janhardo

@lucaud What do you want to achiev...

Answered: janhardo 700

7 hours ago

0 0

@lucaud
What do you want to achieve now?

The system of odes is now for first orde...

Answered: janhardo 700

Yesterday at 10:15 AM

0 0

The system of odes is now for first order solved and it seems be better for handling?

>

$restart; with(plots); with(DEtools); M := 2; m := 1; R := 1; r := .3; g := 9.81; sys := {diff(x1(t), t) = x2(t), diff(x2(t), t) = (-g*(M+m)*sin(x1(t))-m*r*x4(t)^2*sin(x1(t)-x3(t))+m*(g*sin(x3(t))-x2(t)^2*(R-r)*sin(x1(t)-x3(t)))*cos(x1(t)-x3(t)))/(2*M*R-2*M*r-R*m*cos(x1(t)-x3(t))^2+R*m+m*r*cos(x1(t)-x3(t))^2-m*r), diff(x3(t), t) = x4(t), diff(x4(t), t) = (-(2*M+m)*(g*sin(x3(t))-x2(t)^2*(R-r)*sin(x1(t)-x3(t)))+(g*(M+m)*sin(x1(t))+m*r*x4(t)^2*sin(x1(t)-x3(t)))*cos(x1(t)-x3(t)))/(r*(2*M-m*cos(x1(t)-x3(t))^2+m))}; ics := x1(0) = (1/6)*Pi, x2(0) = 0, x3(0) = (1/2)*Pi, x4(0) = 0; sol := dsolve(`union`(sys, {ics}), numeric, output = listprocedure, abserr = 0.1e-7, relerr = 0.1e-7); theta_plot := odeplot(sol, [t, x1(t)], 0 .. 10, color = blue, title = "θ(t)", thickness = 2); phi_plot := odeplot(sol, [t, x3(t)], 0 .. 10, color = red, title = "φ(t)", thickness = 2); display([theta_plot, phi_plot], title = "Hoekbeweging in de tijd", legend = ["θ(t)", "φ(t)"])$

${diff(x1(t), t) = x2(t), diff(x2(t), t) = (-29.43*sin(x1(t))-.3*x4(t)^2*sin(x1(t)-x3(t))+(9.81*sin(x3(t))-.7*x2(t)^2*sin(x1(t)-x3(t)))*cos(x1(t)-x3(t)))/(3.5-.7*cos(x1(t)-x3(t))^2), diff(x3(t), t) = x4(t), diff(x4(t), t) = 3.333333333*(-49.05*sin(x3(t))+3.5*x2(t)^2*sin(x1(t)-x3(t))+(29.43*sin(x1(t))+.3*x4(t)^2*sin(x1(t)-x3(t)))*cos(x1(t)-x3(t)))/(5-cos(x1(t)-x3(t))^2)}$

>

Download lagragne_popgave_wiel_omgezet_naar_1e_orde_systeem_ode_30-7-2025.mw

@lucaud Has to do with Maple, so ...

Answered: janhardo 700

Yesterday at 6:38 AM

0 0

@lucaud

Has to do with Maple, so a investigation on its own ?

@lucaud They are coupled diff...

Answered: janhardo 700

Yesterday at 3:34 AM

0 0

@lucaud

They are coupled differential equations.

I don't know anything about Lagrange problems ( via my ai source, there is a lot to find out ) , but there is something to calculate.

>

restart;
with(plots):
with(DEtools):

# Define constants
M := 2;
m := 1;
R := 1;
r := 0.3;
g := 9.81;

# Define the functions (DO NOT redefine theta := t -> theta(t)!)
# Let theta(t) and phi(t) be treated as symbols from the beginning ( for the odes )

eq1 := (2*M + m)*(R - r)*diff(theta(t), t $ 2) + m*r*diff(phi(t), t $ 2)*cos(theta(t) - phi(t)) + m*r*diff(phi(t), t)^2*sin(theta(t) - phi(t)) = -(M + m)*g*sin(theta(t));
eq2 := m*r^2*diff(phi(t), t $ 2) + m*r*(R - r)*diff(theta(t), t $ 2)*cos(theta(t) - phi(t)) - m*r*(R - r)*diff(theta(t), t)^2*sin(theta(t) - phi(t)) = -m*g*r*sin(phi(t));

