Wikipedia:Reference desk/Archives/Mathematics/2008 November 14

= November 14 =

Fractal circle
Sorry if this is a dumb question, my math is old and crackin'. Ever since encountering Mandelbrot in the library, I've been wondering if one could describe circles with a fractal equation instead of Pi. The obvious advantage of such an application would be that it would be scalable and should thus be superior to Pi. Just to clarify I'm not looking for a fractal circle as in add smaller and smaller circles, but something that would have a circle as result (sort of like:for radius r element size z add another element x with displacement y, repeat till you get a circle}. If this could be done, what would it look like and if yes, why should we not replace existing "pi equations" with it? 76.97.245.5 (talk) 00:43, 14 November 2008 (UTC)


 * Are you acquainted with complex numbers and the exponential definition of trig functions? That might give you some ideas. —Tamfang (talk) 07:25, 14 November 2008 (UTC)


 * Thanks that will give me a starting point. I vaguely recall thinking "Why didn't they teach us this to begin with" when we got to "higher math" way back when.  Unfortunately they didn't go into great detail and I've forgotten most of that since then.  I have a suspicion that I will come back and ask "Why do we bother with pi." after I've read up on the subject (again).  Thks. for your help.76.97.245.5 (talk) 10:03, 14 November 2008 (UTC)


 * Although a circle is not, strictly speaking, a fractal, it can be viewed as the limiting curve of a series of regular polygons. Archimedes obtained an approximate value for &pi; in Measurement of a Circle by calculating the perimeter of a 96-sided regular polygon, and Chinese mathematician Liu Hui obtained a more accurate value using a 192-sided polygon. Our article on Liu Hui's π algorithm describes how the method can be extended to approximate &pi; as closely as you like by calculting the perimier of a polygon with 3(2n) sides, which only involves taking repeated square roots. Gandalf61 (talk) 10:54, 14 November 2008 (UTC)
 * I was not looking for a way to calculate &pi; or calculate it more precisely. The error using a certain value of pi is not related to the size of the circle in question. I was rather trying to replace &pi; with a way of describing circles for which any errors introduced are related to the size of the object.  I haven't had time to look into the complex number avenue suggested by tamnfang in detail, but it looks promising.  76.97.245.5 (talk) 05:44, 16 November 2008 (UTC)


 * I'm not sure I understand. You can describe a circle using x2+y2=R2 (where R is the radius).  No 'pi' anywhere in sight.  Pi only crops up when you want the circumpherence or area of the circle. SteveBaker (talk) 06:28, 16 November 2008 (UTC)
 * Or "all points p such that d(p,0)=R" (where d is the distance function, or metric), you don't even need coordinates. --Tango (talk) 13:57, 16 November 2008 (UTC)


 * If your question is, "are there any fractally things that can output a circle", take a look at Indra's Pearls. Black Carrot (talk) 21:18, 20 November 2008 (UTC)

what kinds of math can and can't you do orally (no symbols)
what kinds of math can and can't a person of exceptional intelligence understand all orally (no symbols)?

relatedly, are there any blind mathematicians? —Preceding unsigned comment added by 83.199.126.76 (talk) 03:40, 14 November 2008 (UTC)


 * Wikipedia has a very incomplete list of blind mathematicians and scientists. Leonhard Euler (not on the list) was not blind from birth, but was nearly blind in his right eye by the time he was in his early thirties and later lost sight in his other eye due to a cataract.  Bernard Morin (on the list) is a topologist who lost his sight as a child.  He helped exhibit an eversion of the sphere (Smale had earlier proved existence).  If you would like more information, the article The World of Blind Mathematicians in the Notices of the AMS is worth reading. Michael Slone (talk) 04:26, 14 November 2008 (UTC)


 * Let's separate a few concepts and misconceptions here:
 * Blind mathematicians are not restricted to verbal communication. There are mathematical "dialects" of Braille, such as Nemeth Braille and GS8 Braille, which, I believe, use systems of markup similar to LaTex.
 * Verbal communication involves symbols just as much as written communication - the spoken phrase "one hundred and twenty three" and the written string of digits "123" are both symbols for the number 123.
 * If the question is really "what kinds of math can a blind person do", then I believe the answer must be any kind, at any level - a mathematician's most productive activity is thinking, not reading or writing. The AMS article mentioned above makes a good case for its claim that mathematics is, if anything, more accessible for blind people than other professions. Gandalf61 (talk) 09:57, 14 November 2008 (UTC)


 * You might be interested in Learning styles Different people think in different ways. Despite what lots of people write one can still think without words, I can see it might be a primary means of thought for most people who write books though :) Dmcq (talk) 13:50, 14 November 2008 (UTC)

anything ever proved true that was almost certainly false (or vice versa?)
Have there ever been a proofs of the truth or falseness of a conjecture that through empirical testing had been almost certain to have the opposite truth value? —Preceding unsigned comment added by 83.199.126.76 (talk) 14:28, 14 November 2008 (UTC)


