Conference rusure::math

Bypass this if you want to try it yourself...

   4      2                 2       2
  x  - 13x  + 36 <= 0 --> (x  - 4)(x  - 9) <= 0 --> (x-3)(x-2)(x+2)(x+3) <= 0

This will be true when any of the terms are 0 or when an odd number of them
are negative; i.e. the set of real numbers that satisfies the inequality is

			[-3,-2] U [2,3].

		   3	      2	
The derivative of x  - 3x = 3x  - 3 which is positive throughout both
subranges, so the maximum must be at an righthand endpoint; trying -2 and 3
shows the max is at 3 and it is 18.

Bypass this if you want to try it yourself...


    20000     100          19900            -100
  10     / (10    + 3) = 10      / (1 + 3*10    )

     19900         1    -100    2     -200    3     -300
 = 10      * (1 - 3 * 10     + 3  * 10     - 3  * 10     + ...

	       198     -19800    199     -19900    200     -20000
	... + 3    * 10       - 3    * 10       + 3    * 10       - ...)

		      100    199     200      100
 = (some integer) * 10    - 3    + (3    / (10    + 3))

				    200    100     100
Now, the last term is less than 1 (3    = 9    < 10   ), so the rightmost
digit of the quotient will be expressed as

		 199		     49
	nnnn0 - 3     = nnnn0 - ((81)   * 27) = nnnnn0 - mmmmm7 = 3

Bypass this if you want to try it...


			  -1  2		  -1
Lemma: sigma(i=0 to n) cot  (i +i+1) = cot  (1/(n+1))

						  -1	     -1
Proof by induction: 1)	its true for n=0, i.e. cot  (1) = cot  (1)

2)	If its true for i=0 through n-1, then

			   -1  2	   -1           -1  2	
	sigma(i=0 to n) cot  (i +i+1) = cot  (1/n) + cot  (n +n+1)

	Using the formula cot(x+y) = (cot(x)cot(y)-1)/(cot(x)+cot(y))
	this yields the desired result and we have a lemma.

Therefore,
			   -1  2			     -1
 sigma(i=0 to infinity) cot  (i +i+1) = lim(n-->infinity) cot  (1/(n+1))

			= cot (0) = pi/2

    finally, one I CAN solve! (it IS the easiest one of the lot)  
    
              .
             / \
        ............. ____
        |  /     \  |
        | /   +   \ |   h
    	|/_________\| ____

                    
    	"+" :== center of circle
    
    when the height of the triangle that sticks up above the rectangle
    is equal to "h", the two areas will be the same.
    
    thus h + (1/2)h = r = 1
    
    		h = 2/3
    
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

Bypass this if you want to try it yourself. Its not a very pretty
solution - is there a better one?


Since each transversal contains exactly one element from each row and column,
if you pick two elements a(i,j) and a(k,l) in a transversal and substitute
the "other corners" a(i,l) and a(k,j) for them you still have a transversal.
If the sum of every transversal in A is the same, it follows that

	a(i,j) + a(k,l) = a(i,l) + a(k,j) for all i,j,k and l

which implies a(i,j) - a(k,j) = a(i,l) - a(k,l), which implies that each
row of the matrix A differs from each other row, and specifically from the
first row, by a constant; i.e.

(1)	a(i,j) = a(1,j) + d(i) for all i and j.

Conversely, any matrix that obeys formula (1) will have the sum of all
its transversals equal to sigma(j=1 to n)a(1,j) + sigma(i=2 to n)d(i), and
is thus the kind of matrix we are looking for. So, the set of matrices we
are looking for is exactly the set of matrices that obey (1).

Now, how many matrices of order n with elements in {-1, 0, 1} satisfy (1)?

Well, there are 3^n possible "first rows" of such a matrix. They fall into
three categories:

(a) First rows where max(a(1,j))-min(a(1,j)) = 0 - there are three such first
	rows, namely all 0, all 1, and all -1.

(b) First rows where max(a(1,j))-min(a(1,j)) = 1 - there are 2^(n+1)-4 such
	first rows, namely all rows consisting of just {0,1} except for all 0
	and all 1 (2^n-2 of these), and all rows consisting of just {0, -1}
	except for all 0 and all -1 (another 2^n-2 of these).

(c) First rows where max(a(1,j))-min(a(1,j)) = 2 - all the remaining first
	rows are of this type, namely 3^n-2^(n+1)+1.

Now:
For first rows of type (a) there are 3 choices of d(i) for every other row
For first rows of type (b) there are 2 choices of d(i) for every other row
For first rows of type (c) there are 1 choices of d(i) for every other row

So, the total number of matrices of order n is:

3 * 3^(n-1)  +  (2^(n+1)-4) * 2^(n-1)  +  (3^n-2^(n+1)+1) * 1^(n-1)

which in the form needed by the problem is:

	2 * 3^n  +  1 * 4^n  +  (-4) * 2^n  +  1 * 1^n

Whew!

    B-4 solution:
    
Show that G goes to zero by picking a particular m and n for each r.

