MaplePrimes Questions

I was trying to get an aproximation of an expression, but it doesnt seem to provide the correct answer. for the following

((12*sqrt(13))/13 - 3/4)/(1 + ((12*sqrt(13))/13)(3/4))

 maple provides 0.59568 and the actual answer is 0.73744

I dont know what I am doing wrong

It was interesting not to miss the solutions of such a test equation  x * (1 + 1.0001 * sin (x ^ 2)) = 0 on a fairly large arbitrary segment using NextZero.
It turned out that NextZero skips solutions.
For example, solutions in the range from 7.5 to 10 are skipped.  And two more specific solutions skipped 21.232154952 and 21.232821004.
I took Digits to help, but it didn't affect anything. Most likely the reason is that I am a bad user.  The example seems to be a good one.

#Digits := 30; 
f := proc (x) options operator, arrow; x*(1+1.0001*sin(x^2)) end proc; 
RootFinding:-NextZero(f, 1.0): 
for j to 154 do RootFinding:-NextZero(f, %) end do


Let L be list say



The list is of this type it can be any list type and can be of any size it should be exported to excel like a table like this

A        B

(1,2)    3

(1,3)    4

(1,4)    4

(1,7)    7


First is in column A

and second is in column B


See brackets changed

kind help with the needul

Hi, dear community,
I have a Maple package "rath" for finding the traveling wave solution of differential equations. But I am not able to loading in Maple 18. Please see this attached zip folder and need help in this regard.
Kind regard




Anyone else experiencing problems uploading Workbooks as MaplePackage to MapleCloud.

Very often it is not possible to upload a workbook as a package, but just as a normal Maple file. Takes a couple of tries with opening the file until it is possible.

Hi dear community!


The following code produces a table, however it always has the text "Tabulate0" as an output as well. Is it possible to supress that? Ordinary : dont work unfortunately.


InitialisierungDF:=Vector[column](nOben-nUnten+1, fill=oE): #Erstellen der auszugebenden Tabelle
InitialisierungSpalte:=Vector[row](nOben-nUnten+1, i->n=nUnten-1+i):
DF:= DataFrame( Concatenate( 2, InitialisierungDF $ 10),
                   columns = [ GKAbs, GKRel, PZPAbs, PZPRel, PZMAbs, PZMRel, PYAbs, PYRel, DAbs, DRel],
                   rows = InitialisierungSpalte);


Thank you very much!

Hi there!

The first time I compile the following code, I get the error message

"Error, cannot split rhs into multiple assignment."

when trying to solve an issue with the procedure. I then have to compile the procedure over and over again, until it finally works (which it does eventually, without changing the code.) The problematic line is

Knoten, Eigenvektoren := Eigenvectors(evalf[15](M));

it is one of the last lines within the code below. Is it possible to get rid of that issue? It is annoying and unprofessional to have to compile a code over and over again until it finally works.




GaußKronrodQuadraturKurz:= proc(Unten, Oben, f,G,n)::real;
  #Unten:= Untere Intervallgrenze; Oben:= Obere Intervallgrenze; G:= Gewicht;
  #f:= zu untersuchende Funktion; n:= Berechnung der Knotenanzahl mittels 2*n+1
A,B,P,S,T, #Listen
a,b,p,s,t, #Listenelemente
i,j,k, #Laufvariablen
M, #werdende Gauss-Kronrod-Jacobi-Matrix
m, #Matrixeinträge
u,l, #Hilfsvariablen Gemischte Momente

A := [seq(a[i], i = 0 .. n)];
B := [seq(b[i], i = 0 .. n)];
P := [seq(p[i], i = -1 .. ceil(3*n/2)+1)];  
S := [seq(s[i], i = -1 .. floor(n/2))];
T := [seq(t[i], i = -1 .. floor(n/2))];
p[-1]:= 0;
for i from -1 to floor(n/2) do
end do;
for j from 1 to 2*n+1 do
  RekursivesZwischenergebnis:= x^j;
  for i from 0 to j-1 do
    RekursivesZwischenergebnis:= RekursivesZwischenergebnis -
    (int(x^j*p[i],x=Unten..Oben)/int(p[i]*p[i],x=Unten..Oben))*p[i]                  #Gram-Schmidt algorithm
  end do;
end do;

  #p[0+1]=(x-a[0])*p[0]-b[0]*p[0-1] -> p[1]=x*p[0]-a[0]*p[0]-b[0]*p[-1] ->
  #p[1]=x*1-a[0]*1-0 -> a[0]=x-p[1] -> a[0]= -coeff(p[1],x,0), da p[1] monisch ist und von Grad 1    #ist
b[0]:=int(p[0]^2, x=Unten..Oben); #by definition
for j from 1 to ceil(3*n/2) do
  #Genau genommen muss a nur bis floor(3/(2*n)) initialisiert werden, allerdings wird der Wert       #ohnehin für die Berechnung von b gebraucht. Die Initialisierung schadet nicht.
  a[j]:= coeff(p[j],x,j-1)- coeff(p[j+1],x,j);
    #p[j+1]=(x-a[j])*p[j]-b[j]*p[j-1] -> p[j+1]=x*p[j]-a[j]*p[j]-b[j]p[j-1] ->
      #(da b[j]*p[j-1] vom Grad j-1 ist) ->
    #coeff(p[j+1],x,j)=coeff(x*p[j],x,j)-a[j], da p[j] monisch ist ->
  b[j]:=  quo((x-a[j])*p[j]-p[j+1],p[j-1],x);

