MaplePrimes Questions

If I have checked the Editable button just below the working window, then the temperature would be very high in the next time when I start Maple 2019. I do not what is going on. But when I unchecked the Editable button, and wait for several seconds, then the temperature and the load of my laptop are on the normal state.  Is this a bug for Maple 2019? My OS is Debian Stretch, that is,

$ uname -a
Linux debian 4.9.0-9-amd64 #1 SMP Debian 4.9.168-1 (2019-04-12) x86_64 GNU/Linux

 

Hi

I try to solve the laplace equation with some special boundary conditions.

But, i get the follwoing error

Error, (in pdsolve/sys) the given system is not polynomial in the variables {f}

 

 laplace_equation.mw

Thank you for any help

 

This worksheet is a modification to Kitonum's excellent http://www.mapleprimes.com/posts/202222-Contour-Curves-With-Labels.

The mod adds the ability to display labelled contours for expressions in x and y defined parametrically.

Your comments are welcome.

Contourplot_with_labels.mw


 

with(LinearAlgebra)

A := Matrix(4, 4, {(1, 1) = 1, (1, 3) = -1, (1, 4) = 3, (2, 2) = 2, (2, 3) = 1, (3, 1) = -1, (3, 2) = 1, (3, 3) = 6, (3, 4) = -1, (4, 1) = 3, (4, 3) = -1, (4, 4) = 10}, fill = 0)

Matrix(%id = 18446746512315154430)

(1)

b := Matrix(4, 1, {(1, 1) = 0, (2, 1) = -2, (3, 1) = -1, (4, 1) = -1})

Matrix(%id = 18446746512315153574)

(2)

x := Matrix([[x1], [x2], [x3], [x4]])

Matrix(%id = 18446746512315146838)

(3)

f := proc (x) options operator, arrow; (1/2)*Transpose(x).A.x+Transpose(b).x end proc

proc (x) options operator, arrow; Typesetting:-delayDotProduct(Typesetting:-delayDotProduct((1/2)*LinearAlgebra:-Transpose(x), A), x)+Typesetting:-delayDotProduct(LinearAlgebra:-Transpose(b), x) end proc

(4)

(1/2)*Transpose(x).A.x+Transpose(b).x

Matrix(%id = 18446746512315132494)

(5)

while g(vk) < 10^(-6) do k end do

Error, cannot determine if this expression is true or false: (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -1, (1, 4) = 3, (2, 1) = 0, (2, 2) = 2, (2, 3) = 1, (2, 4) = 0, (3, 1) = -1, (3, 2) = 1, (3, 3) = 6, (3, 4) = -1, (4, 1) = 3, (4, 2) = 0, (4, 3) = -1, (4, 4) = 10})) . vk+(Matrix(4, 1, {(1, 1) = 0, (2, 1) = -2, (3, 1) = -1, (4, 1) = -1})) < 1/1000000

 

v0 := Matrix([[0], [1], [0], [0]])

Matrix(%id = 18446746512315172734)

(6)

g := proc (x) options operator, arrow; A.x+b end proc

proc (x) options operator, arrow; Typesetting:-delayDotProduct(A, x)+b end proc

(7)

alpha0 := solve(diff(f(v0-g(v0)*alpha)[1, 1], alpha) = 0)

1/10

(8)

v1 := v0-alpha0*g(v0)

Matrix(%id = 18446746512315163582)

(9)

v(k+1) := vk-`&alpha;k`*g(vk)

vk-`&alpha;k`*(Matrix(%id = 18446746512315154430).vk+Matrix(%id = 18446746512315153574))

(10)

`&alpha;k` := solve(diff(f(vk-g(vk)*alpha)[1, 1], alpha) = 0)

Hi, 
Essentially i am trying to programe an iterative loop where v(K+1) can be found fro v(K), I'm not sure how to programe a loop but I know this is not a hard thing to do so I am struggling, any help would be appreciated! Thanks. 
Edit: Also alpha K must be found at each stage by optimising f(v(k+1))
 

Download Optimisation_coursework.mw

Hello everyone!

I'm having some problem with this equation:

solve(0.1 = 23.714*(-0.93205)^2/(20.3+61.4*.884^x), x)

I'm trying to solve for x, but i keeps saying "Warning, solutions may have been lost."

Any ideas?
 

Hello,

What is the minimum period of the following equation.


