MaplePrimes Questions

When print a formula , if variable has power, the power will print in another row

how can it print like a^2 in one row 

Dear all

I would like to minimuze the following function 4 x ^2 + 4 x y  under constraint  16=x^2 y  and both x , y  nonnegative real number 

this is my code

with(Optimization)

Minimize(4*x^2+4*x*y, {x^2*y = 16}, assume = nonnegative)

 

The result obtained from this code is strange

because when i do directly the computation without maple by substituting y =16/x^2 and simple derivation i get the minimum is at x=2 and so y=4 and therefore the minimum is 48

But as i say using my code I obtained a different solution

whats is the problem occurs in this situation

Many thanks

 

I am using MAPLE 2016.1.

I have created an animation of points on a polarplot (2D), but would like it to be projected into 3D, with time t being the 3rd axis.  Here is my code for the polarplot animation:

with(plots);
Repltlist := proc (t) options operator, arrow; [[sin(t), 0], [cos(t), (1/3)*Pi], [cos(2*t), (2/3)*Pi]] end proc;
Impltlist := proc (t) options operator, arrow; [[-sin(3*t), 0], [-2*cos(t), (1/3)*Pi], [-cos(4*t), (2/3)*Pi]] end proc;
titleName := "Polar representation of time-dependence of degrees of freedom in configuration space"; captionName := "Re - red, Im - blue";
P := animate(polarplot, [Repltlist(t), color = "Red"], t = -Pi .. Pi, symbol = solidcircle, style = point, color = red, symbolsize = 12, frames = 100, gridlines = true);
Q := animate(polarplot, [Impltlist(t), color = "Blue"], t = -Pi .. Pi, symbol = solidcircle, style = point, color = blue, symbolsize = 12, frames = 100, gridlines = true);
R := display([P, Q], title = titleName, caption = captionName); R;

I have been trying use the plots[transform] to 'lift' this polarplot into 3D: 

with(plottools); with(plots);

plrPt(2); f := proc (tt) options operator, arrow; plottools:-transform(proc (r, theta) options operator, arrow; [r, theta, tt] end proc) end proc; display((f(2))(plrPt(2)));

but the last command gives no output.

Can anyone help?

MRB
PS:  I am now attemptiing to do the lift to 3D by using a cylinderplot but would like know why the transform function is not able able to lift the polarplot into 3D.

Hello; I found another ODE which Maple gives division by zero on.  Is this also a bug? 

dsolve(x*(a^2*x+(x^2-y(x)^2)*y(x))*diff(y(x),x)^2-(2*a^2*x*y(x)-(x^2-y(x)^2)^2)*diff(y(x),x)+a^2*y(x)^2-x*y(x)*(x^2-y(x)^2) = 0, y(x));

Error, (in dsolve) numeric exception: division by zero

This is from a book. Using Maple 2016.1 on windows.

Please Sir/Ma, I'm trying to generate a recurrent relations of this series and I try to use "if" "else" condition but I didn't get it right. Any one with useful suggestions. Appreciate 

 

restart;
Y[0] := A; Y[1] := B;
if k = a then delta(k-a) := 1 else 0 end if;
                               0
for k from 0 to 10 do Y[k+2] := solve(add(delta(i-1)*(k-i+1)*(k-i+2)*Y[k-i+2], i = 0 .. k)+add((delta(i)-delta(i-1))*(k-i+1)*Y[k-i+1], i = 0 .. k)+lambda*Y[k] = 0, Y[k+2]) end do;
y := sum(Y[j]*x^j, j = 0 .. 10);
 

I use maple 12 on my dual core laptop and i plan to buy maple 2016 and an 8 core system to get advandage of multiple processors. 

Does maple 2016 functions get advandage of multiprocessor systems or it will be the same as having one processor?

Maple 2016.1 on windows. This ODE from a book, and Maple gives division by zero. Is this a bug or expected?

ode:=(a^2*x+(x^2-y(x)^2)*y(x))*diff(y(x),x)+x*(x^2-y(x)^2) = a^2*y(x);
dsolve(ode, y(x));

Error, (in dsolve) numeric exception: division by zero


Mathematica gives this to same ODE, but no division by zero.

