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Answer...

vv 12453
August 21 2020
1 1

Your problem is equivalent to the generalized eigenvalue problem A = lambda * A^+
(A^+ is the transpose)
This problem is solved very efficiently by Maple. Try:

with(LinearAlgebra):
A:=<2,3;4,5>;
Eigenvectors((A^(-1))^+ . A);
Eigenvectors(A , A^+);
A:=RandomMatrix(18, generator=-5..5, datatype=float[8]);
Eigenvectors(A , A^+);
Digits:=200;
A:=RandomMatrix(18, generator=-5..5, datatype=float);
Eigenvectors(A , A^+);

 

Ad hoc solution (brute-force)...

vv 12453
August 20 2020
2 0 
IsEulerPath:=proc(G::list(set), excludecirc:=true)
  local e,  a, a1, ok;
  for a1 in G[1] do  # a1 = first vertex
    ok:=true;
    a:=(G[1] minus {a1})[];
    for e in G[2..] do
      if   a = e[1] then a := e[2]
      elif a = e[2] then a := e[1]
      else ok:=false; break  fi;
    od;
    if ok then break fi 
  od;  
  if not ok then return false fi;
  if excludecirc and (a = a1) then return false fi;
  true
end proc:

G:=[{1,2}, {1,3}, {1,4}, {3,4}$2, {2,4}]:
select(IsEulerPath, combinat:-permute(G)); nops(%)
      [[{1, 2}, {2, 4}, {1, 4}, {1, 3}, {3, 4}, {3, 4}],
        [{1, 2}, {2, 4}, {3, 4}, {1, 3}, {1, 4}, {3, 4}],
        [{1, 2}, {2, 4}, {3, 4}, {3, 4}, {1, 4}, {1, 3}],
        [{1, 3}, {1, 2}, {2, 4}, {3, 4}, {3, 4}, {1, 4}],
        [{1, 3}, {1, 4}, {3, 4}, {3, 4}, {2, 4}, {1, 2}],
        [{1, 3}, {3, 4}, {1, 4}, {1, 2}, {2, 4}, {3, 4}],
        [{1, 3}, {3, 4}, {2, 4}, {1, 2}, {1, 4}, {3, 4}],
        [{1, 4}, {3, 4}, {1, 3}, {1, 2}, {2, 4}, {3, 4}],
        [{1, 4}, {3, 4}, {3, 4}, {2, 4}, {1, 2}, {1, 3}],
        [{1, 4}, {2, 4}, {1, 2}, {1, 3}, {3, 4}, {3, 4}],
        [{3, 4}, {1, 4}, {1, 2}, {2, 4}, {3, 4}, {1, 3}],
        [{3, 4}, {1, 4}, {1, 3}, {3, 4}, {2, 4}, {1, 2}],
        [{3, 4}, {3, 4}, {1, 3}, {1, 2}, {2, 4}, {1, 4}],
        [{3, 4}, {3, 4}, {1, 3}, {1, 4}, {2, 4}, {1, 2}],
        [{3, 4}, {2, 4}, {1, 2}, {1, 3}, {3, 4}, {1, 4}],
        [{3, 4}, {2, 4}, {1, 2}, {1, 4}, {3, 4}, {1, 3}]]

                               16
 
Edit: a bug corrected

solvable...

vv 12453
August 18 2020
4 5

The equation is solvable [in terms of radicals] only for some values of k.

For example, for k=2,
galois(6*w^5 - 15*w^4 + 10*w^3 - 2, w)
returns the group "5T5"  which is not solvable.

But for k=1, solve finds the roots.

In https://en.wikipedia.org/wiki/Quintic_function  there are other methods to check the solvability.

bug...

vv 12453
August 16 2020
1 4

FindAngle(A, T) computes the angle between the sides passing thru A (which is always <= Pi/2) and not the angle of the triangle.

Maybe the bug is in the documentation (which says "angle between two given objects" but then "angle of T at A").

