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dharr

Dr. David Harrington

7720 Reputation

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20 years, 238 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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Contact Dr. David Harrington
I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are answers submitted by dharr

tiled hexagons...

dharr 7720
April 16 2025
2 0

Not optimum. Adapted from some old code I had.

> 

restart;
> 

chexagons:=proc(n::posint,a::positive)
   # n = lattice size, a = hexagon size
   local i, j, imax, hexesA, hexesB, hexesC, hex;
   uses plottools;
   # hex returns coords of hexagon of edge length a centred at x,y.
   hex := (x,y,a) -> [[x+a,y],[x+a/2,y+a*sqrt(3)/2],[x-a/2,y+a*sqrt(3)/2],[x-a,y],[x-a/2,y-a*sqrt(3)/2],[x+a/2,y-a*sqrt(3)/2]];
   hexesA := NULL; hexesB := NULL; hexesC := NULL;
   for i to n do for j to n do
     if j-i mod 3 =0 then
       if i mod 3 = 0 then hexesA := hexesA, hex((j-1)*a+(i-1)*a/2,-(i-1)*a*sqrt(3)/2,a) end if;
       if i mod 3 = 1 then hexesB := hexesB, hex((j-1)*a+(i-1)*a/2,-(i-1)*a*sqrt(3)/2,a) end if;
       if i mod 3 = 2 then hexesC := hexesC, hex((j-1)*a+(i-1)*a/2,-(i-1)*a*sqrt(3)/2,a) end if;
     end if;
   end do end do;
   plots:-display(polygon([hexesA],color=red), polygon([hexesB],color=blue), polygon([hexesC],color=green),
           scaling=constrained,axes=none)
end proc:
> 

chexagons(10,1);

> 

 

Download hexagons.mw

naming state variable...

dharr 7720
April 14 2025
3 4

This is an answer for (1). For (2), would you please upload your worksheet (using green up-arrow) so we can understand more clearly what you want.

p.s. q_d seems to be ignored here; not sure what that means or what you wanted.

> 

restart
> 

with(DynamicSystems)

The default discretetimevar is q, so to use n

> 

SystemOptions(discretetimevar = n)

q

> 

sys2 := StateSpace([u(n) = K_virt*(q_a(n)-q_d)+D_virt*(q_a(n)-q_a(n-1))/T], inputvariable = [q_a(n)], outputvariable = [u(n)], discrete = true, sampletime = T)

Look at the matrices. Since we have c <> 0, the state variable is being used and is implicitly the delayed value - note that c is the coefficient of q_a(n-1)

> 

PrintSystem(sys2)

"[[[`State Space`],[`discrete; sampletime = T`],[`1 output(s); 1 input(s); 1 state(s)`],[inputvariable=[q_a(n)]],[outputvariable=[u(n)]],[statevariable=[x1(n)]],[a=[[[0]]]],[b=[[[1]]]],[c=[[[-D_virt/T]]]],[d=[[[(K_virt T+D_virt)/T]]]]]"

If you want to call it q_a_delayed, you have to explicitly invoke it in the difference equations

> 

sys3 := StateSpace([u(n) = K_virt*(q_a(n)-q_d)+D_virt*(q_a(n)-q_a_delayed(n))/T, q_a_delayed(n) = q_a(n-1)], inputvariable = [q_a(n)], outputvariable = [u(n)], statevariable = [q_a_delayed(n)], discrete = true, sampletime = T)

Same matrices

> 

PrintSystem(sys3)

"[[[`State Space`],[`discrete; sampletime = T`],[`1 output(s); 1 input(s); 1 state(s)`],[inputvariable=[q_a(n)]],[outputvariable=[u(n)]],[statevariable=[q_a_delayed(n)]],[a=[[[0]]]],[b=[[[1]]]],[c=[[[-D_virt/T]]]],[d=[[[(K_virt T+D_virt)/T]]]]]"

> 

NULL

Download DynamicSystems.mw

missing =...

dharr 7720
April 14 2025
2 1

In your params definition, you have varphi__8.33, which should instead be something like varphi__X=8.33.

braces...

dharr 7720
April 12 2025
0 0

Solve has an alternative way of working with braces, with slightly different syntax. 

braces.mw

Same problem...

dharr 7720
April 10 2025
1 3

Get rid of =0.

 F-P-O-W.mw

expression form of PDE...

dharr 7720
April 09 2025
4 1

Both methods work if you remove the "=0" on the pde. I always recommend this, for reasons I've given before.

