## Solution - partial differential equation...

I'm still a maple beginner and looking for a workable solution to the following problem. I have already solved these "by hand" and would now like to control them.

The following equation should be solved:

`diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x))`

The approach function would be for example:

`u := a(x)*sin(k1*(-c*t+x))`

It would be great if maple could give me the solution again in this example:

`solution = -c^2*(d^2*a*sin(k1*(-c*t+x))/dx^2+2*d*a*k1*cos(k1*(-c*t+x))/dx)`

How exactly do I do this best?
Thanks a lot!
Frank

## How to simulate Roll Dice ?...

Hi,

How to simulate 1000 times the Dice1 experiment and put the results in a list to search for the percentils of A=[A1,..,A1000]

( See attachment)

Thanks

QuestionSimulation.mw

## how can solve ordinary differenatial equation?...

how can solve ordinary differenatial equation in maple?

how can use laplce transformation for this equation in maple?

U0 := proc (x) options operator, arrow; cosh(sqrt(2)*x)-1 end proc;
sys_ode := diff(Uc(x), x, x)-(2*(1+U0(x)))*Uc(x) = 0;

## Solving a System over a Restricted Domain...

Hi,

Persay I have a sinusodal function or dampened sinusodial; it is understand that the domain is from negative infinity to infinity.  Say I wish to find the zeros or a certain y value of a sinusodial on a restricted domain - say from A to B. How would I tell maple -syntax wise- to solve on that restricted domain? Using the solve feature on maple, it only yields 1 value that satisfies the condition I give it .

Thank you all for the support; you al l are allowing to me develop a better understanding for maple.

say I have the functions:

y=e^x*sin(x)

y=e

i wish to find all the x values that satisfy that system on the x domain of 0 to 100

## Problem with solving for C1 and C2...

ode_problem.mw Can someone please look at this? I am having a problem with the ordinary differential equations.

## 'fsolve' statement is only rewritting the equation...

Hello there,

I'm pretty new to MAPLE so this may probably an "easy" mistake, but I don't know what's the problem anyway...

I'd like to find a solution to a system of four equations with four unknowns:

fsolve({(T[1]-T[0])/(10000-T[1]+T[0]) = -2.000000000, (T[2]-T[0])/(20000-T[2]+T[0]) = 0, (T[3]-T[0])/(50000-T[3]+T[0]) = 50, .1*T[0]+.3*T[1]+.55*T[2]+0.5e-1*T[3]-5000 = 0}, {T[0], T[1], T[2], T[3]})

Problem: Instead of providing the single solutions, MAPLE is simply just rewriting the fsolve-statement and does not solve. There is no error-message.

Does anybody know, what the problem is here? Is there no solution after all?

## How to check that an n*n matrix is an MDS matrx wh...

An n*n matrix A is called an MDS matrix over an arbitrary field if all determinant of square sub-matrices of A are non-zero over the field. It is not difficult to prove that the number of all square sub-matrices of A is binomial(2*n, n)-1. The code that I use to check whether A is an MDS matrix is in the following form

u := 1;
for k to n do
P := choose(n, k);
for i to nops(P) do
for j to nops(P) do
F := A(P[i], P[j]);
r := Determinant(F);
if r = 0 then

u := 0; k:=n+1;

i := nops(P)+1; j := nops(P)+1;

end if;

end do;

end do;
if u = 1 then

print(A is an MDS Matrix)

end if;
end do:

When I run the mentioned code for n=16, it takes long time since we need to check binomial(32, 16)-1=601080389 cases to verify that A is an MDS matrix or not.

My Question: Is there a modified procedure which can be used to check that an n*n matrix is  whether an MDS matrix for n>=16.

