Please show us how the limit concept is rigorous

[Organic]

Zurtex,

I gave this example:
http://phys23p.sl.psu.edu/~mrg3/mat..._I/Newtons.html
before I wrote something about it.

This kind of a curve 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.

By "does not change its direction" I mean that the tangent line stays in one and only one side of the curve, when N-R is used.

Shortly speaking, it is not switching sides.

If you can't understand all this by looking at the graphic example then it is your problem not mines.

another problem is you don't actually seem to be proving anything...

I do more than that, I make a paradigm shift in the infinity concept.

Please read this:
http://www.geocities.com/complementarytheory/NewDiagonalView.pdf

 

[matt grime]

Organic, how on Earth can you expect people to know exactly what you mean just by posting one example without explanation. How can you not have realized yet that this is physicsforums.net not psychicforums.net. The confusion is entirely your causing by not explaining what you want. From one picture we are supposed to understand that you only want curves with THIS set of properties. Well, that curve in that link is also convex, should the curves only be convex? It only has one root if we carry it on in a naive smooth fashion, it cuts the x-axis at positive x, must the curve always do this? The tangent has positive slope at all points, must this be true as well?

 

[Organic]

MVT or Rolle's theorem plus a little thinking

Examples of it can be found here:http://www.ies.co.jp/math/java/calc/rolhei/rolhei.html

But this is not the case that I show here:
http://phys23p.sl.psu.edu/~mrg3/mathanim/calc_I/Newtons.html

I gave this example, and by this I mean that I am talking only about this example.

If instead you want to speack about another types of curevs, then you are talking to yourself, not to me.

 

[matt grime]

Erm, thanks, Organic, I do know what the MVT and Rolle's theorem state, and, shockingly, I know how to prove them, amazing, someone might think I was a mathematician or something.

WHy didn't you say you only cared about that one example.

 

[Organic]

If I am not mistake we are in a thread that dealing with the limit problem, where there are infinitely many steps that cannot reach the limit point.

I gave this N-R example as something which is another example that is different from the epsilon-deltha method.

Is it not understood?

 

[matt grime]

You are making a mistake. This thread didn't start about that. You hijacked it so that you could talk about your interests again. And I sadly came along too.

Different from? As you've not proven anything to do with convergence that's a little rich.

 

[Organic]

If you look at the previous thread will see the you started this:

But is still makes no sense. x/0 is not a well-defined symbol in the real number system that one can manipulate like this.

 

[matt grime]

And what's that got to do with Newton-Raphson iteration? We explained what the more formal interpretation of your beloved Wolfram definition of infinity is.

 

[Organic]

x/0 deeply connected to the limit problem.

 

[HallsofIvy]

x/0 deeply connected to the limit problem.

Only in the minds of those who do not understand what a limit is.

"x/0" is "undefined" if x is a non-zero constant and "undetermined" if x is zero.

If you mean that x is a variable, then "x/0" makes no sense at all.

If you mean "lim as a-> 0 of x/a" then you should say that: the whole point of the theory of limits is that "x/0" will tell you nothing about the limit.

 

[Organic]

1/0 is the same as 1*oo and in both cases we are no longer in a finite system.

The whole idea of the interesting point of view of the limit concept is that no infinitely many elements can reach the limit itself.

This unclosed gap which is > 0 cannot be closed by infinitely many elements.

therefore the sum of .999... or the intervals of N-R is undefined by definition.

For example, Cauchy method only forcing the impossible to be possible by "raping" infinitely many ... to have a sum.

 

[matt grime]

You're wrong and veering off topic again with you own personal incorrect view of mathematics. Please stay on topic.

Counter examples: let x_n=0 if n is even 1/n n odd. this sequence converges to zero, adn reaches 0 infinitely often. OR let x_n^{M} be the sequence define to be 1 for n<M 0 other wise - this sequence converges to zero and is zero for all n>M. Demonstrate a non-zero real number between 0.9999.. and 1. Hint: can't be done. The infinite sum os defined. It is the limit of the partial sums. (N-R, or Newton Raphson, has no need to be here). I presume Chausy is Cauchy. I don't think you understand enough of the mathematics to be able to form an opinion about completions wrt norms. So, this is mathematics, in the real numbers in decimal notation 0.9999.. is the same as 1. It has been proven many times. If you're going to tell us we're wrong then please don't do so in this thread. Start another one and attempt to understand the answers that will be given. Don't hijack this one please - I've answered your post and told you where you're wrong conceptually as well as physically. If you don't accept that then you aren't using the mathematics correctly and you aren't adding to this thread's worth. Start one in TD say, but this topic has been done to death and that you cannot accept the PROOF is a reflection on you not the mathematics.

 

[Organic]

Another example:

PI exact plece in the real line is unknown, because in any representation method of it we have to use infinitely many elements to define it.

Demonstrate a non-zero real number between 0.9999.. and 1

Demonstrate a zero gap between 0.999... and 1

 

[matt grime]

Again? There are at least 2 proofs of this fact in this thread alone. Let x_n be the n'th partial sum 0f 0.9+.009+.0009...

|1-x_n|= 1/10^n

0.999.. =lim x_n

hence |1-0.999...| =0 as the difference with the limit tends to zero, ie can be made of arbitrarily small absolute value.

If you disagree with that then you are disagreeing with the definition of the real numbers. Got it? If you want to work in a different number system then start a different thread or something.

Just realized this isn't in the thread I thought it was in (new page carry error, bane of mathematics) so rant away in your own private language at will.

