Alfred_F

Mr. Alfred Flaßhaar

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1 years, 267 days
Brandenburg, Germany
As a retired individual with degrees from German universities in mathematics/analysis and structural engineering, I spent my professional life in responsible positions in research, teaching, and practical application, working on the mathematical modeling of states and processes in real-world systems. Now I have the time to explore interesting mathematical problems using Maple. It is my professional curiosity that drives me.

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These are questions asked by Alfred_F

For quite some time, I have wanted to solve the system attached in "test" using Maple. The smallest solution in natural numbers x, y, and z test.mw

restart

kernelopts(version)

`Maple 2026.0, X86 64 WINDOWS, Apr 28 2026, Build ID 2011354`

(1)

interface(version)

`Standard Worksheet Interface, Maple 2026.1, Windows 11, April 28 2026 Build ID 2011354`

(2)

with(NumberTheory)

isolve({x*y*z = w^2, x+y+z = u^2, x*y+x*z+y*z = v^2})

{u = _Z1, v = 0, w = 0, x = _Z1^2, y = 0, z = 0}

(3)

"(->)"

{u = _Z1, v = 0, w = 0, x = _Z1^2, y = 0, z = 0}

(4)

``

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is known, and all these numbers are less than 4 × 10¹². Is this possible in Maple?

(x=1633780814400; y=252782198228; z=3474741058973)

...the essence of plane geometry is hidden within the following puzzle:
Given is a closed curve C. It is assumed to be non-self-intersecting, convex, and continuously differentiable everywhere (a closed Jordan curve). Let line segment AB be a chord of this curve, having a fixed length l. A point P lies on this chord at a fixed distance a from A and b from B, such that l = a + b. An orientation (or direction of circulation) is now assigned to the curve. The chord is then moved continuously along the closed curve in this assigned direction of circulation. As it moves, point P traces out a so-called locus curve O, which—upon completion of one full revolution of the chord—also forms a closed curve lying entirely within C.
The task is to calculate the area of ​​the region between C and O (i.e., the area lying inside C but outside O). Divide the result by the product of a and b, and then apply the "identify" function to the outcome.

To get better acquainted with "plot3d," I browsed through the help documentation. I was surprised to discover not only "Polyhedraplot," but—via "Polyhedra Sets" and "DualSet"—the very tools needed to tackle a classic subject: The duality of Platonic solids.
This reminded me of the following puzzle:
Consider the unit cube (edge ​​length = 1) and, situated within it, its dual polyhedron (the octahedron). As is well known, the dual polyhedron of the octahedron is, in turn, a cube. Now, let us continue this dual construction indefinitely, starting from the unit cube (cube containing octahedron containing cube containing octahedron...). This generates a sequence of nested polyhedra—alternating between cubes and octahedra.
1.) For each of the infinite subsequences—that of the cubes and that of the octahedra—calculate the sum of all volumes and surface areas.
2.) What is the maximum volume an octahedron (regardless of its orientation) can have while being completely contained within the unit cube?

In the attached file, I would like to practice calculating distances between the exact and approximate functions using the "Minimize" function. However, the approach using the integral metric according to (4) produces an error message. I am unable to correct this and would appreciate some advice.test.mw

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with(Optimization)

assume(0 <= r and r <= 100*Pi)

"term(a,b,c,d,r):=(exp(sin(r))-(a&lowast;sin(b&lowast;r+c)+d));"

proc (a, b, c, d, r) options operator, arrow, function_assign; exp(sin(r))-a*sin(b*r+c)-d end proc

(1)

Minimize(term(s, t, u, v, r)^2, {-100 <= s, -100 <= t, -100 <= u, -100 <= v})

[0.443734259186819141e-30, [r = HFloat(0.8573907336951694), s = HFloat(1.0651237756780674), t = HFloat(0.9373750940497176), u = HFloat(0.9414213728272717), v = HFloat(1.0810520913486565)]]

(2)

Minimize(abs(term(s, t, u, v, r)), {-100 <= s, -100 <= t, -100 <= u, -100 <= v})

[0.206279437975354085e-12, [r = HFloat(0.8598182185084339), s = HFloat(1.0763500233776817), t = HFloat(0.9650602699090793), u = HFloat(0.9650599306574976), v = HFloat(1.0839652198942415)]]

(3)

"term1(a,b,c,d):=(&int;)[0]^(2&lowast;Pi)(term(a,b,c,d,r))^(2)&DifferentialD;r;"

proc (a, b, c, d) options operator, arrow, function_assign; int(term(a, b, c, d, r)^2, r = 0 .. 2*Pi) end proc

(4)

Minimize(term1(s, t, u, v), {-100 <= s, -100 <= t, -100 <= u, -100 <= v})

Error, (in Optimization:-NLPSolve) integration variable previously assigned, 2nd argument evaluates to 1. = 0 .. 2*Pi

