The following problems have been sent by Richard Pavelle <rp@XN.LL.MIT.EDU>
to the USENET sci.math.symbolic bulletin board.  They represent
interesting challenges for algebra systems as Dr. Pavelle explained in
his introduction to the series:

        I shall begin, with this message, to send problems or ideas to
        users and implementors of CA systems.  The computations will
        generally be those which occur in applying CA systems to problems
        from industry or government sponsored r&d.  I am selecting those
        problems which can be stated without much verbiage.  I hope they
        will prove interesting.

        I will try to give some background or motivation to the problems.
        However, this will not always be possible.

        I will use MACSYMA input and output since it is the system most
        readily available to me.  I hope that users of other systems will
        post their methods and comments so we can have an exchange of
        ideas.

We shall include the best REDUCE solutions along with the problems as we
receive them.  If there is no solution proposed for a problem, it means we
don't yet have one!

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 1

Problem1: This problem came to me from Xerox and is related to the
frictional forces on a sheet of paper sliding over a metallic surface.
There are two parts. Derive the integral shown and then prove the
answer. Differentiating the expression below introduces some nasty
terms.

'integrate(x^2/(sqrt(x^2+d^2)+c),x) = 
-c*sqrt(d^2-c^2)*asin((c*sqrt(x^2+d^2)+d^2)/(d*(sqrt(x^2+d^2)+c)))-
(d^2-2*c^2)*log(2*(sqrt(x^2+d^2)+x))/2+x*(sqrt(x^2+d^2)-2*c)/2;


  /         2                                             2    2     2
  [        x                         2    2       c sqrt(x  + d ) + d
  I ----------------- dx = - c sqrt(d  - c ) asin(---------------------)
  ]       2    2                                           2    2
  / sqrt(x  + d ) + c                             d (sqrt(x  + d ) + c)


       2      2               2    2                   2    2
     (d  - 2 c ) log(2 (sqrt(x  + d ) + x))   x (sqrt(x  + d ) - 2 c)
   - -------------------------------------- + -----------------------
                       2                                 2


          ------------------ REDUCE SOLUTIONS ------------------

The following script shows a simple solution to the problem running on an
HP 9000/350.  Note however that the answer currently comes out in clumsy
logs.  We are hoping to improve on this in future releases.

load algint;

int(x^2/(sqrt(x^2+d^2)+c),x);

         2    2       2    2    2              2    2               2
(2*SQRT(C  - D )*LOG(C  - D  - X )*C + 2*SQRT(C  - D )*LOG( - SQRT(C

     2                   2    2            2    2   2  2           2
  - D ) - X)*C - 2*SQRT(C  - D )*LOG(SQRT(C  - D )*C *D  + 2*SQRT(C

    2   2  2         2    2   4         2    2   2  2           2
 - D )*C *X  - SQRT(C  - D )*D  - SQRT(C  - D )*D *X  + 2*SQRT(D  +

  2   3             2    2     2                2    2            2
 X )*C *X - 2*SQRT(D  + X )*C*D *X)*C - 2*SQRT(C  - D )*LOG(SQRT(C  -

   2                   2    2                      2    2
  D ) - X)*C - 2*SQRT(C  - D )*LOG(-1)*C + 2*SQRT(D  + X )*X - 2*LOG(

       2    2        2             2    2        2               2
 SQRT(D  + X ) - X)*C  + LOG(SQRT(D  + X ) - X)*D  + 2*LOG(SQRT(D  +

  2        2             2    2        2
 X ) + X)*C  - LOG(SQRT(D  + X ) + X)*D  - 4*C*X)/4


on gcd, rationalize;

% These switch settings are needed to make the derivative come out nicely.

df(ws,x);

     2        2    2
    X *(SQRT(D  + X ) - C)
 - ------------------------
          2    2    2
         C  - D  - X

showtime;

Time: 20247 ms

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 2

The object is prove the exactness of a given solution of the 4th order
PDE op1(f)+op2(f)=0 where

      op1(f):=diff(diff(f,x,2)+diff(f,y,2)+diff(f,z,2),x,2)
      op2(f):=n^2*(diff(f,x,2)+diff(f,y,2))

... It is found that
                                        2    2    2
                              n z sqrt(z  + y  + x )
                          sin(----------------------)
                                        2    2
                                  sqrt(z  + y )
                  f =     ---------------------------
                                    2    2    2
                              sqrt(z  + y  + x )

is an exact solution. Proving this with MACSYMA is time consuming
requiring between 1 and 5 CPU hrs on a VAX 780 depending upon
the approach.

I believe this example provides a rigorous test for the simplification
procedures of any CA system. Would anyone care to try it on another
system and comment on their approach?

