Maple 2016 Questions and Posts

These are Posts and Questions associated with the product, Maple 2016

The worksheet below rolls an ellipse along the y axis with constant energy.

How can the physics be enhanced to roll the ellipse along a non-linear curve (e.g. a sine curve) with constant energy?

EllipseRoll.mw

Help create file Excel in ExcelTools, but error row 564?

thu_file.mw

Please help me? 

Good day everyone.

Please can help me with this code on the taylor series expansion involving Fractional Differential Equation (FDE)? Particularlly, the lines highlighted blue and green respectively in relation to FDE.

Thank you and kind regards

#k=2
restart:q:=n*h:
P:=sum((a[k]*x^(k))/GAMMA(k+1-alpha), k=0..3):
assume(alpha>0,alpha < 1):
Q:=fracdiff(P,x,alpha):
e1:=simplify(eval(P, x=q+h))=y[n+1]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h))=f[n+1]:
e4:=simplify(eval(Q,x=q+2*h))=f[n+2]:
var:=seq(a[i], i=0..3):
M:=e||(1..4):

Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 4 do
	a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
print():
s2:=y[n+2]=simplify(eval(Cf, x=q+2*h)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):

s1:=y[n]=simplify(eval(Cf, x=q)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):

Y[n+1]:=convert(taylor(y(x+h),h=0,12),polynom):
F[n]:=convert(taylor((D(y))(x), h = 0,12), polynom):
F[n+1]:=convert(taylor((D(y))(x+h), h = 0,12), polynom):
F[n+2]:=convert(taylor((D(y))(x+2*h), h = 0,12), polynom):


W:=asympt(expand(eval(rhs(s2),[y[n+1]=Y[n+1],f[n]=F[n],f[n+1]=F[n+1],f[n+2]=F[n+2]])),h,6);
X:=convert(taylor(y(x+2*h),h=0,12),polynom)-W;
lte:=convert(asympt(X,h,8),polynom);

 

 

I have a physics question I need to program it by Maple , Can you help me to solve ? Its Problem 9.34 from Griffiths

Hey there,

I'm trying to numerically intergrate a function s(K,i,j) dK using runge kutta over a 2D grid of i,j values. Essentialy, performing the same sort of integral many different times for slightly different combinations of i and j. The function is more or less gaussian, and so the bulk of the result will come from the values of K around the peak of said gaussian. For some combinations of i and j, the function seems to have a singularity on the right edge of the gaussian peak, which causes my script to spit out an error, telling me the calculation cannot be performed further to the right past the singularity. Now, like i said before, the singularity is on the very edge of the gaussian and therefore I am perfectly happy to stop the integration before the singularity, because anything past it wont contribute very much to the result.

How can I use dsolve events to halt my integration just before hitting the singularity?

RK := (i, j) -> dsolve({diff(n(K), K) = K*s(K, a[i], b[j]), n(0) = 0}, numeric, method = rkf45)

 

Hello, 

I am fairly new to using the Maple software, so I apologize if my question is completely idiotic. Apologies, also, because I could not manage to enter my code as code. When I pressed the button it made the whole text as a code. 

I run the following code to seek -if there are any- analytic solutions for the following differential equation.

odeplus := (r^2+L^2)^(5/2)*(diff(f(r), `$`(r, 2)))+((15/4)*r*(r^2+L^2)^(1/2)+3*(r^2+L^2)^(5/2)/r)*(diff(f(r), r))+M^2*f(r)/(r^2+L^2)^(5/2)-((5/2)*((r^2+L^2)^(1/2))(l-1)+(55/64)*r^2/(r^2+L^2)^(3/2)+(r^2+L^2)^(5/2)*(l^2+3*l+3/2)/r^2)*f(r)+(((r^2+L^2)^(1/2))(5+(5/2)*l)+(5/8)*r^2/(r^2+L^2)^(3/2)-(r^2+L^2)^(5/2)*(3/2+l)/r^2)*f(r) = 0

and then I do 

dsolve(odeplus, f(r))

The solutions that Maple returns is given in terms of DESol. Could anyone try and break it down for me? What is this telling me and if I can indeed from the output obtain analytic solutions? Is this some sort of operator acting on something? 

Thank you in advance. 

A catenoid is the minimal surface between two 3D circles which are co-axial and parallel.

Is there a technique for finding the formula for the minimal surface if the circles are "stretched" into ellipses with proportional major and minor axes?

