Can Someone Explain 'Smoothness' Concisely to Me?

[metacogitans]

I've been working on a solution for one of the millennium prize problems (the Navier-Stokes Equations and Smoothness problem), but one of the finalizing things I need is a formal definition of 'smoothness'.

 

The problem asks for proof which involves a smooth, divergent free vector field, a smooth function for a force, and a smooth function for pressure.

[Strange]

http://mathworld.wolfram.com/SmoothFunction.html

https://en.wikipedia.org/wiki/Smoothness

https://en.wikipedia.org/wiki/Vector_field#Vector_fields_on_manifolds

[Velocity_Boy]

LOL....The Queen of Link Posting! linkogenic teen warhead?

I've been working on a solution for one of the millennium prize problems (the Navier-Stokes Equations and Smoothness problem), but one of the finalizing things I need is a formal definition of 'smoothness'.

 

The problem asks for proof which involves a smooth, divergent free vector field, a smooth function for a force, and a smooth function for pressure.

 

Isn't a smooth function just a function that has continual derivatives? That will extend to up to whatever the specific domain order is?

[studiot]

I'm glad to see strange mentiones manifolds, though I haven't looked into his references.

 

This is because you can't have an everywhere smooth vector field on a spherical shell. (The Hairy Ball Theorem)

 

This is important for instance in applying the NS equations to the Atmosphere.

 

The reason for the ban on divergent (and convergent) fields is that there must be a singularity at the point of divergence (convergence) ie a sink or source

[wtf]

Isn't a smooth function just a function that has continual derivatives?

Of all orders. [math]C^\infty[/math]. Differentiable once, twice, thrice, etc. Even Xerxes and I agree on that

 

But note that there is a [math]C^\infty[/math] function that is not real analytic. In other words it has derivatives of all orders at some point, and we can form its Taylor series at that point, but the function is not equal to the Taylor series. So if you need smoothness it's good to say whether you mean [math]C^\infty[/math] or real or complex analytic.

 

https://en.wikipedia.org/wiki/Non-analytic_smooth_function

Edited April 2, 2017 by wtf

[metacogitans]

I'd really like to talk about my idea for the solution, but I guess this is actually something that if I think I have it, I should publish it first (and make sure the proof is kosher).

 

That makes it hard for me to ask the questions I need to ask, but let me ask this:

If a function ends abruptly, or has any sudden increases/decreases or 'sharp angles', it ceases to be smooth, correct?

[wtf]

If a function ends abruptly, or has any sudden increases/decreases or 'sharp angles', it ceases to be smooth, correct?

That seems a little vague. If a function is [math]C^{10000}[/math] but not [math]C^{10001}[/math] then it's not smooth, but we'd be hard pressed to say it has corners or changes abruptly. Of course a function with a big obvious corner like the absolute value is not differentiable at zero, but infinite differentiability is much stronger than mere differentiability. And the difference is not always visible to the naked eye.

Edited April 2, 2017 by wtf

[Xerxes]

I'd really like to talk about my idea for the solution, but I guess this is actually something that if I think I have it, I should publish it first (and make sure the proof is kosher).

So why are you on this site if you have nothing to share?

 

That makes it hard for me to ask the questions I need to ask, but let me ask this:

With this attitude, as far as I am concerned, I would be disinclined to answer any questions from you.

 

You are abusing the forum and insulting its members

[Strange]

I'd really like to talk about my idea for the solution, but I guess this is actually something that if I think I have it, I should publish it first (and make sure the proof is kosher).

 

 

Why not present your proof and let people here (not me!) review it. That seems like a good way of making sure it is valid.

LOL....The Queen of Link Posting! linkogenic teen warhead?

 

You know me. I am just sceptical that someone who is not familiar with a concept could produce a proof related to it.

 

"I have a proof that Pi is normal but I just need to know what 'normal' means..."

[metacogitans]

 

 

Why not present your proof and let people here (not me!) review it. That seems like a good way of making sure it is valid.

 

You know me. I am just sceptical that someone who is not familiar with a concept could produce a proof related to it.

 

"I have a proof that Pi is normal but I just need to know what 'normal' means..."

I was familiar with smoothness - I just needed a more concise definition for it.

 

As soon as I am closer to a final edit of the solution, I will post it here along with having submitted it to a peer reviewed journal of physics/mathematics.

[1911]

Have you previously worked on any math problems? You cannot prove a theorem to a bunch of mathematicians if you don't speak their language, even if your proof works. I suggest you study mathematics. There are many hobbyist Mathematicians who've claimed lots of things all the time. Just be prepared to be humbled by fellow math enthusiasts.