          ------------------ REDUCE SOLUTIONS ------------------

The following solution was submitted by Kamal Abdali <kabdali@note.nsf.gov>.
It takes about 12 seconds in a PSL-based system on a Sun 3/260:

on time;

f:=sin(n*z*sqrt(z^2+y^2+x^2)/sqrt(z^2+y^2))/sqrt(z^2+y^2+x^2);

fx2:=df(f,x,2)$

fy2:=df(f,y,2)$

fz2:=df(f,z,2)$

op1:=df(fx2+fy2+fz2,x,2)$

op2:=n^2*(fx2+fy2)$

op1+op2;

end;


 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 3

Toeplitz matrices are used, among other areas, in signal processing in
the sampling of stationary data. Here is an example of a 7x7 Toeplitz
matrix:

(c1) toeplitz(7);
                            [ a  b  c  d  e  f  g ]
                            [                     ]
                            [ b  a  b  c  d  e  f ]
                            [                     ]
                            [ c  b  a  b  c  d  e ]
                            [                     ]
(d1)                        [ d  c  b  a  b  c  d ]
                            [                     ]
                            [ e  d  c  b  a  b  c ]
                            [                     ]
                            [ f  e  d  c  b  a  b ]
                            [                     ]
                            [ g  f  e  d  c  b  a ]

Recently I found an application of these matrices which yields a nice
test for factorizing. I believe it is true, although I have not found
a proof, that the characteristic polynomial of a Toeplitz matrix
factors, over the integers, into two parts. As an example, for a
Toeplitz matrix of order 3,

(c2) factor(determinant(toeplitz(3)-L*ident(3)));

                             2                          2    2
(d2)         - (L + c - a) (L  - c L - 2 a L + a c - 2 b  + a )

Here the characteristic polynomial of a 3x3 factors into parts
containing 3 and 6 terms respectively. (MACSYMA does not, as MAPLE
does, have a canned definition of a Toeplitz matrix so I wrote one for
this example. I also edited the "L" above to distinguish it from the
integer 1).

Here is a table of the number of terms in each of the factored parts
as far as I have attempted to proceed with MACSYMA on a VAX 780. 

order   terms in 1st factor     terms in 2nd factor

2               3                       3
3               3                       6
4               11                      11
5               11                      24
6               51                      51
7               49                      107

The characteristic polynomials become large very quickly, and I would
like to know how far users can extend this list. What is the largest
characteristic polynomial of a Toeplitz matrix which can be factored
by a CA system? My second question is whether someone can prove or
find a proof of the conjectured factorization into two parts? 

Richard Pavelle

          ------------------ REDUCE SOLUTIONS ------------------

Using the following program

     operator f;

     procedure settoeplitz n;
        begin
           m:= -1;
           matrix ident(n,n),toeplitz(n,n);
           for i:= 1:n do
             <<ident(i,i) := 1;
               m := m+2;
               for j:= i:n do
                 <<toeplitz(i,j) := f(i+j-m);
                   toeplitz(j,i) := toeplitz(i,j)>>>>;
       end;

     for i := 2:10 do
        <<showtime;
          settoeplitz i;
          x := factorize det(toeplitz - l*ident);
          write "order ",i,": ",length x," ",
                length(y := first x)," ",
                length(y := second x)>>;

the following results were computed on an HP 9000 Series 350 in a 1.65
Mbyte heap (the times include garbage collector time):

        order 2: 2 3 3

        Time: 6205 ms

        order 3: 2 3 6

        Time: 238 ms

        order 4: 2 11 11

        Time: 612 ms

        order 5: 2 11 24

        Time: 1462 ms

        order 6: 2 51 51

        Time: 10166 ms

        order 7: 2 49 107

        Time: 89828 ms

        order 8: 2 270 270

        Time: 1230409 ms

The job then ran out of heap space before order 9 could be computed.

                 ----------------------------------------

Using a similar program, Herbert Melenk <zb6260@db0zib21.bitnet> and
Winfried Neun obtained the following times running on a Cray X-MP with a
Y-MP memory at Cray Research, which allowed 16 million items in the heap:

        order 4: 2 11 11

        Time: 46 ms

        order 5: 2 11 24

        Time: 142 ms

        order 6: 2 51 51

        Time: 879 ms

        order 7: 2 49 107

        Time: 6625 ms

        order 8: 2 270 270

        Time: 87142 ms  plus GC time: 22010 ms

        order 9: 2 248 549

        Time: 1078558 ms  plus GC time: 594765 ms

They started the order 10 computation, but stopped it after they concluded
that they would need an even larger memory for this.


 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 4

One of the difficulties novice users (perhaps all users) encounter
with CA systems is that they will often pose an innocent looking
problem and receive a horrible looking answer or no answer at all. I
have noticed that many researchers routinely derive integrals or ODEs
in their work, but do not remember very much about the analytic
techniques for solving them. I believe these users want a CA system to
give an answer or tell them that it cannot solve the problem. They
are not pleased when a system poses a different problem or asks a
multitude of questions which may be more difficult to answer than the
initial problem. While no system may ever supply all the wants or
needs of a user, I feel that novice users need far more information
about the true capabilities of CA systems and the ease in using them
to solve routine or difficult problems.

