Re(c^n - 1)
quasi
be welcome.
Question:
Let c in C with |c| > 1, and for n in N, let x_n = |Re(c^n - 1)|. Must
x_n approach infinity as n approaches infinity?
quasi
Butch Malahide
>
> Let c in C with |c| > 1, and for n in N, let x_n = |Re(c^n - 1)|. Must
> x_n approach infinity as n approaches infinity?
If c = 2i, isn't x_n = 1 for odd n?
quasi
Of course! (I knew it was easy).
Thanks.
There's another question lurking somewhere, but I'll need to think
about how to phrase it.
quasi
quasi
Here's a quick attempt at a revision ...
Let c in C with |c| > 1, and for n in N, let x_n = |Re(c^n - 1)|. If
c^n is non-real for all n in N, must x_n approach infinity as n
approaches infinity?
quasi
Sorry, let me revise the wording. I think the following restriction is
more natural ...
Let c in C with |c| > 1, and for n in N, let x_n = |Re(c^n - 1)|. If
Re(c^n) is nonzero for all n in N, must x_n approach infinity as n
approaches infinity?
quasi
achille
False, let (s_k), k=1..oo be the sequence:
s_1 = 1
s_k = 2^(s_(k-1)) for k > 1.
and let s be the Liouville number defined by
s = sum_{k=1..oo} 2^{-s_k},
then for c = sqrt(2) exp( i s pi/2 ), we have:
| Re(c^{2^{s_k}}) |
~= 2^{2^{s_k}/2} (pi/2) 2^{s_k - s_(k+1)}
= (pi/2) 2^{s_k - s_(k+1)/2}
As this converges to 0 as k -> oo. x_n has a
sub sequence converges to 1.
BTW, I think the statement is true for almost
all arg(c) because the set of Liouville's number
has measure zero.
quasi
Nice counterexample.
>BTW, I think the statement is true for almost
>all arg(c) because the set of Liouville's number
>has measure zero.
So then, if c is further restricted to be an algebraic number, would
the specified limit then have to approach infinity?
quasi
achille
Not sure, I bet it will but I don't have a proof.
I only know if s is an algebraic irrational, then
for c of the form |c| exp( i pi s ), the
corresponding x_n will approach infinity.
quasi
>> >False, let (s_k), k=1..oo be the sequence:
>> >
>> > s_1 = 1
>> > s_k = 2^(s_(k-1)) for k > 1.
>> >
>> >and let s be the Liouville number defined by
>> >
>> > s = sum_{k=1..oo} 2^{-s_k},
>> >
>> >then for c = sqrt(2) exp( i s pi/2 ), we have:
>> >
>> > | Re(c^{2^{s_k}}) |
>> >~= 2^{2^{s_k}/2} (pi/2) 2^{s_k - s_(k+1)}
>> > = (pi/2) 2^{s_k - s_(k+1)/2}
>> >
>> >As this converges to 0 as k -> oo. x_n has a
>> >sub sequence converges to 1.
>>
>> Nice counterexample.
>>
>> >BTW, I think the statement is true for almost
>> >all arg(c) because the set of Liouville's number
>> >has measure zero.
>>
>> So then, if c is further restricted to be an algebraic number, would
>> the specified limit then have to approach infinity?
>
>Not sure, I bet it will but I don't have a proof.
>I only know if s is an algebraic irrational, then
>for c of the form |c| exp( i pi s ), the
>corresponding x_n will approach infinity.
Well, if you would bet on it, then I'll bet with you.
I'll revise the statement slightly, removing the -1 term (there was
never any need for it, but that's how it came up as part of the
problem I was working on). So here's the (tentative) claim:
Conjecture:
If c is an algebraic number with |c| > 1 such that Re(c^n) is nonzero
for all n in N, then |Re(c^n)| approaches infinity as n approaches
infinity.
quasi
achille
> On Fri, 24 Sep 2010 01:46:35 -0700 (PDT), achille
>
>
>
I think the conjecture is true. Here is my
argument and you need to verify it:
Since the statement is trivially true when c
is real, we will assume Im(c) != 0 from now on:
Let A be the set of algebraic numbers. For any
a in A, let H(a) be its naive height. Since c
is in A\R we can split it as
c = |c| a with |c| in A and a in A\R, |a| = 1.
Now Re(c^k) != 0 for k > 0 => a^k != i or -i for k > 0.
This implies for all b_1, b_2 in Z\{0}, the linear
form of logarithm:
L(b_1,b_2) = b_1 log(a) + b_2 log(i) != 0.
In 1993, Baker and Wustholz has proved following theorem:
For any a_1, ... a_n in A, let L : Z^n -> C be
the linear form of logarithm
L(b_1,..) = b_1 log(a_1) + ... + b_n log(a_n)
L is either 0 or bounded away from 0 by
|L| > exp( -(16nd)^(2n+4) log(A_1)..log(A_n) log(B) )
where d = [Q(a_1,..a_n):Q],
A_i = max( H(a_i), e ),
B = max( b_1, .. b_n, e ).
If we apply this theorem to our linear form, we get
| b_1 log(a) + b_2 log(i) | > B^{-mu} for some const mu.
This implies for large n,
| a^n +/- i | > ( const . n)^{-mu}
=> | Re( c^n ) | > ( const . n)^{-mu} |c|^n
=> | Re( c^n ) | -> oo as n -> oo.
achille
Oops, forget to quote the original reference which
I don't have access. You need to check whether there
are other limitations in using the theorem there:
REF: A. Baker and G. Wustholz, "Logarithmic forms and group
varieties",
J. Reine Angew. Math. 442 (1993) 19-62
quasi
I'll try to verify the details, but my first impression is that you
nailed it. Nice going!
quasi
Axel Vogt
> REF: A. Baker and G. Wustholz, "Logarithmic forms and group
> varieties",
> J. Reine Angew. Math. 442 (1993) 19-62
http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN002210908
quasi
Hmmm ...
Actually, for this problem, after some reflection, I think the
Baker-Wustholz results are massive overkill. This is just an innocent
little problem -- I'm almost certain that a much more elementary proof
is within reach. I can already see two fairly promising lines of
attack, but I thought I'd reopen the problem for others as well.
quasi