Can someone help with the simplification of the result of this code? I am sure the "qs" in the final result should not appear.

Thanking you in anticipation of your positive responses

#k=1
restart:
P:=sum(a[k]*x^k, k=0..2):
assume(alpha>0,alpha <= 1):
Q:=fracdiff(P,x,alpha);
e1:=simplify(eval(P, x=q))=y[n]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h^alpha))=f[n+1]:
var:=seq(a[i], i=0..2):
M:=e||(1..3):

Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 3 do
	a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n], f[n], f[n+1]], recursive):
print():
#y[n+1]=collect(simplify(simplify(expand(eval(Cf,x=q+h^alpha)),size)), [y[n],f[n],f[n+1]], factor);
y[n+1]=simplify(eval(Cf, x=q+h^alpha)):
collect(%, [y[n], f[n], f[n+1]], recursive);

 

Hello,

I have a very simple problem. When Maple displays long outputs I can only see a part of them. Here there is an example

https://www.dropbox.com/s/ymp1vdsg80ewu1s/Untitled.jpeg?dl=0

On my previous versions of Maple I had a slider on the bottom of the page. How can I activate it in Maple 2016?

Thanks, Nicola

Can we use this euler lagrange command for the system of pde's?

If we have a system without lagrangian then euler opertor is applied corresponding to three dependent variables.

so we have three equations but in this euler lagrange command no dependent variable is mentioned.   

I have several functional equations in equally many unknown functions of at least two variables, plus parameters.  ("collect" works just for single equations, right?)

I know that for certain parameter ranges, all the functions involved will be quadratic, and I know some coefficients are zero.  That gives me some  coefficients to determine.  I want to

  1. specify the functional equations [done in a very primitive low-tech way in the attachment, using atomic variables rather than indices ... have I done correctly?!?] 
  2. get Maple to collect coefficients (the K's and the L's in the attachment; the variables are (y,z))
  3. get Maple to state an equation system these coefficients have to satisfy (these will unfortunately be coupled quadratics)
  4. get Maple to solve that equation system if possible, and if not: to tell me when (= for what parameter values, parameters being the "remaining letters" in the attachment) I have specified enough coefficients
  5. in case of a solution, get Maple to tell me which coefficients are real and positive (for those that are solution of quadratic eq's: whether a positive solution exists)

Phew. I am still a complete newbie. Edit: Attachment link: STcoeff2match.mw where the equations themselves are EQ0, EQ1 and EQ2 at the bottom. Copying and pasting them, they look like this (download STcoeff2pastedEQs.mw)

0 = -r__0*(K__011*y^2+K__022*z^2-K__012*(y-L__1)*(z-L__2)-K__01*(y-L__1)+K__02*(z-L__2))+(-2*K__011*y+m__1+K__012*(z-L__2)+K__01)*((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)+(20/9)*K__011*y-(10/9)*K__012*(z-L__2)-(10/9)*K__01-(10/9)*m__1-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(16/9)*K__022*z+(8/9)*K__012*(y-L__1)-(8/9)*K__02+(8/9)*m__2)+(-2*K__022*z+m__2+K__012*(y-L__1)-K__02)*((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)+(20/9)*K__022*z-(10/9)*K__012*(y-L__1)+(10/9)*K__02-(10/9)*m__2-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(16/9)*K__011*y+(8/9)*K__012*(z-L__2)+(8/9)*K__01+(8/9)*m__1)+(-(4/3)*K__011*y+(2/3)*K__022*z+(2/3)*K__012*(z-L__2)-(1/3)*K__012*(y-L__1)-(1/3)*m__2+(2/3)*m__1+(1/3)*K__02+(2/3)*K__01)^2+((2/3)*K__011*y-(4/3)*K__022*z-(1/3)*K__012*(z-L__2)+(2/3)*K__012*(y-L__1)+(2/3)*m__2-(1/3)*m__1-(2/3)*K__02-(1/3)*K__01)^2:

``

0 = -r__1*(K__11*y^2-`K__12 `*y*(z-L__2))+`K__12 `*y*((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)+(20/9)*K__022*z-(10/9)*K__012*(y-L__1)+(10/9)*K__02-(10/9)*m__2-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(16/9)*K__011*y+(8/9)*K__012*(z-L__2)+(8/9)*K__01+(8/9)*m__1)+((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(10/9)*K__022*z+(5/9)*K__012*(y-L__1)-(5/9)*K__02+(5/9)*m__2+(8/9)*K__011*y-(4/9)*K__012*(z-L__2)-(4/9)*K__01-(4/9)*m__1)^2:

