This is my attempt to produce a subplot within a larger plot for magnifying data in a small region, and putting that subplot into the white space of the figure.
Based on the questions: How to insert a plot into another plot? and Inset figure in Maple, I wrote a couple of procedures that create sub-plots and allow the user to place the subplot window as he/she chooses. This avoids the graininess issues mentioned by acer in the second link (and experienced by me).

So far, I only have this completed for point plots, but using acer's method of piecewise functions posted in the plotin2b.mw of the second article, with the subplot function being defined only if it satisfies your conditions, would allow the subplot generating procedure to be generalized easily enough. But the data I'm working with all point plots, so that's the example here.

The basic idea  is to use one procedure to create boxes, make tickmarks on the expanded region, and make tickmark labels, combine all of those into one graph. Then create scaled and shifted versions of the data series, then make graphs of those. Lastly, combine them all into one picture.

Hope this helps someone who has to do the same.

Mapleprimes isn't inserting the contents, but here is the download of the file: SubPlotBoxesandVectorDataSeries.mw

Hello, I was given the problem "Set N:=100. (i) Form four lists L[1], L[3], L[5], L[7] where L[r] contains all primes 2< p < N such that p mod 8 = r ." and came up with the following code:

N := 100;
List1 := [];
List3 := [];
List5 := [];
List7 := [];
for p from 2 to N do
    if isprime(p) and p mod 8 = 1 then List1 := [op(List1), p]; end if;
    if isprime(p) and p mod 8 = 3 then
        List3 := [op(List3), p];
    end if;
    if isprime(p) and p mod 8 = 5 then
        List5 := [op(List5), p];
    end if;
    if isprime(p) and p mod 8 = 7 then
        List7 := [op(List7), p];
    end if;
end do;
List1;
List3;
List5;
List7;
 

this gave me the answer I wanted, however using the above code I have to answer this second question: 

In the notation of Problem 1, make a procedure with input = arbitrary  positive integer N and output = the list [nops(L[1]), nops(L[3]), nops(L[5]), nops(L[7])]. Do some experiments to see for which (if any) r, L[r] is largest. 

I am unsure of how to create a procedure out of the code I already have. I created this: 
 

restart;
F := proc(n)

local N, p, List1, List3, List5, List7;

List1 := [];

List3 := [];

List5 := [];

List7 := [];

for p from N do

if isprime(p) and p mod 8 = 1 then List1 := [op(List1), p]; end if;

if isprime(p) and p mod 8 = 3 then List3 := [op(List3), p]; end if;

if isprime(p) and p mod 8 = 5 then List5 := [op(List5), p]; end if;

if isprime(p) and p mod 8 = 7 then List7 := [op(List7), p]; end if;

return [nops(List1), nops(List3), nops(List5), nops(List7)];

end do;

end proc;

however, when I input F(100) or any other value of N, I am receiving the error message "Error, (in F) initial value in for loop must be numeric or character"

Any ideas on how to improve my program to get the output I desire?

thank you 

Download 23May(1).mw

Seems to make no difference with or without RETURN() for the procedure ?

vraag_op_forum_gesteld_over_boek_vb_binomium.mw

For to know what each graph is standing for in the plot legenda

Could not yet get it  in y , y' , y '' , to 5th derative 

Better is perhaps to have vertical table with the names and function expression together , but that could be difficult betounes_ex_set_task_4def.mw  

f := x -> exp(-x)*sin(x); intvx:= 0..3;
f := proc (x) options operator, arrow; exp(-x)*sin(x) end proc
intvx := 0 .. 3
 

And this one below ( i prefer this one ) , but got in worksheet now the one above
Probably a option issue ?

2 determinants equal, solve for the unknowns in one determinant/Linear Algebra

Hi guys, thank you for reading my question.

A is one determinant "Matrix(2, 3, [[1, 2, 3], [3, 1, 2]])".B is" Matrix(2, 3, [[1, x, 3], [y, 1, z]])"

I wrote 2 matrice in the maple

As I try to solve the unknowns,I wrote:

A:= B, solve(x, y, z) It doesn't work

I switch another phrase,

A = B;solve(B:x,y,z)    It doesn't work as well

Could you help me with it ?

Thank you

Dear All.

Please kindly help to correct the attached code on discretization of fourth order PDE using method of line.
Thank you and kind regards.

