Conference rusure::math

My first guess is a set of fifty twos.

I looked at the following:

 50   25   20    10              50   25   20    10
2  , 4  , 5  , 10  .  Turns out 2  = 4  > 5  > 10  >...

            2    3   32    48                             33
But, since 3  > 2 , 3   > 2  .  So next "guess" would be 3   * 1.

But that "1" doesn't increase the product any, so lets get rid of it.

The expression is ..*3*3*3*1, if we change the last two factors to 2*2, we
get a larger product since 2*2 > 3*1  So my guess is
 32            32
3  * 2 * 2 or 3  * 4.

-Mike

    Re .1, .2:
    
    Sets don't contain multiple copies of anything.
    
    
    				-- edp

I guess {2,3,5,6,7,8,9,10,11,12,13,14}.

    As the submitter of .0, I intended multiples to be allowed.
    
    So, either we replace the word "set of integers" with something
    else, or we announce that {3,3,3...} doesn't mean multiple COPIES
    of 3, but distinct 3's.  Hence we can continue to use the word "set".
    
    Anyway, the question about a set with non-duplicates is an interesting
    one too.
    
    By the way, all 3's except for two 2's is what I believe the answer
    to the original question is.  The "reason" the answer only includes
    2's and 3's is related to the fact that the number "e" (natural
    logarithm base) falls between 2 and 3.
    
    The dependence on "e" can be more clearly seen if you attempt
    to find a collection of positive REALS that add to 100 and multiply
    to a maximum.
    
    /Eric

A multiset is analogous to a set, but elements may appear more than once.
Duplicate elements are indicated with count�element.  For example, the
multiset {1,1,1,1,2,3,3} can be written {4�1,2,2�3}.

Lemma 1:  A multiset with sum = N that maximizes the product of elements
	  contains no elements > 4.

Proof:    Assume m > 4 is an element of a multiset that maximizes the
	  product.  But m can be replaced by the two elements 2 and m-2,
	  yielding the same sum, and a larger product.  Thus, m cannot
	  be an element of the multiset, and the lemma is proved.

Similarly, we can easily show that any multiset that maximizes the product
may contain at most one 3 (since 2*4 > 3*3), need contain no 4s (since 4 = 2*2),
and that, if the sum > 1, there are no 1s (either 2 > 1*1 or 2*(m-1) > 1*m).

Thus, a maximal multiset is either {1} (if N = 1), or several 2s and possibly
a 3.  Finally, we note that the 3 must be present iff N is odd.  Altogether,

For N = 1, the multiset is {1},
for N even, the multiset is {(N/2)�2},
and for N odd and > 1, the multiset is {((N-1)/2)�2,3}.

Other maximal multisets result from substituting 4 for pairs of 2s.

    Re .6:
    
    > Similarly, we can easily show that any multiset that maximizes the product
    > may contain at most one 3 (since 2*4 > 3*3), need contain no 4s (since 4 = 2*2),

    2*4 > 3*3 ?
    
    	-Neil

Oh dear.  Let's pick up 513.6 where it started to go afoul.

Similarly, we can easily show that any multiset that maximizes the product
may contain at most two 2s (since 3*3 > 2*2*2), need contain no 4s (since
4 = 2*2), and that, if the sum > 1, there are no 1s (either 2 > 1*1 or
2*(m-1) > 1*m).

Thus, a maximal multiset is either {1} (if N = 1), or several 3s and two,
one or no 2s.  Finally, we note that the number of 2s can be determined
as a function of N mod 3.  Altogether,

For N = 1, the multiset is {1},
otherwise, a maximal multiset is
	{y�3},                with y = N/3, if N mod 3 = 0,
	{2,2,y�3} or {y�3,4}, with y = (N-4)/3, if N mod 3 = 1, or
	{2,y�3},              with y = (N-2)/3, if N mod 3 = 2.

Or simply, {x�2,y�3}, where x = 2 - (N+2) mod 3, and y = (N-2x)/3.

re 513.5 (same problem, but with a set of positive reals)

	Let us find a set of real positive numbers, with sum S,
	for which the product is maximized.  Now, all the numbers
	must be equal; otherwise, we could take two different numbers
	a-b and a+b and replace them by a and a, giving the same sum,
	but a larger product, since a*a > (a-b)(a+b) = a^2 - b^2.

	Given S, the only freedom we have in maximizing the product
	is the choice of N, the number of numbers.  Each number is S/N
	(since N of them sum to S), and the product is (S/N)^N.  We
	note that the continuous function f(N) = (S/N)^N is concave
	downward, with a maximum at N = S/e.  To maximize f(N) with an
	integer N, we have either N = floor(S/e) or N = ceiling(S/e).

	Does anyone have an idea of how to determine, in general, whether
	the floor or ceiling should be used? (yeah, I know, try 'em both).
	For example, will S/e rounded to the nearest integer always maximize
	the product for integer S?

    Well, you were so quick at comparing e**pi and pi**e (q.v.) without
    "trying them both", so I'd think you could just as effectively
    compare ceiling and floor of (S/e) ! :-)
    
    Anyway, perhaps a small amount of analysis of the derivative will
    reveal which is better, floor or ceiling.  Perhaps we might
    even cheat a little to help the analysis by using the fact that
    e is closer to 3 than 2.  Would this help ?
    
    /Eric

Sometimes S/e should be rounded down, sometimes up.  Let's find values of S
that are 'borderline': rounding either way is just as good.  These S values
have:
	 /   S   \ [S/e]      /    S    \ [S/e]+1
	(  -----  )       =  (  -------  )
	 \ [S/e] /            \ [S/e]+1 /

where [x] is the floor function.  Rearranging the above, we have:

	                      1    [S/e]
	S = ([S/e]+1) ( 1 + ----- )
	                    [S/e]

Now as S (and [S/e]) increases, the expression (1 + 1/[S/e])^[S/e] approaches
e from below.  This makes the above equation for S rather interesting:

	S = ([S/e]+1) (e-delta)

Substituting this into itself gives:

	S = ([ ([S/e]+1)(e-delta)/e ]+1) (e-delta)
	  = ([ ([S/e]+1)(1-delta/e) ]+1) (e-delta)
	  = (   [S/e]                +1) (e-delta)