`c₁₁`, `c₁₂`, `c₁₃`, `e₃₁`, `c₆₆`, `c₄₄`, `e₁₅`, rho, `ϵ₁₁`, `ϵ₃₃` = constants;
`U₁` := unapply(`U₁`(t, x, y, z), x, y, z, t);
`U₂` := unapply(`U₂`(t, x, y, z), x, y, x, t);
`U₃` := unapply(`U₃`(t, x, y, z), x, y, z, t);
phi := unapply(phi(t, x, y, z), x, y, z, t);

PDE1 := `c₁₁`*Diff(`U₁`, y, y) + `c₁₂`*Diff(`U₂`, x, y) + `c₁₃`*Diff(`U₃`, x, z) + `e₃₁`*Diff(phi, x, z) + `c₆₆`*Diff(`U₂`, x, y) + `c₆₆`*Diff(`U₁`, y, y) + `c₄₄`*Diff(`U₃`, x, z) + `c₄₄`*Diff(`U₁`, z, z) + `e₁₅`*Diff(phi, x, y) = rho*Diff(`U₁`, t, t);

PDE2 := `c₆₆`*Diff(`U₂`, x, y) + `c₆₆`*Diff(`U₁`, y, x) + `c₁₂`*Diff(`U₁`, x, y) + `c₁₁`*Diff(`U₂`, y, y) + `c₁₃`*Diff(`U₃`, y, z) + `e₃₁`*Diff(phi, z, y) + `c₄₄`*Diff(`U₃`, y, z) + `c₄₄`*Diff(`U₂`, y, z) + `e₁₅`*Diff(phi, y, z) = rho*Diff(`U₂`, t, t);

PDE3 := `c₄₄`*Diff(`U₃`, x, x) + `c₄₄`*Diff(`U₁`, z, x) + `e₁₅`*Diff(phi, y, x) + `c₄₄`*Diff(`U₃`, y, y) + `c₄₄`*Diff(`U₂`, z, y) + `e₁₅`*Diff(phi, y, y) + `c₁₃`*Diff(`U₁`, x, z) + `c₁₃`*Diff(`U₂`, y, z) + `c₃₃`*Diff(`U₃`, z, z) + `e₃₃`*Diff(phi, z, z) = rho*Diff(`U₃`, t, t);

PDE4 := `e₁₅`*Diff(`U₃`, x, x) + `e₁₅`*Diff(`U₁`, y, x) - `ϵ₁₁`*Diff(phi, x, x) + `e₁₅`*Diff(`U₃`, yx y) + `e₁₅`*Diff(`U₂`, z, y) - `ϵ₁₁`*Diff(phi, y, y) + `e₃₁`*Diff(`U₁`, y, z) + `e₃₁`*Diff(`U₂`, y, z) + `e₃₃`*Diff(`U₃`, x, z) - `ϵ₃₃`*Diff(phi, z, z) = 0;

pds := [PDE1, PDE2, PDE3, PDE4];
sol := pdsolve(pds);

i was solving the above set of pde. but it was showing the following errors,

Error, (in U₁) too many levels of recursion
Error, (in unapply) variables must be unique and of type name
Error, (in U₃) too many levels of recursion
Error, (in phi) too many levels of recursion
Error, (in pdsolve/sys/info) required an indication of the solving variables for the given system

can anyone help me?

In help(EllipticF): Why is the parameter k not called “the modulus of the elliptic function” as in the definition of the inverse Jacobi functions in help(InverseJacobiPQ)?

Instead, it is called “the parameter” which can be confused with the parameter m=k^2 used in other notations (which is refered to "a parameter m" in the EllipticF help page).

Is there a reason for this, or should the parameter definitions of the first, second and thrid elliptic integrals not be aligned with the parameter definitions of the Jacobi functions and their inverses?

DLMF for example defines k as modulus for both, the Elliptic Integrals and the Jacobian Elliptic Functions.

A user who wants to transfer an expression from a different notation to Maple might misinterpret parameters.