 

[Zurtex]

Another example:

PI exact plece in the real line is unknown, because in any representation method of it we have to use infinitely many elements to define it.

Demonstrate a zero gap between 0.999... and 1

How do you mean it is unknown?

I'm fairly sure it is at $\pi$... If you let $\pi$ be your base unit then it is really easy to mark it on.

Or do you just mean there is no given ratio between 1 and $\pi$ in terms of decimals?

 

[matt grime]

Here's a little thing you need that you don't seem to know, Organic.

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

 

[Organic]

Suppose a and b are different real numbers and for any e>0 we know |a-b|<e then a not= b.

proof:

If a is not b then |a-b|>0.

Let d be the difference.

Let e = d/2 then |a-b|=d and |a-b| < d/2 > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.

 

[matt grime]

But your hypothesis is false: if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e. You do understand what the quantifier for all means?

 

[matt grime]

Actually that 'proof' of yours should go down in history: you assume a and b are distinct numbers, make a false claim about them and use that false claim to prove that a and b a different, which is part of the hypothesis... fantastic

 

[matt grime]

What are you trying to prove there anyway, now you've edited it? cos looking at it you can't really tell.

 

[Organic]

How do you mean it is unknown?

I'm fairly sure it is at ... If you let be your base unit then it is really easy to mark it on.

Or do you just mean there is no given ratio between 1 and in terms of decimals?

What I say is very simple: Pi is a notation an element, which its exact place in the real line is unknown.

More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example:

3/9 place is well-known 3/10 place is unknown.

 

[matt grime]

Why is 1/3's place known? How do you know where 1 is? Or zero? The real numbers aren't actually physically a line, Organic. You are confusing the representation of something with the something... Oh, no, you're going to talk about x and model(x) again aren't you?

Actually the statement above is trivially true because it is of the form A=>B whre A is false...

 

[Organic]

if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e.

The two different a and b are both < e.

Therefore |a-b| = d < e, but both d and e > 0.

 

[Zurtex]

What I say is very simple: Pi is a notation an element, which its exact place in the real line is unknown.

More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example:

3/9 place is well-known 3/10 place is unknown.

What are you on about?

Do you know what a number line is? It is not something physical...

 

[matt grime]

The two different a and b are both < e.

Therefore |a-b| = d < e, but both d and e > 0.

but that isn't deducible from your hypothesis: just because |a-b|<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption.

Try writing out the statement of the lemma again, and its proof making sure all the hypotheses are written correctly and that it is not vacuous (which it was first time)

and seeing as the statement was for all e, then you've just shown a=b=0

 

[Organic]

Oh, no, you're going to talk about x and model(x) again aren't you?

Yes exactly, Math is only a theory therefore x-itself does not exist is its scope, only x-model can be used by Math language.

 

[matt grime]

good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.

 

[Organic]

but that isn't deducible from your hypothesis: just because |a-b|<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption.

Thank you for this correction you are right.

When we writing |a-b| < e we mean that d < e.

if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

 

[Organic]

good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.

No, cauchy sequences do not prove that 0.999... = 1 without breaking infinitely many elements to become finitely many elements, by reaching the limit.

 

[matt grime]

Thank you for this correction you are right.

When we writing |a-b| < e we mean that d < e.

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

let e=d>2, then you have

d/2>d and d>0. now think for a second.

d/2>d => d>2d => 0>d ,yet d>0.

Want to rethink that at all.

 

[Organic]

let e=d>2,

This in not the case because e>d always.

 

[matt grime]

No, cauchy sequences do not prove that 0.999... = 1 without breaking infinitely many elements to become finitely many elements, by reaching the limit.

As real numbers are defined... oh look, circles. Tell you what, why don't you tell us what you think the real numbers are? Since your definition must be equivalent to the one using cauchy sequences where 0.9999 =1 by definition you are in trouble. I think this is because when mathematicians speak of a model, in the sense of something satisfying the axioms, an example, they don't mean what you think the mean. ie a model in the sense of a model of turbulence, or something, which is only an approximation (at the moment). There is no approximation; you are confusing the concrete and the abstract. The Cauchy sequence argument is not some "best approximation" mathematically to the "physical" real numbers, they are the real numbers, in and of themselves, it is the things that you draw on the page using axes that are the approximation, not the other way round.

breaking, infinitely many, finite, that's you wishing something to be true that isn't, you are thinking unmathematically (perhaps intuiitive and physically in your opinion, but that isnt' mathematics).

 

[matt grime]

This in not the case because e>d always.

then your initial quantifier, for all e>0, is not correct is it?


Here is your initial post:

Suppose a and b are different real numbers and for any e>0 we know |a-b|<e then a not= b.


No restriction on e>d. at all and it says for any e>0 doesn't it?

Once more you move the goal posts half way through your argument when someone points out where it's gone wrong.

So want to start from the beginning and clearly write out what it is you are trying to prove again?

Because so far you're not doing very well. I mean what was the point of it anyway?

 

[Organic]

you are thinking unmathematically

There is no objective thing like mathematics that we can compare our way of thoughts to it.

Math language is only a rigorous agreement between people, no more no less.

I have found that the current agreement includes lot of weak point in it, where one of them is the infinity concept.

 

[Organic]

Suppose a and b are distinct real numbers and for any e>0 we know |a-b|<e then a not= b.

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

If a is not b then |a-b|>0.

Let d be the difference.

Let e = d/2 then |a-b|=d or |a-b| = d/2 > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.