Maple Questions and Posts

These are Posts and Questions associated with the product, Maple
I have a small Maple library (created with march and savelib) that I want to distribute to some people. It seems to work fine, but if I zip it (with the idea of mailing it), the resulting zipped directory under windows seems to throw up security warnings and gets rejected by my outgoing mail client for being "oversize" although it is in fact small. Are Maple libraries so created just standard files or are they special in some way? I'd be grateful for any insight. Toby
Hello ! I need to create a graph of f(a,b,c) = 1+b/a*c, where "b" goes from 30 to 120, "c" from 0 and 30, and "a" would be from "b" to 220. (but if that part is too complicated, "a" going from 120 to 220 would be okay) The way I visualize this graph is like a typical 3d graph, with no animation, but with a special color mapping. Let's say my variable "a" would be on the x (horizontal) axis variable "b" on the y (depth) axis variable "c" on the z (height) axis then color of each point would be mapped to the result, f(a,b,c) How can this be done ? Thanks you very much !
I need help... I'm running the code attached below which has procedure sdef(). After obtaining results from it, I do a couple of substitutions Esc:=subs(b=b(v),Esc); Esc:=subs(v=w,Esc); at the very end -- but it turns out that NOT ALL b is substituted for b(v) and not all v is substituted for w in Esc. Been trying to understand what's going on for a day and could not... The code: > restart; > interface(showassumed=0); > > sdef:=proc(x) > global b,v; > local m, S, S1, S2, EStmp1, EStmp2, betasol, beta; > m:=1; > > S2:=t*(x+v)*beta; assume(beta>=0,beta
Hello, perhaps my question is really stupid but I cannot find any hints on the help pages (Maple 10): How can I find the lower and upper bound of a range, i.e, "a" and "b" from "a..b"? The following does work lower:=proc(r::range) min($r) end proc ; but is, of course, very inefficient for large ranges and only works for integer bounds.
Considering updating my old package COSVAM, which treats complex-octonionic-valued scalars, vectors, and matrices, first of all I want a much better type check of whether an expression is indeed a complex-octonionic scalar. A complex-octonionic-valued scalar is a linear combination over the complex numbers of the eight units of the octonions 1,e1,...,e7, say. So all the following expressions are valid examples of complex-octonionic-valued scalars
expr1 := 10;
expr2 := e1;
expr3 := e1*I;
expr4 := e1*cos(t)+e2*sin(t);
I have one awful, long way to solve the problem below, but I am sure there is also an elegant solution; unfortunately, it currently escapes me. Consider a rigid sphere of radisu 1 and center at 0 that rotates about its center. The angular velocity is omega(t) = Vector([ cos(t), sin(t), sqrt(3) ]). Does the path of the point starting at (0,0,1) ever reach the same point at a later time?
is there a way to get maple to perform simple logarithm manipulations i.e. ln(a*b^c)=ln(a)+c*ln(b) ? I've noodled around with the convert utility with no luck. I am trying to do a power law regression via the normal equation "showing my work". I have the answer using CurveFitting so I'm just trying to fill in the steps thanks
From an exercise in "Game Physics", we have the following piecewise position vector r. We wish to show that it is well-defined (as well as twice differentiable) at 0. The goal of the exercise is then to show that the Normal vector is not even continuous. So we try: p1,p2 := <t,t^3,0>, <t,0,t^3>; r := piecewise( t<0, p1, p2); limit(r,t=0,left); and unfortunately that limit gives RTABLE(149559852,MATRIX([[t], [t^3], [0]]),Vector[column]) as a result! The obvious thing to try next is to "push in" the piecewise into the components. But the
Hi, I have a small problem. I want to export code to Matlab. The code is mostly a long expression with some named constants. The constants are set to some values in Maple and in Maple only the x-variables are therefore unknown in the expression. However, in the exported code, some of the constants are not replaced by their values. Does someone have a clue what this depends on? Best regards Johan
Suppose the adjacency matrix M is given. How we can get the graph of it from Maple? Can we draw the graph now? Thanx.
can I change the behavior of maple to not represent 1/sqrt(2) as sqrt(2)/2? I guess in a mathmatician's view this is a simplification, but not from an engineering standpoint. also, if I perform combine(sqrt(2)/(sqrt(m)*sqrt(k)),radical,symbolic) I get sqrt(2)*sqrt(1/m k) instead of the desired sqrt(2/(k*m)) this related? I'm sure there is a nuance that I am not getting...
i am trying to plot a moving particle on two separate sphers of different radii. i have the code: restart: > with(plots): > with(plottools): > p1:=plot3d([3*sin(u)*cos(v),3*sin(u)*sin(v),3*cos(u)], u=0..Pi, v=0..2*Pi,color=red, > style=wireframe,axes=normal,scaling=constrained): > p2:=plot3d ([2.9*sin(u)*cos(v),2.9*sin(u)*sin(v),2.9*cos(u)], u=0..Pi, v=0..2*Pi,color=blue,axes=normal,scaling=constrained): > p3:=display(p1,[seq( pointplot( { [3*sin(u)*cos(v)(k/50)(n/50),3*sin(u)*sin(v)(k/50)(n/50),3*cos(u)(k/50)(n/50)],[2.9*sin(u)*cos(v)(k/50)(n/50), 2.9*sin(u)*sin(v)(k/50)(n/50),2.9*cos(u)(k/50)(n/50) ] } ) , (k,n)=(0..314,0..314) ) ],insequence=true):
If i have a sequence of r[i+1] = 2r[i-1] + 3r[i] (where [i] represents subscript i) how do I use the 'do' function to go from the point where initially r[i-1] is, for example, 3 and r[i] is 4 to the point at which r[i-1] is first greater than 300 and find r[i+1] at this point. I hope this question makes sense! Thankyou
Have spent much of the weekend reading the maple help files and googling but still haven't found a current way to do this on XP with v9.5. Tried editing the maple9.5.ini file to point to the folder, but it doesn't work and seems to behave more like a log file. I can use libname as a line in the worksheet but would prefer to do without as I use the modules a lot. Any suggestions would be welcome. Neil
How do I plot f(x) in the following question? f:=x->sum(h(2^n*x)/(2^n),n = 0 .. 20) h(x)={ 2x , 0<><>
First 1991 1992 1993 1994 1995 1996 1997 Last Page 1993 of 2119