Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I would like to perform the following expansion up to an arbitrary order. The dots refer to higher-order terms. All the coefficients are symmetric except for the first one.

Thank you very much!

To Maple support:

I was investigating this pde from a different forum.

I noticed that when using an expanded version of the pde, Maple hangs. Without expanding the PDE, Maple gives an answer in 2 seconds. 

Why does expanding the PDE makes a difference? I do not have an earlier version of Maple on my new PC to check if this is a new issue or not.
 

interface(version);

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

Physics:-Version()

`The "Physics Updates" version in the MapleCloud is 1230 and is the same as the version installed in this computer, created 2022, April 21, 9:8 hours Pacific Time.`

restart;
pde1:=VectorCalculus:-Laplacian(u(r,theta),'polar'[r,theta]);
pde1_expanded:=expand(pde1);
bc  := u(1,theta)=sin(theta)^4,u(3,theta)=1;
pdsolve([pde1=0,bc],u(r,theta))
 

(diff(u(r, theta), r)+r*(diff(diff(u(r, theta), r), r))+(diff(diff(u(r, theta), theta), theta))/r)/r

(diff(u(r, theta), r))/r+diff(diff(u(r, theta), r), r)+(diff(diff(u(r, theta), theta), theta))/r^2

u(1, theta) = sin(theta)^4, u(3, theta) = 1

u(r, theta) = (1/52480)*((328*r^6-26568*r^2)*ln(3)*cos(2*theta)+(-r^8+6561)*ln(3)*cos(4*theta)+19680*(ln(3)+(5/3)*ln(r))*r^4)/(ln(3)*r^4)

pdsolve([pde1_expanded=0,bc],u(r,theta)); #HANGS, Waited more than 40 minutes.

 


 

Download hangs_pde.mw

DirectSearch finds nonexisting roots....     DirectSearch_finds_nonexisting_roots.mw

 

restart

plot([cos(x), 0.001*x*x],x=-40..40,y=0..2)     ### test function plot

 

eq:= cos(x)= 0.001*x*x;     ###   float

cos(x) = 0.1e-2*x^2

(1)

use RealDomain in solve(eq,x,explicit) end use   ## lacking some of the roots

-7.793210110, 7.793210110, -11.11953559, 11.11953559, -4.734809278, 4.734809278, -1.568336644, 1.568336644

(2)

use RealDomain in solve(eq,x) end use

-7.793210110, 7.793210110, -11.11953559, 11.11953559, -4.734809278, 4.734809278, -1.568336644, 1.568336644

(3)

fsolve(eq,x=-30)

-31.17938383

(4)

plot([cos(x), 0.001*x*x],x=-32..-31,y=0.9..1)

 

fsolve(eq,x=-32..-30);  fsolve(eq,x=-31.7..-31.4)

-31.17938383

 

-31.52634294

(5)

with(DirectSearch)

[BoundedObjective, CompromiseProgramming, DataFit, ExponentialWeightedSum, GlobalOptima, GlobalSearch, Minimax, ModifiedTchebycheff, Search, SolveEquations, WeightedProduct, WeightedSum]

(6)

eq;
SolveEquations(eq,AllSolutions):

cos(x) = 0.1e-2*x^2

(7)

interface(rtablesize=90);
SolveEquations(eq, AllSolutions)

90

 

Matrix(%id = 4800942722)

(8)

DirectSearch finds nonexisting roots!!!

Dr. Ali GÜZEL

 

Download DirectSearch_finds_nonexisting_roots.mw

Hello!

How do I get the plot of P(t) and C(t) and also a table with the values of P(t) for each t? Itried but I couldn't...

``

restart; beta := 0.7e-2; Lambda := 0.2e-4; `ρ__o` := 0.3e-2; lambda := 0.8e-1; h := 0.1e-1; n := 100

Error,

restart; beta := 0.7e-2; Lambda := 0.2e-4; `ρ__o` := 0.3e-2; lambda := 0.8e-1; h := 0.1e-1; n := 100

 

for i from 0 to n-1 do P[i+1] := (1+h*(`ρ__o`-beta)/Lambda)*P(i)+h*lambda*C(i); C[i+1] := h*beta*P(i)/Lambda+(-h*lambda+1)*C(i) end do:

seq(i, i = 0 .. 30); seq(P[i], i = 0 .. 30)

Error,

seq(i, i = 0 .. 30); seq(P[i], i = 0 .. 30)

 

with(plots); with(DEtools); p1 := plot([P(t)], t = 0 .. 100, [[P(0) = 1]], scene = [t, P(t)], thickness = 2, linecolor = red, stepsize = .1); p2 := plot([C(T)], t = 0 .. 100, [[C(0) = beta/(Lambda*lambda)]], scene = [t, C(t)], thickness = 2, linecolor = red, stepsize = .1); display([p1, p2])

``

Download System_Recursive_Equations.mw

Thank!!!

Dear all

How can I add more terms, in taylor approximation of odes. 

Attached the code well written for only second order approximatiom, how can i get the fourth order approximation in taylor expansion to approximate an IVP

taylor_fourth_order.mw

thank you

This worksheet displays an intersection between two spheres based on a test which seems unrelated to the display.

How can this be explained?

Intersecting_spheres.mw

Hello,

I am writing an arrow procedure and will like to know if there is a way to implement the following 
 

bj := (G, y) ->  (`@`(seq((t -> u -> v -> G(u, t) - v)(args[1 + nargs - i]), i = 1 .. nargs - 2)))(y)
bj(F, y, a, b, c, d);
v:=[p, q, r, s]
the output

My question is, how can I replace the v with each element of the list to get the following as output
F(F(F(F(y,a)-p),b)-q,c)-r,d)-s

Aany suggestion will be highly appreciated

 

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);
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