## Taylor Series Remainder

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Suppose I have a $C^\infty$ smooth function $f$ defined on the reals.

I can apply Taylor's formula and get the local expression

$$f(x) = \sum_{i=0}^l\frac{f^{(i)}(0)}{l!}x^i+ f^{(l+1)}(\xi(x))x^{l+1}.$$

Question: Is the function $\xi$ smooth? The function $f$ can in principle be as nice as you want.

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2
This is a classical exercise in calculus classes. It is inappropriate to MO. – Denis Serre Sep 2 2011 at 13:14
Completely agree with @Denis – Igor Rivin Sep 2 2011 at 13:42

## 1 Answer

Note that the point $\xi$ in the expression of the remainder is not unique in general (as it is clear already for $l=0$). According to a common phenomenon, lack of unicity may cause a lack of continuity. For an example where there is no continuous $\xi$ (again for $l=0$) think of a smooth function $f$ which is positive and concave on $I:=(0,1)$; with $f(0)=f(1)=0$, with $f'(1) < 1$, and which is flat on an interval $J:=\{f'(x)=0\}\subset I$. Crossing the point $x_0=1$, the point $\xi(x)$ has to jump the interval $J$, causing a discontinuity at $x=1$. Note that in this example $f^{l+2}(\xi(x_0))=0$.

On the other hand, going back to the general situation, if you have a point $\xi_0$ for the expression of the remainder corresponding to $x_0\neq0$, and if $f^{(l+2)}(\xi_0)\neq0$, then the implicit function theorem applies, giving a smooth function $\xi$ in a nbd of $x_0$.

Finally note that any such function $\xi$ is certainly continuous at $x=0$, but it may have discontinuities in any nbd of $0$ even if $f$ is smooth (think of a proper version of the first example, with flat intervals accumulating at $0$).

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$f(x) = x (1-x)^2$ is a very simple example with $l=0$. $\lim_{x \to \infty} \frac{f(x)}{x} = \infty$ so you need $\xi(x) \to \infty$ as $x \to \infty$, but since $\frac{f(x)}{x} \ge 0$ for $x \ge 0$ it has to jump the interval $(1/3, 1)$ where $f' < 0$. – Robert Israel Sep 2 2011 at 19:44
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