DSolve[x*(x^2 - y[x]^2) + (a^2*x + y[x]*(x^2 - y[x]^2))*y'[x] == a^2*y[x], y[x], x]

Solve[x^2/2 - (1/2)*a^2*Log[x - y[x]] + (1/2)*a^2*Log[x + y[x]] +y[x]^2/2 == C[1], y[x]]

Where is the division by zero coming from in Maple?

2016.1 on windows.

Please, anybody with a good answer. Thanks in advance

restart;
U[0] := 1; U[1] := a; U[2] := 3/2; U[3] := (1/6)*b;
for k from 0 to 20 do U[k+4] := (4/factorial(k)+U[k])/((k+1)*(k+2)*(k+3)*(k+4)) end do;
u := sum(U[j]*y^j, j = 0 .. 20);
           3  2   1    3   5   4   /1     1   \  5    7   6
 1 + a y + - y  + - b y  + -- y  + |-- + --- a| y  + --- y 
           2      6        24      \30   120  /      720   

      / 1      1    \  7    1    8   /  1       1     \  9
    + |---- + ---- b| y  + ---- y  + |----- + ------ a| y 
      \1260   5040  /      4480      \45360   362880  /   

        11     10   /   1         1      \  11      13      12
    + ------- y   + |------- + -------- b| y   + --------- y  
      3628800       \4989600   39916800  /       479001600    

      /    1           1       \  13       1       14
    + |--------- + ---------- a| y   + ---------- y  
      \518918400   6227020800  /       5811886080    

      /     1               1        \  15         17        16
    + |------------ + ------------- b| y   + -------------- y  
      \108972864000   1307674368000  /       20922789888000    

      /      1                 1         \  17
    + |-------------- + --------------- a| y  
      \22230464256000   355687428096000  /    

             19         18
    + ---------------- y  
      6402373705728000    

      /       1                   1           \  19
    + |---------------- + ------------------ b| y  
      \7602818775552000   121645100408832000  /    

              1           20
    + ------------------ y  
      115852476579840000    
solve({(sum(U[j]*y^j, j = 0 .. 20), y = 1) = 2*exp(1), (sum(j*(j-1)*U[j]*y^j, j = 0 .. 20), y = 1) = 4*exp(1)}, [a, b]);
 

Here is one that the students generated which caused confusion. 

a := 0.76;
eq1 := 2*cot(a*sqrt(2*E)) = (2*E-5.4)/(sqrt(E*(5.4-E));
solve(eq1, E)

And the results are: 0., 4.411954070, 2.423743792

The problem is with the second answer because it does not exist. If we plot the LHS and the RHS of eq1 vs E where E=0..5.4

plot([2*cot(0.76*sqrt(2*E)), (2*E-5.4)/sqrt(E*(5.4-E))], E=0..5.4, -3..3)

And it gets more interesting, if we calculate:

solve(evalf(eq1), E)

The answers are: 0., 2.423743793, 14.33807304+27.39159712*I

where the 3rd answer is again incorrect.

Finally, if a = 0.8 or larger, the incorrect answers disappear. 

Note - fsolve does handle this problem correctly. And despite my attempts to remind them to use fsolve, they see the solve command as the universal truth. Apparently this will be another teaching moment for next year.

So any thoughts about why this happens and why there is a difference in the outcomes between 0.76 and 0.8 for the value of a?

 

hi .how i can fsolve 8 equations, in which obtained from

for k from 2 to m do eq1[k_] end do

AND

for k from 2 to m do eq2[k_] end do  and so on........

