This showed up on /pol/ a few minutes ago but was quickly deleted.

>>8083709
It has been answered on /pol/ already.

>>8083719
t'wasn't

>>8083720
Yes, it was. You were just too dumb to understand the hint.

f(x) = constant

>Is smooth
>Has *a* solution in the reals
>any degree of derivates are zero for any X
>Is a degree zero polynomial

Or am I misinterpreting something?

>>8083756
you know that at each point some order of derivative vanishes. but that order can vary from point to point.

I was thinking that you could let C_n = {x\in [0,1] | f^{(n)}(x)=0}, then the union of the C_n would be [0,1]. So at least one of the C_n would be uncountable, thus have a limit point in [0,1]. Now if f were analytic, that would be enough, no? But it only says f is C^\infty.

>>8083756
Real rigorous proof your wrote there, fagzoid.

>>8083756
The function doesn't need to have a root anywhere. The point is that if you keep differentiating f you should eventually get a function which is 0 everywhere, so f is a polynomial.

So proving the negation would involve finding a C^inf function from I to R such that at some point x in I there exists no integer a such that f^{a}(x)=0?

I'm pretty sure that's clearly impossible but it's been a long time since I've done any analysis.

so f = 0 (constant)

>>8083846
No, it's true in that case but that's not the only case.

>>8083773
I'm not doing proof by exhastion on every other possible function, and how else would you prove that is has to be a polynomial?

>>8083770
>you know that at each point some order of derivative vanishes. but that order can vary from point to point.

But in which way could it be allowed to vary between points? There can not be a function that associates finite natural number values to the irrational numbers between 0 and 1, as in associating an irrational x with an order of derivate.

Does that not mean the order of derivate would have to be assigned to intervals of x, which makes it a constant function within these intervals?

suppose f is not polynomial

Consider the Taylor series expansion of f at 0 (exists cause f is C infinity)

since f is not polynomial, for all n integer, there is a x_n in I such that f^(n)(x_n)=/=0 (because if it wasn't the case, take such a n and f^(n)(x) is identically 0 and solve a differential equation)

take X = {x_i} i belongs to N

and Y = {n_x in OP's image}, x spanning I

X belongs to I and is non empty therefore there is a nonempty subset of Y (by taking the restriction of Y to X) in which elements verifies both : f^(n_x)(x_n_x)=/=0 (by definition of X) and f^(n_x)(x_n_x)=0 (by definition of Y)

contradiction

>>8083877
There are smooth functions that are not equal to their Taylor expansions though. There are even functions that are smooth but cannot be written as their taylor series at any point on the interval.

If the function was analytic the result would indeed follow.

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

Nice exercise senpai

>>8083944
I was thinking of using Baire category, and then got stuck, couldn't rule out stuff defined polinomially outside stuff like cantor's, set. The point is to use Baire category in the exceptional set X itself. Pretty nice.

>>8083869
Sure, you can have a mapping between [0,1] and the positive integers. At least one integer must be mapped to by uncountably many elements of [0,1] though.

Couldn't you just integrate the constant 0 n times and end up with an n-1 degree polynomial by default?

>>8083709
wow this really looks useful

>>8084098
What does that have to do with this problem?

>>8084133
Isn't [math]f^{(n_x)}(x)[/math] a derivative?

>>8084143
You don't know a priori that any particular derivative is 0.

Ha but everybody know that N* is just {1}

>>8083709
super hard, 0/10 would not try to solve

>>8083709
(I assume N* is the set of natural numbers, so fuck me if it isn't)
My reading of this problem seems to imply that for a SINGLE n_x, the n_x-th derivative is 0 for ALL x on the interval [0, 1] (which is literally what it says, so)
Induction on [eqn] n_x [/eqn]
for [eqn] n_x = 0 [/eqn] this statement is tautological, as f(x) = 0;
so for [eqn] n_x = n - 1 \forall n \in N* \\
f^{(n-1)}(x) = 0 \implies f = a_0x^{n-2} + ... + a_{n-1}
[/eqn]
therefore, for [eqn] n_x = n \\
f^{(n)} = \frac{d}{dx} f^{(n-1)} = \frac{d}{dx} 0 = 0
[/eqn]

>>8084598
your reading's super wrong, it's NOT what it says

it says For all x in I,
there is a n in N > 0
such that f^(n)(x) = 0

>>8083709
Let P(n, x) be a polynomial in x of degree n. Then every polynomial satisfies your definition if we just differentiate P wrt x n+1 times

literally in calc textbooks

>>8084625
0/10 see me after class

File: nice thread.png (14KB, 220x157px) Image search: [Google]

14KB, 220x157px

>>8083709

>>8083709

It's literally not true.

>>8084664

To expand on this, there is an infinitely differentiable function with 0 derivative which is not 0.

It's the standard counterexample to the taylor series working for all functions, I forgot the exact definition but it involved e^x.

>>8084668

Oops, I may have misread the question, nevermind.