Conference rusure::math

Following the form-feed is a solution to the first problem.  This is
a freebie, just to ensure that the problem is understood.

    1.  t = a + b*c


Let f(a,t) = (t-a)*a.  Note that this must be expressed without b's or c's.
Then f = ((a+b*c)-a)*a = a*b*c, which is symmetric in a, b, and c, since
f(a,b,c) = f(a,c,b) = f(b,a,c) = f(c,b,a).

    For t = 1/(b+c), an f is a+1/t.
    
    For t = (1+a^2)/(b+c)^2, an f is {a+sqrt([1+a^2]/t)}^2 -- the last
    square is to make the signs work.
    
    
    				-- edp 

     Re .-1
    
>>    For t = (1+a^2)/(b+c)^2, an f is {a+sqrt([1+a^2]/t)}^2 -- the last
>>    square is to make the signs work.

     With the proposed f(a,t) you get

          f(a,t) = {a + sqrt((1 + a^2)/t)}^2
                 = {a + sqrt((b+c)^2)}^2
                 = (a + |b + c|)^2

     which is not symmetrical:

          f(a=3, b=-1, c=-1) = 25            (This replaces an earlier
          f(a=-1, b=3, c=-1) =  1             reply where I had these
                                              wrong)

     Dan

    Items 3 and 4 are still unsolved.  Given t as a function of a, b, and c
    that is symmetric in b and c, find a non-trivial function f(a,t) such
    that f(a,t) is symmetric in a, b, and c.
    
    	3.  t = (1+a^2)/(b+c)^2       
    
    	4.  t = b + c + 1/(b*c)
    
    
    				-- edp

3.  t = (1+a^2)/(b+c)^2

This one is not too hard.  Suppose / let ...

	f1(a,t) = (1+a^2)/t = (b+c)^2

So far so good.  Let's guess that the desired f is:

	f(a,t) = (a+b+c)^2

In other words, we'll try f = f1 + 2ab + 2ac + a^2.
Now we massage those last terms: f = f1 + 2a(b+c) + a^2.
This works, since we have b+c = sqrt(f1)!

Solution:

	f(a,t) = (1+a^2)/t + 2 a sqrt((1+a^2)/t) + a^2

	       = (a+b+c)^2

    Re .6:
    
    The problem described in .4 still applies -- sqrt((1+a^2)/t) =
    sqrt((b+c)^2) = |b+c|, not b+c.
    
    
    				-- edp

>>    Re .6:
>>    
>>    The problem described in .4 still applies -- sqrt((1+a^2)/t) =
>>    sqrt((b+c)^2) = |b+c|, not b+c.

	Maybe we just have to restrict b+c to be nonnegative in
	order to solve #3.

	Dan

    3.	t = (1+a^2)/(b+c)^2

We can easily (and reversibly) change this problem to:

    3a.	t = (b+c)^2

We want F (t,a) = a symmetric function of a, b, and c.
         1

Let's suppose that F (t,a) takes the form of a polynomial in t, where each
                    1
coefficient in the polynomial is itself a polynomial in a.  We will below

that this cannot yield a symmetric function.


Suppose
	F(b+c,a) = P (a) + P (a) (b+c) + P (a) (b+c)^2 + P (a) (b+c)^3 + ...
	            0       1             2               3
where
	P (a) = p   + p  a + p  a^2 + p  a^3 + ...
	 i       i0    i1     i2       i3

Then the symmetry constraint

	F(b+c,a) = F(a+b,c)

greatly constrains the p  , so that F can be expressed as (below, p  = p  ).
                        ij                                         j    j0

	F(b+c,a) =
		(p  + p  a + p  a^2 + p  a^3 + p  a^4 + ...) +
		  0    1      2        3        4

		(p  + 2 p  a + 3 p  a^2 + 4 p  a^3 + 5 p  a^4 + ...) (b+c) +
		  1      2        3          4          5

		(p  + 3 p  a + 6 p  a^2 + 10 p  a^3 + 15 p  a^4 + ...) (b+c)^2 +
		  2      3        4           5           6

