Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

can maple code saved into database for self modifying easily?

self modifying , metaprogramming , database, snapshot, blockchain for code integrity etc

For my own use, I am attempting to port Joe Riel’s glyph package for geometric algebra into a module more compatible with recent versions of Maple. To this end, I have been testing individual procedures extracted from the package into Maple 2016, both to understand the algorithms and to check for glitches caused by the code running in more current Maple 2016. The procedure for carrying out the exterior multiplication of blades does not seem to work reliably, and I haven’t the necessary knowledge of Maple language to determine whether this is due to an error on my part or a feature of Maple V that no longer works.  I have attached a worksheet,tablemultiplyexample.mw,  that includes the procedures necessary for the multiplication routine to work, but I can’t get any consistency in the results.  Can anyone advise me what is the problem?  

As I understand the routine, setup defines a anti-symmetric root blade table with an indexing function that precludes assignment to the table. Clifford blades are then represented as indexed variables using the root table. The process is as follows see worksheet for actual code):

initialize := proc ()
 global _e, tableroot;
tableroot := table(antisymmetric, blade);
tableroot[] := 1;
_e := tableroot;
end proc:
#The index function `blade` is as follows:

`index/blade` := proc (Indices, tableau)
 if nargs = 2 then if Indices = [] then 1
          else tableau[op(checkindices(Indices))] end if
elif Indices = [] then tableau[Indices[]] := 1
 else ERROR("cannot assign to a blade", Indices) end if
end proc;

#Exterior multiplication is performed by the following routine.
b_exteriorp := proc (u, v)
option remember;
 if  u = 1 or v = 1 then u*v
else _e[op(u), op(v)] end if
end proc:

As near as I understand, the procedure joins the lists representing the two input blade into a single list that is processed by the antisymmetric indexing function and outputs the indexes as the product blade. I don’t understand how the case of duplicate indexes (which should return 0) is supposed to be handled by the procedure.  What the procedure usually returns is simply the appended list of the two blades without modification by the indexing function.

Can anyone give me a hint about how to fix this procedure?

tablemultiplyexample.mw

Hi
sry guys i have a problem with my maple when i use my fsolve sometimes it gave me no solution but i khow we have a solution ill use simplify to gave it a simplified input but also no awenser is there any other command yo make my problem easier for maple and also is there anyway that i can force my cpu and memory to focus more on solving problems

and at last what is default method for fsolve ( i mean is it newton method or something else and can i change it )

tnQ for your attentions
thats my maple file problem.mw

 

How do I animate a matrixplot frame by frame so it displayes the highest and lowest value first and then sort of starts "relaxing" to the boundary conditions that will be set to 0? Thank you.

I want to plot contour for piecewise  function , here is my code

with(plots):
d1:=0.1:eta:=0.01:k:=2:L:=1:
mu:=0.01:delta1:=0.1:delta2:=0.2:
d1:=0.2:d2:=0.2:L1:=0.2: L2:=0.2:
h:=z->piecewise( z<=d1,    1,
                 z<=d1+L1,   1-(delta1/(2))*(1 + cos(2*(Pi/L1)*(z - d1 - L1/2))), 
                        z<=d1+L1+d2,  1 ,          
                    z<=d1+L1+d2+L2,  1-(delta2/(2))*(1 + cos(2*(Pi/L2)*(z - d1 - d2-L1-L2/2))),
                 z<=L,    1):
A1:=((k+2)*(k+3)*(1-mu)^(k+1)-2*(1-mu)^(k+2)*(k+2+mu))^(1/k):
A2:=(A1*r/h(z)^(1+3/k)):
contourplot(A2,z=0..1,r=-h(z)..h(z),colour=black,axes=boxed);

I have upgraded my monitor  to a 4K monitor. I know how to enlarge the fonts in work-sheets and thow to increase the icons size. Is it possible to  alsoincrease the font size in the side and horizontal bars? They are ridicously small. The zoom factor facility in Tools-> Options does noit work, May this is has to be done  by some guid setting. I w\ould appreciate any help.

