MaplePrimes Questions


How I can remove these errors?





restart; eq31g := diff(u(t), `$`(t, 2))+u(t)+mu[s]*(diff(u(t), t))^3 = (1-mu[s])*(diff(u(t), t))*(diff(u(t), `$`(t, 2)))

diff(diff(u(t), t), t)+u(t)+mu[s]*(diff(u(t), t))^3 = (1-mu[s])*(diff(u(t), t))*(diff(diff(u(t), t), t))






diff(diff(u(tau), tau/omega), tau/omega)+u(tau)+mu[s]*(diff(u(tau), tau/omega))^3 = (1-mu[s])*(diff(u(tau), tau/omega))*(diff(diff(u(tau), tau/omega), tau/omega))



omega = epsilon^2*omega[2]+epsilon*omega[1]+1



u(tau) = epsilon*u[1](tau)+epsilon^2*u[2](tau)+epsilon^3*u[3](tau)



Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument



Error, invalid input: lhs received temp, which is not valid for its 1st argument, expr



Error, invalid input: lhs received eq33b, which is not valid for its 1st argument, expr


The general solution of the first-order equation, eqEps[1], can be expressed as


Error, (in dsolve) not a system with respect to the unknowns [u[1](tau)]



Error, invalid input: lhs received eqEps[1], which is not valid for its 1st argument, expr


Expanding the right-hand side of eq33c in a Fourier series using trigonometric identities yields


Error, invalid input: rhs received eq33c, which is not valid for its 1st argument, expr


Eliminating the terms,  and , demands that . Then, the particular solution of eq33c can be expressed as


Error, (in dsolve) expecting an ODE or a set or list of ODEs. Received eq33c


Substituting sol1 and sol2 into the third-order equation, eqEps[3], and using the fact that , we obtain


Error, invalid input: lhs received eqEps[1], which is not valid for its 1st argument, expr


Expanding the right-hand side of eq33d in a Fourier series using trigonometric identities, we have


Error, invalid input: rhs received eq33d, which is not valid for its 1st argument, expr


Eliminating the terms that lead to secular terms from eq33d_RHS demands that


omega[2] = omega[2]


As discussed above, for a second-order uniform expansion, we do not need to solve for . Combining the first- and second-order solutions, we obtain, to the second approximation, that


Error, invalid input: subs received sol1, which is not valid for its 1st argument




tau = (epsilon^2*omega[2]+1)*t






How do I take the square root of :

    X:= r2 (R + r cos(v))2

with the condition that R and r are positive and R>r and get:

r (R+ r cos(v))

I assume I need to use simplify with some assumptions, but I cannot figure it out.  

browsing on math.stackexchange I came across this maple code:

asympt(int(sqrt(-k^2+1)*exp(I*k*x), k = -1 .. 1), x, 2)

that gave the correct answer:

so I wondered would it not be great if maple done asymptotic expansion of integrals in general.

say as an example Li(x).  Can maple expand Li(x) and how?


I am perplexed about the geometry and plots packages.  When starting a programme involving geometric shapes, how does one decide which to use?  ..Or can the two packages be used in the same programme?  To iiiustrate my question there is a text program below which draws a diagram of a large square, with a circle inscribed tangentially to the square.  There is then a further smaller concentric circle, with a square drawn inside; the vertices touching this circle.  (It isdesigned as a problem suitable for young high-school teenagers.)

   I wanted to fill in various regions of this diagram in colors red and green.  I believe the with(geometry) package does this easily.  However, as I'm more used to the with(plots) package I stuck to this.  To get the coloured regions I drew a sequence of colored lines.  I also drew a colored square (See    SmallSq:=[op(SmallSq),sq_small||ii]:   ) by drawing a seq of concentric squares.  At the end of my program there are two almost identical plots[display] commands.   The difference is that the first does not incude the SmallSq.  I was expecting the two displays to be identical, since I'd specified the color white.  The first display has a dark colored square.  Why?? .  

   .   The program is very klutsy!  I'm embarrassed to publish it:-(  The SmallSq is adequate for my purposes, but has several white spaces - almost dots - on the dark version.  The thickness and style of the lines was the widest possible  - perhaps I should have set the larger rarius R, to a bigger value?

   And I have a feeling this diagram could have been constructed ib the geometry package in far fewer lines of code!

As always, any comments, suggestions would be most appreciated.




