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Solve Linearly System in explicit form...

Asked by: PhearunSeng 15
September 14 2022
0  19

Hello dear

please guide me how to solve linearly system in explicit form.

In this purpose of this problem is to find  w1, w2 w3  as symblic solution in Matrix form. and I hope that in the futher the vairiables consist of w1 ... wn and will still be applied in those simplied form or explicit form.

Thank you

_____________code________________________________________________________________

Vector[column](3, [0.85*((phi__cs*beta__s*f__con) . (D__pile*w__1)) + Vector[column](1, [-5/6*H - 5/8*P]), 0.85*((phi__cs*beta__s*f__con) . (D__pile*w__2)) + Vector[column](1, [5/6*H - 5/8*P]), 0.85*((phi__cs*beta__s*f__con) . (D__pile*w__3)) + Vector[column](1, [1/2*H - 3/8*P])])

New display of arbitrary constants and functions ...

by: ecterrab 14540 Maple
typesetting differential-equations
September 13 2022
11  8


 

New display of arbitrary constants and functions

 

When using computer algebra, first we want results. Right. And textbook-like typesetting was not fully developed 20+ years ago. So, in the name of getting those results, people somehow got used to the idea of "give up textbook-quality computer algebra display". But computers keep evolving, and nowadays textbook typesetting is fully developed, so we have better typesetting in place. For example, consider this differential equation:

 

Download New_arbitrary_constants_and_functions.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Special Theory of Relativity: Uniformly accelerated...

by: Freddy Baudine 70 Maple 2022
tensor physics special-relativity
September 13 2022
5  0

Problem statement:
Determine the relativistic uniformly accelerated motion, i.e. the rectilinear motion for which the acceleration w in the proper reference frame (at each instant of time) remains constant.

As an application of the post presented by Dr Cheb Terrab in MaplePrimes on the principle of relativity ( found here ), we solve the problem stated on page 24 of Landau & Lifshitz book [1], which makes use of the relativistic invariant condition of the constancy of a four-scalar, viz., `w__μ`*w^mu where w^mu is the four-acceleration. This little problem exemplify beautifully how to use invariance in relativity. This is the so-called hyperbolic motion and we explain why at the end of this worksheet.

NULL

let's introduce the coordinate system, X = (x, y, z, tau)with tau = c*t 

with(Physics)

Setup(coordinates = [X = (x, y, z, tau)])

[coordinatesystems = {X}]

 (1)

%d_(s)^2 = g_[lineelement]

%d_(s)^2 = -Physics:-d_(x)^2-Physics:-d_(y)^2-Physics:-d_(z)^2+Physics:-d_(tau)^2

 (2)

NULL

Four-velocity

 

The four-velocity is defined by  u^mu = dx^mu/ds and dx^mu/ds = dx^mu/(c*sqrt(1-v^2/c^2)*dt) 

Define this quantity as a tensor.

Define(u[mu], quiet)

The four velocity can therefore be computing using

u[`~mu`] = d_(X[`~mu`])/%d_(s(tau))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/%d_(s(tau))

(1.1)

NULL

As to the interval d(s(tau)), it is easily obtained from (2) . See Equation (4.1.5)  here with d(diff(tau(x), x)) = d(s(tau)) for in the moving reference frame we have that d(diff(x, x)) = d(diff(y(x), x)) and d(diff(y(x), x)) = d(diff(z(x), x)) and d(diff(z(x), x)) = 0.

 Thus, remembering that the velocity is a function of the time and hence of tau, set

%d_(s(tau)) = d(tau)*sqrt(1-v(tau)^2/c^2)

%d_(s(tau)) = Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2)

(1.2)

subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/%d_(s(tau)))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))

(1.3)

Rewriting the right-hand side in components,

lhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = Library:-TensorComponents(rhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

u[`~mu`] = [Physics:-d_(x)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(y)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(z)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]

(1.4)

Next we introduce explicitly the 3D velocity components while remembering that the moving reference frame travels along the positive x-axis

NULL

simplify(u[`~mu`] = [Physics[d_](x)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](y)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](z)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)], {d_(x)/d_(tau) = v(tau)/c, d_(y)/d_(tau) = 0, d_(z)/d_(tau) = 0}, {d_(x), d_(y), d_(z)})

u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]

(1.5)

Introduce now this explicit definition into the system

Define(u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)])

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], u[mu], w[`~mu`], w__o[`~mu`], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(1.6)

NULL

Computing the four-acceleration

 

This quantity is defined by the second derivative w^mu = d^2*x^mu/ds^2 and d^2*x^mu/ds^2 = du^mu/ds and du^mu/ds = du^mu/(c*sqrt(1-v^2/c^2)*dt)

Define this quantity as a tensor.

