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dharr

Dr. David Harrington

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15 years, 220 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a professor of chemistry at the University of Victoria, BC, Canada, where 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

symbolic eigenval...

dharr 1018
Yesterday at 11:21 AM
0 0

A.B-1 is rank 1 so there is only one nonzero eigenvalue. If this eigenvalue is found symbolically, then parameter values may be varied in plots as shown.
 

> 

restart
> 

with(LinearAlgebra); A := Matrix(6, 6, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = alpha[1]*S[p], (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = beta, (4, 4) = 0, (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 0, (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = 0, (6, 5) = 0, (6, 6) = 0}); B := Matrix(6, 6, {(1, 1) = mu, (1, 2) = tau, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = e[o], (2, 1) = 0, (2, 2) = tau+mu, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = beta+eta[1]+sigma+mu, (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = (1-delta)*alpha[2]-mu, (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = alpha[2]*delta, (5, 5) = mu+eta[2], (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = 0, (6, 5) = 0, (6, 6) = mu+e[o]}); C := A.(1/B); Rank(C); evs := Eigenvalues(C); eig := op(`minus`({entries(evs, nolist)}, {0}))

A := Matrix(6, 6, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = alpha[1]*S[p], (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = beta, (4, 4) = 0, (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 0, (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = 0, (6, 5) = 0, (6, 6) = 0})

B := Matrix(6, 6, {(1, 1) = mu, (1, 2) = tau, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = e[o], (2, 1) = 0, (2, 2) = tau+mu, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = beta+eta[1]+sigma+mu, (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = (1-delta)*alpha[2]-mu, (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = alpha[2]*delta, (5, 5) = mu+eta[2], (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = 0, (6, 5) = 0, (6, 6) = mu+e[o]})

C := Matrix(6, 6, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = alpha[1]*S[p]/(beta+eta[1]+sigma+mu), (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = beta/(beta+eta[1]+sigma+mu), (4, 4) = 0, (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 0, (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = 0, (6, 5) = 0, (6, 6) = 0})

1

evs := Vector(6, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = alpha[1]*S[p]/(beta+eta[1]+sigma+mu)})

alpha[1]*S[p]/(beta+eta[1]+sigma+mu)

> 

params := {beta = .5, delta = .2115, mu = 0.2041e-1, sigma = .9, tau = .33, S[p] = 34.722406639004, alpha[1] = 0.2e-3, alpha[2] = .2, e[o] = .33, eta[1] = 0.96e-1, eta[2] = 0.2485e-2}

{beta = .5, delta = .2115, mu = 0.2041e-1, sigma = .9, tau = .33, S[p] = 34.722406639004, alpha[1] = 0.2e-3, alpha[2] = .2, e[o] = .33, eta[1] = 0.96e-1, eta[2] = 0.2485e-2}

Plot vs mu, for example

> 

plot(eval(eig, remove(has, params, mu)), mu = 0 .. 1)

> 

 


 

Download eval.mw

cursor and rectangle...

dharr 1018
April 02 2020
2 1

For regular 1-D input, a rectangle surrounds the parenthesis that matches the one next to the cursor (vertical line here)

getdata is in plottools package...

dharr 1018
March 28 2020
0 2

If you don't try to see R0 (terminate with colon), then implicitplot is fine with gridrefine=3 (gridrefine=10 runs into a memory allocation on my machine), and the data is extracted with plottools:-getdata without issue.

implictplot.mw

isosurface...

dharr 1018
March 19 2020
1 1

The ISOSURFACE plot structure (see ?plot,structure) is used internally by implicitplot3d to draw a surface where a function is zero over an evenly spaced grid of values, so it is close to your dataarray, assuming your x,y,z values are evenly spaced. So to get a red surface enclosing -1 values can be done as below. Not sure it makes sense to color the rest of space green since then you couldn't see anything.

