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subsindets...

vv 12453
October 07 2018
4 0

subsindets(p, [0,algebraic], u -> NULL);

Simplify first...

vv 12453
October 06 2018
3 2

You have an inequality in the system, so the variables will be assumed to be real.
Your system is hence polynomial and Maple should use SolveTools:-SemiAlgebraic, which usually gives correct answers.
Unfortunately, due to the presence of abs, the system is not recognized as polynomial.
So, use simplify first.
 

eq:=[a>0,b*d*(abs(c)^2-abs(a)^2) = 0, -a*c*(abs(b)^2-abs(d)^2) = 0, -abs(b)^2*a*d+abs(d)^2*b*c = 0, abs(c)^2*a*d-abs(a)^2*b*c = 0]:
eq1:=simplify(eq) assuming real:
solve(eq1);

{b = -d, c = -a, 0 < a, d < 0}, {b = 0, c = 0, d = d, 0 < a}, {b = d, c = a, 0 < a, d < 0}, {b = 0, d = 0, 0 < a, c < -a}, {b = 0, c = -a, d = 0, 0 < a}, {b = 0, d = 0, 0 < a, c < 0, -a < c}, {c = 0, d = 0, 0 < a, b < 0}, {c = 0, d = 0, 0 < a, 0 < b}, {b = 0, d = 0, 0 < a, 0 < c, c < a}, {b = 0, c = a, d = 0, 0 < a}, {b = 0, d = 0, 0 < a, a < c}, {b = -d, c = -a, 0 < a, 0 < d}, {b = d, c = a, 0 < a, 0 < d}

 

Vector...

vv 12453
October 05 2018
1 0

Replace vector with Vector.
(vector is deprecated).
If you insist with vector then change the last line from out to eval(out).

P.S. base should be close to 1 (e.g. base=1.01), otherwise the sequence increases too fast.

syntactically...

vv 12453
October 05 2018
4 1

evalb checks syntactically and does not simplify, see ?evalb.  Use

is(a * conjugate(a) = abs(a) ^ 2);
or                            
evalb(simplify(a * conjugate(a) = abs(a) ^ 2));
 

 

Infinite loop...

vv 12453
October 03 2018
1 1

It seems to be a bug and Maple enters an infinite loop.
Execute
infolevel[`mod`]:=2;
and see the messages.

Edit. But it might just working hard. Maxima also fails.
Where have you obtained a factorization?

Answer for the Hausdorff measure...

vv 12453
September 30 2018
1 0

 

Here is the answer for the Hausdorff measure (corresponds to the length).

Unfortunately Maple reveals some bugs for symbolic integration!

 

 

> 

restart;
> 

dS:=sqrt(a^2*sin(t)^2+b^2*cos(t)^2):
S :=(t1,t2) -> Int( dS, t=t1..t2 ):
> 

Mean:= (t1,t2) -> Int( sqrt(a^2*cos(t)^2+b^2*sin(t)^2)*dS, t=t1..t2 )/S(t1,t2):

 

Examples

 

> 

### 1 ###
ans:=eval(Mean(0,Pi), [a=2,b=1]);

(Int((4*cos(t)^2+sin(t)^2)^(1/2)*(4*sin(t)^2+cos(t)^2)^(1/2), t = 0 .. Pi))/(Int((4*sin(t)^2+cos(t)^2)^(1/2), t = 0 .. Pi))

 (1)
> 

simplify(value(ans));  # Maple bug

0

 (2)
> 

evalf(ans);

1.463684753

 (3)
> 

### 2 ###
> 

ans:=eval(Mean(Pi/4,3*Pi/2), [a=2,b=1]);

(Int((4*cos(t)^2+sin(t)^2)^(1/2)*(4*sin(t)^2+cos(t)^2)^(1/2), t = (1/4)*Pi .. (3/2)*Pi))/(Int((4*sin(t)^2+cos(t)^2)^(1/2), t = (1/4)*Pi .. (3/2)*Pi))

 (4)
> 

simplify(value(ans)); evalf(%);  # another bug!!!