Define(w[mu], quiet)

Applying the definition just given,

w[`~mu`] = d_(u[`~mu`])/%d_(s(tau))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/%d_(s(tau))

(2.1)

Substituting for d_(s(tau))from (1.2) above

subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/%d_(s(tau)))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))

(2.2)

Introducing now this definition (2.2)  into the system,

Define(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2)), quiet)

lhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = TensorArray(rhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

w[`~mu`] = Array(%id = 36893488148327765764)

(2.3)

Recalling that tau = c*t, we get

"PDETools:-dchange([tau=c*t],?,[t],params=c)"

w[`~mu`] = Array(%id = 36893488148324030572)

(2.4)

Introducing anew this definition (2.4)  into the system,

"Define(w[~mu]=rhs(?),redo,quiet):"

NULL

In the proper referential, the velocity of the particle vanishes and the tridimensional acceleration is directed along the positive x-axis, denote its value by `#msub(mi("w"),mn("0"))`

Hence, proceeding to the relevant substitutions and introducing the corresponding definition into the system, the four-acceleration in the proper referential reads

  "Define(`w__o`[~mu]= subs(v(t)=`w__0`, v(t)=0,rhs(?)),quiet):"

w__o[`~mu`] = TensorArray(w__o[`~mu`])

w__o[`~mu`] = Array(%id = 36893488148076604940)

(2.5)

NULL

The differential equation solving the problem

 

NULL``

Everything is now set up for us to establish the differential equation that will solve our problem. It is at this juncture that we make use of the invariant condition stated in the introduction.

The relativistic invariant condition of uniform acceleration must lie in the constancy of a 4-scalar coinciding with `w__μ`*w^mu  in the proper reference frame.

We simply write the stated invariance of the four scalar (d*u^mu*(1/(d*s)))^2 thus:

w[mu]^2 = w__o[mu]^2

w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`]

(3.1)

TensorArray(w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`])

(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4

(3.2)

NULL

This gives us a first order differential equation for the velocity.

 

Solving the differential equation for the velocity and computation of the distance travelled

 

NULL

Assuming the proper reference frame is starting from rest, with its origin at that instant coinciding with the origin of the fixed reference frame, and travelling along the positive x-axis, we get successively,

NULL

dsolve({(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4, v(0) = 0})

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)

(4.1)

NULL

As just explained, the motion being along the positive x-axis, we take the first expression.

[v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)][1]

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)

(4.2)

This can be rewritten thus

v(t) = w__0*t/sqrt(1+w__0^2*t^2/c^2)

v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)

(4.3)

It is interesting to note that the ultimate speed reached is the speed of light, as it should be.

`assuming`([limit(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = infinity)], [w__0 > 0, c > 0])

limit(v(t), t = infinity) = c

(4.4)

NULL

The space travelled is simply

x(t) = Int(rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)), t = 0 .. t)

x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t)

(4.5)

`assuming`([value(x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t))], [c > 0])

x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0

(4.6)

expand(x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0)

x(t) = c*(t^2*w__0^2+c^2)^(1/2)/w__0-c^2/w__0

(4.7)

This can be rewritten in the form

x(t) = c^2*(sqrt(1+w__0^2*t^2/c^2)-1)/w__0

x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0

(4.8)

NULL

The classical limit corresponds to an infinite velocity of light; this entails an instantaneous propagation of the interactions, as is conjectured in Newtonian mechanics.
The asymptotic development gives,

lhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0) = asympt(rhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0), c, 4)

x(t) = (1/2)*w__0*t^2+O(1/c^2)

(4.9)

As for the velocity, we get

lhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)) = asympt(rhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)), c, 2)

v(t) = t*w__0+O(1/c^2)

(4.10)

Thus, the classical laws are recovered.

NULL

Proper time

 

NULL

This quantity is given by "t'= ∫ dt sqrt(1-(v^(2))/(c^(2)))" the integral being  taken between the initial and final improper instants of time

Here the initial instant is the origin and we denote the final instant of time t.

