MaplePrimes Questions

Dear friends:

I have a long expression having a/some RootOf(something..) inside it. 

Any way to get just that "something..", i.e., the argument of RootOf() from that expression?

 

Thank you very much.

César Lozada

 

 

Hi all,

 

Sorry if the question of the title wasn't very clear. I am not sure how to express it correctly.

But here is what I do for linux,

and I get

 

So when I submitted a lot of jobs, I can see the arguments that was passed to it. I can easily kill a process.

 

How do I do that in windows if anyone knows?

Here is a screenshot from windows task manager:

 

That does not give informative details, just like the 'top' command in linux.

 

Thanks,

How to make Decimal number in maple by default? 

How to make radian to degrees? Think It is like this "degcos" but I cant see the correct result, because it is not showed in decimals

Why do I have to active all my varibles when I open the document after I have saved it?

 

Regards

Østerbro

Does anyone have JavaViewLib running on Maple 18?  Could you help me out?  I can't get it running and I really need a way to export some 3d plots to a web page where a colleague can view them (rotate and such).

Thanks, Chad.

Hello everyone,

I'm trying to do some fitting using NonlinearFit, for the coefficients I know in advance, that they have to fulfill a condition (a+b>c+1). I couldn't find a way to make Maple take this condition into consideration while fitting my data. I tried to use Parameterrange to make the difference a+b-c+1 positive, this works for linear conditions like mine but leads to computational difficulties and errors like "no improved point could be found".

Thanks in advance,

Sören

Dear.All

 

I'm a beginner of Maple 13. 

 

In MATLAB, there is a command 'sigmaplot' which draw the 'SINGULAR VALUE OF SYSTEM H(jw)' over all frequency range. 

I want to obtain the function of the 'sigmaplot' graph about frequency variable 'w'.

 

so I defined matrix A,B,C,D in Maple13. 

 

and specify H like following

 

H:= Multiply(C,Multiply(Inversematrix(s*IdentityMatrix(8)-A),B))+D              

 

          --> It is to express H=C*((sI-A)^-1)*B+D that is the state-space matrix of frequency domain form.

 

and Maple 13 gave me a very long formular expressed by 's'.

 

Then I try to 'SingularValues' command 

 

but there came up 'Error Message : Error. (in content/polynom) general case of floats not handled'

 

I couldn't resolve this problem.... 

 

So I'm requesting your advice like this.

 

Could you give some advice ? 


Dear users,

In my attached file I have two PDES, (PDE1 and PDE2). PDE1 is a function of v(t) and w(x,t) and PDE2 is also a function of v(t) and w(x,t). I can solve PDE2 if I say v(t) is 1 for example and you can see the plot. But what if I put v(t) back in PDE2 and want to find v(t) and w(x,t) from PDE1 and PDE2 together? 

Many Thanks,

Baharm31

 

Define PDE Euler-Bernoulli Beam

 

NULL

restart:

Parametrs of piezoelectric and cantilever beam

 

``

Ys := 70*10^9: # Young's Modulus structure

Yp := 11.1*10^10: # Young's Modulus pieazo

ha := -0.00125: # Position

hb := 0.001: # Position

hc := 0.0015: # Position

d31 := -180*10^(-12): # Piezoelectric constant

b := 0.01: #Width of the beam

tb := 0.002:

epsilon33 := 15.92*10^(-9):

hp :=0.00025: # Position

hpc := 0.00125: # Position

YI := b*(Ys*(hb^3- ha^3)+Yp*(hc^3-hb^3))/3: # Bending stiffness of the composit cross section

cs := 0.564: # The equivqlent coefficient of strain rate damping

ca := 0: # Viscous air damping coefficient

Ibeam := (b * tb^3 )/12: # The equivalent moment of inertia

m := 0.101: # Mass of the structure

upsilon := - Yp*d31*b*(hc^2-hb^2)/(2*hp): # Coupling term

lb := 0.57:# Length of the structure (Cantilever Beam)

lp := 0.05:# Length of the Piezoelectric

R:= 10000: # Shunted resistor

Electrical circuit equation

 