> 

restart;with(plots):

Generate dataarray with -1 values within an ellipse, +1 outside

> 

ellipse:=(x,y,z)->if x^2/0.2+y^2/0.8+z^2/0.3 < 1 then -1 else 1 end if;

proc (x, y, z) options operator, arrow; if x^2/.2+y^2/.8+z^2/.3 < 1 then -1 else 1 end if end proc

> 

x:=Vector(1..10,i->(i-5.5)/5):
y:=Vector(1..10,i->(i-5.5)/5):
z:=Vector(1..10,i->(i-5.5)/5):
dataarray:=Array(1..10,1..10,1..10,(i,j,k)->ellipse(x[i],y[j],z[k]),datatype=float[8],order=C_order):

Fill array required for ISOSURFACE structure

> 

isoarray:=Array(1..10,1..10,1..10,1..4,datatype=float[8],order=C_order):
> 

for ix to 10 do
  for iy to 10 do
    for iz to 10 do
      isoarray[ix,iy,iz,1]:=x[ix];
      isoarray[ix,iy,iz,2]:=y[iy];
      isoarray[ix,iy,iz,3]:=z[iz];
      isoarray[ix,iy,iz,4]:=dataarray[ix,iy,iz];
   end do;
 end do;
end do:
> 

display(PLOT3D(ISOSURFACE(isoarray)),view=[-1..1,-1..1,-1..1],color=red);

> 

 


 

Download isosurface2.mw

 

2 argument arctan...

dharr 1018
March 12 2020
3 1

Try arctan(-diff(fy,t),diff(fx,t)), or perhaps arctan(diff(fy,t),diff(fx,t)) I didn't follow the details, but it removes the discontinuity.

slow...

dharr 1018
March 10 2020
0 1

It's just very slow, especially for small and large times. You are doing an integral within an integral. If you change the plot range to 1..1000 then you do get a plot in a few seconds.

and a few more...

dharr 1018
March 09 2020
0 0

In addition to @tomleslie 's solutions there is one real one between 0 and 1. I have also made the mininum number of changes to convert from the old linalg package to the LinearAlgebra package. To more consistently find the roots, you could use the NextZero routine in the RootFinding package.

I can't see any typos, so I don't know where the 0.5 root could come from.


 

> 

restart
> 

with(LinearAlgebra)
> 

V0 := 1

1

> 

a := 2

2

> 

b := 1

1

> 

assume(En > 0)
> 

ks := sqrt(2*En)

2^(1/2)*En^(1/2)

> 

qs := sqrt(2*(V0-En))

(2-2*En)^(1/2)

> 

psi11 := proc (x) options operator, arrow; A1*exp(qs*x)+B1*exp(-qs*x) end proc

proc (x) options operator, arrow; A1*exp(qs*x)+B1*exp(-qs*x) end proc

> 

psi21 := proc (x) options operator, arrow; A2*exp(I*ks*x)+B2*exp(-I*ks*x) end proc

proc (x) options operator, arrow; A2*exp(I*ks*x)+B2*exp(-I*ks*x) end proc

> 

lp := a+b

3

> 

psi12 := proc (x) options operator, arrow; exp(I*Kp*lp)*(A1*exp(qs*(x-lp))+B1*exp(-qs*(x-lp))) end proc

proc (x) options operator, arrow; exp(I*Kp*lp)*(A1*exp(qs*(x-lp))+B1*exp(-qs*(x-lp))) end proc

> 

psi22 := proc (x) options operator, arrow; exp(I*Kp*lp)*(A2*exp(I*ks*(x-lp))+B2*exp(-I*ks*(x-lp))) end proc

proc (x) options operator, arrow; exp(I*Kp*lp)*(A2*exp(I*ks*(x-lp))+B2*exp(-I*ks*(x-lp))) end proc

> 

eq1 := psi11(0) = psi21(0)

A1+B1 = A2+B2

> 

eq2 := (D(psi11))(0)-(D(psi21))(0)

A1*(2-2*En)^(1/2)-B1*(2-2*En)^(1/2)-I*A2*2^(1/2)*En^(1/2)+I*B2*2^(1/2)*En^(1/2)

> 

eq3 := expand(psi21(a)-psi12(a))

A2*(exp(I*2^(1/2)*En^(1/2)))^2+B2/(exp(I*2^(1/2)*En^(1/2)))^2-(exp(I*Kp))^3*A1/exp((2-2*En)^(1/2))-(exp(I*Kp))^3*B1*exp((2-2*En)^(1/2))

> 

eq4 := expand((D(psi21))(a)-(D(psi12))(a))

