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

I am creating an user interface with embedded components (Labels, Buttons, plotwindow ...).

All works well.

Now I need to include a combobox selection in that code. I can't get any further with that.

The specific question is what to type here (this is "edit selected code", in the combobox options)

use DocumentTools in 

end use; 

to get the choosen value as a variable in the rest of the code (startup code and other code in buttons with edit click code).

A simple example explaining that would be very helpfull:)

When I try to solve the determinant the system hangs and Maple doesn't give any result. Here $\lambda_1,2$ is not a function of 'x' and 't'.

>  >  >  >  >  >  (1) >  >  (2) >  >  >  >  >  >  >  >

Download determinant.mw

Hi everyone, Is it possible to extract data by using Maple from any figure published by any author in different papers and then ussed that to reprouced same graph in Maple?

Question about using the command „ExtremPoints“

I am using Maple 2019.

Using the command ExtremPoints I got different list when defining the function over a closed intervall piecewise or with f(x), x=a..b.

Maple desciption states:
ExtremePoints(f(x), x = a..b) command returns all extreme points of f(x) in the interval [a,b] as a list of values.

An extreme point is defined as any point which is a local minimum or maximum, which includes any finite end points.

So I expected, Maple returns the same list, independend of how the same function is defined (see example below). Instead: with the piecewise definition Maple returns as extrempoints only the local extrempoints without the finite endpoints.
Defining the same function with f(x),x=a..b Maple returns the list with local minimum or maximum, which includes any finite end points.

Example:

f := x -> piecewise(-1 <= x and x <= 2, x^2, undefined);

Return:  ExtremePoints(f(x));

                              [0] 

g := x -> x^2

ExtremePoints(g(x), x = -1 .. 2);

                           [-1, 0, 2]

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.):
M1:=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):

In a dataframe you can easily replace 'undefined' by another value using FillMissing. However, if you want to remove the row in which a cell contained 'undefined' this seems not to be possible. 'DropMissing' will remove the complete column in which one of the cells contains 'undefined'. Is there another command that removes the row of the dataframe as a column contains an 'undefined'? Sorry for the simplistic question but I am still in my learning curve. 

Hello. I am trying to solve the following polynomial system. Maple solution is empty but when I add the polynomial X4_4-2^(1/2)/10 to the list then maple gives me a set of solutions. What could I be doing wrong?

with(SolveTools):
F:=[
10*X4_4 + Y1_1,
10*X4_4*Y1_1 + 2,
20*X4_4*Y3_1,
2*Y3_1,
10*X4_4*Y1_2,
20*X4_4*Y3_2 + 2,
10*X4_4 + 2*Y3_2,
4*L2 - 7*L1 + 5*L3 - 6*L4 + 5*L5 + 9*L6 - 6*L7 - 1,
L1 + 10*L2*X4_4,
10*L5*X4_4,
2*L4 + 20*L3*X4_4,
2*L7 + 20*L6*X4_4
]:
V:=[X4_4, Y1_1, Y1_2, Y3_1, Y3_2, L1, L2, L3, L4, L5, L6, L7]:

Sols := PolynomialSystem(F, V);

Dear all

I have a system that I want to solve it, 
But I get 

''too many levels of recursion''

code_solve_system_of_equations.mw

Thank you for your help 

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)]);
 

I have written a program that checks many possible cases of something. In a line of the code I have programmed  that it prints the number of cases checked every 100 cases checked. However when the number is 200, 300, 400, etc, the number is printed in a new line, and I would like just to replace the precceeding number with the new one. Is it possible? How?

*** Edited to make more clear ***

The function algcurves.puiseux creates generator series for an algebraic function at a point.  The code below generates a few terms of each of the five generator series for the 1,2,3,4 and 5-cycle branch of  w(z) at the origin written implicitly as f in the code below.  Consider the 3-cycle branch:  I extract the 3-cycle generator and plot the imaginary sheet.  Note that's only one surface of a 3-valued function..  In order to plot the other two surfaces the generator series has to be conjugated.   The documentation doesn't appear to have a option to do this.   

If there's no procedure for doing this already and some are interested in this, we need to conjugate each individual monomial separately and  multiply each power of z expressed in terms of p/3 by the root of unity raised to the power p.   Not hard to do manually  for a few terms (see updated notebook below)  but I would like to generates several hundred terms.  Note in my updated notebook below how each monomial is multiplied by (exp(2k Pi i/3))^p (for k=1,2) where p is the numerator of the exponent in this case over 3 for the associated power of z.  We can graphically check if this is correct by stitching the sheets together as a Riemann surface via plots:-display(convert({sheet1Plot, sheet2Plot, sheet3Plot}, list)) which shows a 3-valued analytically-continuous surface wraping  around the z-plane three times.   I'll try to write a procedure but I'm brand new to Maple so might take a while.  Maybe call it as

                         conjugatedSeries=conjugatePuiseux(generator,k) 

for sheet k.

Download puiseuxPlotsVer2.mw 

I have tried to translate the Mathematica-Code of OEIS (A219954):

with MmaTranslator. In the translator occurs an error message with the IF-Statement ...

I will be happy to have this in Maple. I am interested in the digitCount in the above Code.

Sequence is:

Auto_regne_dokument.mw

My maple froze while it was running. I saved and closed it and now it gives me this message when i try to open it "There was a problem in the loading process, you worksheet may be incomplete.". 

There is only one backup file and it is corrupted as well. I tried to see if i could work it out in the text file, but im not very good at doing this.

If anyone knows how to uncorrupt it pls help me