NULL

`#mrow(mi("t"),mo("′"))` = Int(sqrt(1-rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2))^2/c^2), t = 0 .. t)

`#mrow(mi("t"),mo("′"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t)

(5.1)

Finally the proper time reads

`assuming`([value(`#mrow(mi("t"),mo("′"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t))], [w__0 > 0, c > 0, t > 0])

`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0

(5.2)

When proc (t) options operator, arrow; infinity end proc, the proper time grows much more slowly than t according to the law

`assuming`([lhs(`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0) = asympt(rhs(`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0), t, 1)], [w__0 > 0, c > 0])

`#mrow(mi("t"),mo("′"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2)

(5.3)

combine(`#mrow(mi("t"),mo("′"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2), ln, symbolic)

`#mrow(mi("t"),mo("′"))` = ln(2*t*w__0/c)*c/w__0+O(1/t^2)

(5.4)

NULL

Evolution of the four-acceleration of the moving frame as observed from the fixed reference frame

 

NULL

To obtain the four-acceleration as a function of time, simply substitute for the 3-velocity (4.3)  in the 4-acceleration (2.4)

" simplify(subs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2),?),symbolic)"

w[`~mu`] = Array(%id = 36893488148142539108)

(6.1)

" w[t->infinity]^( mu)=map(limit,rhs(?),t=infinity) assuming `w__0`>0,c>0"

`#msubsup(mi("w"),mrow(mi("t"),mo("→"),mo("∞")),mrow(mo("⁢"),mo("⁢"),mi("μ",fontstyle = "normal")))` = Array(%id = 36893488148142506460)

(6.2)

We observe that the non-vanishing components of the four-acceleration of the accelerating reference frame get infinite while those components in the moving reference frame keep their constant values . (2.5)

NULL

Evolution of the three-acceleration as observed from the fixed reference frame

 

NULL

This quantity is obtained simply by differentiating the velocity v(t)given by  with respect to the time t.

 

simplify(diff(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t), size)

diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)

(7.1)

Here also, it is interesting to note that the three-acceleration tends to zero. This fact was somewhat unexpected.

map(limit, diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2), t = infinity)

limit(diff(v(t), t), t = infinity) = 0

(7.2)

NULL

At the beginning of the motion, the acceleration should be w__0, as Newton's mechanics applies then

NULL

`assuming`([lhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)) = series(rhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)), t = 0, 2)], [c > 0])

diff(v(t), t) = series(w__0+O(t^2),t,2)

(7.3)

NULL

Justification of the name hyperbolic motion

 

NULL

Recall the expressions for x and diff(t(x), x)and obtain a parametric description of a curve, with diff(t(x), x)as parameter. This curve will turn out to be a hyperbola.

subs(x(t) = x, x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0)

x = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0

(8.1)

`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0

`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0

(8.2)

The idea is to express the variables x and t in terms of diff(t(x), x).

 

isolate(`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0, t)

t = sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)*c/w__0

(8.3)

subs(t = sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)*c/w__0, x = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0)

x = c^2*((1+sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2)^(1/2)-1)/w__0

(8.4)

`assuming`([simplify(x = c^2*((1+sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2)^(1/2)-1)/w__0)], [positive])

x = c^2*(cosh(`#mrow(mi("t"),mo("′"))`*w__0/c)-1)/w__0

(8.5)

We now show that the equations (8.3) and (8.5) are parametric equations of a hyperbola with parameter the proper time diff(t(x), x)

 

Recall the hyperbolic trigonometric identity

cosh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2 = 1

cosh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2 = 1

(8.6)

Then isolating the sinh and the cosh from equations (8.3) and (8.5),

NULL

isolate(t = sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)*c/w__0, sinh(`#mrow(mi("t"),mo("′"))`*w__0/c))

sinh(`#mrow(mi("t"),mo("′"))`*w__0/c) = t*w__0/c

(8.7)

isolate(x = c^2*(cosh(`#mrow(mi("t"),mo("′"))`*w__0/c)-1)/w__0, cosh(`#mrow(mi("t"),mo("′"))`*w__0/c))

cosh(`#mrow(mi("t"),mo("′"))`*w__0/c) = x*w__0/c^2+1

(8.8)

and substituting these in (8.6) , we get the looked-for Cartesian equation

 

subs(sinh(`#mrow(mi("t"),mo("′"))`*w__0/c) = t*w__0/c, cosh(`#mrow(mi("t"),mo("′"))`*w__0/c) = x*w__0/c^2+1, cosh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("′"))`*w__0/c)^2 = 1)