#Initial conditions
ics := theta(0) = Pi/6, D(theta)(0) = 0, phi(0) = Pi/2, D(phi)(0) = 0:

#Solve the system numerically
sol := dsolve({eq1, eq2, ics}, {phi(t), theta(t)}, numeric, output = listprocedure);

#Plot theta(t) and phi(t)
theta_plot := odeplot(sol, [t, theta(t)], 0 .. 10, color = blue, title = "θ(t)", thickness = 2):
phi_plot := odeplot(sol, [t, phi(t)], 0 .. 10, color = red, title = "φ(t)", thickness = 2):

#Display combined plot
display([theta_plot, phi_plot], title = "Angular Motion Over Time", legend = ["θ(t)", "φ(t)"]);

3.5*(diff(diff(theta(t), t), t))+.3*(diff(diff(phi(t), t), t))*cos(-theta(t)+phi(t))-.3*(diff(phi(t), t))^2*sin(-theta(t)+phi(t)) = -29.43*sin(theta(t))

0.9e-1*(diff(diff(phi(t), t), t))+.21*(diff(diff(theta(t), t), t))*cos(-theta(t)+phi(t))+.21*(diff(theta(t), t))^2*sin(-theta(t)+phi(t)) = -2.943*sin(phi(t))

Download best_way_to_solve_langrange_problem_30-7-2025_in_worksheet_format_.mw

@Carl Love Thanks, yes, it's 2d...

Answered: janhardo 700

July 25 2025

0 0

@Carl Love
Thanks, yes, it's 2d format what i am using these days mostly, and 1d less
The 1d worksheet code can be added in a readable format
"Unfortunately, the conversion from 2d to 1d input removes all formatting such as indentation and line breaks"
No way to fix this in the editor ?., probably impossible , because there is 1d and 2d code what can be used

@Alfred_F If you want to comp...

Answered: janhardo 700

July 21 2025

0 0

@Alfred_F

If you want to compare further terms for equality , you can use this procedure.
It can use two check methods: is command method or via difference method.

>

term1 := proc (n) options operator, arrow; (2*cos(2^n*x)+1)/(2*cos(x)+1) end proc; term2 := proc (n) options operator, arrow; product(2*cos(2^k*x)-1, k = 0 .. n-1) end proc; InductieControleHybride(term1(n), term2(n), n, optie = 'product')

proc (n) options operator, arrow; product(2*cos(2^k*x)-1, k = 0 .. n-1) end proc

### Step 1: Base case check (n = 1):

is(%):

✅ Base case correct.

### Induction hypothesis:

product(2*cos(2^k*x)-1, k = 0 .. n-1) = (2*cos(2^n*x)+1)/(2*cos(x)+1)

### Product form: try `is(...)` method first:

is(%):

✅ Inductive step valid via `is(...)`

(1)

>

term1 := proc (n) options operator, arrow; (1/6)*n*(n+1)*(2*n+1) end proc; term2 := proc (n) options operator, arrow; sum(k^2, k = 1 .. n) end proc; InductieControleHybride(term1(n), term2(n), n, optie = 'sum')

proc (n) options operator, arrow; sum(k^2, k = 1 .. n) end proc

### Step 1: Base case check (n = 1):

is(%):

✅ Base case correct.

### Induction hypothesis:

### Product form: try `is(...)` method first:

(1/6)*n*(n+1)*(2*n+1)*(2*n^2+7*n+6)/(2*n^2+n) = (1/6)*(n+1)*(n+2)*(2*n+3)

is(%):

✅ Inductive step valid via `is(...)`

(2)

>

term1 := proc (n) options operator, arrow; (1/2)*n*(n+1) end proc; term2 := proc (n) options operator, arrow; sum(k, k = 1 .. n) end proc; InductieControleHybride(term1(n), term2(n), n, optie = 'sum')

proc (n) options operator, arrow; sum(k, k = 1 .. n) end proc

### Step 1: Base case check (n = 1):

is(%):

✅ Base case correct.

