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Weierstrass's Monster
- graycat 2y agoA favorite on my bookshelf: Bernard R.\ Gelbaum and John M.\ H.\ Olmsted, {\it Counterexamples in Analysis,\/} Holden-Day, San Francisco, 1964.\ \
- thomassmith65 2y agoHere's Weierstrass in Python: https://scipython.com/blog/the-weierstrass-function https://scipython.com/blog/the-weierstrass-function I wonder what they sound like as audio. Edit: apparently it sounds like this https://youtu.be/vWmZnlPQK14 https://youtu.be/vWmZnlPQK14
- waveBidder 2y agohonestly less terrible than I was expecting
- thomassmith65 2y agoWhat piqued my interest is that the illustration in the article looks a lot like a typical organic audio waveform. That's unusual. At a lower pitch, with some variation in pitch and amplitude, it probably sounds nicer.
- emmanueloga_ 2y agoAt some frequencies sounds like a more (or less) chip-crushed church organ.
- praptak 2y agoAudio spectrum is bounded, so we are only hearing the first few terms of the series that make up the function (the coefficient under sine grows exponentially with n). You could even replace the audible part with Chopin (or anything else, even f(x)=0) to get a Weierstrass-like function which sounds like Chopin (or silence) but is still not differentiable anywhere.
- emmanueloga_ 2y agoMy attempt at nice looking TS version: function sum(f: (x: number) => number, upper = 500) { let sum = 0; for (let n = 0; n < upper; n++) sum += f(n); return sum; } const weierstrass = (x: number, a = 0.5, b = 3) => sum(n => a ** n * Math.cos(b ** n * x));
- cyberax 2y agoMy favorite counter-example in math is https://en.wikipedia.org/wiki/Schwarz_lantern https://en.wikipedia.org/wiki/Schwarz_lantern Intuitively, you'd think that the surface area in 3D can be approximated by using a series of converging polyhedra, but that's not the case.
- rtpg 2y agoThere's a wonderful book on Mathematical counterexamples in French[0], meant for undergrad/engineering school hopefuls. So many of the continuity counter-examples are throwing Weierstrass at the wall and getting something to stick. It's fun but also feels a bit like cheating. I do recommend this book for any french-speaking mathematician-adjacent person though. Real great dictionary for remembering why certain things only work in one direction. [0] Les contre-exemples en mathématiques: https://www.editions-ellipses.fr/accueil/5328-les-contre-exemples-en-mathematiques-9782729834180.html https://www.editions-ellipses.fr/accueil/5328-les-contre-exe...
- daquisu 2y agoIt is interesting how France became so focused on analysis and properly proving theorems and stuff, while the applications don't have the same highlight in prépa. One professor of mine commented that most French engineers are better mathematicians than most mathematicians in Brazil. It is the opposite of what the linked article mentions that was happening in Weierstrass' time.
- mbeex 2y agoCannot compare, because I own only the following ones: https://www.amazon.com/Counterexamples-Analysis-Dover-Books-Mathematics/dp/0486428753 https://www.amazon.com/Counterexamples-Analysis-Dover-Books-... Weierstrass function is also there, but exactly this one contains an error... :) https://math.stackexchange.com/questions/1507399/is-the-weierstrass-function-given-in-counterexamples-in-analysis-a-typo https://math.stackexchange.com/questions/1507399/is-the-weie... There is also https://en.wikipedia.org/wiki/Counterexamples_in_Topology https://en.wikipedia.org/wiki/Counterexamples_in_Topology
- croes 2y agoDid I miss the mentioning of the actual function?
- twiceaday 2y agohttps://en.wikipedia.org/wiki/Weierstrass_function#Construction https://en.wikipedia.org/wiki/Weierstrass_function#Construct...
- emmanueloga_ 2y ago+1, it the obvious thing missing in the article!
- rmrfchik 2y agoThey even don't let you select and search the word. What a ... strange choice of ui/ux.
- seanhunter 2y agoThe Weierstrass function is cool but the undisputed champion of calculus counterexamples has to be the Dirichlet function[1] f(x) = 1 if x is rational, 0 otherwise. It is defined over all real numbers but continuous nowhere. Also if you take the Dirichlet function and multiply it by x so you get g(x) = x if x is rational, 0 otherwise …then you have something that is continuous at exactly one place (0) and nowhere else, which also is pretty spectacular. [1] https://mathworld.wolfram.com/DirichletFunction.html https://mathworld.wolfram.com/DirichletFunction.html
- philomath_mn 2y agoNice. Reminds me of the discrete metric in my Real Analysis course: distance is 0 if it is the same point, 1 otherwise. We used to joke that it was the perfect metric for finding weird counter examples
- unkulunkulu 2y agothere is a whole book “counterexamples in analysis”, basically for each theorem they find why every condition is required. these functions and ideas based on them are indispensable. this weierstrass function is definitely a mind bender I remember from high school.
