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mehdi jafari

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you are vrey welcome...

mehdi jafari 769
April 08 2014
0 0

@lemelinm u can also try this, which is so clearer than the maple defualt .
go to tools,options, go to display tab, change input display to maple notation, i think it is some what more clear . good luck

why u use matlab ?!...

mehdi jafari 769
April 08 2014
0 0

@sharena2 i think u can do whatever u want with maple ! 

yess !...

mehdi jafari 769
April 08 2014
0 10

@Markiyan Hirnyk  actually i can edit  ! by i can not delet ! :D lol

i can not delete my answer !...

mehdi jafari 769
April 08 2014
0 0

there is no deletion option !

you are very welcome...

mehdi jafari 769
April 08 2014
0 1

@sharena2 good luck

on the link provieded...

mehdi jafari 769
April 07 2014
0 0

@Carl Love Robert isreal has said:

Note:  This code is experimental, and is not guaranteed to work.  
# In particular, "allsolve" uses "evalr" to do interval arithmetic, and is
# therefore subject to the weaknesses of "evalr".  Among the bugs and
# weaknesses I know about:
# 1) It doesn't work with "GAMMA".
# 2) It doesn't work with the two-variable version of "arctan".

at the end of the help page related to ?...

mehdi jafari 769
April 07 2014
0 0

@Carl Love at the end of this help page there is :

See Also
evalr, operators/precedence, ScientificErrorAnalysis, Units

actually the first related argument to the topic is this according to maple help.

according to maple help page :

evalr - evaluate an expression using range arithmetic

and i actually guess evalr can be uesd  do u think it is incorrect ?.

Reboert Isreal has some topic about this at 
http://www.math.ubc.ca/~israel/advisor/advisor4/allsolve.txt

i emailed him to help us here, i do not know whether he has time or not.

actually purpose is this page,...

mehdi jafari 769
April 07 2014
0 0

@Preben Alsholm 
http://www.mapleprimes.com/questions/201301-Factory-Simulation-Question

none of your uploaded worksheet is downl...

mehdi jafari 769
April 07 2014
0 0

@sharena2  none of your uploaded worksheet is downlodable , please uplaod again and make sure that your file is downloadable . tnx

please upload your worksheet,...

mehdi jafari 769
April 07 2014
0 0

please upload your worksheet here . tnx

yes...

mehdi jafari 769
April 06 2014
0 0

@Aroob2012 PDE and BCs is just name i have assing your pde equation and your boundary conditions to them,respectively. u can use pdsolve so solve your pde equation, for more information, u should see ?pdsolve 

pdsolve - find solutions for partial differential equations (PDEs) and systems of PDEs

i upload two pdfs of maple help page here, but please see them yourself on the maple help. good luck!

pdsolve,numeric_(pds.pdf

pdsolve.pdf

yes...

mehdi jafari 769
April 05 2014
0 0

@Carl Love yes Sir,you are right, i thought iindex notation is better, and as you said,i shoud say about that. thank u for your remark.

use eval...

mehdi jafari 769
April 05 2014
0 0

the file is not uploading correctly, u can do like this ( at the end of your code ) :

x(t):=add(a[i]*f[i](t),i=1..numelems(C));u(t):=add(b[i]*f[i](t),i=1..numelems(C));

eval(x(t),op(2,MM));eval(u(t),op(2,MM));


> 

restart:
with(Optimization):
with(LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= DiagonalMatrix(P20):
P3:= DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Matrix(%id = 18446744073930868126)

Matrix(%id = 18446744073930869806)

Matrix(%id = 18446744073934930878)

Matrix(%id = 18446744073934931118)

Vector[row](%id = 18446744073934931478)

Vector[row](%id = 18446744073934931598)

Vector[row](%id = 18446744073934931718)

Vector[row](%id = 18446744073934931838)

Vector[row](%id = 18446744073934934014)

 (1)
> 

J:=X.E.Transpose(X)+U.E.Transpose(U);

a[1]^2+(1/2)*a[2]^2+(1/2)*a[3]^2+(1/2)*a[4]^2+(1/2)*a[5]^2+b[1]^2+(1/2)*b[2]^2+(1/2)*b[3]^2+(1/2)*b[4]^2+(1/2)*b[5]^2

 (2)
> 

indets(J); indets(C)

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

 (3)
> 

MM:=Minimize(J,{seq(C[i]=0,i=1..numelems(C))});

[1.52587355672056, [a[1] = HFloat(1.1444051675322553), a[2] = HFloat(0.07357433785547854), a[3] = HFloat(0.01908371625198537), a[4] = HFloat(-0.23114059946679755), a[5] = HFloat(-0.11990652558360362), b[1] = HFloat(-0.3814683891883076), b[2] = HFloat(1.0769416608491511e-11), b[3] = HFloat(-2.602282105079734e-12), b[4] = HFloat(-0.23114059939913112), b[5] = HFloat(-0.11990652561630648)]]

