Mathematics - the language of a deterministic Universe?

[studiot]

 

Actually that's the beauty of mathematics, there could be millions of function pairs f(x) and g(x) for which g(x) equals f( f(x) ). For any of the pair if it is proven that g(x) equals f( f(x) ) then that proof must of course always hold for those function pairs.

 

All mathematical proofs if valid must always hold, otherwise they're not valid. I think here we can agree?

 

 

That didn't answer (=avoided) the question of what you mean by the statement

 

f(x) = g(x)

 

So how can anyone agree or disagree if they don't know exactly what you mean?

 

(hint the above equality statement (equation) is incomplete as it stands)

 

I will be helpful this once, although the onus is on you since you are the one trying to prove something.

 

The functions g(x) = sin (x) and f(x) = tan(x)

 

are equal for an infinite number of values of x

 

Unfortunately they are also not equal for an infinite number of values of x

 

They are also very nearly equal for an infinite number of values of x

 

They are also as different as it is possible to be ( that is f(x) - g(x) = [math]\infty [/math] for an infinite number of values of x

 

If h(x) = x then h(x) = f(x) = g(x) for one value only of x.

 

So what exactly do you mean?

 

And you have still avoided my example points.

 

Edited February 6, 2015 by studiot

[quanta'namo nay!]

 

That didn't answer (=avoided) the question of what you mean by the statement

 

f(x) = g(x)

 

So how can anyone agree or disagree if they don't know exactly what you mean?

 

(hint the above equality statement (equation) is incomplete as it stands)

 

Perhaps you misread, I wrote

 

g(x) equals f( f(x) )

not

g(x) equals f(x)

 

It is not a statement but is a requirement for statements where the first result of f(x) is fed once more into the same function.

[studiot]

 

Perhaps you misread, I wrote

 

g(x) equals f( f(x) )

not

g(x) equals f(x)

 

It is not a statement but is a requirement for statements where the first result of f(x) is fed once more into the same function.

 

 

You did indeed write that, but did not explain why, so I took it as a typo. Sorry.

 

It seems a bit pointless since a simple substitution reduced f(f(x)) to q(x) and we have the situation I have described (in post 51) and you need to elaborate on.

Edited February 6, 2015 by studiot

[quanta'namo nay!]

The functions g(x) = sin (x) and f(x) = tan(x)

 

Here we can see that for

g(x) equals f( f(x) )

 

you get:

sin(x) = tan( tan(x) )

 

It it obvious that this does not hold, so there is no proof (for particular function pair). My point was that there was some proven function pair that fulfils

g(x) equals f( f(x) )

 

If the proof is valid, it must always hold.

 

You did indeed write that, but did not explain why, so I took it as a typo. Sorry.

 

It seems a bit pointless since a simple substitution reduced f(f(x)) to q(x) and we have the situation I have described (in post 51) and you need to elaborate on.

 

See above.

Edited February 6, 2015 by quanta'namo nay!

[studiot]

Since I am addressing your points, either you address my points as well or we will need to discontinue this conversation as one sided.

 

I said

 

For an infinite number of x, sin(x) = tan(x).

 

It is also true that for all these x, sinx = tan(tan(x))

 

So your claim

sin(x) = tan( tan(x) )

 

It it obvious that this does not hold, so there is no proof (for particular function pair). My point was that there was some proven function pair that fulfils

g(x) equals f( f(x) )

 

 

is nonsense.

[quanta'namo nay!]

Since I am addressing your points, either you address my points as well or we will need to discontinue this conversation as one sided.

 

I said

 

For an infinite number of x, sin(x) = tan(x).

 

It is also true that for all these x, sinx = tan(tan(x))

 

So your claim

 

is nonsense.

 

I explicitly referred to any pair of functions f(x) and g(x) that fulfil the requirement of g(x) equals f( f(x) ) and that being proven for that pair of functions.

 

Why do you offer a pair of functions for which this has not been proven, but also obviously do not fulfil the requirement?

 

For any pair that fulfils the requirement it is not nonsense, instead it would be a proof that holds always, deterministically without exception.

[Strange]

These are the two last categories of the three you listed. I was trying to ask you how you could distinguish between them.

 

I thought I had answered that.