- JJMcJ 2y agoAlso "Counterexamples In Topology" - same but for topology.
- shwouchk 2y agoI don’t know if they are the same books as the names are quite generic, but i picked up both this one and the topology book sibling suggested at the library during my undergrad, and read both cover to cover. Both slim books and very clear. My favorite counterexamples in analysis were based on the cantor set as it allowed counerexamples for many different scenarios while still producing functions that were “nice” in other respects
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- t234oi324324 2y agoI've increasingly come to doubt the utility of all these things. Real numbers and Lesbesgue integration that stem out of 19th c. analysis came before Computability was a thing... and it turns out that all but a measure 0 set of real-numbers are actually computable. So all these things that we worry about don't even "exist" in a computational sense. Mathematicians haven't come to terms with this (barring non-standard analysis etc.), and I suspect this is one of the reasons it's losing mind-share.
- tsimionescu 2y agoNote that "computable" just means "computable by a Turing machine or equivalent system", at this time. We don't know if we are missing some other form of computation that goes beyond the abilities of a Turing machine, that could upend the whole concept. And given that we haven't proven that general human thinking is a computable process, there is at least one significant candidate for possible computation beyond Turing machines. Note that I don't personally think it's likely that our thinking goes beyond the capabilities of a Turing machine, not at all. But I also think it would be premature to throw out all mathematics that doesn't conform to Turing machine computability before we are more sure that this is the best possible model. And even if it were, there is also the question of whether physical processes are computable or not. Right now there are plenty of physical processes where our best and only reliable models for predicting their behavior require assumptions from calculus, like the existence of all real numbers. For example, there is no successful formulation of quantum mechanics where the distance between two particles moving relative to each other can be constrained to any subset of the [minDistance, maxDistance] interval of the real number line. Which means that, as the particles move away from each other, at some times the distance between them will have to be an uncomputable number (and given their density in the real number line, this will be approximately all the time).
- magneticnorth 2y ago>> and it turns out that all but a measure 0 set of real-numbers are actually computable. This is the wrong way around. The computable numbers have measure 0. In fact, they are countable. The non-computable real numbers have full measure. (proof: by definition, computable numbers are generated by a program, which can be represented by a finite number of bits. The set of finite numbers is countable).
- andrewhalle 2y agoHere’s a really cool thing about continuous-everywhere differentiable-nowhere functions. A consequence of the Baire Category Theorem is that most continuous functions are nowhere-differentiable (in the same sense that “most” real numbers are irrational). The familiar functions from calculus are the vanishing minority of continuous functions.
- coke12 2y agoAre there any hierarchies in mathematics that aren't like this? It seems like most hierarchies are constructed such that going up one "level" expands your scope so much that the previous rung looks like a drop of water in the ocean. Examples include, for example: the Chomsky hierarchy of languages, where most context-free languages are not regular; Turing computability/solvability; and so on.
- awanderingmind 2y agoMany mathematicians dismissed the function. It was an anomaly, they said — the work of a pedant, mathematically useless. This is low-key hilarious given how much of mathematics seems like pedantry to an outsider.
- staunton 2y agoIt used to be less so
- Sharlin 2y agoIndeed, Weierstrass was one of the people thanks to whom math became the pedantic endeavor we know it as today.
- pfdietz 2y agoWay back when, the rigor (or lack thereof) of calculus was a culture war issue. Bishop Berkeley famously argued the dubious foundations of calculus rendered it no more trustworthy than religion. So, there was external as well as internal motivation for mathematics to clean up its act. https://en.wikipedia.org/wiki/The_Analyst https://en.wikipedia.org/wiki/The_Analyst
- Sharlin 2y agoOh wow, hadn't heard of this before!
- constantcrying 2y agoThese examples are what caused them to be this way. Hand waving was a lot more acceptable in mathematics while Weierstrass was alive. The discovery of clear counter examples to hand waving arguments lead to the desire to put mathematics on the strong footing it is on today. It may annoy students today as there is seemingly little utility in these distinctions, but they actually are important. The more complex mathematics becomes, the more important it is to actually be on solid footing.
- xelxebar 2y agoGood memories of my Real Analysis course in undergrad. I'm also a fan of the Devil's Staircase [0], a function that is both continuous everywhere and constant almost everywhere, while still managing to climb from 0 to 1. [0]:https://en.wikipedia.org/wiki/Cantor_function https://en.wikipedia.org/wiki/Cantor_function
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- aa-jv 2y agoI wonder if there are any synthesizers using this in wavetable generation, somehow .. ?
- synergy7 2y agoWeierstrass function is used prominently in Abbot's (2015) "Understanding Analysis" book. Abbott also relies heavily on three other mind-bending functions - Dirichlet (nowhere continuous), Thomae (discontinuous at every rational and continuous at every irrational point), and Cantor (increasing and continuous on [0, 1], yet constant at [0, 1]\C. where C is the Cantor set that is of measure zero). Dirichlet, Thomae, and Cantor functions are central in Abbott to introduction and exercises on continuity, differentiation, and integration. I thought that was an interesting pedagogical choice for an undergraduate book, especially when it is used for the very first course in mathematical analysis as in Princeton’s MATH215 (I do think it is a really nice book).