(5*EllipticK((1/4)*3^(1/2)*5^(1/2))+(25*I)*EllipticK(1/4)-(16*I)*EllipticE(1/4)+(4*I)*EllipticE(4)-(4*I)*EllipticE((1/2)*I, 4)+(20*I)*EllipticF((1/2)*I, 4)-10)/(16*EllipticE((1/2)*2^(1/2), (1/2)*3^(1/2))+32*EllipticE((1/2)*3^(1/2)))

  

-.2882716323+.3166809263*I

 (5)
> 

evalf(ans);

1.406678625

 (6)
> 

### 3 ###
> 

ans:=eval(Mean(Pi/4,3*Pi/2), [a=1,b=1]);  # circle, finally bug-free

(Int(cos(t)^2+sin(t)^2, t = (1/4)*Pi .. (3/2)*Pi))/(Int((cos(t)^2+sin(t)^2)^(1/2), t = (1/4)*Pi .. (3/2)*Pi))

 (7)
> 

simplify(value(ans)); evalf(%);

1

  

1.

 (8)
> 

evalf(ans);

1.000000000

 (9)
> 

 

> 

 

Nice recurrence...

vv 12453
September 30 2018
3 1
> 

ans:=rsolve({f(1) = 1, f(n+1) = 1/2*(f(n)+9/f(n))}, {f});

3*coth((1/2)*arccoth(1/3)*2^n)

 (1)
> 

# It's nice that Maple did it!
> 

ans1:=simplify(convert(ans,compose,ln,exp)) assuming posint, n>1;

(3*2^(2^(-1+n))+3)/(2^(2^(-1+n))-1)

 (2)
> 

#(Not valid for n=1)

> 

 

Edit for a simpler version

# Check
F:=n -> 1/2*(F(n-1)+9/F(n-1)): F(1):=1:
seq([F(n), ans1],n=1..7);

[1, 9], [5, 5], [17/5, 17/5], [257/85, 257/85], [65537/21845, 65537/21845], [4294967297/1431655765, 4294967297/1431655765], [18446744073709551617/6148914691236517205, 18446744073709551617/6148914691236517205]

 

GL2...

vv 12453
September 27 2018
3 5

Find the group generated in GL(2,Q) by the matrices A, B:

> 

restart;
> 

A :=  Matrix(2, 2, [[2, 0], [0, 1]]) ;

B := Matrix(2, 2, [[1, 1], [0, 1]]) ;

Matrix(2, 2, {(1, 1) = 2, (1, 2) = 0, (2, 1) = 0, (2, 2) = 1})

  

Matrix(%id = 18446744074411464574)

 (1)
> 

pow:=LinearAlgebra:-MatrixPower:
> 

pow(A,n);

Matrix(2, 2, {(1, 1) = 2^n, (1, 2) = 0, (2, 1) = 0, (2, 2) = 1})

 (2)
> 

pow(B,n);

Matrix(2, 2, {(1, 1) = 1, (1, 2) = n, (2, 1) = 0, (2, 2) = 1})

 (3)
> 

pow(A,m).pow(B,n).pow(A,p).pow(B,q);

Matrix(2, 2, {(1, 1) = 2^m*2^p, (1, 2) = 2^m*2^p*q+2^m*n, (2, 1) = 0, (2, 2) = 1})

 (4)

It will be easy now to see that the group generated by A and B is  { X(a,b,c) : a,b,c in Z }, where

 

 

> 

X(a,b,c) =  <2^a, 2^b*c; 0, 1>;

X(a, b, c) = (Matrix(2, 2, {(1, 1) = 2^a, (1, 2) = 2^b*c, (2, 1) = 0, (2, 2) = 1}))

 (5)

posint...

vv 12453
September 26 2018
3 8

In Maple permutations have posint entries. You must shift them by 1.

restart;
with(group):
a:=[[seq(i,i=1..6)],[seq(i,i=7..9)]];
              a := [[1, 2, 3, 4, 5, 6], [7, 8, 9]]
mulperms(a,a);
               [[1, 3, 5], [2, 4, 6], [7, 9, 8]]
b:=a:
for i to 6 do
  lprint('a'^i=b);
  b:=mulperms(b,a);
od:
a = [[1, 2, 3, 4, 5, 6], [7, 8, 9]]
a^2 = [[1, 3, 5], [2, 4, 6], [7, 9, 8]]
a^3 = [[1, 4], [2, 5], [3, 6]]
a^4 = [[1, 5, 3], [2, 6, 4], [7, 8, 9]]
a^5 = [[1, 6, 5, 4, 3, 2], [7, 9, 8]]
a^6 = []