I*A2*2^(1/2)*En^(1/2)*(exp(I*2^(1/2)*En^(1/2)))^2-I*B2*2^(1/2)*En^(1/2)/(exp(I*2^(1/2)*En^(1/2)))^2-(exp(I*Kp))^3*A1*(2-2*En)^(1/2)/exp((2-2*En)^(1/2))+(exp(I*Kp))^3*B1*(2-2*En)^(1/2)*exp((2-2*En)^(1/2))

> 

row1 := Vector[row]([1, 1, -1, -1])

row1 := Vector[row](4, {(1) = 1, (2) = 1, (3) = -1, (4) = -1})

> 

row2 := Vector[row]([coeff(eq2, A1), coeff(eq2, B1), coeff(eq2, A2), coeff(eq2, B2)])

row2 := Vector[row](4, {(1) = (2-2*En)^(1/2), (2) = -(2-2*En)^(1/2), (3) = -I*2^(1/2)*En^(1/2), (4) = I*2^(1/2)*En^(1/2)})

> 

row3 := Vector[row]([coeff(eq3, A1), coeff(eq3, B1), coeff(eq3, A2), coeff(eq3, B2)])

row3 := Vector[row](4, {(1) = -(exp(I*Kp))^3/exp((2-2*En)^(1/2)), (2) = -(exp(I*Kp))^3*exp((2-2*En)^(1/2)), (3) = (exp(I*2^(1/2)*En^(1/2)))^2, (4) = 1/(exp(I*2^(1/2)*En^(1/2)))^2})

> 

row4 := Vector[row]([coeff(eq4, A1), coeff(eq4, B1), coeff(eq4, A2), coeff(eq4, B2)])

row4 := Vector[row](4, {(1) = -(exp(I*Kp))^3*(2-2*En)^(1/2)/exp((2-2*En)^(1/2)), (2) = (exp(I*Kp))^3*(2-2*En)^(1/2)*exp((2-2*En)^(1/2)), (3) = I*2^(1/2)*En^(1/2)*(exp(I*2^(1/2)*En^(1/2)))^2, (4) = -I*2^(1/2)*En^(1/2)/(exp(I*2^(1/2)*En^(1/2)))^2})

> 

C := Matrix(4, 4, 0)

C := Matrix(4, 4, {(1, 1) = 1, (1, 2) = 1, (1, 3) = -1, (1, 4) = -1, (2, 1) = (2-2*En)^(1/2), (2, 2) = -(2-2*En)^(1/2), (2, 3) = -I*2^(1/2)*En^(1/2), (2, 4) = I*2^(1/2)*En^(1/2), (3, 1) = -(exp(I*Kp))^3/exp((2-2*En)^(1/2)), (3, 2) = -(exp(I*Kp))^3*exp((2-2*En)^(1/2)), (3, 3) = (exp(I*2^(1/2)*En^(1/2)))^2, (3, 4) = 1/(exp(I*2^(1/2)*En^(1/2)))^2, (4, 1) = -(exp(I*Kp))^3*(2-2*En)^(1/2)/exp((2-2*En)^(1/2)), (4, 2) = (exp(I*Kp))^3*(2-2*En)^(1/2)*exp((2-2*En)^(1/2)), (4, 3) = I*2^(1/2)*En^(1/2)*(exp(I*2^(1/2)*En^(1/2)))^2, (4, 4) = -I*2^(1/2)*En^(1/2)/(exp(I*2^(1/2)*En^(1/2)))^2})

> 

for i to 4 do C[1, i] := row1[i]; C[2, i] := row2[i]; C[3, i] := row3[i]; C[4, i] := row4[i] end do
> 

C

Matrix([[1, 1, -1, -1], [(2-2*En)^(1/2), -(2-2*En)^(1/2), -I*2^(1/2)*En^(1/2), I*2^(1/2)*En^(1/2)], [-(exp(I*Kp))^3/exp((2-2*En)^(1/2)), -(exp(I*Kp))^3*exp((2-2*En)^(1/2)), (exp(I*2^(1/2)*En^(1/2)))^2, 1/(exp(I*2^(1/2)*En^(1/2)))^2], [-(exp(I*Kp))^3*(2-2*En)^(1/2)/exp((2-2*En)^(1/2)), (exp(I*Kp))^3*(2-2*En)^(1/2)*exp((2-2*En)^(1/2)), I*2^(1/2)*En^(1/2)*(exp(I*2^(1/2)*En^(1/2)))^2, -I*2^(1/2)*En^(1/2)/(exp(I*2^(1/2)*En^(1/2)))^2]])

