/sci/ - Science & Math » Thread #13264178

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Prove or disprove

Anonymous Fri 11 Jun 03:13:01 2021 No.13264178 View Reply Original Report

Quoted By: >>13264201 >>13264223 >>13264594 >>13265248 >>13265856

I literally can't approach this one, does it have something to do with with uncountability of the reals?

Anonymous

Anonymous Fri 11 Jun 2021 03:19:32 No.13264201 Report

Quoted By:

>>13264178
I would think disprove, but I dont remember a lot about calculus. My reasoning is the difference between
1) for every real number x there is a natural number n such that (condition)
Compared to
2) there is some natural number n such that (condition) holds for all real numbers x
Intuitively those seem like very different meanings to me.

El Arcón

El Arcón Fri 11 Jun 2021 03:23:29 No.13264213 Report

Quoted By: >>13264252 >>13264311

>Sometimes I go to the store
>I go to the store, sometimes

Anonymous

Anonymous Fri 11 Jun 2021 03:26:27 No.13264223 Report

Quoted By: >>13264239

>>13264178
I'm pretty sure the second condition implies f is a polynomial. So maybe construct something that isn't a polynomial that has property 1.

Anonymous

Anonymous Fri 11 Jun 2021 03:29:54 No.13264239 Report

Quoted By: >>13264246

>>13264223
Was about to say this, the second condition just says the nth derivative is the zero function, integrate that n times to get a polynomial

Anonymous

Anonymous Fri 11 Jun 2021 03:32:40 No.13264246 Report

Quoted By:

>>13264239
Yeah, and I think you can construct something with fourier series that has property 1 buy isn't a polynomial. I'm sure there is an easier way, though.

Anonymous

Anonymous Fri 11 Jun 2021 03:34:58 No.13264252 Report

Quoted By:

>>13264213
That analogy doesn't line up though, does it?

Anonymous

Anonymous Fri 11 Jun 2021 03:52:28 No.13264311 Report

Quoted By:

>>13264213
Doesn't make any sense in this context

Anonymous

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Anonymous Fri 11 Jun 2021 04:59:59 No.13264558 Report

Quoted By: >>13264561 >>13264781 >>13265077

It's obvious that 2 implies 1, so assume 1 and let $Z_n = (f^{n})^{-1}(0)$ . Then by 1, for any $A \subseteq \mathbb{R}$ , $A = \cup_n (Z_n \cap A)$ . Also note that each $Z_n$ is a closed set by smoothness of $f$ . So by the Baire theorem, we see that the set $S$ of points such that $f$ is a polynomial on a neighborhood of the point is open and dense in $\mathbb{R}$ and it's also not difficult to see that the complement of $S$ has no isolated points and hence is a perfect set. Now all that remains is to show that $S$ is also closed and for that I have no ideas.

Anonymous

Anonymous Fri 11 Jun 2021 05:01:24 No.13264561 Report

Quoted By:

>>13264558
By f^n I meant the n-th derivative

Anonymous

Anonymous Fri 11 Jun 2021 05:16:16 No.13264594 Report

Quoted By:

>>13264178
the second statement supposes that at some point the n-th derivative is the zero function

the only class of function this applies to is a polynomial, i think

the first statement is trickier though

Anonymous

Anonymous Fri 11 Jun 2021 06:06:52 No.13264730 Report

Quoted By:

there must be an n for which an uncountable number of points the derivative is 0

Anonymous

Anonymous Fri 11 Jun 2021 06:28:46 No.13264781 Report

Quoted By: >>13264799 >>13264875

>>13264558
Yeah, I've been thinking BCT as well. Obviously, the idea can be extended that for each $x \in \mathbb{R}$ that there is some neighbourhood $x \in U_x \subseteq \mathbb{R} [/maath] for which<br /><script type="math/tex; mode=display"> f^{(n_x)}|_{U_x} = 0.$
Perhaps partitions of unity can then show that this function must be identically zero everywhere eventually for some finite $n$ .

Anonymous

Anonymous Fri 11 Jun 2021 06:37:31 No.13264799 Report

Quoted By:

>>13264781
>maath
what did he mean by this

and also, evidently you can make shit as wide as you want in LaTeX
$8======================================================================================================================================================================================================================================================================================================================================================================================================================================================================================D~$

Anonymous

Anonymous Fri 11 Jun 2021 08:10:33 No.13264875 Report

Quoted By: >>13266179

>>13264781
The set S in my post is actually exactly the set of points for which there is some neighborhood of the point where the function is polynomial, lol. That S is open and dense follows by Baire, what's not clear is whether it's equal to R. I tried thinking about partitions of unity too but they're an absolute nightmare to do calculations with.