Thank you

Fernando

I am attempting to write a series representation of a general integral of a function from a to b as follows:

int(f(x), x = a..b)= h*sum((c_k)*f(a+kh))+O(h^p),k=1..N;

where h:=(b-a)/(N+1), p(N) is greater than or equal to N + 1 and c_k are coefficients.  I then need to write procedures with Maple to evalue c_k from 1,..,N and also to evaluate P(N) for any N.  If I take the case for N = 3 and N = 6 I have to use those procedures to prove that:

int(f(x), x = a..b)=(4h/3)*(2*f_1 - f_2 +2*f_3) + O(h^5) = (7*h/1440)*(611*(f_1 + f_6) - 453*(f_2 + f_5) + 562*(f_3 + f_4)) + O(h^7) 

where f_k = f(a + kh).  I am really at a loss as to how to write this procedure, although I may have used something similar before:

P:=proc(p) add((1/k^(1/10))*(sin(1/k)-1/k), k=1..10^p) end proc;
seq( evalhf(P(p)), p = 1 .. 5 );
 

 



I am using MAPLE for quantum computation.
It appears to me that the Physics[Expand] and Physics[Symplify] functions do not operate as I expected on
Kets with multiple quantum numbers. The functions no not consider the non commutative nature of tensor
products of Kets. (neither of bras incidently). It would be very useful if thos could be adjusted in some way.

Thank you for your help

LL

restart;

with(Physics):

Setup(mathematicalnotation=true,
      noncommutativecolor=black,
      quantumoperators=q,
      noncommutativeprefix={q,psi,beta});

[mathematicalnotation = true, noncommutativecolor = black, noncommutativeprefix = {beta, psi, q}, quantumoperators = {q}]

(1)

Ket(psi):=Ket(q,-1)*Ket(q,1)*Ket(q,-1);
Ket(psi):=Ket(q,-1,1,-1);
'q[1].Ket(psi)'=q[1].Ket(psi);
'q[2].Ket(psi)'=q[2].Ket(psi);
'q[3].Ket(psi)'=q[3].Ket(psi);

Physics:-`*`(Physics:-Ket(q, -1), Physics:-Ket(q, 1), Physics:-Ket(q, -1))

 

Physics:-Ket(q, -1, 1, -1)

 

Physics:-`.`(q[1], Physics:-Ket(psi)) = -Physics:-Ket(q, -1, 1, -1)

 

Physics:-`.`(q[2], Physics:-Ket(psi)) = Physics:-Ket(q, -1, 1, -1)

 

Physics:-`.`(q[3], Physics:-Ket(psi)) = -Physics:-Ket(q, -1, 1, -1)

(2)

Ket(beta,-1,-1):=(1/2)*(sqrt(2)*(Ket(q,-1)*Ket(q,-1)));
Ket(beta,-1,-1):=(1/2)*(sqrt(2)*(Ket(q,-1,-1)));

(1/2)*2^(1/2)*Physics:-`*`(Physics:-Ket(q, -1), Physics:-Ket(q, -1))

 

(1/2)*2^(1/2)*Physics:-Ket(q, -1, -1)

(3)

Ket(Prod):='Ket(psi)'*'Ket(beta,-1,-1)';

Physics:-`*`(Physics:-Ket(psi), Physics:-Ket(beta, -1, -1))

(4)

Ket(Prod):=Ket(psi)*Ket(beta,-1,-1);
'Expand(Ket(Prod))'=Expand(Ket(Prod));
'Expand(Ket(Prod))'<>'(Ket(Prod))';

'Simplify(Ket(Prod))'=Simplify(Ket(Prod));
'Simplify(Ket(Prod))'<>'(Ket(Prod))'

(1/2)*2^(1/2)*Physics:-`*`(Physics:-Ket(q, -1, 1, -1), Physics:-Ket(q, -1, -1))

 

Physics:-Expand(Physics:-Ket(Prod)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Ket(q, -1, -1), Physics:-Ket(q, -1, 1, -1))

 

Physics:-Expand(Physics:-Ket(Prod)) <> Physics:-Ket(Prod)

 

Physics:-Simplify(Physics:-Ket(Prod)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Ket(q, -1, -1), Physics:-Ket(q, -1, 1, -1))

 

Physics:-Simplify(Physics:-Ket(Prod)) <> Physics:-Ket(Prod)

