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I find mapleform software   but I dont know how to use , and does anyone know another method to convert.  I know use 

> with(MmaTranslator);
> MmaToMaple();  

I can  automatically translated .nb files.  but this have  

Error, (in readline) file or directory does not exist
Error, (in readline) file or directory does not exist
Error, (in readline) file or directory does not exist
Error, (in readline) file or directory does not exist

another Error, missing operator or `;`

also I try to use 

> with(MmaTranslator);
> FromMmaNotebook(Mma_notebook_filename, options);

I still dont know how is works?  can you explains for me and show me some example  , here is my example


hi friends

I encountered a problem and I can not draw the plot of this code

> sol := fsolve({diff(S, x) = 0, diff(S, y) = 0}, {x, y});

> with(VectorCalculus);
> with(linalg);
> s1 := evalf(subs(sol, linalg[grad](S, [x, y])));

> with(VectorCalculus);
> with(LinearAlgebra);
> s2 := evalf(subs(sol, linalg[hessian](S, [x, y]))); pmp0 := [x-subs(sol, x), y-subs(sol, y)]; sapprox := s0+evalm(`&*`(`&*`(transpose(pmp0), s2), pmp0));
> with(Statistics);
> with(stats); statevalf[icdf, chisquare[4]](.95);

> with(VectorCalculus);
> with(plottools);
> with(plots);
> with(linalg); ellips := {seq(stats*([statevalf[icdf, chisquare[4]]])(c) = sapprox, c = [.5, .95, .999])};
> plots(ellips(x, y), x = 950 .. 1000, y = 700 .. 750, grid = [50, 50], view = [950 .. 1000, 700 .. 750]);



can you helpe me?Thank you

Dear Friends,

I am solving 6 ODEs using maple15. then i got this error. anyone know abou this? thank you.



restart:with (plots): B:=1:M:=1:Gr:=0.5:Pr:=3:w:=0.02:blt:=5:Bi:=10:


diff(diff(diff(f(eta), eta), eta), eta)-(diff(f(eta), eta))^2+f(eta)*(diff(diff(f(eta), eta), eta))+H(eta)*(F(eta)-(diff(f(eta), eta)))-(diff(f(eta), eta))+.5*theta(eta) = 0



(1+Nr)*(diff(diff(theta(eta), eta), eta))+3*f(eta)*(diff(theta(eta), eta))+(2/3)*H(eta)*(theta1(eta)-theta(eta)) = 0



H(eta)*F(eta)+H(eta)*(diff(G(eta), eta))+G(eta)*(diff(H(eta), eta)) = 0



F(eta)^2+G(eta)*(diff(F(eta), eta))+F(eta)-(diff(f(eta), eta)) = 0



G(eta)*(diff(G(eta), eta))+f(eta)+G(eta) = 0



G(eta)*(diff(theta1(eta), eta))+l*(theta1(eta)-theta(eta)) = 0



f(0) = 0, (D(f))(0) = 1, (D(theta))(0) = -10+10*theta(0), (D(f))(5) = 0, F(5) = 0, G(5) = -f(5), H(eta) = 0.2e-1, theta(5) = 0, theta1(5) = 0



[.5, 1, 1.5, 2]


for k from 1 to 4 do p:=dsolve(eval({Eq1,Eq2,Eq3,Eq4,Eq5,Eq6,bcs},Nr=L[k]),[f(eta),F(eta),G(eta),H(eta),theta(eta),theta1(eta)],numeric,output=listprocedure);end do:

Error, (in dsolve/numeric/bvp) unevaluated names in system not allowed: {Y[9], Y[10]}







I would like to pick out specific christoffel symbols once maple calculates them. Right now I am using CoefficientList, but I would like a more direct way, as I am working with someone using Maple 17, and they are put in a different order in the CoefficientList. Here is the bit of my code I would like help with.

