Please show us how the limit concept is rigorous

[Organic]

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

Please show us how the limit concept is rigorous.

 

[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?

 

[Organic]

Yes Matt,

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

Thank you.

 

[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.

 

[Organic]

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.

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

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

 

[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]

 

[Organic]

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]

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

    |                 |
    - 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?

 

[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.

 

[Organic]

...for any e>0 there is an m in N such that |x_n-x| < e for all n>m.

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

Right?

 

[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.

 

[chroot]

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

- Warren

 

[Organic]

Hi ShawnD,

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.

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

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

 

[matt grime]

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

Right?

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?

 

[Organic]

Matt,

Please look at this:

http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/Newtons.html

If you translate Newton's method to the limit of a sequence, do you get delta-epsilon definition?

 

[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.

 

[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?

Generally N-R converges very rapidly as can generally be indicated graphically

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?

 

[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.

 

[Organic]

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

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.

 

[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.

 

[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?

 

[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.

 

[Hurkyl]

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

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 - x_m| < &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.

 

[Organic]

Hi Hurkyl,

...neighborhoods or nearness.

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

 

[Organic]

Matt,

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

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?

 

[matt grime]

Hi Hurkyl,

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

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?

 

[Hurkyl]

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

No...

The normal topological definition says that L is the limit of <x> iff:

For every neighborhood U of L, there is a natural number N such that if m > N, then x_m is in U.


Anyways, Organic, have you considered the sequence that is always 1? Shouldn't it's limit be 1?

 

[Organic]

What do you mean by 'change direction of curvature'?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

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?

 

[matt grime]

Then my counter example possibly doesn't cut it. However as the way to prove you claim has been merely to elminate all counter examples offered, why don't you offer a proof with all your many hypotheses, to demonstrate that the N-R method is only terminal for certain kinds of functions.

What that has to do with the defintion of limit is beyond a bear of little brain like me.

 

[Organic]

Let me ask this:

Is the case of abs(x_n-x)=0 (which means that x_n=x) exists, according to what I wrote?

 

[matt grime]

What you wrote when? If you'll excuse the ungrammatical paraphrasing.

Here's a reason why you won't get s generalized notion of seqences converging from N-R for a very good reason, and bearing in mind we're having to guess what it is that you mean, AGAIN.

The seqeunce x_n =1 n even, x_n=1+1/n n odd cannot arise from any Newton Raphson method given that if the iteration sends a to b, it always sends a to b, so there's no way 1 can go to 1+1/3 and 1+1/5 and 1+1/7 and...

 

[Organic]

What do you mean by 'change direction of curvature'?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

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.

Let me ask this:

Is the case of abs(x_n-x)=0 (which means that x_n=x) exists, according to the curve that was described by me?

 

[matt grime]

I don't know. It's your conjecture, why don't you try to prove it in the light of all your artificial hypotheses which are solely to exclude counter examples that you didn't exclude originally?

If I may rephrase your question so that it makes sense to others who haven't read your stuff before - is it possible that there is some curve satisfying various criteria which is differentiable, and such that the Newton Raphson method does not converge to the exact root in a finite number of steps. The criteria are that the curve is nowhere linear, and that "the tangent line always stays on the same side of the curve". Note, that although he hasn't excluded it, one must presume that Organic also means that we do not choose the root as the initial input.

I don't know the answer, nor do I care really - my geometric intuition tells me it is probably true as one isn't allowed to do something bizarre like 1-sided derivatives.

Does that answer your query?

 

[Organic]

Matt,

What is "probably true"?

By your point of view x_n=x?

Please answer by yes or no.

Thank you.

 

[matt grime]

My view is that the conditions you've decided to impose on the type of curve probably imply that any non root initail choice for a N-R iteration will not reach the root in a finite number of steps. Probably means I can see how that proof would go (MVT or Rolle's theorem plus a little thinking).. I cannot answer yes or no because I have not proved it and have no proof in mind, only a possible proof. But it isn't my conjecture, it is yours, and I have no real desire to make it rigorous, try proving it yourself.

 

[Zurtex]

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.

What?

You just said it was a curve with non-zero curvature at all points... therefore it is always changing its direction. How can anyone take you seriously when you make such obvious contradictions in your own argument?

I mean that the tangent line stays in one and only ony side of the curve, when N-R is used.

I'll write it again:

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.

Well why don't you actually say what you mean rather than just writing nonsense. Half the problems with your posts is you say one thing and mean something else all together, another problem is you don't actually seem to be proving anything...