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Suppose $ f(x)$ that is infinitely differentiable in $[a,b]$.

For every $c\in[a,b] $ the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial.

Is true that $f(x)$ is a polynomial?

I can show it is true if for every $c\in [a,b]$, there exists a neighborhood $U_c$ of $c$, such that
$$f(x)=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n\quad\text{for every }x\in U_c,$$ but, this equality is not always true.

What can I do when $f(x)\not=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n$?

  • 9
    Two solutions starting from weaker assumptions are given [in this MO thread](http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial)2011-12-22
  • 5
    Put $F_n:=\bigcap_{k\geq n}\{x\in [a,b], f^{(k)}(x)=0\}$ and apply Baire's category theorem.2011-12-22
  • 6
    I'm left wondering if the stronger assumptions here permit some more elementary proof.2011-12-22
  • 2
    @t.b. Would you (or @Davide) mind typing up a correct answer (possible just taken from MO), perhaps as community wiki? (Or, I can do it if no one else wants to). There are currently 10 incorrect answers (some deleted), and no correct answers.2012-10-22

2 Answers 2

1

As I confirmed here, if for every $c\in[a,b] $, the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial, then for every $c\in[a,b]$ there exists a $k_c$ such that $f^{(n)}(c)=0$ for $n>k_c$.

If $\max(k_c)$ is finite, we're done: $f(x)$ is a polynomial of degree $\le\max(k_c)$.

If $\max(k_c)=\infty$ it means there is an infinite number of unbounded $k_c$'s, but $f$ is infinitely differentiable, so (hand waving) the $c$'s can't have a limit point, i.e. although $\max(k_c)=\infty$ it can't be $\lim_{c\to c_\infty}k_c=\infty$ for some $c_\infty\in[a,b]$ because that would mean $k_{c_\infty}=\infty$, i.e. not a polynomial.

So the infinite number of unbounded $k_c$'s need to be spread apart, e.g. like a Cantor set.

Does this suggest a counterexample or can a Cantor-like distribution of $k_c$'s never be infinitely differentiable?

-1
  1. All polynomials are Power Series but not all Power Series are not polynomials. For a certain Power Series $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k = a_0 + a_1 (x-c)^1 + a_2 (x-c)^2 + a_3 (x-c)^3 + \cdots$ to be a Polynomial of degree $n$, then for all $k>n$, $a_k = 0$.

  2. If $ f(x)$ is infinitely differentiable in the interval $[a,b]$, then for every $k \in \mathbb{N}$, $f^{(k)}(x) \in \mathbb{R}$ i.e. exists as a finite number. The Taylor Series of $f(x)$ in the neighbourhood of $c$ is $\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $ and

  3. If the remainder, $R_N(x) = f(x) - \sum\limits_{k=0}^N \cfrac{f^{(k)}(c)}{k!}(x-c)^k $ for a certain $N \in \mathbb N$, converges to $0$ then $f(x) = \sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $
  4. Taylor's inequality: If $|f^{(N+1)}(x)|≤ B$ for all $x$ in the interval $[a, b]$, then the remainder $R_N(x)$ (for the Taylor polynomial to $f(x)$ at $x = c$) satisfies the inequality $$|R_N(x)|≤ \cfrac {B}{(N+ 1)!}|x − c|^{N+1}$$ for all $x$ in $[c − d, c + d]$ and if the right hand side of this inequality converges to $0$ then $R_N(x)$ also converges to $0$.

According to your question, supposing that $\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $, $\forall c \in [a,b]$ is a polynomial which translates to $$\text{given } c\in[a,b],\ \ \exists n_c\in \mathbb N \ (\text{ $n_c$ depends on c}) \quad|\quad\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k=P_{n_c}(x)$$ $$\quad \quad \quad\quad \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \text { and} \ \forall k>n_c, \ k\in \mathbb N, \ {f^{(k)}(c)}=0$$

This is true because if one looks at the finite sum $N\ge n_c$, $$\displaystyle\sum^N_{k=0} a_k(x-c)^k=\sum^N_{k=0}\sum^k_{i=0}a_k\binom ki(-1)^{k-i} c^{k-i}x^{i}=\sum^N_{i=0}x^{i}\sum^N_{k=i}a_k\binom ki(-1)^{k-i} c^{k-i}$$ if this is a polynomial $P_{n_c}(x)$ of degree $n_c$, then $$\forall i>n_c,\ \ \displaystyle \sum^N_{k=i}a_k\binom ki(-1)^{k-i} c^{k-i}=0$$ Solving this system of equations gives that $\forall n_c

$$a_k=\cfrac{f^{(k)}(c)}{k!}=0\implies f^{(k)}(c)=0, \ \ \forall k>n_c$$ This holds when $N\rightarrow \infty$

Since $n_c$ depends on each $c\in[a,b]$, it is sufficient to take $\displaystyle n=\max_{c\in[a,b]} (n_c)$ such that for any $c\in [a,b]$ and for any $k>n,\ \ k\in \mathbb N$, we have $f^{(k)}(c)=0$.