The interrupt button is great for stopping long calculations. But if there is an accidental calculation of a long list that has a mistype for example and the computer is busy calculating behind the scenes getting ready to output to the screen (evaluating icon has a heartbeat) there is no interrupt for that.  You either wait for the output or kill maple and start again (you may be able to save the worksheet before you close it - that might be an option).

I am reading (and enjoying) this:

"On the computation of the nth decimal digit of various transcendental numbers" by

Simon Plouffe, 1996 (2009)

On the third page Simon calculates this:

a := 1/binomial(100, 50);
                                 1               
              a := ------------------------------
                   100891344545564193334812497256

ifactor(denom(a));
                                                           
(8)  (81)  (11) (13) (17) (19) (29) (31) (53) (59) (61) (67) (71) (73) (79) (83) (89) (97)

and builds a sum of fractions where the above primefactors are the denominators in this sum.

5/8 + 20/81 + 10/11 + 2/13 + 13/17 + 10/19 + 4/29 + 5/31 + 23/53 + 41/59 + 29/61 + 37/67 + 33/71 +19/73 + 36/79 + 7/83 + 13/89 + 88/97

Solving a diophantine equation and using continued fractions are the steps to get that fractions (see link above).

May someone explain these steps to me ?

Thanks

with(GraphTheory):
s:=DrawGraph(CompleteGraph(5),size=[250,250])
Export("D:\\s1.jpg",s)

But I would like to export it using PDF format. However, the modified code below seems to be quite unsuccessful. I am aware that Maple has export options in the front end, but I still prefer to use code for this purpose.

Export("D:\\s1.pdf",s)

Error, (in Export) invalid input: member received _ImportExport:-InfoTable["PDF"][4], which is not valid for its 2nd argument, s

There are two reasons for this.

• One is for batch conversion;
•  the second is the previously mentioned issue (see https://www.mapleprimes.com/questions/236142-How-To-Remove-The-Mosaic-Of-Vertices) that has not been resolved. I'm thinking that perhaps converting Tabulate to PDFs using codes  might remove the mosaic issue.

with(GraphTheory):
Graphs:=[NonIsomorphicGraphs(6,8,output=graphs,outputform = graph)]:
num_g:=nops(Graphs):
num:=ceil((num_g)/5.):
Matrix (num,5,(i,j)->`if`((i-1)*5+j<=num_g, DrawGraph(Graphs[(i-1)*5+j],size=[250,250] ,overrideoptions ,showlabels=false,style=planar, stylesheet =  [
 vertexcolor     = orange
,vertexfontcolor = black
,vertexborder    = false
,edgethickness   = 0.6
,edgecolor       = MidnightBlue
,vertexshape     =  "circle"
,vertexfont      = [Arial, 4],
vertexthickness=5], caption = cat(H__,5*(i-1)+j),captionfont=["ROMAN",7]),plot(x = 0 .. 1, axes = none))):
DocumentTools:-Tabulate (M1[1..5,.. ],widthmode=percentage ,width=80 , exterior =all):

Since strings are not mutable objects in Maple, the package provides two procedures, StringTools:-OldStringBuffer and StringTools:-StringBuffer, which appear heavily correlated with Java's  and . 

The help page of StringBuffer claims that use of a is much more efficient than the naive approach: 

(*
`G` and `F` are taken from the link above.
*)
G := proc()
   description "extremely inefficient string concatenator";
   local   r;
   r := proc()
       if nargs = 0 then
           ""
       elif nargs = 1 then
           args[ 1 ]
       else
           cat( args[ 1 ], procname( args[ 2 .. -1 ] ) )
       end if
   end proc;
   r( args )
end proc:
# # This can be transformed into an O(1) algorithm by passing a string buffer to the recursive calls.
F := proc()
   description "efficient version of G";
   local    b, r;
   b := StringTools:-StringBuffer();
   r := proc()
       if nargs = 1 then
           b:-append( args[ 1 ] )
       else
           b:-append( args[ 1 ] );
           procname( args[ 2 .. -1 ] )
       end if
   end proc;
   r( args ):-value()
end proc:
s := 'StringTools:-Random(10, print)' $ 1e4:
NULL;
time(G(s));
                             5.375

time(F(s));
                             1.125

But why not use the built-in cat directly? 

time(cat(s));
                               0.

time(StringTools:-Join([s], ""));
                               0.