An integral I encountered in a study of optical discrimination addresses
some of these issues:

'integrate(sin(x)*acos(b^2/sin(x)),x)=b^2*atan(cos(x)/sqrt(sin(x)^2-b^4))-
asin(b^2*cot(x)/sqrt(1-b^4))-cos(x)*acos(b^2/sin(x));

      /                2
      [               b           2            cos(x)
      I sin(x) acos(------) dx = b  atan(------------------)
      ]             sin(x)                       2       4
      /                                  sqrt(sin (x) - b )

                                  2             2
                                 b             b  cot(x)
                 - cos(x) acos(------) - asin(------------)
                               sin(x)                   4
                                              sqrt(1 - b )

If you present this integral (the left hand side) to MACSYMA (versions
412.45 and 309.1), it does not return anything useful. MACSYMA gives a
different looking, very messy integral. To coax MACSYMA to give an
answer it is necessary to change variables. Then one encounters an
expression which requires some extensive manipulation to get it into
the form I have shown above.

Let me add that integrands of the form trig(x)*inverse_trig(k/trig(x))
are fairly common in problems involving spherical geometry and
projected surfaces.

My queries are the following: 

1) Can any CA system perform this integration without a change of 
   variable or trickery? That is, can a novice use ANY system to
   perform this integration?
2) If not, would the implementors explain why their systems are
   unable to handle this class of integrals?
3) If a change of variable is needed, how little work would be 
   necessary to get the answer into a useful form, similar to the one 
   I wrote above?
4) Would some users care to tell us how often they find a CA system to
   be of limited use in their work because it cannot solve problems 
   which you believe it should be able to solve? -- 

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 5

Prove the following identity:

               3 %pi          2 %pi
           tan(-----) + 4 sin(-----) = sqrt(11)
                11             11

This identity is given by Hobson (without proof) in his 1891 "A Treatise
on Plane and Advanced Trigonometry".  The identity is the next to last in
Chapter 7.

With MACSYMA I have found many more similar identities for 
foo(m,n,r):=tan(m*%pi/11)+n*sin(r*%pi/11)-sqrt(11);

foo(1, 4, 3)=0 
foo(1, 4, 8)=0 
foo(3, 4, 2)=0 /* Hobson's */
foo(3, 4, 9)=0 
foo(4, 4, 1)=0 
foo(4, 4, 10)=0
foo(9, 4, 5)=0 
foo(9, 4, 6)=0 

For 7 replacing 11 there are several:
foo(m,n,r):=tan(m*%pi/7)+n*sin(r*%pi/7)-sqrt(7);
foo(3, 4, 8)=0 
foo(5, 4, 3)=0 
foo(5, 4, 4)=0 
foo(6, 4, 2)=0 
foo(6, 4, 5)=0 
foo(10, 4, 8)=0 

For 13 replacing 11 I find none. 
foo(m,n,r):=tan(m*%pi/13)+n*sin(r*%pi/13)-sqrt(13)=0

In each case I let only let m,n,r run thru 10 so I have certainly
missed many. There are three questions which I pose:

1) Why does the integer 4 appear to play a role in these identities?
2) Is there some general scheme for generating these integer 
   identities?
3) Is there some simple underlying mathematics behind these
   identities?

              ------------------ SOLUTIONS ------------------

% From Hans-Gert Graebe <graebe@informatik.uni-leipzig.dbp.de>.

% Define trigonometric simplification rules :

for all x let tan x = sin x / cos x,cot x = cos x/sin x;

for all x,y let sin(x+y)=sin x * cos y+ cos x* sin y;

for all x,y let cos(x+y)=cos x * cos y - sin x * sin y;

for all x,k such that fixp k and k>1 let
  sin(k*x)=sin((k-1)*x)*cos x+cos((k-1)*x)*sin x,
  cos(k*x)=cos((k-1)*x)*cos x-sin((k-1)*x)*sin x;
                
for all x let sin(x)^2=1-cos(x)^2;

% Input the example :

off allfac;

f:=tan(3x)+4sin(2x);

f:=sub(cos x=y,f^2-11);

% Compute the minimal polynomial of y=cos(pi/11) :

sub(cos x=y,cos(11x)+1);

factorize ws;

% This is the char. polynomial :
second ws; 

% Define the correct rule for y :
let ws=0;  

% And then f is zero:
f;         

end;


 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 6

Finding reasonable looking solutions to systems of non-linear equations is
unusual.  In the example below I had the feeling that the solution(s) must
be simple owing to the physical nature of the problem.  This feeling was
strengthened by assigning numerical values to the constants and solving
the far easier problem.

