MaplePrimes Questions

What Maple15 commands will display a Rouleaux tetrahedron or a Meissner tetrahedron?

Hi all,

I'm having a challenge solving an integration that has some error functions associated with it.

Description:

I am trying to find the Riemann midpoint of the integral. All the parameters are known except the u and du terms which came into the expression following a laplace transform.

the output I get from the above input is shown below, and as it can be seen, the u and du terms coupled with a complimentary error function still exist. I have tried the "simplify" function and it returns similar output expression.

 

Any help will really be appreciated.

Thanks :)

 

 

 


plot(x^2, x = -2 .. 2)

Error, (in plot) incorrect first argument x^2

 

plot(2);

 

solve(x-1 = 0, x)

Error, (in coulditbe) invalid input: `coulditbe/internal` uses a 1st argument, obj, which is missing

 

``


Download Maple_Worksheet.mw

Hi,

I'm using Maple 18 and none of my worksheets that I've developed on earlier versions work on it. I tried starting a new worksheet with a very simple command (attached) and even that's giving me problems. As you can see, I have no trouble plotting constant functions, but as soon as I put in a variable it breaks down. It won't solve equations either.

FYI I'm using an HP G72 laptop with Windows 7 and no installation issues came up.

Can anyone tell me what's going on here?

Thanks,

Tom

Hi all,

I'm trying to use logplot but it's very time consuming. Any suggestion please!

Cheers!

restart:with(plots):

S1:=x+x^2*Sum(t^(k*alpha+1)/GAMMA(k*alpha+2),k=0..infinity);

S2:=x+x^2*Sum(t^(k*alpha+1)/GAMMA(k*alpha+2),k=0..n);

alpha:=0.75:x:=0.5:n:=10:

logplot(abs(S1-S2),t=0..1);

 

 

So here is the issue: I have a 50 by 50 tridiagonal matrix. The entries in the first row, first column are -i*x and the last row last column is -i*x; these are along the main diagonal, where i is complex and x is a variable. Everything in between these two entries is 0. Above and below the main diagonal the entries are -1. My issue is that I have to find a conditon on x that makes the eigenvalues real. I am completely new to maple and have no programming experience.. Can someone show me how to this?

Using Maple 18, I solved for minimum and maximum price. Instead of using fsolve I wanna use procedure programming structure in order to get the same results. How can I do it?

min_sol := fsolve([bc_cond, slope_cond, x[G, 1] = w[aggr, 1]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_min := subs(min_sol, p); max_sol := fsolve([bc_cond, slope_cond, x[G, 2] = w[aggr, 2]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_max := subs(max_sol, p);
{p = 0.3857139820, x[G, 1] = 127.8000000, x[G, 2] = 38.99045418}
0.3857139820
{p = 0.8841007104, x[G, 1] = 44.30160890, x[G, 2] = 164.2000000}
0.8841007104

Hi,

I am trying to realize the following calculation in Maple.

$
  \left[\sum_{i=0}^n y_i(x) \partial_x^i , \sum_{j=0}^m z_j(x) \partial_x^j \right]  \\
=   \sum_{i=0}^n \sum_{j=0}^m \sum_{l=0}^i  \binom il y_i(x) \left( \partial_x^{i-l} z_j(x)\right) \partial_x^{l+j} \\
- \sum_{j=0}^m \sum_{i=0}^n \sum_{l=0}^j  \binom jl z_j(x) \left( \partial_x^{j-l} z_i(x)\right) \partial_x^{l+i} \ .

$

 

Is there a way to make maple understand d/dx as a differential opperator and calculate with it? When i for example try to calculate diff(d/dx, x) it should give me d^2/dx^2 as a result. Unfortunately i don't know how to realize this.

Basic problem is i don't know how to realize operator expressions in maple like for example:

f(x) d/dx      ( f(x) is a smooth function of x here )

where when applied to a function h(x) it should result in f(x) d/dx h(x) .

 

Is that possible?

 

Thank you very much in advance.

I am trying to expand out the terms  of equation 13.  The expand command causes the lhs to be zero?


