Polynomial approximations to $e^{f(x)}$

Question

Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is a bounded function on some compact subset of the real numbers, i.e. $|f(x)|\leq B$ for every $x$ in the domain $D = [-L,L]\cap \mathbb{R}$. For simplicity, it is sufficient to assume $f(x)$ is some degree $d$ polynomial.

There exist polynomial approximations to $e^{-\beta(1+x)}$ that provide an $\epsilon$-accurate approximation in the $L^2$ sense that scale as $d' = O(\sqrt{\beta\log(1/\epsilon)})$ where $d'$ is the degree of the approximating polynomial. How does $d'$ change if I replace $\beta(1+x)$ with $f(x)$?

Answer 1

Since $f(x)$ is a bounded polynomial in $[-L,L]$ with $|f(x)|\le B$, $f$ is analytic, and $e^{f(x)}$ is also analytic in $[-L,L]$, meaning it is equal to its Taylor expansion.

To see this, let $g(x)=e^{f(x)}$. Then $\forall x\in[-L,L]$: $|g(x)|\le e^B < \infty$. Moreover, if $f$ is a polynomial, then the $n$'th derivative of $g$ will look something like

$$g^{(n)} = e^f p(f^{(n)},f^{(n-1)}f',f^{(n-2)}f^{(2)},\dots,f')$$

Where $p$ is a polynomial such that the sum of derivatives in each term is $n$. For example

$$g^{(5)} = e^f \left(f^{(5)}+10f^{(3)}f^{(2)}+10f^{(3)}f'^2+10f'^3f^{(2)}+5f^{(4)}f'+15{f^{(2)}}^2f'+f'^5\right)$$

Since $f$ is a polynomial, eventually, most of these derivatives will be $0$. So $g^{(n)}$ is also bounded by some number $A^n$ on $[-L,L]$, let's say

$$\sup_{x\in[-L,L]} |g^{(n)}(x)| \le A^n$$

So the convergence radius of $g = e^f$ is

$$r = \frac{1}{\limsup_{n\to\infty}\sqrt[n]{\left|\frac{g^{(n)}(x_0)}{n!}(x-x_0)^n\right|}} \ge \frac{1}{\limsup_{n\to\infty}\sqrt[n]{\left|\frac{A^n}{n!}(2L)^n\right|}} \stackrel{\frac{(2AL)^n}{n!}\to 0}{=} +\infty$$

Therefore

$$e^{f(x)} = g(x) = \sum_{n=0}^\infty \frac{\frac{d^n}{dx}(e^{f(x_0)})}{n!}(x-x_0)^n$$

for any $x_0 \in [-L,L]$.

So the idea is to choose an $x_0 \in [-L,L]$ such that $e^{f(x_0)}$ is easily calculatable. For example

• if $f(x_0)=0$, then $e^{f(x_0)}=1$, or
• if $f(x_0)=1$, then $e^{f(x_0)}=e$, or
• if $f(x_0)=\ln(2)$, then $e^{f(x_0)}=2$, etc.

With the choice of $x_0 \in [-L,L]$ made, now we just need to caculate the individual derivatives of $e^{f(x)}$. WolframAlpha helps in this, for example here is the $5$'th derivative of $e^{f(x)}$.

The first few derivatives are: $$\begin{align} g(x)&=e^{f(x)} \\ g'(x)&=e^{f(x)}(f'(x)) \\ g^{(2)}(x)&=e^{f(x)}(f^{(2)}(x)+f'(x)^2) \\ g^{(3)}(x)&=e^{f(x)}(f^{(3)}(x)+3f'(x)f^{(2)}(x)+f'(x)^3) \\ g^{(4)}(x)&=e^{f(x)}(f^{(4)}(x)+4f^{(3)}(x)f'(x)+6f'(x)^2f^{(2)}(x)+3f^{(2)}(x)^2+f'(x)^4) \\ g^{(5)}(x)&=e^{f(x)}(f^{(5)}(x)+10f^{(3)}(x)f^{(2)}(x)+10f^{(3)}(x)f'(x)^2+10f(x)'^3f^{(2)}(x)+5f^{(4)}(x)f'(x)+15f^{(2)}(x)^2f'(x)+f'(x)^5) \end{align}$$

So for example, the $5$'th order polynomial approximation of $e^{f(x)}$ is the following:

• Choose $x_0\in[-L,L]$, such that $e^{f(x_0)}=c$ is easy to calculate.
• Then $$\begin{align} e^{f(x)} &\approx c[1+f'(x_0)(x-x_0)+\\ &\frac{1}{2}\left(f^{(2)}(x_0)+f'(x_0)^2\right)(x-x_0)^2+ \\ &\frac{1}{6}\left(f^{(3)}(x_0)+3f'(x_0)f^{(2)}(x_0)+f'(x_0)^3\right)(x-x_0)^3+ \\ &\frac{1}{24}\left(f^{(4)}(x_0)+4f^{(3)}(x_0)f'(x_0)+6f'(x_0)^2f^{(2)}(x_0)+3f^{(2)}(x_0)^2+f'(x_0)^4\right)(x-x_0)^4+ \\ &\frac{1}{120}\left(f^{(5)}(x_0)+10f^{(3)}(x_0)f^{(2)}(x_0)+10f^{(3)}(x_0)f'(x_0)^2+10f(x_0)'^3f^{(2)}(x_0)+5f^{(4)}(x_0)f'(x_0)+15f^{(2)}(x_0)^2f'(x_0)+f'(x_0)^5\right)(x-x_0)^5] \\ \end{align}$$