MaplePrimes Questions

Here is a Maple 2020 worksheet that ran fine on Maple 2020, but runs slower on Maple 2022, especially when plots[display] is used it seems to take much longer?

with(NumberTheory);
with(plots);
NULL;
NULL;
theta := [14.134725, 21.022039, 25.010858, 30.424876, 32.935062, 37.586178, 40.918719, 43.327073, 48.00515, 49.773832, 52.970321, 56.446248, 59.347044, 60.831779, 65.112544, 67.079811, 69.546402, 72.067158, 75.704691, 77.144840, 79.337375, 82.91038, 84.735493, 87.425273, 88.809111, 92.491899, 94.651344, 95.870634, 98.831194];
theta := [14.134725, 21.022039, 25.010858, 30.424876, 32.935062, 

  37.586178, 40.918719, 43.327073, 48.00515, 49.773832, 

  52.970321, 56.446248, 59.347044, 60.831779, 65.112544, 

  67.079811, 69.546402, 72.067158, 75.704691, 77.144840, 

  79.337375, 82.91038, 84.735493, 87.425273, 88.809111, 

  92.491899, 94.651344, 95.870634, 98.831194]

y[1] := x -> -2*sqrt(x)*cos(theta[1]*ln(x) - argument(0.5 + theta[1]*I))/(abs(0.5 + theta[1]*I)*ln(x));
y[1] := proc (x) options operator, arrow; -2*sqrt(x)*cos(theta[1\

  ]*ln(x)-argument(.5+I*theta[1]))/(abs(.5+I*theta[1])*ln(x)) 

   end proc

plot(y[1](x), x = 20 .. 100, title = 'Fig1*(S &G theta) = 1/2 + 14.134725*i');

y[2] := x -> -2*sqrt(x)*cos(theta[2]*ln(x) - argument(0.5 + theta[2]*I))/(abs(0.5 + theta[2]*I)*ln(x));
y[2] := proc (x) options operator, arrow; -2*sqrt(x)*cos(theta[2\

  ]*ln(x)-argument(.5+I*theta[2]))/(abs(.5+I*theta[2])*ln(x)) 

   end proc

plot(y[2](x), x = 20 .. 100, title = 'Fig1*(S &G theta) = 1/2 + 21.022040*i');

y[3] := x -> -2*sqrt(x)*cos(theta[3]*ln(x) - argument(0.5 + theta[3]*I))/(abs(0.5 + theta[3]*I)*ln(x));
y[3] := proc (x) options operator, arrow; -2*sqrt(x)*cos(theta[3\

  ]*ln(x)-argument(.5+I*theta[3]))/(abs(.5+I*theta[3])*ln(x)) 

   end proc

plot(y[3](x), x = 20 .. 100, title = 'Fig1*(S &G theta) = 1/2 + 25.00858*i');

y[4] := x -> -2*sqrt(x)*cos(theta[4]*ln(x) - argument(0.5 + theta[4]*I))/(abs(0.5 + theta[4]*I)*ln(x));
y[4] := proc (x) options operator, arrow; -2*sqrt(x)*cos(theta[4\

  ]*ln(x)-argument(.5+I*theta[4]))/(abs(.5+I*theta[4])*ln(x)) 

   end proc

plot(y[4](x), x = 20 .. 100, title = 'Fig1*(S &G theta) = 1/2 + 30.424876*i');

y[5] := x -> -2*sqrt(x)*cos(theta[5]*ln(x) - argument(0.5 + theta[5]*I))/(abs(0.5 + theta[5]*I)*ln(x));
y[5] := proc (x) options operator, arrow; -2*sqrt(x)*cos(theta[5\

  ]*ln(x)-argument(.5+I*theta[5]))/(abs(.5+I*theta[5])*ln(x)) 

   end proc

plot(y[5](x), x = 20 .. 100, title = 'Fig1*(S &G theta) = 1/2 + 32.93502*i');

T[1] := x -> -2*sum(Moebius(n)*Re(Ei((0.5 + theta[1]*I)*ln(x)))/n, n = 1 .. trunc(ln(100)/ln(2)) + 1);
T[1] := proc (x) options operator, arrow; -2*(sum(NumberTheory:-\

