# On limits

1. Apr 3, 2004

### Organic

This is the reason why I used the word "General".

Please show us how the limit concept is rigorous.

Last edited: Apr 3, 2004
2. Apr 3, 2004

### matt grime

Oh heck, not again. The limit concept is a definition, and therefore by, um, definition it is rigorous. Do you need to see the definition of limits?

3. Apr 3, 2004

### Organic

Yes Matt,

Please write a formal definition of it, and then please explain it in plain English.

Thank you.

Last edited: Apr 3, 2004
4. Apr 3, 2004

### matt grime

Ok, for real numbers we say that a sequence x_n indexed by the natural numbers tends to a limit x if for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

A sequence that does not converge is said to diverge. In particular, if x_n is a sequence such that for every X in R x_n>X for all n sufficiently large then x_n is said to diverge to infinity. Note this does not tell you waht infinity is. It tells you what the phrase tends to infinity means.

This idea that infinity is a quantity larger than any real numbers is a short hand and unmathematical way of stating that fact about divergent series. The more rigorous way of saying 1/0 is infinity is to say that if x_n is any sequnce of (positive) real numbers converging to zero then 1/x_n diverges to infinity.

OK? Infinity isn't there, that is the easiest way of thinking about it, it isn't part of the real numbers.

5. Apr 3, 2004

### Organic

Please tell me if this is a correct picture of this definition:

Code (Text):

|                 |
- e               - e
|                 |
- |x_m-x|  <--->  - m<n --> |x_n-x|
|                 |           |
|                 - <---------'
|                 |
- 0               - 0

Last edited: Apr 4, 2004
6. Apr 3, 2004

### matt grime

Perhaps if by that you mean that if you plotted all the points |x_n-x| on the real axis, then all of the ones you plot after the m'th lie in the interval [0,e]

7. Apr 3, 2004

### Organic

Yes, |x_n-x| is for (what you call) all points and also |x_m-x| and e.

Code (Text):

|                 |
- e               - e
|                 |
- |x_m-x|  <--->  - m<n --> |x_n-x|
|                 |           |
|                 - <---------'
|                 |
- 0               - 0

What a show is the invariant state that stands in the basis of the rigorous definition, or more to the point this invariant picture is the reason that we call it rigorous.

Do you agree?

Last edited: Apr 4, 2004
8. Apr 3, 2004

### matt grime

I cannot agree because I cannot decide what invariant state means. Seeing as this needs to be true for every e>0 and each time the m is different I see nothing very invariant. Moreover if you were to offer that as the definition of convergence of a sequence you'd be once more not providing enough information.

Last edited: Apr 3, 2004
9. Apr 3, 2004

### Organic

This is the invariant state that for any given n there is m<n in N.

Right?

10. Apr 3, 2004

### ShawnD

Organic, the difference between a limit and normal numbers is that a limit is what the number gets close to.
the limit of x -->0 for 1/x is infinity. the value of 1/0 is undefined.

Thanks for answering the question guys.

11. Apr 3, 2004

### chroot

Staff Emeritus
The limit of 1/x with x from the right is infinity; from the left it's negative infinity.

- Warren

12. Apr 3, 2004

### Organic

Hi ShawnD,
|x_n-x| is for (what you call) all points and also |x_m-x| and e.
Code (Text):

|                 |
- e               - e
|                 |
- |x_m-x|  <--->  - m<n --> |x_n-x|
|                 |           |
|                 - <---------'
|                 |
- 0               - 0 is the limit x in this case

Last edited: Apr 4, 2004
13. Apr 4, 2004

### matt grime

No, get your quantifiers in the right order: for any e>0, there is an m in N (dependent on e) such that for all n>m....

It is not the trivial true or false assertion (dependent on n being 1 or not) that for any given n there is m<n at all. That deosn't even mention x_i, x or e, does it?

14. Apr 4, 2004

### Organic

Last edited: Apr 4, 2004
15. Apr 4, 2004

### matt grime

I don't understand what you want. If you mean: in order to make Newton-Raphson's iterative method rigorous in the sense that we wish to prove the iterated sequence converges must we use epsilons? Yes, and no. There are other equivalent definitions of convergence using nets and ultrafilters and topologies, but here I imagine one could prove that given certain constraints the algorithm converges in the epsilon definition. Generally N-R converges very rapidly as can generally be indicated graphically. One may prove for instance the that x^2-p yields a recurrence formual x_n = (p + x_(n-1)2)2/x_(n-1) or something which gives a monotone decreasing sequence bounded below which thus convrgese and to sqrt(p) without using an epsilon at all.

16. Apr 4, 2004

### Organic

So, can we define a rigorous definition to a limit of a sequence without using an epsilon?