Clearly, this is even more efficient. 

Here is the last example in that link. 

FilterFile := proc( fname::string, filter )
   local   b, line;
   b := StringTools:-StringBuffer();
   do
       line := readline( fname );
       if line = 0 then break end if;
       b:-append( filter( line ) )
   end do;
   b:-value()
end proc: # verbatim 
filename__0 := FileTools:-JoinPath(["example", "odyssey.txt"], 'base' = 'datadir'):
filename__1 := URL:-Download("https://gutenberg.org/ebooks\
/2600.txt.utf-8", "War-and-Peace.txt"):

fclose(filename__0):
    time[real]((rawRes0 := FilterFile(filename__0, StringTools:-Unique)));
                             0.223

fclose(filename__1):
    time[real]((rawRes1 := FilterFile(filename__1, StringTools:-Unique)));
                             1.097

Nevertheless, 

close(filename__0):
use StringTools, FileTools:-Text in
	time[real]((newRes0 := String(Support~(fscanf(filename__0, Repeat("%[^\n]%*c", CountLines(filename__0))))[])))
end;
                             0.118

close(filename__1):
use StringTools, FileTools:-Text in
	time[real]((newRes1 := String(Support~(fscanf(filename__1, Repeat("%[^\n]%*c", CountLines(filename__1))))[])))
end;
                             0.580

evalb(newRes0 = rawRes0 and newRes1 = rawRes1);
                              true

As you can see, these experiments just tell an opposite story. Isn't the so-called "StringBuffer" obsolete today? 

restart:
Digits:=30;

h0:=0.156;
d:=0.32*h0;
l:=1;
h1:=h0-d;
h2:=h0+d;
h3:=0.5*h0;
g:=9.8;
d1:=1;
Term:=5;
Num:=150:
n:=1:
l1:=l/n;
p:=2;

for N from 1 to Num do
lambda:=2*n*Pi/l:## N1 wei sha ba tiao shu 
k0:=evalf(0.5*Pi+2*(N-1)*Pi/(Num-1)):
tau0:=evalf(k0*h0):
omega:=evalf((g*k0*tanh(k0*h0))^(1/2)):
E:=g/(omega)^2:
k1:=abs(fsolve(omega^2=g*k*tanh(k*h1),k)):
tau1:=evalf(k1*h1):
k2:=abs(fsolve(omega^2=g*k*tanh(k*h2),k)):
tau2:=evalf(k2*h2):
k3:=abs(fsolve(omega^2=g*k*tanh(k*h3),k)):
tau3:=evalf(k3*h3):

F:=tau->tanh(tau1)+tau*add((eval(diff(tanh(tau),tau$s),tau=tau1))/s!*(tau-tau1)^(s-1),s=1..10);##F1(K)
T:=tau->tanh(tau2)+tau*add((eval(diff(tanh(tau),tau$s),tau=tau2))/s!*(tau-tau2)^(s-1),s=1..10);##F2(K)

P:=tau->(sinh(2*tau)-2*tau*cosh(2*tau))/(2*tau+sinh(2*tau))^2;##P
Q:=tau->(16*tau^4+32*tau^3*sinh(2*tau)-9*sinh(2*tau)*sinh(4*tau)+12*tau*(tau+sinh(2*tau))*((cosh(2*tau))^2-2*cosh(2*tau)+3))/3/(2*tau+sinh(2*tau))^4; ##Q
A:=unapply(taylor(2*(tau-tau1)/sinh(2*tau)-tanh(tau)*(h0-E*tau*tanh(tau))*(2*tau+sinh(2*tau))/((E*tau*tanh(tau)-h2)*F(tau)*sinh(2*tau)),tau=tau1,Term+1),tau);##A(K)
B:=unapply(taylor((1+2*tau/sinh(2*tau))^2*(-(tau-tau1)/lambda^2/E/(E*tau*tanh(tau)-h2)/F(tau)-(h0-E*tau*tanh(tau))*(tau-tau1)*tanh(tau)/(E*tau*tanh(tau)-h2)/F(tau)*P(tau)+(tau-tau1)^2*Q(tau)),tau=tau1,Term+1),tau);##B(K)