- JKCalhoun 2y agoWhen I first learned about limits (my senior year in high school), I visualized a "step" shape graphed (0, 0) to (0, 1) and then to (1, 1). Up one unit, right one unit. (I know, not technically a function since it has a vertical component.) But now you subdivide both segments and put in an extra step. Up 0.5, right 0.5, up 0.5, right 0.5. Same starting point, same ending point. Even the same length of line segments as before — but the area bounded beneath is 3/4 the original. I reasoned that if you continued this subdivision, as you approach infinity the area under the "curve" approaches that of a half-unit right triangle (i.e. root-two). How is it, I asked my math teacher, that this shape can have the perimeter of a square but the area of a triangle? I don't recall the answer from the teacher. Maybe somewhat hand-wavy; ultimately unsatisfactory to me. I still don't really know the answer. I suppose the answer is that it is not in fact ever a triangle — just a thing with a fat edge.
- aqme28 2y agoReminds me of Gabriel’s Horn, which has an infinite perimeter but finite area.
- wat10000 2y agoThe answer is that the perimeter “at infinity” matches the triangle. There’s nothing that says the perimeter of the final “infinite” shape has to equal the limit of the perimeters as they go to infinity. In order for this to work, you have to prove that they’re equal. Typically you’d come up with two limits, one which you prove is always greater than the value in question, and one which is always less. Then you show that they’re equal in the limit. That squeezes the number you’re interested in, proving that it’s also equal to that number. For example, the typical proof of the circumference of a circle involves making two polygons, one inside the circle and one outside. The perimeter of the one inside is always less than the circumference, and the one outside is always greater. Then you take the limit of the polygon perimeters as the number of sides goes to infinity. They converge on the same number, 2 pi r.
- pif 2y agoSorry, you are completely wrong. The perimeter is equal to the square one at every iteration, thus its limit towards infinity is still the square one.
- calebm 2y agoI just made a similar function (except with only 2 waves) just recently: https://gods.art/math_videos/convolv_waves.html https://gods.art/math_videos/convolv_waves.html
- waynecochran 2y agoThere are also functions that have a first derivate, but no second derivative. Much of my graduate research involved studying these type of functions. Many of the original ideas came from the paper "The calculus of fractal interpolation functions" https://www.sciencedirect.com/science/article/pii/0021904589900804 https://www.sciencedirect.com/science/article/pii/0021904589... I wrote a paper on how to compute the surface normal (for rendering) of related functions: https://link.springer.com/article/10.1007/PL00013408 https://link.springer.com/article/10.1007/PL00013408 Interestingly enough, while you can not differentiate the Weierstrass function, you can integrate it -- i.e. you can treat it like a differential equation that has a set of well defined solutions.
- sfink 2y agoI'll throw in my funky function (which is probably similar to a well-known one, but I'm no mathematician so I know not what is well known): f(x) = a random number uniformly selected from [-1, 1]. (So every real number maps to a distinct random number.) It's nowhere continuous, nowhere differentiable, yet its integral over any interval is exactly zero. Which is kind of weird if you think of integrals and derivatives as inverses, because as I understand it the derivative of a constant g(x) = c is g'(x) = 0, yet the integral of the above function is F(x) = 0 yet F'(x) = f(x) = the random nonsense monster. So the derivative of zero can be either zero or random noise? (And while f(x) is not differentiable, its integral surely is.)
- eafer 2y ago> yet its integral over any interval is exactly zero That function doesn't actually have an integral in any sense that I know of. The Riemann integral doesn't work because, for every interval in every possible partition, the function's upper bound will be 1 and the lower bound will be -1.
- sfink 2y agoOh. Er... I didn't really consider whether it could be proven to have an integral, I just went off of intuition. After an additional lifetime of study, magically compressed into 15 minutes of skimming Wikipedia articles, I'm going to (tenuously) claim that the function is Lebesgue integrable. Heck, if it works for the {0 if irrational, 1 if rational} function, this should be much easier. I hope? My original intuition is that every value in [0, 1] will be cancelled by a value in [-1, 0]. My updated intuition to make it more Lebesgue-compatible would be to sort the (infinite) set of values in the interval. This will give the everywhere continuous, linear function from -1 at the minimum coordinate of the interval to 1 at the maximum coordinate. Hm... I notice that when applying Lebesgue to Dirichlet, it uses countability, and reals aren't countable. I have no clue whether that's a problem here or not.
- eafer 2y agoIt won't be Lebesgue integrable because it won't be measurable. Your function is actually a good (though not rigorous) argument for the existence of non-measurable sets.
- nejsjsjsbsb 2y agoDo you need axiom of choice for the Weierstrass function?