 

A plausible (?) answer...

vv 12453
September 24 2018
1 8

 The behavior seems to be intentional (but not documented) being useful for boolean expressions containing variables.
Such expressions are simplified generically for boolean operators  but are left unsimplified for the corresponding procedures.

restart;
x := a=0:  y := b=0:
eval(`or`(x,y), [a=0]);   evalb(%);
                         0 = 0 or b = 0
                              true
eval(x or y, [a=0]);   # generically false, so the substitution is useless
                             false

 

indets...

vv 12453
September 20 2018
1 1

A "sequential order" does not have much sense in Maple because e.g. a unique ordering of the operands in sum or product is not possible.
For the set of all functions use:

map2(op,0,indets(expr,function));

 

Maple complete solution...

vv 12453
September 18 2018
2 3

I had the impression that posting the complete Maple solution is not very interesting (the 13 regions were already listed).

The idea of the solution is simple: obtain the explicit values for the six floors (after some simplifications).

However it seems that there are interested users here, so here it is. I have inserted a minimum of comments

 

 

> 

restart;
> 

S0:=[4*x1+7*x2+6*x3 - 186, floor((1/2)*x1)+floor((1/5)*x2)+floor((1/3)*x3) - 18,
floor((1/5)*x1)+floor((1/2)*x2)+floor((1/4)*x3) - 21];

[4*x1+7*x2+6*x3-186, floor((1/2)*x1)+floor((1/5)*x2)+floor((1/3)*x3)-18, floor((1/5)*x1)+floor((1/2)*x2)+floor((1/4)*x3)-21]

 (1)
> 

id:=t->t:
> 

Suv:=eval(S0,floor=id)-[0,3*u,3*v];

[4*x1+7*x2+6*x3-186, -3*u+(1/2)*x1+(1/5)*x2+(1/3)*x3-18, -3*v+(1/5)*x1+(1/2)*x2+(1/4)*x3-21]

 (2)

3u and 3v  are the sum of the fractional parts in the 2nd and 3rd equation;  0 <= u,v < 1

 

> 

#Suv:=[4*x1+7*x2+6*x3-186, (1/2)*x1+(1/5)*x2+(1/3)*x3-18-3*u, (1/5)*x1+(1/2)*x2+(1/4)*x3-21-3*v]
> 

xuv:=solve(Suv,[x1,x2,x3])[];

[x1 = 4920/77+(750/77)*u+(680/77)*v, x2 = -(120/77)*u+4110/77+(1000/77)*v, x3 = -(1620/77)*v-5688/77-(360/77)*u]

 (3)
> 

Suv:=eval(S0,xuv)[2..3];

[floor(73/77+(375/77)*u+(340/77)*v)-2+floor(-(24/77)*u+52/77+(200/77)*v)+floor(-(540/77)*v+29/77-(120/77)*u), floor(60/77+(150/77)*u+(136/77)*v)-2+floor(-(60/77)*u+53/77+(500/77)*v)+floor(-(405/77)*v+41/77-(90/77)*u)]

 (4)

Suv  is the new system in the unknowns u,v.  The relations with the initial system is below (uvx).

> 

uvx:=solve(xuv[1..2],{u,v});

{u = -(17/270)*x2-23/9+(5/54)*x1, v = (1/90)*x1-53/12+(5/72)*x2}

 (5)

Now Suv can be solved easily because the floors may take only a few values.

> 

A:=eval(indets(Suv[1])minus{u,v},floor=id)[];
B:=eval(indets(Suv[2])minus{u,v},floor=id)[];

73/77+(375/77)*u+(340/77)*v, -(24/77)*u+52/77+(200/77)*v, -(540/77)*v+29/77-(120/77)*u

  

60/77+(150/77)*u+(136/77)*v, -(60/77)*u+53/77+(500/77)*v, -(405/77)*v+41/77-(90/77)*u

 (6)
> 

Min:=f -> min(eval(f,[u=0,v=0]), eval(f,[u=0,v=1]), eval(f,[u=1,v=0]), eval(f,[u=1,v=1]));
Max:=f -> max(eval(f,[u=0,v=0]), eval(f,[u=0,v=1]), eval(f,[u=1,v=0]), eval(f,[u=1,v=1]));

proc (f) options operator, arrow; min(eval(f, [u = 0, v = 0]), eval(f, [u = 0, v = 1]), eval(f, [u = 1, v = 0]), eval(f, [u = 1, v = 1])) end proc

