Maple Questions and Posts

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Hello

1)I need some guides to solve my problem. I have attached my file. Could you please help me to understand what's wrong in my code?

2) Is there any option to pause "Loops" and watching results, something like break point.

THank you.

 

 

restart

 

with(ImageTools):``

with(FileTools):````

Input Section

 

NULL

filelocation := "C:\\Users\\Mohammad\\Desktop\\1.jpg":

k := 1:NULL

 

zimage := Read(filelocation):

zwidth := Width(zimage):NULL

kernel__length := VectorCalculus:-`+`(VectorCalculus:-`*`(2, k), 1):````

kernel__data := `~`[`*`](Matrix(1 .. kernel__length, 1 .. kernel__length, 1), 1/kernel__length^2):

imheight := Height(zimage):``

imwidth := Width(zimage):NULLNULL

Width(zimage, lower):

Width(zimage, upper):

View(zimage):``

new1zpic := Convolution(zimage, kernel__data):

View(new1zpic)``NULL

new2zpic := zimage:

dummy := 0:

"for i from (k+1)  to (Width(zimage,upper)-k) do     for j from (k+1)  to (Height(zimage,upper)-k) do           for m from 1 to kernel[length]  do               for n from 1 to kernel[length]  do                         dummy:=dummy+kernel[data] (m, n)*new2zpic(i+m-2, j+n-2)                                         end do;              end do;                      new2zpic(i,j):=dummy:                   dummy:=0:   end do; end do;"

Error, final value in for loop must be numeric or character

 

 

 

 

NULL

 

Download Exercise_II-1.0.mw

restart

 

with(ImageTools):``

with(FileTools):````

Input Section

 

NULL

filelocation := "C:\\Users\\Mohammad\\Desktop\\1.jpg":

k := 1:NULL

 

zimage := Read(filelocation):

zwidth := Width(zimage):NULL

kernel__length := VectorCalculus:-`+`(VectorCalculus:-`*`(2, k), 1):````

kernel__data := `~`[`*`](Matrix(1 .. kernel__length, 1 .. kernel__length, 1), 1/kernel__length^2):

imheight := Height(zimage):``

imwidth := Width(zimage):NULLNULL

Width(zimage, lower):

Width(zimage, upper):

View(zimage):``

new1zpic := Convolution(zimage, kernel__data):

View(new1zpic)``NULL

new2zpic := zimage:

dummy := 0:

"for i from (k+1)  to (Width(zimage,upper)-k) do     for j from (k+1)  to (Height(zimage,upper)-k) do           for m from 1 to kernel[length]  do               for n from 1 to kernel[length]  do                         dummy:=dummy+kernel[data] (m, n)*new2zpic(i+m-2, j+n-2)                                         end do;              end do;                      new2zpic(i,j):=dummy:                   dummy:=0:   end do; end do;"

Error, final value in for loop must be numeric or character

 

 

 

 

NULL

 

Download Exercise_II-1.0.mw

f:=sin((x-1)*(y+3))/(exp((x-1)^2+(y+3)^2)-1);
limit(f,{x=1,y=-3});


On running the above code, I don't get a result. I don't understand why, and what are some of the underlying math principles that I would have to use to evaluate the limit of f at (1,-3) if I were to solve this question on paper?

I need to find where the limit of the function:

{1-cos(x*y^3)}/(x^2+y^6)^(1+a) exists/does not exists for different values of a, given 0<=a<=1 and as (x,y)->(0,0).

I've tried writing a procedure that changes a in particular increments but that is clearly not the most efficient way. Are there any rules of limits that I should be using?

 

I've got a set

E:={(x,y,z): x^2+y^2=-2*z-x, z^2+y^2=1} and need to find points of E which have minimal or maximal distance from (0,0,0). I've set up the Lagrangian as F:=sqrt(x^2+y^2+z^2) + L1(x^2+y^2+2z+x)+L2(z^2+y^2-1)

and consequently obtained the equations:

x/sqrt(x^2+y^2+z^2) + 2*x*L1+L1=0

y/sqrt(x^2+y^2+z^2) + 2*y*L1+2*y*L2=0

z/sqrt(x^2+y^2+z^2)+2*L1+2*L2*z=0

for which I've set up
eqn1,eqn2,eqn3 as the three equations and vars:=x,y,z

and used solve() but I'm not getting the right answer( I need to first express x,y,z in terms of L1, L2 and then get values for L1 and L2 by substituting in the constraints and eventually get values of x,y,z.)

How should I implement that?

For example, given a 3d point p(x,y,z), with (x,y,z) as its coordinates. Then it is transformed by rotation and translation, as 

p'=R(p)*p+t(p), where R(p) is a 3x3 rotation matrix that is a matrix of functions of p, and t(p) is a 3x1 vector function of p. 

My question is how to derive dp'/d(as a 3x3 matrix) using maple? 

To make it clear,I want to do it in a way that dp'/dp = ∂p'/∂p + ∂p'/∂R*∂R/∂p +  ∂p'/∂t*∂t/∂p

And I'd like to know each intermediate quantity, such as p'/∂R, R/∂p.


Anyone can help?

Thanks a lot. 

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


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

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

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

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.

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