(5)

q[1]*q[2]-q[2]*q[1]<>0;
q[1]*q[3]-q[3]*q[1]<>0;
q[2]*q[3]-q[3]*q[2]<>0;

Physics:-`*`(q[1], q[2])-Physics:-`*`(q[2], q[1]) <> 0

 

Physics:-`*`(q[1], q[3])-Physics:-`*`(q[3], q[1]) <> 0

 

Physics:-`*`(q[2], q[3])-Physics:-`*`(q[3], q[2]) <> 0

(6)

 


 

Download Expand_Simplify-of-Kets.mw


 

 

 

Dear Friends
Is there a way to solve a complicated integration in less possible time?

Thanks

_________________________________________________________________________________
 

restart;
Digits := 100:
tm := time():
with(LinearAlgebra):

m := 6:
a := 0.1:
b := 10*a:
E := 1:
h := 1:
nu := 0.3:

w := (r-b)^2*(r-a)^2*add(add(W[n, i]*r^n*t^(i-n), n = 0 .. i), i = 0 .. m):
ur := -z*(diff(w, r)):
ut := -z*(diff(w, t))/r:
er := diff(ur, r)+(1/2)*(diff(w, r))^2:
et := ur/r+(diff(ut, t))/r+(diff(w, t))^2/(2*r^2):
grt := diff(ut, r)-ut/r+(diff(ur, t))/r+(diff(diff(w, t), r))/r:
u := -(1/2)*E*(2*er*et*nu+er^2+et^2)/(nu^2-1)+(1/2)*E*grt^2/(2*(1+nu)):

PI := int(int(int(u*r, z = -(1/2)*h .. (1/2)*h), t = 0 .. 2*Pi), r = a .. b)-0.5*P*(int(int(r*(diff(w, r))^2, r = a .. b), t = 0 .. 2*Pi)):

Time = time()-tm;

I am looking for a window bat file to run my maple input file.mw

Does anyone know how to calculate basic reproduction number using maple coding? Or by any chance, anyone know how to solve it by hand with this complicated equations?
 

restart

interface(imaginaryunit = j)

I

(1)

lambda := k*tau*(C*Upsilon+I)/N

k*tau*(C*Upsilon+I)/N

(2)

eqn1 := (1-p)*Pi+phi*V+delta*R-(mu+lambda+`&vartheta;`)*S

(1-p)*Pi+phi*V+delta*R-(mu+k*tau*(C*Upsilon+I)/N+vartheta)*S

(3)

eqn2 := p*Pi+`&vartheta;`*S-(lambda*`&epsilon;`+mu+phi)*V

p*Pi+vartheta*S-(epsilon*k*tau*(C*Upsilon+I)/N+mu+phi)*V

(4)

eqn3 := rho*lambda*S+rho*`&epsilon;`*lambda*V+I*(1-q)*eta-(mu+beta+chi)*C

rho*k*tau*(C*Upsilon+I)*S/N+rho*epsilon*k*tau*(C*Upsilon+I)*V/N+(1-q)*eta*I-(mu+beta+chi)*C

(5)

eqn4 := (1-rho)*lambda*S+(1-rho)*`&epsilon;`*lambda*V+chi*C-I*(mu+alpha+eta)

(1-rho)*k*tau*(C*Upsilon+I)*S/N+(1-rho)*epsilon*k*tau*(C*Upsilon+I)*V/N+chi*C-(mu+alpha+eta)*I

(6)

eqn5 := beta*C+I*q*eta-(mu+delta)*R

beta*C+q*eta*I-(mu+delta)*R

(7)

``


 

Download Equation_for_basic_reproduction_number.mwEquation_for_basic_reproduction_number.mw

I am using the Physics package for quantum mechanic.

Ket product are supposed to be noncommutative and the Simplify function
appears to ignore the propety.

I must be doing someting wrong.

Thank you for your help

LL

 

Please, where are the statistical tolerance intervals in Maple, similar to those we have in MiniTab and in R ?  maybe in statistical quality control ?

