Wikipedia:Reference desk/Archives/Mathematics/2009 January 30

= January 30 =

Calc 1 Question
Ok, although this was a homework question its already been handed in and was someone else's and has now gained the intrigue of two math majors. The question was to algebraically solve 0=e^x-3x At the calc one level (i.e. derivatives is as far as they have gone). As far as we can tell, it cannot be solved for algebraically, but it can be solved for graphically. How would you solve it algebraically? --Omnipotence407 (talk) 00:57, 30 January 2009 (UTC)
 * I'm not quite sure what you mean by solving it algebraically, but I don't think it's possible. Certainly the solution is solutions are not an algebraic number algebraic numbers (by the Lindemann–Weierstrass theorem say). It They can be expressed in terms of the Lambert W function, as &minus;W(&minus;1/3), but I doubt that counts as an algebraic solution. Algebraist 01:06, 30 January 2009 (UTC)

There are two solutions, so the expression in terms of Lambert's W is not the end of the story. Michael Hardy (talk) 01:14, 30 January 2009 (UTC)
 * Oops, sorry. I wrote W0 (the single-valued version) when I should have written W for the multi-valued version. The solutions of the given equation are the two real values of &minus;W(&minus;1/3). Algebraist 01:22, 30 January 2009 (UTC)
 * Sorry, forgot this piece. 0≤x≤1 --omnipotence407 (talk) 01:25, 30 January 2009 (UTC)
 * Then you do want W0, for what it's worth. Algebraist 01:29, 30 January 2009 (UTC)

The equation has two real and infinitely many nonreal complex solutions. A J expression for the solutions is this: p.(^-3&*)t.i.9. The transcendental function ^-3&* is expanded as a taylor series, t.. The first 9 coefficients, i.9, are used for an approximation, and 8 roots are computed, p.. The real solutions are approximately 0.6190 and 1.512. Bo Jacoby (talk) 12:39, 30 January 2009 (UTC).

Common name for spheres and circles.
I could loosely describe either a rectangle or a cuboid as a 'box'. But I'm looking for an analogous word that could be used to describe both circles and spheres (and maybe also 4D hyperspheres). Our Sphere article offers 'n-Sphere' which is kinda nasty - is there anything better than that? SteveBaker (talk) 14:33, 30 January 2009 (UTC)


 * How about Ball? Strictly speaking it should be 'n-ball', but plain 'ball' should be clear enough in context, and at least it's easier to pronounce than 'sphere'. I suppose it depends who your audience is... AndrewWTaylor (talk) 16:44, 30 January 2009 (UTC)


 * If you're talking to mathematicians, though, remember that to us, balls and spheres are completely different things. Algebraist 16:46, 30 January 2009 (UTC)


 * Yes, as a (former) mathematician I realise that, but it looks as if Steve wants a fairly informal term. AndrewWTaylor (talk) 18:05, 30 January 2009 (UTC)


 * "Round" can mean circular, spherical, or many other shapes which lack sharp corners, edges, and planar faces (such as ellipses and ovals). StuRat (talk) 18:00, 30 January 2009 (UTC)


 * Yeah - this is for naming a type in a computer program - so precision of language isn't everything - I just know that if I start talking about 2D spheres it's going to mess with everyone's head. I'm aware of the distinction between a ball and a sphere (the former being the solid inside and the latter being just the surface) - analogous to disk and circle.  But my data structure is a center coordinate and a radius and a whole bunch of tests for overlaps and inclusions and such - I'm being deliberately vague about whether it's solid or not because it's going to be used in both situations.  So what I REALLY STRICTLY want is a common name for disk,circle,ball,sphere,hyperball,hypersphere and perhaps even the 1D analogs of those things.  I guess "Round thingy" doesn't cut it mathematically then? SteveBaker (talk) 18:55, 30 January 2009 (UTC)


 * I can't help but be reminded of this - "Hey! What's this thing suddenly coming towards me very fast? Very very fast. So big and flat and round, it needs a big wide sounding name like ... ow ... ound ... round ... ground! That's it! That's a good name - ground! I wonder if it will be friends with me?" -- LarryMac  | Talk  20:55, 30 January 2009 (UTC)


 * Both "disk" and "ball" are used in any dimension and norm. To be more precise one can say: "euclidean ball" or also " n-dimensional euclidean ball". PS: LarryMac: aren't you talking about The Prisoner, with Patrik McGoohan, are you? --pma (talk) 21:46, 30 January 2009 (UTC)


 * Not The Prisoner, sorry, I should have put a link in there - List of minor characters from The Hitchhiker's Guide to the Galaxy -- LarryMac | Talk  21:50, 30 January 2009 (UTC)


 * In computereeze, how about "round object" ? StuRat (talk) 15:52, 31 January 2009 (UTC)
 * By the way, while today "sphere" sounds quite more technical than "ball", let's recall that the Greek term just coincides with "ball" both in the ordinary sense and in the mathematical one. pma (talk) 17:50, 31 January 2009 (UTC)


 * The mathematical terms are "ball" and "sphere", you're not going to get any better answer than that on the maths desk. Try the language one! --Tango (talk) 23:41, 31 January 2009 (UTC)


 * The computing desk might also be appropriate. If you're designing a program to simulate moving objects, represented by circles and spheres, I'd call them "balls". In three dimensions both the surface and the solid can be reasonably called that, and in two dimensions (thanks to cartoons probably) it's natural to extend the terminology downward. If I were programming a sidescroller game, for instance, nobody would complain about a two-dimensional representation of a baseball being a "ball". Natural language hasn't yet produced a word that applies to four dimensions, but I don't think it would hurt to use it there as well. All that matters is that it's readily comprehensible to someone reading your code. Black Carrot (talk) 17:22, 1 February 2009 (UTC)

