Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12

i use

A -> B

B -> C

would like to find A -> C directly from A -> B and B -> C

but logic table show that AND(A->B, B->C) is not equal to A -> C

and (A->B) -> (B->C) is not equal to A -> C too

which logic operations can do this?

because do not want to use result to verify and because i do not know the result in the beginning

I could keep on using the commands 

with(GraphTheory):

with(RandomGraphs):

DrawGraph(RandomTree(7))

but then this is not helpful because I do not get non-isomorphic graph each time and there are repetitions. So any other suggestions would be very helpful. 

for example

func1 := proc(system1)

for i from 1 to 100 do

solve([system1[1], system1[2]],[x,y]);

od:

end proc:

 

func1([diff(y,t) = data[i+t+1], diff(x,t) = data[i+t+1]])

i is depend on the for loop inside a function, but woud like to pass this system into a function with i

this will cause error

how to write better for passing a system as parameter using variable inside a function?

indets(AA(MM(AA(X1,X2),B2),MM(AA(B2,X3),X5)));
this can get a list of X1,X2,B2,X3,X5

propositionsentence := MM(AA("it", "run"), "maple"):
indets(propositionsentence);

but this can not get "it", "run", "maple"

how to do this in case this in maple 12?
 

i use optimization package with constraint hello >= 0

Minimize(xx=0, {hello >= 0})

but solution only return the case when hello = 0

how about hello > 0?

i would like to find all possible set of solutions using this constraint

do i need to set upper bound, such as {hello <= 7, hello >=0}

can it return solution when hello = 1.1, 1.2, ...2, 2.1, 2.2, 2.3, ....7

How to create a hyperplane which perpendicular to groebner basis

tord := plex(x, y, z);
G := Basis([hello1, hello2, hello3], tord);
ns, rv := NormalSet(G, tord);
Error, (in Groebner:-NormalSet) The case of non-zero-dimensional varieties is not handled
is this error due to version of maple?
which version do not have this error?
 

A lemniscate is a polar curve of the form r^2=a^2*cos(2*theta) or r^2=a^2*sin(2*theta). I have just started using Maple and I wrote the following commands: 

> with(plots):
> polarplot(2*sqrt(cos(2*t)), axiscoordinates = cartesian, angularunit = radian, color = "Black");
 

But I am getting the following graph 

which is not satisfactory since some points are missing. I know that using the square root may have caused this, but I am not sure as to how should I resolve this issue. I used plus/minus symbol before the expression 2(cos(2t))^(0.5) but there was an error and the discontinuity still persisits. Kindly help me in plotting this curve. 

Thank you.

Hi everyone!

I would really appreciate if someone could give me a hand on telling me what is wrong with this problem! pdsolve gives the error: Error, (in pdsolve/sys/info) found functions with same name but depending on different arguments in the given DE system: {f(0, y), f(x, 0), f(x, y), (D[2](f))(0, y), (D[2](f))(x, 0)}.


Thanks in advance!!! 




 

 

Dear All,

I am plotting the following function using implicitplot command.:


plots[implicitplot3d]((17.31626331*M^3-(4*(z[1]-z[2])^2*M^2-1.171300684*(z[1]+z[2])^2)*(1.082266457-2*M)*(1.082266457-3*M))^2 = 4.598621420*(z[1]+z[2])^2*M*(1.082266457-2*M)^3*(4*(z[1]-z[2])^2*M^2-1.171300684*(z[1]+z[2])^2), M = 0 .. 1, z[1] = 0 .. 10, z[2] = -10 .. 0);

How can I extract data points from the plot obtained

with(DEtools):
rho := 0.1;
w0 := 2;
sys := {diff(x(t),t) = y(t),diff(y(t),t) = -2*rho*y - w0^2*x};
DEplot(sys, [x(t), y(t)], t = 0 .. 12, [[x(0) = 1, y(0) = 0]]);

i use flow above, would like to plot a circle move from right hand side to left hand side

and see how a circle influence the flow diagram in animation like weather diagram

