Is there a way to track a thread without responding to it?

I have two small questions: 1. Given a list L: L:=[false, false, true, true, FAIL,false]; and I want to get the positions of "FAIL" (in this case 5) and false (1,2,6). HOW could I get that? 2. Now, given another list Z with the same number of elements: Z:=[z1,z2,z3,z4,z5,z6]; I would like to replace the values of FAIL by 0 (zero). (in the way that Z became: Z:=[z1,z2,z3,z4,0,z6];) It is possible to do that without using "for do" commands? Thanks in advance, JJacques

Does anyone know if it's possible to discover the largest known prime number just using Maple 10 on a stand alone machine (PC)? Or does Maple 10 and or all of it's current additions / upgrades lack the computing power required to attempt this possible recreational task ? v/r, Dan

Hi dear Maple's gurus. I have the lists a and b: a:=[A, B, C] b:= [[K1,L1][K2,L2][K3,L3]] I want to add the each term of "a" in the intervals given in "b": sum(A from L1 to K1)+ ..+sum(C from L3 to K3). I tried with this procedure: > sums:= proc(a,b) local j,tot; tot:=0; for j from 1 to nops(a) do tot:=tot+sum(a[j], k=b[j,2]..b[j,1]); end do: end proc: >sums([a1,a2,a3],[[ku1,kl1],[ku2,kl2],[ku3,kl3]]); 1. Is there some more efficient way to do the same? 2. I would like also that the sums on a[j] under the following 2 join conditions: 2.1. assuming that a[j] is real and positive, and

Hi there! Assuming that I have the following fraction: fr1:= (x+1)/(x-1); I have the following 2 questions: 1. with which command could I take only the numerator "(x+1)"? (and for the denominator?) 2. Is it normal that whattype(fr1); returns a star * Thanks in advance for your answers, Jean Jacques

Simple question: I can't seem to use a negative number as a minimum value for a slider control (in Maplet Builder)... Is this not possible? Thanks! Bryan

Hello, I am relatively new to Maple and using version 9.5. Does anyone out there know of a way for symbolic characters such as Pi to appear in a plot title? This is the input I am working with. f:=x->piecewise(x < Pi/3,4*sin(x),Pi/3 <= x,x^3-x+1); plot (f(x),x=-Pi..Pi,discont=true,title="f(x) [-Pi, Pi]",titlefont=[TIMES,BOLD,16]); Any help would be greatly appreciated. Thanks

i got the solution of a ODE system (dsn), the results are time dependent variable theta[i](t),and its deriavtives diff(theta[i](t),t). and i have one set of expressions, which are explicit form of theta[i](t),and diff(theta[i](t),t), such as ss[1]:=fx[1]=sin(theta[1](t))*3+cos(theta[1](t))*diff(theta[2](t),t). I want to put the solution i got into the explicit expression of fx[i], and then polt them, how can I do it? i use eval(ss[1],dsn)but it does not work. here is the code: > restart: interface(warnlevel=0): with(SolveTools):with( plots ): with( plottools ): with( DEtools ): with( PDEtools ): g:=9.81: x[0][2]:=0: y[0][2]:=0: #initialization function of each beam,in this function, we should give 3 variables, n=the number of this #beam, la=the length of this beam, ma=the mass of this beam beams:=proc(n,la,ma) global l,m,x,y: l[n]:=la: #length of beam n m[n]:=ma: #mass of beam n #initialize the positions of each beam x[n][2]:=0: #the x position of the end point of each beam y[n][2]:=0: #the x position of the end point of each beam x[n][1]:=0: #the y position of the center point of each beam y[n][1]:=0: #the y position of the center point of each beam end proc; #define the forces that applied on the end of beam n Force:=proc(n,fm,fn) global fx,fy: fx[n]:=fm: fy[n]:=fn: end proc:

when i do this loop for i from 0 to 10 do f(i) mod 11; end do; it only outputs the last value, ie f(10) mod 11 i want it to output all of them, what is going on here?

Hello, contourplot can be given a list of contours levels or a integer specifying the number of evenly spaced contour levels it should plot. In the second case, or even in the default case, is there a way to know what these contour levels are? The legend only shows Curve1, Curve2, ... but to which contour levels do these curves correspond? -- Thanks in advance, Franky.

Hello, I was wondering just how evaluation to n levels works as compared to the default eval statement (Maple 10). As an example, consider this: a := 1; whattype(''a''); // uneval whattype(eval(''a'')); // symbol whattype(eval(eval(''a''))); // integer The same information can be retrieved by adding a number to eval, indicating the level. as in: a := 1; whattype(eval(''a'',1)); // uneval whattype(eval(''a'',2)); // symbol whattype(eval(''a'',3)); // integer So I would expect the same behavior for the following example: whattype(''sin(2.0)''); // uneval whattype(eval(''sin(2.0)'')); // function

Hi, I think I will go crazy if I have to type

maple("printf(MathML:-ExportPresentation( ... ))");

one more time. It would be great to have an abbreviate version of this like

mapleml("...");

or something. Or how about a function available in the question bank editor to translate short LaTeX phrases into mathml? This would solve a huge number of my Maple T.A. issues. Anyway, along these lines, what I'd really like to do right now is ask my students about double integrals and display something like $\int\int_R f(x,y)\ dA$ Any suggestions? Can I use mathml() to display symbols like the integral?

Given an algebraic expression p(x,y)..it is possible to factorize assuming that some terms are involved?. For example if I have poly1 and I expand it, I obtain a poly2 (with several terms).. now, if I factor(poly3) I dont recover the exact expression poly1. But at least in this case, it would be possible to factorize (or collect) poly3 asumming that in numerator and denominator we have the term(x+y)?. Thanks in advance, JJacques. poly1:= ((1-x-y)^2+ (2+x+y)^3)/((1+x+y)^2+ (4-x-y)^2); poly2:= expand(%); poly3:=factor(%);

Hi, I cannot understand why IS fails below. Am I doing something wrong?. Maybe I generated a new variable when I wrote 2*L/(1-x-y).. Thanks a lot for reading this message, > restart; > assume(L=1/2); > assume(x>0,y>0); > additionally(x+y <> is(x <>"true" > is(2*L/(1-x-y)<>"FAILS" JJacques,

can i use maple to solve 15 nonlinear equations in 15 unknowns in the following form: T1^2 + 5(T1T2) + 6(T1T3)+ 8 (T1T4) + 7 (T1T5) + ....... -5 (T1T15) = 0 T2^2 + 3(T2T1) + 6(T2T3) -3 (T2T4) + 5 (T2T5) + ....... -3 (T2T15) - (T1T3) = 0 where T1 + T2 + T3 + T4 + ... + T15 =1