### Induction hypothesis:

### Product form: try `is(...)` method first:

is(%):

✅ Inductive step valid via `is(...)`

(3)

Download proc_Inductiecontrolhybride_-via_is_of_verschil_check_mprimes_21-7-2025.mw

@Alfred_F The prove idea is a teles...

Answered: janhardo 700

July 20 2025

0 0

@Alfred_F
The prove idea is a telescopical product of fractions

>

$restart; telescopic_product_full_output := proc (x, n) local k, fractions, product_fractions, factors, factor_product, equation; print(" Goal: Prove the identity:"); print(" (2*cos(2^n*x)+1)/(2*cos(x)+1) = ∏_{k=0}^{n-1} (2*cos(2^k*x) - 1)"); print("\n Step 1: Telescoping fractions:"); fractions := [seq((2*cos(2^(k+1)*x)+1)/(2*cos(2^k*x)+1), k = 0 .. n-1)]; for k to n do print(fractions[k]) end do; print("\n Step 2: Telescoping product (still as fractions):"); product_fractions := mul(fractions[k], k = 1 .. n); print(product_fractions); print("\n Telescoping effect:"); print(" Each denominator cancels with the numerator of the next fraction. What remains:"); print((2*cos(2^n*x)+1)/(2*cos(x)+1)); print("\n Step 3: Rewrite each fraction:"); factors := [seq(simplify(fractions[k]), k = 1 .. n)]; for k to n do print(fractions[k] = factors[k]) end do; print("\n Step 4: Right-hand side as product form:"); factor_product := mul(factors[k], k = 1 .. n); print(factor_product); print("\nConclusion:"); equation := (2*cos(2^n*x)+1)/(2*cos(x)+1) = factor_product; print(equation) end proc$

>

$" (2*cos(2^n*x)+1)/(2*cos(x)+1) = ∏_{k=0}^{n-1} (2*cos(2^k*x) - 1)"$

$" Step 1: Telescoping fractions:"$

$" Step 2: Telescoping product (still as fractions):"$

$" Each denominator cancels with the numerator of the next fraction. What remains:"$

$" Step 3: Rewrite each fraction:"$

(2*cos(4096*x)+1)/(2*cos(x)+1) = (2*cos(x)-1)*(2*cos(2*x)-1)*(2*cos(4*x)-1)*(2*cos(8*x)-1)*(2*cos(16*x)-1)*(2*cos(32*x)-1)*(2*cos(64*x)-1)*(2*cos(128*x)-1)*(2*cos(256*x)-1)*(2*cos(512*x)-1)*(2*cos(1024*x)-1)*(2*cos(2048*x)-1)

(1)

>

using this procedure command : ( 12 ... now, it is 12, but you can choose any number n )

Download telescopische_term_1_i=_term_2_mprimes_20-7-2025.mw

@Alfred_F Good workWas the statemen...

Answered: janhardo 700

July 20 2025

0 0

@Alfred_F
Lol..at your age ,.good work
Was the statement : term 1= term 2 ?

L := (2*cos(2*x) + 1)/(2*cos(x) + 1);
R := 2*cos(x) - 1;
L_simpl := simplify(L, trig);
R_simpl := simplify(R, trig);
simplify(L_simpl - R_simpl);

this is recursive identity for n = 1 derived from the general identity

Faster proc and with fixed colors for sp...

Answered: janhardo 700

July 01 2025

0 0

Faster proc and with fixed colors for spirals now and choose any system
Download analyse_faseplots_via_polar_eq_vaste_kleuren_31-6-2025DEFA.mw

@lucaud Glad i could help you..so...

Answered: janhardo 700

June 27 2025

0 0

@lucaud

Glad i could help you..sorry?

@lucaud ......