## how to fix bug on DirectSearch package?...

i'm doing an optimaztion for my final project. when i execute the coding, it says:

Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)].

here my full coding:

restart;
with(linalg);
with(Optimization);
with(DirectSearch);
[BoundedObjective, CompromiseProgramming, DataFit,

ExponentialWeightedSum, GlobalOptima, GlobalSearch, Minimax,

ModifiedTchebycheff, Search, SolveEquations, WeightedProduct,

WeightedSum]
q := Array(0 .. .111, [0.586e-2, 0.67475e-3, 0.52476e-3, 0.419785e-3, 0.354814e-3, 0.324859e-3, 0.319948e-3, 0.31e-3, 0.295013e-3, 0.28e-3, 0.259974e-3, 0.25496e-3, 0.26e-3, 0.270027e-3, 0.280026e-3, 0.284987e-3, 0.274934e-3, 0.279893e-3, 0.294824e-3, 0.324765e-3, 0.374672e-3, 0.439555e-3, 0.509439e-3, 0.56934e-3, 0.60923e-3, 0.634201e-3, 0.634233e-3, 0.624319e-3, 0.61442e-3, 0.624491e-3, 0.6495e-3, 0.684465e-3, 0.714452e-3, 0.729463e-3, 0.749451e-3, 0.789387e-3, 0.864347e-3, 0.964331e-3, 0.1064261e-2, 0.119413e-2, 0.1333876e-2, 0.150354e-2, 0.1683293e-2, 0.1883023e-2, 0.2102591e-2, 0.2356929e-2, 0.2656062e-2, 0.3004981e-2, 0.3403637e-2, 0.3841874e-2, 0.4349482e-2, 0.4931339e-2, 0.5587622e-2, 0.6293742e-2, 0.7044467e-2, 0.780967e-2, 0.8563865e-2, 0.9291539e-2, 0.9987678e-2, 0.10677379e-1, 0.1139506e-1, 0.12195833e-1, 0.13108819e-1, 0.14262384e-1, 0.15557439e-1, 0.17002799e-1, 0.18574592e-1, 0.20263971e-1, 0.22103523e-1, 0.24043337e-1, 0.26165465e-1, 0.28548531e-1, 0.31489568e-1, 0.3458584e-1, 0.37891201e-1, 0.41511365e-1, 0.45444649e-1, 0.49867283e-1, 0.54877269e-1, 0.60478547e-1, 0.66532745e-1, 0.7397857e-1, 0.81900977e-1, 0.90034141e-1, 0.98921462e-1, .108746024, .11839162, .130752255, .14368791, .157201822, .169770195, .177233464, .188823516, .204733363, .224639756, .249617627, .273409998, .298674064, .304378882, .319592743, .342821691, .367512872, .395257421, .426270115, .460795318, .498484563, .537417055, .577542345, .618093683, .660441764, .703856817, 1]);
variables = [m1, m2, m3, sigma1, sigma2, sigma3];
variables = [m1, m2, m3, sigma1, sigma2, sigma3]
psi1 := 0.11480601e-1; psi2 := 0.8890123e-2; psi3 := .979629276;
0.011480601
0.008890123
0.979629276
s1(x):=exp(-(x/(m1))^((m1/(sigma1)))):  s1(x+1):=exp(-((x+1)/(m1))^((m1/(sigma1)))):  s2(x):=1-exp(-(x/(m2))^((-m2/(sigma2)))):  s2(x+1):=1-exp(-((x+1)/(m2))^((-m2/(sigma2)))):  s3(x):=exp(exp(-m3/(sigma3))-exp((x-m3)/(sigma3))): s3(x+1):=exp(exp(-m3/(sigma3))-exp((x+1-m3)/(sigma3)))  :  s(x):=(psi1*s1(x)+psi2*s2(x)+psi3*s3(x)): s(x+1):=(psi1*s1(x+1)+psi2*s2(x+1)+psi3*s3(x+1)):   qtopi(x):=1-((s(x+1))/(s(x))):
fungsikerugian := add((1-qtopi(x)/q[x])^2, x = 0 .. .110);
for x from 0 to 111 do c1[x] := 0 <= s1(x+1); c2[x] := 1 >= s1(x+1); c3[x] := 0 <= s2(x+1); c4[x] := 1 >= s2(x+1); c5[x] := 0 <= s3(x+1); c6[x] := 1 >= s3(x+1); c7[x] := 0 <= s1(x); c8[x] := 1 >= s1(x); c9[x] := 0 <= s2(x); c10[x] := 1 >= s2(x); c11[x] := 0 <= s3(x); c12[x] := 1 >= s3(x); c13[x] := 0 <= s(x); c14[x] := 1 >= s(x); c15[x] := 0 <= s(x+1); c16[x] := 1 >= s(x+1); c17[x] := 0 <= qtopi(x); c18[x] := 1 >= qtopi(x) end do;