		(p  + 4 p  a + 10 p  a^2 + 20 p  a^3 + 35 p  a^4 + ...) (b+c)^3
		  3      4         5           6           7

		+ ...

		          j              i
	    =	Sum  (b+c)  Sum  C(i,j) a
		j>=0        i>=i

where C(i,j) is `i choose j', the binomial coefficient.  For any choice of the
p  values in the above equation for F, we have a symmetric function in a, b, c.
 j


Now for our problem at hand, we want the odd terms (those (b+c) terms with odd
exponent) to be identically zero.  For the first odd (b+c) term, this forces

	p  = p  = p  = ... = 0.
	 1    2    3

So the only polynomial function F ((b+c)^2,a) that is symmetric in a, b, and c
                                 1
is the trivial one -- a constant value.


A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
it's implausible.

        re .9
        
>>	A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
>>	it's implausible.
        
        I agree most rigorously!
        
        Dan

	I haven't worked through the details, but yours seems to be a
very plausible approach.   Tell me, why did you select the four example
polynomials that you did?   Is this a puzzle with a known solution?

	A restatement of the problem which formalizes the definition of the
problem, but which doesn't seem to help to reach any solutions is the following:

	Let K be a field.   For which t lying in K(a,b,c) [the field of
fractions in three indeterminates, a, b & c] is the intersection of
the two fields K(a,t) & K(a+b+c, ab+ac+bc, abc) strictly greater than K?

Regards,
Andrew.

> Tell me, why did you select the four example polynomials that you did?
> Is this a puzzle with a known solution?

I got the problem from Stan Rabinowitz.  I'll ask him.

    I'm not comfortable with the statement that a non-polynomial function
    of (b+c)^2 and a could not be symmetric.
    
    Consider F(t,a) =
    	0 for t= 4,a=1
    	1 for t= 9,a=1 and t=4,a=2
    	2 for t=16,a=1 and t=9,a=2
    	3 for t=16,a=2
    	4 otherwise
    
    Now for f(a,b,c) = F( (b+c)^2, a) =
    	0 when all of a, b, and c are 1.
    	1 when one of a, b, and c is 2 and the others are 1.
    	2 when two of a, b, and c are 2 and the other is 1.
    	3 when all of a, b, and c are 2.
    	4 otherwise.
    
    This is symmetric.  It's also not "trivial", although it's close.  It
    demonstrates symmetry can be achieved without a polynomial.  Certainly
    we could add as many more values to F's range as we please.  Can it be
    done in a more aesthetic manner?
    
    
    				-- edp

	re .9 (I think)

>>	A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
>>	it's implausible.

	re .13

>>    I'm not comfortable with the statement that a non-polynomial function
>>    of (b+c)^2 and a could not be symmetric.
    
	Well, for a, b, and c nonnegative reals, we have already
	seen that F(a, t) = a + sqrt(t) yields F(a, (b+c)^2) = a+b+c.
	So for some domains nonpolynomial functions of a and (b+c)^2
	can be symmetric.  Over the whole real line it just becomes
	harder.

	Dan

re .13:

    No.  Actually, given your definition of F(t,a),

    f(a,b,c) = F((b+c)^2, a) =
	0 for |b+c|=2, a=1
	1 for |b+c|=3, a=1, or |b+c|=2, a=2
	2 for |b+c|=4, a=1, or |b+c|=3, a=2
	3 for |b+c|=4, a=2
	4 otherwise

    f(a,b,c) is not symmetric, since f(1,1/2,3/2) = 0,
    but f(1/2,3/2,1) = 4.  Or using positive integers,
    f(1,1,3) = 2, while f(3,1,1) = 4.

.15:

    Thanks.  I was looking for a proof like this, but
    couldn't find it.  Can you find one for (4)?

	Knowing that it is a quadratic doesn't imply real solutions.

	Dan

.21> Knowing that it is a quadratic doesn't imply real solutions.

True, but it doesn't matter -- only existence matters.

