a:=n->((-1)^(n+1)+4*n-1)/2:

seq(a(n), n=1..8);

           2, 3, 6, 7, 10, 11, 14, 15

eliminate({x-lambda*p=0, y-lambda=0}, lambda):

p=solve(op(%[2]), p);

           p=x/y

Example:

S:={[1, 1], [2, 1], [2, 2], [3, 1], [3,2]}:

f:=proc(x,y)

x^2+y^2:

end proc:

seq(f(op(S[i])), i=1..nops(S));

              2, 5, 8, 10, 13

If you frequently use the symbol perpendicularity it makes sense to write a special procedure. Formal arguments: L - a list of the coordinates of the center of the character, l - length of a stick, th - thickness, c - color, alpha - angle of the vertical stick from the positive direction of the horizontal axis.

PerpSymb:=proc(L, l, th, c, alpha)

local A, B;

uses plottools;

A:=line(L,[L[1]+l*cos(alpha), L[2]+l*sin(alpha)], color=c, thickness=th);

B:=line([L[1]-l*sin(alpha)/2, L[2]+l*cos(alpha)/2], [L[1]+l*sin(alpha)/2, L[2]-l*cos(alpha)/2], color=c, thickness=th);

A, B;

end proc;

Example:

plots[display](PerpSymb([1,0.7], 0.15, 2, red, 2*Pi/3), PerpSymb([1.5,1], 0.2, 4, black, Pi/2), PerpSymb([2,0.7], 0.25, 6, blue, Pi/3), view=[0..3, 0..2], scaling=constrained);

with(LinearAlgebra):

do M:=RandomMatrix(3, 3, generator=1..9):

if Determinant(M)<>0 then break: fi:

od:

v:=RandomVector(3, generator=1..9):

'M'=M, 'v'=v;

x=LinearSolve(M, v);

The easiest way to solve this problem is by generating functions, noting that the number of integer and positive roots of the equation is the number of non-negative integer roots of the equation

2a +3 b +10 c = 2010 - (2 +3 +10) = 1995

coeff(convert(series(1/(1-x^2)/(1-x^3)/(1-x^10),  x=0,  1996), polynom),  x^1995);

                                                              33367

The problem can be also solved programmatically as follows:

restart:

n:=0:

for a from 1 to floor(2010/2) do

for b from 1 to floor((2010-2*a)/3) do

for c from 1 to floor((2010-2*a-3*b)/10) do

if 2*a+3*b+10*c=2010 then n:=n+1: fi:

od: od: od:

n;

                                                       33367

Maple incorrectly decided your example 3. This equation has only one root, and this root is positive.

In accordance with your wishes

Newton := proc( f, x0, n, tol)

local x, g, k;

g := D(f);

x[0] := evalf(x0);

for k from 1 to n do

while evalf(abs(g(x[k-1])))<=10^(-8) do

x[k-1]:=x[k-1]+0.00001; od;

x[k] := evalf( x[k-1] - f(x[k-1])/g(x[k-1]) );

if abs(x[k]-x[k-1]) < tol then

return x[k]; fi;

od;

x[k-1];

end proc;

Examples:

Newton(x->1-x^2, 0, 20, 10^(-6));

             1.000000002

Newton(x->x^2-2, 0, 20, 10^(-6));

              1.414214589

You can build any number of triangles for which the coordinates of the vertices and the coordinates of the center of circle circumscribed  are integers as follows:

1) Take  any three points with integer coordinates that do not lie on a straight line. Let be points A, B, C.

2) Find the coordinates of the center of circle circumscribed about triangle ABC as the point of intersection of three planes. The coordinates of this point will be rational numbers.

3) Find the least common multiple of the denominators of the fractions.

4) Multiply the coordinates of all the points on this number.

As it should be |sin(x)|<=1, then we are looking for solutions only in the range x = -8 .. 8 .

The code:

RootFinding[Analytic](sin(x)=x/8, re=-8..8, im=-1..1);

Add the condition

(y1*z2-z1*y2)^2+(z1*x2-x1*z2)^2+(x1*y2-y1*x2)^2<>0

Topological structure is the same as in the figure. The form can be changed.

tubeplot({[(cos(5*t/2)+13/5)*cos(t), (cos(5*t/2)+13/5)*sin(t), 3*sin(5*t/2)/5],[(2/5*cos(7*t/3)+12/5)*cos(t), (2/5*cos(7*t/3)+12/5)*sin(t), 2*sin(7*t/3)/7]}, t=0..6*Pi, radius=1/5, style=surface,numpoints=2500, scaling=constrained, lightmodel=light4, orientation=[50, 20]);

Include your assumptions into the system.

Example:

solve(x^2=1, x>0});

        {x=1}

See this link http://www.mapleprimes.com/questions/136257-Symbolic-Operations-With-Arbitrary-Matrices

Two drawbacks:

1) The perimeter of a triangle is equal to 2p, not p.

2) If S>0, then all the vertices of the triangle are the different points.

Fixed lines:

if type(2*p, integer) and type(S, posint)

then L:=[op(L), [[0, 0, 0], [x1, y1, z1], [x2, y2, z2]]]: fi:

From a geometrical point of view is actually found  only one triangle with sides 3, 4, 5. To find other such triangles  not looking for them in space but on the plane, because search range can be substantially increased.