Advanced Search

mehdi jafari

769 Reputation

13 Badges

11 years, 237 days
Contact mehdi jafari

MaplePrimes Activity


These are replies submitted by mehdi jafari

how these commands can be used?...

mehdi jafari 769
December 02 2024
0 0

@Carl Love could please introduce the syntax of using this commands?

exactly...

mehdi jafari 769
November 07 2024
0 0

@Alfred_F thank you, i solved the problem as you commented

what is the relation between left and ri...

mehdi jafari 769
March 12 2024
0 0

@vv what is the relation between left and right eigenvectors? should multiplication of them give the IdentityMatrix? but in maple i cannot achive this? how can i find diagonal form of the matrix using these eigenvectors? thank you for the help.

exactly...

mehdi jafari 769
December 20 2023
0 0

@mmcdara as u mentioned above, what is the idea for solving my problem? what should i use ?

some points...

mehdi jafari 769
October 16 2022
0 0

@mmcdara @Carl Love @acer 

thank you for the responses. there are some points which i explain in the following:
1. @mmcdara when you do not set any method for the optimization, it means that maple itself will search for the proper method according to the help. only when the method is set modifiednewton maple can find an answer and for other methods such as sqp, maple can not find an answer and we got divison by zero error. 
2. is there any reason that you choosed those ranges for "x"s;(i.e  7..10, 5..10, 1..10). since for other ranges we can not reach an answer except with modifiednewton method.
3. dear @acer , actually we do NOT have any atitude on the integration's value, but on the modified code by @mmcdara the result is 2.204832770*10^6 which is positive. Could you please explain me why those values are wrong and what values for the integration are expected? actually i want to use your inforamtion for deeper understanding of the problem for further problems. 


 

> 

restart:with(LinearAlgebra):
> 

N:=3:
> 

m:=Vector([ 1 , log(x+b3) , b2/(x+b3) ]):
> 

A:=m.m^+:
> 

for i to N do
m||i:=eval(A,[x=x||i]);
od:
> 

M:=add(w||i*m||i,i=1..N-1)+(1-add(w||i,i=1..N-1))*m||N:
> 

MM:=( LinearAlgebra:-Trace(MatrixInverse(M)) ):
> 

IF1 := proc(__w1, __w2, __x1, __x2, __x3)
  local J:
  J := evalf(
         Int(
           eval(MM, [w1=__w1, w2=__w2, x1=__x1, x2=__x2, x3=__x3])
           , [b2=1..2,b3=1..2]
           , method = _d01ajc
           , epsilon=0.001
         )
       ):
   #print([_passed], J);
  return J:
end proc:
> 

# check

IF1(0.6, .1, 8, 7, 5);

2204832.770

 (1)
> 

s:= Optimization:-NLPSolve(
  IF1
  , 0..1, 0..1, 1..10, 1..10, 1..10
  , initialpoint=[0.6, .1, 8, 7, 5]
  , maximize=false
  , method=modifiednewton
)

Warning, convergence is not assured; examine Hessian values, or consider raising tolerance

  

[1077.30026359427006, Vector(5, {(1) = .2390983436806223, (2) = .5071151402152273, (3) = 10.0, (4) = 3.400128412709433, (5) = 1.0})]

 (2)
> 

 

 

 
> 

 

> 

 


 

LinearLog-A-Bayesian_1_mmcdara_moodified_newton.mw

LinearLog-A-Bayesian_mmcdara3_sqp_method.mw


thank you for your patience and attention.
sincerely yours.

some points...

mehdi jafari 769
October 15 2022
0 0

@acer Actually, we are looking for a design for the model which finds minimum for the Information Matrix. The design points are x1,x2 and x3 and the weights are w1,w2 and 1-(w1+w2).
The information matrix is M. 
According to the Beysian approach, ranges of b1 and b2 are considered. 
we are looking for a global optimum; although in convex optimization problems, a locally optimal solution is also globally optimal.
The ranges for "w"s are necessary, but for "x"s we can consider identical ranges for them.
we thought that the integration is complicated so that we used this method, but if another simple way can be adopted, we gratefully will use that. 
Thanks for your attention and help.

yes, use restart...

mehdi jafari 769
October 14 2022
0 0

@Miguel_Vega as you mentioned, just supress that line using # in front of the command and just restart the code.

actually you are right, but ......

mehdi jafari 769
August 03 2022
0 0

@tomleslie actually you are right, but the problem is the same pde with replacing z^2 with z, can be solved with these boundary/initial conditions. thus the analytical solution can be found, but maple can not do it for this problem.
thank you for your answer.

 

> 

restart
> 

pde__1 := diff(z(x, t), `$`(t, 2))-(diff(z(x, t), `$`(x, 2)))+z(x, t)^2 = 6*x*t*(-t^2+x^2)+x^6*t^6

diff(diff(z(x, t), t), t)-(diff(diff(z(x, t), x), x))+z(x, t)^2 = 6*x*t*(-t^2+x^2)+x^6*t^6

 (1)
> 

conds__1 := z(x, 0) = 0, z(0, t) = 0, z(1, t) = t^3
> 

pdsolve([pde__1, conds__1])
> 

pde__2 := diff(z(x, t), `$`(t, 2))-(diff(z(x, t), `$`(x, 2)))+z(x, t) = 6*x*t*(-t^2+x^2)+x^6*t^6

diff(diff(z(x, t), t), t)-(diff(diff(z(x, t), x), x))+z(x, t) = 6*x*t*(-t^2+x^2)+x^6*t^6

