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Science Forum Index » Mathematics Forum » Periodic sequences--is this obvious?
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| Kevin Searle |
 Posted: Wed Dec 17, 2003 7:24 pm |
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I have been told the following fact by someone who didn't know
where they knew it from:
Every periodic sequence can be expressed as the sum of a
constant sequence and an irreducible sequence;
where an "irreducible sequence" is one such that if you take
successive differences iteratively, you eventually end up with
the sequence you started with;
a constant sequence is one whose terms are all equal;
and the sum of two sequences is the sequence whose nth term
is the sum of the nth terms of the two summand sequences.
In other words, and stated even less rigorously than the
above: every periodic sequence is offset by some constant
amount from an irreducible sequence.
Does this seem obvious? Does it even seem true? Is it a result
that you've seen published anywhere? I find it surprising but
my intuition is not to be trusted.
Thanks --Kev |
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| Larry Hammick |
| Posted: Thu Dec 18, 2003 12:41 am |
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"Kevin Searle"
Quote: I have been told the following fact by someone who didn't know
where they knew it from:
Every periodic sequence can be expressed as the sum of a
constant sequence and an irreducible sequence;
where an "irreducible sequence" is one such that if you take
successive differences iteratively, you eventually end up with
the sequence you started with;
a constant sequence is one whose terms are all equal;
and the sum of two sequences is the sequence whose nth term
is the sum of the nth terms of the two summand sequences.
In other words, and stated even less rigorously than the
above: every periodic sequence is offset by some constant
amount from an irreducible sequence.
Something's wrong with the statement, or maybe I misunderstand it.
If a periodic sequence is irreducible, then the sum of the terms in any
period must be zero (because that is true of any derived sequence). So, if
this periodic sequence:
2 3 -5 2 3 -5 ...
(period=3) is the sum of a constant and an irreducible, then the constant is
zero. But here's what happens when we take the first few "derivatives":
2 3 -5 ...
7 1 -8 ...
15 -6 -9 ...
etc.
Hereafter, all the terms are multiples of 3, and therefore we never get back
the original sequence.
LH |
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| Kevin Searle |
| Posted: Fri Dec 19, 2003 9:35 am |
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Larry Hammick <larryhammick@telus.net> wrote:
Quote: "Kevin Searle"
I have been told the following fact by someone who didn't know
where they knew it from:
Every periodic sequence can be expressed as the sum of a
constant sequence and an irreducible sequence;
where an "irreducible sequence" is one such that if you take
successive differences iteratively, you eventually end up with
the sequence you started with;
a constant sequence is one whose terms are all equal;
and the sum of two sequences is the sequence whose nth term
is the sum of the nth terms of the two summand sequences.
In other words, and stated even less rigorously than the
above: every periodic sequence is offset by some constant
amount from an irreducible sequence.
Something's wrong with the statement, or maybe I misunderstand it.
Sorry, something was wrong with the statement.
It should be about periodic sequences of residues mod n. For example,
consider this sequence of period 6:
11 10 11 7 2 7 11 10 11 7 2 7...
first differences: 4 11 1 8 7 5 4 11 1 8 7 5 ...
2nd differences: 11 7 2 7 11 0 11 7 2 7 11 0 ...
3rd differences: 1 8 7 5 4 11 1 8 7 5 4 11 ...
4th differences: 2 7 11 10 11 7 2 7 11 10 11 7...
and we're back where we started--this is an "irreducible" sequence. The
statement should have been that every periodic sequence of residues mod n
can be expressed as the sum (mod n) of a constant sequence and an
irreducible sequence (where differences are taken mod n).
Does this sound any more convincing?
Thanks --Kevin |
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