Download Discretization_of_PDE_Order_4.mw

restart;

ii := 50;

seq(ii, ii = 0 .. 5);

evalf(int(ii, ii = 0 .. 5))

For sequence the syntax is correct i.e the output is 0,1,2,3,4,5. is ii inside sequence command takes the value 5 assigned it then changes when ii ranges from 0 to 5?

restart;

ii:=50:
seq(ii)

output: 50

restart;

ii:=50:

seq(seq(ii),ii=1..5)

expected answer: 50,50,50,50,50

obtained: 1,2,3,4,5

please explain how seq command works

Trebuchet, Phase I, 2020-05-27 Ki restart; with(RealDomain); with(SolveTools); assume(h < r1); additionally(h < r2); [Im, Re, ^, arccos, arccosh, arccot, arccoth, arccsc, arccsch, arcsec, arcsech, arcsin, arcsinh, arctan, arctanh, cos, cosh, cot, coth, csc, csch, eval, exp, expand, limit, ln, log, sec, sech, signum, simplify, sin, sinh, solve, sqrt, surd, tan, tanh ] [AbstractRootOfSolution, Basis, CancelInverses, Combine, Complexity, Engine, GreaterComplexity, Identity, Inequality, Linear, Parametric, Polynomial, PolynomialSystem, RationalCoefficients, SemiAlgebraic, SortByComplexity] hz1 := g*(r3*m3*cos(phi1(t))-r1*m2*sin(phi2(t))/sin(phi2(t)-phi1(t)))/theta3; whattype(phi1(t)); whattype(phi2(t)); / r1 m2 sin(phi2(t)) \ g |r3 m3 cos(phi1(t)) + -----------------------| \ sin(-phi2(t) + phi1(t))/ hz1 := ------------------------------------------------ theta3 function function hz2 := phi1(t)+arcsin((h+r1*sin(phi1(t)))/r2); /h + r1 sin(phi1(t))\ hz2 := phi1(t) + arcsin|-------------------| \ r2 / subs(phi2 = hz2, hz1); / / | | 1 | | ------ |g |r3 m3 cos(phi1(t)) theta3 | | | | \ \ // /h + r1 sin(phi1(t))\\ \ \\ r1 m2 sin||phi1(t) + arcsin|-------------------||(t)| || \\ \ r2 // / || + ----------------------------------------------------------|| / / /h + r1 sin(phi1(t))\\ \|| sin|-|phi1(t) + arcsin|-------------------||(t) + phi1(t)||| \ \ \ r2 // /// deq := diff(phi1(t), t, t)-% = 0; / / / 2 \ | | | d | 1 | | deq := |---- phi1(t)| - ------ |g |r3 m3 cos(phi1(t)) | 2 | theta3 | | \ dt / | | \ \ / /h + r1 sin(phi1(t))\ \ \ r1 m2 sin|phi1(t)(t) + arcsin|-------------------|(t)| | \ \ r2 / / | + -----------------------------------------------------------| / /h + r1 sin(phi1(t))\ \| sin|-phi1(t)(t) - arcsin|-------------------|(t) + phi1(t)|| \ \ r2 / // \ | | | = 0 | | / ics := phi1(0) = -arcsin(h/r1), (D(phi1))(0) = 0; /h \ ics := phi1(0) = -arcsin|--|, D(phi1)(0) = 0 \r1/ dsolve({deq, ics}, phi1(t)); dsolve(deq); deq_numeric := subs(r1 = 8, r2 = 8, r3 = 1, h = 5, m2 = 1, m3 = 20, theta3 = 20, deq); / | / d / d \\ 1 | |--- |--- phi1(t)|| - -- g |20 cos(phi1(t)) \ dt \ dt // 20 | | \ / /5 \ \ \ 8 sin|phi1(t)(t) + arcsin|- + sin(phi1(t))|(t)| | \ \8 / / | + --------------------------------------------------------| = 0 / /5 \ \| sin|-phi1(t)(t) - arcsin|- + sin(phi1(t))|(t) + phi1(t)|| \ \8 / // ics_numeric := phi1(0) = 0, (D(phi1))(0) = -.63; ics_numeric := phi1(0) = 0, D(phi1)(0) = -0.63 hz1 := dsolve({ics, deq_numeric}, phi1(t), numeric); Error, (in dsolve/numeric/process_input) unknown arcsin(5/8+sin(phi1(t))) present in ODE system is not a specified dependent variable or evaluatable procedure sin(-phi1(t)-arcsin((h+r1*sin(phi1(t)))/r2)); / /h + r1 sin(phi1(t))\\ -sin|phi1(t) + arcsin|-------------------|| \ \ r2 //

How to convert expression below to LaTex:  $\lim _{x \rightarrow 2} \frac{2}{x+3}$

And How to convert expression below to LaTex:  $\int_{2}^{3} \frac{1}{x^{2}+2} d x$

file test: newLaTex.mw

ode3 := diff(P(t), t) = 0.2*P(t) - 300;
                        d                       
              

ICs := [[0, 1300], [0, 1800]];
DEplot(ode3, p(t), p = -50 .. 2000, t = 0 .. 2000, ICs, arrows = LINE, color = RED, linecolor = BLUE, title = "Solution Curves ode3");
Error, (in DEtools/DEplot) invalid variable in specification of independent variable range
 

I want to plot y=sec(x) and x=-pi/4, x= pi/4. I also want the bounded region shaded. I have been able to graph sec(x) but I can't figure out how to get the two vertical lines and the bounded region shaded.

