vv

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@janhardo You should be aware that Maple can help (in Calculus here) only at a formal level. To understand things you must study carefully a good textbook and be able to solve at least the simple exercises by hand. Not doing so, you are wasting your time.

@janhardo Is this a question, whether f(x) and f(y) are two functions?
You have only one f, in two variables. y is a variable, no y(x) here! 
Try to use proper terms and notations, otherwise everything becomes a mess.

@mthkvv I'd consider writing a procedure P(m) using some generated polynomials (preferably with known results).
Then call P(m) for an increasing sequence of m's inside CodeTools:-Usage and try to figure out the amount of needed memory as a function of m. [It probably also depends of how dense the polynomials are].

@Pascal4QM `--` is an assignment operator in Maple 2020 (2019?)

I don't think that a teacher would be very happy to see one of his students writing: -3 - -5  (unless he missed the class about parantheses).
The next level would be:  --5+-7-+2++3*-1

 

@mmcdara Unfortunately it is easy to find modern "applied" math formulations: "The function y = f(x) ..." or "Let f(x) be the function ...".
Even Maple is not very careful with the math terminology. E.g. a better name for the type function would have been funcall.

@Carl Love It should be better:

proc(L::{list, set}, $) option overload; ilcm__orig(L[]) end proc

 

@acer  Let's consider just polynomials in one variable over Q.
Is there a good simplifier for them? compoly could be a candidate but it most often fails.
ex := -10*x^9 + 45*x^8 - 120*x^7 + 210*x^6 - 252*x^5 + 210*x^4 - 120*x^3 + 45*x^2 - 11*x + 1;

@nm If you need all the chains for each eigenvalue then this is equivalent with finding the Jordan form and the transition matrix. So practically you will have to program a new JordanForm. It remains to see whether it will be better than Maple's version.

@dharr  Maple computes diff(HeunB(a1,a2,a3,a4, u(z)), z,z);  with a formula having u(z) at the denominator.
 

@janhardo  The geometric idea is simple. A path integral equals the length of the curve for f = 1.

E.g. if the curve is the unit circle in the xy plane and S is the right circular cylinder based on this circle and the height is 1, then the area of the cylinder equals (obviously) the length of the circle:
plots:-display(
plot3d([cos(t), sin(t), z], t=0..2*Pi, z=0..1, color=pink, style=surface, transparency=0.25),
plots:-spacecurve([cos(t), sin(t), 0], t=0..2*Pi, color=blue, thickness=3)
);

@Carl Love The replacement is necessary for r>0 too.

@Carl Love 
Thanks. I wonder why the formula was not included somewhere (e.g. in `simplify/Psi` or some convert); maybe it was forgotten.

Your extension is wrong for Psi(r) when r is a negative rational; probably because the original Psi has 2 arguments (or recursive overload problem); it would be interesting to investigate.

Edit. The mistake: (n-1)*(q+1) must be replaced with (n-1)*q+p

Just curious: what's wrong with a .mla update? It used to work well in the past. Is it not the case in "modern Maple"?

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