     #p[j+1]=(x-a[j])*p[j]-b[j]*p[j-1] -> -p[j+1]+(x-a[j])*p[j]= b[j]*p[j-1]

end do;    
t[0]:=b[n+1]; #t[0]:= /hat{b}[0], Beginn der ostwärtigen Phase
for i from 0 to n-2 do # n-2 ist die Anzahl der zu berechnenden Diagonalen
  for k from floor((i+1)/2) to 0 by -1 do # aufgrund des diagonalen Vorgehens ist nur bei jedem
                                          # zweiten Schleifendurchlauf eine Inkrementierung
                                          # vorzunehmen
    u:=u+(a[k+n+1]-a[l])*t[k]+b[k+n+1]*s[k-1]-b[l]*s[k]; # Ausrechnen gemischter Momente über die
                                                         # fünfgliedrige Rekursion
  end do;
  for j from -1 to floor(n/2) do  # Durchrotieren der Werte der gemischten Momente, da ein                                           # jeweiliges gemischtes Moment beim zweiten auf die Generierung                                    # folgenden
                                  # Schleifendurchlauf das letzte mal benötigt und danach über-
                                  # schrieben wird. Die am Ende vorliegenden Werte sind gerade
                                  # die, die bei der südwärtigen Phase benötigt werden.
  end do;
end do;
for j from floor(n/2) to 0 by -1 do
end do;
for i from n-1 to 2*n-3 do #entspricht der Anzahl der restlichen Diagonalen
  for k from i+1-n to floor((i-1)/2) do #berechnet die gemischten Momente innerhalb einer
                                        #Diagonalen, von oben rechts nach unten links.
  end do;
  if i mod 2 = 0 then #Ausrechnen eines fehlenden Koeffizienzen über die fünfgliedrige Rekursion                         #am Eintrag (k,k)
    k:= i/2;
  else                #Ausrechnen eines fehlenden Koeffizienzen über die fünfgliedrige Rekursion                         #am Eintrag (k,k-1)
  end if;
  for j from -1 to floor(n/2) do #Erneutes Durchrotieren der Werte der gemischten Momente
  end do;
end do;
a[2*n]:=a[n-1]-b[2*n]*s[0]/t[0]; #Berechnung des letzten fehlenden Koeffizienten über die                                           #fünfgliedrige Rekursion am Eintrag (n-1,n-1)

M:=Matrix(2*n+1, shape=symmetric);#definieren der werdenden Gauß-Krondrod-Matrix
for m from 2 to (2*n+1) do #generieren der Gauss-Kronrod-Matrix
  M(m-1,m):= sqrt(b[m-1]);
  M(m,m-1):= sqrt(b[m-1]);
  M(m,m):= a[m-1];
end do;
Knoten, Eigenvektoren := Eigenvectors(evalf[15](M));# "Die gesuchten Knoten sind die Eigenwerte #dieser Matrix, und die Gewichte sind proportional zu den ersten Komponenten der normalisierten #Eigenvektoren"


for m from 1 to 2*n+1 do
  AktuellerNormierterVektor:= Normalize(Column(Eigenvektoren,m),Euclidean);

end do;



end proc


An example of an application of the procedure is


GaußKronrodQuadraturKurz(-2, 1, 3*x*3*x^2*sin(x),x,3)


Thank you very much!



How to add a moving texte ( for example  notation of vector v and vector a ) to my animation ?


I am looking to have Maple run the algorithm I am given:

function Crunch (x∈ R)

if x ≥ 100 then

return x/100


return x+Crunch(10*x)

I am looking to be able to input 4 different numbers for x and have the algorithm execute. I understand how to arrive at the answers manually but as algorithms get more complex, I won't be able to. Thank you!

I want to find a weak form to Navier equations and obtain a solution formulation . I am interested in solving this problem using a finite element solver for which we need to introduce these equations in weak form. Can anyone help me in this regard? What is a higher-order continuity in the FEM approximation. This is challenge with the FEM which is based the Lagrange basis functions. To overcome the shortcomings, should we use the isogemetric analysis being based on the NURBS basis functions? How FEM cover this shortage for solving this type of equations?

Thanks in advance for your guidance.

HI all, I obtained a solution to a system solved by Maple but it seems not to be correct. And I couldn't detect where the mistake lies!

I find some book named automatic construction of graph for CAD

Then searched some paper related with it about geometric constraint solver

then find YouTube has AutoCAD can automatic design

Do maple has these things?

I have no idea in design, which input should be for these automatic design from nothing?

is there any mechanical knowledge as constraint by default in this automatic design?

We can write a list, set, MutableSet and array type together with their entries types. For example list(polynom) or array(array(integer)). But what about a table? For example how can I emphasize a type table with indices of the type integer and entries a list of integers? I was guessing table(integer,list(integer)) which is not working, so my guess is not correct. I tried some other combinations which they didn't work too. I can't see anything in the programming guide and the help of Maple or a post here that is addressing this question.


type(test,table(integer,list(integer))): # which of course is not working.



HazardRate seems not to work for a convolution - it always returns zero.

X1 := RandomVariable(Uniform(0, 1));
X2 := RandomVariable(Uniform(0, 1));
HazardRate(1/2*X1 + 1/2*X2, 0.4); #returns zero
plot(HazardRate(1/2*X1 + 1/2*X2, t), t = 0 .. 0.9, legend = "Hazard Rate"); #plots zero
#writing hazard rate explicitly works fine
plot(PDF(1/2*X1 + 1/2*X2, t)/(1 - CDF(1/2*X1 + 1/2*X2, t)), t = 0 .. 0.9, legend = "Hazard Rate explicitly");

Thank you!


Let $\triangle ABC$ be an equilateral triangle and $\angle BAC$=$20^\circ$, take point $D$ is on $AC$ such that $AD$ $=$ $BC$. Find $\angle ABD$= ?

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