 

We have models of deterministic systems (that may or may not be predictable - i.e. may or may not be chaotic) and we have models of probabilistic systems.

 

Do you think it is possible that absolutely all events (macro and quantum) in the universe fall into the domain of causality?

 

It is, of course, possible. But there is absolutely no reason to think it is likely and a lot of evidence to suggest that it is so.

Actually that's the beauty of mathematics, there could be millions of function pairs f(x) and g(x) for which g(x) equals f( f(x) ). For any of the pair if it is proven that g(x) equals f( f(x) ) then that proof must of course always hold for those function pairs.

 

All mathematical proofs if valid must always hold, otherwise they're not valid. I think here we can agree?

That is true but has nothing to do with determinism.

That didn't answer (=avoided) the question of what you mean by the statement

 

f(x) = g(x)

 

So how can anyone agree or disagree if they don't know exactly what you mean?

You are being ridiculous again.

 

IF it is shown that f(x) = g(x) THEN:

a) It doesn't matter what f() and (g) are; and

b) it will always be true.

[studiot]

 

Why do you offer a pair of functions for which this has not been proven, but also obviously do not fulfil the requirement?

 

I asked you (politely) to explain in more detail you meant by your "requirement" and gave an example of why I consider your statement

f(f(x)) = g(x) incomplete.

 

You have made no attempt to do so that I can see.

 

So I offered an example for which there are an infinite number of values of x for which

 

f(f(x)) = g(x)

 

Do you deny that this is so and if you deny it can you prove it mathematically?

[quanta'namo nay!]

I thought I had answered that.

 

We have models of deterministic systems (that may or may not be predictable - i.e. may or may not be chaotic) and we have models of probabilistic systems.

 

 

It is, of course, possible. But there is absolutely no reason to think it is likely and a lot of evidence to suggest that it is so.

 

Suppose there are two competing mathematical models in physics supported by the exact same data (the evidence) and no data against. Would you use Occam's razor to decide which one of these models is better?

 

That is true but has nothing to do with determinism.

 

Well determinism means predictability without exception. Mathematical proofs when valid are predictability without exception, they will always hold. No matter what numeric values or other substitutions you test them with they always hold.

 

I asked you (politely) to explain in more detail you meant by your "requirement" and gave an example of why I consider your statement

f(f(x)) = g(x) incomplete.

 

You have made no attempt to do so that I can see.

 

So I offered an example for which there are an infinite number of values of x for which

 

f(f(x)) = g(x)

 

Do you deny that this is so and if you deny it can you prove it mathematically?

 

In mathematics if there is any x for which

g(x) = f( f(x) )

 

does not hold, it means that there is no proof. If there is only a specific range of x for which the above holds true, then you need to specify that range. Limited to the range then again the proof may hold. In that case of course always. What I defined was not for any limited range, but any.

Edited February 6, 2015 by quanta'namo nay!

[studiot]

 

What I defined was not for any limited range, but any.

 

 

But you didn't say so.

 

Which is why I called your statement incomplete.

 

Fo one function to exactly match another they must share the same domain and codomain.

 

You haven't specified this either.

 

How will you make that work for f(f(X)) ?

[quanta'namo nay!]

 

But you didn't say so.

 

Which is why I called your statement incomplete.

 

In mathematics the default is no range unless specified. Have you seen anyone assume a range without specification?

 

It is merely a problem without a presented solution. The problem is:

Which functions f(x) and g(x) will hold true for g(x) = f( f(x) )?

 

You might find some function pair that will always hold true for the above.

 

 

Fo one function to exactly match another they must share the same domain and codomain.

 

You haven't specified this either.

 

How will you make that work for f(f(X)) ?

No restrictions on domains, they can be any.

If codomains of f( f(x) ) and g(x) are contradictory then of course the statement g(x) = f( f(x) ) is false and there is no proof for that particular function pair.

[imatfaal]

 

 

In mathematics the default is no range unless specified. Have you seen anyone assume a range without specification?

 

It is merely a problem without a presented solution. The problem is:

Which functions f(x) and g(x) will hold true for g(x) = f( f(x) )?