  

proc (f) options operator, arrow; max(eval(f, [u = 0, v = 0]), eval(f, [u = 0, v = 1]), eval(f, [u = 1, v = 0]), eval(f, [u = 1, v = 1])) end proc

 (7)

SA, SB  will contain the possible floors

> 

SA:=NULL:
for i from floor(Min(A[1])) to  floor(Max(A[1])) do
for j from floor(Min(A[2])) to  floor(Max(A[2])) do
for k from floor(Min(A[3])) to  floor(Max(A[3])) do
if i+j+k=2 then SA:=SA,[i,j,k] fi
od od od:
nops([SA]);

38

 (8)
> 

SB:=NULL:
for i from floor(Min(B[1])) to  floor(Max(B[1])) do
for j from floor(Min(B[2])) to  floor(Max(B[2])) do
for k from floor(Min(B[3])) to  floor(Max(B[3])) do
if i+j+k=2 then SB:=SB,[i,j,k] fi
od od od:
nops([SB]);

33

 (9)
> 

NR:=0:
for sa in [SA] do for sb in [SB] do
sa[1]<=A[1],A[1]<sa[1]+1, sa[2]<=A[2],A[2]<sa[2]+1, sa[3]<=A[3],A[3]<sa[3]+1,
sb[1]<=B[1],B[1]<sb[1]+1, sb[2]<=B[2],B[2]<sb[2]+1, sb[3]<=B[3],B[3]<sb[3]+1;
ss:=solve(eval({%},uvx));    # nu e nevoie sa includem  0<=u<1, 0<=v<1  caci le-am folosit deja in Min.Max
if [ss]<>[] then NR:=NR+1;R[NR]:=ss; fi
od od:  NR;

13

 (10)

We have ontained the NR=13 regions.  Here is the first one

> 

R[1];

{x2 = 56, 68 <= x1, x1 <= 137/2}, {x1 = 68, x2 <= 394/7, 56 < x2}, {56 < x2, 68 < x1, x1 < 333/2-(7/4)*x2, x2 < 394/7}, {x1 = 333/2-(7/4)*x2, 56 < x2, x2 < 394/7}

 (11)

Now  plots:-inequal is able to plot the regions (projected on the (x1,x2) plane).

> 

proj:=plots:-display( seq(  plots:-inequal({R[r]},x1=67..81,x2=53..65), r=1..NR), size=[800,800] );

 
> 

#proj:=plots:-display( seq(  plots:-inequal({R[r]},x1=67..81,x2=53..65,
#optionsclosed=[color=red,thickness=4], optionsopen=[color=white,thickness=0]), r=1..NR), size=[800,800] );
> 

 

> 

# List all 13 regions
for r to NR do lprint(Region||r=[R[r]]) od:

Region1 = [{x2 = 56, 68 <= x1, x1 <= 137/2}, {x1 = 68, x2 <= 394/7, 56 < x2}, {56 < x2, 68 < x1, x1 < 333/2-(7/4)*x2, x2 < 394/7}, {x1 = 333/2-(7/4)*x2, 56 < x2, x2 < 394/7}]

