N(e) tells you how far into the sequence you must go so that all the terms after the returned value of N(e) are at most e away from the limit of the sequence. That means that N(e) also depends somewhat on a, the limit of the sequence.
For any arbitrary epsilon, can N(e) return only a singular value?
No. N(e) can return a range of values. Let’s restrict N(e) to returning the _least_ of the values it could return.
So let’s now redefine N(e):
Exactly how far you must go into the sequence so that every value after N(e) is at most e away from the limit of the sequence.
Now if you plug in any epsilon, you’ll return the first term that satisfies |(term-limit)| <epsilon. This is a problem because the term could be below or above the limit and this does not tell us which one it is. To remedy this, we would simply first filter functions into two categories, monotone increasing and monotone decreasing. For functions that are not either, I am not sure how we would handle.
We can eliminate the absolute value to give term-limit <epsilon. This now eliminates us being unsure as to which side of the limit the value of the sequence returned by N(e) is on.
sequence = a_n
limit = a
First, let N(e) be equal to the index of the last term of the sequence a_n such that all terms in the sequence after this term are less than a+e.
As we have a strictly less than, when N(e) changes value, we know that N(e) must be equal to a term in the sequence, specifically, the N(e)th term. We even know the value of this term; it’s a+e. Therefore, we can construct the sequence a_n from its N(e) and limit using the following provided that it is monotonic decreasing (the monotonic increasing case should follow simply):
When N(e) changes value, let a_N(e) = a+e.
The question is, how do we determine when N(e) changes value over the domain of epsilon.
This means that if you have a change of epsilon and your N(e) changes then your past N(e)