The problem comes from a study in quantum chemistry where we are given six
equations

equations:[de3*m3-de2*m2+de1*m1,   -ep3*m3+ep2*m2+ep1*m1,
 -r1^2+(ep1-ep2)^2+(de2+de1)^2,   -r2^2+(ep3+ep2)^2+(de3+de2)^2,
 de1*m1*(2*m3+m2)-de2*m2*(m3-m1)-de3*(m2+2*m1)*m3,
 -r1*r2*sin(th)+(de2+de1)*ep3+(de1-de3)*ep2+(de3+de2)*ep1]$

to be solved for ep1,ep2,ep3,de1,de2,de3.  The solution must be physical,
namely real and positive.  The ep(j) and de(j) are coordinates of a
triatomic molecule in two different systems, m1,m2,and m3 are atomic
masses, and r1 and r2 are bond lengths.  Therefore, m1+m2+m3 = m (total
mass) is a natural relation in the solution shown below although it does
not appear with regularity.

solution:
[de1 = m2*m3*r1*r2*sin(th)/sqrt(-2*h1*h2*cos(th)+h2^2+h1^2),
 de2 = 2*m1*m3*r1*r2*sin(th)/sqrt(-2*h1*h2*cos(th)+h2^2+h1^2),
 de3 = m1*m2*r1*r2*sin(th)/sqrt(-2*h1*h2*cos(th)+h2^2+h1^2),
 ep1 = ((h1-h2)*(m3*r2-m3*r1-m2*r1)-(h2*m3*r2+h1*m3*r1+h1*m2*r1)*(cos(th)-1))
     /(m*sqrt(-2*h1*h2*cos(th)+h2^2+h1^2)),
 ep2 = ((h1-h2)*(m3*r2+m1*r1)-(h2*m3*r2-h1*m1*r1)*(cos(th)-1))
     /(m*sqrt(-2*h1*h2*cos(th)+h2^2+h1^2)),
 ep3 = ((h1-h2)*(m2*r2+m1*r2-m1*r1)-(h2*m2*r2+h2*m1*r2+h1*m1*r1)*(cos(th)-1))
     /(m*sqrt(-2*h1*h2*cos(th)+h2^2+h1^2))]$

where
[h1 = (m2+2*m1)*m3*r2,h2 = m1*(2*m3+m2)*r1]$

When I turned MACSYMA loose on the full set of equations, asking it to
solve them, it ground away for hours and returned a horrendous mess of
thousands of useless terms.  This was not very surprising since it was
using brute force.  Eventually I found, over several days, the solution
above by eliminating particular variables from the equations and dealing
interactively with them.

Most systems claim the ability to solve systems of non-linear equations,
but finding a solution and finding a reasonable looking solution are not
equivalent.

The question I should like to pose is whether any system can get close,
automatically, to the solution I have shown above.  In addition perhaps
others would share their experiences or tricks used in solving systems of
non-linear equations.

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 7

The following is the seventh in a series of problems which I am sending to
readers of sci.math.symbolic.  While I was able to solve the second
problem at the bottom of this message with MACSYMA (with some difficulty),
its generalizations may be too hard or impossible for any CA system.
Therefore I am also sending the message to sci.math to see whether the
human touch can beat the computer (as it often does).

Background:  Multidimensional normal integrals are generally difficult to
evaluate numerically, much less in closed form, over any range of
integration.  Closed forms are known for a very limited class of
covariance matrices.  However, I have recently found two closed forms for
particular integrals whose covariance matrix is of a the most elementary
Toeplitz form.  The answers are so simple in appearance that extensions to
higher dimensions may be possible.  However, I have been unable to go
above three dimensions.

Here are the MACSYMA statements which generate the 3D and 4D integrands
for 3x3 and 4x4 Toeplitz matrices:

mat3:matrix([1,p,p^2],[p,1,p],[p^2,p,1])$
vec3:matrix([x,y,z])$
-factor(ratsimp(transpose(vec3).mat3^^(-1).vec3))/2;
kernal3:exp(%)/((2*%pi)^(3/2)*sqrt(factor(determinant(mat3))));

mat4:matrix([1,p,p^2,p^3],[p,1,p,p^2],[p^2,p,1,p],[p^3,p^2,p,1])$
vec4:matrix([x,y,z,w])$ 
-factor(ratsimp(transpose(vec4).mat4^^(-1).vec4))/2;
kernal4:exp(%)/((2*%pi)^(4/2)*sqrt(factor(determinant(mat4))));

The extensions to higher dimensions should be obvious.

Here are the integrals and answers for the first three dimensions. The
first is trivial. The second is obscure but known in the literature.
The third may be new. 

                                   2
                          inf     x
                         /      - --
                         [        2
                         I    %e     dx
                         ]
                         /
                          0                 1
                        ----------------- = -
                        sqrt(2) sqrt(%pi)   2


                     2              2
         inf  inf   y  - 2 p x y + x
        /    /      -----------------
        [    [      2 (p - 1) (p + 1)
        I    I    %e                  dx dy
        ]    ]
        /    /
         0    0                               2 asin(p) + %pi
        ----------------------------------- = ---------------
                                2                  4 %pi
                2 %pi sqrt(1 - p )


                  2              2  2    2              2
 inf  inf  inf   z  - 2 p y z + p  y  + y  - 2 p x y + x
/    /    /      ----------------------------------------
[    [    [                 2 (p - 1) (p + 1)
I    I    I    %e                                         dx dy dz
]    ]    ]
/    /    /
 0    0    0
------------------------------------------------------------------ = 
                              3/2
                 2 sqrt(2) %pi    (p - 1) (p + 1)

        2
2 asin(p ) + 4 asin(p) + %pi
----------------------------
           8 %pi

My queries to readers are the following:

1) Can anyone find a closed form for the 4D integral and possibly
   extend it to higher dimensions? 
2) Can any using a Computer Algebra system verify or derive the second
   and third integrals above? If so, how?-- 

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 8

The following is the eighth in a series of problems which I am sending
to readers of this newsgroup: The problems below interest me because
they came from two different, unrelated people in the same week and
exhibit a weakness in CA systems. 