 (4)
> 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):
> 

 

> 

op(2,MM);

[a[1] = HFloat(1.1444051675322553), a[2] = HFloat(0.07357433785547854), a[3] = HFloat(0.01908371625198537), a[4] = HFloat(-0.23114059946679755), a[5] = HFloat(-0.11990652558360362), b[1] = HFloat(-0.3814683891883076), b[2] = HFloat(1.0769416608491511e-11), b[3] = HFloat(-2.602282105079734e-12), b[4] = HFloat(-0.23114059939913112), b[5] = HFloat(-0.11990652561630648)]

 (5)
> 

x(t):=add(a[i]*f[i](t),i=1..numelems(C));u(t):=add(b[i]*f[i](t),i=1..numelems(C));

a[1]*f[1](t)+a[2]*f[2](t)+a[3]*f[3](t)+a[4]*f[4](t)+a[5]*f[5](t)

b[1]*f[1](t)+b[2]*f[2](t)+b[3]*f[3](t)+b[4]*f[4](t)+b[5]*f[5](t)

 (6)
> 

eval(x(t),op(2,MM));eval(u(t),op(2,MM));

HFloat(1.1444051675322553)*f[1](t)+HFloat(0.07357433785547854)*f[2](t)+HFloat(0.01908371625198537)*f[3](t)-HFloat(0.23114059946679755)*f[4](t)-HFloat(0.11990652558360362)*f[5](t)

-HFloat(0.3814683891883076)*f[1](t)+HFloat(1.0769416608491511e-11)*f[2](t)-HFloat(2.602282105079734e-12)*f[3](t)-HFloat(0.23114059939913112)*f[4](t)-HFloat(0.11990652561630648)*f[5](t)

 (7)
> 

 

good luck!

Download solved.mws



@mahmood180 

tnx for your remark...

mehdi jafari 769
April 04 2014
0 0

@Carl Love i thought nops(c) has returned 5 not 3 . tnx for your remark.
Mr. Dadkhah you are very welcome. as my friend Carl love says your corrected form of answer is :

> 

restart:
with(Optimization):
with(LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= DiagonalMatrix(P20):
P3:= DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Matrix(%id = 18446744073931261342)

Matrix(%id = 18446744073931263022)

Matrix(%id = 18446744073935324094)

Matrix(%id = 18446744073935324334)

Vector[row](%id = 18446744073935324694)

Vector[row](%id = 18446744073935324814)

Vector[row](%id = 18446744073935324934)

Vector[row](%id = 18446744073935325054)

Vector[row](%id = 18446744073935327230)

 (1)
> 

J:=X.E.Transpose(X)+U.E.Transpose(U);

a[1]^2+(1/2)*a[2]^2+(1/2)*a[3]^2+(1/2)*a[4]^2+(1/2)*a[5]^2+b[1]^2+(1/2)*b[2]^2+(1/2)*b[3]^2+(1/2)*b[4]^2+(1/2)*b[5]^2

 (2)
> 

indets(J); indets(C)

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

 (3)
> 

Minimize(J,{seq(C[i]=0,i=1..numelems(C))});

[1.52587355672056, [a[1] = HFloat(1.1444051675322553), a[2] = HFloat(0.07357433785547854), a[3] = HFloat(0.01908371625198537), a[4] = HFloat(-0.23114059946679755), a[5] = HFloat(-0.11990652558360362), b[1] = HFloat(-0.3814683891883076), b[2] = HFloat(1.0769416608491511e-11), b[3] = HFloat(-2.602282105079734e-12), b[4] = HFloat(-0.23114059939913112), b[5] = HFloat(-0.11990652561630648)]]

 (4)
> 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):
> 

 

> 

 

> 

 

> 

 

> 

 



Download solved.mws




could please write down the matrix inste...

mehdi jafari 769
April 04 2014
0 0

@upaska could please write down the matrix instead of image ? maybe you need to see ?Eigenvalue. i guess your problem is with datatype = float option.

as maple help page :

The Eigenvalues(A) command solves the simple eigenvalue problem by returning the eigenvalues of Matrix A in a column Vector.
In the simple floating-point eigenvalue problem, if A has either the symmetric or the hermitian indexing function then the returned object has float[8] or sfloat datatype. Otherwise the returned object has complex[8] or  complex(sfloat) datatype.

 

If the implicit option (imp) is included in the calling sequence as just the symbol implicit or in the form implicit=true, then the eigenvalues are expressed by using Maple's RootOf notation for algebraic extensions or by expressing the eigenvalues in terms of exact radicals (if possible).

if you write down your matrix here, we can try possible options which can solve your problem.


i guess set the option implicit=true will solve the problem,but not sure of,good luck.

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