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

_local(gamma)
> 

NULL
> 

declare(u(x, y, z, w, t))

u(x, y, z, w, t)*`will now be displayed as`*u

> 

declare(f(x, y, z, w, t))

f(x, y, z, w, t)*`will now be displayed as`*f

> 

pde := 4*(diff(u(x, y, z, w, t), x, t))-(diff(u(x, y, z, w, t), `$`(x, 3), y))+diff(u(x, y, z, w, t), x, `$`(y, 3))+12*(diff(u(x, y, z, w, t), x))*(diff(u(x, y, z, w, t), y))+12*u(x, y, z, w, t)*(diff(u(x, y, z, w, t), x, y))-6*(diff(u(x, y, z, w, t), z, w))

4*(diff(diff(u(x, y, z, w, t), t), x))-(diff(diff(diff(diff(u(x, y, z, w, t), x), x), x), y))+diff(diff(diff(diff(u(x, y, z, w, t), x), y), y), y)+12*(diff(u(x, y, z, w, t), x))*(diff(u(x, y, z, w, t), y))+12*u(x, y, z, w, t)*(diff(diff(u(x, y, z, w, t), x), y))-6*(diff(diff(u(x, y, z, w, t), w), z))

> 

oppde := [op(expand(pde))]; u_occurrences := map(proc (i) options operator, arrow; numelems(select(has, [op([op(i)])], u)) end proc, oppde); linear_op_indices := ListTools:-SearchAll(1, u_occurrences); pde_linear := add(oppde[[linear_op_indices]]); pde_nonlinear := expand(simplify(expand(pde)-pde_linear))

4*(diff(diff(u(x, y, z, w, t), t), x))-(diff(diff(diff(diff(u(x, y, z, w, t), x), x), x), y))+diff(diff(diff(diff(u(x, y, z, w, t), x), y), y), y)-6*(diff(diff(u(x, y, z, w, t), w), z))

12*(diff(u(x, y, z, w, t), x))*(diff(u(x, y, z, w, t), y))+12*u(x, y, z, w, t)*(diff(diff(u(x, y, z, w, t), x), y))

> 

pde_linear, pde_nonlinear := selectremove(proc (term) options operator, arrow; not has((eval(term, u(x, y, z, w, t) = a*u(x, y, z, w, t)))/a, a) end proc, expand(pde))

4*(diff(diff(u(x, y, z, w, t), t), x))-(diff(diff(diff(diff(u(x, y, z, w, t), x), x), x), y))+diff(diff(diff(diff(u(x, y, z, w, t), x), y), y), y)-6*(diff(diff(u(x, y, z, w, t), w), z)), 12*(diff(u(x, y, z, w, t), x))*(diff(u(x, y, z, w, t), y))+12*u(x, y, z, w, t)*(diff(diff(u(x, y, z, w, t), x), y))

> 

NULL

Download seperate_L-NL.mw

RootOf...

dharr 7720
April 07 2025
4 0

Well a lot depends on your objective, but here are some comments. 

(i) If the argument of RootOf is not a polynomial in _Z, then Maple has given up, and although you might be able to do some things with it, such as differentiate wrt a parameter, allvalues and convert radical won't do anything with it. I'm assuming remove_RootOf will put it back into unsolved form.

(ii) remove_RootOf just generates an implicit equation for which solve would give the RootOf solution; in some sense that is going backwards, but if you don't like RootOfs then it may be for you. On the other hand it is implicit, which you may not like.

(iii) If the argument of RootOf is a polynomial in _Z of degree n and it can be converted to radical form, then allvalues gives the n solutions in radical form. convert(..,radical) gives only the first of these; it is equivalent to convert(RootOf(..., index=1), radical).

(iv)  If the argument of RootOf is a polynomial in _Z of degree n and it can't be converted to radical form, which is the case for most high-degree polynomials, then convert(..,radical) won't do anything. allvalues will just return the n values still with RootOfs:

RootOf(..., index=1), RootOf(..., index=2), ... RootOf(..., index=n).

Ellipse example...

dharr 7720
April 06 2025
1 2

Here some code which should apply to any parametric curve. I tried to do it analytically, which works for the ellipse example, but in the general case will depend on the capabilities of solve.

> 

restart

Curve in a parametric form (here an ellipse)

> 

curve := proc (t) options operator, arrow; [2*sin(t), cos(t)] end proc

proc (t) options operator, arrow; [2*sin(t), cos(t)] end proc

> 

pcurve := plot([curve(t)[], t = 0 .. 2*Pi], scaling = constrained, color = red)

procedure to plot tangent line at parameter value t.  y=y0 + m*(x-x0), [x0,y0] on curve

> 

pT:=proc(t) local x0,y0,Dc,m,x;
x0,y0:=curve(t)[];
Dc:=D(curve)(t);
m:=Dc[2]/Dc[1];
plot(m*x+y0-m*x0,x,color=blue);
end proc:

Procedure to calculate curvature

> 

C:=proc(t) local dx,dy,d2x,d2y;
dx,dy:=D(curve)(t)[];
d2x,d2y:=(D@@2)(curve)(t)[];
abs(dx*d2y-dy*d2x)/(dx^2+dy^2)^(3/2)
end proc:

Slope of tangent``(non-normalized tangent vector)

> 

Dcurve := D(curve)

proc (z) options operator, arrow; [2*cos(z), -sin(z)] end proc

Find two points on curve with the same tangent slope and curvature

> 

T1 := (D(curve))(t__1); T2 := (D(curve))(t__2); equalslopes := T1[1]*T2[2] = T1[2]*T2[1]

[2*cos(t__1), -sin(t__1)]

[2*cos(t__2), -sin(t__2)]

-2*cos(t__1)*sin(t__2) = -2*sin(t__1)*cos(t__2)

For t spaced multiples of Pi apart the slopes are the same

> 

t2 := solve(equalslopes, t__2, allsolutions)

arctan(sin(t__1)/cos(t__1))+Pi*_Z1

> 

simplify(t2); simplify(t2, symbolic)

arctan(tan(t__1))+Pi*_Z1

Pi*_Z1+t__1

And the curvatures are the same under this condition

> 

C(.2); C(.2+Pi)``

.2615262308

.2615262308

And for the general case.

> 

simplify(C(t)-C(t+Pi))

0

> 

plots:-display(pcurve, pT(.2), pT(.2+Pi), view = [-2 .. 2, -2 .. 2])

> 

NULL

Download Blaschke.mw

Features...

dharr 7720
April 06 2025
1 0

Generally, Maple works identically across platforms. The help page ?platform lists platform-specific issues, but only Mac OS has small differences from the other versions. I'm assuming by features you mean the capabilities of Maple itself. (For the GUI, the Java interface is also intended to be platform independent.) 

simplify is best...

dharr 7720
April 06 2025
1 3

I would have said that s is already in factored form. Simplify is the correct way here, I think, though the answer appears to be 2^m not 2^(m+1). Maple's commands for working with polynomials and rational functions such as factor don't work for generic/unspecified exponents/degrees. Even if given a specific m, the default field for the coefficients is the rationals, so you will have to work more specifically on them to determine their factors, as I did here.

> 

restart

Factored form

> 

s := (1+sqrt(3))^(2*m+1)

(1+3^(1/2))^(2*m+1)

Example for m=6

> 

s6 := eval(s, m = 6)

(1+3^(1/2))^13

Expanded form

> 

e6 := expand(s6)

236224+136384*3^(1/2)

Extracting parts is ugly; then find gcd

> 

op(1, e6), op([2, 1], e6); igcd(%); 2^6

236224, 136384

64

64

> 

simplify(s/2^m)

(1+3^(1/2))*(2+3^(1/2))^m

> 

NULL

Download test2.mw

 

Uploading on Mapleprimes...

dharr 7720
April 05 2025
1 0

Click on the green up-arrow in the Mapleprimes editor, click on choose file and browse to the file on your computer you want to upload and select it. Then click upload, and then click one of "insert link" or "insert contents" buttons. This definitely works for worksheet (*.mw) files. It doesn't work directly for *.mpl files, but if it contains only text then just rename it to *.txt and upload that. It works for some other types of files, but if it doesn't work then you can convert it to a .zip file and upload that.

Inserting contents of a worksheet sometimes fails to display the contents, but gives a link that works anyway.

evaluation...

dharr 7720
April 04 2025
3 1

Full evaluation is occuring in the first case, but not the second. If you change the second to eval(cat(x_,i-1));  it becomes the same as the first.

So in the first case, for i = 2, x_2 := x_1, but x_1 is already x_0, so you get x_2 := x_0.

Needs assumptions...

dharr 7720
April 02 2025
1 1

I don't have Maple 2023, but for 2025 you need assumptions to get a result. I assumed everything positive, but you may have more specific requirements. For future reference. please upload a worksheet so we don't have to type in from the image.

Download Fourier.mw

complexplot...

dharr 7720
March 31 2025
3 0 
plots:-complexplot([allvalues(RootOf(x^6 - 1))], style = point,
     symbol = solidcircle, color = red, symbolsize = 15, scaling = constrained);

The same result can be obtained with fsolve

plots:-complexplot([fsolve(x^6 - 1, complex)], style = point,
     symbol = solidcircle, color = red, symbolsize = 15, scaling = constrained);

allvalues(RootOf(x^6 - 1)) gives the symbolic solutions and  fsolve(x^6 - 1, complex) gives the numeric solutions.

plottools:-polygon...

dharr 7720
March 31 2025
0 0

Here's an example.

> 

restart
> 

with(plots); with(plottools)
> 

display(polygon([[0, 0], [3, 4], [3, 1]], color = "Plasma 212"), polygon([[-1, 1], [-2, 4], [-5, 1]], color = red))

> 

NULL

Download polygons.mw

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