 * Borsuk's conjecture...? --CiaPan (talk) 15:14, 14 November 2008 (UTC)


 * Gödel's Incompleteness Theorem was a surprise. Bo Jacoby (talk) 15:17, 14 November 2008 (UTC).
 * Skewes' numbers are linked to such kind of conjectures.--Pokipsy76 (talk) 16:44, 14 November 2008 (UTC)

first mathematician. left-handed mathematician.
who is the first mathematician known by name?

of the truly famous mathematicians, are any left-handed? —Preceding unsigned comment added by 83.199.126.76 (talk) 18:06, 14 November 2008 (UTC)


 * The first name I see mentioned in History of mathematics is the Greek mathematician (and philosopher), Thales. --Tango (talk) 19:38, 14 November 2008 (UTC)


 * Wikipedia is quite keen on purging lists of left handed people so its hard to find this info here. Anyway Charles Dodgson aka (Lewis Carroll) Alan Turing, and Nicole Oresme were lefthanded mathematicians. There is speculation that Pythagoras was a lefty. There is some very speculative research disproportionate number of people with high mathematical ability are left handed.--Salix (talk): 23:47, 14 November 2008 (UTC) (a lefty)


 * What?! Why would it purge this, it's way more relevant, notable and incontrovertible, not to mention well-documented with copious references, than ethnicity or nationality or....   How do they possibly justify this?  —Preceding unsigned comment added by 83.199.126.76 (talk) 01:08, 15 November 2008 (UTC)


 * I suspect that they have some sinister motive. :-) StuRat (talk) 19:00, 16 November 2008 (UTC)


 * C'mon Stu, lets call a spade a spade ;-) -hydnjo talk 03:09, 17 November 2008 (UTC)


 * it's just that when they made sure all the information was right, nothing was left.Gzuckier (talk) 19:18, 19 November 2008 (UTC)


 * I heard it was all due to the actions of one evil deletionist named Dexter. StuRat (talk) 00:58, 20 November 2008 (UTC)

O(n log n) Polynomial Derivative Evaluation?
I was wondering if there's an algorithm for calculating the value of a degree-n polynomial and its n derivatives at a given point in O(n log n) time. The extended Horner's rule does it in O(n^2) time, but there are a large number of algorithms that are nominally O(n^2) which can be evaluated recursively to get O(n log n) time. I've tried making algorithms that subdivide high and low terms and that subdivide even and odd terms, and I can't get it to work. --Zemylat 23:48, 14 November 2008 (UTC)


 * Yesterday, reading your post, I doubted it, but not anymore. If the point you're considering is zero, we get the answer immediately, looking at the polynomial as a Taylor series,
 * $$p(x) = \sum_{j=0}^n p_j x^j \Rightarrow p^{(j)}(0) = p_j j! \text{.}$$
 * Now let's look at the problem not just at zero, but at a constant $$a$$ instead. That is, we're interested in $$p^{(j)}(a)\ \forall j \in \{0, 1, \dots, n\} \text{.}$$ $$p(a)$$ can be expressed as a polynomial at zero,
 * $$p(a) = \left. \sum_{j=0}^n p_j (x+a)^j \right|_{x=0} \text{.}$$
 * If we knew the coefficients before $$x^j$$ in that expression, we would immediately know the $$p^{(j)}(a)$$ we're looking for. Let's expand it using the Binomial theorem and the Heaviside step function ($$H(x \ge 0) \equiv 1,\ H(x<0) \equiv 0$$):
 * $$\sum_{j=0}^n p_j (x+a)^j = \sum_{j=0}^n p_j \sum_{k=0}^j x^k a^{j-k} {j \choose k} = \sum_{j=0}^n p_j \sum_{k=0}^n x^k a^{j-k} {j \choose k} H(j-k)=$$
 * $$= \sum_{j=0}^n p_j \sum_{k=0}^n x^k a^{j-k} \frac{j!}{k!(j-k)!} H(j-k) = \sum_{k=0}^n \frac{1}{k!} x^k \sum_{j=0}^n (j! p_j) \frac{a^{j-k} H(j-k)}{(j-k)!} \text{.}$$
 * We've implicitly defined
 * $$\frac{H(x)}{x!} \equiv 0\ \forall x<0 \text{.}$$
 * Defining $$f_j = j! p_j$$ and $$g_j = \frac{a^{-j} H(-j)}{(-j)!}$$ it's clear that $$p^{(j)}(a)$$ is calculable as the discrete convolution $$(f*g)_j \text{.}$$ Both $$f$$'s and $$g$$'s relevant parts are sequences of length $$n+1$$ and computable in $$O(n)$$ time. The convolution, on the other hand, can be calculated by some fast convolution algorithm in $$O(n \log n)$$ time, the goal sought! In particular, both sequences can be fast-Fourier-transformed in $$O(n \log n)$$, the transformed ones multiplied together in $$O(n)$$ and finally inverse-transformed in $$O(n \log n)$$ again.undefined&mdash;undefinedPt(T) 13:16, 18 November 2008 (UTC)