Dear All, If I have a square with lenght b and width 2a

.The question is how can I make Maple plot this square as points

.I want to define a function f(x_[i],y_[i]) for all sides.

Thanks

Hi, I'm a new user of Maple 2016.  I'm plotting a surface using plot3d and I'd like to add a line on the surface. You can see my unsatisfying attempt to do this below.  The line that shows in the third figure is basically what I want to show but I don't want the rest of the second surface at all, just the line visible on the first surface.  

thanks

Tom 
 

Download surface_plot_question.mw
 

Download surface_plot_question.mw

restart;
with(PDEtools);
assume(k::real, x::real, omega::real, t::real, theta::real, c::real);
tr0 := c*(t*upsilon+x) = xi;
tr1 := I*(k*x+omega*t+theta);
tr2 := phi(lhs(tr0))*exp(tr1);
PDE := proc (u) options operator, arrow; I*(diff(u, t))+diff(u, x, x)-I*sigma*u*(conjugate(u)*(diff(u, x))-u*(diff(conjugate(u), x))) end proc;
Eq1 := PDE(tr2);
Eq2 := simplify(convert(expand(subs(tr0, Eq1)), diff));

for_maple_prime.mw

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

I wrote this simple set of instructions  (Maple 2015 and Maple 2016)

restart:
p := x -> sum('a__||k' * x^k, k=0..5):
p(x);    # returns a₀+ a₁x +  ... + a₅x⁵

p(1);   # returns a₀+ ... + a₅

p(0);   # returns 0 ...  not a₀

Probably so huge a mistake that I can't see it !?!?!?

Could you please help me to fit it ?

TIA

I have a problem with writing this equation in maple.

the equation is: 

I want to write this equation in maple, so I write it in this way:

for k from 1 to 7 do tau[k] := add*(d(k, j)*(diff(theta[j](t), t, t)), j = 1 .. 7)+add(add(c[i, j, k]*(diff(theta[i](t), t))*(diff(theta[j](t), t)), j = 1 .. 7), i = 1 .. 7)+phi[k] end do:

but I see this error: Error, index out of bounds 

I appriciate if you help me find the problem.

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);

> S:=-1/2*((-30*sqrt(3)+81*I)^(2/3)+21+2*sqrt(3)*(-30*sqrt(3)+81*I)^(1/3)-6*(-30*sqrt(3)+81*I)^(1/3)+I*sqrt(3)*(-30*sqrt(3)+81*I)^(2/3)-21*I*sqrt(3))/(-30*sqrt(3)+81*I)^(1/3);

When I enter

simplify(S);

a not much simpler expression involving arctan is output
but when I enter

factor(S);

the expected simplification to the number

3

is output.

This is not a but report, just a report of curious behavior.

I am using Maple 2016 on a Windows 10 PC.

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

Near the top of the large loop at the bottom of the uploaded worksheet IntSect is copied to IntSectRoll.

IntSectRoll is then modified several times within the loop. However IntSect also changes content even though there are no commands within the loop to cause this.

To avoid this is the reason IntSect is recreated from previous data each time before it is copied.

Is this a bug in my worksheet or in Maple 2016?

Constant_Diameter.mw

The theory behind the worksheet is described, without attribution, in the book The Penguin Dictionary of Curious and Interesting Geometry by David G. Wells.

Does this theory have a name and a discoverer?

I am puzzled by the strange behaviour of the following exampleSectionaken directly from the Programming Guide section 8.6. When I copy the example from the help to a worksheet as follows, I get an error stating that module m does not evaluate to a module in the use statement. 

> m := module() export a, b; a := 2; b := 3; end module:
use m in a + b end use;

Error: `m` does not evaluateto a module.

However, after a restart, when I execute the two statements as separate statements (separate prompts), the module does evaluate properly.

My question is, what is the safe workaround in coding to avoid this situation.

I have attached the worksheet.moduleeval.mw

In the following codes I am very sure exp(-3*q) is a factor both at numerator and denominator. However, I dont know how to annihilate this factor. Can someone kindly help to ensure it cancels out? Thanks and kind regards.

restart:
s:=(sum(a[j]*x^j,j=0..2)+sum(a[j]*exp(-(j-2)*x),j=3..4)):
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))=f[n+1]:
p4:=simplify(eval(F,x=q+3*h/2))=f[n+3/2]:
p5:=simplify(eval(F,x=q+2*h))=f[n+2]:

vars:= seq(a[i],i=0..4):
Cc:=eval(<vars>, solve({p||(1..5)}, {vars})):
for i from 1 to 5 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
T:y[n+2]=collect(simplify(eval(Cf,x=q+2*h)), [y[n],f[n],f[n+1],f[n+3/2],f[n+2]], recursive);

hello,

this is my first post here so sorry if it's done wrong...

I'm having difficulties with the following command because maple only returns a solution if the first two boundary conditions of the "if" arguments are true. If the the first or the second condition is false then maple doesn't give a solution solution.

for i to n do if H[i] < 2.7 then if A[i, f] < 12 then if A[i, o] < 1.2 then Q[i, foo] := evalf(610*(A[i, o]*sqrt(H[i])*h[k]*A[i, T])^(1/2)) else Q[i, foo] := evalf(7.8*A[i, t]+378*A[i, o]*sqrt(H[i])) end if end if end if; print(Q[i, foo]); end do;

I've also tried with the elif command but it gives a similar problem

Anyone knows how to solve this?