  Moebius(n)*Re(Ei((.5+I*theta[1])*ln(x)))/n, n = 1 .. 

   trunc(ln(100)/ln(2))+1)) end proc

plot(T[1](x), x = 20 .. 100, title = 'T[1]');

T[2] := x -> -2*sum(Moebius(n)*Re(Ei((0.5 + theta[2]*I)*ln(x)))/n, n = 1 .. trunc(ln(100)/ln(2)) + 1);
T[2] := proc (x) options operator, arrow; -2*(sum(NumberTheory:-\

  Moebius(n)*Re(Ei((.5+I*theta[2])*ln(x)))/n, n = 1 .. 

   trunc(ln(100)/ln(2))+1)) end proc

plot(T[2](x), x = 20 .. 100, title = 'T[2]');

T[3] := x -> -2*sum(Moebius(n)*Re(Ei((0.5 + theta[3]*I)*ln(x)))/n, n = 1 .. trunc(ln(100)/ln(2)) + 1);
T[3] := proc (x) options operator, arrow; -2*(sum(NumberTheory:-\

  Moebius(n)*Re(Ei((.5+I*theta[3])*ln(x)))/n, n = 1 .. 

   trunc(ln(100)/ln(2))+1)) end proc

plot(T[3](x), x = 20 .. 100, title = 'T[3]');

T[4] := x -> -2*sum(Moebius(n)*Re(Ei((0.5 + theta[3]*I)*ln(x)))/n, n = 1 .. trunc(ln(100)/ln(2)) + 1);
T[4] := proc (x) options operator, arrow; -2*(sum(NumberTheory:-\

  Moebius(n)*Re(Ei((.5+I*theta[3])*ln(x)))/n, n = 1 .. 

   trunc(ln(100)/ln(2))+1)) end proc

plot(T[4](x), x = 20 .. 100, title = 'T[4]');

T[5] := x -> -2*sum(Moebius(n)*Re(Ei((0.5 + theta[5]*I)*ln(x)))/n, n = 1 .. trunc(ln(100)/ln(2)) + 1);
T[5] := proc (x) options operator, arrow; -2*(sum(NumberTheory:-\

  Moebius(n)*Re(Ei((.5+I*theta[5])*ln(x)))/n, n = 1 .. 

   trunc(ln(100)/ln(2))+1)) end proc

plot(T[5](x), x = 20 .. 100, title = 'T[5]');

f10 := x -> Li(x) - 2*sum(Re(Ei((1/2 + theta[n]*I)*ln(x))), n = 1 .. 10) - ln(2) + int(1/(t*(t^2 - 1)*ln(t)), t = x .. infinity);
f10 := proc (x) options operator, arrow; Li(x)-2*(sum(Re(Ei((1/2\

  +I*theta[n])*ln(x))), n = 1 .. 10))-ln(2)+int(1/(t*(t^2-1)*ln(\

  t)), t = x .. infinity) end proc

R10 := x -> sum(Moebius(l)*f10(x^(1/l))/l, l = 1 .. 8);
R10 := proc (x) options operator, arrow; sum(NumberTheory:-Moebi\

  us(l)*f10(x^(1/l))/l, l = 1 .. 8) end proc

plot1 := plot(R10(x), x = 2 .. 100);

plot2 := plot(pi(x), x = 2 .. 100);

display([plot1, plot2]);

f29 := x -> Li(x) - 2*sum(Re(Ei((1/2 + theta[n]*I)*ln(x))), n = 1 .. 29) - ln(2) + int(1/(t*(t^2 - 1)*ln(t)), t = x .. infinity);
f29 := proc (x) options operator, arrow; Li(x)-2*(sum(Re(Ei((1/2\

  +I*theta[n])*ln(x))), n = 1 .. 29))-ln(2)+int(1/(t*(t^2-1)*ln(\

  t)), t = x .. infinity) end proc

R29 := x -> sum(Moebius(l)*f29(x^(1/l))/l, l = 1 .. 8);
R29 := proc (x) options operator, arrow; sum(NumberTheory:-Moebi\

  us(l)*f29(x^(1/l))/l, l = 1 .. 8) end proc

plot3 := plot(R29(x), x = 2 .. 100);

NULL;
display([plot1, plot2, plot3]);