Initialize the metric and tetrad

 

restart; with(Physics); with(Tetrads); with(PDETools)

0, "%1 is not a command in the %2 package", Tetrads, Physics

(1.1)

X = [zetabar, zeta, v, u]

X = [zetabar, zeta, v, u]

(1.2)

ds2 := Physics:-`*`(Physics:-`*`(2, dzeta), dzetabar)+Physics:-`*`(Physics:-`*`(2, du), dv)+Physics:-`*`(Physics:-`*`(2, H(zetabar, zeta, v, u)), (du+Physics:-`*`(Ybar(zetabar, zeta, v, u), dzeta)+Physics:-`*`(Y(zetabar, zeta, v, u), dzetabar)-Physics:-`*`(Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), dv))^2)

2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2

(1.3)

declare(ds2)

Ybar(zetabar, zeta, v, u)*`will now be displayed as`*Ybar

(1.4)

NULL

vierbien = Matrix([[1, 0, -Ybar(zetabar, zeta, v, u), 0], [0, 1, -Y(zetabar, zeta, v, u), 0], [Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Physics:-`*`(H(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1-Physics:-`*`(Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Ybar(zetabar, zeta, v, u)), H(zetabar, zeta, v, u)], [Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u), -Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1]])

vierbien = (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -Ybar(zetabar, Zeta, v, u), (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = -Y(zetabar, Zeta, v, u), (2, 4) = 0, (3, 1) = H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u), (3, 2) = H(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 3) = 1-H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 4) = H(zetabar, Zeta, v, u), (4, 1) = Y(zetabar, Zeta, v, u), (4, 2) = Ybar(zetabar, Zeta, v, u), (4, 3) = -Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (4, 4) = 1}))

(1.5)

``

NULL

Setup(tetrad = rhs(vierbien = Matrix(%id = 18446744078213056502)), metric = ds2, mathematicalnotation = true, automaticsimplification = true, coordinatesystems = (X = [zetabar, zeta, v, u]), signature = "+++-")

[automaticsimplification = true, coordinatesystems = {X}, mathematicalnotation = true, metric = {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}, signature = `+ + + -`, tetrad = {(1, 1) = 1, (1, 3) = -Ybar(X), (2, 2) = 1, (2, 3) = -Y(X), (3, 1) = H(X)*Y(X), (3, 2) = H(X)*Ybar(X), (3, 3) = 1-H(X)*Y(X)*Ybar(X), (3, 4) = H(X), (4, 1) = Y(X), (4, 2) = Ybar(X), (4, 3) = -Y(X)*Ybar(X), (4, 4) = 1}]

(1.6)

gamma_[4, 1, 1] = 0

diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0

(1)

gamma_[4, 2, 2] = 0

diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0

(2)

gamma_[1, 4, 4] = 0

(diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0

(3)

gamma_[2, 4, 4] = 0

(diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0

(4)

gamma_[3, 4, 4] = 0

0 = 0

(5)

gamma_[4, 4, 4] = 0

0 = 0

(6)

shearconditions := {diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0, diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0, (diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0, (diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0}:

 

 

RicciT := proc (a, b) options operator, arrow; SumOverRepeatedIndices(Ricci[mu, nu]*e_[a, `~mu`]*e_[b, `~nu`]) end proc

proc (a, b) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(Physics:-Ricci[mu, nu], Physics:-Tetrads:-e_[a, `~mu`]), Physics:-Tetrads:-e_[b, `~nu`])) end proc

(7)

SlashD := proc (f, a) options operator, arrow; SumOverRepeatedIndices(D_[b](f)*e_[a, `~b`]) end proc

proc (f, a) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-D_[b](f), Physics:-Tetrads:-e_[a, `~b`])) end proc

(8)

SlashD(f(X), 1)

diff(f(X), zeta)-Ybar(X)*(diff(f(X), u))

(9)

SlashD(f(X), 2)

diff(f(X), zetabar)-Y(X)*(diff(f(X), u))

(10)

SlashD(f(X), 3)

(1+H(X)*Y(X)*Ybar(X))*(diff(f(X), u))-H(X)*((diff(f(X), zeta))*Y(X)+Ybar(X)*(diff(f(X), zetabar))+diff(f(X), v))

(11)

SlashD(f(X), 4)

-Y(X)*Ybar(X)*(diff(f(X), u))+Ybar(X)*(diff(f(X), zetabar))+(diff(f(X), zeta))*Y(X)+diff(f(X), v)

(12)