constr := {m2 > sigma2, m3 > sigma3, seq(c1[x], x = 0 .. .110), seq(c10[x], x = 0 .. .111), seq(c11[x], x = 0 .. .111), seq(c12[x], x = 0 .. .111), seq(c13[x], x = 0 .. .111), seq(c14[x], x = 0 .. .111), seq(c15[x], x = 0 .. .110), seq(c16[x], x = 0 .. .110), seq(c17[x], x = 0 .. .110), seq(c18[x], x = 0 .. .110), seq(c2[x], x = 0 .. .110), seq(c3[x], x = 0 .. .110), seq(c4[x], x = 0 .. .110), seq(c5[x], x = 0 .. .110), seq(c6[x], x = 0 .. .110), seq(c7[x], x = 0 .. .111), seq(c8[x], x = 0 .. .111), seq(c9[x], x = 0 .. .111), m1 < sigma1, 0 <= m1 and m1 < 17, 17 <= m2 and m2 < 33, 33 <= m3 and m3 < 111};
solusi := DirectSearch:-Search(fungsikerugian, constr, assume = positive, evaluationlimit = 5555);
Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)]

thanks before

## taylor( (2 * pi * n) ^ (1/(2n), n = infinity, 2)...

The command

taylor( (2 * pi * n) ^ (1/(2n), n = 2)

return   1+((1/2)*ln(2*Pi)+(1/2)*ln(n))/n+O(1/n^2)

I think, this is incorrect, in my oppinion must be:

1+((1/2)*ln(2*Pi)+(1/2)*ln(n))/n+O(ln(n)^2/n^2)

Is this error in maple?

## A command to evaluate the non-negative integer sol...

Assume that a[1],a[2],..a[k] are positive integer numbers. Let n be a positive integer number.

Suppose that igcd(a[1],a[2],..a[k])=1.

My question: Is there a command in Maple such that the output of the this command be true provided that there are "non-negative integer numbers" x[1],x[2],..x[k] which satisfy the following condition:

a1*x1+a2*x2+...+ak*xk=n

## How to let sqrt(5*x+5+y) become sqrt(5*(x+1)+y) au...

How to let sqrt(5*x+5+y) become sqrt(5*(x+1)+y) automatically?

## derivative of an implicit function...

Dear all,

Following the comments I am editing this post:

I have a function F of variables (r1,r2,theta1,theta2,r,theta,a). r1, r2,theta1,theta2 are function of r,  theta and a. I want to take derivative of F with respect to a. r and theta are independent of a . I expressed everything in terms of 'a' as a function of 'a' at first. Then I use diff(F, a). I see there is an error in the final expression G .There is a restriction that theta1 should lie between -Pi to Pi and theta2 between 0 to 2*Pi. I speculate this is the source of error. Work sheet is attached. Reason: value of G: integration in 0 to pi/4 gives some  value but for 0 to pi it evaluates to zero and so is the case with 0 to 2*Pi. As "G "physically represents energy it must be a positive value.

