dharr

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Name: Dr. David Harrington

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Member for: 20 years, 336 days

Company/Institution: University of Victoria

Title: Professor or university staff

Location: Victoria, British Columbia, Canada

Website: http://web.uvic.ca/~dharr

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Contact Dr. David Harrington

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

assign...

Answered: dharr 8205

March 30 2019

0 0

assign(solution) will do it.

explicit...

Answered: dharr 8205

March 30 2019

1 1

use the option explicit=true after the set of variables.

tickmarks...

Answered: dharr 8205

March 29 2019

1 2

(No, really, it looks right in the worksheet.)

Download costicks.mw

Hint...

Answered: dharr 8205

March 29 2019

0 1

Plotting to look is always a first step.

Looks to be between 2 and 4.

Download hint.mw

fsolve is a numerical solver and Digits controls accuracy... Good luck.

You need the two equations in a set (or list), you had a Y instead of y, and the derivatives as entered weren't right, as can be seen by converting to 1-D input. Here is the worksheet with the right syntax (please check). dsolve ran for about 3 ks in Maple 2017 before I gave up, so in your later version you may (eventually) get a solution.

Ric_Eqns.mw

partial fraction...

Answered: dharr 8205

March 15 2019

0 4

I added some explicit multiplications, so hope that gives what is intended.

Download parfrac.mw

change default behaviour...

Answered: dharr 8205

February 24 2019

1 0

To change the default behaviour, set:

_EnvUseHeavisideAsUnitStep := true;

(see the help page for Heaviside)

Or just make a piecewise function

MyStep:=x->piecewise(x<0,0,0<=x,1);

not scanned...

Answered: dharr 8205

February 17 2019

0 3

I don't see a problem here - my interpretation is that seq returns the default value without accessing ("scanning") the entry in the rtable, just as V[6] returns 0.

exhaustive test...

Answered: dharr 8205

February 01 2019

2 4

I think the following logic works for mod 2 as well are regular matrices. We want to test all nonsingular matrices without too much duplication. Below shows that we need only test all perturbation matrices (n! of them) plus an extra 1 (n^2-n possibilities) and another extra 1 (n^2-n-1 possibilities). This is possible in about an hour on my laptop. If produces many duplicates and I show only one answer. It would be possible to eliminate duplicates with a bit more effort.

Of the 8 columns, none have only zeros so six of then have one 1 and two have two 1s. For the six columns with one 1, there will be two row positions, say i and j that have no 1s for any of the six columns. Now consider the 7th column, with two 1s.

If neither of these two 1s are in row i or j then this column is a sum of two of the six columns and so the matrix would be singular. So the 7th column has one or two 1's in row i or j. If it has only one, say in i, then for the 8th column there must be a 1 in row j, otherwise row j would be zero and the matrix would be singular. In this case, every column has a 1 in a different row (a permutation matrix) and there are two extras.

If the 1s in column 7 are in row i and j, then consider column 8. If its two 1s avoided row i and j, it would be a sum of the six columns. So it must have one 1 in i or j. Then column 7 has a 1 in the other or i or j and again we have a perturbation matrix with two extra 1s.

The following tests this.

restart;

>	with(LinearAlgebra[Modular]):with(Iterator):

>	n:=8; #s = n+2 f:=x^8+x^7+x^5+x+1;#f:=x^8+x^4+x^3+x^2+1; perms:=Permute(n): Number(perms)(n^2-n)(n^2-n-1); # matrices to check

st:=time():
p:=0:          # count matrices checked
nsucc:=0:      # count matrices with correct char. poly.
succ:=table(): # collect succesful ones
for perm in perms do
  A:=Create(2,n,n,integer[]):
  for i to n do
    A[perm[i],i]:=1:
  end do:
  for k to n^2 do
    if A(k)=1 then next else A(k):=1 end if;
    for m to n^2 do
      if A(m)=1 then next else A(m):=1 end if;
      p:=p+1;
      #if p mod 10000000 =0 then print(p,(time()-st)/60*min) end if;
      if CharacteristicPolynomial(2,A,x)=f then nsucc:=nsucc+1; succ[nsucc]:=copy(A) end if;
      A(m):=0;
    end do;
    A(k):=0;
  end do;
end do:
print(p,`matrices tested.`,nsucc,`successes.`,`cpu time:`,(time()-st)/60*min);

>	B:=succ[1];add(B);CharacteristicPolynomial(2,B,x);

_rtable[18446744628225754526]

Download Modular5.mw

Edit: improved with modifications suggested by @vv :

Modular6.mw

parametric plots...

Answered: dharr 8205

January 29 2019

0 1

A bit contrived but this does it:

Download plots.mw

maybe not faster...

Answered: dharr 8205

January 29 2019

1 3

[Edit: see improved answer below]

Still takes a long time for 8x8 so I didn't wait. Below is just a different approach, FYI.

restart;

>	with(LinearAlgebra[Modular]):with(Iterator):

>	n:=4;s:=3; f:=x^4+x; perms:=Permute([1$s,0$(n^2-s)]): Number(perms);

>	st:=time():i:=0: for perm in perms do i:=i+1; lst:=convert(perm[],list); A:=Matrix(n,n,lst); cp:=CharacteristicPolynomial(2,A,x); if cp=f then print(time()-st,A,cp) end if; end do: i; time()-st;

$.281, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 1, (3, 1) = 0, (3, 2) = 1, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0}), x^4+x$

$.328, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 1, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 1, (4, 1) = 0, (4, 2) = 1, (4, 3) = 0, (4, 4) = 0}), x^4+x$

$.375, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 1, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 1, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0}), x^4+x$

$.406, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 1, (2, 1) = 1, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 1, (4, 3) = 0, (4, 4) = 0}), x^4+x$

$.484, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 1, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 1, (4, 1) = 1, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0}), x^4+x$

$.547, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 1, (1, 4) = 0, (2, 1) = 1, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 1, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0}), x^4+x$

$.578, Matrix(4, 4, {(1, 1) = 0, (1, 2) = 1, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 1, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 1, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0}), x^4+x$

.640, Matrix(%id = 18446744237432444678), x^4+x

Download Modular2.mw

Edit: since you wanted nonsingular I should have chosen x^4+x+1 (degree 4 without zero root) and s=4 (>=4 or there will be a zero row/column)

Edit: OP has changed algorithm so above is no longer comparable.

fsolve needs help...

Answered: dharr 8205

January 21 2019

0 0

I think you will have to give some ranges for the variables to make any progress here. These can probably be quite wide, e.g., 0..infinity if you know a parameter is positive and don't wan't to be too restrictive.