 

 

f(x) = -x

[studiot]

Let f(x) = a [math]\forall [/math]x

 

Then f(a) = a

 

thus f(f(x)) = a [math]\forall [/math]x

 

Let g(x) = a for x=a and be undefined otherwise.

 

then g(x) = f(f)x)).

But this still avoids the two points I asked you directly about.

 

Are you refusing to answer them?

 

Nice one imatfaal +1

Edited February 6, 2015 by studiot

[Strange]

Suppose there are two competing mathematical models in physics supported by the exact same data (the evidence) and no data against. Would you use Occam's razor to decide which one of these models is better?

 

Probably. But it is more likely that the two models overlap and we will choose to use the simpler one when it is good enough. For example, Newton's theory of gravity is very often accurate enough and way simpler than GR. And there are many classical (deterministic) approximations that can be used in place of full quantum formalizations (e.g. Maxwell's equations instead of QED).

[ajb]

Assuming that we are discussing functions on the real line then we know what is going on here. As long as [math]f\circ f[/math] makes sense then we can simply define [math]g= f\circ f[/math]. We have to be a little careful with the domain and range for the composition to be okay, but mod that there is nothing in this question.

Edited February 7, 2015 by ajb

[quanta'namo nay!]

 

f(x) = -x

 

Yes, indeed. The simplest solution with unrestricted domain is f(x)=-x and g(x)=x. And this here will hold always. This is the determinism I was referring to. As said If it was not deterministic it would be random, and if it was random it mostly would not hold (could hold extremely rarely when one is lucky).

 

Let f(x) = a [math]\forall [/math]x

 

Then f(a) = a

 

thus f(f(x)) = a [math]\forall [/math]x

 

Let g(x) = a for x=a and be undefined otherwise.

 

then g(x) = f(f)x)).

But this still avoids the two points I asked you directly about.

 

Are you refusing to answer them?

 

Nice one imatfaal +1

 

If you are referring to your f(x) = tan(x) and g(x) = sin(x) example, that was already addressed. Strange also addressed this.

 

sin(x) = tan( tan(x) ) does not hold always, and as such is false as a statement.

sin(x) = tan( tan(x) ) will hold when you say x=0. Then you have reduced your equation to 0=0.

 

As this is also not random but deterministic, that will hold always when you let x=0.

 

 

Probably. But it is more likely that the two models overlap and we will choose to use the simpler one when it is good enough. For example, Newton's theory of gravity is very often accurate enough and way simpler than GR. And there are many classical (deterministic) approximations that can be used in place of full quantum formalizations (e.g. Maxwell's equations instead of QED).

 

Yes. Newton's model for gravity is also known to be false. We can of course still use it for approximations.

 

Science searches for the best models. In that sense a classical model always has the upper hand over any nonclassical models that produce logical paradoxes. Occam's razor says that the paradoxical model is the worse model.

[ajb]

Yes, indeed. The simplest solution with unrestricted domain is f(x)=-x and g(x)=x.

What about just using the identity mapping. That is even simpler, but not so interesting. You could also consider the obvious pairs of linear maps.

 

 

 

This is the determinism I was referring to.

But this is because you are using functions. In the sense you describe, we have 'determinism' which just means that we have expressions that are unambiguous.

 

What about mappings that are one-to-many? Are they 'deterministic' by your definition?

[quanta'namo nay!]

But this is because you are using functions. In the sense you describe, we have 'determinism' which just means that we have expressions that are unambiguous.

 

What about mappings that are one-to-many? Are they 'deterministic' by your definition?

 

Yes. One-to-many and many-to-one mappings are of course deterministic, you can not produce random results via any function. It would mean your results change even though what you substitute into the function does not.

[ajb]

Yes. One-to-many and many-to-one mappings are of course deterministic, you can not produce random results via any function. It would mean your results change even though what you substitute into the function does not.

Functions are understood as many-to-one, usually.

 

Okay, you don't produce random results as such, but you are happy if the result is not unique?

 

Anyway, in some sense what you have been saying is true. I would also add that we cannot truly generate random numbers on a computer, we come close for many practical purposes, but I don't know of any algorithm that is truly random.