You may use: LinearAlgebra:-VectorAngle(B-A, C-A); A,B,C being lists or vectors of any dimension.

 

Answer...

vv 12453
August 13 2020
2 0 
restart;
em:=eulermac(1/(x^k-1), k = 1 .. infinity, 3) assuming x>1;
answer:=series(em, x=1);

Note that series uses a weaker big-O notation (up to a sub-polynomial growth), so  O(x - 1) =O((x - 1) *ln(x-1)).
 

Answer...

vv 12453
August 10 2020
1 4

Let's compute the ranks for some integer random values for the symbols.

 

> 

CO := Matrix(28, 28, {(1, 2) = s1, (1, 3) = s2, (1, 4) = s3, (1, 5) = s4, (1, 6
) = s5, (1, 7) = s6, (1, 8) = s7, (1, 9) = s8, (1, 10) = s9, (1, 11) = s10, (1,
12) = s11, (1, 13) = s12, (1, 14) = s13, (1, 15) = s14, (1, 16) = s15, (1, 17)
= s16, (1, 18) = s17, (1, 19) = s18, (1, 20) = s19, (1, 21) = s20, (1, 22) =
s21, (1, 23) = s22, (1, 24) = s23, (1, 25) = s24, (1, 26) = s25, (1, 27) = s26,
(1, 28) = s27, (2, 7) = s6, (2, 8) = s7, (2, 9) = s8, (2, 10) = s9, (2, 11) = 2
*s10, (2, 12) = -s11, (2, 16) = s15, (2, 17) = -s16, (2, 21) = s20, (2, 22) = -
s21, (2, 26) = s25, (2, 27) = s26, (2, 28) = -s27, (3, 7) = -s6, (3, 11) = s10,
(3, 12) = s11, (3, 13) = s12, (3, 14) = s13, (3, 15) = s14, (3, 16) = 2*s15, (3
, 18) = -s17, (3, 21) = s20, (3, 23) = -s22, (3, 26) = s25, (3, 27) = s26, (3,
28) = -s27, (4, 8) = -s7, (4, 11) = s10, (4, 13) = -s12, (4, 16) = s15, (4, 17)
= s16, (4, 18) = s17, (4, 19) = s18, (4, 20) = s19, (4, 21) = 2*s20, (4, 24) =
-s23, (4, 26) = s25, (4, 27) = s26, (4, 28) = -s27, (5, 9) = -s8, (5, 11) = s10
, (5, 14) = -s13, (5, 16) = s15, (5, 19) = -s18, (5, 21) = s20, (5, 22) = s21,
(5, 23) = s22, (5, 24) = s23, (5, 25) = s24, (5, 26) = 2*s25, (5, 27) = s26, (5
, 28) = -s27, (6, 10) = -s9, (6, 11) = s10, (6, 15) = -s14, (6, 16) = s15, (6,
20) = -s19, (6, 21) = s20, (6, 25) = -s24, (6, 26) = s25, (6, 27) = 2*s26, (6,
28) = -2*s27, (7, 12) = s1-s2, (7, 13) = s7, (7, 14) = s8, (7, 15) = s9, (7, 16
) = s10, (7, 17) = -s17, (7, 22) = -s22, (8, 12) = -s12, (8, 17) = s1-s3, (8,
18) = s6, (8, 19) = s8, (8, 20) = s9, (8, 21) = s10, (8, 22) = -s23, (9, 12) =
-s13, (9, 17) = -s18, (9, 22) = s1-s4, (9, 23) = s6, (9, 24) = s7, (9, 25) = s9
, (9, 26) = s10, (10, 12) = -s14, (10, 17) = -s19, (10, 22) = -s24, (10, 27) =
s10, (11, 12) = -s15, (11, 17) = -s20, (11, 22) = -s25, (11, 28) = s9, (12, 18)
= -s16, (12, 23) = -s21, (13, 17) = s11, (13, 18) = s2-s3, (13, 19) = s13, (13,
20) = s14, (13, 21) = s15, (13, 23) = -s23, (14, 18) = -s18, (14, 22) = s11, (
14, 23) = s2-s4, (14, 24) = s12, (14, 25) = s14, (14, 26) = s15, (15, 18) = -
s19, (15, 23) = -s24, (15, 27) = s15, (16, 18) = -s20, (16, 23) = -s25, (16, 28
) = s14, (17, 24) = -s21, (18, 24) = -s22, (19, 22) = s16, (19, 23) = s17, (19,
24) = s3-s4, (19, 25) = s19, (19, 26) = s20, (20, 24) = -s24, (20, 27) = s20, (
21, 24) = -s25, (21, 28) = s19, (25, 27) = s25, (26, 28) = s24, (27, 28) = s5},
storage = triangular[upper,strict], shape = [skewsymmetric]):
> 