> 

chareq := Determinant(C)

2*((2*I)*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*(exp(I*Kp))^6*exp((2-2*En)^(1/2))*(exp(I*2^(1/2)*En^(1/2)))^2-I*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2*(exp(I*2^(1/2)*En^(1/2)))^4-I*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*(exp(I*Kp))^3*(exp(I*2^(1/2)*En^(1/2)))^4+2*En*(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2*(exp(I*2^(1/2)*En^(1/2)))^4-I*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2-(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2*(exp(I*2^(1/2)*En^(1/2)))^4-2*En*(exp(I*Kp))^3*(exp(I*2^(1/2)*En^(1/2)))^4-I*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*(exp(I*Kp))^3+(2*I)*2^(1/2)*En^(1/2)*(2-2*En)^(1/2)*exp((2-2*En)^(1/2))*(exp(I*2^(1/2)*En^(1/2)))^2+(exp(I*Kp))^3*(exp(I*2^(1/2)*En^(1/2)))^4-2*En*(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2+(exp(I*Kp))^3*(exp((2-2*En)^(1/2)))^2+2*En*(exp(I*Kp))^3-(exp(I*Kp))^3)/(exp((2-2*En)^(1/2))*(exp(I*2^(1/2)*En^(1/2)))^2)

> 

chareq0 := simplify(subs(Kp = 0, chareq))

(-(2*I)*(2-2*En)^(1/2)*2^(1/2)*En^(1/2)+4*En-2)*exp(-(2-2*En)^(1/2)-(2*I)*2^(1/2)*En^(1/2))+(-(2*I)*(2-2*En)^(1/2)*2^(1/2)*En^(1/2)-4*En+2)*exp(-(2-2*En)^(1/2)+(2*I)*2^(1/2)*En^(1/2))+(-(2*I)*(2-2*En)^(1/2)*2^(1/2)*En^(1/2)-4*En+2)*exp((2-2*En)^(1/2)-(2*I)*2^(1/2)*En^(1/2))+(-(2*I)*(2-2*En)^(1/2)*2^(1/2)*En^(1/2)+4*En-2)*exp((2-2*En)^(1/2)+(2*I)*2^(1/2)*En^(1/2))+(8*I)*(2-2*En)^(1/2)*2^(1/2)*En^(1/2)

Real parts are all zero for En<1

> 

plot(Re(chareq0), En = 0 .. 3, color = red, thickness = 3)

Imaginary parts are all zero for En>1

> 

plot(Im(chareq0), En = 0 .. 3, color = red, thickness = 3)

> 

sol1 := fsolve(chareq0, En = .2 .. .4); sol2 := fsolve(chareq0, En = 2 .. 2.5)

.2669813621+0.*I

2.375133245-0.*I

> 

 


 

Download qm_maple_-_periodic_potentials.mw
 

different compressions and formats...

dharr 1018
March 08 2020
1 2

As @acer suggested, they have different compressions (and formats). Corel PhotoPaint reports the original 125 KB file is 1 bit B&W, 300 dpi, CCITT group 3 subformat (compression), and the 133 KB one written by Maple is 72 dpi, 8 bit grayscale and LZW compression. LZW is a lossless compression, but it is interesting that the output resolution is lower.

Edit: aside from the resolution issue, the above is as indicated on the ImageTools[Formats] help page.

 

Enter after prompt...

dharr 1018
March 08 2020
2 0

There was nothing wrong with  the actual code, but it hadn't been entered after a prompt (in a worksheet) and was somehow two separate regions. If you enter 1-D code as you did, you need to format it yourself. You get some nicer formatting with a code edit region if you are going to enter a lot of code.

For this site, uploading a worksheet as you did is probably best.

proc.mw

Edit: Using a startup code edit region for initialization and procedure definitions can be a nice way to do it.

Matrix of integers...

dharr 1018
March 07 2020
2 0

The help for CayleyTableGroup specifies: The single required argument is the Cayley table (or multiplication table): an n by n
 Matrix or Array, m, describing the group with n elements, where the [i, j] entry contains the positive integer k such that the product of the ith and j h group elements is the kth group element.

