You've disproven the Roundamajig Theorem!

Not that I disagree with the conclusion, but do you have a more rigorous proof that the limit curve is simply f_inf(x) = 0 for all x? It's pretty clear that that's the key step in the proof, and you hand-wave past it pretty quickly.

Um, shouldn't the case for floor(nx) congruent to 1 mod 2 be 1/n + (1/n)floor(nx) - x? It doesn't match up otherwise. However, the rest of the math checks out. I think.

Now, when I read "pointwise limit", I immediately thought, "Hey, is this a uniform convergence question?" After some reading, I believe I now understand what's going on here. f_n is actually uniformly convergent to its pointwise limit function f_inf. We see this because the distance between f_n(x) and f_inf(x) is, if I've done it correctly, the {sup(|f_n(x) - f_inf(x)|) over x in [0,1]} = 1/n -> 0 as n -> infinity. So f_n converges uniformly to f_inf = 0. I believe we get the pointwise limit by observing that the maximum of the function is given by 1/n, and as n -> infinity, since f_n is non-negative for all x in [0,1], the result follows by Squeeze Theorem.

Because of this, we get a couple neat properties. Ie, the limit function is continuous (which in this case is trivial), and the integral of f_n over [0,1] does indeed converge to the integral of the limit function, ie the area under the curve is 0.

However! We're looking for the arc length of the curve, which involves the derivative of f_n. Since f_n is NOT continuously differentiable (in fact, f_n isn't even C1 on [0,1]), we don't have that the derivative of f_n converges uniformly to the derivative of the limit. This is clear. Thus, since the derivative of f_n does not converge uniformly to the derivative of the limit, we have that the integral of the arc lengths of f_n does not converge to the integral of the arc lengths of the limit function, assuming that we can apply the same idea to a function of the limit function (... which is probably not a safe bet, but it still makes sense).

Cool. First time poster, btw. I couldn't help myself.

Easy way to add TeX formatted equations to your blog: use a personal user page on Wikipedia as a scratchpad, with Wikitext. Save (or just preview), then grab the PNG.
Example: http://en.wikipedia.org/wiki/User:Rpresser/fnord

Fixed the 1/n error in the definition of f_n(x).

Made the uniform convergence proof much more explicit.

Last time I tried to sketch out mathematical equations on my Wikipedia user page somebody just came along and blanked it because you're not supposed to do that.

A quick note on the convergence: I'm being a bit lazy on typing out the details of the proof (which is similar to one for the squeeze theorem), but it seems to me that any time you have a function sandwiched between two functions that uniformly converge to "f", the sandwiched function converges to f. That is, if for all (relevant) x and n, f_n(x)<=g_n(x)<=h_n(x), and (f_n) and (h_n) converge uniformly to f, then (g_n) converges uniformly to f. (Similarly, if they both converge pointwise, g converges pointwise.)

Taking f_n and h_n to be horizontal lines at the infimum and supremum of the range of g_n (assuming such exist), you then have that any time the sequence of infima and the sequence of suprema of the range approach a constant value, the function uniformly converges to that value.

Or, to put it in a less jargon-y way, when a function is sandwiched between horizonal lines that squeeze into a single limit, they squeeze the function into the same limit (since the convergence of the lines is obviously uniform, so is the convergence of the squeezed function).

Sadly, this was the most obvious proof that sprang to mind when I was thinking of the uniform convergence of these sawtooth functions.

http://www.math.union.edu/~dpvc/jsMath/
jsMath is what my uni uses on its maths student forum. It's usually pretty reliable in my experience. No idea about setting it up though.

And then there's always MathML, but it still only seems to work in Firefox (although Webkit support is underway)

If you want TeX on your webpage, use MathJax. http://www.mathjax.org/

This what both mathoverflow.net and math.stackexchange.com run on.

>Last time I tried to sketch out mathematical equations on my Wikipedia user page somebody just came along and blanked it because you're not supposed to do that.

Sam, you *could* just try having a blank LaTeX document around to recompile and take PNG snapshots of... not all that much more work than the Wikipedia solution.

Of course this is a terrible kludge and I'm sure that, long-term, you'd be much happier with mathjax or something along those lines (and you've shown before that you're willing to invest time in getting this website to run).

This also presumes that you've already installed LaTeX and have an easy way to compile documents. But that isn't too much of a stretch, I hope...?

