# A question on Taylor Series and polynomial

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

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Two solutions starting from weaker assumptions are given in this MO thread –  t.b. Dec 22 '11 at 13:39
Put $F_n:=\bigcap_{k\geq n}\{x\in [a,b], f^{(k)}(x)=0\}$ and apply Baire's category theorem. –  Davide Giraudo Dec 22 '11 at 13:39
I'm left wondering if the stronger assumptions here permit some more elementary proof. –  leonbloy Dec 22 '11 at 14:44
@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. –  Jason DeVito Oct 22 '12 at 19:44

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 $\forall n_c<k\le N, \ \ a_k=0$ and

$$a_k=\cfrac{f^{(k)}(c)}{k!}=0\implies f^{(k)}(c)=0, \ \ \forall k>n_c$$ 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.

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The point is that $k$ is allowed to depend on $c$, whereas our $k$ is independent of $c$. –  Jason DeVito Oct 22 '12 at 19:42
why the down-vote? –  user31280 Nov 11 '12 at 16:33
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 $c$. –  Jason DeVito Nov 11 '12 at 18:21
@JasonDeVito I didn't get your point the first time. I'll fix the answer after working on it. Thanks. –  user31280 Nov 11 '12 at 18:32
Well, I didn't make my point very well the first time - sorry about that! –  Jason DeVito Nov 11 '12 at 19:01
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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?

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At first , Taylor Series is an approximation of any polynomial for a value between a and b , given the polynomial is differentiable in the closed interval between a and b .

If the equality is not always true , that exists some neighborhood of c inside a and b is not differentiable for the polynomial.

In this sense, the assumption that polynomial is differentiable in the closed interval between a and b fails. And hence, the coefficient of the terms in taylor series and the coucannot be found.

Simply not using this formula in the case is the solution.

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Unless I'm missing something why doesn't the following work.

Pick a $c\in[a,b]$. By assumption $g(x)=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!} (x-c)^n$ is a polynomial (I'm assuming this is suppose to mean that the series converges to a polynomial function on some nonzero size interval around $c$). This says that $g^{(k)}(c) = 0$ for $k>d$ where $d$ is the degree of that polynomial. The taylor series $g(x)=\sum\limits_{n=0}^d \cfrac{g^{(n)}(c)}{n!} (x-c)^n$, which is just a polynomial, has the same value as $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!} (x-c)^n$ on some interval and therefore the coefficients are equal. We conclude that $f^{(k)}(c)=g^{(k)}(c) = 0$ for $k>d$.

To show that $f(x)$ agrees with its expansion around $c$ consider the lagrange form of the remainder. $$f(x) = \sum_{n=0}^k \frac{f^{(n)}(c)}{n!} (x-c)^n + f^{(k+1)}(h)\frac{(x-c)^{(k+1)}}{(k+1)!}$$ where $h$ lies between $c$ and $x$ and $x\in [a,b]$. That equality holds for $x\in[a,b]$ is guaranteed since $f$ is $k+1$ times differentiable on $[a,b]$.

We choose $k$ so that $k+1>d$, this guarantees $f^{(k+1)}=0$ and simplifies the above to $$f(x) = \sum_{n=0}^k \frac{f^{(n)}(c)}{n!} (x-c)^n$$ where $x\in [a,b]$. Or in other words f is a polynomial.

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The problem is the assumption that $f^{(k+1)}(h)=0$, when you actually only know that $f^{(k+1)}(c)=0$. Note that $d$ depends on $c$. (Or rather, you have to prove that $d$ can be chosen independently of $c$.) (By the way: the series being a polynomial just means what you concluded, that for all $c$ there exists $d$ such that $f^{(k)}(c)=0$ for all $k>d$. Note that polynomials converge everywhere, so reference to "nonzero size interval" is unnecessary.) –  Jonas Meyer Dec 24 '11 at 3:56

The Taylor theorem states that every function which is differentiable over a space [a,b] can be rewritten as a polynomial. It doesn't mean the function was a polynomial at first place. Consider Sin(x), which is uniformly differentiable. The Taylor polynomial (for any space (a,b)) is actually the function itself, but sin(x) is not a polynomial.

