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 




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]);


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?

this can get a list of X1,X2,B2,X3,X5

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

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

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  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.

Thanks 4 any help you can provide.



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


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


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


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


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


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


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


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)







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:


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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