hi friends

After this cods i see very error

 > restart;

read(orbit.sav ): whit(plots):
ax := -G*Mz*x/(x^2+y^2)^(3/2);
ay := -G*Mz*y/(x^2+y^2)^(3/2);
i := 'i'; j := i+1;
for k from 0 to 3 do
x := 7*10^6; Vx := 0;
y := 0; Vy := 9000;
dt := evalf(1/2^k);
for i from 0 to 328 do
X[i] := evalf(x); Y[i] := evalf(y);
for n to 40*2^k do
x := evalf((1/2)*ax*dt^2+Vx*dt+x); y := evalf((1/2)*ay*dt^2+Vy*dt+y);
Vx := evalf(ax*dt+Vx); Vy := evalf(ay*dt+Vy)
if i mod 41= 0 then
dX[k, i] := X[i]-XS[j]; dY[k, i] := Y[i]-YS[j]
p[k] := plot([seq([(X[i]-XS[j])*(1/1000), (Y[i]-YS[j])*(1/1000)], i = 0 .. 328)], color = green) end do;
p1 := display({seq(p[k], k = 0 .. 3)}, thickness = 3)
SI := [seq(41*i, i = 0 .. 8)]
p2 := plot({seq([seq([(1/1000)*dX[k, i], (1/1000)*dY[k, i]], k = 0 .. 1), [0, 0]], i = SI)}, color = black)
display({p1, p2}, scaling = constrained, labels = ['dx', 'dy'])
display({p1, p2}, view = [-.1 .. .5, -.4 .. .2], scaling = constrained, labels = ['dx', 'dy'])

can you help me Please?

Thank you





     Maple is seriously used in my article Approximation of subharmonic functions in the half-plane by the logarithm of the modulus of an analytic function. Math. Notes 78, No 4, 447-455 in two places. The purpose of this post is to present these applications.                                                                                                 First, I needed to prove the elementary inequality (related to the properties of the minimal harmonic majorant of the function 1/Im z in a certain strip)                                                                                                    2R+sqrt(R)-R(R+sqrt(R))y - 1/y   1/4                                                                                                  for    y ≥ 1/(R+sqrt(R)) and  y ≤ 1/R, the parameter R is greater than or equal to 1.   The artless attemt                                                                          
restart; `assuming`([maximize(2*R+sqrt(R)-R*(R+sqrt(R))*y-1/y, y = 1/(R+sqrt(R)) .. 1/R)], [R >= 1])

maximize(2*R+R^(1/2)-R*(R+R^(1/2))*y-1/y, y = 1/(R+R^(1/2)) .. 1/R)


fails. The second (and successful) try consists in the use of optimizers:

F := proc (R) options operator, arrow; evalf(maximize(2*R+sqrt(R)-R*(R+sqrt(R))*y-1/y, y = 1/(R+sqrt(R)) .. 1/R)) end proc:





Optimization:-Minimize('F(R)', {R >= 1})

[.171572875253809986, [R = HFloat(1.0)]]


To be sure ,
DirectSearch:-Search(proc (R) options operator, arrow; F(R) end proc, {R >= 1})

[.171572875745665, Vector(1, {(1) = 1.0000000195752754}, datatype = float[8]), 11]


Because 0.17
"158 < 0.25, the inequality is  proved.   "
Now we establish this  by the use of the derivative. 

solve(diff(2*R+sqrt(R)-R*(R+sqrt(R))*y-1/y, y) = 0, y, explicit)

1/(R^(3/2)+R^2)^(1/2), -1/(R^(3/2)+R^2)^(1/2)


maximize(1/sqrt(R^(3/2)+R^2)-1/(R+sqrt(R)), R = 1 .. infinity, location)

(1/2)*2^(1/2)-1/2, {[{R = 1}, (1/2)*2^(1/2)-1/2]}


minimize(eval(2*R+sqrt(R)-R*(R+sqrt(R))*y-1/y, y = 1/sqrt(R^(3/2)+R^2)), R = 1 .. infinity, location)

3-2*2^(1/2), {[{R = 1}, 3-2*2^(1/2)]}





The second use of Maple was the calculation of the asymptotics of the following integral (This is the double integral of the Laplacian of 1/Im z over the domain {z: |z-iR/2| < R/2} \ {z: |z| ≤ 1}.). That place is the key point of the proof. Its direct calculation in the polar coordinates fails.