Example 1 :  Package « tolerance » in R (https://cran.r-project.org/web/packages/tolerance/tolerance.pdf) : 

 

« Description :  Statistical tolerance limits provide the limits between which we can expect to find a specified proportion of a sampled population with a given level of confidence. This package provides functions for estimating tolerance limits (intervals) for various univariate distributions (binomial, Cauchy, discrete Pareto, exponential, two-parameter exponential, extreme value, hypergeometric, Laplace, logistic, negative binomial, negative hypergeometric, normal, Pareto, Poisson-Lindley, Poisson, uniform, and Zipf-Mandelbrot), Bayesian normal tolerance limits, multivariate normal tolerance regions, nonparametric tolerance intervals, tolerance bands for regression settings (linear regression, nonlinear regression, nonparametric regression, and multivariate regression), and analysis of variance tolerance intervals. Visualizations are also available for most of these settings. »

 

Example 2 : 

https://support.minitab.com/en-us/minitab/18/help-and-how-to/quality-and-process-improvement/quality-tools/how-to/tolerance-intervals-normal-distribution/methods-and-formulas/methods-and-formulas/

https://support.minitab.com/fr-fr/minitab/18/help-and-how-to/quality-and-process-improvement/quality-tools/how-to/tolerance-intervals-nonnormal-distribution/methods-and-formulas/tolerance-intervals/   

Hi, i want to investigate  chaos for the problem , cantilever beam under random narro band excitation, but the code has errors .the code is this:

      restart:with(plots):      h:=1: Omega:=(0..376):alpha1:=617.2:alpha2:=1.02*10^(8): c:=.002:k:=18.4:  step:=0.1:imax:=376:  for i from 0 to imax do;  Omega[i]:=i*step:   f:=evalf(solve({((-a*Omega[i]^(2)+alpha1*a+3/(4)*alpha2*a^(3)+1/(4)*k*Omega[i]^(2)*a^(3)-(3)/(4)*k*Omega[i]^(2)*a^(3))^(2)+(c*Omega[i]*a^())^(2))=h^(2),a>0}));  ff[i]:=((rhs(f[1]))^(2))/(2):  end do:   l1:=[[Omega[n],ff[n]] $n=0..imax]:  p1:=plot(l1, x=0..3,y=0..1,  style=point,symbol=solidcircle,symbolsize=4,color=red):    jmax:=914: f1:=array(377..914):f2:=array(377..914):f3:=array(377..914):Omega1:=array(377..914):  for j from 377to jmax do;  Omega1[j]:=j*step:   fff:=evalf(solve({((-a*Omega1[j]^(2)+alpha1*a+3/(4)*alpha2*a^(3)+1/(4)*k*Omega1[j]^(2)*a^(3)-(3)/(4)*k*Omega1[j]^(2)*a^(3))^(2)+(c*Omega1[j]*a^())^(2))=h^(2),a>0}));  f1[j]:=((rhs(fff[1,1]))^(2))/(2):f2[j]:=((rhs(fff[2,1]))^(2))/(2):f3[j]:=((rhs(fff[3,1]))^(2))/(2):  end do:   ll1:=[[Omega1[n],f1[n]] $n=377..jmax]:  pp1:=plot(ll1, x=0..10,y=0..1,  style=point,symbol=solidcircle,symbolsize=4,color=red):    ll2:=[[Omega1[n],f2[n]] $n=377..jmax]:  pp2:=plot(ll2, x=0..10,y=0..1,  style=point,symbol=solidcircle,symbolsize=4,color=red):    ll3:=[[Omega1[n],f3[n]] $n=377..jmax]:  pp3:=plot(ll3, x=0..15,y=0..1,  style=point,symbol=solidcircle,symbolsize=4,color=red):       plot({  seq(seq(p1), seq(seq(pp1),seq(seq(pp2),seq(seq(pp3))  },style=point,title=`Pitchfork Diagram`);  Thanks for your help

We conjecture that the polynomial h(n) = n^2 + n + 41 is prime for an infinite number of values n.
We furthur conjecture that p(n) = n^2 + 1 is prime an infinite number of times.

I have shown that the set (x,y) with h(y) mod x is congruent to 0 can be written down.  It is p(x,y).  p(x,y) is the set of all divisors of h(n).  See

https://sites.google.com/site/primeproducingpolynomial/

landau.mw

Regards,

Matt

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