Maple Questions and Posts

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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

I'm trying to generate 100 equation to solve them numerically , buy maple won't let me , here's the code :

for i by 0 to 99 do
a := a[i]+a[i+1] = i^2
end do ;

Parametric equation of a circle in 3d by three points. Draghilev method.

CIRCLE_3_POINTS_geom3d_2.mw

I bought a new macbook pro and want to move my maple2015 software and licence to this new platform?  Simply copy my folder to the new application folder? How do I reinstall the necessary licence ?

Thx for your help

Dear all:

I have used the "diff" command in Maple to help me derive a huge and very long function, and now I want to convert this huge expression from Maple to Matlab format, for example, into a Matlab .m file. The format of this expression in Maple is very different from Matlab.

So could you help me with this problem?

Thank you all.

with(DEtools):
rho := 0.1;
w0 := 2;
sys := {diff(x(t),t) = y(t),diff(y(t),t) = -2*rho*y - w0^2*x};
DEplot(sys, [x(t), y(t)], t = 0 .. 12, [[x(0) = 1, y(0) = 0]]);

i use flow above, would like to plot a circle move from right hand side to left hand side

and see how a circle influence the flow diagram in animation like weather diagram

Hello,

In my worksheet (dedicated to the calculations of mechanism geometry), i'm trying to have a more "POO" approach (Programming Oriented Object).

In this sense, I begin to get more familiar with module and package. Attached, you can find a worksheet which gives the idea of the structure of my code. With this code, the module is very useful for me to call some results (attributes in terms of POO).

My way of doing is :
1) at the beginning, i use Maple as a calculator that is to say that at each prompt, i calculate expressions.
2) i try to make modules so as to have a easier access to the data that I calculated.

However, as the definition of the expressions that I calculated are outside my module (for example, in the worksheet attached, i calculate the expression of Gamma and psi before in my worksheet and i retrieve these expressions in the module with "global"), i can not use this module independently of this worksheet and namely in a Excel Addin.

In fact, the module works if the worksheet has been launched one time and consequently the expression of psi and Gamma are in the kernel (or in memory i want to say).

Question:

How can I do to transform my module in a independent and thus portable module ?

One idea will be to copy/ paste all the geometric calculations which have been done line by line inside the module.
However, i have many lines in separated prompts and moreover, it is always convenient to keep the line by line approach in the worksheet to make some tests.

I look forward to hearing your tips.

Thanks a lot for your help

MonPackage_ProcModule_variante4.mw

 


 

Is it possible to plot chart like this in Maple?

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