 > restart;
 > theta1 := unapply(arctan(r*sin(theta)/(r*cos(theta)-a)), a);
 (1)
 >
 > ## theta1 -->[-Pi,Pi] and theta2-->[0,2*Pi]
 >
 > theta2 := unapply(arctan(r*sin(theta)/(r*cos(theta)+a)), a);
 (2)
 >
 > r1:=unapply(sqrt((r*cos(theta)-a)^2+r^2*(sin(theta))^2),a);r2:=unapply(sqrt((r*cos(theta)+a)^2+r^2*(sin(theta))^2),a);
 (3)
 > sigma12:=0;sigma22:=sigma;
 (4)
 >
 > ## I have to use a constraint that
 > assume(theta1(a) < Pi, theta1(a) > -Pi, theta2(a) > 0, theta2(a) < 2*Pi, a>0,r>0)
 > u1:=(1+nu)*sigma22*sqrt(r1(a)*r2(a))*(4*(1-2*nu)*cos((theta1(a)+(theta2(a)))/2)-4*r*(1-nu)*cos(theta)/sqrt(r1(a)*r2(a))-2*r^2/(r1(a)*r2(a))*(cos((theta1(a)+(theta2(a)))/2)-cos(2*theta-theta1(a)/2-(theta2(a))/2)))/(4*E)+(1+nu)*sigma12*sqrt(r1(a)*r2(a))*(2*(1-2*nu)*sin((theta1(a)+(theta2(a)))/2)-2*r*(1-nu)*sin(theta)/sqrt(r1(a)*r2(a))+1*r^2/(r1(a)*r2(a))*sin(theta)*cos(theta-theta1(a)/2-(theta2(a))/2))/(E);
 >
 (5)
 >
 > u2:=(1+nu)*sigma*sqrt(r1(a)*r2(a))*(8*(1-nu)*sin((theta1(a)+(theta2(a)))/2)-4*r*(nu)*sin(theta)/sqrt(r1(a)*r2(a))-2*r^2/(r1(a)*r2(a))*(sin((theta1(a)+(theta2(a)))/2)+sin(2*theta-theta1(a)/2-(theta2(a))/2)))/(4*E)+(1+nu)*sigma12*sqrt(r1(a)*r2(a))*((1-2*nu)*cos((theta1(a)+theta2(a))/2)+2*r*(1-nu)*cos(theta)/sqrt(r1(a)*r2(a))-1*r^2/(r1(a)*r2(a))*sin(theta)*sin(theta-theta1(a)/2-theta2(a)/2))/(E);
 >
 (6)
 >
 > ## get u_r and u_theta as u[1] and u[2]
 > u[1] := u1*cos(theta)+u2*sin(theta);
 (7)
 > u[2] := -sin(theta)*u1+cos(theta)*u2;
 (8)
 > Diff_ur := simplify(diff(u[1], a));
 (9)
 >
 >
 > Diff_ut := simplify(diff(u[2], a));
 (10)
 >
 > # find the limiting case
 > Att := limit(Diff_ut*r*sin(2*theta), r = infinity);
 (11)
 > Arr := limit(Diff_ur*r*(1-cos(2*theta)), r = infinity);
 (12)
 > G := (1/8)*(int(Arr+Att, theta = 0 .. Pi/2))*sigma*4;
 (13)
 > simplify(G)
 (14)
 >

Thanks,

## Adding Arrows to Parametrics to Show Direction...

Hi,

Is there any way to add arrows to the parametrics plot to show the direction they are going? Thanks.

## Finding Intersection of Parametrics;...

Hi,

How would I tell maple to solve for the variable t? The answer should be 4.

[t-2,(t-2)^2]

[(3t/2)-4,(3t/2)-2]

I need to set them equal to eachother and then solve for t.

Thanks!

## How to create recursive loops in procedure ...

Suppose that n is a positive integer number and [k1,k2,...,kn] is a list of non-negative integer numbers.

My Question: How to create the procedure proc([k1,k2,...,kn]) such that the output of this procedure is in the following form:

For i1 from 0 to k1 do

for i2 from 0 to k2 do

......................

for in from 0 to kn do

i1+i2+...+in

end do; end do;...end do;

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