for j from 0 to Term do
a[j]:=coeff(A(tau),tau-tau1,j):
b[j]:=coeff(B(tau),tau-tau1,j):
end do;

for m from 1 to Term do
f1[0]:=1;
f2[0]:=1;
f1[m]:=-(add(f1[m-i]*((m-i)*a[i]+b[i]),i=1..m))/(m*(m-1+1/2));##pm(z1)
f2[m]:=-(add(f2[m-i]*((m-i+1/2)*a[i]+b[i]),i=1..m))/((m+1/2)*(m+1/2-1+1/2));##qm(Z2)
end do;

CC:=unapply(taylor(2*(tau-tau2)/sinh(2*tau)-tanh(tau)*(h0-E*tau*tanh(tau))*(2*tau+sinh(2*tau))/(E*tau*tanh(tau)-h1)/T(tau)/sinh(2*tau),tau=tau2,Term+1),tau);
DD:=unapply(taylor((1+2*tau/sinh(2*tau))^2*(-(tau-tau2)/lambda^2/E/(E*tau*tanh(tau)-h1)/T(tau)-(h0-E*tau*tanh(tau))*(tau-tau2)*tanh(tau)/(E*tau*tanh(tau)-h1)/T(tau)*P(tau)+(tau-tau2)^2*Q(tau)),tau=tau2,Term+1),tau);

j:='j':
for j from 0 to Term do
cc[j]:=coeff(CC(tau),tau-tau2,j):
dd[j]:=coeff(DD(tau),tau-tau2,j):
end do;

i:='i':
f3[0]:=1;
f4[0]:=1;
m:='m':
for m from 1 to Term do
f3[m]:=-(add(f3[m-i]*((m-i)*cc[i]+dd[i]),i=1..m))/(m*(m-1+1/2));
f4[m]:=-(add(f4[m-i]*((m-i+1/2)*cc[i]+dd[i]),i=1..m))/((m+1/2)*(m+1/2-1+1/2));
end do:

xi11:=unapply(add(f1[j1]*(tau-tau1)^(j1),j1=0..m-1),tau);
xi12:=unapply(add(f2[j2]*(tau-tau1)^(j2+1/2),j2=0..m-1),tau);
xi21:=unapply(add(f3[j3]*(tau-tau2)^(j3),j3=0..m-1),tau);
xi22:=unapply(add(f4[j4]*(tau-tau2)^(j4+1/2),j4=0..m-1),tau);

u0:=evalf(g*tanh(tau0)*(1+2*tau0/(sinh(2*tau0)))/(2*k0));
u01:=evalf(g*tanh(tau3)*(1+2*tau3/(sinh(2*tau3)))/(2*k3));
u1:=evalf(g*(1-(tanh(tau0))^2)*(sinh(2*tau0)-2*tau0*cosh(2*tau0))/(4*(2*tau0+sinh(2*tau0))));
H:=evalf((1+2*tau0/sinh(2*tau0))/(-lambda*k0*d));
delta00:=evalf((lambda*d*u1/u0+I*k0)*H);
delta01:=evalf((lambda*d*u1/u0-I*k0)*H);
delta11:=evalf((lambda*d*u1/u0+I*k0)*H*exp(I*k0*l));
delta12:=evalf((lambda*d*u1/u0-I*k0)*H*exp(-I*k0*l));
delta21:=evalf(exp(I*k0*(l1)));
delta22:=evalf(u0*k0*exp(I*k0*(l1)));
delta31:=evalf(exp(-I*k0*(l1)));
delta32:=evalf(-u0*k0*exp(-I*k0*(l1)));
delta41:=evalf(exp(I*k3*(l1)));
delta42:=evalf(u01*k3*exp(I*k3*(l1)));
delta51:=evalf(exp(-I*k3*(l1)));
delta52:=evalf(-u01*k3*exp(-I*k3*(l1)));
delta61:=evalf(exp(I*k3*(l1+d1)));
delta62:=evalf(u01*k3*exp(I*k3*(l1+d1)));
delta71:=evalf(exp(-I*k3*(l1+d1)));
delta72:=evalf(-u01*k3*exp(-I*k3*(l1+d1)));
Y11 := evalf(exp(k0*(l1 + d1)*I));
Y12 := evalf(-exp(-k0*(l1 + d1)*I));
Y21 := evalf(u0*k0*exp(k0*(l1 + d1)*I));
Y22 := evalf(-u0*k0*exp(-k0*(l1 + d1)*I));

