[seqfan] Re: Help with proof strategy needed.

Brad Klee bradklee at gmail.com
Thu Jul 16 17:01:39 CEST 2015


Hi Ed,

Can't really see all the details on the proof, but maybe I can help you
with the log expression.

Instead of your form, I take:

x = log(B1) log(A2) - log(A1) log(B2) .

Using vector notation with "*" the inner product,

x = { log( A2 ) , log( A1 )  } * { log( B1 ) , - log( B2 )  } .

Inserting identity,

2 x = { log( A2 ) , log( A1 )  } * T^{-1} * T * { log( B1 ) , - log( B2 )
} .

with,

T = {
{ 1 , 1 },
{ 1 , -1 }
} .

Then,

2 x = { a , b } * { -c, d } = - a c + b d,

with,

a = log ( A2 A1 ) ,
b = log ( A2 / A1 ) ,
c = log( B2 / B1 ) ,
d = log ( B2 B1 ) .

Assuming Ai, Bi in Naturals and A2 > A1, B2 > B1 all quantities { a, b, c,
d } are positive. Assuming Bi > Ai imples d > a. The same can be said if
you exchange B1 & A2.


Hope this helps.

Thanks,

Brad

On Tue, Jul 14, 2015 at 2:27 PM, L. Edson Jeffery <lejeffery2 at gmail.com>
wrote:

> I am trying to prove a conjecture related to the 3x+1 problem which arises
> from the definition of a new sequence. I want to include the proof in a PDF
> in the links section on the sequence page. I know of only one way to
> structure the proof but have encountered a technical problem which I do not
> know how to solve. I would of course like to complete the proof myself, so
> in the following I generalize the problem and ask for your help with how to
> proceed (questions are at the end of this note).
>
> Let A and B be rectangular arrays (of natural numbers) with entries in row
> n and column k denoted by A(n,k) and B(n,k), respectively, n,k >= 1. Let L
> be the array with entries
>
> L(n,k) = log(B(n,k))/log(A(n,k)).
>
> We are given the following properties of the arrays.
>
> 1. A(n,k) and B(n,k) are defined as functions of n and k.
>
> 2. The rows {A(n,k)} and {B(n,k)}, k=1,2,..., are strictly monotonically
> increasing sequences, and each is an arithmetic progression.
>
> 3. The column bisections {A(2*n-1,k)}, {A(2*n,k)}, {B(2*n-1,k)} and
> {B(2*n,k)}, n=1,2,..., are strictly monotonically increasing sequences, and
> each sequence of first differences is a geometric progression.
>
> 4. There exists an r such that B(n,k) < A(n,k), for all n <= r, and such
> that B(n,k) > A(n,k), for all n > r.
>
> 5. For the rows of L, we have the repeated limits
>
>    lim_{n->infinity} lim_{k->infinity} L(n,k) = 1.
>
> 6. For c a known constant, for the column bisections of L we have the
> repeated limits
>
> (i) lim_{k->infinity} lim_{n->infinity} L(2*n-1,k)) = c;
>
> (ii) lim_{k->infinity} lim_{n->infinity} L(2*n,k) = c.
>
> I also proved for the main diagonal of L that
>
> 7. lim_{n->infinity} L(n,n)) = c.
>
> This leads to the following
>
> Conjecture. L(n,k) < c, for each pair n,k.
>
> I tried structuring the proof as follows.
>
> For the r of property 4, for each n <= r, L(n,k) < 1 and the conjecture is
> true. Suppose that n > r, so L(n,k) > 1. We must first prove:
>
> (I) For each n > r, row n of L is a strictly monotonically decreasing
> sequence.
>
> If (I) is true, then it follows that
>
> (II) For each n > r, row n of L contains a maximal element, namely L(n,1).
>
> We must then prove:
>
> (III) For column 1 of L, each of the bisections {L(2*n-1,1)} and {L(2*n,1)}
> is a strictly monotonically increasing sequence.
>
> If (III) is true, then it follows that
>
> (IV) L(1,1) is the least element of sequence {L(2*n-1,1)}, and L(2,1) is
> the least element of sequence {L(2*n,1)}.
>
> To complete the proof, it remains to show that
>
> (V) L(1,1) < c, and L(2,1) < c.
>
> But (V) is easily verified numerically, so the result should follow.
>
>
> Now, to prove (I) it must be shown that L(n,k) - L(n,k+1) > 0, for each k.
> However, the expression
>
> (1)  log(B(n,k))/log(A(n,k)) - log(B(n,k+1))/log(A(n,k+1))
>
> seems to defy algebraic evaluation, as do the expressions L(n,1) -
> L(n+2,1). (It seems that properties 2 and 3 should be useful here.) Could
> someone please tell me how to evaluate the expression (1)? Or is there some
> other approach to the proof that avoids this problem altogether?
>
> Thank you for any help,
>
> Ed Jeffery
>
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>
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>



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