CI := Matrix(27, 27, {(1, 6) = s6, (1, 7) = s7, (1, 8) = s8, (1, 9) = s9, (1,
10) = 2*s10, (1, 11) = -s11, (1, 15) = s15, (1, 16) = -s16, (1, 20) = s20, (1,
21) = -s21, (1, 25) = s25, (1, 26) = s26, (1, 27) = -s27, (2, 6) = -s6, (2, 10)
= s10, (2, 11) = s11, (2, 12) = s12, (2, 13) = s13, (2, 14) = s14, (2, 15) = 2*
s15, (2, 17) = -s17, (2, 20) = s20, (2, 22) = -s22, (2, 25) = s25, (2, 26) =
s26, (2, 27) = -s27, (3, 7) = -s7, (3, 10) = s10, (3, 12) = -s12, (3, 15) = s15
, (3, 16) = s16, (3, 17) = s17, (3, 18) = s18, (3, 19) = s19, (3, 20) = 2*s20,
(3, 23) = -s23, (3, 25) = s25, (3, 26) = s26, (3, 27) = -s27, (4, 8) = -s8, (4,
10) = s10, (4, 13) = -s13, (4, 15) = s15, (4, 18) = -s18, (4, 20) = s20, (4, 21
) = s21, (4, 22) = s22, (4, 23) = s23, (4, 24) = s24, (4, 25) = 2*s25, (4, 26)
= s26, (4, 27) = -s27, (5, 9) = -s9, (5, 10) = s10, (5, 14) = -s14, (5, 15) =
s15, (5, 19) = -s19, (5, 20) = s20, (5, 24) = -s24, (5, 25) = s25, (5, 26) = 2*
s26, (5, 27) = -2*s27, (6, 11) = s1-s2, (6, 12) = s7, (6, 13) = s8, (6, 14) =
s9, (6, 15) = s10, (6, 16) = -s17, (6, 21) = -s22, (7, 11) = -s12, (7, 16) = s1
-s3, (7, 17) = s6, (7, 18) = s8, (7, 19) = s9, (7, 20) = s10, (7, 21) = -s23, (
8, 11) = -s13, (8, 16) = -s18, (8, 21) = s1-s4, (8, 22) = s6, (8, 23) = s7, (8,
24) = s9, (8, 25) = s10, (9, 11) = -s14, (9, 16) = -s19, (9, 21) = -s24, (9, 26
) = s10, (10, 11) = -s15, (10, 16) = -s20, (10, 21) = -s25, (10, 27) = s9, (11,
17) = -s16, (11, 22) = -s21, (12, 16) = s11, (12, 17) = s2-s3, (12, 18) = s13,
(12, 19) = s14, (12, 20) = s15, (12, 22) = -s23, (13, 17) = -s18, (13, 21) =
s11, (13, 22) = s2-s4, (13, 23) = s12, (13, 24) = s14, (13, 25) = s15, (14, 17)
= -s19, (14, 22) = -s24, (14, 26) = s15, (15, 17) = -s20, (15, 22) = -s25, (15,
27) = s14, (16, 23) = -s21, (17, 23) = -s22, (18, 21) = s16, (18, 22) = s17, (
18, 23) = s3-s4, (18, 24) = s19, (18, 25) = s20, (19, 23) = -s24, (19, 26) =
s20, (20, 23) = -s25, (20, 27) = s19, (24, 26) = s25, (25, 27) = s24, (26, 27)
= s5}, storage = triangular[upper,strict], shape = [skewsymmetric]):
> 