`assuming`([(int(int(2/(r^2*sin(phi)^3), r = 1 .. R*sin(phi)), phi = arcsin(1/R) .. Pi-arcsin(1/R)))/(2*Pi)], [R >= 1])

(1/2)*(int(int(2/(r^2*sin(phi)^3), r = 1 .. R*sin(phi)), phi = arcsin(1/R) .. Pi-arcsin(1/R)))/Pi


In order to overcome the difficulty, we find the inner integral

`assuming`([(int(2/(r^2*sin(phi)^3), r = 1 .. R*sin(phi)))/(2*Pi)], [R*sin(phi) >= 1])



and then we find the outer integral. Because
`assuming`([int((R*sin(phi)-1)/(sin(phi)^4*R*Pi), phi = arcsin(1/R) .. Pi-arcsin(1/R))], [R >= 1])

int((R*sin(phi)-1)/(sin(phi)^4*R*Pi), phi = arcsin(1/R) .. Pi-arcsin(1/R))


is not successful, we find the indefinite integral  

J := int((R*sin(phi)-1)/(sin(phi)^4*R*Pi), phi)



We verify that  the domain of the antiderivative includes the range of the integration.
plot(-cos(phi)/sin(phi)^2+ln(csc(phi)-cot(phi)), phi = 0 .. Pi)


plot((2/3)*cos(phi)/sin(phi)^3+(4/3)*cos(phi)/sin(phi), phi = 0 .. Pi)


    That's all right. By the Newton-Leibnitz formula,

eval(J, phi = Pi-arcsin(1/R))-(eval(J, phi = arcsin(1/R)));



Finally, the*asymptotics*is found by

asympt(eval(J, phi = Pi-arcsin(1/R))-(eval(J, phi = arcsin(1/R))), R, 3)



      It should be noted that a somewhat different expression is written in the article. My inaccuracy, as far as I remember it, consisted in the integration over the whole disk {z: |z-iR/2| < R/2} instead of {z: |z-iR/2| < R/2} \ {z: |z| ≤ 1}. Because only the form of the asymptotics const*R^2 + remainder is used in the article, the exact value of this non-zero constant is of no importance.

       It would be nice if somebody else presents similar examples here or elsewhere.



hi friends

i have a problem in maple with an error

local i,j,n;
global C1,C2,C3,V1,V2,V3;
print("Order of the variables:");
for i from 2 to 2*n +1 do
for j from 1 to n do
select(has,numpr,diff(coor[j](t),t)) then
print(coor[j],C[j]," ",diff(coor[j](t),t),V[j]);
end :

IniC:=z(0)=0.75,D(z)(0)=0, x(0)=1,D(x)(0)=0,y(0)=0,D(y)(0)=1:



for i from 0 to 1000 do ;

X[i]:=rhs(NsT[C1]); Vx[i]:=rhs(NsT[V1]);
Y[i]:=rhs(NsT[C2]); Vy[i]:=rhs(NsT[V2]);
Z[i]:=rhs(NsT[C3]); Vz[i]:=rhs(NsT[V3]);

KepVec[i]:=convert(crossprod([X[i],Y[i], Z[i]],[Vx[i],Vy[i], Vz[i]]),list);

but i see this error and I can't draw PLOT:

Error, invalid input: rhs received Ns(0)[C1], which is not valid for its 1st argument, expr

this cods is for draw plot:

spacecurve({[seq([X[i], Y[i], Z[i]], i = 0 .. 1000)], [[-1/2, 0, 0], [1/2, 0, 0]]}, labels = ['x', 'y', 'z']);

spacecurve([seq(KepVec[i], i = 0 .. 1000)], orientation = [0, 90], labels = ['x', 'y', 'z'])

plot([seq([(1/25)*i, KepAbs[i]], i = 0 .. 1000)], labels = ['t', 'MofI'])

can you helpe me?Thank you

I’ve been having some issues working with large datasets / matrixes in maple 17.02 and 2015. My data consists of a 10^7 x 14 csv file with several lines of header information. Attached is a small sample. The ImportData assistant hangs while importing said file. The javaw process stops responding for a period of time then stops consuming cpu time. I’ve have successfully imported a file of the same format but reduced in size (10^6 x 14) with this same function. So I don’t believe it’s a formatting issue but rather its size.

Are there size limitations to the ImportData function?

The attached maple file has a test case in which the data set (sans header info) is created and exported as a csv file. The export time took longer than I expected (~2 hrs). I then attempted to import the file using two different functions. The ImportMatrix function successfully imported the test case file in approximately 20 minutes, however the ImportData functions seems to fail in the same way as it does importing my actual dataset. I haven’t successfully used the ImportMatrix function on my actual dataset; I’m assuming the header information is the source of the problem.