And gain unknown functions as f3[2], f4[4],....

thanks?

fdm-maple.mw
 

 

 ############################Define some parameters

 

 
restart; Digits := 15; n := 1; m := 3; len := 1; h := len/m; nn := m+1
 ############################Define some equation

eq1[k_] := -3.0*h*(-f2[k]*f1[k-1]+f2[k]*f1[k+1]+f1[k]*(-f2[k+1]+f2[k-1]))*f4[k]^2+((-8.0*f1[k]+4.0*f1[k-1]+4.0*f1[k+1])*f3[k]+(-f1[k+1]+f1[k-1])*(-f3[k+1]+f3[k-1]))*f4[k]-f3[k]*(-f1[k+1]+f1[k-1])*(-f4[k+1]+f4[k-1]):

 

 

 

 

                                     ######################################  APPLY BOUNDARY CONDITIONS


f2[0] := f2[2];

1.0

(1)


``for k from 2 to m do eq1[k_] end do

-1.00000000000000*(-f1[2]*f2[3]+f1[3]*(-f2[4]+f2[2]))*f4[3]^2+((-8.0*f1[3]+4.0*f1[2])*f3[3]+f1[2]*(-f3[4]+f3[2]))*f4[3]-f3[3]*f1[2]*(-f4[4]+f4[2])

(2)

for k from 2 to m do eq2[k_] end do

(-4.16666666666667*f2[2]^2+(50.0000000000000*f2[3]+8.33333333333333*f2[4])*f2[2]-100.000000000000*f2[3]^2+50.0000000000000*f2[3]*f2[4]-4.16666666666667*f2[4]^2+1/27)*f4[3]^2+((f2[2]-1.*f2[4]-.5*f2[3])*f3[3]+(-50.*f3[2]+50.*f3[4])*f2[2]+(-50.*f3[2]+50.*f3[4])*f2[4]+(100.*f3[2]-100.*f3[4])*f2[3])*f4[3]+((50.*f4[2]-50.*f4[4])*f2[2]+(50.*f4[2]-50.0*f4[4])*f2[4]+(-100.0*f4[2]+100.*f4[4])*f2[3])*f3[3]

(3)

for k from 2 to m do eq3[k_] end do

-0.123456790123457e-1*f4[3]^3*f3[3]+((1/27)*(.5*f2[2]-.5*f2[4])*f3[3]-0.277777777777778e-1*f2[3]*f3[2]+0.277777777777778e-1*f2[3]*f3[4])*f4[3]^2+(-.222222222222222*f3[3]^2+((1/9)*f3[2]+(1/9)*f3[4]+4.*f2[3]^2+(-4.*f2[2]-4.*f2[4])*f2[3]+f2[2]^2+2.*f2[4]*f2[2]+f2[4]^2)*f3[3]+0.277777777777778e-1*f3[2]^2-0.555555555555556e-1*f3[2]*f3[4]+0.277777777777778e-1*f3[4]^2)*f4[3]+(1/9)*((-.25*f4[2]+.25*f4[4])*f3[2]+(.25*f4[2]-.25*f4[4])*f3[4])*f3[3]

(4)

for k from 2 to m do eq4[k_] end do

-0.493827160493827e-1*f4[3]^4-(1/27)*(-f2[4]+f2[2])*f4[3]^3+(.111111111111111*f2[3]*(f4[4]-f4[2])-.888888888888889*f3[3]+16.0*(f2[3]-.500000000000000*f2[2]-.500000000000000*f2[4])^2)*f4[3]^2+.444444444444444*((f3[3]+.250000000000000*f3[2]-.250000000000000*f3[4])*f4[2]+f4[4]*(f3[3]-.250000000000000*f3[2]+.250000000000000*f3[4]))*f4[3]-(1/9)*f3[3]*(f4[4]-f4[2])^2

(5)

``


 

Download fdm-maple.mw

 

hi.

how i can dsolve this differential equations?

