MaplePrimes Questions

Hi Maple friends. :)

I would like to graph and solve: 2*sin(2 x)=1 for the interval 0<=x<=360. My textbook gives the answers of x=15, 75, 195 and 225.

But I get an error messages:

plot(2*sin(2*x) = 1, x = 0 .. 360);
Error, invalid input: plot expects its 1st argument, p, to be of type {array, list, rtable, set, algebraic, procedure, And(`module`, appliable)}, but received 2*sin(2*x) = 1

solve(2*sin(2*x) = 1, x = 0 .. 360);
Error, invalid input: too many and/or wrong type of arguments passed to solve; first unused argument is x = 0 .. 360

Any help would be appreciated! Thanks in advance.

 

x: =Matrix([[a1,a2],[a3,a4]])

after some calculation,

assign(%)

a1 etc have value,

how to make a1,a2,a3,a4 back to variable in maple 12?

How to calculate floor(10^(10^(10^(10^(10^(-10^10))))))? My simpleminded try is

Error, numeric exception: underflow
PS. I think that is about 10^(10^10).

m := Matrix([[a1,a2,a3],[a4,a5,a6],[a7,a8,a9]]);

m2 := Determinant(m-Matrix([[1,0,0],[0,1,0],[0,0,1]]));

1. which family of polynomials do m2 belong to?

2. how to analyze m2?

What´s the error here??????

Solução do Sistema de EDO's por Transformada de Laplace

 

restart

with(inttrans):

eq1 := diff(x(t), t)+(2.2*x(t)*y(t)+0.5e-1*x(t)/(5+0.5e-1*t)) = 0;

diff(x(t), t)+2.2*x(t)*y(t)+0.5e-1*x(t)/(5+0.5e-1*t) = 0

(1)

eq2 := diff(y(t), t)+(2.2*x(t)*y(t)-(0.5e-1*(0.25e-1-y(t)))/(5+0.5e-1*t)) = 0;

diff(y(t), t)+2.2*x(t)*y(t)-0.5e-1*(0.25e-1-y(t))/(5+0.5e-1*t) = 0

(2)

eq3 := diff(z(t), t)-2.2*x(t)*y(t)+0.5e-1*z(t)/(5+0.5e-1*t) = 0;

diff(z(t), t)-2.2*x(t)*y(t)+0.5e-1*z(t)/(5+0.5e-1*t) = 0

(3)

EQ := [eq1, eq2, eq3]:

for i to 3 do La[i] := laplace(EQ[i], t, s) end do;

s*laplace(x(t), t, s)-1.*x(0.)+2.200000000*laplace(x(t)*y(t), t, s)+laplace(x(t)/(100.+t)^1., t, s) = 0.

 

-1.*y(0.)-0.2500000000e-1*(exp(100.*s))^1.*Ei(1., 100.*s)^1.+1.*s^1.*laplace(y(t), t, s)^1.+2.200000000*laplace(x(t)^1.*y(t)^1., t, s)+1.*laplace(y(t)^1./(100.+1.*t)^1., t, s) = 0.

 

s*laplace(z(t), t, s)-1.*z(0.)-2.200000000*laplace(x(t)*y(t), t, s)+laplace(z(t)/(100.+t)^1., t, s) = 0.

(4)

LL := subs({laplace(x(t), t, s) = X, laplace(y(t), t, s) = Y, laplace(z(t), t, s) = Z}, [La[1], La[2], La[3]]);

[s*X-1.*x(0.)+2.200000000*laplace(x(t)*y(t), t, s)+laplace(x(t)/(100.+t)^1., t, s) = 0., -1.*y(0.)-0.2500000000e-1*(exp(100.*s))^1.*Ei(1., 100.*s)^1.+1.*s^1.*Y^1.+2.200000000*laplace(x(t)*y(t), t, s)+1.*laplace(y(t)/(100.+t)^1., t, s) = 0., s*Z-1.*z(0.)-2.200000000*laplace(x(t)*y(t), t, s)+laplace(z(t)/(100.+t)^1., t, s) = 0.]