Answered: janhardo 700

June 26 2025

0 0

@lucaud
It is not in the right Physics notation style in order to learn serious Physics with Maple

>

$restart; m := m; M := M; a := a; k := k; g := g; y := 'y'; theta := 'theta'; dy := 'dy'; dtheta := 'dtheta'; py := 'py'; ptheta := 'ptheta'; T := (1/2)*m*dy^2+(1/6)*M*a^2*dtheta^2; U := m*g*y+(1/2)*k*(3*y^2+2*a*y*sin(theta)+3*a^2); L := T-U; p_y_expr := m*dy; p_theta_expr := (1/3)*M*a^2*dtheta; dy_sol := solve(py = protected, dy); dtheta_sol := solve(ptheta = p_theta_expr, dtheta); L_sub := subs({dtheta = dtheta_sol, dy = dy_sol}, L); H := simplify(dy_sol*py+ptheta*dtheta_sol-L_sub); dq1 := simplify(diff(H, py)); dq2 := simplify(diff(H, ptheta)); dp1 := simplify(-(diff(H, y))); dp2 := simplify(-(diff(H, theta))); printf("? Hamiltoniaan:\n\n6"F[q-"); "\n? Hamilton-vergelijkingen:\n\nFas-F^"; "dy/dt = "; dq1; printf("\ndtheta/dt = "); dq2; printf("\ndpy/dt = "); Typesetting; printf("\ndptheta/dt = "); dp2;$

✅ Hamiltoniaan:

(1/2)*py^2/m+(3/2)*ptheta^2/(M*a^2)+m*g*y+(1/2)*k*(3*y^2+2*a*y*sin(theta)+3*a^2)

✅ Hamilton-vergelijkingen:

dy/dt =

dtheta/dt =

dpy/dt =

dptheta/dt =

(1)

>

Download hamilton_systeem_animatie_26-6-2025.mw

The LAD info versionChoose infolevel fro...

Commented: janhardo 700

June 25 2025

The LAD info version
Choose infolevel from 1 to 4

>

Example from A-M. Wazwaz, Applied Mathematics and Computation 111 (2000) 53. Sec. 4.

>

(1)

>

LADinfo: [Structure] original PDE (expanded form): diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

(2)

>

LADinfo: [Structure] original PDE (expanded form): diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] PDE after rational conversion and conjugate substitution: diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] L = U[t]; R = 0; N = -U^2*U[x]; G = 0.

LADinfo: [Structure] nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

LADinfo: [Init] component 0 = 3*x

LADinfo: [Iteration] component 1 = -27*x^2*t

LADinfo: [Iteration] component 2 = 486*x^3*t^2

LADinfo: [Iteration] component 3 = -10935*x^4*t^3

LADinfo: [Iteration] component 4 = 275562*x^5*t^4

LADinfo: [Iteration] component 5 = -7440174*x^6*t^5

LADinfo: [Iteration] component 6 = 210450636*x^7*t^6

LADinfo: [Iteration] component 7 = -6155681103*x^8*t^7

LADinfo: [Iteration] component 8 = 184670433090*x^9*t^8

LADinfo: [Iteration] component 9 = -5650915252554*x^10*t^9

LADinfo: [Iteration] component 10 = 175692092397588*x^11*t^10

(3)

>

LADinfo: [Structure] original PDE (expanded form): diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] PDE after rational conversion and conjugate substitution: diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] L = U[t]; R = 0; N = -U^2*U[x]; G = 0.

LADinfo: [Structure] time-dependent term L = U[t]; derivative order m = 1; scale factor = 1

LADinfo: [Structure] nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

LADinfo: [Init] component 0 = 3*x

LADinfo: [Iteration] computing component 1 at t = 0.

LADinfo: [Iteration] component 1 = -27*x^2*t

LADinfo: [Symbolic] evaluated nonlinear variables for component 1: [-27*x^2*t, -54*x*t]

LADinfo: [Iteration] computing component 2 at t = .459

LADinfo: [Iteration] component 2 = 486*x^3*t^2

LADinfo: [Symbolic] evaluated nonlinear variables for component 2: [486*x^3*t^2, 1458*x^2*t^2]

LADinfo: [Iteration] computing component 3 at t = .873

LADinfo: [Iteration] component 3 = -10935*x^4*t^3

LADinfo: [Symbolic] evaluated nonlinear variables for component 3: [-10935*x^4*t^3, -43740*x^3*t^3]