After some easy searching, I found a nice mo post having the complete solutions to this problem and every solution there uses Baire (the top solution applies it to the complement of my set S and then uses the fact that it's a perfect set in a very clever way).

Anonymous

Anonymous Fri 11 Jun 2021 08:13:27 No.13264885 Report

Quoted By: >>13264888 >>13265874

What's crazy is these are the first math posts I've seen in forever. Maybe it's my fault for only clicking sensational threads

$we have truly lost sci$

Anonymous

Anonymous Fri 11 Jun 2021 08:14:26 No.13264888 Report

Quoted By: >>13265874

>>13264885
Yeah, it's been a long time since I've seen a good math problem on /sci/ lol

Anonymous

Anonymous Fri 11 Jun 2021 09:41:10 No.13265077 Report

Quoted By: >>13265103

>>13264558
just consider compact subsets [-k,k] of R.
you can create an open cover of [-k,k] by taking open epsilon balls around points of S, such that f is polynomial on each epsilon ball. then use a finite subcover, since [-k,k] is compact to show that f is a polynomial on [-k,k]. From this it can be concluded that f is a polynomial on R since the polynomial representation of f is the same for all k,k'>0 since they have to agre on the intersection of [-k',k'] and [-k,k]

Anonymous

Anonymous Fri 11 Jun 2021 09:50:14 No.13265103 Report

Quoted By:

>>13265077
>you can create an open cover of [-k,k] by taking open epsilon balls around points of S
How exactly would you do this? I've only shown that S is open and dense. Such a set can have arbitrarily small measure, so there's no guarantee that it can cover [-k, k].

Anonymous

Anonymous Fri 11 Jun 2021 11:11:08 No.13265248 Report

Quoted By:

>>13264178
This is a very hard exercise actually, don't beat yourself up over that. It does have to do with countability (in some sense)

Anonymous

Anonymous Fri 11 Jun 2021 14:43:16 No.13265832 Report

Quoted By: >>13265844 >>13266179

What's wrong with this:

Condition 2 says that f is a polynomial of degree at most n. This implies condition 1.
Condition 1 says that at every point in IR, the function behaves like a polynomial of finite degree.
Take the maximum of these degrees, and the function that satisfies 1 also satisfies 2 for n equal to this maximum.

Anonymous

Anonymous Fri 11 Jun 2021 14:45:37 No.13265844 Report

Quoted By: >>13265865

>>13265832
>Take the maximum of these degrees,
You get infinity. What now?

Anonymous

Anonymous Fri 11 Jun 2021 14:47:46 No.13265856 Report

Quoted By:

>>13264178
for f(x) = sin(x) there is no n : f^n(x) = 0, x in R

Anonymous

Anonymous Fri 11 Jun 2021 14:50:10 No.13265865 Report

Quoted By: >>13266179

>>13265844
Doesn't the function then not satisfy the first condition

Anonymous

Anonymous Fri 11 Jun 2021 14:51:33 No.13265874 Report

Quoted By: >>13266161

>>13264885
>>13264888
/sci/ has gotten pretty shit lately but even before that the math threads on /b/, on the rare occasion they would appear, were always better.

Anonymous

Anonymous Fri 11 Jun 2021 16:33:33 No.13266147 Report

Quoted By: >>13266179

You need to find a function that has property 1 but the n in 1 must be unbounded so that no maximum exists

Anonymous

Anonymous Fri 11 Jun 2021 16:36:42 No.13266161 Report

Quoted By:

>>13265874
I'll have a hard time believing that unless you post examples

Anonymous

Anonymous Fri 11 Jun 2021 16:41:14 No.13266179 Report

Quoted By: >>13266190

>>13266147
>>13265865
>>13265832
I posted this earlier >>13264875, but I'll mention it again since everyone seems to have missed the post. The two conditions are equivalent and here's a post which has multiple proofs of the result. All of them use Baire's theorem

https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial

Anonymous

Anonymous Fri 11 Jun 2021 16:42:39 No.13266190 Report

Quoted By:

>>13266179
And don't be mistaken by the title of the post which has an obvious error

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