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mehdi jafari

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11 years, 167 days
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These are replies submitted by mehdi jafari

maple 2018.1 windows 10 x64...

mehdi jafari 754
August 19 2018
0 0

@tomleslie thnx for the quick answer
the versions are maple 2018.1 windows 10 x64.
i push all the arrows, Pd Dn, Pg Up , ...
but nothing happens and i can not see other pages

maybe it is related to your version of w...

mehdi jafari 754
July 17 2018
0 0

i don't think maple has any options related to sound. maybe you want to check your windows setting for this.
how about trying this : open Control panel, on the search box of right top corner, write "sound" . open the sound option and go to sound tab. and change sound scheme to nothing. but it will mute all sounds related to windows events and programs. maybe it solves the problem.

fsolve...

mehdi jafari 754
July 13 2018
0 0

@samen 
 

> 

restart; Digits := 5; with(plots); with(LinearAlgebra)
> 

NULL
> 

a := 0; b := 1; N := 9; h := (b-a)/(N+1); k := 1/1000; phi := .5; K := 10^(-6); mu := 1.67; alpha := K/(phi*mu); lambda := alpha*k/h^2

0.11976e-6

 (1)
> 

NULL
> 

for i from 0 while i <= N do u[i, 0] := h*i+1 end do
> 

NULL
> 

for j from 0 while j < N+1 do u[0, j] := .1; u[N+1, j] := .5 end do
> 

NULL
> 

printlevel := 2; for i while i <= N do for j from 0 while j <= N do eq[i, j] := lambda*u[i-1, j]+(2-2*lambda)*u[i, j]+lambda*u[i+1, j] = -lambda*u[i-1, j+1]+(2+2*lambda)*u[i, j+1]-lambda*u[i+1, j+1] end do end do
> 

sys := ([seq])(seq(eq[i, j], j = 0 .. 9), i = 1 .. 9); indets(sys)
> 

fsolve(sys)

{u[0, 10] = u[0, 10], u[1, 1] = 1.1000, u[1, 2] = 1.1000, u[1, 3] = 1.1000, u[1, 4] = 1.1000, u[1, 5] = 1.1000, u[1, 6] = 1.1000, u[1, 7] = 1.1000, u[1, 8] = 1.1000, u[1, 9] = 1.1000, u[1, 10] = 1.1000+0.59880e-7*u[0, 10]+0.98982e-65*u[10, 10], u[2, 1] = 1.2000, u[2, 2] = 1.2000, u[2, 3] = 1.2000, u[2, 4] = 1.2000, u[2, 5] = 1.2000, u[2, 6] = 1.2000, u[2, 7] = 1.2000, u[2, 8] = 1.2000, u[2, 9] = 1.2000, u[2, 10] = 1.2000+0.35856e-14*u[0, 10]+0.16530e-57*u[10, 10], u[3, 1] = 1.3000, u[3, 2] = 1.3000, u[3, 3] = 1.3000, u[3, 4] = 1.3000, u[3, 5] = 1.3000, u[3, 6] = 1.3000, u[3, 7] = 1.3000, u[3, 8] = 1.3000, u[3, 9] = 1.3000, u[3, 10] = 1.3000+0.21470e-21*u[0, 10]+0.27605e-50*u[10, 10], u[4, 1] = 1.4000, u[4, 2] = 1.4000, u[4, 3] = 1.4000, u[4, 4] = 1.4000, u[4, 5] = 1.4000, u[4, 6] = 1.4000, u[4, 7] = 1.4000, u[4, 8] = 1.4000, u[4, 9] = 1.4000, u[4, 10] = 1.4000+0.12856e-28*u[0, 10]+0.46100e-43*u[10, 10], u[5, 1] = 1.5000, u[5, 2] = 1.5000, u[5, 3] = 1.5000, u[5, 4] = 1.5000, u[5, 5] = 1.5000, u[5, 6] = 1.5000, u[5, 7] = 1.5000, u[5, 8] = 1.5000, u[5, 9] = 1.5000, u[5, 10] = 1.5000+0.76980e-36*u[0, 10]+0.76988e-36*u[10, 10], u[6, 1] = 1.6000, u[6, 2] = 1.6000, u[6, 3] = 1.6000, u[6, 4] = 1.6000, u[6, 5] = 1.6000, u[6, 6] = 1.6000, u[6, 7] = 1.6000, u[6, 8] = 1.6000, u[6, 9] = 1.6000, u[6, 10] = 1.6000+0.46096e-43*u[0, 10]+0.12857e-28*u[10, 10], u[7, 1] = 1.7000, u[7, 2] = 1.7000, u[7, 3] = 1.7000, u[7, 4] = 1.7000, u[7, 5] = 1.7000, u[7, 6] = 1.7000, u[7, 7] = 1.7000, u[7, 8] = 1.7000, u[7, 9] = 1.7000, u[7, 10] = 1.7000+0.27602e-50*u[0, 10]+0.21471e-21*u[10, 10], u[8, 1] = 1.8000, u[8, 2] = 1.8000, u[8, 3] = 1.8000, u[8, 4] = 1.8000, u[8, 5] = 1.8000, u[8, 6] = 1.8000, u[8, 7] = 1.8000, u[8, 8] = 1.8000, u[8, 9] = 1.8000, u[8, 10] = 1.8000+0.16528e-57*u[0, 10]+0.35856e-14*u[10, 10], u[9, 1] = 1.9000, u[9, 2] = 1.9000, u[9, 3] = 1.9000, u[9, 4] = 1.9000, u[9, 5] = 1.9000, u[9, 6] = 1.9000, u[9, 7] = 1.9000, u[9, 8] = 1.9000, u[9, 9] = 1.9000, u[9, 10] = 1.9000+0.98970e-65*u[0, 10]+0.59880e-7*u[10, 10], u[10, 10] = u[10, 10]}

 (2)
> 

 


 

Download Crank_scheme.mw

a comment...

mehdi jafari 754
July 13 2018
0 0

as a comment, can the space between a cube with corners (-1,-1,-1,) , (1,1,1) , and this sphere be filled? or can the volumes be subtracted from each other in plotting? 