        But if the only roots that exist are complex, then it
        says nothing about whether there is a nontrivial F(a,t)
        over the reals that satisfies the symmetry constraint.
        
        Dan

>>	The fool who wrote this should have written there is a path of
>>	length at most 1 + ceil(||b|-|c||/k).

	You should be careful ... not everyone who reads these
	realizes that you are referring to a note that you wrote.
	So they go away thinking that this is a hostile conference.

	Dan

	The much-admired person who wrote .30 inquired as to the fate of
t = b + c + 1/bc when M = R+.   In fact, using a (hopefully familiar) result,
F(1,1) = 1, otherwise F(x,y) = 0, gives Ft as a 3-way symmetric function.

Regards,
Andrew.

.30> Now ~ connects M

	It took me a while to understand this.  In the case of Example2, it
	means that

		F(b,k) = F(c,k), for any bc < 0 and k in M.

	But now we "connect".  For b > 0,

		F(b,k) = F(-1,k) = F(1,k).

	And for b < 0,

		F(b,k) = F(1,k).

    5.  t(b,c) = ( b + c + bc )/( 1 + b + c )

Enjoy.

.37> That's neat.   What made it occur to you as a function to suggest?

I was looking at functions t(b,c)=N(b,c)/D(b,c) that had a minimum (or maximum).
If M=t(z,z) is an extrema, then F(a,k) = { K1 if a=z and k=M; else K2 }.

Actually, (b+c+bc)/(b+c+1) has a far less trivial solution!

.37> Have you had any feedback from Stan the Man?

Yes, I saw him this weekend.  The problem is his own, and I saw the solutions
given in `Crux' -- these were for t=(b^2+c^2) and t=bc(b+c), and the solutions
were very problem-specific.  The second half of the problem asks for a t(b,c)
that permits a symmetric F(a,t).  It's still open in `Crux'.  I'm still working
on it.

What's Stan doing?  He's still at Avid, and...

Stan is working on a concordance of math problems that have been published in
math magazines over the past 5 (or ten) years.  If you're interested in some
extra cash, by typing some in, or by translating Polish, Rumanian, ..., or if
you know someone who is, then call Stan at (617) 272-1680.

(1) Theory

	I like Peter's new idea, which I tidy slightly below.

	Essentially, this *entire* problem is all about connectivity of a 
certain graph, which is infinite for most of the problem we want to consider.  
Connectivity being what it is, there is unlikely to be any all-encompassing 
necessary & sufficient condition for the graph to be connected.   However, we 
have been successful in identifying some good general *sufficient* conditions.

	We can view the vertices of our graph as an array of elements (a,k), 
where a lies in M, and k runs over N.   We can view the edges of the graph as
lying in two families:
		(i) (a,t(a,b)) <-> (b,t(a,a))			[a <> b]
		(ii) (a,t(b,c)) <-> (b,t(a,c)) <-> (c,t(a,b))   [a<>b<>c<>a]

	The earlier algorithm was interested in showing that any one row was 
connected, (ie: (b,k) = (c,k) for all b,c,k) since from that the connectivity
of the entire graph follows trivially.

	Peter's new algorithm takes a different approach.   He asks, given 
a,b & c can we find x1,x2,y1,y2, such that the following hold true:
		(1) t(x1,x2) = t(b,c)
		(2) t(x2,a) = t(y1,y2)
	If we can, then we know that (a,t(b,c)) is connected to (y2,t(x1,y1)).
Now suppose that:
		(3) y2, & t(y1,x1) are symmetric in a,b,c.   
	Then the two families of edges listed above, can give us no more 
connections than we already have deduced.   So, F(a,k) = (y2, t(x1,y1)) is a 
symmetric function of a & t and as Peter says any other symmetric function of 
a & t(b,c) will be of the form QF, where Q is composed with F.

Notes:
(o)	Yes, F is a mapping from M x N -> M x N (but not nec. surjective!)
(i)	(1), (2) & (3) are not *necessary* for F to exist.
(ii)	x1,x2 & y1 can be non-symmetrical without disrupting the algorithm.
(iii)   (1) & (2) are widely satisfied, they are related to Peter's "step 2".
	(3) is rather tougher to satisfy.