(5)

sol := solve(LL, [X, Y, Z]):

assign(sol):

SOLS[X, Y, Z]:

SOLT := map(invlaplace, [X, Y, Z], s, t);

[-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+x(0), -1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))+y(0)+0.2500000000e-1*ln(1.+0.1000000000e-1*t), 2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))+z(0)]

(6)

SOLTT := evalf(subs({x(0) = 0.5e-1, y(0) = 0, z(0) = 0}, SOLT));

[-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+0.5e-1, -1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))+0.2500000000e-1*ln(1.+0.1000000000e-1*t), 2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))]

(7)

xx := evalc(Re(SOLTT[1]));

-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+0.5e-1

(8)

yy := evalc(Re(SOLTT[2]));

0.2500000000e-1*ln(abs(1.+0.1000000000e-1*t))-1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))

(9)

zz := evalc(Re(SOLTT[3]));

2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))

(10)

plot([xx, yy, zz], t = 0 .. 500, legend = [x, y, z]);

Warning, expecting only range variable t in expression -2.200000000*int(x(_U1)*y(_U1),_U1 = 0. .. t)-1.*int(x(_U1)/(100.+_U1),_U1 = 0. .. t)+.5e-1 to be plotted but found names [_U1, x, y]

 

 

NULL

NULL


Download laplace.mw

Dear Friends

I have a problem in CPU time in MAPLE.

I write the codes in maple related to the nonlinear heat conduction problem in one dimension by Collocation method, but after 30 minutes no solution has been observed!!!

My codes are for N=4!, i.e., I have 25 equations with 25 unknowns!!!

If MAPLE can not solve this simple system, How can I solve 3 dimensional pdes by N=9,

In this case, I have 1000 equations with 1000 unknowns!!!

please help me and suggest me a fast iterative solver.

I should remark that my problem is stated in this paper

http://www.sciencedirect.com/science/article/pii/S1018364713000025

If there exist any other suitable method, I will be happy to receive any support.

 

With kind regards,

Emran Tohidi.

 

> restart;
> Digits := 20; N := 4; st := time(); u := sum(sum(a[m, n]*x^m*t^n, m = 0 .. N), n = 0 .. N); u := unapply(u, x, t); ut := diff(u(x, t), `$`(t, 1)); ut := unapply(ut, x, t); ku := simplify(1+u(x, t)^2); ku := unapply(ku, x, t); ux := diff(u(x, t), `$`(x, 1)); ux := unapply(ux, x, t); K := ku(x, t)*ux(x, t); K := unapply(K, x, t); Kx := diff(K(x, t), `$`(x, 1)); Kx := unapply(Kx, x, t); f := proc (x, t) options operator, arrow; x*exp(t)*(1-2*exp(2*t)) end proc;
print(`output redirected...`); # input placeholder
> S1 := {seq(u(i/N, 0)-i/N = 0, i = 0 .. N)}; S2 := {seq(u(0, j/N) = 0, j = 1 .. N)}; S3 := {seq(u(1, j/N)+ux(1, j/N)-2*exp(j/N) = 0, j = 1 .. N)}; S4 := {seq(seq(Kx(i/N, j/N)+f(i/N, j/N)-ut(i/N, j/N) = 0, i = 1 .. N-1), j = 1 .. N)}; S := `union`(`union`(`union`(S1, S2), S3), S4); sol := DirectSearch:-SolveEquations([op(S)], tolerances = 10^(-4), evaluationlimit = 1000000);
print(`output redirected...`); # input placeholder
> assign(sol);
%;
> u(x, t);
> CPUTIME := time()-st;
plot3d(u(x, t) - x exp(t), x = 0 .. 1, t = 0 .. 1)

Hello,

I just bought and installed "The mathematical Survival Kit" but I can't figure out how does it work

Anybody can help?

 

Thanaks

Martina

EDIT:
Let f:X->f(X)  be a polynomial function from C^n to C^p. Let r(X) be the rank of the Jacobian matrix of f in X. What is the maximal value of r(X) when X goes throught C^n ?

In other words, I'd want to obtain the maximal dimension of the components of im(f). 

How to proceed ?

Thanks in advance.

