I would like to understand how approximations by monomials and approximations by other kinds of functions are related which I illustrate with an example.

Consider the interval $[-\pi,\pi]$ let's say.

The Stone Weierstrass theorem tells us that the linear hull of functions

$$f_n(x)=(\arctan(x))^n$$ for $n \in \mathbb N_0$ is dense in the continuous functions on $[-\pi,\pi].$ So it makes sense to use these functions to approximate continuous functions on this interval.

On the other hand, we know that for the smaller class of $C^{\infty}$ functions on $(-\pi,\pi)$ Taylor's formula holds, i.e. for $f\in C^{\infty}$

$$ f(x) = \sum_{i=0}^k a_i x^i + M_k$$ and $M_k$ can be explicitly estimated using the mean value theorem.

Now I would like to understand whether one can get also a Taylor-type formula

$$f(x) = \sum_{i=0}^k b_i f_i(x) + N_k$$

such that $N_k$ can be explicitly estimated?

I figured out already a way to compute the coefficients $b_i$ in that case.

Namely, since $f_n$ has a Taylor expansion $$\arctan(x)^n=x^n(1-\frac{nx^2}{3}+\mathcal O(x^4))$$

you can recursively express the $a_i$ in terms of $b_i.$ The problem is that I do not know how to estimate the error $N_k$ this way.

Thus, I would like to understand whether one can get for $f \in C^{\infty}[-\pi,\pi]$ a nice explicit power series expansion in terms of functions $f_i$ with full control on the error.

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    $\begingroup$ The Taylor polynomials are not guaranteed to approximate $f$. $\endgroup$ Nov 9, 2018 at 16:22
  • $\begingroup$ no, but you can explicitly give an error bound. $\endgroup$
    – Sascha
    Nov 9, 2018 at 16:57
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    $\begingroup$ Isn't what you are doing, the standard Taylor expansion of $f(\tan s)$, say $f(\tan s)=\sum_{i=0}^k b_is^i+M_k(s)$, where we then put $\tan s=x$ (So $f(x)=\sum_{i=0}^k b_i(\arctan x)^i+M_k(\arctan x)$ and $N_k(x)=M_k(\arctan x)$ ). $\endgroup$ Nov 9, 2018 at 17:32

1 Answer 1


Suppose you want to expand the function $f$ on an interval $I$ into powers of another function $h$ on $I$, which is invertible and maps $I$ onto an interval $J$. Let $H:=h^{-1}$ and $g:=f\circ H$, so that $f=g\circ h$. Thus, you want to expand $g(u)$ into powers of $u-u_0$ for some $u_0$ in the interior of $J$. The problem is to estimate the remainder, which reduces to bounding the $n$th derivative of $g$. If $g$ is analytic in an open disc $D\subset\mathbb C$ centered at $u_0$ and we can bound $g$ on $D$, then we can use the Cauchy integral formula $$ g^{(n)}(u)=\frac{n!}{2\pi i}\int_\gamma\frac{g(z)\,dz}{(z-u)^{n+1}} $$ for (say) a circle $\gamma\subset D$ to bound $g^{(n)}$ in a neighborhood of $u_0$, as desired.


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