathematically communicate I often end up assuming everyone follows my thought process completely without taking time to explain fully out of not being accustomed to it. Meh its something I need to and am working on. </p>
<p>I had considered the rational endpoint idea very briefly actually, however at the time (Around 3:30 am) it did not seem like a viable approach for whatever reason.</p>
<p>Thank you for the advice.</p>
<p>Basically bringing this up; it has some pretty good advice about the interview process. Found it on good 'ol google. Here’s the link to Stu Schmill’s blog entry on the new blog site: <a href=“http://www.mitadmissions.org/topics/apply/interviews_educational_counselors_ecs/advice_on_how_to_approach_your.shtml[/url]”>http://www.mitadmissions.org/topics/apply/interviews_educational_counselors_ecs/advice_on_how_to_approach_your.shtml</a></p>
<p>Good luck, everyone! My interview is on Thursday. ^^;</p>