Rotation along a certain point...
What's the formula used to rotate any given point (2,3) for example, around another point (5,4) any number of degrees in any direction???? Thanks, Saebjorn (talk) 14:47, 30 January 2009 (UTC)


 * Translate your center of rotation to the origin, rotate around the origin using the formula in rotation (mathematics), and translate it back. — Emil J. 15:02, 30 January 2009 (UTC)


 * You can compose this (in fact any) sequence of translations and rotations into a single matrix using affine transformations. This is very widespread in 3d computer graphics, and also works fine in 2d spaces. —Preceding unsigned comment added by 84.187.97.24 (talk) 01:28, 31 January 2009 (UTC)


 * Of course. But in order to derive what the matrix looks like you need to performs the steps I wrote above. Anyway, the explicit solution is
 * $$x'=a+(x-a)\cos\theta-(y-b)\sin\theta,$$
 * $$y'=b+(x-a)\sin\theta+(y-b)\cos\theta,$$
 * or
 * $$z'=c+(z-c)e^{i\theta}$$
 * using complex numbers, where c = a + ib is the centre of rotation, and $$\theta$$ is the angle. — Emil J. 12:11, 3 February 2009 (UTC)

Curiosity has revived the calculus - A math homework from yesteryear
208.92.185.250 (talk) 21:32, 30 January 2009 (UTC) This is a nostalgic trip back to a math problem presented in a graduate applied math course in my youth, forty years ago. My professor for that course had a had a lecturing style which can only be be described as succinct. At the front of our lecture hall, there was a stage in front of which were we students, arrayed in orderly rows to the back of the hall. Behind the stage, the wall was a continuous twenty feet of blackboard, and, on either end, there was a door to the exterior hall. The professor would arrive precisely on the hour, walk through the left-hand door, pick up his chalk and begin lecturing (which invariably, was the detailed and annotated solution of a  rather mundane math problem but always with an engaging unexpected, even ironic, twist). He would develop his problem and its solution - at all times addressing the blackboard - never acknowledging us, his students. At the conclusion of the lecture, invariably 47 or 48 minutes after his entry, he would lay down the chalk, open the right-hand door and disappear into the hall, not to be seen until the next class.

One particular lecture lives forever in my memory, but I can no longer locate the notebook from that class, and I cannot reproduce the solution - a lapse which continues to haunts me. Perhaps someone here can take the time to outline an approach or provide a solution.

Statement: If a known mass is elevated above a elastic surface and let to drop a known distance, the mass will, of course, acquire momentum during its fall: upon impact with the resilient surface, it will decelerate to zero velocity as a function of the momentum thus absorbed and by the “stiffness” of the surface (expressed as Young's modulus, or, maybe, bulk modulus). The value for the momentum can be calculated from the mass' initial height and from the local gravitational constant.

Further, the interaction with the rebounding surface will cause the mass to decelerate rapidly as the surface distorts to absorb the the transferred momentum until the masses velocity is reduced to zero. The compression caused to the surface during this process is, also, easily calculable.

The question poised, that day forty years ago, was – how much is the apparent weight of the mass at the moment that its velocity is reduced to zero. The answer is to be found in the force that it takes to deflect the elastic material a distance equal to that which it takes the mass to decelerate to zero velocity.

As an extension and through intuition, it is obvious that the apparent weight will be greater if the weight is elevated, say, a meter above the surface compared to the apparent weight when the mass is elevated a centimeter. The “hooker” that day was that, as the professor exited the lecture hall, his last remark was, to the effect -”Gentlemen, please submit as your homework assignment, a solution to this problem when the distance the mass falls is set to zero.”

Thus, this was a pretty standard applied materials problem until the introduction of the “hooker”. In the ensuing discussion, we, the students, reformulated the problem by recasting of the rebounding surface to that of the deck supported by a coiled spring, the deck in turn being “geared” to a pointer to register the degree of spring compression, (analogous to, say, a weighing scale such as typically found in  a grocer's shop). Then, if the mass is released above the deck, the momentum it acquires will be “absorbed” in the compression of the spring of the scale mechanism, and when the mass reaches minimum velocity, the scale's pointer will register maximum “weight”. Within the bounds of this model, we then asked the question - if the mass were “raised” off the deck until the “scale” registered zero weight, but remains in contact with the surface, then released, what is the maximum weight the pointer will register.

I dimly recall that the solution was found in taking the limit of the function controlling the deceleration and that the answer was that the apparent “weight” of the mass was precisely twice the rest weight, but I can no longer reproduce the steps to arrive at this conclusion.

Any suggestion or comment would be appreciated. TIA —Preceding unsigned comment added by 208.92.185.250 (talk) 17:58, 30 January 2009 (UTC)


 * Initially the particle is at rest at height z0>0. The particle hits the spring at z=0. The particle is again at rest when it has compressed the spring to the maximum at z1<0. The spring force on the particle, F = &minus;kz1, is the 'weight'. The conserved energy of the particle is  mgz0 = mgz1+(1/2)kz12. The negative solution defines z1 as a function of m, g, z0, and k. For z0=0 you get z1=&minus;2mg/k and F=2mg. This is independent of k, which is nice, because the value of k was not provided. What was the name of that professor? Bo Jacoby (talk) 21:00, 30 January 2009 (UTC).

Thanks Bo Jacoby. I do not remember the professor's name but have contacted the university with a request for help in his identification. He was a fully tenured professor in the math dept at Rutgers Univ (New Brunswick) in the early 1960's. I believe that he retired in 1967 or 1968. It was widely rumoured that he often collaborated with Einstein, so teaching us first year graduate students must not have been very challenging.208.92.185.250 (talk) 21:32, 30 January 2009 (UTC)