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What is the number of all the solutions of the equation frac(x*floor(x)) = 1/2 belonging to RealRange(1,100)? How to count it with Maple?


I'm trying to solve system of linear and nonlinear equations with inequalities and it looks like this:


    SX := solve(
                       seq(diff(EX,WX[k+1])=0, k = 0..m), # these are linear

x = a*(3*cos(t) - cos(3*t))
y = a*(3*sin(t) - sin(3*t))

is there a library or function to convert above parametric equations into one equation in terms of x and y?

sys := x = a*(3*cos(t) - cos(3*t));
k := solve(sys, t);
simplify(subs(t=k[1], y = a*(3*sin(t) - sin(3*t))));

any simpler form?


I have this matrix with coefficients that I need to estimate;

I get there after some simple calculations and I know that my matrix it´s equal to the zero matrix; I have something like:


A:=Matrix(3, 3, [5*a-4,  5*sqrt(a) *sqrt(b)-5, 7*sqrt(a) *sqrt(c)-6,  

        8*sqrt(a) *sqrt(b)-5, 8*b-2,  8*sqrt(b) *sqrt(c)-9,   


I am using the function solve() to find roots of a trig. equation. Such as for sin(k*x) = 0, Maple retuns x = 0, whereas I expecting to get  x = n ∏/k, for n = 0,1,2,.... I am sorry I am new to Maple, can anyone help me get what I am looking for?

Thank you


Has anyone seen this form of a nonlinear equation with respect to X, but linear with respect to Y & Z?  I provided a contour plot within the region for all 3 variables between -2 & 2.  The plot is actually Z*conjugate(Z) so that the magnitude is above ZERO.  If I am correct I may have seen this before in orbital mechanics, but my undergrad years are a bit...

As many of you know now the MRB constant = sum((-1)^n*(n^(1/n)-1),n=1..infinity).

Here are some equations involving various forms of that summation.

The first one involves convergent series and is too obvious. The others involve divergent series.

The last two, however, are new!


Let c=MRB constant and a, c~, x, and y = any number.


sum((-1)^n*(c~*n^(1/n)-c~),n=1..infinity)= c*c~.

evalf(sum((-1)^n*(n^(1/n)-a),n=1..infinity)) gives c-1/2*(1-a).

evalf(sum((-1)^n*(x*n^(1/n)+y*n),n=1..infinity)) gives (c-1/2)*x-1/4*y.

And it appears that

evalf(sum((-1)^n*(x*n^(1/n)-a),n=1..infinity)) gives (c - 1/2)*x + 1/2*a.

I would like to know if there is any way to coax maple into spitting out the solution (1,1,1) for the system of equations {x2+y2+z2=3, x+y+z=3}. So far all I have come up with is


This produces


as desired, but I would like to accomplish this without specifying the polynomial...

I'm new to using Maple. I'm trying to evaluate an equation where all unknowns are given. So to say use Maple

as a kind of desktop calculator with the possibility to change certain input variables and see what the outcome is.


I'm wondering why the trigonometric functions are not evaluated.


I want to solve the equation (1 + 1/x)*(1 + 1/y)*(1 + 1/z)=2 with integer solutions. I tried 

isolve((1 + 1/x)*(1 + 1/y)*(1 + 1/z)=2);

and I get {x = -1, y = -2, z = 0}. Is this a bug?

I have a differential equation having a parameter say 'a'.

If I increase 'a' by 10%, got change in the solution as well.

How to calculate change in percentage in the solution?.


 I am conveting ordinary differntial equations into polynomial using differential tranformed method and also after using pade arroximants for finding converge point at infinity. because differential transform cannot satisfy boundary at infinity. EQ1 and EQ2 does not built up fromt he preceeding iterations. its pade approximants diagonal value results like pade[3,3],[4,4],[5,5] and so on. i can solve this type for single equations but when  i solve coupled equations...


recently I was helped on this forum when I asked how to work with exterior products:

Now I would like to know how I can efficiently let Maple solve systems of equations between wedge products.


For instance, I would ask Maple to solve 



I'm currently working on the modelling of a thermodynamic process.

Briefly, I cool down a solution (water + polymer) from -5°C to -15°C to induce a phase separation. At the end (and after removing of the water by lyophilisation) I obtain a porous sponge like material.

The process uses a home made cooling system which can be described like this:

- A Peltier module

- An aluminium layer recovered by teflon (And also a layer of ethanol)

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