LADinfo: [Iteration] computing component 4 at t = 1.312

LADinfo: [Iteration] component 4 = 275562*x^5*t^4

LADinfo: [Symbolic] evaluated nonlinear variables for component 4: [275562*x^5*t^4, 1377810*x^4*t^4]

LADinfo: [Iteration] computing component 5 at t = 1.742

LADinfo: [Iteration] component 5 = -7440174*x^6*t^5

LADinfo: [Symbolic] evaluated nonlinear variables for component 5: [-7440174*x^6*t^5, -44641044*x^5*t^5]

LADinfo: [Iteration] computing component 6 at t = 2.169

LADinfo: [Iteration] component 6 = 210450636*x^7*t^6

LADinfo: [Symbolic] evaluated nonlinear variables for component 6: [210450636*x^7*t^6, 1473154452*x^6*t^6]

LADinfo: [Iteration] computing component 7 at t = 2.583

LADinfo: [Iteration] component 7 = -6155681103*x^8*t^7

LADinfo: [Symbolic] evaluated nonlinear variables for component 7: [-6155681103*x^8*t^7, -49245448824*x^7*t^7]

LADinfo: [Iteration] computing component 8 at t = 3.006

LADinfo: [Iteration] component 8 = 184670433090*x^9*t^8

LADinfo: [Symbolic] evaluated nonlinear variables for component 8: [184670433090*x^9*t^8, 1662033897810*x^8*t^8]

LADinfo: [Iteration] computing component 9 at t = 3.435

LADinfo: [Iteration] component 9 = -5650915252554*x^10*t^9

LADinfo: [Symbolic] evaluated nonlinear variables for component 9: [-5650915252554*x^10*t^9, -56509152525540*x^9*t^9]

LADinfo: [Iteration] computing component 10 at t = 3.857

LADinfo: [Iteration] component 10 = 175692092397588*x^11*t^10

LADinfo: [Iteration] completed at time 4.159

(4)

>

LADinfo: [Structure] original PDE (expanded form): diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] PDE after rational conversion and conjugate substitution: diff(U(x,t),t)+U(x,t)^2*diff(U(x,t),x)

LADinfo: [Structure] L = U[t]; R = 0; N = -U^2*U[x]; G = 0.

LADinfo: [Structure] time-dependent term L = U[t]; derivative order m = 1; scale factor = 1

LADinfo: [Structure] nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

LADinfo: [Init] component 0 = 3*x

LADinfo: [Iteration] computing component 1 at t = 0.

LADinfo: [Symbolic] Laplace-transformed sum: -27*x^2/s

LADinfo: [Iteration] component 1 = -27*x^2*t

LADinfo: [Symbolic] evaluated nonlinear variables for component 1: [-27*x^2*t, -54*x*t]

LADinfo: [Symbolic] derived nonlinear term A[1] = 972*x^3*t

LADinfo: [Iteration] computing component 2 at t = 1.020

LADinfo: [Symbolic] Laplace-transformed sum: 972*x^3/s^2

LADinfo: [Iteration] component 2 = 486*x^3*t^2

LADinfo: [Symbolic] evaluated nonlinear variables for component 2: [486*x^3*t^2, 1458*x^2*t^2]

LADinfo: [Symbolic] derived nonlinear term A[2] = -32805*x^4*t^2

LADinfo: [Iteration] computing component 3 at t = 2.044

LADinfo: [Symbolic] Laplace-transformed sum: -65610*x^4/s^3

LADinfo: [Iteration] component 3 = -10935*x^4*t^3

LADinfo: [Symbolic] evaluated nonlinear variables for component 3: [-10935*x^4*t^3, -43740*x^3*t^3]

LADinfo: [Symbolic] derived nonlinear term A[3] = 1102248*x^5*t^3

LADinfo: [Iteration] computing component 4 at t = 3.077

LADinfo: [Symbolic] Laplace-transformed sum: 6613488*x^5/s^4

LADinfo: [Iteration] component 4 = 275562*x^5*t^4

LADinfo: [Symbolic] evaluated nonlinear variables for component 4: [275562*x^5*t^4, 1377810*x^4*t^4]