So this needs to be a matrix of integers.


 

> 

restart;interface(rtablesize=12);

10

> 

with(GroupTheory):
> 

ct := <<e | p | q | r | s | t | u | v | w | x | y | z>, <p | q | e | y | u | w | z | r | x | t | v | s>, <q | e | p | v | z | x | s | y | t | w | r | u>, <r | z | t | s | e | y | v | x | p | u | q | w>, <s | w | y | e | r | q | x | u | z | v | t | p>, <t | r | z | x | w | u | e | q | y | p | s | v>, <u | x | v | p | y | e | t | z | s | r | w | q>, <v | u | x | z | q | r | y | w | e | s | p | t>, <w | y | s | t | x | z | p | e | v | q | u | r>, <x | v | u | w | t | s | q | p | r | e | z | y>, <y | s | w | u | p | v | r | t | q | z | e | x>, <z | t | r | q | v | p | w | s | u | y | x | e>>;

ct := Matrix(12, 12, {(1, 1) = e, (1, 2) = p, (1, 3) = q, (1, 4) = r, (1, 5) = s, (1, 6) = t, (1, 7) = u, (1, 8) = v, (1, 9) = w, (1, 10) = x, (1, 11) = y, (1, 12) = z, (2, 1) = p, (2, 2) = q, (2, 3) = e, (2, 4) = y, (2, 5) = u, (2, 6) = w, (2, 7) = z, (2, 8) = r, (2, 9) = x, (2, 10) = t, (2, 11) = v, (2, 12) = s, (3, 1) = q, (3, 2) = e, (3, 3) = p, (3, 4) = v, (3, 5) = z, (3, 6) = x, (3, 7) = s, (3, 8) = y, (3, 9) = t, (3, 10) = w, (3, 11) = r, (3, 12) = u, (4, 1) = r, (4, 2) = z, (4, 3) = t, (4, 4) = s, (4, 5) = e, (4, 6) = y, (4, 7) = v, (4, 8) = x, (4, 9) = p, (4, 10) = u, (4, 11) = q, (4, 12) = w, (5, 1) = s, (5, 2) = w, (5, 3) = y, (5, 4) = e, (5, 5) = r, (5, 6) = q, (5, 7) = x, (5, 8) = u, (5, 9) = z, (5, 10) = v, (5, 11) = t, (5, 12) = p, (6, 1) = t, (6, 2) = r, (6, 3) = z, (6, 4) = x, (6, 5) = w, (6, 6) = u, (6, 7) = e, (6, 8) = q, (6, 9) = y, (6, 10) = p, (6, 11) = s, (6, 12) = v, (7, 1) = u, (7, 2) = x, (7, 3) = v, (7, 4) = p, (7, 5) = y, (7, 6) = e, (7, 7) = t, (7, 8) = z, (7, 9) = s, (7, 10) = r, (7, 11) = w, (7, 12) = q, (8, 1) = v, (8, 2) = u, (8, 3) = x, (8, 4) = z, (8, 5) = q, (8, 6) = r, (8, 7) = y, (8, 8) = w, (8, 9) = e, (8, 10) = s, (8, 11) = p, (8, 12) = t, (9, 1) = w, (9, 2) = y, (9, 3) = s, (9, 4) = t, (9, 5) = x, (9, 6) = z, (9, 7) = p, (9, 8) = e, (9, 9) = v, (9, 10) = q, (9, 11) = u, (9, 12) = r, (10, 1) = x, (10, 2) = v, (10, 3) = u, (10, 4) = w, (10, 5) = t, (10, 6) = s, (10, 7) = q, (10, 8) = p, (10, 9) = r, (10, 10) = e, (10, 11) = z, (10, 12) = y, (11, 1) = y, (11, 2) = s, (11, 3) = w, (11, 4) = u, (11, 5) = p, (11, 6) = v, (11, 7) = r, (11, 8) = t, (11, 9) = q, (11, 10) = z, (11, 11) = e, (11, 12) = x, (12, 1) = z, (12, 2) = t, (12, 3) = r, (12, 4) = q, (12, 5) = v, (12, 6) = p, (12, 7) = w, (12, 8) = s, (12, 9) = u, (12, 10) = y, (12, 11) = x, (12, 12) = e})