(x*w__0/c^2+1)^2-w__0^2*t^2/c^2 = 1

(8.9)

NULL

This is the Cartesian equation of a hyperbola, hence the name hyperbolic motion

NULL

Reference

 

[1] Landau, L.D., and Lifshitz, E.M. The Classical Theory of Fields, Course of Theoretical Physics Volume 2, fourth revised English edition. Elsevier, 1975.

NULL

Download Uniformly_accelerated_motion.mw

Maple Conference 2022 Creative Works Deadline...

by: John May 3036 Maple , Maple Learn
September 13 2022
2  0

This is a friendly reminder that the deadline for submissions for this year's Maple Conference Creative Works Exhibit is fast approaching!

If you are looking for inspiration, you can take a look at the writeup of the works that were featured last year in this write up in the most recent issue of Maple Transations.

Also, don't forget that you can also submit art made in Maple Learn for a special exhibit alongside the main gallery.

Cannot find the solution of a non-linear equation ...

Asked by: lemelinm 1535
solve brachistochrone
September 13 2022
1  1

In solving the brachistochrone for a fine string of length L under the pull of g passing through the point (0,0) and (a,b) using the euler-lagange method,I stumble on this non-linear relation:

C*sinh(mu*g*a/C)=L

which I need to solve for C.

Maple give me the famous RootOf:

RootOf(A*exp(_Z)^2 - 2*_Z*L*exp(_Z) - A)

where A = mu*g*a

Can it be solve for C or am I force to use numeric method?

Thank you in advance for your help.

Mario

Error for doing For Loop ...

Asked by: PhearunSeng 15
matrix syntax loop
September 13 2022
2  10

hello Dear

I am new here I am not clear about Loop in Maple, I am stuck here.

Please correct and guide me 

thank you

input

 

Loop

Expect result

-------------------------------------------Code--------------------------------

KTe := Matrix(3, 1, [[Matrix(4, 4, [[216, -288, -216, 288], [-288, 384, 288, -384], [-216, 288, 216, -288], [288, -384, -288, 384]])], [Matrix(4, 4, [[216, 288, -216, -288], [288, 384, -288, -384], [-216, -288, 216, 288], [-288, -384, 288, 384]])], [Matrix(4, 4, [[500, 0, -500, 0], [0, 0, 0, 0], [-500, 0, 500, 0], [0, 0, 0, 0]])]]);
DOFe := Matrix(3, 4, [[1, 2, 3, 4], [3, 4, 5, 6], [5, 6, 1, 2]]);
with(ListTools);


with(LinearAlgebra);
nn1 := Row(DOFe, 1);
                      nn1 := [1, 2, 3, 4]

KG1 := Matrix(6, 6);


KG := Matrix(3, 1);

for k from 1 to 3  for i from 1 to 4 do      for j from 1 to 4 do    nn:=Row(DOFe,k):   KG1[nn1[i],nn1[j]]:=(KTe[k[],1])[i,j]    end do  end do   KG[k,1]:=KG1[k] end do;
 

Print points with records...

Asked by: Stretto 260
points plotting
September 13 2022
0  5

Is there a way to print points that have additional information that can be probed when mousing over the points?

I have an array of arrays with information in each element including the point: [x,y1,y2,R1, R2, R3, R4]

Where y1 and y2 are different values to graph per x(different plots)  and Rk are records for the data(other info that goes along with them such as day of week, rainfall on that day, etc). The different graphs might have different scales so it would be nice if they handled scaling effectively(e.g., |y1| < 1 and y2 > 3430).

I have about 50k points so it has to be relatively fast... as it's already taking a few minutes to process the data in to the array from a file using fopen and readline(not sure why so slow as I'm just using a few parses and cats to put it in an array but probably is being slowed by not being able to pre-allocate the array.

Translation fron matemática to maple ...

Asked by: emersondiaz 60
mathematica import
September 12 2022
1  1

Hello every one, i am New using maple and I am trying to translate This code from matemática to maple, someone can help me please? 

how to find how many occurances a function shows u...