Are there other methods to import this data?

As stated above, I’m tried both maple 17 and 2015 both 64 bit versions running on an Intel i7 M620 @ 2.67Ghz, 8 GB ram (~ 6 GB avail), sata 2 ssd.

Thank you,

Ron  Sample.txt



is there a way I can use data (variables) from Maple environment in the Maplesim environment. 

I have a scirpt in maple that generates the robots joints angles and need to use them in the 3D robot built in maplesim. I know I can export/Import data, but this sounds redundant. Is there a way to simply use an input block as a source of the data in maplesim and have the variable name generated in maple used int. Similar to what Matlab/Simulink does.. 




I'm attempting to plot the Nyquist plot for a complex system such that Maple cannot determine the frequency limits automatically, and suggested I use 'range' to specify the frequency limits. How do you do this? For example if you have a system G=1/(s+1), how do you plot the Nyquist plot in the frequency range 0 to 1rad/s?  Thanks

hi friends

i have a problem in maple with an error


for i from 0 to 1000 do;
X[i]:=rhs(NsT[C1]); Vx[i]:=rhs(NsT[V1]);
Y[i]:=rhs(NsT[C2]); Vy[i]:=rhs(NsT[V2]);
Z[i]:=rhs(NsT[C3]); Vz[i]:=rhs(NsT[V3]);
KepVec[i]:=convert(crossprod([X[i],Y[i], Z[i]],[Vx[i],Vy[i], Vz[i]]),list);


but i see:

Error, invalid input: seq expects 2 arguments, but received 4

can you helpe me?


Maplesim 7.0(1) installer looks for Maple 18 installation.  Will Maplesim work with Maple 2015?

If not, what is the expected date of making both products compatible?



Here is a sample code:


N := 3;

for i to 2 do

   f[i] := unapply(a[0][i]+Sum(a[j][i]*t^j, j = 1 .. N), t) end


sys : = {diff(x(t),t) = y(t), diff(y(t),t) = x(t) }

syssub := subs([x(t) = f[1](t), y(t) = f[2](t)],sys)


   solve(simplify(syssub) union {simplify(f[1](0))=2, simplify(f[2](0))=1}),
     {a[0][1], a[0][2], a[1][1], a[1][2], a[2][1], a[2][2]}


Now for some reason solve returns nothing. This doesn't make sense to me since the system is determinable.

Here is a proof,

 ans: = dsolve(sys union {x(0) = 2, y(0) = 1},{x(t), y(t)}, type = series)


will actually return the result.


The reason I am doing this is because I am currently working on a bigger nonlinear DE system and it couldn't be done with just dsolve and since there isn't a package in Maple that makes a series representation and plug and chug, I have to write out the necessary steps.


Something tells me that Maple has a lot of trouble getting around the variable "t". But it still makes no sense why it can't even return a[0][1] = 1,a[0][2]=2

hi friends

i have a problem in maple with an error

> Us:=subs(G=1,m=1,L=1,U):
> D2r:=[diff(x(t),t,t),diff(y(t),t,t),diff(z(t),t,t)]:
> g:=subs(x=x(t),y=y(t),z=z(t),grad(Us,[x,y,z])):
> IniC:=x(0)=1,D(x)(0)=0,y(0)=0,D(y)(0)=1,z(0)=3/4,D(z)(0)=0:
> Ns:=dsolve({seq(D2r[i]=g[i],i=1..3),IniC},{x(t),y(t),z(t)},numeric);

after this i see :

Error, (in f) unable to store 'grad(U, [1., 0., .750000000000000000])[1]' when datatype=float[8] 

meaning it can only store 8 byte hardware floats.
Is this statement true? And how can i solve this problem? Or could/should i use a different data type?