R29(100);
R10(100);
pi(100);
                          25.25165721

                          25.28503922

                               25

RR10 := x -> sum(Moebius(l)*f10(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2)) + 1);
RR10 := proc (x) options operator, arrow; sum(NumberTheory:-Moeb\

  ius(l)*f10(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2))+1) end 

   proc

RR10(1000);
pi(1000);
                          168.1328341

                              168

RR29 := x -> sum(Moebius(l)*f29(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2)) + 1);
RR29 := proc (x) options operator, arrow; sum(NumberTheory:-Moeb\

  ius(l)*f29(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2))+1) end 

   proc

RR29(1000);
                          167.6113955

P1 := plot(RR29(x), x = 880 .. 930);

P2 := plot(pi(x), x = 880 .. 930);

display([P1, P2]);

f0 := x -> Li(x) - ln(2) + int(1/(t*(t^2 + 1)*ln(t)), t = x .. infinity);
f0 := proc (x) options operator, arrow; Li(x)-ln(2)+int(1/(t*(t^\

  2+1)*ln(t)), t = x .. infinity) end proc

RR0 := x -> sum(Moebius(l)*f0(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2)) + 1);
RR0 := proc (x) options operator, arrow; sum(NumberTheory:-Moebi\

  us(l)*f0(x^(1/l))/l, l = 1 .. trunc(ln(1000)/ln(2))+1) end proc

P3 := plot(RR0(x), x = 880 .. 930);

NULL;
display([P1, P2, P3]);

P4 := plot(RR10(x), x = 880 .. 930);

display([P1, P2, P3, P4], color = [green, blue, purple, yellow]);

evalf(Li(2)), evalf(ln(2));
                   1.045163780, 0.6931471806

evalf(li(2));
                             li(2)

evalf(Ei(2));
                          4.954234356

evalf(Int(1/ln(t), t = 0 .. 2));
                        Float(undefined)

evalf(Ei(ln(2)));
                          1.045163780

Li(1000.);
                          177.6096580

isprime, [$ (1 .. 100)];
isprime, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 

  17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 

  33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 

  49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 

  65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 

  81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 

  97, 98, 99, 100]

nops(select(isprime, [$ (1 .. 100)]));
                               25

theta[1];
                           14.134725

evalf(Ei((1/2 + theta[1]*I)*ln(x)));
             Ei((0.5000000000 + 14.134725 I) ln(x))

evalf(Ei(ln(1/2 + theta[1]*I)));
                  4.386989035 + 6.632175089 I

plot(Li(x), x = 0 .. 5);

Let 
                              "a"

 and 
                              "b"

 be real numbers and 

       
"A = Matrix(3, 3, [[a, a - 1, -b], [a - 1, a, -b], [b, b, 2*a - 

   1]])"


,  
                              "B="


 "Matrix(5, 5, [[0, a, 3, 0, a], [3, 0, 0, b, 0], [0, 1, b, 0, 

    1], [b, 0, 0, 1, 0], [0, a, 1, 0, b]])"


(a) Show that if 
                            "0 <= a"


                            "a <= 1"

 and 
                      "b^2 = 2*a*(1 - a)"

, then A is an orthogonal matrix with determinant equal to one. 
(b) For what values of a and b is the matrix B singular? Determine the inverse of B (for those values of a and b for which B is invertible).
 

Find all rational function solutions to the Kadomtsev-Petviashvili equation 
                         
(&PartialD;)/(&PartialD;x);

diff(u, t) + 6*u*diff(u, x) + diff(u, x, x, x) - diff(u, y, y) = 0;

by u = 2 
diff(ln, x, x)*f;

=(2 (((&PartialD;)^2)/(&PartialD;x^2) f) f-2 ((&PartialD;)/(&PartialD;x) f)^2)/(f^2);
 with 
f;
  =
(a[1 ]x+a[2] y+a[3] t+a[4])^2+(a[5] x+ a[6] y+a[7] t+a[8])^2+a[9], ;
where 
a[i], i=1..9, ;
are real constants.
 