Asked by: nm 11353
programming indets
September 12 2022
1  3

I need to count how many times a special function shows up in an expression.

The problem is that indets returns a set. So if the same function shows up more than one time in the original expression, with same arguments, only one of these will show up in the result. So I would not know if there were mmore than one of these.

Here is a simple example, using sin(x) here.

restart;
expr:=sin(x)+3*cos(x)*sin(x)+1/sin(2*x);
indets(expr,'specfunc(anything,sin)')

#gives
#   {sin(x), sin(2*x)}

So when I do nops() on the above, it gives 2 and not 3.

How to obtain number of times a function shows in an expression, even it if is repeated?

Complex Plots without a variable made me wonder. ...

Asked by: The function 110
complex plotting
September 12 2022
0  5

Hello everybody, im at it again. Math with maple. I should be picking up some speed again to plow through this Dutch math book that explains Maple. It creeps me out.. But hey, im learning Maple in the process, and that is what its all about!

The example shows what is done. I made question a. happen, and the answer was right. The thing is with question b. they ask to plot the phase vectors alpha of f(t), g(t), and s(t), although there is no variable t in the phase vector. So how on earth will i plot it in the complex plane?

The literal translation of quesion b is: "check graphically the answer of part a. by drawing the phase vectors f(t), g(t), and s(t) in the complex plane."

I cant get it done. 

Would anyone know the right question. There were no graphs displayed at the answers in the back of the book. 

Thank you!

Greetings,

The Function

> 

Opdracht 2
> 

a.
> 

"f(t):=3*cos(2*t-Pi/(4))"

proc (t) options operator, arrow, function_assign; 3*cos(2*t-(1/4)*Pi) end proc

 (1)
> 

"g(t):=4*cos(2*t+Pi/(6))"

proc (t) options operator, arrow, function_assign; 4*cos(2*t+(1/6)*Pi) end proc

 (2)
> 

f(t)+g(t)

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

 (3)
> 

smartplot(3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi))

 
> 

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

 (4)
> 

3*exp(I*(0-(1/4)*Pi))

(3/2)*2^(1/2)-((3/2)*I)*2^(1/2)

 (5)
> 

4*exp(I*((1/6)*Pi))

2*3^(1/2)+2*I

 (6)
> 

NULL
> 

3*sqrt(2)*(1/2)-(1/2)*(3*I)*sqrt(2)+2*sqrt(3)+2*I

(3/2)*2^(1/2)-((3/2)*I)*2^(1/2)+2*3^(1/2)+2*I

 (7)
> 

evalf(%)

5.585421959-.121320343*I

 (8)
> 

arctan((2-3*sqrt(2)*(1/2))/(3*sqrt(2)*(1/2)+2*sqrt(3)))

arctan((-(3/2)*2^(1/2)+2)/((3/2)*2^(1/2)+2*3^(1/2)))

 (9)
> 

evalf(%)

-0.2171747628e-1

 (10)
> 

5.585421959*cos(2*t-0.2171747628e-1)

5.585421959*cos(2*t-0.2171747628e-1)

 (11)
> 

smartplot(5.585421959*cos(2*t-0.2171747628e-1))

 
> 

b.
> 

"fc(t):=3*(e)^(I*(0-Pi/(4)))"

proc (t) options operator, arrow, function_assign; 3*exp(-((1/4)*I)*Pi) end proc

 (12)
> 

"gc(t):=4*(e)^((I*Pi)/(6))"

4*exp(((1/6)*I)*Pi)

 (13)
> 

"sc(t):=(3 sqrt(2))/2-(3 &ImaginaryI; sqrt(2))/2+2 sqrt(3)+2 &ImaginaryI;"

proc (t) options operator, arrow, function_assign; (3/2)*sqrt(2)-((3/2)*I)*sqrt(2)+2*sqrt(3)+2*I end proc

 (14)
> 

``

Download Mapleprimes_Question_Book_2_Paragraph_3.9_Question_2_b.mw

ArrayInterpolation in contourplot...

Asked by: Dkunb 70
duplicate-question arrayinterpolation incomplete-question
September 12 2022
0  6

I wonder if there is any way to use ArrayInterpolation with contourplot or similar effect?