 

d := evalf(expand((100+100*cos(6*t)+200*cos(12*sqrt(2)*t))^2))

40000.-2580480000.*cos(t)^2*cos(1.414213562*t)^6+604800000.*cos(t)^2*cos(1.414213562*t)^4-51840000.*cos(t)^2*cos(1.414213562*t)^2+2621440000.*cos(t)^6*cos(1.414213562*t)^12-7864320000.*cos(t)^6*cos(1.414213562*t)^10+8847360000.*cos(t)^6*cos(1.414213562*t)^8-4587520000.*cos(t)^6*cos(1.414213562*t)^6+1075200000.*cos(t)^6*cos(1.414213562*t)^4-92160000.*cos(t)^6*cos(1.414213562*t)^2-3932160000.*cos(t)^4*cos(1.414213562*t)^12+0.1179648000e11*cos(t)^4*cos(1.414213562*t)^10-0.1327104000e11*cos(t)^4*cos(1.414213562*t)^8+6881280000.*cos(t)^4*cos(1.414213562*t)^6-1612800000.*cos(t)^4*cos(1.414213562*t)^4+138240000.*cos(t)^4*cos(1.414213562*t)^2+1474560000.*cos(t)^2*cos(1.414213562*t)^12-4423680000.*cos(t)^2*cos(1.414213562*t)^10+4976640000.*cos(t)^2*cos(1.414213562*t)^8+720000.*cos(t)^2+274560000.*cos(1.414213562*t)^4-5125120000.*cos(1.414213562*t)^6+0.4942080000e11*cos(1.414213562*t)^8-0.2811494400e12*cos(1.414213562*t)^10+0.1013841920e13*cos(1.414213562*t)^12+0.1677721600e12*cos(1.414213562*t)^24-0.1006632960e13*cos(1.414213562*t)^22+0.2642411520e13*cos(1.414213562*t)^20-0.3984588800e13*cos(1.414213562*t)^18+0.3810263040e13*cos(1.414213562*t)^16-0.2406481920e13*cos(1.414213562*t)^14-5760000.*cos(1.414213562*t)^2+10240000.*cos(t)^12-30720000.*cos(t)^10+34560000.*cos(t)^8+1320000.*cos(t)^4-16000000.*cos(t)^6

(1)

``


 

Download period

 

 

When typing 

z:=exp(I*2*Pi/3);
convert(z,'sincos')

Maple evaluates the intermediate result which is cos(2*Pi/3)+I*sin(2*Pi/3) and  gives

Is there a way to tell it not to do this? I'd like to see the result as when typing

'cos(2*Pi/3)+I*sin(2*Pi/3)'

Is there an option or method to tell Maple not to immediate evaluation in the above? it can do evaluate next time the expression is used.

 


How to solve a system of partial equations with boundary conditions.I used this formula

restart;
sys_Pde := diff(V(x, t), x, x) = 0, diff(T(x, t), x, x) = 0;
          
BC := eval(diff(V(x, t), x), x = 1) = 0, eval(V(x, t), x = 0) = 1+cos(w*t), eval(T(x, t), x = 0) = 0, eval(T(x, t), x = 1) = 1;
pdsolve({sys_Pde});
   {T(x, t) = _F1(t) x + _F2(t), V(x, t) = _F3(t) x + _F4(t)}
pdsolve({BC, sys_Pde});


Error, (in PDEtools:-ToJet) found functions to be rewritten in jet notation, {V(1, t)}, having different dependency than the indicated in [V(x, t)]

 

I

have the following function phi(x,y) which depends on two unknown functions F1 and F2 which are differentiable.

Phi(x,y)=y sin(x y) + F1(y) + y^2 F2(x y)

Note that F1 and F2 are obtained by hand when I solve a partial differential equation

I would like to find  phi(x,y)  that satisfies phi=1 and diff(phi(x,y),y)=1-x^2*cos(x^2) on line y=x

 

constraint.mw

Thanks for your help

 

Hi, given the function f(t)=t^2. Determine the Fourier coefficients a_0, a_1 and b_1 for f(t). Plot the graphs for f(t) and also plot the graphs for the Fourier sum F(t) with N=1 in the same coordinate system. Let the t-axis go from 0 to 2*Pi.

Can someone help with this?

Thanks

I have a system of linear equations with a fixed number of equations, but the stracture of the system varies in T. I have solved the system for a fixed T and have to run for each value of T, which is quite boaring. I want to solve the system by a single run for multiple T values using a for loop. When I tried with the code, I got the following error.