 

> 

restart
> 

A1:=plots:-implicitplot3d(x^2+y^2+z^2=1,x=-1..1,y=-1..1,z=-1..1,grid= [10$3], style= wireframe, scaling=constrained, axes=boxed, orientation=[25,60]):
> 

A2:=plots:-implicitplot3d(x^2+y^2+z^2=1,x=-1..1,y=-1..1,z=-1..1,grid= [10$3], style= patchnogrid, scaling=constrained, axes=boxed, orientation=[25,60]):
> 

P0:=plots:-display(A1,A2):
> 

A:=plots:-display(plottools:-cuboid([-1,-1,-1], [1,1,1]),scaling=constrained, colorscheme = ["xyzcoloring", proc (x, y, z) options operator, arrow; x+y-z^2 end proc], style = surface,style= wireframe,axes=none):
> 

plots:-display(A,P0)

 
> 

 

> 

 

> 

 

> 

 

> 

 

NULL

NULL


 

Download comment.mw

the problem is function f is complex !...

mehdi jafari 754
April 23 2018
0 0

@mehran rajabi actually i don't know maple can find maximium value of a complex function or not ! as far as i know , it can't !

suggestion...

mehdi jafari 754
April 23 2018
0 0

as a suggestion, your file name can be dependent to the loop numerator. so it can be done without over writing, but as @Carl Love said, is there need to store the jacobian every loop ?! 

Steady state...

mehdi jafari 754
April 23 2018
0 0

steady state is a state in which variables are independent of time and actucally the "state" of the system is "steady". thus if differential equations are available, the differentiation with respect to the time becomes zero. ( see: https://en.wikipedia.org/wiki/Steady_state). 
then i think you are maximizing your function with respect to the varibale "x" , and your function is complex, is that right?

tnx for the answer but......

mehdi jafari 754
April 01 2018
0 0

@Preben Alsholm @Kitonum 
tnx for the answer but as i already said, it is in the loop and the loop is updating the term f,
for example in the next loop f is as follows :


 

> 

restart
> 

f2:=-2*(sqrt(6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)*(-sigma[11]^2-3*sigma[12]^2-3/2*(beta[11]^2)+2*beta[11]*sigma[11]+6*beta[12]*sigma[12]-3*beta[12]^2)+E*delta_gamma*omega*(beta[11]^2+6*beta[12]^2-12*beta[12]*sigma[12]+6*sigma[12]^2))*(1/sqrt(6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2))*(1/(3*beta[11]^2-4*beta[11]*sigma[11]+6*beta[12]^2-12*beta[12]*sigma[12]+2*sigma[11]^2+6*sigma[12]^2))

-2*((6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)^(1/2)*(-sigma[11]^2-3*sigma[12]^2-(3/2)*beta[11]^2+2*beta[11]*sigma[11]+6*beta[12]*sigma[12]-3*beta[12]^2)+E*delta_gamma*omega*(beta[11]^2+6*beta[12]^2-12*beta[12]*sigma[12]+6*sigma[12]^2))/((6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)^(1/2)*(3*beta[11]^2-4*beta[11]*sigma[11]+6*beta[12]^2-12*beta[12]*sigma[12]+2*sigma[11]^2+6*sigma[12]^2))

 (1)
> 

 

> 

 

NULL


isn't there a direct substitution way?

Download problem2.mw

a comment...

mehdi jafari 754
February 02 2018
0 0

@minhhieuh2003 maybe u want to try this :

 

> 

restart:
> 

ans:=int((a*x+c)/(b*x+d), x);

a*x/b-ln(b*x+d)*a*d/b^2+ln(b*x+d)*c/b

 (1)
> 

eval(ans,{a=2,b=1,c=1,d=1,x=6})-eval(ans,{a=2,b=1,c=1,d=1,x=2})

8-ln(7)+ln(3)

 (2)
> 

 


now , if u change parameters, the answer will be change.
also maybe u want to try these commands : subs,eval,simplify...

Download parametric.mw

Tnx...

mehdi jafari 754
January 25 2018
0 0

@tomleslie thank you, you are great.

another qeustion...

mehdi jafari 754
January 25 2018
0 0

@tomleslie in order to use the results in matlab, the command 'double' is used ( in order to convert symbolic result to floating point number used in matlab), but when there is more than one unkown in the system, it returns some errors, what should i do ? could pls help? tnx

yes, they all work, tnx for your comment...

mehdi jafari 754
January 23 2018
0 0

@tomleslie yes, they all work, tnx for your comments, they were useful

what about introducing more than one var...

mehdi jafari 754
January 23 2018
0 0

@acer thnx for quick answer, it solved my problem, can i introduce more than one variable simultaneously?
for example sth like this :
setmaple('a,b',5,6) ? ( it returns errors)
or i should define them seprately ? 
with Regards, M.Jafari

tnx...

mehdi jafari 754
December 08 2017
0 0

@tomleslie tnx for your compelete answer

you are right,...

mehdi jafari 754
December 07 2017
0 0

@tomleslie sorry for the inconvenience, you are right, but a1 is not constant and is a function of time, it is int(a1(t)*A(t),t)...
actually a1=(t)->(wl/gamma1)*cos(w*t);
can it be solved?

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