N_data.xlsx 

Thank you in advance,

> 

restart;
> 

with(CurveFitting)

[ArrayInterpolation, BSpline, BSplineCurve, Interactive, LeastSquares, Lowess, PolynomialInterpolation, RationalInterpolation, Spline, ThieleInterpolation]

 (1)
> 

with(plots);

[animate, animate3d, animatecurve, arrow, changecoords, complexplot, complexplot3d, conformal, conformal3d, contourplot, contourplot3d, coordplot, coordplot3d, densityplot, display, dualaxisplot, fieldplot, fieldplot3d, gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, interactive, interactiveparams, intersectplot, listcontplot, listcontplot3d, listdensityplot, listplot, listplot3d, loglogplot, logplot, matrixplot, multiple, odeplot, pareto, plotcompare, pointplot, pointplot3d, polarplot, polygonplot, polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, semilogplot, setcolors, setoptions, setoptions3d, shadebetween, spacecurve, sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]

 (2)
> 

alpha := <seq(0..10,evalf(10/50))>:
beta := <seq(0..10,evalf(10/50))>:
> 

excelfile:= FileTools:-JoinPath(["C:","Users","aimer","OneDrive","Desktop","Msc Thesis","Maple ref","N_data.xlsx"]);

"C:\Users\aimer\OneDrive\Desktop\Msc Thesis\Maple ref\N_data.xlsx"

 (3)
> 

NN:=ImportMatrix(excelfile,source=Excel):

_rtable[36893489576445216036]

 (4)
> 

#?ImportMatrix;
> 

#NN:=ImportMatrix(matlabData, source=MATLAB);
> 

#currentdir();

"C:\Users\aimer\OneDrive\Desktop\Msc Thesis\Maple ref"

 (5)
> 

 

> 

contourplot(ArrayInterpolation([beta,alpha],NN,[x,y]),x=0..10,y=0..10,contours=[0]);

Error, (in CurveFitting:-ArrayInterpolation) invalid input: xvalues are not specified correctly

  
> 

#?listcontplot
> 

 

Download test1.mw

ImportMatrix problem...

Asked by: Dkunb 70
importmatrix
September 12 2022
0  2

I do not know what is the problem with Using ImportMatrix. N_data.xlsx is in the same directory.

Any comment would be appreciated.

> 

restart;
> 

with(CurveFitting)

[ArrayInterpolation, BSpline, BSplineCurve, Interactive, LeastSquares, Lowess, PolynomialInterpolation, RationalInterpolation, Spline, ThieleInterpolation]

 (1)
> 

with(plots);

[animate, animate3d, animatecurve, arrow, changecoords, complexplot, complexplot3d, conformal, conformal3d, contourplot, contourplot3d, coordplot, coordplot3d, densityplot, display, dualaxisplot, fieldplot, fieldplot3d, gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, interactive, interactiveparams, intersectplot, listcontplot, listcontplot3d, listdensityplot, listplot, listplot3d, loglogplot, logplot, matrixplot, multiple, odeplot, pareto, plotcompare, pointplot, pointplot3d, polarplot, polygonplot, polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, semilogplot, setcolors, setoptions, setoptions3d, shadebetween, spacecurve, sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]

 (2)
> 

alpha := <seq(0..10,evalf(10/50))>:
beta := <seq(0..10,evalf(10/50))>:
> 

excelfile:= FileTools:-JoinPath(["C: ","Users","aimer","OneDrive","Desktop","Msc Thesis","Maple ref","N_data.xlsx"]);

"C: \Users\aimer\OneDrive\Desktop\Msc Thesis\Maple ref\N_data.xlsx"

 (3)
> 

NN:=ImportMatrix(excelfile,source=Excel);

Error, (in ImportMatrix) file or directory does not exist: C: \Users\aimer\OneDrive\Desktop\Msc Thesis\Maple ref\N_data.xlsx

  
> 

?ImportMatrix;
> 

#NN:=ImportMatrix(matlabData, source=MATLAB);
> 

currentdir();

"C:\Users\aimer\OneDrive\Desktop\Msc Thesis\Maple ref"

 (4)
> 

?Joinpath
> 

 

Download test1.mw

dsolve, numeric: events...

Asked by: Hnx 20
ode events dsolve
September 12 2022
0  3

I have a non-linear deq of 2nd order which I want to solve numerically. The integration should stop if the integrated variable exceeds a certain value, i.e. phi(t)>phi_ end or equivalently phi(t)-phi_end>0. Maple doesn't accept this type of event and the test phi(t)=phi_end doesn't catch (obviously). How to work around?

