593 Reputation

8 Badges

16 years, 238 days

MaplePrimes Activity

These are answers submitted by resolvent

I am keeping things incredibly simple. The problem is - if things are too simple, classical mechanics predicts electrons and protons collapse. Of course, we know by quantum mechanics that they do not - that there exist minimal distances between them. I crudely use this as a boundary condition on the distances between particles.

I am making a static model - as if the electrons and atomic nuclei simply "hang there" in space, with absolute knowledge of their positions - in complete violation of Heisenberg's Uncertainty Principle.  So, no time derivative.

I am working on 2 electrons and 2 atomic nuclei right now: a tetrahedron. I am trying to prove that their configuration ought to be planar - in order to keep the 2 electrons as far apart from one another as possible and the atomic nuclei as far apart from one another.

On wikipedia, I found the determinantal formula for the volume of a tetrahedron - given the lengths of its 6 edges.  If one chooses the 6 lengths arbitrarily in such a way that this formula yields a negative value, then no such tetrahedron with those edge lengths exist.  The volume =0 is the degenerate / critical / boundary case.

I cannot really make use of calculus (derivative = 0) to find the minimum value of the total potential energy of this system

A point in my network is either an electron or an atomic nucleus.

E = sum over i<j of Zi*Zj*ei*ej/rij where Zi = atomic number (Zi=1 for an electron), ei=e if an electron, ei=-e if an atomic nucleus


Thanks, Alex Smith! But, the output of your code seems to be just the code itself. However, thank you for introducing me to a new set of commands Procedures. I will have many questions very soon about that.
I always had Document Mode and Worksheet Mode backwards. I had always assumed I was working in Worksheet Mode when I found out I had been in Document Mode all along.
My program - the uploaded one - has added comments which have disabled my program. Why?
Please replace this text with the link to your file. The link can be found in the File Manager View 5348_Maple Flow Control on MapleNet or Download 5348_Maple Flow Control
View file details
How were you able to write code with line breaks something 1; something 2; something 3; without setting off the execution mode in Maple after hitting the Enter key? Please let me know, as this will be very useful to me. Also, where is the Units template? I use the Help menu all the time. But, to assign units to a variable, they ask me to go to the Units template. But, I do not see the Units template on the Worksheet. Thanks.
That's always the best way to get a problem solved, anyway. Too bad (yes, I am being superstitious) that figuring out a problem on my own comes only after I beg the entire world for an answer. Turns out I was simply making wrong assumptions. Given a partition A of N, for most (random) choices of another partition B of N, there does NOT exist a domino tabloid of shape A of type B. The parts of B have to be parts of a (sub)partitions of each part of A. Plus, I was confusing a "tabloid" with a "tableaux". My bad. Ian Grant Macdonald is the only person on earth, it seems, who can help me with this stuff. I wish I could become an expert on his book all by myself, just by doing the examples. I'm just not self-disciplined enough, so it seems. I need to use all this theory all the time. In case anyone else is curious, here's what I got. Let z^n + (-1)^(n-s)*e(n-s)*z^s + (-1)^n*e(n) be trinomial in z. e(n-s) = the n-s-th elementary symmetric function in this trinomial's roots. e(n)= the n-th elementary symmetric function of the roots. Let p(m)= the m-the powersum of the roots. m is a positive integer, with p(0)=n. Then p(m)= the sum of (((a+b)-choose-a)*n - ((a+b-1)-choose-b)*s) times ((e(n-s))^a )*((e(n))^b) * ((-1)^(m-a-b)) over all nonnegative integers a and b such that a*(n-s)+b*n = m. This is a "one-dimensional" sum, due to the condition a*(n-s)+b*n = m on a and b. One can replace b with (m - a*(n-s))/n and add the condition "sum over all nonnegative integers, a, such that (m - a*(n-s))/n is a nonnegative integer".
To Robert Israel and John Fredsted and other interested parties: Of course, I need to edit out even MORE junk from it just when I thought I HAD something new (that Corollary 2.3 is nothing new nor applicable just to resolvents). Plus, I need to keep my paper as short as possible because I will submit it to the International Journal of Mathematics & Mathematical Sciences which charges authors by the page to publish. That's ok, though. I've become pretty emotionally inured to editing - from editing math papers to editing videos - having done a lifetime of it. I can be pretty "cruel/cool" and "objective" about my own work (and the work of others). (At least I like to think I am.) I've got a whole bunch more truly original ideas (math and applied math) stacked up for me to work on and perhaps someday publish.
Robert Israel: (y'(1) - 9)^(5/2) = -5 does have a real solution, since y'(1)=9 + (-5)^(2/5) = 9 - 5^(2/5) power I do not have any need a solution for this particular equation. I deliberately through a lot in there just to test Maple's capabilities. So I can type this, I will pick a simpler ODE (and one without the functional term!), say, (y')^2 + y^3 = x given y(0) = a < 0 So y'(0)= - abs(a)^(3/2) (I would settle for not even caring about Maple expressing the multiplicity of initial conditions when they appear algebraically. If y'(0) is complex, let it be.) What I was hoping Maple would have built-in was this feature. 2y'y" + 3y^2*y' = 1 => y"(0) = (1 - 3*y(0)^2*y'(0))/(2y'(0)) 2(y")^2 + 2y'y''' + 6*y*(y')^2 + 3y^2*y" = 0 => y'''(x) = rational function of y, y',y" so compute y'''(0) So we can compute the first three terms of a Taylor series expansion y(x) = y(0) + y'(0)*x + y"(0)*x^2/2! + y'''(0)*x^3/3! I'll worry about convergence of anything later.
1 2 Page 2 of 2