## 19069 Reputation

14 years, 145 days

## Solving of cryptarithms with Maple...

Maple

A kryptarithm - this is an example of arithmetic, in which all or some of the digits are replaced by letters. The rule must be satisfied: the different letters represent different digits, the identical letters represent the identical digits. See the link  http://en.wikipedia.org/wiki/Verbal_arithmetic.

The following procedure, called  Ksolve , solves kryptarithms in which are used four operations ...

## Explicit equation of arbitrary continuou...

Maple

It's interesting that every continuous piecewise linear function can be specified by one explicit equation with absolute values​​. The procedure JoggedLine carries out such conversion.

Formal arguments of the procedure:

A - a list of the coordinates of the vertices of the polyline or the continuous piecewise linear expression defined on the entire real axis.

B (optional) - a point on the left "tail"...

## Perimeter, area and visualization of a p...

Given a figure in the plane bounded by the non-selfintersecting piecewise smooth curve. Each segment in the border defined by the list in the following format (variable names  in expressions can be arbitrary):

1) If this segment is given by an explicit equation, then  [f(x), x=x1..x2)]

2) If it is given in polar coordinates, then  [f(phi), phi=phi1..phi2, polar] , phi is polar angle

3) If the segment is given parametrically, then  [[f(t), g(t)], t=t1..t2]

4) If several consecutive segments or entire border is a broken line, then it is sufficient to set vertices the broken line [ [x1,y1], [x2,y2], .., [xn,yn]]

The first procedure symbolically finds perimeter of the figure. Global variable  Q  saves the lengths of all segments.

Perimeter := proc (L) #  L is the list of all segments of the border

local i, var, var1, var2, e, e1, e2, P;

global Q;

for i to nops(L) do if type(L[i], listlist(algebraic)) then P[i] := seq(simplify(sqrt((L[i, j, 1]-L[i, j+1, 1])^2+(L[i, j, 2]-L[i, j+1, 2])^2)), j = 1 .. nops(L[i])-1) else

var := lhs(L[i, 2]); var1 := min(lhs(rhs(L[i, 2])), rhs(rhs(L[i, 2]))); var2 := max(lhs(rhs(L[i, 2])), rhs(rhs(L[i, 2])));

if type(L[i, 1], algebraic) then e := L[i, 1]; if nops(L[i]) = 3 then P[i] := simplify(int(sqrt(e^2+(diff(e, var))^2), var = var1 .. var2)) else

P[i] := simplify(int(sqrt(1+(diff(e, var))^2), var = var1 .. var2)) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := abs(simplify(int(sqrt((diff(e1, var))^2+(diff(e2, var))^2), var = var1 .. var2))) end if end if end do;

Q := [seq(P[i], i = 1 .. nops(L))];

add(Q[i], i = 1 .. nops(Q));

end proc:

The second procedure symbolically finds the area of the figure. For correct work of the procedure, all the segments in the list L  of border must pass sequentially in clockwise or counter-clockwise direction.

Area := proc (L)

local i, var, e, e1, e2, P;

for i to nops(L) do

if type(L[i], listlist(algebraic)) then P[i] := (1/2)*add(L[i, j, 1]*L[i, j+1, 2]-L[i, j, 2]*L[i, j+1, 1], j = 1 .. nops(L[i])-1) else

var := lhs(L[i, 2]);

if type(L[i, 1], algebraic) then e := L[i, 1];

if nops(L[i]) = 3 then P[i] := (1/2)*(int(e^2, L[i, 2])) else

P[i] := (1/2)*simplify(int(var*(diff(e, var))-e, L[i, 2])) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := (1/2)*simplify(int(e1*(diff(e2, var))-e2*(diff(e1, var)), L[i, 2])) end if end if

end do;

abs(add(P[i], i = 1 .. nops(L)));

end proc:

The third procedure shows this figure. To paint the interior of the boundary polyline approximation is used. Required parameters: L - a list of all segments of the border and C - the color of the interior of the figure in the format color = color of the figure. Optional parameters: N - the number of parts for the approximation of each segment (default N = 100) and Boundary is defined by a list for special design of the figure's border (the default border is drawed by a thin black line). The border of the figure can be drawn separately without filling the interior by the global variable Border.