Thus, the Taylor series is of $f$ is a polynomial of degree $\displaystyle n=\max_{c\in[a,b]} (n_c)$ because $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k=P_n(x)$.

At this point it is sufficient to prove that $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k=P_n(x)$ using the Taylor Remainder Theorem (#4).

We've already found out that $f^{(k)}(c) = 0,\space \forall k>n$, thus $ f^{(n+1)}(x) = 0$ or simply $ f^{(n+1)}(x) \le 0$ (to work with inequalities) which implies that $B = 0$. At this point it is clear that $|R_N(x)|≤ \cfrac {B}{(N+ 1)!}|x − c|^{N+1} = 0$ and we can conclude that $R_N(x)$ converges to $0$ and that $f(x) = \sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k = P_n(x)$.

$f$ is a polynomial.

  • 0
    The point is that $k$ is allowed to depend on $c$, whereas our $k$ is independent of $c$.2012-10-22
  • 0
    why the down-vote?2012-11-11
  • 1
    I didn't downvote, but as I said in my comment, this doesn't answer the OPs question. The OPs question is this: "We know that for each point $c$, the Taylor series at $c$ is a polynomial. Why is the original function a polynomial?" We do **not** know that at each point $c$, the taylor series is a polynomial of degree $n$ for some $n$, because $n$ can vary as $c$ varies. In particular, your use of point $4$ is invalid because it's possible that at some point $c$, there is no universal $n$ that works any interval containing $c$.2012-11-11
  • 0
    @JasonDeVito I didn't get your point the first time. I'll fix the answer after working on it. Thanks.2012-11-11
  • 0
    Well, I didn't make my point very well the first time - sorry about that!2012-11-11
  • 0
    Is there something I missed?2012-11-21
  • 0
    Unfortunately, I think there is - why is $\max(n_c)$ a finite number?2012-11-22
  • 0
    @JasonDeVito a polynomial is always finite, so $\forall c \in [a,b]$, $n_c$ is always finite or else it is not the degree of a polynomial.2012-11-22
  • 1
    I agree that each $n_c$ is finite, but the maximum of an infinite number of finite numbers need not be. For example, what if $n_{1/k} = k$?2012-11-22
  • 0
    @JasonDeVito I see no reason why it shouldn't be finite. Enlighten me more clearly please.2012-11-22
  • 0
    Well, look at $c = 1$, $c=1/2$, $c=1/3$, $c=1/4$, etc. For each of these, there is a corresponding $n_c$. What stops the following from occuring? $n_1 = 1$, $n_{1/2} = 2$, $n_{1/3} = 3$, $n_{1/4} = 4$, etc. If this does occur, then there is no (finite) maximum of all the $n_c$ because if you say "the max is 10000000000!" then I say, "well, what about $n_{1/10000000001}$?" (and clearly, by a similar method, I can beat any supposed maximum.)2012-11-22
  • 0
    @JasonDeVito just because it can't be determined doesn't mean it isn't finite OR does it?2012-11-22
  • 0
    Well, if for a particular $f(x)$ you choose it happens to be finite, I agree with your proof. But the point of a proof is that it must handle all cases. Thus, you either need to prove that even if the $\max$ is $\infty$, $f(x)$ is still a polynomial OR prove that the $\max$ can't be $\infty$. (As a hint, the first statment is false - if $\max = \infty$ then there's no way $f(x)$ was a polynomial. So, you need to prove the $\max$ is finite. In my own opinion this is the crux of the overall proof and by far the hardest part.)2012-11-22
  • 1
    @JasonDeVito I do get your point now.2012-11-22
  • 1
    I did downvote. In addition to $\max\limits_{[a,b]}n_c$ being assumed finite without justification, point #3 (in which I assume "as $x$ approaches $c$" is implied in the statement of convergence to zero) is incorrect. In particular, it would prevent the existence of [smooth non-analytic functions](http://en.wikipedia.org/wiki/Non-analytic_smooth_function).2012-11-24