Assume we have a Taylor polynomial approximation to a smooth $n$ times differentiable function around $x_0$. Can we show any result for how the coefficients of the polynomial will change as $x_0$ changes?

We have:

$$P_{x_0}(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

But can we somehow measure how well and smoothly the coefficients $c_k$ change in

$$\sum_{k=0}^{n}c_kx^k$$ as $x_0$ starts moving around?

The only thing I have thought to attack the problem is to consider $x_0$ the variable and to do, maybe say, a Taylor polynomial on it?! But I think it would get rather convoluted to say the least.


It suffices to consider an example to illustrate the general rule.

Consider the function $$ f(x) = \begin{cases} 0 & x\leq 1 \\ (x-1)^3 & > 1 \end{cases} $$ This function is twice continuously differentiable, but its third derivative has a jump discontinuity at $x = 1$. Let us consider:

$P^{(0)}[f; x_0] = f(x_0)\cdot x^0$: the coefficient $c_0$ is twice continuously differentiable as a function of $x_0$.

$P^{(1)}[f; x_0] = f(x_0) \cdot x^0 + f'(x_0) \cdot (x^1 - x_0)$: in this case the coefficient $c_1 = f'(x_0)$ is only once continuously differentiable, and $$ c_0 = f(x_0) - x_0 f'(x_0) = \begin{cases} 0 & x_0 \leq 1 \\ - (1 + 2 x_0) (x_0 - 1)^2 & x_0 > 1\end{cases} $$ is also only once continuously differentiable, with a jump discontinuity in the second derivative when $x_0 = 1$.

$P^{(2)}[f; x_0] = f(x_0) \cdot x^0 + f'(x_0) \cdot (x^1 - x_0) + \frac12 f''(x_0) \cdot (x^2 - 2 x_0 x^1 + (x_0)^2)$: The coefficients are

  • $c_0 = f(x_0) - f'(x_0) x_0 + \frac12 f''(x_0) x_0^2$
  • $c_1 = f'(x_0) - f''(x_0) x_0$
  • $c_2 = \frac12 f''(x_0)$

and can be checked fairly easily that all three functions are merely continuous and not differentiable (as functions of $x_0$) at $x_0 = 1$.

In general: let $f$ be a function such that it is $n$ times differentiable at a point $x_0$ and not $n+1$ times differentiable there. Consider for some $k \leq n$ its Taylor polynomial $P^{(k)}[f; y]$. Then it is relatively easy to check (by induction) that the coefficients of the polynomial takes the form $$ c_\ell(y) = \sum_{m = \ell}^{k} \alpha_{k,\ell,m} f^{(m)}(y) y^{m - \ell} $$ where $\alpha_{k,\ell,m}$ are universal constants.

This shows immediately that,

  • $c_\ell(y)$ must be at least $(n - k)$ times differentiable as a function of $y$, at the point $x_0$, and
  • provided that $x_0 \neq 0$, $c_\ell(y)$ is exactly $(n-k)$-times differentiable at $x_0$ and no better.
  • if $x_0 = 0$, then we have that $c_\ell(y)$ is in fact $(n - \ell)$-times differentiable at $x_0$, but no better.

We are given an explicit formula for the $k$th coefficient when expanded at $y$, namely $f^{(k)}(y) / k!$. So it appears that you are asking: how does $f^{(k)}(y)/k!$ behave as a function of $y$?

The answer is that the smoothness of these coefficients is directly related to the smoothness of the function, and bounds on the rate of change of these coefficients amount to bounds on the derivatives of $f$. If the derivatives are wild and difficult to understand, then the behavior of the coefficients will be wild and difficult to understand.

If $f$ is only finitely many times, say $n$ times, differentiable, then the behavior of the $n$th coefficient is challenging to understand, as one can't understand its rate of change by analyzing $f^{(n+1)}(y)$.

  • $\begingroup$ Yes partly that is what I am asking but also some things happen as we do the $(x-x_0)^k$. It will distribute among the coefficients differently as $x_0$ changes. $\endgroup$ – mathreadler Oct 21 '17 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.