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Even Great Mathematicians Guess Wrong
- iwwr 16y agoAnother fascinating idea is to consider the split-complex numbers (and their split-quaternion cousins) as ways to express hyperbolic geometry (in 2 or 3/4 dimensions respectively).
- aksbhat 16y agoComplex numbers are a powerful tool for studying the two-dimensional plane: each point corresponds to a unique complex number . The beauty of this correspondence is that it allows you to add, subtract, and multiply points in the plane I wish someone would have told me this five years ago, it is such a neat way of thinking!
- cdavid 16y agoif you are interested in new insights about complex numbers and analysis from a geometric POV, this book is wonderful: http://www.amazon.com/Visual-Complex-Analysis-Tristan-Needham/dp/0198534469/ref=pd_sim_b_1 http://www.amazon.com/Visual-Complex-Analysis-Tristan-Needha... and quite accessible.
- happy4crazy 16y agoMy introductory physics class in college included a problem that investigated the effects of "2d" gravity (a generic attractive force proportional to 1/r instead of 1/r^2) on a point mass in the plane. When the professor mentioned that I could write all the vectors as complex numbers, and do regular old algebra with them, my brain almost melted.
- oikjhgbpokj 16y agoThat's why it's important to wait until college to teach kids anything like mathematics. By ensuring that all they learn in school is just arithmatic and keeping the debate to "are calculators good/bad" we stop all the kids being interested in math and so keep it all to ourselves ....evil genius laugh ......
- xyzzyz 16y agoWell, his argument is unconvincing -- it only proves that there is no normed division algebra on R^3. Division algebra on R^3 would be interesting and maybe useful even without norm. However, its existence is still impossible. One can prove it for instance by noticing that if it existed, then {1, v, v^2} for any non-scalar v would span whole R^3 (the hardest part of proof), so v^3 would be a linear combination of them, say v^3 = a 1 + b v + c v^2 for some a, b, c \in R. Then v is the zero of polynomial f(x) = x^3 - c x^2 - b^x - a. But deg f = 3, so it has real zero, so for some q, w, p \in R, f(x) = (x-q)(x^2 + w x + p), and 0 = f(v) = (v-q)(v^2 + w v + p 1), and factors on the right side are non-zero, so we have non-trivial zero divisors which is imposible in division algebra. There is also a theorem due to H. Hopf, stating that the only finite dimensional, commutative division algebras over R are C and R itself, but the only proof I know requires pretty heavy topological machinery.
- happy4crazy 16y agoIf you enjoy playing around with quaternions, you may also have fun with geometric algebra. Geometric Algebra for Physicists is very clear and gently paced. http://www.amazon.com/Geometric-Algebra-Physicists-Chris-Doran/dp/0521715954/ http://www.amazon.com/Geometric-Algebra-Physicists-Chris-Dor... This one also looks good: http://www.amazon.com/Linear-Geometric-Algebra-Alan-Macdonald/dp/1453854932/ http://www.amazon.com/Linear-Geometric-Algebra-Alan-Macdonal...
- yummyfajitas 16y agoAnother great example (technically a physicist, not a mathematician) is Einstein and his quest for a local theory of quantum mechanics. In his original paper deriving EPR ( http://en.wikipedia.org/wiki/EPR_paradox http://en.wikipedia.org/wiki/EPR_paradox ), he believed it was a reducto ad absurdum which invalidated configuration-space based quantum mechanical theories: We are thus forced to conclude that the quantum mechanical description of physical reality given by wavefunctions is not complete [...] No reasonable definition of reality could be expected to permit this. http://prola.aps.org/pdf/PR/v47/i10/p777_1 http://prola.aps.org/pdf/PR/v47/i10/p777_1 Nevertheless, he thought a local and complete theory of QM was possible and spent many years searching. Bell showed this to be impossible (after Einstein's death). Eventually experiments showed the predictions of the EPR paper to hold, thereby implying that the definition of reality is not "reasonable". If a scientist never gets things wrong, they aren't doing anything interesting.
- ArbitraryLimits 16y agoI think it's an even better story how he "guessed wrong" with Hamiltonian mechanics, also; the only difference is that he never fixed the mistake and it wasn't really fixed until the early 20th century with the development of optimal control. See "The Brachistochrone Problem and Modern Control Theory" (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.26.825&rep=rep1&type=pdf http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.26....) The TL;DR is that classical mechanics can be formulated as finding a time history of position and velocity which minimizes a certain function of them both. Hamiltonian mechanics essentially optimizes the function over position and velocity as if position and velocity were independent, then adds the constraint that velocity be the time derivative of position via a Lagrange multiplier. This Lagrange multiplier is the momentum. Of course, no one in a physics course actually learns it that way; undergraduates just memorize some equations and graduate students, if they are lucky, learn if from the Hamilton-Jacobi equation (which is how Hamilton developed it in the first place) where the momenta become the spatial gradient of the function to be minimized when evaluated at all positions. (It's exactly the same as the Bellman return function in dynamic programming; hence the name "Hamilton-Jacobi-Bellman equation.") IMHO, neither of these other developments is as intuitive as the Lagrange multipler one; but because Hamilton's original formulation obscures it, no one in physics learns it that way. A fun postscript: when you consider the equations of classical mechanics in terms of position and momenta (this combination is what physicists call "phase space") they form a manifold with a special property called "symplectic"; symplectic geometry can be formulated in terms of matrices over the field of quaternions. According to some people, this idea was a big breakthrough, but I don't believe Hamilton himself, despite having formulated both, ever noticed.
- starwed 16y ago"IMHO, neither of these other developments is as intuitive as the Lagrange multipler one; but because Hamilton's original formulation obscures it, no one in physics learns it that way." I'm pretty sure learning a bit about Lagrange multipliers is standard for a grad classical mechanics class. Granted, I've forgotten all I learned back then, b/c it has never come up in anything else I do, but it was definitely part of the course.