Picture := proc (L, C, N::posint := 100, Boundary::list := [linestyle = 1])

local i, var, var1, var2, e, e1, e2, P, Q, h;

global Border;

for i to nops(L) do

if type(L[i], listlist(algebraic)) then P[i] := op(L[i]) else

var := lhs(L[i, 2]); var1 := lhs(rhs(L[i, 2])); var2 := rhs(rhs(L[i, 2])); h := (var2-var1)/N;

if type(L[i, 1], algebraic) then e := L[i, 1];

if nops(L[i]) = 3 then P[i] := seq(subs(var = var1+h*i, [e*cos(var), e*sin(var)]), i = 0 .. N) else

P[i] := seq([var1+h*i, subs(var = var1+h*i, e)], i = 0 .. N) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := seq(subs(var = var1+h*i, [e1, e2]), i = 0 .. N) end if end if

end do;

Q := [seq(P[i], i = 1 .. nops(L))];

Border := plottools[curve]([op(Q), Q[1]], op(Boundary));

[plottools[polygon](Q, C), Border];

end proc:

Examples of works:

Example 1.

L := [[sqrt(-x), x = -1 .. 0], [2*cos(t), t = -(1/2)*Pi .. (1/4)*Pi, polar], [[1, 1], [1/2, 0], [0, 3/2]], [[-1+cos(t), 3/2+(1/2)*sin(t)], t = 0 .. -(1/2)*Pi]];

Perimeter(L); Q; evalf(`%%`); evalf(`%%`); Area(L);

plots[display](Picture(L, color = grey, [color = "DarkGreen", thickness = 4]), scaling = constrained);

plots[display](Border, scaling = constrained);

Example 2.

The easiest way to use this  procedures for polygons.

L := [[[3, -1], [-2, 2], [5, 6], [2, 3/2], [3, -1]]];

Perimeter(L), Q;

Area(L);

plots[display](Picture(L, color = pink, [color = red, thickness = 3]));

Example 3 (more complicated )

3 circles on the plane C1, C2 and C3 defined by the parametric equations  of their borders. We want to find the perimeter, area, and paint the figure  C3 minus (C1 union C2) . For details see attached file.

C1 := {x = -sqrt(7)+4*cos(t), y = 4*sin(t)};

C2 := {x = 3*cos(s), y = 3+3*sin(s)};

C3 := {x = 4+5*cos(u), y = 5*sin(u)};

L := [[[-sqrt(7)+4*cos(t), 4*sin(t)], t = -arccos((1/4)*(7+4*sqrt(7))/(sqrt(7)+4)) .. -arctan((3*(-23+sqrt(7)*sqrt(55)))/(23*sqrt(7)+9*sqrt(55)))], [[3*cos(s), 3+3*sin(s)], s = -arctan((1/3)*(9+sqrt(7)*sqrt(55))/(-sqrt(7)+sqrt(55))) .. arctan((1/3)*(-9+4*sqrt(91))/(4+sqrt(91)))], [[4+5*cos(u), 5*sin(u)], u = arctan((3*(41+4*sqrt(91)))/(-164+9*sqrt(91)))+Pi .. arctan(3/4)-Pi]];

Perimeter(L), Q; evalf(%);

Area(L); evalf(%)

A := plot([[rhs(C1[1]), rhs(C1[2]), t = 0 .. 2*Pi], [rhs(C2[1]), rhs(C2[2]), s = 0 .. 2*Pi], [rhs(C3[1]), rhs(C3[2]), u = 0 .. 2*Pi]], color = black);

B := Picture(L, color = green, [color = black, thickness = 4]);

plots[display](A, B, scaling = constrained);

More examples and all codes see in attached file

Plane_figure.mw

## Another proof of stunningly beautiful id...

I propose a different proof of this remarkable identity (see  http://www.mapleprimes.com/posts/144499-Stunningly-Beautiful-Identity-Proved ) in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals.

For the proof, we need three simple identities with cubic roots (a, b, c -any real numbers):