Region2 = [{x2 = 54, 70 <= x1, x1 < 72}, {x1 = 70, 54 < x2, x2 < 55}, {54 < x2, 70 < x1, x1 < 333/2-(7/4)*x2, x2 < 55}, {x1 = 333/2-(7/4)*x2, 54 < x2, x2 < 55}]
Region3 = [{56 <= x2, x2 <= 402/7, 70 < x1, x1 < 72}, {x1 = 70, 56 < x2, x2 < 58}, {70 < x1, 402/7 < x2, x1 < 345/2-(7/4)*x2, x2 < 58}, {x1 = 345/2-(7/4)*x2, 402/7 < x2, x2 < 58}]
Region4 = [{58 <= x2, 72 <= x1, x2 <= 412/7, x1 < 74}, {x1 = 72, 412/7 < x2, x2 < 60}, {72 < x1, 412/7 < x2, x1 < 177-(7/4)*x2, x2 < 60}, {x1 = 177-(7/4)*x2, 412/7 < x2, x2 < 60}]
Region5 = [{x2 = 60, 147/2 < x1, x1 < 74}, {60 < x2, x1 < 74, x2 < 426/7, 357/2-(7/4)*x2 < x1}, {426/7 <= x2, x2 <= 430/7, 72 < x1, x1 < 74}, {x1 = 72, 426/7 < x2, x2 < 62}, {72 < x1, 430/7 < x2, x1 < 363/2-(7/4)*x2, x2 < 62}, {x1 = 363/2-(7/4)*x2, 430/7 < x2, x2 < 62}]
Region6 = [{408/7 < x2, x1 < 75, x2 < 412/7, 177-(7/4)*x2 < x1}, {412/7 <= x2, x2 <= 414/7, 74 < x1, x1 < 75}, {x1 = 74, x2 <= 418/7, 412/7 < x2}, {74 < x1, 414/7 < x2, x1 < 357/2-(7/4)*x2, x2 < 418/7}, {x1 = 357/2-(7/4)*x2, 414/7 < x2, x2 < 418/7}]
Region7 = [{404/7 < x2, x1 < 76, x2 < 58, 177-(7/4)*x2 < x1}]
Region8 = [{410/7 < x2, x1 < 76, x2 < 414/7, 357/2-(7/4)*x2 < x1}, {414/7 <= x2, 75 < x1, x1 < 76, x2 < 60}, {x1 = 75, 414/7 < x2, x2 < 60}]
Region9 = [{426/7 < x2, x1 < 75, x2 < 430/7, 363/2-(7/4)*x2 < x1}, {430/7 <= x2, 74 < x1, x1 < 75, x2 < 62}, {x1 = 74, 430/7 < x2, x2 < 62}]
Region10 = [{438/7 < x2, x1 < 75, x2 < 442/7, 369/2-(7/4)*x2 < x1}, {442/7 <= x2, x2 <= 444/7, 74 < x1, x1 < 75}, {x1 = 74, 442/7 < x2, x2 < 64}, {74 < x1, 444/7 < x2, x1 < 186-(7/4)*x2, x2 < 64}, {x1 = 186-(7/4)*x2, 444/7 < x2, x2 < 64}]
Region11 = [{414/7 < x2, x1 < 78, x2 < 60, 363/2-(7/4)*x2 < x1}]
Region12 = [{432/7 < x2, x1 < 78, x2 < 62, 186-(7/4)*x2 < x1}]
Region13 = [{442/7 < x2, x1 < 80, x2 < 64, 381/2-(7/4)*x2 < x1}]

  

Plotting the 3d regions is now easy.

> 

f23 := plottools:-transform((x1,x2) -> [x1,x2,-2*x1*(1/3)-7*x2*(1/6)+31]):
> 

plots:-display(f23(proj));

 
> 

 


 

Download sysfloor-sent.mw

2*Pi...

vv 12453
September 17 2018
1 0

The integral is indeed 2*Pi for any complex a. This follows from the Residue theorem.

Maple finds the antiderivative Ei(1, exp(I*x))  and using Newton-Leibniz concludes that the integral is 0.
Unfortunately the antiderivative is discontinuous but Maple cannot check this (at least for a symbolic a).
Such situations are common and will not disappear soon (most of them).

Graphics only...

vv 12453
September 15 2018
2 14

The system can be reduced to 2 equations eliminating e.g. x3.
It is easy to obtain  60 < x1 < 85,  50 < x2 < 70.

S12:=[floor((1/2)*x1)+floor((1/5)*x2)+floor(-(2/9)*x1-(7/18)*x2+1/3)-8,
      floor((1/5)*x1)+floor((1/2)*x2)+floor(-(1/6)*x1-(7/24)*x2+3/4)-14]:
f:=(i,j) -> eval( S12,[x1=60+i/500*25,x2=50+j/500*20]): 
M:=Matrix(500, (i,j) -> `if`(f(i,j)=[0,0],1,0)):
plots:-sparsematrixplot(M, size=[500,500],
  labels=["",""],tickmarks=[[1=60,500=85], [1=50,500=70]]);

It is possible to write a code and obtain all these regions, but probably a motivation would be necessary
(without it the problem does not seem to be interesting enough).

Round-off errors...

vv 12453
September 15 2018
0 0

Increasing Digits:=15 (which is always a good idea for numerics) is enough in this case.

(Note that the summation order is not necessarily the same in I1 and I2 and `+` is not associative for floats.)

See also here.

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