The problems are unusual owing to their simplicity. I cannot recall
seeing a "reasonable problem for a CA system", which I could do by
hand in seconds, for which CA systems fail as in Part1 below. For fun
I shall go out on a limb by stating that I believe no Computer Algebra
system can do either of these problems directly, and I challenge the
community to prove me wrong subject to certain guidelines. For example,
it would not be fair to insert a numerical integration routine for
erf(x) as a plotting fix.

Part1: Find the limit as x --> infinity of exp(x^2)*(1-erf(x)). 
Part2: Plot the above function from x=0 to x=12.

Comments: Part1 is trivial if l'Hopital's rule is applied but calling
"limit" directly fails in the CA systems I tried. Part2 seems to fail
because the numerical precision of erf(x) for large values of x is not
sufficiently accurate and roundoff errors occur. I was, however, able
to plot the function from 0 to 12 using Fortran on an Amdahl to
quadruple precision. 

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 9

The following is the ninth in a series of problems which I am sending
to readers of this newsgroup. In August, 1988, I posted Problem3
concerning the factorization of the determinant of symmetric Toeplitz
matrices. This provided a test of factorization algorithms and placed
some limits on computing power. The following problem has some
similarities because the expression we seek also factors, but it is a
perfect square. Therefore a different routine may be used. However,
the growth in the size of the expression for increasing order is far
faster than the example in Problem3. By way of historical interest,
this problem was posed by Greg Fee at the University of Waterloo about
8 years ago. It was shown to be valid for the 4x4 and 5x5 cases using
MACSYMA. It circulated among users of the MIT-MC machine for several
weeks until someone proved it with elementary methods and a written 
proof was found in a book. I do not have access to those records at
this time so perhaps someone will be able to come up with a reference.

The problem(theorem) may be stated as follows:
If M and N are antisymmetric matrices then the determinant 
DET(I-M.N) is a perfect square where I is the identity matrix. 

Here is MACSYMA output for the 3x3 case. The MACSYMA command SQRF
finds polynomial factors which are perfect squares. I do not show the
user function which generates the antisymmetric matrices.

(c2) mat(x,3);
                             [   0     x12   x13 ]
                             [                   ]
(d2)                         [ - x12    0    x23 ]
                             [                   ]
                             [ - x13  - x23   0  ]

(c3) mat(y,3);
                             [   0     y12   y13 ]
                             [                   ]
(d3)                         [ - y12    0    y23 ]
                             [                   ]
                             [ - y13  - y23   0  ]

(c4) sqfr(determinant(ident(3)-mat(x,3).mat(y,3)));

                                                       2
(d4)                  (x23 y23 + x13 y13 + x12 y12 + 1)

Below is a table of the cases which I have been able to run on a VAX
780 using MACSYMA 309.1. I have not been able to get beyond the 5x5
case on a VAX or on a Lisp Machine. In the 6x6 case the determinant,
before simplification, has more than 4 million terms and the VAX runs
out of space at this stage while the lispm ran for 6 hours without
returning an answer. In the following LDET is the number of terms in
the simplified determinant and LSQR is the number of terms in square
free term.

Order   LDET    LSQR   Time (sec)

  3      10       4       1 
  4     124      16      32
  5    1506      56     700

My queries are:

1) Can anyone extend this table?
2) Can anyone provide us with a proof or reference?-- 

 ----------------------------------------------------------------------------

From: smith@joshua.math.ucla.edu (Douglas Smythe)

Here is a simple proof (which works over any field not of characteristic 2):

I will base it on some simple steps, each with proofs or hints:

(1) An odd-sized antisymmetric matrix has determinant 0.

    This is routine to verify.

(2) The determinant of an even-sized antisymmetric matrix  A  (whose
    entries above the diagonal are indeterminates), is non-zero
    and it is a perfect square (of a polynomial in its entries)

    Proof:   Suppose  A  is  n  by  n. The result is clear for  n = 2. 
    For larger  n  expand  A  using Sylvester's identity:  If  B  is the
    (n-2)  by  (n-2)  matrix obtained by omitting the first and last rows
    and columns of  A, then Sylvester's identity states that

      det(A) det(B) = det(A[1,1]) det(A[n,n]) - det(A[1,n]) det(A[n,1])

    where  A[i,j]  denotes the  (n-1) by (n-1)  matrix obtained by
    putting back the  ith  row  and  jth  column.   In the above formula
    A[1,1]  and  A[n,n]  are odd-sized anti-symmetric matrices and hence
    by (1) have determinant 0.   The other two terms, when written out
    are negatives of each other and are easily seen, by induction, to be
    non-zero.  This shows that  det(A)  is non-zero and a square of a
    rational function of its entries.  Since  det(A)  is a polynomial in
    its entries, this rational function must also be a polynomial in the
    entries of  A.