 

However, as I have said before we do have mathematical methods for dealing with truly random things. Thus mathematics can handle more than what is usually meant by deterministic. The expressions you get will be 'deterministic' in your sense. By this I mean you may calculate a probability, an average value, an expectation value etc of something exactly, but you cannot interpret the meaning in a truly deterministic way.

Edited February 7, 2015 by ajb

[quanta'namo nay!]

Uniqueness is not a requirement of determinism. Only that the result is not random but repeatable and predictable.

[ajb]

Uniqueness is not a requirement of determinism. Only that the result is not random but repeatable and predictable.

This is interesting. By determinism I would have in mind some dynamical system, not simply a function on the real line or two. For example, if we were thinking of the dynamics of a classical particle and the equations of motion for given initial condition were not at least locally unique then what path would the particle take? How would it decide where to go?

 

p.s.

 

I edited my comment above, sorry of you missed that.

Edited February 7, 2015 by ajb

[quanta'namo nay!]

Anyway, in some sense what you have been saying is true. I would also add that we cannot truly generate random numbers on a computer, we come close for many practical purposes, but I don't know of any algorithm that is truly random.

 

It often pains me that it is so difficult to generate "random" numbers. I usually just end up using the time a program is executed as an easy "random" input to the program as it is not in control of the program. That generates adequate variation but still the linear progression is seen as some kind of pattern.

 

However, as I have said before we do have mathematical methods for dealing with truly random things. Thus mathematics can handle more than what is usually meant by deterministic. The expressions you get will be 'deterministic' in your sense. By this I mean you may calculate a probability, an average value, an expectation value etc of something exactly, but you cannot interpret the meaning in a truly deterministic way.

 

Interpret the meaning in what sense? As applied to the natural world for physics? I'd say that without physics mathematics is just abstractions absent any meaning.

 

This is interesting. By determinism I would have in mind some dynamical system, not simply a function on the real line or two. For example, if we were thinking of the dynamics of a classical particle and the equations of motion for given initial condition were not at least locally unique then what path would the particle take? How would it decide where to go?

 

p.s.

 

I edited my comment above, sorry of you missed that.

 

Suppose you want to model the path with a plotted curve. Can you find a function f(x) that gives you a "two way" parallel path?

No worries about the edit..

[ajb]

Interpret the meaning in what sense? As applied to the natural world for physics? I'd say that without physics mathematics is just abstractions absent any meaning.

 

Even mathematically interpreting a result. I may have some mathematical system for which the states evolve with some parameter, details don't matter here. Let us suppose that I want to know a state at some given parameter. I may only be to calculate probabilities etc. I will have to interpret this result using probability theory even though my calculation is 'deterministic' in your sense.

 

Suppose you want to model the path with a plotted curve. Can you find a function f(x) that gives you a "two way" parallel path?

It cannot be a simple function on the real line, by definition. We need something that has different branches and we won't make a 'branch cut'. What about taking the square root on the positive line and allowing both positive and negative roots? The domain here will be the positive line but the range will be the whole real line.

[quanta'namo nay!]

Even mathematically interpreting a result. I may have some mathematical system for which the states evolve with some parameter, details don't matter here. Let us suppose that I want to know a state at some given parameter. I may only be to calculate probabilities etc. I will have to interpret this result using probability theory even though my calculation is 'deterministic' in your sense.

 

You could of course see from the mathematical system which was the previous state of the parameter, and observe in simulation how the parameter changes its state from one to another whenever there is any change (maybe program the simulation to detect any change in the given parameter). That allows you to also plot the state of the parameter with respect to simulated time. This would give you more insight to the behaviour of the parameter than just calculating probabilities.

 

It cannot be a simple function on the real line, by definition. We need something that has different branches and we won't make a 'branch cut'. What about taking the square root on the positive line and allowing both positive and negative roots? The domain here will be the positive line but the range will be the whole real line.

 

Wouldn't that be the same as just plotting two functions instead of one?

It is interesting that the notion is the "real" line as opposed to something that is not real.

[Strange]

Yes. Newton's model for gravity is also known to be false. We can of course still use it for approximations.

 

I wouldn't say it is false. It is correct within its domain of application. The same is true of any scientific theory. There are very few theories that have been shown to be completely wrong.

 

Occam's razor says that the paradoxical model is the worse model.

 

That isn't at all what it says.