LADinfo: [Symbolic] derived nonlinear term A[4] = -37200870*x^6*t^4

LADinfo: [Iteration] computing component 5 at t = 4.110

LADinfo: [Symbolic] Laplace-transformed sum: -892820880*x^6/s^5

LADinfo: [Iteration] component 5 = -7440174*x^6*t^5

LADinfo: [Symbolic] evaluated nonlinear variables for component 5: [-7440174*x^6*t^5, -44641044*x^5*t^5]

LADinfo: [Symbolic] derived nonlinear term A[5] = 1262703816*x^7*t^5

LADinfo: [Iteration] computing component 6 at t = 5.170

LADinfo: [Symbolic] Laplace-transformed sum: 151524457920*x^7/s^6

LADinfo: [Iteration] component 6 = 210450636*x^7*t^6

LADinfo: [Symbolic] evaluated nonlinear variables for component 6: [210450636*x^7*t^6, 1473154452*x^6*t^6]

LADinfo: [Symbolic] derived nonlinear term A[6] = -43089767721*x^8*t^6

LADinfo: [Iteration] computing component 7 at t = 6.195

LADinfo: [Symbolic] Laplace-transformed sum: -31024632759120*x^8/s^7

LADinfo: [Iteration] component 7 = -6155681103*x^8*t^7

LADinfo: [Symbolic] evaluated nonlinear variables for component 7: [-6155681103*x^8*t^7, -49245448824*x^7*t^7]

LADinfo: [Symbolic] derived nonlinear term A[7] = 1477363464720*x^9*t^7

LADinfo: [Iteration] computing component 8 at t = 7.218

LADinfo: [Symbolic] Laplace-transformed sum: 7445911862188800*x^9/s^8

LADinfo: [Iteration] component 8 = 184670433090*x^9*t^8

LADinfo: [Symbolic] evaluated nonlinear variables for component 8: [184670433090*x^9*t^8, 1662033897810*x^8*t^8]

LADinfo: [Symbolic] derived nonlinear term A[8] = -50858237272986*x^10*t^8

LADinfo: [Iteration] computing component 9 at t = 8.252

LADinfo: [Symbolic] Laplace-transformed sum: -2050604126846795520*x^10/s^9

LADinfo: [Iteration] component 9 = -5650915252554*x^10*t^9

LADinfo: [Symbolic] evaluated nonlinear variables for component 9: [-5650915252554*x^10*t^9, -56509152525540*x^9*t^9]

LADinfo: [Symbolic] derived nonlinear term A[9] = 1756920923975880*x^11*t^9

LADinfo: [Iteration] computing component 10 at t = 9.298

LADinfo: [Symbolic] Laplace-transformed sum: 637551464892367334400*x^11/s^10

LADinfo: [Iteration] component 10 = 175692092397588*x^11*t^10

LADinfo: [Iteration] completed at time 9.916

(5)

Download LADinfo_versie_25--6-2025.mw

@Rouben Rostamian ...

Commented: janhardo 700

June 23 2025

@Rouben Rostamian

@dharr Could the procedure be furt...

Commented: janhardo 700

June 23 2025

@dharr
Could the procedure be further extended for new types of pdes or does that make no sense ?

on Windows 11 : Shift + Microssoft + S...

Answered: janhardo 700

June 23 2025

0 0

on Windows 11 : Shift + Microssoft + S : a screenshot

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@lucaud What do you want to achiev...

The system of odes is now for first orde...

@lucaud Has to do with Maple, so ...

@lucaud They are coupled diff...

@Carl Love Thanks, yes, it's 2d...

@Alfred_F If you want to comp...

@Alfred_F The prove idea is a teles...

@Alfred_F Good workWas the statemen...

Faster proc and with fixed colors for sp...

@lucaud Glad i could help you..so...

@lucaud ......

The LAD info versionChoose infolevel fro...

@Rouben Rostamian ...

@dharr Could the procedure be furt...

on Windows 11 : Shift + Microssoft + S...

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