(3) The claimed theorem is true for even-sized matrices.

    Proof:  Write
                                    -1
        det(I - A B)  = det(A) det(A   -  B).

         -1                                    -1
    Now A   is antisymmetric and hence so is  A   -  B.

    It is immediate that both determinants on the right hand side are
    squares, hence is so  det(I - AB)  and, using the same argument as
    above it is the square of a polynomial in the entries of  A  and  B.

(4) If the theorem is true for an  n by  n  matrix  A  it is
    true for an  (n-1)  by  (n-1)  matrix.

    This is seen by replacing the entries in the last row and column of 
    A  by 0.

This completes the proof.  Note that it would be quite easy to give
explicit formulas for the desired squareroot.

Comment:  I don't think this problem is a good test for a symbolic
manipulation routine.  The problem shows exponential explosion and is
not the kind of thing that someone using symbolic manipulation routines
would normally do.

 ----------------------------------------------------------------------------

                             PROBLEM NUMBER 10


The following is the tenth in a series of problems which I am sending
to readers of this newsgroup (following a pause of a few months). This
problem interests me because it looks simple, and yet I question
whether any CA system can perform it without a fair amount of user
interaction. If I am wrong then users of various systems will be able
to correct me.  

It comes from a problem, recently solved, of obtaining a closed form
for the radiation from microstrip bends with arbitrary angle. I may
send a second (far more difficult) related integral later. 

The problem is to derive the right hand side of the following where 
0 < g < 1 .

     %pi
     ---
      2
    /             4                                  2     2
    [          cos (x)            3 %pi (2 sqrt(1 - g ) + g  - 2)
    I      --------------- dx = - -------------------------------
    ]                    2                        4
    /      (g sin(x) + 1)                      2 g
       %pi
     - ---
        2

I used MACSYMA 309.1 for this. 

--------------------------------------------------------------------------

                             PROBLEM NUMBER 11

Prof. Knuth posed a problem in the June-July American Mathematical Monthly 
which I thought would be easy with MACSYMA. WRONG!!!"!!!

Here it is as stated on Page 525:

Solve the recurrence

X  = A, X  = B, X    = X    + X  /(N+1) for N = 0,1,2,.....
 0       1       N+2    N+1    N

both exactly (in terms of familiar functions of N) and asymptotically.


--------------------------------------------------------------------------

                             PROBLEM NUMBER 12

The following is the 12th in a series of problems which I am sending
to readers of this newsgroup. The following correspondence in the
newsgroup rec.puzzles appeared yesterday:

Newsgroups: rec.puzzles
Subject: Re: 2 problems   (PARTIAL SPOILER)
Date: 2 Feb 90 13:11:30 GMT
References: <1043@acf5.NYU.EDU>
Organization: Penn State University

In article <1043@acf5.NYU.EDU>, ambati@acf5.NYU.EDU (FJLevM{n[]Balamurali
Ambati) says:
>>1) How many squares are composed of the digits 1,4, and 9 alone.
>>For example (144, 49).

>My guess is that there are only 7: 1^2=1, 2^2=4, 3^2=9, 7^2=49, 12^2=144,
>21^2=441, and 38^2=1444.  But I can't prove it.  The obvious approach is
>to show that there's a bound on the number of digits at either the front
>or the end which can be 1, 4, or 9.  But there isn't; for any N, there
>exist squares which begin with N 1's and end with N 1's or 4's.

As users of Computer Algebra systems know, while the systems are not
designed for numeric problems, they can often be used effectively
for this purpose. The above is a good example of such usage. It took me
a few minutes of coding MACSYMA and two hours of run-time on a VAX780
to produce these ten examples, thereby extending the set of seven examples
mentioned above. I would be interesting to see how users of other CA
systems can handle this problem with the tools provided.

1^2     = 1
2^2     = 4
3^2     = 9
7^2     = 49
12^2    = 144
21^2    = 441
38^2    = 1444
107^2   = 11449 
212^2   = 44944
31488^2 = 991494144


              ------------------ SOLUTIONS ------------------

In the absence so far of a REDUCE solution, I include a Maple solution:

found := 4:
pairs := [[1,1],[2,4],[3,9],[7,49]]:
settime := time():
for i from 10 by 10 while found < 11 do
    candidates := [i+1,i+2,i+3,i+7,i+8,i+9];
    for j in candidates do   
        square := convert(j^2,'name');
        digits := {'substring(square,k..k)' $ 'k'=1..length(square)};
        if digits intersect {`1`,`4`,`9`} = digits then 
           pairs := [ op(pairs), [j,j^2] ]:
           found := found + 1:
        fi
    od
od:
cpu_time := time()-settime:
print(pairs);
print('cpu_time'=cpu_time);
done