TO:=[indets(CO)[]]: nops(%);

27

 (1)
> 

VO:=[seq( rand(-1000..1000)(), 1..%)]:
> 

LinearAlgebra:-Rank(eval(CO, TO=~VO));

28

 (2)
> 

TI:=[indets(CI)[]]: nops(%);

27

 (3)
> 

VI:=[seq( rand(-1000..1000)(), 1..%)]:
> 

LinearAlgebra:-Rank(eval(CI, TI=~VI));

26

 (4)

Both results are exact, being maximal
(Rank(CI)<27  because CI is antisymmetric of odd order).

> 

 

> 

 

 

Change the color...

vv 12453
August 07 2020
1 0 
foo:=proc(ode::`=`,y,x)
local t; #hoping this will not be the same looking as actual x
PDEtools:-dchange({x=t^2},ode,t);  
eval(%, t=cat(`#mi(`,x,`,mathcolor=red)`));
end proc:
ode:=diff(y(t),t)=sin(t):
foo(ode,y,t)

1 operand?...

vv 12453
August 06 2020
0 0

It seems that Grid:-Map does not work correctly when the expression has a single operand. Of couse it does not make sense to use Map in such a case, but anyway it's a bug.
If you have >1 operands, it is OK.

map(taylor, [sin(x),cos(x)],x=0,10);
Grid:-Map(taylor, [sin(x), cos(x)],x=0,10);

 

Convert to sums...

vv 12453
August 03 2020
4 0 
restart;
ex:=product(q^(n -2)- q^i, i = 0 .. r )/product(q^(n-2) - q^(i+1), i = 0 .. r):
subsindets(ex, specfunc(product), u -> exp(Sum(ln(op(1,u)), op(2,u)))):
simplify(value(combine(%)));

              

Golden ratio...

vv 12453
August 03 2020
1 0

For f : (0, oo) --> (0, oo), a solution is:

phi := sqrt(5)/2 + 1/2:
f := x -> phi^(1-phi)*x^phi;

Try to check it.

Another workaround: rtf or latex first...

vv 12453
August 03 2020
0 1

Save the worksheet as rtf (or LaTeX) and then produce the pdf from here.
A pdf obtained from rtf is very small!

 

f(p)+g(p)=h(p)...

vv 12453
August 02 2020
0 8

1.  y = f(p) + g(p)   means actually  y = h(p)

so, it's in the d'Alembert pattern y = x*f(p) + g(p)   for f=0, g=h.

2. dsolve(ode) solves the ode using the y = g(y') pattern [independent of x], so, a RootOf form.

The situation is almost the same as in y...

vv 12453
August 01 2020
0 0

The situation is almost the same as in your previous question except that now Expand works properly in Maple 2015
(Maple 2020 works directly -- your "Method 1").

Expand...

vv 12453
July 31 2020
1 6

In Maple 2020 the result is directly   -4.211444465*10^(-43)*y + 4.534504493*10^(-43)

In earlier versions try:

Int(q*AAA/sqrt(p^2 + q^2 + 1), [p = 0 .. L, q = 0 .. L]):
evalf(IntegrationTools:-Expand(%));

 

mpl...

vv 12453
July 31 2020
0 0

The simplest solution is to open the worksheet in Maple 2020 and then export it as mpl (i.e. text file).
Now you can open the .mpl file. You may want to split it into several execution groups (1D input).

Another solution (since you are already using 1D input)  is to simply remove the output from worksheet (Ctrl + D) and reexecute.

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