 **  **  **  **  **  **  **  **  **  **  **  **  **  **  **  ** 

Let  H(r)  =  | r - sqrt(M^2 + 2*N^2) |

where   M = floor (r)
and     N = floor ( sqrt( (r-M)*M ) )

Then G(r) <= H(r), and H(r) ---> zero

(1)  Example:	r = 1234.5
		M = 1234
		sqrt (617) = 24.8
		N = 24
		sqrt (1234^2 + 2 * 24^2) = sqrt (1523908) = 1234.467
		H(1234.5) = | 1234.5 - 1234.467 | = .033

                | r^2 - M^2 - 2*N^2 |         | r^2 - M^2 - 2*N^2 |
(2)  H(r) =  ---------------------------  <  -----------------------
              | r + sqrt(M^2 + 2*N^2) |                 r

(3)  Work on the numerator:

	Let   X = r - M   and   Y = sqrt( (r-M)*M ) - N

	| r^2 - M^2 - 2*N^2 |

	=  | r^2 - (r-X)^2 - 2*(sqrt( (r-M)*M - Y)^2 |

	=  | r^2 - r^2 + 2*r*X - X^2 - 2*(sqrt( X*M - Y)^2 |

	=  | 2*r*X - X^2 - 2*( X*M - 2*Y*sqrt( X*M ) + Y^2 ) |

	=  | 2*r*X - X^2 - 2*X*M + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | 2*r*X - X^2 - 2*X*(r-X) + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | 2*r*X - X^2 - 2*r*X + 2*X^2 + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | X^2 + 4*Y*sqrt( X*M ) - 2*Y^2 |

	<= | X^2 | + | 4*Y*sqrt( X*M ) | + | 2*Y^2 |

	<= 1  +  4 * sqrt (M)  +  2

	<= 4 * sqrt(r)  +  3

                  4 * sqrt(r)  +  3
(4)  So H(r)  <  -------------------
                          r

	which ---> zero as r ---> infinity.
    
    
    John

    re .7:
    
    So what is "h"? 
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

    Re:             -< after all that, what's the answer? >-

    Oh well, maybe I'll learn to read someday.

    - Jim

Solution for B-2 follows


The simultaneous equations

		x (x-1) + 2 y z = y (y-1) + 2 z x = z (z-1) + 2 x y,

are equivalent to

		(x-y) (x+y-1-2z) = 0
		(y-z) (y+z-1-2x) = 0

The solutions to these are:

		((x = y) or (x+y = 1+2z)) and ((y = z) or (y+z = 1+2x))

Going through the 4 possibilities, the solutions are:

		x = y = z
		x = y, z = 1+x
		y = z, x = 1+y
		z = x, y = 1+z

In these cases, the ordered triple T = (x-y, y-z, z-x) takes values:

		( 0,  0,  0)
		( 0, -1,  1)
		( 1,  0, -1)
		(-1,  1,  0)

I agree with .12.

The (0,1,1) triple can't be right because the components
have to sum to (x-y)+(y-z)+(z-x) = 0.

I 'found' what appears to be a solution to problem A-6, but as yet have
no idea of why it turns out the way it does, or even whether it is the
general solution.  It follows the <FF>.


	Let a(1), ..., a(n) be real numbers, and let b(1), ..., b(n) be
distinct positive integers.  Suppose there is a polynomial f(x) satisfying 
the identity (1 - x)^n f(x) = 1 + sum(a(i) x^b(i)), where the sum is taken
from 1 to n.  Find a simple expression (not involving sums) for f(1) in
terms of b(1), ..., b(n) (but independent of a(1), ..., a(n).)

	f(1) = product(b(i)) / n!, where the product is taken from 1 to n.

I can't resist trying yet another one.

    Instead of summing the angles we can consider instead the problem
    of rotating in the complex plane and recover the resulting angle.

    arccot(k^2+k+1) = the included angle of the complex number (k^2+k+1) + j.

    Form the product PROD (k>=0) [(k^2+k+1) + j] = (1+j)*(3+j)*(7+j)*...

    The first few partial products z[k] = x[k] + j y[k] are

	z[0] = 1 + j
	z[1] = 2 + j 4
	z[2] = 10 + j 30
	z[3] = 100 + j 400

    Now assume that in the k'th product y[k-1] = k*x[k+1]
    and write out the recurrance

	x[k] = x[k-1]*(k^2+k+1) - y[k-1] = x[k-1]*(k^2+k+1) - k*x[k-1]
	y[k] = y[k-1]*(k^2+k+1) + x[k-1] = k*x[k-1]*(k^2+k+1) + x[k-1]

	x[k] = x[k-1]*(k^2+1)
	y[k] = x[k-1]*(k^3+k^2+k+1)

    But k^3+k^2+k+1 = (k+1)*(k^2+1) and so y[k] = (k+1)*x[k], and the
    included angle must therefore approach pi/2 as k approaches oo.

    - Jim

>	f(1) = product(b(i)) / n!, where the product is taken from 1 to n.

Yeah, I suspected that as the answer by plugging in

		f(x) = 1 + x + x^2 + ... + x^n

which makes the a(i) and b(i) terms easier to figure.

One approach that looks real promising, but has me pretty well bogged down
in the details, is to make use of the following properties of polynomials
with repeated roots, which can be derived by expressing the polynomial in
product-of-roots form and differentiating it repeatedly:

1)	If a polynomial P has an n'th order repeated root r, then

			     2			n-1
		dP       ,  d P       ,  ...   d   P
	P(r),	---(r)	    ---(r)	       -----(r) are all zero
		dx	      2			 n-1
			    dx		       dx

2)	for that same polynomial P, the value of the reduced polynomial
	Q = P / (r-x)^n at r is given by

	    n	n
	(-1)   d P
	---- * ---(r)
		 n
	 n!    dx

In this case, property (1) with r=1 gives a set of n simultaneous equations
in the a(i) and b(i), which can be used to express the a(i) in terms of the
b(i), and property (2) gives the value of f(1).

The term prod(b(i)) showed up real early when I tried to solve the set of
equations, but the complexity quickly became more than I could afford.....

There must be an easier way!
							Richie