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I think the OP means that the Taylor series is finite. –  Javier Badia Aug 23 '12 at 14:33
You're using the word polynomial with two different meanings, saying that $\sin x$ can be written as a polynomial, but saying it is not a polynomial. If something can be written as a polynomial, then it is a polynomial. If you want to distinguish, you could say $\sin x$ can be written as an infinite polynomial, but not a finite polynomial. But, without the adjective in front, polynomial will almost always be taken to mean finite polynomial. So, I don't think your answer adds anything. –  Graphth Sep 28 '12 at 20:20

The situation is not as complex as it seems.

Let $n$ be an nonnegative integer.

Let Coo denote infinitely differentiable.

For starters if $f(x)$ is a polynomial of degree $n$ then the expansion around every $c$ (or $x$) must be of degree at most $n$.

Since $f(x)$ is Coo everywhere taking the $n+1$ th derivative of $f$ at $c$ ( or $x$ ) must always give $0$.

For the same reason : taking the $n$ th derivative of $f$ at $x$ must always give the same value $a$.

Hence we get $f(x) = 0 + a + ...$

Analogue taking the $n+2$ th deriv must give $bx$ agreed upon for all $c$ because $f(x)$ is Coo.

Hence we get $f(x) = 0 + a + bx + ...$

(By induction) We can continue this and then end up with $f(x) = 0 + a + bx + cx^2 + ... + £ x^n$ or in other words the polynomial that equals $f(x)$.

Conclusion : same upperbound degree $n$ for every polynomial being the taylor expansion around any $c$ (or $x$) and $f$ being Coo (everywhere) is both sufficient and almost neccesary.

Actually it is only necc to be $C^{n+1}$ however when polynomial it means also Coo of course so actually these conditions are both sufficient and neccesary.

Thus the situation is completely understood.

QED.

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I didn't downvote, but I responded in a similar way on FOla Yinka's answer: The point is that the $n$ that works at each point is allowed to vary as $c$ varies. So it's not "same degree $n$ from every $c$", but rather, "at each $c$, some $n$". Further, there are plenty of $C^{\infty}$ functions which are not equal to their Taylor series at any point, so saying $f(x) = 0+a+....$ is not necessarily valid. –  Jason DeVito Nov 11 '12 at 18:24
Let me look at your argument line by line. First, the "if $f$ is a polynomial" line is irrelevant because it's assuming the conclusion. What you say in the next line, "Since $f$ is $C^\infty$ everywhere, taking the $n+1$st derivative must give $0$" is incomplete. What's assumed is that at each point $c$, there is an $n_c$ such that taking the derivative $n_c$ or more times, then plugging in $c$ gives $0$. What's not necessarily true is that $n_c =n_{d}$ for two different choices $c$ and $d$. This fact needs to be proven. (Continued) –  Jason DeVito Nov 17 '12 at 19:45
What could go wrong (before you prove the fact) is that at each $c$, $n_c$ is finite, but if you take the supremum of the $n_c$ over all $c$ you get $\infty$. Said another way, it could be that $f$ looks a like a degree one polynomial (in terms of derivatives at one point), a degree $2$ polynomial at another point, a degree $3$ at another point,...., so that overall, it doesn't look like it has finite degree. –  Jason DeVito Nov 17 '12 at 19:47
Could you point out exactly where you prove it? It looks like you just assume it starting with your third line. –  Jason DeVito Nov 17 '12 at 21:36
"if a function is a polynomial in an interval": this isn't relevant since we don't know $f$ is a polynomial in any interval whatsoever. (We cannot assume $f$ is analytic anywhere in its domain, so local Taylor polynomials needn't necessarily agree with $f$ beyond a single point.) This answer shows that if $f$ is a polynomial then its local Taylor polynomials have lesser-or-equal degrees and in fact are equal to $f$. However, this line of reasoning says nothing about what happens if we do not assume $f$ is a polynomial, which is what you would need to do to answer the original question. –  anon Nov 24 '12 at 12:47
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