How to find the values of X(1),X(2),..&Y(1),Y(2)...Plese help .

restart;

 

for k from 0 to 5 do
X(k+1):=solve(2*(k+1)*X(k+1)+(k+1)*Y(k+1)-X(k)-Y(k)+(1)/k!,X(k+1));
Y(k+1):=solve((k+1)*X(k+1)+(k+1)*Y(k+1)+2*X(k)+Y(k)+(1)/k!,Y(k+1)); od;


 

2

 

1

 

Warning, solving for expressions other than names or functions is not recommended.

 

Error, (in solve) a constant is invalid as a variable, -(1/2)*Y(1)+1

 

``

Download DE_Using_DTM-Ex-3(1).mw

 

The serie is :Sum(-5*3^(-k - 1)*(x - 2)^k, k = 0 .. infinity)

How to simplify (with collect ? with convert ?...) this expression to get this more "traditionnal" writing :

-5/3*sum(((x - 2)/3^k)^k, k = 0 .. infinity)

Thank you for your help.

To Maple support:

I see 2 problems here. Maple solves the ode using series method.

First problem: Using odetest shows the syntax according to help does not work. Which is

           odetest(sol, ODE, series, point = 0);

The above gives internal error.

When changing to the following syntax

         odetest(sol,ODE,type='series',point=0); 

No internal error.

So help page should be corrected.

The second problem is that Maple odetest does not return 0 on its own solution. I verified manually that the solution is correct actually. So I do not know why maple does not return zero here. Simplfication does not help. 
 

interface(version);

`Standard Worksheet Interface, Maple 2022.0, Windows 10, March 8 2022 Build ID 1599809`

restart;

Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+x^2*diff(y(x),x)+y(x) = 0;
maple_sol:=dsolve(ode,y(x),type='series',x=0):
odetest(maple_sol,ode,series,point=0);
odetest(maple_sol,ode,'series',point=0);

6

x^2*(diff(diff(y(x), x), x))+x^2*(diff(y(x), x))+y(x) = 0

Error, (in odetest/series) complex argument to max/min: 13/2-1/2*I*3^(1/2)

Error, (in odetest/series) complex argument to max/min: 13/2-1/2*I*3^(1/2)

odetest(maple_sol,ode,type='series',point=0); #This should return zero, but it does not.

-I*3^(1/2)*x^(3/2-((1/2)*I)*3^(1/2))*(series(-1/2-(I*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2)))*x-((1/4)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^2-((1/6)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3-((7/16)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^4+O(x^5),x,5))*_C1+((1/2)*I)*3^(1/2)*x^(3/2+((1/2)*I)*3^(1/2))*(series(1-(1/2)*x+(((1/2)*I)*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2)))*x^2-((1/12)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x^3+((1/24)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^4-((7/80)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^5+O(x^6),x,6))*_C2+_C1*x^(5/2-((1/2)*I)*3^(1/2))*(series(-I*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2))-((1/2)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x-((1/2)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^2-((7/4)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3+O(x^4),x,4))+_C2*x^(5/2+((1/2)*I)*3^(1/2))*(series(I*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2))-((1/2)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x+((1/2)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^2-((7/4)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^3+O(x^4),x,4))+_C1*x^(3/2-((1/2)*I)*3^(1/2))*(series(-1/2-(I*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2)))*x-((1/4)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^2-((1/6)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3-((7/16)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^4+O(x^5),x,5))+_C2*x^(3/2+((1/2)*I)*3^(1/2))*(series(-1/2+(I*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2)))*x-((1/4)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x^2+((1/6)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^3-((7/16)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^4+O(x^5),x,5))-((1/2)*I)*3^(1/2)*x^(3/2-((1/2)*I)*3^(1/2))*(series(1-(1/2)*x-(((1/2)*I)*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2)))*x^2-((1/12)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3-((1/24)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^4-((7/80)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^5+O(x^6),x,6))*_C1+I*3^(1/2)*x^(3/2+((1/2)*I)*3^(1/2))*(series(-1/2+(I*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2)))*x-((1/4)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x^2+((1/6)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^3-((7/16)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^4+O(x^5),x,5))*_C2+x^(5/2-((1/2)*I)*3^(1/2))*(series(-1/2-(I*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2)))*x-((1/4)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^2-((1/6)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3-((7/16)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^4+O(x^5),x,5))*_C1+x^(5/2+((1/2)*I)*3^(1/2))*(series(-1/2+(I*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2)))*x-((1/4)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x^2+((1/6)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^3-((7/16)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^4+O(x^5),x,5))*_C2+(1/2)*x^(3/2-((1/2)*I)*3^(1/2))*(series(1-(1/2)*x-(((1/2)*I)*3^(1/2)/((I*3^(1/2)-1)*(I*3^(1/2)-2)))*x^2-((1/12)*((5*I)*3^(1/2)+3)/((I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^3-((1/24)*((8*I)*3^(1/2)+9)/((I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^4-((7/80)*((3*I)*3^(1/2)+5)/((I*3^(1/2)-5)*(I*3^(1/2)-4)*(I*3^(1/2)-3)*(I*3^(1/2)-2)*(I*3^(1/2)-1)))*x^5+O(x^6),x,6))*_C1+(1/2)*x^(3/2+((1/2)*I)*3^(1/2))*(series(1-(1/2)*x+(((1/2)*I)*3^(1/2)/((1+I*3^(1/2))*(I*3^(1/2)+2)))*x^2-((1/12)*((5*I)*3^(1/2)-3)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)))*x^3+((1/24)*(-9+(8*I)*3^(1/2))/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)))*x^4-((7/80)*((3*I)*3^(1/2)-5)/((1+I*3^(1/2))*(I*3^(1/2)+2)*(I*3^(1/2)+3)*(I*3^(1/2)+4)*(I*3^(1/2)+5)))*x^5+O(x^6),x,6))*_C2