Maple is unusable...

Asked by: AdamJHK 0
September 12 2022
0  2

Hi

just downloaded maple on my windows computer with AMD processor but I can't get it to work!!!!

The program opens up but as soon as I start to write, the program freezes. When the program is frozen there is nothing I can do I can't even close it.

I have tried just about everything ( antivirus, deleting the program, updating the computer) u name it.

if anyone has any suggestions, please write to me!!!

Error, (in dsolve/numeric/process_input) system mu...

Asked by: Naveshin Jawaheer 15
ode syntax homework
September 11 2022
1  8

Hi everyone, I have copied soem code from a paper. I hope to try and manupiluate some of the varables. However it seems while I am coping the code I get this error Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations .I have checked online but it seems that there is nothing I can find to fix this problem. My code is posted below 

Thanks for your time and help!

restart

with(VariationalCalculus)

with(ODEtools)

Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received ODEtools

  

with(DEtools)

with(plots); with(plottools); PDEtools[declare]((theta, phi, psi)(t), prime = t)

`derivatives with respect to`*t*`of functions of one variable will now be displayed with '`

 (1)

with(linalg)

We will now declare our first equations

 

 

x[1] := l[1]*sin(theta(t))

y[1] := -l[1]*cos(theta(t))``

x[2] := -l[2]*sin(theta(t)); y[2] := l[2]*cos(theta(t))

x[4] := x[1]-l[4]*sin(theta(t)+phi(t)); y[4] := y[1]+l[4]*cos(theta(t)+phi(t)); x[3] := x[2]+l[3]*sin(theta(t)-psi(t)); y[3] := y[2]-l[3]*cos(theta(t)-psi(t))

R[1] := vector(2, [x[1], y[1]]); R[2] := vector(2, [x[2], y[2]]); R[4] := vector(2, [x[4], y[4]]); R[3] := vector(2, [x[3], y[3]])

`\`R`[1]*` &Assign; map`(diff, R[1], t); `\`R`[2]*` &Assign; map`(diff, R[2], t); `\`R`[4]*` &Assign; map`(diff, R[4], t); `\`R`[3]*`&Assign;map`(diff, R[3], t)

NULL

NULL

T := combine(simplify(collect((1/2)*m[1]*innerprod(`\`R`[4]*`,`*R[4]*`)+m[2]/2* innerprod(\`R[3]\`,\`R[3]`), [l[1], l[2], l[3], l[4], m[1], m[2], diff(theta(t), t), diff(phi(t), t), diff(psi(t), t), cos(psi(t)), sin(psi(t))])), trig)

NULL

U := collect(simplify(expand(g*m[1]*y[4]+g*m[2]*y[3])), [l[1], l[2], l[3], l[4], m[1], m[2], diff(theta(t), t), diff(phi(t), t), diff(psi(t), t), cos(psi(t)), sin(psi(t)), g])

NULL

L := simplify(collect(T-U, [l[1], l[2], l[3], l[4], m[1], m[2], diff(theta(t), t), diff(phi(t), t), diff(psi(t), t), cos(psi(t)), sin(psi(t))]))

NULL

NULL

NULL

Eul := remove(has, EulerLagrange(L, t, [theta(t), phi(t), psi(t)]), K[1]); eq1 := op(select(has, Eul, (diff(psi(t), t))^2)); eq2 := op(select(has, remove(has, Eul, (diff(psi(t), t))^2), sin(psi(t)))); eq3 := op(remove(has, remove(has, Eul, (diff(psi(t), t))^2), sin(psi(t))))

NULL

NULL

INITS := {phi(0) = (1/4)*Pi, psi(0) = (1/4)*Pi, theta(0) = 3*Pi*(1/4), (D(phi))(0) = 0, (D(psi))(0) = 0, (D(theta))(0) = 0}

NULL

PARAM := [g = 9.8, l[1] = 10, l[2] = 100, l[3] = 100, l[4] = 21, m[1] = 1000, m[2] = 1]

NULL

sys := eval([eq1, eq2, eq3], PARAM)

NULL

sol := dsolve([op(sys), op(INITS)], numeric, output = listprocedure)

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

  

NULL``

 

 

NULL

NULL

``

Download Test_1.mw

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