Error, (in LinearAlgebra:-GenerateMatrix) invalid subscript selector


t.mw

L:=[1,2,0,4]:
s := sum(1/L[i], i=1..nops(L));

Error, (in limit/mrv/limsimpl) too many levels of recursion

So, this error cannot be caught by try

try    # bug
  s := sum(1/L[i], i=1..nops(L));
  catch:  s:=infinity
end try;

Error, (in limit/mrv/limsimpl) too many levels of recursion

Strangely, for L:=[1, 2, 0.0, 4]  it's OK.

Everything works with add instead of sum, but this is another thing.

 

  I want list all  diameter-2  Nonisomorphismgraphs  of order n . I use the following code, but its running speed  is slow with  order of graph gradually increasing . (for example,n=10)
 Are there other ways to Improve it? 
restart:
with(GraphTheory):
Graphs_data:=[NonIsomorphicGraphs(4,restrictto =[connected], output = graphs, outputform =graph)]:
Diameter_select:=select[flatten](t->Diameter(t)=2,Graphs_data):
DrawGraph(Diameter_select,size=[50,50]);

  By the way,  Why doesn't size=[50,50] work ?

 
                       
                   
               
               
           
           
               
                   
                 

Hi,

How can find the range of  the parameter k, such that f(x)=b ( for a given b in the set of real number R) we have a unique solution of the equation f(x)=b 

 

Many thanks for your help

Hello,

I'm trying to solve a system of ODE in a model of infectious disease. But, unfortunately, when I try to plot the curves, I get the following error message: Error, (in fproc) unable to store '-1.*HFloat(0.0)[1]' when datatype=float[8]. The code is here

