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

How I can plot the answers obtained from solving two differential equations?

Thanks

SAL.mw
 

Download SAL.mw

I am writing a question in Mobius that requires students to enter the union and infinity symbols. I am using a Maple graded answer, and I am able to get it to grade infinity properly by typing the word "infinity" in the answer and using Maple syntax with symbolic entry, so that they have access to the infinity symbol in the equation editor. There is a union symbol in the equation editor, but I am not sure how to code the answer to make it accept the union symbol as correct.

Thanks