 

 

restart;

with(combinat):

symMonomial := proc(test)

h := 0;

for i from 1 to nops(test) do

                h[i] := choose(test,i);

od;

 

c := copy(test);

k := 0;

for k from 1 to nops(test) do

                c[k] := 0;

                for i from 1 to nops(h[k]) do

                                ki := 1;

                                for j from 1 to nops(h[k][i]) do

                                                ki := ki*h[k][i,j];

                                od;

                                c[k] := c[k] + ki;

                od;

od;

return c;

end proc;

 

 

sympoly := proc(test, number_of_roots)

with(combinat):

h := 0;

for i from 1 to nops(test) do

                h[i] := choose(z,i);

od;

 

c := 0;

for k from 1 to nops(test) do

                c[k] := 0;

                for i from 1 to nops(h[k]) do

                                ki := 1;

                                for j from 1 to nops(h[k][i]) do

                                                ki := ki*h[k][i,j];

                                od;

                                c[k] := c[k] + ki;

                od;

od;

poly := x^number_of_roots;

for k from 1 to number_of_roots do

                poly := poly + c[k]*x^(number_of_roots-k);

od;

end proc;

 

z := [x1, x2, x3, x4, x5, x6, x7, x8, x9, x10];

sigma := symMonomial(z);

f := expand(sigma[1]*sigma[2]);

f := expand(sigma[1]*sigma[1]); # two lambda value are the same

f := expand(sigma[1]*sigma[1]*sigma[1]); # three lambda value are the same

 

f := expand(sigma[1]^2+2*sigma[1]*sigma[2]+sigma[2]);

f := subs(x1=3, f);

f := subs(x2=2, f);

f := subs(x3=3, f);

f := subs(x4=4, f);

f := subs(x5=5, f);

f := subs(x6=6, f);

f := subs(x7=7, f);

f := subs(x8=8, f);

f := subs(x9=9, f);

f := subs(x10=lambda, f);

evalf(solve(f, lambda));

if degree(f) = 2 then

                f := f + lambda^3;

                evalf(solve(f, lambda));

end if:

 

how to convert above f back to matrix form such as

 

m := Matrix([[a1,a2,a3],[a4,a5,a6],[a7,a8,a9]]);

m-Matrix([[lambda,0,0],[0,lambda,0],[0,0,lambda]]);

m2 := Determinant(m-Matrix([[lambda,0,0],[0,lambda,0],[0,0,lambda]]));

 

after tested m2 can not be expressed in terms of shell like polynomial,

it seems that it is from symmetric polynomial and it is from a non-homogenous polynomial which homogenize with a lambda

 

if solve f for new eigenvalue,

can traditional eignvector method calculate new eigenvector for these new kind of eigenvalues method?

A*x = lambda*x

hi ,

 

i solve my odes with ode analyzer and obtained h(t) , u(t) and s(t) ... i wana put u(t) and s(t) in another equation to obtain T(t,r) , i did as follow but i have an error :

 

400*(s(t)*(diff(T(t, r), r))+u(t)*(diff(T(t, r), t))/r) = 0.3e-1*((diff(T(t, r), r)+r*(diff(T(t, t), r, r)))/r+(diff(T(t, r), r))/r^2)+0.1e-2*(2*((diff(u(t), t))/r+s(t)/r)^2+((diff(s(t), t))/r-u(t)/r)^2)

 

4.00000000000000*u(t)^2*h(t)*(diff(y(t), t))-8.00000000000000*h(t)^2*x(t)^2+16.0000000000000*u(t)*h(t)*x(t)*(diff(h(t), t))+(1367.18750000000*(-29.8200000000000+u(t)))*h(t)^2*x(t)+363.281250000000*u(t)*h(t)*y(t)-1367.18750000000*h(t)^2*(30-u(t))*x(t)+u(t)^2*h(t)-0.225000000000000e-1+0.225000000000000e-1*(0.749297072727423e-1+1.49578322716831*sin(t-0.75e-1))^2+4.00000000000000*(u(t)*y(t)+h(t)*x(t))^2 = 0

 

diff(s(t), t) = 8.00000000000000*(diff(x(t), t))*h(t)+16.0000000000000*y(t)*x(t)+8.00000000000000*u(t)*(diff(y(t), t))

 

diff(h(t), t) = y(t)

 

diff(u(t), t) = x(t)

 

x(t) = 427.2460938*u(t)+385620.1174/u(t)-25671.38673

 

h(0) = 0.1e-10, s(0) = .6, u(0) = .62, x(0) = 596561.43714, y(0) = .12096, T(0, r) = 0, t = 0 .. .18, r = 0 .. .125