 (2)
> 

conds__2 := z(x, 0) = 0, z(0, t) = 0, z(1, t) = t^3
> 

pdsolve([pde__2, conds__2], z(x, t))

z(x, t) = -(Sum(12*sin(n*Pi*x)*((1/12)*n^4*(Pi^2*n^2+1)*Pi^4*((-12*Pi^2*n^2+72)*(-1)^n*(Pi^2*n^2+1)^(1/2)+_C1[n]*n^3*Pi^3*(Pi^2*n^2+1)^2)*cos((Pi^2*n^2+1)^(1/2)*t)+(-86400+(-120*Pi^6*n^6+3600*Pi^4*n^4-43200*Pi^2*n^2+86400)*(-1)^n)*sin((Pi^2*n^2+1)^(1/2)*t)+(((Pi^6*n^6-30*Pi^4*n^4+360*Pi^2*n^2-720)*(Pi^2*n^2+1)^2*t^5+(-20*Pi^8*n^8+580*Pi^6*n^6-6600*Pi^4*n^4+7200*Pi^2*n^2+14400)*t^3-(7/2)*Pi^6*n^6*(Pi^2*n^2+1)^2*t^2+(120*Pi^6*n^6-3600*Pi^4*n^4+43200*Pi^2*n^2-86400)*t+n^8*Pi^8-5*Pi^6*n^6-6*Pi^4*n^4)*(-1)^n+720*(120+(Pi^2*n^2+1)^2*t^4+(-20*Pi^2*n^2-20)*t^2)*t)*(Pi^2*n^2+1)^(1/2)-(1/12)*_C1[n]*n^7*Pi^7*(Pi^2*n^2+1)^3)/((Pi^2*n^2+1)^(7/2)*n^7*Pi^7), n = 1 .. infinity))+t^3*x

 (3)
> 

NULL


 

Download pde_(2).mw

right...

mehdi jafari 769
July 16 2022
0 0

@tomleslie thank you for your explanation. i think i have made a mistake, tnx alot.

there are some problems...

mehdi jafari 769
April 02 2022
0 0

@Carl Love thank you for your answer, but therr is a problem. 

The polynomials are correct  but betas are not correct and are different from the betas in equations above. 

Are the betas in this procedure identical with the ones in the equation?

 

is seq[reduce= `+`] equivalent to `+`(se...

mehdi jafari 769
March 29 2022
0 0

@Carl Love is seq[reduce= `+`] equivalent to `+`(seq) ? it seems the latter is more efficient. Am i wrong?
on my question on page https://www.mapleprimes.com/questions/233919-Repated-Summetion, i reffered here. but i couldn't find out how to code nested loops. are nested loops explaind here?

> 

restart:
Iterator:-CartesianProduct([1,1]): #Force compilation

P:= 2..4: #suffixes of m and c variables
N:= [$0..4]: #evaluation values of m variables

B:= subs(
    {_C= [$P], _V= [c||P]},
    proc(M)
    local r:= add(M*~_C), s:= 1+r, t:= s-add(M);
        r!*c0^t*mul(_V^~M)/t!/mul(M!~)/c1^s
    end proc
):
CodeTools:-Usage(
    seq[reduce= `+`](B(v), v= Iterator:-CartesianProduct(N$(rhs(P)-1)))
):

memory used=0.69MiB, alloc change=0 bytes, cpu time=16.00ms, real time=14.00ms, gc time=0ns

  
> 

restart:
Iterator:-CartesianProduct([1,1]): #Force compilation

P:= 2..4: #suffixes of m and c variables
N:= [$0..4]: #evaluation values of m variables

B:= subs(
    {_C= [$P], _V= [c||P]},
    proc(M)
    local r:= add(M*~_C), s:= 1+r, t:= s-add(M);
        r!*c0^t*mul(_V^~M)/t!/mul(M!~)/c1^s
    end proc
):
CodeTools:-Usage(
    `+`(seq(B(v), v= Iterator:-CartesianProduct(N$(rhs(P)-1))))
):

memory used=395.63KiB, alloc change=0 bytes, cpu time=0ns, real time=7.00ms, gc time=0ns

  
> 

 

Download seq.mw

yes you are right...

mehdi jafari 769
March 28 2022
0 0

@Carl Love yes as you said,  t[j] and t[i] are symbolic variables independent of t .the definiton of t[j] is as follows:

which if not of importance to use it here. i just want to know whether these summations and product can be coded?
tnx for the help

thank you...

mehdi jafari 769
September 09 2021
0 0

@mmcdara thank you for you comprehensive explanations. interesting code it was.

equation 1 in terms of variables other t...

mehdi jafari 769
September 08 2021
0 0

@tomleslie tnx for your attention. 

Actually i want equation one in terms of other functions and variables and differential opertors so that equation 1 doesn't contain the function psi(x,t).

Actually if i could write equation 2 in terms differential operators and then factor psi(x,t) (sth like isolating psi(x,t) in terms of other functions), i could replace it in equation 1 so that equation 1 does NOT contain any psi(x,t).

is it clear now?

can it be done by factoring differential operators and NOT directly using dsolve?

Both qualitative and quantitative insigh...

mehdi jafari 769
May 17 2021
0 0

@acer Thank you for clarifying explanations, actually we want to compare the results both qualitatively and quantitatively with the results of an article. But when mu=0 or mu=1 , there is an unwanted pick in the plot a shown blow:

For mu=8, qualitative asnwer is correct.

Here is the article:

2_5348170783959026711.pdf

1 2 3 4 5 6 7 Last Page 1 of 23