NULL

  simplify(RicciT(1, 2), shearconditions) = 0

H(X)*(diff(diff(Y(X), zeta), zetabar))*Ybar(X)-H(X)*Ybar(X)*Y(X)*(diff(diff(Ybar(X), u), zetabar))-H(X)*Ybar(X)^2*(diff(diff(Y(X), u), zetabar))-H(X)*Y(X)^2*(diff(diff(Ybar(X), u), zeta))-2*H(X)*Y(X)*Ybar(X)*(diff(diff(Y(X), u), zeta))+H(X)*Y(X)^2*Ybar(X)*(diff(diff(Ybar(X), u), u))-H(X)*Y(X)*(diff(diff(Ybar(X), u), v))+H(X)*Y(X)*Ybar(X)^2*(diff(diff(Y(X), u), u))-H(X)*(diff(diff(Y(X), u), v))*Ybar(X)+H(X)*(diff(Ybar(X), zetabar))^2+(-3*H(X)*Y(X)*(diff(Ybar(X), u))-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), zetabar))+H(X)*(diff(Y(X), zeta))^2+(-4*H(X)*(diff(Y(X), u))*Ybar(X)-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Y(X), zeta))+2*H(X)*Y(X)^2*(diff(Ybar(X), u))^2-Y(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), u))+2*(diff(Y(X), u))*Ybar(X)*(H(X)*(diff(Y(X), u))*Ybar(X)+(1/2)*(diff(H(X), u))*Y(X)*Ybar(X)-(1/2)*(diff(H(X), zeta))*Y(X)-(1/2)*(diff(H(X), zetabar))*Ybar(X)-(1/2)*(diff(H(X), v))) = 0

(13)

``

0 = 0

0 = 0

(14)

``

Why does the expand command cause the lhs to be zero?

NULL


Download Question_R12.mw

I want to know with what x,y, z,  function f is minimum, whereas function g is constant.

 

regards

Hi

I want to know with what x/y, z,  function f is minimum, whereas function g is constant.

regards

 

Hello every one,

Is any one knows how to solve the following inequality with assumptions that all parameters are real positive and k<1 and delta > c*alpha

(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))<0

I tried the following code but it  dosn't make sense:

u:=(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))

solve({u < 0,alpha > 0, beta > 0, c > 0, delta > 0, delta > c*alpha, k > 0, k < 1, })

In fact I want to know under which circumastances the above inequality is negative.

THX

Is there a command in maple that shows which files were read during start up? I am confused as to which maple init file is being read on my PC.  From http://www.maplesoft.com/support/help/Maple/view.aspx?path=worksheet/reference/initialization

it says

"Under Windows, the initialization file is called maple.ini.
If <Maple>\lib\maple.ini exists, it is loaded first (where <Maple> is your Maple installation directory). With a  network installation of Maple, the commands in this initialization file will be executed by all users on the network.
To execute a user's personal set of commands, only the first initialization file in one of the following paths will be loaded.
1) The binary directory of your current working directory (for example, "c:\Program_Files\Maple\bin.win\maple.ini")
2) The <Maple>\Users directory (for example, "c:\Program_Files\Maple\Users\maple.ini")
3) The user's personal profile directory (multiuser only) (for example, "c:\Documents_and_Settings\userid\maple.ini")
      
Maple reads and executes the network initialization file before the personal initialization file."

How do I find out, from inside Maple, which file(s) were read? Or make maple shows a trace of the loading process to see what files ini files it is reading?

Maple 18.2 on windows. btw, the above help page seems old. I am on windows 7, and I do not have "c:\Program_Files\Maple\" folder. And do not have "c:\Documents_and_Settings" folder. So the above help is not very useful. May be it was written during windows 95 times?

Hi all,

I have some "boolean variable" constraint equation like this:

a1*x1+a2*x2+...+an*xn>=b1*y1+b2*y2+...+bn*yn

where a1,a2,...,an and b1, b2, ..., bn are 1 or -1

These equations will be used in LPSolve or the other command to find a group of parameters which can fit them.

Now I used for-loop to deal with this kind of question, for example:

But there are more than 10 boolean variables in my case and It's very inefficient. On the other hand, using for-loop to determine the equation we solve in the command will lead to great confusion.

I think there should be some ways able to solve this kind of "boolean variables" question in Maple, such as, through assume command to define the type of "boolean variable".

But I have no idea how to do it.

I have the following construct:
for i from 1 to 10 do
cubeprod:=i^3;
if irem(cubeprod,3)=0  and if(modp(cubeprod,2)<>0 then
cubesum(cubeprod);// I need both if statements to be true in order to invoke cubesum(). I've noticed that using an and between both if statements is incorrect but :

for i from 1 to 10 do
cubeprod:=i^3;
if irem(cubeprod,3)=0 then 
 if modp(cubeprod,2)<>0 then cubesum(cubeprod) fi;
end if;
end do;

gives me an error as well. What is the right syntax to achieve this?

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