Please how can I define two sequences in a procedure with two arguments?

  ans:= Array([seq( [j, doCalc(j, u)], j=-2..0, 0.006, u=1..334)]):

The procedure is doCalc(j,u)  and I received this error: invalid input: seq expects between 1 and 3 arguments, but received 4

Find attached my complete code.Seq_Proc.mw

Please, how do I find the minimum of the real part of a complex function? I tried min ( ) function it didn't work. 

Find attached the fileFinding_min_zero.mw
 

Download Finding_min_zero.mw

I have tried a code in python for Francis QR algorithm but didn't desire the result. I don't know if it is possible to code in maple.

Given that A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]].

1. Write a little program that computes 1 step of Francis QR and test your program by starting from

A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]]  and compute A^1, A^2 ...A^6.

I expected to get:

A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]], 

A^1 = [[3.5,  -4.264, 0.2688], [-9.206, 1.577, 9.197], [0., -1.41, 1.923]], 

... A^6 = [[8.056,  1.596, 8.584], [0.3596, -2.01, -7.86], [0., 2.576*10^(-16), 0.9542]]. 

But didn't get the same results.

Here is my python code:

# Import packages
import numpy as np
from numpy.linalg import qr # QR from Linear Algebra Library
import scipy.linalg   # SciPy Linear Algebra Library
 

A = np.array([[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]])
p = [1, 2, 3, 4, 5, 6]
for i in range(30):
    q, r = qr(A)
    a = np.dot(r, q)
    if i+1 in p:
        print("Iteration {i+1}")
        print(A)

I would really appreciate your help.

Thank you.

Please, how can I find all the roots  of: (H/(Hc))^2 -(4*q^2)/Pi^2*((tan(q)- q/(1-alpha))/(tan(q)-q)) with q=(i-1)*Pi+Pi/2..i*Pi+Pi/2 for n=20.

See my attempt below:

restart:with(RootFinding):
a[1] := .1093; k[3] := 7.5*10^(-12); k[2] := 3.8*10^(-12); d := 0.2e-3; eta[1] := 0.240e-1; alpha[2] := -.1104; alpha[3] := -0.1104e-2; eta[2] := .1361; xi := 1.219*10^(-6); alpha := 1-alpha[3]^2/(a[1]*eta[1]); theta[0] := 0.5e-1;
Hc := (Pi/d)*sqrt(k[2]/xi):

H := 5.5*Hc;
lambda := a[1]/(xi*H^2);

RootFinding:-NextZero(proc (q) options operator, arrow; (H/(Hc))^2 -(4*q^2)/Pi^2*((tan(q)- q/(1-alpha))/(tan(q)-q)) end proc, 0);

for j to 9 do RootFinding:-NextZero(proc (q) options operator, arrow; H^2/Hc^2-4*q^2*(tan(q)-q/(1-alpha))/(Pi^2*(tan(q)-q)) end proc, %) end do;

Error, invalid input: RootFinding:-NextZero expects its 2nd argument, zz, to be of type numeric, but received FAIL

restart;
with(LinearAlgebra):
a := .1093;
b[3] := -0.1104e-2;
k[1] := 3.8*10^(-12);
d := 0.2e-3;
xi := -1;
theta0 := 0.1e-3;
eta[1] := 0.240e-1;
alpha := 1-b[3]^2/(a*eta[1]);
c := b[3]*xi*alpha/(eta[1]*(4*k[1]*q^2/d^2-b[3]*xi/eta[1]));

f:= q-(1-alpha)*tan(q)- c*tan(q) ;
plot(f,q=0..4*Pi,view=[DEFAULT,-1..10]):

   
n:=3: 
q[1]:=fsolve(f,q=(0..Pi/2)): 
for i from 2 to n do
q[i]:=fsolve(f,q=(2*i-3)*Pi/2..(2*i-1)*Pi/2): ##q[n]
od:

printlevel := 2;
for i from 2 to n+1 do for j from i+1 to n do b[i, j] := int((cos(q[i]-2*q[i]*z/d)-cos(q[i]))*(cos(q[j]-2*q[j]*z/d)-cos(q[j])), z = 0 .. d); b[j, i] := b[i, j] end do end do;

for i from 2 to n do a[i, i] := int((cos(q[i]-2*q[i]*z/d)-cos(q[i]))*(cos(q[i]-2*q[i]*z/d)-cos(q[i])), z = 0 .. d) end do;

for i from 2 to n do f[i] := theta0*(int(cos(q[i]-2*q[i]*z/d)-cos(q[i]), z = 0 .. d)) end do;

## I have generated a matrix using the a[i,i] above
A := Matrix([seq([seq(a[i, j], i = 2 .. n)], j = 2 .. n)]);

b := Vector([seq(f[i], i = 2 .. n)]);
     

#Please, how do I generate an upper triangular matrix using b[i,j] above?