restart;
with(plots);
Q := 1000; a := .9; b := 0.8e-3; mu := 0.247e-2; k := .2; y := 0.2e-1; e := .5; g := .5; T := 8; n := 100;
sigma[1] := 0.5e-1; sigma[2] := 0.9e-1; alpha[1] := 0.52e-2; alpha[2] := 0.52e-2; delta[1] := .8; delta[2] := .904; delta[3] := .8; c[1] := 50; c[2] := 250; c[3] := 50; w[1] := 140; w[2] := 130; w[3] := 150; w[4] := 160;
u[1] := min(max(0, z), 1); z := (a*i[p](t)*(i[p](t)+i[pm](t))*(lambda[4](t)-lambda[3](t))+a*s(t)*(i[p](t)+i[pm](t))*(lambda[1](t)-lambda[2](t)))/(n.w[1]); u[2] := min(max(0, c), 1); c := (b*(i[m](t)+i[pm](t))*s(t)*(lambda[3](t)-lambda[1](t))+b*(i[m](t)+i[pm](t))*i[p](t)*(lambda[4](t)-lambda[2](t)))/(n.w[2]); u[3] := min(max(0, j), 1); j := (i[p](t)*lambda[2](t)+i[pm](t)*(lambda[4](t)-lambda[7](t))-(i[p](t)+i[pm](t))*lambda[5](t))/w[3]; u[4] := min(max(0, o), 1); o := (i[m](t)*lambda[3](t)-i[pm](t)*(lambda[4](t)-lambda[7](t))-(i[m](t)+i[pm](t))*lambda[6](t))/w[4]; u[2] := 0; u[3] := 0;
;
sys := diff(s(t), t) = Q+delta[1]*r[p](t)+delta[2]*r[m](t)+delta[3]*r[pm](t)-(a*(1-u[1])*(i[p](t)+i[pm](t))/n+b*(1-u[2])*(i[m](t)+i[pm](t))/n+mu)*s(t), diff(i[p](t), t) = (1-u[1])*a*(i[p](t)+i[pm](t))*s(t)/n-(1-u[2])*b*(i[m](t)+i[pm](t))*i[p](t)/n-(sigma[1]+u[3])*i[p](t)-(alpha[1]+mu)*i[p](t), diff(i[m](t), t) = (1-u[2])*b*(i[m](t)+i[pm](t))*s(t)/n-(1-u[1])*a*(i[p](t)+i[pm](t))*i[m](t)/n-(sigma[2]+u[4])*i[m](t)-(alpha[2]+mu)*i[m](t), diff(i[pm](t), t) = (1-u[2])*b*(i[m](t)+i[pm](t))*i[p](t)/n+(1-u[1])*a*(i[p](t)+i[pm](t))*i[m](t)/n-(y+u[3]+u[4])*i[pm](t)-(alpha[1]+alpha[2]+mu)*i[pm](t), diff(r[p](t), t) = (sigma[1]+u[3])*i[p](t)+(e*y+u[3])*i[pm](t)-(delta[1]+mu)*r[p](t), diff(r[m](t), t) = (sigma[2]+u[4])*i[m](t)+(y*g*(1-e)+u[4])*i[pm](t)-(delta[2]+mu)*r[m](t), diff(r[pm](t), t) = (y*(1-g)*(1-e)+u[3]+u[4])*i[pm](t)-(delta[3]+mu)*r[pm](t), diff(lambda[1](t), t) = lambda[1](t)*(a*(1-u[1])*(i[p](t)+i[pm](t))/n+b*(1-u[2])*(i[m](t)+i[pm](t))/n+mu)-lambda[2](t)*(1-u[1])*a*(i[p](t)+i[pm](t))/n-lambda[3](t)*(1-u[2])*b*(i[m](t)+i[pm](t))/n, diff(lambda[2](t), t) = -c[1]+lambda[1](t)*(1-u[1])*a*s(t)/n-lambda[2](t)*((1-u[1])*a*s(t)/n+b*(1-u[2])*(i[m](t)+i[pm](t))/n+sigma[1]+u[3]+alpha[1]+mu)-lambda[4](t)*(b*(1-u[2])*(i[m](t)+i[pm](t))/n+(1-u[1])*a*i[m](t)/n)-lambda[5](t)*(sigma[1]+u[3]), diff(lambda[3](t), t) = -c[2]+lambda[1](t)*(1-u[2])*b*s(t)/n+lambda[2](t)*(1-u[2])*b*i[p](t)/n-lambda[3](t)*(u[4]+alpha[2]+sigma[2]+a*(1-u[1])*(i[p](t)+i[pm](t))/n-(1-u[2])*b*s(t)/n)-lambda[4](t)*((1-u[2])*b*i[p](t)/n+a*(1-u[1])*(i[p](t)+i[pm](t))/n)-lambda[6](t)*(sigma[2]+u[4]), diff(lambda[4](t), t) = -c[3]+lambda[1](t)*((1-u[1])*a*s(t)/n+(1-u[2])*b*s(t)/n)-lambda[2](t)*((1-u[2])*b*i[p](t)/n+(1-u[1])*a*s(t)/n)-lambda[2](t)*((1-u[2])*a*i[m](t)/n+b*(1-u[1])*s(t)/n)-lambda[4](t)*(y+u[3]+u[4]+alpha[1]+alpha[2]+mu-(1-u[2])*b*i[p](t)/n-b*(1-u[1])*i[m](t)/n)-lambda[5](t)*(e*y+u[3])-lambda[6](t)*(y*g*(1-e)+u[4])-lambda[7](t)*(y*(1-g)*(1-e)+u[3]+u[4]), diff(lambda[5](t), t) = -lambda[1](t)*delta[1]+lambda[5](t)*(delta[1]+mu), diff(lambda[6](t), t) = -lambda[1](t)*delta[2]+lambda[6](t)*(delta[2]+mu), diff(lambda[7](t), t) = -lambda[1](t)*delta[3]+lambda[7](t)*(delta[3]+mu), i[p](0) = 300, i[m](0) = 200, i[pm](0) = 150, r[p](0) = 200, r[m](0) = 150, r[pm](0) = 150, s(0) = 1000, lambda[1](T) = 0, lambda[2](T) = 0, lambda[3](T) = 0, lambda[4](T) = 0, lambda[5](T) = 0, lambda[6](T) = 0, lambda[7](T) = 0;
p1 := dsolve({sys}, type = numeric, method = bvp[midrich], abserr = 0.1e-5, maxmesh = 2400);
Error, (in fproc) unable to store '-1.*HFloat(0.0)[1]' when datatype=float[8]
p2o := odeplot(p1, [t, i[p](t)], 0 .. 3, numpoints = 100, labels = ["Time (Months)", " Population"], labeldirections = [horizontal, vertical], style = line, color = green, axes = boxed);
 

Can anyone help me please? I read some related problems here, but couldnt find a solution yet.

Thanks for your time

Best regards

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