On the off chance you've not already installed LaTeX, and for the benefit of other people: If you're running Windows, try TeXworks (which comes bundled with MiKTeX); if on Linux, Kile works fairly well but I don't know how easy it is to get working. Both editors support compiling with one keystroke (ctrl+t, alt+6 respectively; at least I think it's alt-6).

For TeX you can use Google Charts API. Enter any LaTeX expression after this:
http://chart.apis.google.com/chart?cht=tx&chl=

Sam, I explicitly stated "GRAB THE IMAGES". After you've got your formulas right you download the images they render to and server them from your own server. Then you can blank your own page because you don't need it anymore.

But the Google Charts option is probably less work.

Would WolframAlpha be of any use here?

I can't believe how intellectual this website is. All the comments are properly formatted with good grammar and good form - no caps lock, no one-liners, no nothing.

My Father offered a simpler explanation; I was wondering if it is correct: if you take the process of corner-concaving and repeat it infinitely on the circle, you will never end up with the curve itself, no matter how infinitesimal you get - you will end up with a series of right angles with legs of infinitesimal size with an arc (the curve of the circle) scrawled across. While the individual difference from arc to right-angle may be tiny, the number of these structures is, as the adjective implies, nearly infinite; and thus you still have a perimeter which properly adds up to four.

Sorry for the bad English. I'm terrible at outputting my mental simulations.

Danny, I'm sorry but your father's argument doesn't make any mathematical sense.

If you repeat the corner-concaving for a finite number of steps you NEVER get a circle and the perimeter is ALWAYS 4. But if you "repeat for an infinite number of steps" i.e. take the limit at infinity, you DO have a circle of perimeter pi.

Doesn't f_n have no derivative at the points where x-1/n[nx] = 1/n +1/n[nx]-x ? If so, doesn't that mean that you cannot use an integral to find the arc length of f_n?

There is a discontinuity, but because the discontinuity is isolated (i.e. there is a positive finite distance between it and the next discontinuity), the function is still integrable. The value of the function at this isolated point has no effect on the integral so the fact that it is undefined doesn't present a problem.

Pi is actually 4, all you need to do is rotate the surrounding square 1/4, within it a perfect circle is inscribed. Also, when you annotated the square you seem to have put '1' on each of it's sides -- this is incorrect. The left and right, and top and bottom, exist as two sets of 0 value pairs. (Left + Right, is the complete package, both cancel each other out to 0, neither holds significance and neither is individually '1', without 'one' the other cannot exist). With this interpretation you can see that the square is 3 dimensional. It has an inside, and an outside. A top and a bottom; and a left and a right = 3 sets of 0 value pairs that are reliant on each other to exist. It's purely down to perception, and this takes into account the symmetrical nature of the universe as oppose to the human created asymmetrical nature-matrix of innovation that we subside in.

Have a good day.

The point of troll science pictures are to get you to waste your time.

You fell for it lol.

another option for equation rendering for the web that doesn't involve slow-loading javascript interpreter is tex4ht
http://www.tug.org/applications/tex4ht/mn.html

Feed it a valid tex file and it will give you back an html file with all the equations converted to pngs.

To each of your right angled triangles surrounding the circle, draw the hypotenuse as well (chord for circle). As you go on making the them smaller and infinite, the sum of all such chords equals perimeter of the circle, and from your arguments sum of other two sides for all such small triangles also equals the perimeter of circle. So in effect "the sum of two smaller sides now become equal to the larger side". This can only be true when your right angle goes on increasing to become obtuse, eventually ending up at angle pi (180 degree). So that these triangles are effectively straight lines tangent to the circle.

a heuristic (not proof) to see where the problem is, is that every time the corners are folded in, the absolute distance from the square to the circle is decreased. This, I assume, is using the euclidean norm, not the 'taxicab norm.' For this reason, a contradiction arises. Using taxicab geometry, the 'circle' is actually a square rotated counterclockwise 45 degrees (which is stil a square, regardless). Now, is this were to use the Minkowski Norm, then the original square would be a ciricle in that defined space. However, seeing as this is trying to prove taht a square is a circle, we can assume that euclidean norms are used, while the implicit starting point is in the Taxicab Vector Space, using the Minkowski Norm. Therefore, three contradictions of space, metrics, and norms are made using this argument and pi does not equal 4

Nobody mentioned taxicab geometry or the Minkowski Norm.

By inspection the shape made from removing squares from the corners would be a diamond (the same square rotated 45') not a circle.

People take Troll circle so seriously....