Thanks for your time

Best regards

Could anyone please hepl me? I have the following system

e1 := exp(F(r)/phi_0)*L*A(r) = (1/2)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))/(B(r)^2*A(r)*r*C(r));
e2 := alpha*(diff(F(r), r, r))+(alpha^2+omega)*(diff(F(r), r))^2+(1/4)*(4*(diff(C(r), r, r))*B(r)*A(r)^2*C(r)*r+2*(diff(A(r), r, r))*A(r)*B(r)*r*C(r)^2-2*B(r)*A(r)^2*(diff(C(r), r))^2*r-(diff(A(r), r))^2*B(r)*r*C(r)^2-2*A(r)^2*C(r)*(diff(C(r), r))*(diff(B(r), r))*r-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)^2+8*B(r)*A(r)^2*C(r)*(diff(C(r), r))-4*A(r)^2*C(r)^2*(diff(B(r), r)))/(r*A(r)^2*B(r)*C(r)^2)-(1/4)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))/(B(r)*A(r)^2*r*C(r)) = 0;
e3 := (1/4)*(-2*(diff(C(r), r, r))*B(r)*A(r)*r^2-B(r)*(diff(A(r), r))*(diff(C(r), r))*r^2+A(r)*(diff(C(r), r))*(diff(B(r), r))*r^2-8*B(r)*A(r)*(diff(C(r), r))*r-2*B(r)*(diff(A(r), r))*C(r)*r+2*A(r)*C(r)*(diff(B(r), r))*r+4*B(r)^2*A(r)-4*B(r)*A(r)*C(r))/(B(r)^2*A(r)) = -(1/4)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))*r/(B(r)^2*A(r)^2);
e4 := -(alpha^2+2*omega)*(diff(F(r), r))*(-(1/2)*(-(diff(A(r), r))*B(r)*r^4*C(r)^2-A(r)*(diff(B(r), r))*r^4*C(r)^2-4*A(r)*B(r)*r^3*C(r)^2-2*A(r)*B(r)*r^4*C(r)*(diff(C(r), r)))/(A(r)*B(r)*r^4*C(r)^2)-(diff(B(r), r))/B(r)+(diff(F(r), r, r))/(diff(F(r), r))+alpha*(diff(F(r), r)))/B(r) = -exp(F(r)/phi_0)*V_0*(alpha-1/phi_0);

phi_0 := -alpha/(2*alpha^2+2*omega); L := V_0*(1-(alpha-1/phi_0)*alpha/(3*alpha^2+2*omega)); V_0 := -lambda*exp(-fc/phi_0); fc := ln((4*alpha^2+2*omega)/(G_0*(3*alpha^2+2*omega)))/alpha; m := (2/(1+g))^(1/2); n := g*(2/(1+g))^(1/2); P := (G_0*(3*alpha^2+2*omega)/(4*alpha^2+2*omega))^(-2*alpha/(n-m)); eta := 1.4*G_0*Ms*(2/(1+g))^(-1/2)/c^2; g := 1-alpha^2/(2*alpha^2+omega);

omega := -10^5; alpha := 1; G_0 := 6.67*10^(-11); lambda := 10^(-52); c := 2.9*10^8; Ms := 1.9*10^30;
ri := evalf(1000*eta);

ics := A(2.109660445*10^6) = 1, (D(A))(2.109660445*10^6) = 2.370091128*10^(-15)*sqrt(2)*sqrt(99998)*sqrt(199997), B(2.109660445*10^6) = 1, C(2.109660445*10^6) = 1, (D(C))(2.109660445*10^6) = 4.740182256*10^(-15)*(1-(99999/19999300006)*sqrt(2)*sqrt(99998)*sqrt(199997))*(1-1.000017501*10^(-8)*sqrt(2)*sqrt(99998)*sqrt(199997))^(-(99999/19999300006)*sqrt(2)*sqrt(99998)*sqrt(199997))*sqrt(2)*sqrt(99998)*sqrt(199997), f(2.109660445*10^6) = 23.43081116, (D(f))(2.109660445*10^6) = 4.749681180*10^(-15):

eta:=2109.660445: sys:=e1,e2,e3,e4; vars:=[A(r),B(r),C(r),F(r)];

dsn3 := dsolve([sys, ics], numeric, vars, range = 3*eta .. 50*eta);

Results in

Warning, cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up

Setting f(r)=Const,V_0=0 which is a physically relevant case, results in

Error, (in simplify/normal) numeric exception: division by zero

I suugest the problem is that the equation contain sqared derivatives, hence there are several solution branches corresponding to different signs of square root. Maple chooses the singular branch. How can I force it to choose another branch or calculete all of them?

Thanks in advance..

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