 

NULL

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...and starting values. I want to reproduce the maximum in Maple using the solution structure in the attached file, similar to the old Mathcad method. Instead of using the Euler-Lagrange equation, I want to separately enter the objective function, then the boundary conditions, and finally the starting values ​​for an iteration (as was done in the old Mathcad solution block). I entered the latter into the "maximize" command using the Maple help text. But nothing is being calculated. What am I doing wrong? As I said, I only want a numerically generated approximate solution.test.mw

restart

kernelopts(version)

`Maple 2026.0, X86 64 WINDOWS, Mar 05 2026, Build ID 2001916`

(1)

interface(version)

`Standard Worksheet Interface, Maple 2026.0, Windows 11, March 05 2026 Build ID 2001916`

(2)

with(VariationalCalculus)

[ConjugateEquation, Convex, EulerLagrange, Jacobi, Weierstrass]

(3)

with(plots, implicitplot)

[implicitplot]

(4)

with(Optimization)

 

assume(x >= 0, y(x) >= 0)

"p(a,b,c,x):=a&lowast;x^(2)+b&lowast;x+c;"

proc (a, b, c, x) options operator, arrow, function_assign; a*x^2+b*x+c end proc

(5)

" q(d,f,x):=d&lowast;x^(2)+f&lowast;x;"

proc (d, f, x) options operator, arrow, function_assign; d*x^2+f*x end proc

(6)

"yn(a,b,c,d,f,g,x):=p(a,b,c,x)&lowast;exp(q(d,f,x))+g;"

proc (a, b, c, d, f, g, x) options operator, arrow, function_assign; p(a, b, c, x)*exp(q(d, f, x))+g end proc

(7)

" wn(a,b,c,d,f,g,xend):=(&int;)[0]^(xend)x&lowast;yn(a,b,c,d,f,g,x)&DifferentialD;x;"

proc (a, b, c, d, f, g, xend) options operator, arrow, function_assign; int(x*yn(a, b, c, d, f, g, x), x = 0 .. xend) end proc

(8)

"constr(a,b,c,d,f,g,xend):=(&int;)[0]^(xend)(sqrt(1+((&DifferentialD;)/(&DifferentialD;x)(yn(a,b,c,d,f,g,x)))^(2))-50)&DifferentialD;x=0;"

proc (a, b, c, d, f, g, xend) options operator, arrow, function_assign; int(sqrt(1+(diff(yn(a, b, c, d, f, g, x), x))^2)-50, x = 0 .. xend) = 0 end proc

(9)

 

maximize(wn(a, b, c, d, f, g, xend), constr(a, b, c, d, f, g, xend), initialpoint = {a = -0.2e-1, b = 1.06, c = 0.14e-2, d = 0.46e-3, f = -0.12e-2, g = -0.14e-2, xend = 30})

maximize(-(1/16)*(8*a*xend^2*(-d)^(11/2)*exp(d*xend^2+f*xend)-8*g*xend^2*(-d)^(13/2)+4*a*f*xend*(-d)^(9/2)*exp(d*xend^2+f*xend)+8*b*xend*(-d)^(11/2)*exp(d*xend^2+f*xend)+2*exp(d*xend^2+f*xend)*(-d)^(7/2)*a*f^2+4*b*f*(-d)^(9/2)*exp(d*xend^2+f*xend)+8*c*(-d)^(11/2)*exp(d*xend^2+f*xend)-erf((1/2)*(2*d*xend+f)/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*d^3*a*f^3+2*erf((1/2)*(2*d*xend+f)/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*d^4*b*f^2-4*erf((1/2)*(2*d*xend+f)/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*d^5*c*f+8*a*(-d)^(9/2)*exp(d*xend^2+f*xend)+erf((1/2)*f/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*a*f^3*d^3-2*erf((1/2)*f/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*b*f^2*d^4+4*erf((1/2)*f/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*c*f*d^5-2*a*f^2*(-d)^(7/2)-4*b*f*(-d)^(9/2)-8*c*(-d)^(11/2)+6*erf((1/2)*(2*d*xend+f)/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*d^4*a*f-4*erf((1/2)*(2*d*xend+f)/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*d^5*b-6*erf((1/2)*f/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*a*f*d^4+4*erf((1/2)*f/(-d)^(1/2))*Pi^(1/2)*exp(-(1/4)*f^2/d)*b*d^5-8*a*(-d)^(9/2))/(-d)^(13/2), int((1+((2*a*x+b)*exp(d*x^2+f*x)+(a*x^2+b*x+c)*(2*d*x+f)*exp(d*x^2+f*x))^2)^(1/2)-50, x = 0 .. xend) = 0, initialpoint = {a = -0.2e-1, b = 1.06, c = 0.14e-2, d = 0.46e-3, f = -0.12e-2, g = -0.14e-2, xend = 30})

(10)

 

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