> 

elems:=convert(ct[1],list);
for i to nops(elems) do f(elems[i]):=i end do: #use remember table

[e, p, q, r, s, t, u, v, w, x, y, z]

> 

M:=map(f,ct);

M := Matrix(12, 12, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (1, 4) = 4, (1, 5) = 5, (1, 6) = 6, (1, 7) = 7, (1, 8) = 8, (1, 9) = 9, (1, 10) = 10, (1, 11) = 11, (1, 12) = 12, (2, 1) = 2, (2, 2) = 3, (2, 3) = 1, (2, 4) = 11, (2, 5) = 7, (2, 6) = 9, (2, 7) = 12, (2, 8) = 4, (2, 9) = 10, (2, 10) = 6, (2, 11) = 8, (2, 12) = 5, (3, 1) = 3, (3, 2) = 1, (3, 3) = 2, (3, 4) = 8, (3, 5) = 12, (3, 6) = 10, (3, 7) = 5, (3, 8) = 11, (3, 9) = 6, (3, 10) = 9, (3, 11) = 4, (3, 12) = 7, (4, 1) = 4, (4, 2) = 12, (4, 3) = 6, (4, 4) = 5, (4, 5) = 1, (4, 6) = 11, (4, 7) = 8, (4, 8) = 10, (4, 9) = 2, (4, 10) = 7, (4, 11) = 3, (4, 12) = 9, (5, 1) = 5, (5, 2) = 9, (5, 3) = 11, (5, 4) = 1, (5, 5) = 4, (5, 6) = 3, (5, 7) = 10, (5, 8) = 7, (5, 9) = 12, (5, 10) = 8, (5, 11) = 6, (5, 12) = 2, (6, 1) = 6, (6, 2) = 4, (6, 3) = 12, (6, 4) = 10, (6, 5) = 9, (6, 6) = 7, (6, 7) = 1, (6, 8) = 3, (6, 9) = 11, (6, 10) = 2, (6, 11) = 5, (6, 12) = 8, (7, 1) = 7, (7, 2) = 10, (7, 3) = 8, (7, 4) = 2, (7, 5) = 11, (7, 6) = 1, (7, 7) = 6, (7, 8) = 12, (7, 9) = 5, (7, 10) = 4, (7, 11) = 9, (7, 12) = 3, (8, 1) = 8, (8, 2) = 7, (8, 3) = 10, (8, 4) = 12, (8, 5) = 3, (8, 6) = 4, (8, 7) = 11, (8, 8) = 9, (8, 9) = 1, (8, 10) = 5, (8, 11) = 2, (8, 12) = 6, (9, 1) = 9, (9, 2) = 11, (9, 3) = 5, (9, 4) = 6, (9, 5) = 10, (9, 6) = 12, (9, 7) = 2, (9, 8) = 1, (9, 9) = 8, (9, 10) = 3, (9, 11) = 7, (9, 12) = 4, (10, 1) = 10, (10, 2) = 8, (10, 3) = 7, (10, 4) = 9, (10, 5) = 6, (10, 6) = 5, (10, 7) = 3, (10, 8) = 2, (10, 9) = 4, (10, 10) = 1, (10, 11) = 12, (10, 12) = 11, (11, 1) = 11, (11, 2) = 5, (11, 3) = 9, (11, 4) = 7, (11, 5) = 2, (11, 6) = 8, (11, 7) = 4, (11, 8) = 6, (11, 9) = 3, (11, 10) = 12, (11, 11) = 1, (11, 12) = 10, (12, 1) = 12, (12, 2) = 6, (12, 3) = 4, (12, 4) = 3, (12, 5) = 8, (12, 6) = 2, (12, 7) = 9, (12, 8) = 5, (12, 9) = 7, (12, 10) = 11, (12, 11) = 10, (12, 12) = 1})

> 

G := Group(M,labels=elems);

_m231969395104

> 

 


 

Download Cayley.mw

large numbers...

dharr 1018
March 04 2020
1 0

I don't get an error message for the plot in Maple 2015, just an empty plot. But eval(S,w=1) gives -1.853089027*10^329399 so your numbers seem to be unreasonably large.

realroot...

dharr 1018
February 28 2020
0 0


 

> 

restart;
> 

A:=(x^3-13*x^2+55*x-73)/(x-1)=0:
p:=numer(lhs(A));
realroot(p)[];

x^3-13*x^2+55*x-73

[169/64, 339/128]


 

Download realroot.mw

avoided error message but still wrong...

dharr 1018
February 25 2020
0 2

[Edited based on Carl's comment]. At least for real parameters (where the Matrix is Hermitian), there must be four eigenvectors.Not setting phi=Pi circumvents the OPs error message, and 4 eigenvectors are found, but then substituting phi=Pi leads to an error.


