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9 years, 310 days
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vv 13805
March 13 2024
0 0

@dharr Just a remark. The eigenvalues are not correctly sorted; the option key=evalf should be added.

example...

vv 13805
March 12 2024
0 0

@mehdi jafari 
 

> 

with(LinearAlgebra):
> 

A := <<-1,-3,-6>|<3,5,6>|<-3,-3,-4>>;

Matrix(3, 3, {(1, 1) = -1, (1, 2) = 3, (1, 3) = -3, (2, 1) = -3, (2, 2) = 5, (2, 3) = -3, (3, 1) = -6, (3, 2) = 6, (3, 3) = -4})

 (1)
> 

(V,Q):=Eigenvectors(A);

V, Q := Vector(3, {(1) = -4, (2) = 2, (3) = 2}), Matrix(3, 3, {(1, 1) = 1/2, (1, 2) = -1, (1, 3) = 1, (2, 1) = 1/2, (2, 2) = 0, (2, 3) = 1, (3, 1) = 1, (3, 2) = 1, (3, 3) = 0})

 (2)
> 

Q^(-1).A.Q = DiagonalMatrix(V);

(Matrix(3, 3, {(1, 1) = -4, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 2})) = (Matrix(3, 3, {(1, 1) = -4, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 2}))

 (3)
> 

(Vt,Qt):=Eigenvectors(A^+); # left eigenvectors; the eigenvalues are always the same

Vt, Qt := Vector(3, {(1) = 2, (2) = 2, (3) = -4}), Matrix(3, 3, {(1, 1) = -2, (1, 2) = -1, (1, 3) = 1, (2, 1) = 0, (2, 2) = 1, (2, 3) = -1, (3, 1) = 1, (3, 2) = 0, (3, 3) = 1})

 (4)
> 

A . Q[..,1] = V[1] * Q[..,1];         # A.v = lambda*v

(Vector(3, {(1) = -2, (2) = -2, (3) = -4})) = (Vector(3, {(1) = -2, (2) = -2, (3) = -4}))

 (5)
> 

Qt[..,1]^+ . A = Qt[..,1]^+ * Vt[1];  # w.A = lambda*w  

(Vector[row](3, {(1) = -4, (2) = 0, (3) = 2})) = (Vector[row](3, {(1) = -4, (2) = 0, (3) = 2}))

 (6)
> 

 

 

 

Funny...

vv 13805
March 06 2024
0 0

@acer However,

restart;
local a; 	
a := 17:
:-a := 18:
p := proc()
  local a;
  a := 25;
  return parse("a",statement);
end proc:

p();  #17

 

Sorry...

vv 13805
March 06 2024
0 0

@mmcdara Sorry, probably I have seen an older version of your answer, or maybe I was not careful enough.

useassumptions...

vv 13805
March 06 2024
0 0

@mmcdara You have ommited useassumptions. So, solve has another task: a single inequality instead of a system.

Digits=10...

vv 13805
February 18 2024
0 0

@Axel Vogt Just replace 10.0 by 10

1/f...

vv 13805
February 18 2024
0 0

@sursumCorda 
I used Maple 2018 on an old Windows tablet  and I obtain 
limit(1/f, x=infinity) = 0.

comment...

vv 13805
February 04 2024
0 0

@petit loup   "#" starts a comment, so actually method='integer'  is not needed when the matrix has integer entries.
For our matrix A,
S,U,V := SmithForm(A, output=['S,U,V'], method='integer');
and
S,U,V := SmithForm(A, output=['S,U,V']);
give the same result.
 

interval...

vv 13805
February 03 2024
0 0

@mmcdara Actually, when limit returns an interval a..b, this means that the limit does not exist; mathematically this is equivalent to the undefined result, but Maple also says that the limit points (aka accumulation points) are contained in the interval a..b.

Note that e.g. 

limit(3*sin(1/x)-4*cos(1/x), x=0);
    -7 .. 7

but actually the set of accumulation points is the interval -5 .. 5, which Maple cannot find.

It would be nice to have in Maple the comands  sup, inf, limsup, liminf for functions and sequences, just like in Mathematica!