On a SUN 4/280 in about 10 minutes:


    [[1, 1], [2, 4], [3, 9], [7, 49], [12, 144], [21, 441], [38, 1444],

        [107, 11449], [212, 44944], [31488, 991494144], [70107, 4914991449]]

                               cpu_time = 355.817

Andre Heck
Expertise Centre CAN
e-mail: heck@cwi.nl

--------------------------------------------------------------------------

                             PROBLEM NUMBER 13

I posted this problem in July 88 to sci.math.symbolic, and many of the
systems were able to solve it.  Mathematica did not have an opportunity to
respond or was not in shape to solve it.  Neither was Derive.  In any case
the problem will now be published as part of a paper, and the referee has
asked that a table of systems and timings be included.  I tried to use
Mathematica on a Mac, but it runs out of space.  I do not have Derive.
Perhaps someone would try these systems or others on different machines
and post the times.  If someone can improve on the timings below with
their system please post it, and I will include it in the table.  Here it
the problem written in MACSYMA:

op1(f):=diff((diff(f,x,2)+diff(f,y,2)+diff(f,z,2)),x,2)$
op2(f):=n^2*(diff(f,x,2)+diff(f,y,2))$
exp:(sin(n*sqrt(x^2+y^2+z^2)*z/sqrt(y^2+z^2)))/sqrt(x^2+y^2+z^2)$
ratsimp(op1(exp)+op2(exp));

This final expression is zero.

The table from 1988 was:
        
user            system          machine         CPU times

Todd McComb     Maple 4.2       Sun-4           61 sec
Kamal Abdali    REDUCE 3.3      SUN 3/160       70 sec
Greg Fee        Maple 4.2       VAX 11/785      80 sec
R. Pavelle      MACSYMA 412.45  Symbolics 3650  1200 sec
R. Pavelle      MACSYMA 309.1   VAX 11/780      10800 sec

              ------------------ SOLUTIONS ------------------

Here's a simple REDUCE solution running on a SPARCstation 1 using the
PSL-based version:

REDUCE 3.3, 15-Jan-88 ...

1: showtime;

Time: 17 ms

operator op1,op2;

for all f let op1(f)=df((df(f,x,2)+df(f,y,2)+df(f,z,2)),x,2),
              op2(f)=n^2*(df(f,x,2)+df(f,y,2));

exp := (sin(n*sqrt(x^2+y^2+z^2)*z/sqrt(y^2+z^2)))/sqrt(x^2+y^2+z^2)$

op1(exp)+op2(exp);

0

showtime;

Time: 4607 ms

--------------------------------------------------------------------------

                             PROBLEM NUMBER 14

Keywords: A difficult matrix inversion and simplification
Date: 10 May 90 18:29:25 GMT

I encountered a problem this week which makes a good candidate for the
14th entry in the series I am submitting to the group.  The problem arises
in photon scattering and involves little more than inverting a particular
4x4 matrix.  But owing to the nature of the matrix the components of the
inverse are huge.  It therefore creates an unusually difficult matrix
inversion problem for my MACSYMA.

I am running Unix version 309.1 and when I try to invert it the MACSYMA
runs out of space.  This may be a difficulty with the underlying Franz,
but it may just be that big.  I had to use some special tricks to invert
and factor it on this machine, and it took more than 4 CPU hrs.  The final
answer when factored is greatly simplified.  So I am curious to know
whether any system can compute the inverse in a direct manner and factor
the components.  Here is the matrix:

s:matrix(
[(er^2+ei^2)/(4*w^2+g^2+2*er^2+2*ei^2),
-(er^2+ei^2)*(2*p1+g)/((p1+g)*(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g)),
-(er^2+ei^2)*(2*p2+g)/((p2+g)*(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g)),
-(er^2+ei^2)*(2*p3+g)/((p3+g)*(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g))],
[-(2*er*w-ei*g)/(4*w^2+g^2+2*er^2+2*ei^2),
(2*er*w-2*ei*p1-ei*g)/(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g),
(2*er*w-2*ei*p2-ei*g)/(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g),
(2*er*w-2*ei*p3-ei*g)/(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g)],
[-(2*ei*w+er*g)/(4*w^2+g^2+2*er^2+2*ei^2),
(2*ei*w+2*er*p1+er*g)/(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g),
(2*ei*w+2*er*p2+er*g)/(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g),
(2*ei*w+2*er*p3+er*g)/(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g)],
[(4*w^2+g^2+er^2+ei^2)/(4*w^2+g^2+2*er^2+2*ei^2),
-(4*p1*w^2+4*g*w^2+4*p1^3+8*g*p1^2+5*g^2*p1+2*er^2*p1+2*ei^2*p1+g^3
+er^2*g+ei^2*g)
 /((p1+g)*(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g)),
-(4*p2*w^2+4*g*w^2+4*p2^3+8*g*p2^2+5*g^2*p2+2*er^2*p2+2*ei^2*p2+g^3
  +er^2*g+ei^2*g)
 /((p2+g)*(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g)),
-(4*p3*w^2+4*g*w^2+4*p3^3+8*g*p3^2+5*g^2*p3+2*er^2*p3+2*ei^2*p3+g^3
  +er^2*g+ei^2*g)
 /((p3+g)*(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g))])$