G11 := evalf(exp(k0*(2*l1 + d1)*I));
G12 := evalf(-exp(-k0*(2*l1 + d1)*I));
G21 := evalf(u0*k0*exp(k0*(2*l1 + d1)*I));
G22 := evalf(-u0*k0*exp(-k0*(2*l1 + d1)*I));

W11 := evalf(exp(k3*(2*l1 + d1)*I));
W12 := evalf(-exp(-k3*(2*l1 + d1)*I));
W21 := evalf(u01*k3*exp(k3*(2*l1 + d1)*I));
W22 := evalf(-u01*k3*exp(-k3*(2*l1 + d1)*I));

Z11 := evalf(exp(k3*(2*l1 + 2*d1)*I));
Z12 := evalf(-exp(-k3*(2*l1 + 2*d1)*I));
Z21 := evalf(u01*k3*exp(k3*(2*l1 + 2*d1)*I));
Z22 := evalf(-u01*k3*exp(-k3*(2*l1 + 2*d1)*I));

V11:=evalf(exp(I*k0*2*(l1+d1)))；
V21:=evalf(u0*k0*exp(I*k0*2*(l1+d1)))；

delta91:=evalf(exp(I*k0*l));
delta92:=evalf(exp(-I*k0*l));

Hi:=(Matrix([[1,1],[delta00,delta01]]))^(-1);
H0:=Matrix([[1,1],[delta00,delta01]]);
H1:=(Matrix([[delta21,delta31],[delta22,delta32]]))^(-1);
H2:=Matrix([[delta41,delta51],[delta42,delta52]]);
H3:=(Matrix([[delta61,delta71],[delta62,delta72]]))^(-1);
H4 := Matrix([[Y11, Y12], [Y21, Y22]]);
H5 :=( Matrix([[G11, G12], [G21, G22]]))^(-1);
H6 := Matrix([[W11, W12], [W21, W22]]);
H7 := (Matrix([[Z11, Z12], [Z21, Z22]]))^(-1);
H8 := Matrix([[V11], [V21]]);

Ht:=Matrix([[1],[delta11]]);

e1:=Matrix([[evalf(xi11(tau0)),evalf(-xi12(tau0))],[evalf(eval(diff(xi11(tau),tau),tau=tau0)),evalf(eval(diff(-xi12(tau),tau),tau=tau0))]]);

e2:=(Matrix([[evalf(xi11(tau0)),evalf(xi12(tau0))],[evalf(eval(diff(xi11(tau),tau),tau=tau0)),evalf(eval(diff(xi12(tau),tau),tau=tau0))]]))^(-1);

EE:=evalm(e1.e2);

ee1:=Matrix([[evalf(xi21(tau0)),evalf(xi22(tau0))],[evalf(eval(diff(xi21(tau),tau),tau=tau0)),evalf(eval(diff(xi22(tau),tau),tau=tau0))]]);

ee2:=(Matrix([[evalf(xi21(tau0)),evalf(-xi22(tau0))],[evalf(eval(diff(xi21(tau),tau),tau=tau0)),evalf(eval(diff(-xi22(tau),tau),tau=tau0))]]))^(-1);
EEE:=evalm(ee1.ee2);

MM:=evalm(Hi.EE.EEE.H0.H1.H2.H3.H4.H5):
R:=evalf(MM[2,1]/MM[1,1]):
Kr:=abs(evalf(R)):
KR[N]:=abs(Kr);

end do:

N:='N':
seq(KR[N],N=1..Num);
with(plots):
listplot([seq([0.5+2*(N2-1)/(Num-1),KR[N2]],N2=1..Num)]);
 

Dear all
I have a boundary value problem, 
How can I solve the problem using maple or maybe we can introduce serie expansion to solve it or something else.