PDE1:=(epsilon33 * b*lp / hp) * diff(v(t), t) + (v(t)/R)+ int(d31*Yp*hpc*b* diff(w(x, t),$(x, 2))*diff(w(x, t), t),x = 0..lp)=0;

0.3184000000e-7*(diff(v(t), t))+(1/10000)*v(t)+int(-0.2497500000e-3*(diff(diff(w(x, t), x), x))*(diff(w(x, t), t)), x = 0 .. 0.5e-1) = 0

(1.1.1.1)

``

 

PDE Equation

 

fn := 3.8:# Direct Excitation frequency;

wb(x,t) := 0.01*sin(fn*2*Pi*t):#Direct Excitation;

plot(wb(x,t),t = 0 .. 0.25*Pi,labels = [t,wb], labeldirections = ["horizontal", "vertical"], labelfont = ["HELVETICA", 15], linestyle = [longdash], axesfont = ["HELVETICA", "ROMAN", 10], legendstyle = [font = ["HELVETICA", 10], location = right],color = black);

 

 

FunctionAdvisor(definition, Dirac(n,x));

[Dirac(n, x) = (1/2)*(Int((I*_k1)^n*exp(I*_k1*x), _k1 = -infinity .. infinity))/Pi, `with no restrictions on `(n, x)]

(1.2.1)

 

PDE2 := YI*diff(w(x, t),$(x, 4))+ cs*Ibeam*diff(w(x, t),$(x, 4))*diff(w(x, t), t)+ ca* diff(w(x, t), t) + m * diff(w(x, t),$(t, 2))+ upsilon*v(t)*(Dirac(1,x) -Dirac(1,x-lp) ) =-m*diff(wb(x, t),$(t, 2))-ca*diff(wb(x, t), t);#PDE

1.567812500*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.3760000000e-11*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(diff(w(x, t), t))+.101*(diff(diff(w(x, t), t), t))+0.4995000000e-3*Dirac(1, x)-0.4995000000e-3*Dirac(1, x-0.5e-1) = 0.583376e-1*sin(7.6*Pi*t)*Pi^2

(1.2.2)

tmax := 0.3:

xmin := 0:

xmax := lb:

N := 20:#NUMBER OF NODE POINT

bc1 := dw(xmin, t) = 0:

bc2 := dw(xmax, t) = 0:

bc3 := w(xmin, t) = 0:

ic1 := wl(x, 0) = 0:

Maple's pdsolve command

 

 

 

bcs := { w(x,0)=0 , D[2](w)(x,0)=0 , w(0, t) = rhs(bc1), D[1](w)(0, t)= rhs(bc1), D[1,1](w)(lb,t) = rhs(bc2), D[1,1,1](w)(lb,t) = rhs(bc2)}; # Boundary conditions for PDE2.

{w(0, t) = 0, w(x, 0) = 0, (D[1](w))(0, t) = 0, (D[2](w))(x, 0) = 0, (D[1, 1](w))(.57, t) = 0, (D[1, 1, 1](w))(.57, t) = 0}

(2.1)

PDES := pdsolve(PDE2, bcs, numeric, time = t, range = 0 .. xmax, indepvars = [x, t], spacestep = (1/1000)*xmax, timestep = (1/1000)*tmax);

 

module () local INFO; export plot, plot3d, animate, value, settings; option `Copyright (c) 2001 by Waterloo Maple Inc. All rights reserved.`; end module

(2.2)

PDES:-plot3d(t = 0 .. tmax, x = 0 .. xmax, axes = boxed, orientation = [-120, 40], shading = zhue, transparency = 0.3);

 

 

NULL


Download Euler-Bernoulli_Beam-last_version.mw

 

I wnat to print the polar that contain a part of loops,but it always contains all loops.

how do I solve this preblem?

Any one can help me,please?

 

 

 

 

Dear Maple users

 

I have a question about applying pdsolve MAPLE for solving two dimensional heat equations:

My codes have been provided but it shows to me this error:

Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}

If kindly is possible, please help me in this case.