 


 

Download problems_with_series_solution.mw

 

I want to use this L3 list as the index of another list. I am trying to create a model for short time electricity load forecasting in Maple. I am fairly new to maple coding structures. Can anybody suggest an easier way for doing this kind of thing in Maple? Can I use matrix generation? Please give me suggestions. 
Note: Currently working in a Doc File. 
The code is pasted below: 

L1 := [seq([seq(seq1[i], i = 1 .. 7)], i = 1 .. 24)];

L2 := [seq(i, i = 1 .. 24)];



local(i, j, L3);
L3 = [];
for i to 24 do
    for j to 7 do if i = 1 then L3[i][j] := L1[i][j]; else L3[i][j] := L1[i][j] + L2[i] - 1; end if; end do;
end do;
print(L3);

How can I found the area of the region bounded by 

y=e^xcos(x), y=o, x=−π/2, x= π/2 ?

I have an example where I try to calculate an infinite sum (G&R1.641.1)  in Maple 2022 (or 2021).  I get a different numerical result when I evaluate the sum if I express the coefficients c(m) in the sum using Maple's doublefactorial function for m!! (c1(m)) or the equivalent expression involving the factorial function (c2(m)).  Only the latter gives the correct value for the sum.  This happens even though Maple gives the same numerical value for the coefficients c1(m) and c2(m).  What's going on?? Please see the attached spreadsheet.

Download doublefactorial_test.mw

Dear all

I solved a linear system using Jacobi method , then I plot the solution, someting is wrong, I can't plot the solution 

jacobi.mw

Thank you for your help 

I have the following expression (result of a calculation):

(1/1296)*cBooP0-(1/1296)*cSRP0-(1/1296)*tStartRamp*f__SR/N

Rather obviously the common factor 1/1296 can be factored out, except I cannot get Maple to factor the 1/1296 without also factoring out N, which I do not want. My desired end result is this:

(1/1296)*(cBooP0-cSRP0-tStartRamp*f__SR/N)

I don't seem to be able to coerce Maple to do this. I can freeze the tStartRamp*f__SR/N term (leaving the 1/1296 unfrozen), but in that case I don't get Maple to pull the 1/1296 factor out at all.

Any hint would be appreciated. I am doing this on Maple 2015. It is really a bit cosmetic, but sometimes I use Maple to write "smart" documentation & then I'd like to end up with a somewhat polished result.

Mac Dude

Dear all,

Please I want to solve the following boundary value problem numerically with the attached code

y''=((y')^2+y^2)/(2*exp(x)),      0<x<1

with boundary conditions as follows

y(0)-y'(0)=0, y(1)+y'(1)=2*exp(1)

The exact solution is y(x)=exp(x).

How do I modify the code to be able to handle it?