	However, before I go into examples, I want Peter to check his working
on the famous (b+c+bc)/(b+c+1).   I don't believe that the y2 he derives is
symmetric, even for x1=y1=0.

Regards,
Andrew.

I've nailed b+c+1/(bc) for the positive reals.  It wasn't easy or pretty,
but it may illustrate some of the problems with these problems.

We're working with R+ throughout.

	t(b,c) = b + c + 1/(bc)

	t(b,c) = t(1/(bc),c) = t(b,1/(bc))

		(aside: F(a,b,c) = F(a,b,1/(bc)) = F(a,b,c(b/a)) is a nice
			transformation, but I couldn't find what to do with it)

	t(b,c) is minimized when b = c = 1; t(1,1) = 3.

	f(b,t(1/(bc),a))
		= f(a,t(1/(bc),b))
		= f(a,t(1/(bc),c))
		= f(c,t(1/(bc),a))

	So f(b,k) = f(c,k) for any k in { t(1/(bc),a) }.  (a,b,c all in R+)

	k in { t(1/(bc),a) }  <=>  k >= t(1/(bc),sqrt(bc)) = 2sqrt(bc) + 1/(bc)
		(since a=sqrt(bc) minimizes t(1/(bc),a)).

	Is f(b,k)=f(c,k)?  They are equal if k-3 >= (p-1)^2 (2p+1) / p^2, where
	p = sqrt(bc); but that's only a small part of the b-c plane.  But for
	a given k > 3, the graph of this inequality looks roughly like a
	hyperbola with some non-zero `width'.  In a vertical slice through
	(the width of) this curve (i.e., with a constant b), f(b,k)=f(c,k),
	where c varies within the bounds allowed by the inequality.  And
	in a horizontal slice, f(b,k)=f(c,k), with b varying.  Within this
	wide hyperbola, we can zig-zag to connect any two values.

		c3	\--\
			 \ |\
			  \| \
		c2	   \--\
			    \ |\
			     \| \
		c1	      \--\
			       \ |\
			  b1 b2 b3

	To visualize this, see the diagram and see that
		f(c1,k)=f(b,k) for b2 <= b <= b3;
	    and	f(b2,k)=f(c,k) for c1 <= c <= c2;
	    and	f(c2,k)=f(b,k) for b1 <= b <= b2;
	    and	f(b1,k)=f(c,k) for c2 <= c <= c3.
	So f(b3,k)=f(c3,k) even though the point (b3,c3) is not on the
	wide hyperbola.

	The following argument extends the equality between k values:
	Since the value of f(a,k) is independent of a, by symmetry it's also
	independent of b and c, and so f(a,k) must be a constant for any k.
	(We could instead find suitable a,b,c so that f(a,t(b,c)) = f(b,t(a,c))
	connects any/all different k values).


	All that remains is to handle k=3.  It can't be handled as above,
	because when k=3, the hyperbola has no width.  We know that t(x,y)=3
	iff x=y=1.  Now f(a,3)=f(a,t(1,1))=f(1,t(a,1)).  And if a<>1, this
	last expression connects to all the other f(a,k) with k > 3.

	And finally, there is f(1,t(1,1)).  Symmetry constrains it to equal
	itself (which it does), and the only other connection involves finding
	other t(x,y) equal to t(1,1), which we can't.  So f(1,3) remains
	disconnnected, and is permitted its own value.

So we have:

    Problem:
	Let t(b,c) = b + c + 1/(bc) be a function R+ x R+ -> R+.
	Find the most general F(a,k) for which F(a,t(b,c)) is symmetric
	in its parameters.

    Solution:
	Let F(a,k) =
		K1	if k>3,
		K1	if k=3 and a<>1,
		K2	if k=3 and a=1,
	     undefined	if k<3.
	
	Where K1 and K2 are arbitrary constants (i.e., expressions that don't
	involve a or k).