> restart:
> # # # # # # # # # # # # # # # # # # # # # # # # # # # #

# Maple 7
> # Area puzzle involving squares & circles
> #To Do:  fill areas with color
> #  Use segment command and seq to make a set of colored lines
> #NB  Numbering for lines L1 is 1,2,3,4 for N,E,S,West
> #    Numbering for lines L2 is 1,2,4,3 for N,E,S,West
> # # # # # # # # # # # # # # # # # # # # # # # # # # # #
> with(plots):
> with(plottools):
> print(`The diagram shows two squares and two circles.`);
> print(`The red and green regions are equal in area.`);
> print(`To find the ratio of the radius of the large circle to the
> smaller.`);

> #Doesn't seem to like a combination of geometry & with plots
> #with(geometry);
> r:=R*sqrt((4-Pi)/(Pi-2));  #for the pretty printout only!
> R:=49:
> r:=R*sqrt((4-Pi)/(Pi-2)):  #give r a numerical value
> c_big := circle([0,0], R, color=red):
> c_small := circle([0,0], r, color=green):
> sq_big := rectangle([-R,R], [R,-R], color=white):
> #sq_small := rectangle([-r/sqrt(2),r/sqrt(2)], [r/sqrt(2),-r/sqrt(2)],
> color=white):
> #y:=R/2:

> SmallSq:=[]:
> for ii from 1 to round(r) do
> sq_small||ii := rectangle([-ii/sqrt(2),ii/sqrt(2)],
> [ii/sqrt(2),-ii/sqrt(2)], color=white,linestyle=1 , thickness=3):
> SmallSq:=[op(SmallSq),sq_small||ii]:
> end do:

> Llines1:=[]:Llines2:=[]:Llines3:=[]:Llines4:=[]:
> for yy from 1 by 1 to R do
> l1||yy := line([-R,yy], [-sqrt(R^2-yy^2),yy], color=red, linestyle=1 ,
> thickness=3):
> l2||yy := line([sqrt(R^2-yy^2),yy], [R,yy], color=red, linestyle=1 ,
> thickness=3):
> l3||yy := line([sqrt(R^2-yy^2),-yy], [R,-yy], color=red, linestyle=1 ,
> thickness=3):
> l4||yy := line([-sqrt(R^2-yy^2),-yy], [-R,-yy], color=red, linestyle=1
> , thickness=3):
> #List of lines is Llines
> Llines1:=[op(Llines1),l1||yy]:
> Llines2:=[op(Llines2),l2||yy]:
> Llines3:=[op(Llines3),l3||yy]:
> Llines4:=[op(Llines4),l4||yy]:
> end do:
> first:=round(r*(sqrt(2))/2)+1:
> last:=round(r):
> L2lines1:=[]:L2lines2:=[]:L2lines2D:=[]:L2lines3:=[]:L2lines3Up:=[]:L2
> lines4:=[]:
> for yy from first by 1 to last do
> l2_1||yy := line([-sqrt(r^2-yy^2),yy], [sqrt(r^2-yy^2),yy],
> color=green, linestyle=1 , thickness=3):
> #l2_2||yy := line([yy,r/sqrt(2)],[yy,sqrt(r^2-yy^2)],  color=green,
> linestyle=1 , thickness=3):
> l2_2||yy := line([yy,sqrt(r^2-yy^2)],[yy,0],  color=green, linestyle=1
> , thickness=3):
> l2_2D||yy := line([yy,-sqrt(r^2-yy^2)],[yy,0],  color=green,
> linestyle=1 , thickness=3):

> l2_3||yy := line([-yy,-sqrt(r^2-yy^2)],[-yy,0],  color=green,
> linestyle=1 , thickness=3):
> l2_3Up||yy := line([-yy,sqrt(r^2-yy^2)],[-yy,0],  color=green,
> linestyle=1 , thickness=3):

> l2_4||yy := line([-sqrt(r^2-yy^2),-yy], [sqrt(r^2-yy^2),-yy],
> color=green, linestyle=1 , thickness=3):
> #List of lines is Llines
> L2lines1:=[op(L2lines1),l2_1||yy]:
> L2lines2:=[op(L2lines2),l2_2||yy]:
> L2lines2D:=[op(L2lines2D),l2_2D||yy]:
> L2lines3:=[op(L2lines3),l2_3||yy]:
> L2lines3Up:=[op(L2lines3Up),l2_3Up||yy]:
> L2lines4:=[op(L2lines4),l2_4||yy]:
> end do:

> plots[display](c_big,c_small,Llines1,Llines2,Llines3,L2lines1,L2lines2
> ,L2lines2D,Llines4, L2lines3,L2lines3Up,L2lines4,
> scaling=constrained);
> plots[display](c_big,c_small,Llines1,Llines2,Llines3,Llines4,L2lines1,
> L2lines2,L2lines2D,L2lines3,L2lines3Up,
> L2lines4,SmallSq,scaling=constrained);

Warning, the name changecoords has been redefined

Warning, the name arrow has been redefined

            The diagram shows two squares and two circles.