thanks

ich.mw
 

restart; Digits := 50; dsol1 := dsolve({diff(F(eta), eta, eta, eta)+.5*H(eta)*((diff(F(eta), eta))^2+F(eta)*(diff(F(eta), eta, eta)))/G(eta)^2+2*(diff(G(eta), eta))*(diff(F(eta), eta, eta))/G(eta)-(diff(H(eta), eta))*(diff(F(eta), eta, eta))/H(eta) = 0, diff(G(eta), eta, eta)+H(eta)*((diff(F(eta), eta))*G(eta)+.5*F(eta)*(diff(eta, eta)))/G(eta)^2+2*(diff(G(eta), eta))^2/G(eta)-((diff(H(eta), eta))*(diff(H(eta), eta)))/H(eta)+(diff(F(eta), eta, eta))^2-(H(eta)/G(eta))^2 = 0, diff(H(eta), eta, eta)+(.5*1.3)*H(eta)*(5*(diff(F(eta), eta))*H(eta)+F(eta)*(diff(H(eta), eta)))/G(eta)^2+2*(diff(G(eta), eta))*(diff(H(eta), eta))/G(eta)-(diff(H(eta), eta))^2/H(eta)+(1.3*1.44)*H(eta)*(diff(F(eta), eta, eta))/G(eta)-(1.3*1.92)*(H(eta)/G(eta))^3 = 0, F(0) = 0, G(0) = 0, H(0) = 0, (D(F))(0) = 1, (D(F))(1) = 0, (D(G))(0) = 0, (D(H))(0) = 0}, 'maxmesh' = 900, numeric, output = listprocedure, method = bvp[middefer], abserr = 0.1e-2); fy := eval(F(eta), dsol1)

Error, invalid input: eval received dsol1, which is not valid for its 2nd argument, eqns

 

 

NULL


 

Download ich.mw

 

In the test package below I want the the Value of Clr to be "B" when the package is loaded. How do I achieve that?
 

restart

mylibdir := cat(kernelopts(homedir), kernelopts(dirsep), "maple", kernelopts(dirsep), "toolbox", kernelopts(dirsep), "personal", kernelopts(dirsep), "lib")

"C:\Users\Ronan.Ronan-PC\maple\toolbox\personal\lib"

(1)

``

libname := mylibdir, libname

"C:\Users\Ronan.Ronan-PC\maple\toolbox\personal\lib", "C:\Program Files\Maple 2016\lib", "C:\Users\Ronan.Ronan-PC\maple\toolbox\personal\lib"

(2)

``

``

``

"RTest:=module()  option package;  export  Sums;    global Clr;  Clr:=B;  Sums:=proc(p1,p2,c:=Clr)  if c=B or c=b  then    p1+p2  elif c=G or c=g then  p1-p2  elif c=R or c=r  then      if p2 =0 then "Not Valid"       else ( p1)/(p2)        end if  else "Error"  end if  end proc;    end module; "

_m655171968

(3)

``

savelib('RTest')

restart

``

with(RTest)

[Sums]

(4)

"At This point I want the value of Clr to be automaticially   B"

Sums(2, 3)

"Error"

(5)

RTest:-Sums(2, 3, b)

5

(6)

``

RTest:-Sums(2, 3, r)

2/3

(7)

RTest:-Sums(2, 3, g)

-1

(8)

Clr := g

g

(9)

RTest:-Sums(2, 3)

-1

(10)

RTest:-Sums(2, 3, b)

5

(11)

RTest:-Sums(2, 3)

-1

(12)

RTest:-Sums(2, 0, r)

"Not Valid"

(13)

``


 

Download Module_Test_global_Var.mw

HI everyone,

As can be seen from the attached file, the first three equations of Eq. (5) will render some of the other equations (and other terms) redundant. How can I obtain a simplified system automatically?

Thanks.

Pdesample.mw

Hello

I know there are other methods to solve this classic probem, but I wanted to cast it as a linear program. I wonder if an expert can look at my IP coding and please correct my error.

TSP_IP.mw

sorry its a mishmash of 1D and 2D inputs

Sorry if this question is trivial, but my search for an answer was not a success.

I get diagnostics [Length of output exceeds limit of 1000000]. This is what is expected, and I expect to get this diagnostics quite often. I know how I can increase the limit (Tools->Options->Precision), but I wonder how I can compute the length of the output using some Maple function, so that I don't need to determine the required limit by the try-and-error method each time or keep the limit very high and get output that I don't need (I know how to switch off output of an operator completely).

Thank you

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