[seqfan] Re: Help with proof strategy needed.

[L. Edson Jeffery]

Hello Brad,

I use your notation in the following and L(x) for log(x).

We are given the conditions 1 < A1 < A2 and 1 < B1 < B2 on adjacent entries
in a row of arrays A and B. From properties of the logarithmic function and
in particular the series

log(x) = 2*sum_{n=1..infinity} (1/(2*n-1))*((x-1)/(x+1))^(2*n-1)  (x>0),

it is easy to see that if A1 > B1 and A2 > B2, then

(1)  L(A2)/L(A1) - L(B2)/L(B1) < 0.

I had already proved that. Now we have to prove that (1) implies

(2)  L(B1)/L(A1) - L(B2)/L(A2) < 0.

But

L(A2)/L(A1) - L(B2)/L(B1) = (L(B1)/L(B1))*(L(A2)/L(A1) - L(B2)/L(B1))

                          = (1/L(B1))*(L(B1)*L(A2)/L(A1) - L(B2))

                          = (L(A2)/(L(B1)*L(A2)))*(L(B1)*L(A2)/L(A1) -
L(B2))

                          = (L(A2)/L(B1))*(L(B1)/L(A1) - L(B2)/L(A2))

from which it follows that L(B1)/L(A1) - L(B2)/L(A2) < 0, since all array
entries are positive. Similarly, it can be shown that if A1 < B1 and A2 <
B2, then the inequalities (1) and (2) are simply reversed. QED

Thanks for your help,


Ed Jeffery

On Thu, Jul 16, 2015 at 10:01 AM, Brad Klee <bradklee at gmail.com> wrote:

> 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
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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