@mmcdara Probably the designers con...

vv 13805
January 08 2024
0 0

@mmcdara Probably the designers consider that <...> is enough, e.g. the first method or

<M[1,1],M[1,2]; M[2,1],M[2,2]>;

 

Example...

vv 13805
December 29 2023
0 0

@Mikey_W 

restart;
m:=1000: n:=1000: A:=LinearAlgebra:-RandomMatrix(m, n):
x:=Pi/77:
    
f:=proc(x,A) local i,j,s:=0; 
for i to m do for j to n do
  s:=s+cos(sin(x+A[i,j]));
od od;
s
end proc:

CodeTools:-Usage( evalf(f(x,A)) );
## memory used=6.11GiB, alloc change=33.00MiB, cpu time=10.89s, real time=10.90s, gc time=1.70s

##                        7.639612369 10^5 

CodeTools:-Usage( evalhf(f(x,A)) );
## memory used=7.63MiB, alloc change=7.63MiB, cpu time=265.00ms, real time=266.00ms, gc time=0ns

##                    7.63961236894650501 10^5

 

@acer  No, please do....

vv 13805
December 21 2023

@acer  No, please do.

typo...

vv 13805
December 21 2023
0 0

@sursumCorda Corrected and .mw inserted.

OK...

vv 13805
December 20 2023
0 0
> 

restart;
> 

SYS := [cos(x)*cos(y)*cos(x + y) - 2*(sin(x)*sin(y) - 1)*2*(sin(x)*sin(x + y) - 1)*2*(sin(y)*sin(x + y) - 1),
       (cos(x + y) - (cos(x) + cos(y)) + 1)**2 + 2*cos(x)*cos(y)*cos(x + y)]:
> 

Sys:=eval(expand~(SYS)):
> 

solve(eval(Sys[2],y=Pi)); #Not real; idem x=Pi and for Sys[1]

arccos(2/3+((1/3)*I)*2^(1/2)), arccos(2/3-((1/3)*I)*2^(1/2))

 (1)
> 

tanhalf := [ cos(x)=(1-t^2)/(1+t^2), sin(x)=2*t/(1+t^2), cos(y)=(1-u^2)/(1+u^2), sin(y)=2*u/(1+u^2) ]:
> 

f:=seq(
  convert( SolveTools:-SemiAlgebraic( {numer(eval(Sys[i], tanhalf))}), radical),
i=1..2);

[[u = -1, t = -1], [u = -1, t = 0], [u = -(1/3)*3^(1/2), t = -(1/3)*3^(1/2)], [u = 0, t = -1], [u = 0, t = 1], [u = (1/3)*3^(1/2), t = (1/3)*3^(1/2)], [u = 1, t = 0], [u = 1, t = 1]], [[u = -1, t = -1], [u = -1, t = 0], [u = -(1/3)*3^(1/2), t = -(1/3)*3^(1/2)], [u = 0, t = -1], [u = 0, t = 1], [u = (1/3)*3^(1/2), t = (1/3)*3^(1/2)], [u = 1, t = 0], [u = 1, t = 1]]

 (2)
> 

evalb({f[1][]} = {f[2][]});   # The equations have the same solutions

true

 (3)
> 

XY:=t -> arctan(2*t/(1+t^2),  (1-t^2)/(1+t^2));

proc (t) options operator, arrow; arctan(2*t/(1+t^2), (1-t^2)/(1+t^2)) end proc

 (4)
> 

map( z -> [x=XY(eval(t,z)), y=XY(eval(u,z))], f[1]);

[[x = -(1/2)*Pi, y = -(1/2)*Pi], [x = 0, y = -(1/2)*Pi], [x = -(1/3)*Pi, y = -(1/3)*Pi], [x = -(1/2)*Pi, y = 0], [x = (1/2)*Pi, y = 0], [x = (1/3)*Pi, y = (1/3)*Pi], [x = 0, y = (1/2)*Pi], [x = (1/2)*Pi, y = (1/2)*Pi]]

 (5)
> 

select(z -> is(eval(x,z)+eval(y,z) >=0), %);

[[x = (1/2)*Pi, y = 0], [x = (1/3)*Pi, y = (1/3)*Pi], [x = 0, y = (1/2)*Pi], [x = (1/2)*Pi, y = (1/2)*Pi]]

 (6)

Download sysok_vv.mw

The only one...

vv 13805
December 20 2023
0 0

@mmcdara If you don't know an IC for some x>1, you cannot find a solution.
For example, 
sol3 := piecewise( x<1, 1/(1-x), 1/(a-x) )

is a solution for any a<1.

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