 

pdsolve(1, 2, 3, 4, 5, 6, {7}, numeric);
%;
Error, (in pdsolve) invalid input: `pdsolve/numeric` expects its 2nd argument, IBCs, to be of type {list, set}, but received 2

Hi,

according to my previous question

http://www.mapleprimes.com/questions/201435-ODE-With-Constraint

I wrote the following code. at first, the code solve the equation for f and when it slves that, I want to solve Theta in such a way that use the values of f in previous calculation. I use the command 'known' but i couldnt find thesolution

I would be most grateful if you could help me in this problem

Thanks for your attentions in advance


restart; # Notice that Restart (capital R) has no effect (to catch that use semicolon, not colon)
a:=0.13:
b:=0.41:
reynolds:=1.125*10^7;  
Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)^2*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)^2*x,x$1);
eq1:=isolate(Eq1,diff(f(x),x,x,x));
eq2:=subs(g=f,isolate(Eq2,diff(g(x),x,x,x)));
EQ:=diff(f(x),x,x,x)=piecewise(x<c*0.1,rhs(eq1),rhs(eq2));


c:=75:
;
Q:=proc(pp2) local res,F0,F1,F2;
print(pp2);
if not type(pp2,numeric) then return 'procname(_passed)' end if:
res:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=pp2},numeric,output=listprocedure);
F0,F1,F2:=op(subs(subs(res),[f(x),diff(f(x),x),diff(f(x),x,x)])):
F1(c)-1;
end proc;


fsolve(Q(pp2)=0,pp2=(0..102));
se:=%;
res2:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=se},numeric,output=listprocedure);
G0,G1,G2:=op(subs(subs(res2),[f(x),diff(f(x),x),diff(f(x),x,x)])):

plots:-odeplot(res2,[seq([x,diff(f(x),[x$i])],i=0..2)],0..2); #This plots from and past 0.1*c
pr:=1;
prt:=0.89;

Eq11:=diff(theta(x),x$2)+pr*diff(theta(x),x$1)*f(x)+pr/prt*b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(theta(x),x$1)*x^2,x$1);
Eq22:=diff(g(x),x$2)+pr*diff(g(x),x$1)*f(x)+pr/prt*a^2*c*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(g(x),x$1)*x^1,x$1);
eq11:=isolate(Eq11,diff(f(x),x,x));
eq22:=subs(g=theta,isolate(Eq22,diff(g(x),x,x)));
EQT:=diff(theta(x),x,x)=piecewise(x<c*0.1,rhs(eq11),rhs(eq22));


QT:=proc(pp3) local res3,theta0,theta1;
print(pp3);
if not type(pp3,numeric) then return 'procname(_passed)' end if:
res3:=dsolve({EQT,theta(0)=1,D(theta)(0)=pp3,known=f},numeric,output=listprocedure);
theta0,theta1:=op(subs(subs(res),[theta(x),diff(theta(x),x)])):
theta0(c);
end proc;

fsolve(QT(pp3)=0,pp3=(0..200));
res3(0);



Amir

in maple 15

https://drive.google.com/file/d/0B8F2D27rfQWgVXE1alN0V3JWU1U/edit?usp=sharing

there are 3 equation to be minimized

and i limit x between x + 5 and x - 5 as constraints

 

though f1 got a error in first line of command,

later i type a correct command for f1 in later part of script

What is the difference between the "old" Maplet format and the "new" Möbius Project App format? Is the Möbius Project just a new name for writing Maplets? Is the programming different; are the mechanics of putting the file in a form to be used by students different; does the file containing the program that can be directly run by students have a different format and extension; are the kinds of things students see different? Is the Maplets function on the way to being discontinued in favor of the Möbius Apps?

I cannot find anything in the documentation for either format that relates it to the other format, although the general description of the output seems entirely the same.

Hi:

how find the coefficients C1,C2?
q(0)=0,D(q)(0)=0

q(T) := sin(T)*_C2+cos(T)*_C1-(1/3000)*Pi*(4012562293500*Pi^3*cos(T)^3-32100498340000*Pi^3*cos(T)^2-3009421720125*Pi^3*cos(T)+16050249170000*Pi^3+435778855000*Pi*cos(T)^2-217889427500*Pi-3539762622):

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