If yes, can you write a rigorous definition, which is based on N-R?

It is a non-linear sequence of x1, x2, x3, ... that converges very slowly to the limit and run away from the limit (diverges to infinity) very rapidly, as we can see here: http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html

But in both directions no sequence (geometrical or arithmetical) can reach the limit, isn't it?

Last edited: Apr 4, 2004
17. Apr 4, 2004

### matt grime

Look, that is one particular example of N-R, you cannot decide how rapidly it converges in general from one example, and I'm not talking temporally to do with the animation anyway.

The rest of what you write is wrong: it is perfectly possible for a sequnce in N-R to attain the limit in a finite number of steps, nor does one reach infinity, in any sense because infinity isn't there, we cannot emphasize that enough to you.

I can see no way of using Newton-Raphson to define the notion of a limit of a sequence, or what it means for a sequence to converge.

18. Apr 4, 2004

### Organic

No, it is possible only if a curve become a straight line before the limit point, or changes its direction before or in the limit point.

I am am talking about a curve that does not become a sraight line or changes its direction before or in the limit point.

Therefore if we reach the limit then we make a phase transition that leave behind (ignore) N-R method.

If you don't think so, please prove That by N-R we are not approaching but reaching the limit in finitely many steps, where a curve does not become a sraight line or changes its direction before or in the limit point.

Last edited: Apr 4, 2004
19. Apr 4, 2004

### matt grime

A straight line is a curve, it has curvature zero at all points. Why is N-R not applicable to straight lines anyway? Moreover your assertion is not true, there are other curves that will reach their root in a finite number of steps (if your initial term is the root for instance), others may even cycle with period. The second sentence makes no sense to me. I don't know exactly what you're asking me to prove in the third step, but I think the trivial counter example you need is to take the initial term to be the root. That converges after 1 step, as well as the other trivial disproof that a straight line is a curve. You have shown a lack of belief of proofs in the past anyway, so what would it do for me to prove a result for you anyway? I've given you an example where it terminates in a finite number of steps, you've given an example of this, apparently, I've not checked the details, which you've dismissed for no good reason (a straight line is a curve, just not a very curvy one). Perhaps there are no other examples but I imagine you might be able to construct something that isn't technically a straight line (adjoin a non-straight line to it).

Ah you've now edited your post to remove the counter examples that you have NOW decided are not allowed for some reason, even though you didn't say they weren't allowed before! Naughty, naughty.

20. Apr 4, 2004

### Janitor

Does this website own the copyright to the ongoing Grime-Organic debate? I'm smelling \$ in the air on this one! How about publishing it in print?

21. Apr 4, 2004

### matt grime

I want to stop, really I do, but I just can't leave ignorance alone. It's pointless, but some of us somehow appear to take it in turns to duke it out. Hurkyl did for a while and now he's stopped. This annoys me though because it is hijcaking another persons thread *yet* *again*. Hopefully it'll be locked again.

22. Apr 4, 2004

### Hurkyl

Staff Emeritus
How about: The sequence <x> converges to the limit L iff:

For every &rho; > 0 there exists a natural number N such that for all m > N:
|L - xm| < &rho;

There are certainly other ways to go about defining limits; they're all equivalent to this one, though, when applied to sequences of real numbers. A common one is to use the topological notions of neighborhoods or nearness.

23. Apr 4, 2004

### Organic

Hi Hurkyl,
When these words are used, do we can understand that no element in the sequence reaching the limit?

24. Apr 4, 2004

### Organic

Matt,
In this case ( http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/newtons.html )

I am talking about a curve that has non-zero curvature at all points, and this non-zero curvature does not changes its direction or become zero curvature at or before the limit point.

Do you think that in this case we get infenitely many x_n values where abs(x_n-x)>0?

Last edited: Apr 4, 2004
25. Apr 4, 2004

### matt grime

It is perfectly possible in some convergent sequence x_n tending to x, for none, finitely many, infinitely many but not all, or all of the x_n to equal x. Why is this a problem?

As for N-R. What do you mean by 'change direction of curvature'?

At this rate you're going to say: ah but it's true except for all the counter examples.

Here's one for you

Consider a curve given by x^(1/2) for x positive, -(-x)^1/2 for x negative. Call this f(x). Pick some point x in (0,1) draw the tangent line in. Suppose just for the sake of the argument that I can tell exactly where that tangent line cuts curve again, call that point t. Then the curve g(x)= f(x)-t has a non-trivial starting point for a N-R sequence that converges in a finite number of steps. Does that have 'curvature that changes direction' though? Perhaps you mean sign?

It seems to me that you're just going to keep inventing restrictions each time to make any counter examples false. Why do you do that? Are those all the criteria you need? Are you sure? Not going to change your mind again?