Please the delta in the code represents y'.

Thank you for your time and best regards

restart;

 

e1:=y[n+2] = (1/12)*h^2*f(n)+(5/6)*h^2*f(n+1)+(1/12)*h^2*f(n+2)+2*y[n+1]-y[n]:

 

NULL

NULL

     h          Num.y          Num.z            Ex.y           Ex.z        Error y        Error z
0.25000      0.702642933  -11.119327426    0.044732488   -9.516563326       0.65791         1.6028
0.50000     -1.480941776    3.706008390   -0.195836551   -1.080050970        1.2851         4.7861
0.75000      2.177304037    3.857405154    1.966274055   -2.618345553       0.21103         6.4758
1.00000     -0.353401232  -12.899134092   -0.541621655  -16.265122833       0.18822          3.366
1.25000      2.205257809   -0.748313267    1.880461001    0.803458961        0.3248         1.5518
1.50000     -0.457853582   -9.260175135    0.888094914  -17.501091793        1.3459         8.2409
1.75000      2.368665639   -0.220734441    0.227799905   -7.823705892        2.1409          7.603
2.00000     -0.556459764  -11.300336415    2.230324739   -5.428300487        2.7868          5.872
2.25000      2.205494666   -0.953807966   -0.582405956  -17.141951993        2.7879         16.188
2.50000     -0.947043873  -11.871777444    1.457323206    3.126163062        2.4044         14.998
2.75000      1.867780693   -1.104185167    0.366014055  -11.889781987        1.5018         10.786
3.00000     -1.448466064  -12.600325627   -0.692660166   -0.858678078       0.75581         11.742
3.25000      1.408438927   -1.076955833    1.243407129    4.765140072       0.16503         5.8421
3.50000     -1.960671678  -12.838644035   -1.682658102   -6.465546461       0.27801         6.3731
3.75000      0.947838606   -0.761420206    0.210882522   14.697480238       0.73696         15.459
4.00000     -2.352223093  -12.652936159   -0.678627396    0.245000716        1.6736         12.898
4.25000      0.596650899   -0.266003839   -1.802692162    9.387643085        2.3993         9.6536
4.50000     -2.528856863  -12.055589881    0.398695396   14.817725265        2.9276         26.873
4.75000      0.441416507    0.296276900   -2.296698552    0.729525901        2.7381        0.43325
5.00000     -2.446901406  -11.201642111   -0.256333100   19.522565217        2.1906         30.724

 

Download K2_Prob_4_direct_2nd_derivative.mw

As the LPsolve package returns the solution points in a list format, how we can analyze the list if we're having a large instance problem? By large instance, I mean that there are around 1500+ variables in the linear programming model. 

Also, I have attached the screenshot of the output in this thread. To perform further analysis, I want to fetch the value of the DQV, QF, and UQT variables. But since the list contains the values as an expression, for example DQV[0, 0, 0, 0] = 0. In such cases, I cannot perform mathematical operations for the solution points results.

Also, if I can delete the DQV[0,0,0,0] from the expression mentioned above, I'll end with the numeric values list, which can then be converted into an array and make things easier for me. But currently, I'm struggling to delete the DQV part of the expression in one go (manually deletion for each expression, it's very time consuming to delete for 1500 variables).

Please suggest to me some approach to deal with this issue.

 

I am thanking you.

Regards

I am having a problem in evaluating a fluid flow problem at a boundary that includes infinity using a semi-analytic method Adomian Decomposition Method.

Find the attached code. Thanks in advance.

I recently upgraded to Maple 2022 just to use the new latex command. It seems I still can't export DataFrame types, I'm very disappointed. What am I doing wrong?

restart:
test:= DataFrame( [[1]], columns = [1] );
latex(test);

returns:

Error, (in DataFrame:-type) invalid input: subtype expects its 2nd argument, _t, to be of type type, but received And(symbol,satisfies(u -> substring(u,1 .. 9) = ('`\\mapleref`')))

Using File > Export As > LaTeX also doesn't work.
Printing the function call using printlevel := 100000: returns the following:

I've attached the worksheet for reference. Demo.mw
Does anyone know a workaround?

-Thanks for the Help

First 26 27 28 29 30 31 32 Last Page 28 of 2141