 

> 

restart;
> 

with(LinearAlgebra):
> 

assume(k1,real,k2,real,m1,real,m2,real,phi,real);

Leave phi unspecified for now

> 

A := <
          <0, 0, exp(I*k1) + m1, exp(I*k2) + m2>|
          <0, 0, exp(I*phi)*(exp(-I*k2) + m2), exp(-I*k1) + m1>|
          <exp(-I*k1) + m1, exp(-I*phi)*(m2 + exp(I*k2)), 0, 0>|
          <exp(-I*k2) + m2, exp(I*k1) + m1, 0, 0>
     >;

A := Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = exp(-I*k1)+m1, (1, 4) = exp(-I*k2)+m2, (2, 1) = 0, (2, 2) = 0, (2, 3) = exp(-I*phi)*(exp(I*k2)+m2), (2, 4) = exp(I*k1)+m1, (3, 1) = exp(I*k1)+m1, (3, 2) = exp(I*phi)*(exp(-I*k2)+m2), (3, 3) = 0, (3, 4) = 0, (4, 1) = exp(I*k2)+m2, (4, 2) = exp(-I*k1)+m1, (4, 3) = 0, (4, 4) = 0})

Under the assumptions that the parameters are real, the Matrix is Hermitian and so must have 4 eigenvectors

> 

simplify(A-A^%H);

Matrix([[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]])

If we specifiy phi=Pi, we get the OP's error

> 

evals,evecs:=LinearAlgebra:-Eigenvectors(eval(A,phi=Pi)):

Error, (in LinearAlgebra:-Eigenvectors) multiplicity mismatch

If we leave phi unspecified, then we don't get the error message, and get 4 eigenvectors

> 

evals,evecs:=LinearAlgebra:-Eigenvectors(A):
nops(evals),nops(evecs);  #nops always reterns 3, as Carl points out
numelems(evals),numelems(evecs);

3, 3

4, 16

Now specify phi=Pi. Works for the eigenvalues, but not for the eigenvectors. Perhaps some manipulation could improve this, but I'm not sure

> 

evals2:=eval(evals,phi=Pi);
eval(evecs,phi=Pi);

evals2 := Vector(4, {(1) = (2+m1^2+m2^2+2*m1*cos(k1)+2*m2*cos(k2))^(1/2), (2) = -(2+m1^2+m2^2+2*m1*cos(k1)+2*m2*cos(k2))^(1/2), (3) = (2+m1^2+m2^2+2*m1*cos(k1)+2*m2*cos(k2))^(1/2), (4) = -(2+m1^2+m2^2+2*m1*cos(k1)+2*m2*cos(k2))^(1/2)})

Error, numeric exception: division by zero

 


 

Download Hermitian2.mw

need multiple curves...

dharr 1018
February 21 2020
1 0

When legend is a list there need to be multiple curves, but you had one curve with 3 points. One way to do this is:

display(pointplot([28,.6481496576],color=blue),pointplot([28,.648149657615473],color=orange),pointplot([28,.6512873548],color=red),style=point,labels = ["k", "y(k)"], legend = ["10-digit precision", "15-digit precision", "Floating-point iteration"] ,legendstyle = [font = ["HELVETICA", 9], location = right]);
 

 

invlaplace...

dharr 1018
February 19 2020
1 12

Not entirely certain I know what you want, but this may be a start.

[Edit: I changed pi (just a symbol) to Pi (the circle constant thing)]

At the beginning I define i to be sqrt(-1) as you want and D to be a regular variable rather than the differential operator.

Maple gives the answer as a sum over roots of a quartic polynomial. To get a simpler formula, you will need to give values to some variables or parhaps tell Maple their signs (using assuming)

Download Linear_System.mw

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