              ------------------ SOLUTIONS ------------------

From: monagan@inf.ethz.ch (Michael)
Date: 11 May 90 16:27:27 GMT

In article <22001@ethz-inf.UUCP>, maeder@inf.ethz.ch (Roman Maeder) writes:
> > From: rp@XN.LL.MIT.EDU (Richard Pavelle)
> > Subject: Problem14
> > Keywords: A difficult matrix inversion and simplification
> > 
> > I encountered a problem this week which makes a good candidate
> > for the 14th entry in the series I am submitting to the
> > group. The problem arises in photon scattering and involves
> > little more than inverting a particular 4x4 matrix. But owing
> > to the nature of the matrix the components of the inverse are
> > huge. It therefore creates an unusually difficult matrix
> > inversion problem for my MACSYMA.
> 
> (all following timings are with Mathematica 1.2 on a Sun 4/330)
> 
> Well, the inversion is no problem, it takes about 14 seconds to
> compute it, but trying to print the result is next to impossible.
> Canceling common terms in the numerators and denominators 
> brings the output down to reasonable size, however (this takes about
> 25 min.). Max memory used was about 4.5MB.
> ... 

Not to be outdone by my new colleague, I ran it in Maple.
Takes only 30 seconds and under 1 MB to compute, simplify
(i.e. cancel common terms in the numerators and denominators
which is done automatically in Maple) and print the inverse
in Maple 4.3 on a DEC station.

Mike
monagan@inf.ethz.ch

    |\^/|
._|\|   |/|_. Licensed by the University of Waterloo
 \  MAPLE  /  Version 4.3 --- Mar 1989
 <____ ____ >  For on-line help, type  help(); 
      |
 > s:=array(
 > [[(er^2+ei^2)/(4*w^2+g^2+2*er^2+2*ei^2),
 > -(er^2+ei^2)*(2*p1+g)/((p1+g)*(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g)),
 > -(er^2+ei^2)*(2*p2+g)/((p2+g)*(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g)),
 > -(er^2+ei^2)*(2*p3+g)/((p3+g)*(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g))],
 > [-(2*er*w-ei*g)/(4*w^2+g^2+2*er^2+2*ei^2),
 > (2*er*w-2*ei*p1-ei*g)/(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g),
 > (2*er*w-2*ei*p2-ei*g)/(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g),
 > (2*er*w-2*ei*p3-ei*g)/(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g)],
 > [-(2*ei*w+er*g)/(4*w^2+g^2+2*er^2+2*ei^2),
 > (2*ei*w+2*er*p1+er*g)/(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g),
 > (2*ei*w+2*er*p2+er*g)/(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g),
 > (2*ei*w+2*er*p3+er*g)/(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g)],
 > [(4*w^2+g^2+er^2+ei^2)/(4*w^2+g^2+2*er^2+2*ei^2),
 > -(4*p1*w^2+4*g*w^2+4*p1^3+8*g*p1^2+5*g^2*p1+2*er^2*p1+2*ei^2*p1+g^3
 > +er^2*g+ei^2*g)
 >  /((p1+g)*(2*er*w+2*ei*w+2*er*p1-2*ei*p1+er*g-ei*g)),
 > -(4*p2*w^2+4*g*w^2+4*p2^3+8*g*p2^2+5*g^2*p2+2*er^2*p2+2*ei^2*p2+g^3
 >   +er^2*g+ei^2*g)
 >  /((p2+g)*(2*er*w+2*ei*w+2*er*p2-2*ei*p2+er*g-ei*g)),
 > -(4*p3*w^2+4*g*w^2+4*p3^3+8*g*p3^2+5*g^2*p3+2*er^2*p3+2*ei^2*p3+g^3
 >   +er^2*g+ei^2*g)
 >  /((p3+g)*(2*er*w+2*ei*w+2*er*p3-2*ei*p3+er*g-ei*g))]]):
 > evalm(1/s);
                                     2                     2
 [1/4 (- 4 p2 p3 p1 - 4 p1 p2 g - 4 g  p1 - 4 p1 g p3 - 4 g  p2 - 4 p2 g p3

         2       2        3      2         2    2       2       2
     + er  g + ei  g - 4 g  - 4 g  p3) (4 w  + g  + 2 er  + 2 ei )

       /     2     2                       2    2       2       2
      /  ((er  + ei ) p2 p3 p1), - 1/8 (4 w  + g  + 2 er  + 2 ei ) (
     /

        ...


           (p3 + g) (2 er w + 2 ei w + 2 er p3 - 2 ei p3 + er g - ei g)
     - 1/4 ------------------------------------------------------------]
                                            2
                         (p1 p2 - p3 p1 + p3  - p2 p3) p3

 > quit
 bytes used=7585924, alloc=925696, time=28.950

--------------------------------------------------------------------------