I resolved the coefficients to a 2nd order diff eq of the form:ay''+by'+cy=f(t)

I have included the .mw file for convenience at the link at the bottom of the page.  I resolved the coefficients in 2 different ways & they do not concur.  The 1st approach used the LaPlace transform & partial fraction decomposition.  The coefficient results are given by equations # 14 & 15.  The 2nd approach used undetermined coefficients where I assumed the particular solution and then applied the initial conditions to resolve the coefficients pertaining to the homogeneous solution which are given in the results listed in equation #23.  Noted in the 1st case the coeff's are A3 & A4 and for the 2nd approach the coeff's are A1 & A2.  I have worked this numerous times & do not understand why they do not concur.  So I thought I should get some fresh eyes on the problem to find where I may have gone wrong.

Any new perspective will be greatly apprecieated.

I had trouble uploading the .mw file so I have included an alternative link to retrieve the file if the code contents is illegible or you cannot dowlad the file drectly from the weblink  Download coeffs_of_homogen_soln_discrepancy.mw.  You should be able to download from the alternative link below once you paste the link into your browser.  If you cannot & wish for me to provide the file in some other fashion respond with some specific instructions & I will attempt to get the file to you.

https://unl.box.com/s/dywe90wwpy0t4ilkuxshkivz2z26mud8

Thanks 4 any help you can provide.

Download coeffs_of_homogen_soln_discrepancy.mw


 

restart; with(plots); beta := 0.1e-1; Bi := 1; Pr := 3.0; L0 := 1; w = 0.2e-1

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2+beta*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2+0.1e-1*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

(1)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+beta*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+0.1e-1*F(eta)-0.1e-1*(diff(f(eta), eta)) = 0

(2)

Eq3 := G(eta)*(diff(G(eta), eta))+beta*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+0.1e-1*f(eta)+0.1e-1*G(eta) = 0

(3)

Eq4 := H(eta)*F(eta)+H(eta)*(diff(G(eta), eta))+G(eta)*(diff(H(eta), eta)) = 0

H(eta)*F(eta)+H(eta)*(diff(G(eta), eta))+G(eta)*(diff(H(eta), eta)) = 0

(4)

Eq5 := (diff(theta(eta), eta, eta))/Pr+f(eta)*(diff(theta(eta), eta))+(2*beta*H(eta)*(1/3))*(theta[p](eta)-theta(eta)) = 0

.3333333333*(diff(diff(theta(eta), eta), eta))+f(eta)*(diff(theta(eta), eta))+0.6666666667e-2*H(eta)*(theta[p](eta)-theta(eta)) = 0

(5)

Eq6 := G(eta)*(diff(theta[p](eta), eta))+L0*beta*(theta[p](eta)-theta(eta)) = 0

G(eta)*(diff(theta[p](eta), eta))+0.1e-1*theta[p](eta)-0.1e-1*theta(eta) = 0

(6)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(theta))(0) = -Bi*(1-theta(0)), (D(f))(5) = 0, F(5) = 0, G(5) = -f(5), H(5) = w, theta(5) = 0, theta[p](5) = 0

f(0) = 0, (D(f))(0) = 1, (D(theta))(0) = -1+theta(0), (D(f))(5) = 0, F(5) = 0, G(5) = -f(5), H(5) = w, theta(5) = 0, theta[p](5) = 0

(7)

p := dsolve({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1}, numeric)

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

 

odeplot(p, [eta, f(eta)], 0 .. 10);

odeplot(p, [eta, f(eta)], 0 .. 10)

(8)

``

 

 


 

Download from_net.mw

I attempted to use convolution in AudioTools to convolve 2 vectors with arbitrary values, but could not since the operation is expecting numeric values.  Can this be done?    Is there not a convolution operastor in the LinearAlgebra package?  See attached:

convolutionWithAu[1].mw

After several attempts on this question,

Int(x*sqrt(2*x^4+3),x) with substitution u = sqrt(2)*x^2,

I don't seem to find the solution. Can you guys help me?

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