# Power series in functions other than monomials

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.

• The Taylor polynomials are not guaranteed to approximate $f$. – Christian Remling Nov 9 '18 at 16:22
• no, but you can explicitly give an error bound. – Sascha Nov 9 '18 at 16:57
• 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 $N_k(x)=M_k(\arctan x)$ ). – Pietro Majer Nov 9 '18 at 17:32

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.