             The red and green regions are equal in area.

  To find the ratio of the radius of the large circle to the small\

                                     4 - Pi
                         r := R sqrt(------)
                                     Pi - 2

I'm a brand-new Maple user, so be gentle!

I was fooling around with a document and manually changed some fonts on various items, by just highlighting them and setting a new font.  I even selected everything and changed the font but not the size.  I didn't actually change any styles. 

I want to reset to the Maple default now, but can't seem to do it.  Format / Manage Style Sets... / Default Maple Style Set / OK doesn't do it, neither does Format / Manage Style Sets... / Load Style Set... navigating to the default stylesets\ and OK / OK.  Nothing seems to change.  I've been through Help:worksheet/documenting/styles and the verbiage doesn't match the dialog box screen shots there.  I'm not really sure what to do with that "help".

I can't seem to switch to a customized style set either.  If you manually change fonts, does that supercede the style?  Is there any way to revert?

Thanks for any help!


I have a list of displayed sequences  

S[j]:=display(seq(R1[i],i=,seq(R[i],i=, scaling = constrained, axes = none);

that can be animated easily in a worksheet using the following command 

display(seq(S[n]$5, n=1..10), insequence=true);

but when I try to embed that in a maplet, it doesnt work. The problem is with insequence. I removed insequence and the maplet showed S[10]. What else can I use?  

What can I do to have the maplet show the sequence of displays? Is there a way to use the animate command here?

Thanks for the help. 


I'm trying to compute the flux over a closed cylinder but can't define the top and bottom.

For the envelope

works fine. But top and bottom?


why does the sum(sin(k), k = -214748364 .. 214748364) not equal 0?

This behavior seems rather odd.  Is it documented anywhere?

v := <a,b>;



a := 3: b := 4:




convert(v, list);

[a, b]


I expected that final result to be [3,4].  Why is it [a,b]?

I know that I can do eval(convert(v, list)) to get the numerical value, but why is the eval necessary?


1. Why doesn't simplify(subs(...)) as a single expression work?

simplify(subs(z = -z, erf(z)));
subs(z = -z, erf(z));

2. This just hangs. Would be nice if simplify could deal with this, but in any case I don't think it should take so long:

timelimit(60, simplify(n <= p*n+sqrt(n)*p*(1-p))) assuming n > p*(n+(1-p)*sqrt(n));
Error, (in assuming) when calling '`property/ObjProperty`'. Received: 'time expired'

3. It seems that assuming doesn't like f(constant) objects. Doesn't seem right, because x(a) would work fine:

simplify(abs(x(0))) assuming x(0) > 0;

simplify(abs(y)) assuming x(0) > 0, y > 0;
Error, (in assuming) when calling 'assume'. Received: 'cannot assume on a constant object'

4. Can simplify be made to work on things like this (since it can handle the coefficient at (t+1))?



Hello everybody, 

I use Maple to obtain the solution of a reccurrence equation :
r := rsolve({u(n)=n*(u(n-1)+1), u(0)=0}, u)

Now I want to define a function s(n) wich returns the integer value of the nth term u(n).
In the attached file three different attempts to compute the numerical value of s(2) (you can replace "2" by any other positive integer) are given.

  • S := unapply(r, n);
  • S := unapply(convert(r, sum), n);
  • S := unapply(simplify(convert(r, sum)), n);

But none of them returns an integer and I'm always forced to apply some operation to "transform" the output into the desired integer.
Could you please help me to understand why it is so ?

Thanks in advance


​​​​​​use epilson- delta definition of a limit to prove that

Limit as x approaches 5 of 1/x =1/5

There seems to be a bug: in some of the cases, the intersection command works, in others it doesn't.

I'd appreciate any hints as to why this is the case.


I think its a simple question and I hope, someone can help me. For Example:

I have a list of numbers ("M") and I want to write these numbers in a new List but with the restriction

if x<Mthen Mi*2
else Mi 

Can someone explain how to solve this in maple?

Thank you very much.


I am looking for a pragmatic solution to the question: Can my rolled carpet fit inside a given cardboard box?

It is the (generally unsolvable) problem "straw in a box":

...with the restriction that the cylinder is aligned along the spatial diagonal.


So the application is this:
Is it possible to put a 120cm long rolled (radius 10cm) carpet inside a cardbord box of (inner) dimensions 118cm x 58cm x 38cm ?

Maple code would be perfect, but I have very little Maple experience so I cannot even provide any attempts/code at this stage.


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