``

0 = -r__2*(K__22*z^2-`K__21 `*(y-L__1)*z)+`K__21 `*z*((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)+(20/9)*K__011*y-(10/9)*K__012*(z-L__2)-(10/9)*K__01-(10/9)*m__1-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(16/9)*K__022*z+(8/9)*K__012*(y-L__1)-(8/9)*K__02+(8/9)*m__2)+((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(10/9)*K__011*y+(5/9)*K__012*(z-L__2)+(5/9)*K__01+(5/9)*m__1+(8/9)*K__022*z-(4/9)*K__012*(y-L__1)+(4/9)*K__02-(4/9)*m__2)^2:

``

 

 

restart;
with(PDEtools);
assume(k::real, x::real, omega::real, t::real, theta::real, c::real);
u := phi(c*(t*upsilon+x))*exp(I*(k*x+omega*t+theta));
PDE := proc (u) options operator, arrow; I*(diff(u, t))+diff(u, x, x)-I*sigma*u*(conjugate(u)*(diff(u, x))-u*conjugate(diff(u, x))) end proc;
Eq1 := PDE(u)

Hi everyone.

I am trying to build a package in Maple out of a bunch of procedures that I have. To create the procedures, I write them on a Vim editor, than copy and paste into the Maple worksheet to test. In this way, it works (I end the procedure with ; and then I see all the text in blue and I afterwords test it).

I then gathered these procedures to a module, option package, and I do the same: edit on Vim, copy paste it to Maple worksheet and press enter to see a blue message "module() ... end module" in blue, and then I start playing.

However, I decided to add some more procedures to the module that was working. I have a new procedure that works (meaning when I copy to the worksheet it gives the blue text and I can call normally), but when I copy the contents of this new procedure inside the package, with all the others, and I copy everything to the worksheet, the package now fails to work. It gives a message "Error," in pink, no more text, and that is it. Does this sort of error sound familiar to anyone? Thanks in Advance,

Marcelo

Good day. Please can someone kindly help to reduce the result of this code. Thank you and kind regards

restart:
s:=(sum(a[j]*x^j,j=0..3)+sum(a[j]*exp(-(j-3)*x),j=4..7)):
F:=diff(s,x):
p1:=simplify(eval(s,x=q))=y[n]:
p2:=simplify(eval(F,x=q))=f[n]:
p3:=simplify(eval(F,x=q+h/3))=f[n+1/3]:
p4:=simplify(eval(F,x=q+h))=f[n+1]:
p5:=simplify(eval(F,x=q+5*h/3))=f[n+5/3]:
p6:=simplify(eval(F,x=q+2*h))=f[n+2]:
p7:=simplify(eval(F,x=q+7*h/3))=f[n+7/3]:
p8:=simplify(eval(F,x=q+3*h))=f[n+3]:


vars:= seq(a[i],i=0..7):
Cc:=eval(<vars>, solve({p||(1..8)}, {vars})):
for i from 1 to 8 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
L:=collect(simplify(simplify(expand(eval(Cf,x=q+3*h)),size)), [y[n],f[n],f[n+1/3],f[n+1],f[n+5/3],f[n+2],f[n+7/3],f[n+3]], factor):
length(L);
H := ee -> collect(numer(ee),[exp],h->simplify(simplify(h),size))/collect(denom(ee),[exp],h->simplify(simplify(h),size)):
M:=y[n+3]=(H@expand)(L);
length(M);

 

Hi, I've been doing a few small explore plots where the ranges for the parameters are something like [-1,1], but 0 is a special value which will often need to be set.  The slider, gauge components, ... are a bit fiddly to set with exactly zero using a mouse to drag the value.  Thought about a few different ways to do this

* Right click and set value from the component properties option - a bit fiddly for the user and the graph doesn't immediately update when OK'ed.

* Extra control with a tick to set a zero value (fiddly programming and possible confusing interpretation)

* Slider component and snaptoticks=true

Last option seems to be the easiest all round, but I was wondering why it is only the slider component that has a snaptoticks property - none of the other components seem to support it and a gauge would probably be a more natural component to use.

Thanks in advance

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