 

With kind regards,

Emran Tohidi.

 

> restart;
> with(plots);
print(??); # input placeholder
> with(PDEtools);
print(??); # input placeholder
> declare(u(x, y, t));
print(`output redirected...`); # input placeholder
                    u(x, y, t) will now be displayed as u
> S := 1/100; tR := 0 .. 1; xR := 0 .. 1; yR := 0 .. 1; NF := 30; NP := 100;
print(??); # input placeholder
> N := 3; L1 := [red, blue, green]; L2 := [0, 1/2, 1]; Ops := spacestep = S, timestep = S;
print(??); # input placeholder
> Op1 := frames = NF, numpoints = NP;
print(??); # input placeholder
> PDE1 := diff(u(x, y, t), t)-(diff(u(x, y, t), `$`(x, 2)))-(diff(u(x, y, t), `$`(y, 2))) = 0;
print(??); # input placeholder
> IC := {u(x, y, 0) = exp(x+y)}; BC := {u(0, y, t) = exp(2*t+y), u(1, y, t) = exp(2*t+y+1), u(x, 0, t) = exp(2*t+x), u(x, 1, t) = exp(2*t+x+1)};
print(??); # input placeholder
> Sol := pdsolve(PDE1, `union`(IC, BC), numeric, u(x, t), Ops);
Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}


A linear system with three variables has augmented matrix that is row-equivalent to the 
following matrix: 

k + 3 2 k − 4 = 3 
0 2 −9 = 5 
0 0 k^2 + k − 2 = k − 1 
Determine the values of k for which the system has: 
(a) exactly one solution, 
(b) infinitely many solutions, 
(c) no solutions.

(a) Write down the augmented matrix corresponding to the linear system: 
4x1 + 2x2 − 7x3 − 11x4 = 5 
2x1 + 1x2 − 2x3 − 4x4 = 2 
4x1 + 2x2 − 10x3 − 14x4 = 6 

(b) Use Gauss-Jordan elimination to reduce this matrix to reduced row-echelon form. 


(c) Solve the linear system.

I am having problems with the curly brackets in math mode. I am using a Danish keyboard, and since changing to Maple 17, the inline evaluation (usually [CTRL]+=) has been placed as [CTRL]+[ALT]+0 which - by Maple - is interpreted the same way as [ALT-GR]+0. Unfortunately [ALT-GR]+0 is the way to get the end-bracket for curly brackets, i.e. }

Therefore, I am not able to write } in math mode. Maple interprets my keystrokes as a wish to do inline evaluation when trying to write }.

Has anyone experienced this issue - and if so: Is there a way to fix it?

Is there a way to change the shortcut keys in Maple (e.g. make an alternative shortcut for inline evaluations)?

 

Hi.

I am new in Maple and I'm trying to get functions from system of equations.

Constants are defined in line 4 and equations are:

eq1 := E2 = fE2(1+(KaE2+Ca)/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsE2*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

eq2 := T = fT(1+KaT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

eq3 := DHT = fDHT(1+KaDHT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsDHT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

KsT = 0.10e11; KaT = 4.6*0.10e6; KsE2 = 3.14*0.10e10; KaE2 = 4.21*0.10e6; KsDHT = 3*0.10e6; KaDHT = 3.5*0.10e6;

fT, fE2 and fDHT are variables, not functions (i.e. fT is not f(T) ) and I am trying to get fT=f(E2,T,DHT,Ca,Cshbg), fE2=f(E2,T,DHT,Ca,Cshbg) and fDHT=f(E2,T,DHT,Ca,Cshbg).

When I type:

eliminate({eq1, eq2, eq3}, {fE2, fT, fDHT})

Maple gives me a blank field. No error, no other comment.

I have no idea where I'm making mistakes.

Any suggestion is appreciated.

 

Thanks in advance.

Could you tell me what the maple player is ?

How can I calculate the residue of a function with complex variable in maple ?

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