Hi there!

Essentially, sum(sqrt(-2),x=2..1) is, as it should be, 0. My issue is, that e.g. sum(sqrt(-2),x=2..0) or sum(sqrt(-2),x=2..-1)  are declared by Maple as -I*sqrt(2) and -(2*I)*sqrt(2), which, to my knowledge, is not true, and should be 0 as well. I need those results for a rather complicated algorithm to be 0, otherwise it doesn't work.

NEUZMinus:= proc(Unten, Oben, f,G,Liste,n)::real;
  #Unten:= Untere Intervallgrenze; Oben:= Obere Intervallgrenze; g:= zu integrierende Funktion;
  #G:= Gewicht; n:= Hinzuzufügende Knoten;
 
  Basenwechsel:=proc(Dividend, m);
 
  print(Anfang,Dividend,p[m]);
  Koeffizient:=quo(Dividend, p[m],x);

  Rest:=rem(Dividend, p[m],x);
 
  if m=0 then
    Basenwechsel:=[Koeffizient];
  else

    Basenwechsel:=[Koeffizient,op(Basenwechsel(Rest,m-1))];
   
  end if;
 
  end proc;
p[-1]:=0;
p[0]:=1;
for i from 1 to max(n,numelems(Liste)) do
  p[i]:=x^i-add(int(x^i*p[j]*diff(G,x),x=Unten..Oben)*p[j]/int(p[j]^2*diff(G,x),x=Unten..Oben),j=0..i-1);
  print(p[i]);
c[i-1]:=lcoeff(p[i],x)/lcoeff(p[i-1],x);
d[i-1]:=coeff(p[i],x,(i-1))/coeff(p[i-1],x,(i-1));
if i <> 1 then
  e[i-1]:=coeff(p[i]-(c[i-1]*x+d[i-1])*p[i-1],x,i-2)/coeff(p[i-2],x,i-2);
else
  e[i-1]:=1;
end if;
end do;
print(Liste[1],numelems(Liste));
Hn:=mul(x-Liste[i],i=1..numelems(Liste));
print(Hn);
 Koeffizienten:=Basenwechsel(Hn,numelems(Liste));
print(Koeffizienten);
for j from 0 to numelems(Liste)-1 do
  a[s][j][j]:=1;
end do;
for s from 1 to numelems(Liste)-1 do
  a[s][0]:=[1];
  a[s][1]:=[-e[s]*c[0]/c[s],d[0]-d[s]*c[0]/c[s],c[0]/c[s]];
  for j from 2 to numelems(Liste)-1 do
    print(1);
    a[s][j][abs(s-j)]:=sum(-c[j-1]*e[i+1]*a[s][j-1][i+1]/c[i+1],i=abs(s-j)..min(abs(s-j),s+j-2));
    print(2);
    a[s][j][abs(s-j)+1]:=sum(d[j-1]-c[j-1]*d[i]/c[i])*a[s][j-1][i],i=abs(s-j)+1..min(abs(s-j)+1,s+j-1)+sum(-c[j-1]*e[i+1]*a[s][j-1][i+1]/c[i+1],         i=abs(s-j)+1..min(abs(s-j)+1,s+j-2));
    print(3);
    for i from abs(s-j)+2 to s+j-2 do
      a[s][j][i]:=c[j-1]*a[s][j-1][i-1]/c[i-1]+(d[j-1]-c[j-1]*d[i]/c[i])*a[s][j-1][i]-c[j-1]*e[i+1]*a[s][j-1][i+1]/c[i+1]+e[j-1]*a[s][j-2][i];

      print(4);
    end do;
    a[s][j][s+j-1]:=sum(c[j-1]*a[s][j-1][i-1]/c[i-1],i=max(s-j+2,s+j-1)..s+j-1)+sum((d[j-1]-c[j-1]*d[i]/c[i])*a[s][j-1][i],i=max(s-j+1,s+j-1));
    print(5);
    a[s][j][s+j]:=sum(c[j-1]*a[s][j-1][i-1]/c[i-1],i=max(s-j+2,s+j)..s+j);
    print(6);
  end do;
end do
 

end proc

I'm having problems with Multiplying Complex Numbers in Maple 2019.

Thanks in Advance.

Download Multiplying_Complex_num.mw