BVP_frac.mw

Thank you for your help 

I have data that I've binned in list. I'd like to plot in as a histogram. 

dataplot sort of does it but doesn't give the x-axis that I used but just the bins.

Yes, I know about Histogram from Statistics. To plot 1000000 values I have to enter them all into a list which seems crazy when all I want is a 100 bin histogram. Doing the binning is trivial, but I can't figure out how to plot it with a sensible x-axis (show range used to define the 'histogram')

Hello,

I am experiencing difficulties using my old Maple programs with the newer version. I tried changing the types of inputs and the typesetting level, but it just doesn't work. I would appreciate it if someone could help me overcome my ignorance.

Some simple input is attached with the output.

Hi,

I use the RamdonGraph function quite a bit and wanted to know if it is possible to generate random graphs with specific properties, beyond the typical order, size, connectednes, etc. Specifically, I am interesting in generating eulerian (containing an eulerian circuit), semi-eulerian (containing an eulerian trail) and hamiltonian (containing a spanning cycle)  graphs. Also, the abiltiy to randomly generate graphs that have none of these properties would be helpful.

Dear all 

I have a system of  second order difference equation.

How, can I update the iterate solution and solve the system

System_of_equations.mw

Thank you

Maple (2023.1) opens regularly but I cannot use "open" or "save" or "save as" and after opening Maple I no longer can close it.

That is a big problem for me.

The issue is on my new laptop Lenovo L13 Yoga with Windows 11.

Any suggestion? Thanks

hi i have a problem where maple dosent have a varible theta inside cos and sin and when i give it a size it dosent solve the equtation 

The function f := x -> (x + 1)*(x^2 + (m - 5)*x - 7*m + 2) satifies
solve(discrim(f(x), x) = 0, m) has three solutions 1, -17, -1
How to find the integer numbers a, b, c, d, k, t so that the function 
f := x -> x^3 + (a*m + b)*x^2 + (c*m + d)*x + k*m+t  satifies the equation 

solve(discrim(f(x), x) = 0, m)  has three  integer numbers m?

Consider the function f:=x-> a*x^2 + b*m*x + x^3 + c*m.
I tried
restart;
f := x -> x^3 + a*x^2 + b*m*x + c*m;
solve(f(x) = 0, m);
g := x -> -x^2*(a + x)/(b*x + c);
solve(diff(g(x), x) = 0, x);

restart;
n := 0;
f := x -> -x^2*(a + x)/(b*x + c);
for a from -10 to 20 do
    for b to 20 do
for c from -10 to 20 do
mydelta := a^2*b^2 - 10*a*b*c + 9*c^2;
if 0 < mydelta and type(mydelta, integer) then
x1 := (-b*a - 3*c + sqrt(a^2*b^2 - 10*a*b*c + 9*c^2))/(4*b):
x2 := -(b*a + sqrt(a^2*b^2 - 10*a*b*c + 9*c^2) + 3*c)/(4*b):
x3 := 0:
if type(x1, integer) and type(x2, integer) and nops({0, x1, x2}) = 3 and type(f(x1), integer) and type(f(x2), integer) then n := n + 1; L[n] := [a, b, c]; end if; end if; end do; end do; end do;
L := convert(L, list);

I get
L := [[-10, 1, 6], [-10, 2, 12], [-9, 3, 5], [-8, 1, 10], [-5, 1, 3], [-4, 1, -6], [4, 1, 6], [5, 1, -3], [6, 3, 20], [7, 1, 15], [8, 1, -10], [8, 1, 12], [9, 3, -5], [10, 1, -6], [10, 1, 12], [12, 1, 18], [14, 1, -4], [15, 1, -9], [18, 1, 20], [18, 3, -10], [18, 3, 4], [20, 1, 2]]

With L[1], solve(discrim(x^3 - 10*x^2 + m*x + 6*m, x) = 0, m) ;
The equation has three integer solutions: 0, 12, -500