# Homework Help: Sequences problem

1. Feb 13, 2013

### trollcast

1. The problem statement, all variables and given/known data
For an increasing sequence of numbers, how many other sequences could this be the average sequence of.

2. Relevant equations

Where the average sequence, a = 0.5( s + s[i+1] )

3. The attempt at a solution

If theres n terms in the original sequence.
The number of differences between consecutive terms is (n - 1)
Find all these, (n-1) and find the lowest difference.

eg.
s = 1, 3, 6, 10, 12
s[i+1] - s = 2, 3, 4, 2

The lowest difference here is 2 so theres 2 possible sequences for which s is the average of?

2. Feb 13, 2013

### I like Serena

Hey trollcast!

Suppose we start with the simplest possible sequence we can think of: 1,2,3,4,...

Now we'd look for another sequence that has the same averages.
To get the same average, if we increase a specific number by some amount ε > 0, the next number must be decreased by that same amount ε.
So we'd add an alternating sequence $(-1)^i ε$.
Since the resulting sequence still has to be increasing, that ε must be less than (1/2).

So we'd start with the sequence specified by $s_i = i$ and we'd end up with the sequence specified by $s_i' = i + (-1)^i ε$ with $0 < ε < \frac 1 2$.

How many of those ε's are there?

3. Feb 13, 2013

### trollcast

Would there be 0 if the sequence has to be integers?

4. Feb 13, 2013

### I like Serena

Yes if we're talking about the sequence 1,2,3,...

But for another sequence like 0,10,20,30,40,...
we can find 1,9,21,29,41,...
or 2,8,22,28,42,...
or ...

5. Feb 13, 2013

### haruspex

Maybe I'm misreading the OP. If a = 0,10,20,30,40,.. is the given sequence, I thought we were looking for other sequences s s.t. a = 0.5( s + s[i+1] ). 1,9,21,29,41,... does not appear to be a solution.
If we start with an arbitrary s[0] then s[i+1] = 2a - s would appear to generate the rest uniquely. Perhaps it is also required that s is increasing (or maybe non-decreasing)?

6. Feb 14, 2013

### trollcast

Yeah thats how the example worked, the new sequence must be non decreasing as well.

7. Feb 16, 2013

### trollcast

Actually I looked at this again and for the easiest one, ie. the average sequence 1, 2, 3, 4, ... , theres 2 possible sequences that could have came from:

0, 2, 2, 4, 4, 6, 6, ...

or

1, 1, 3, 3, 5, 5, ...

I've sat and mucked around with various other sequences and found that something like:

1, 2, 8, 9 has no sequences for which it could be the average but I can't figure out a way to work this out without going through and working out the possible values for each average sequence?