Approximate a known function by a sum of real power function

approximationregression

Given a known function $f(x):[a,b]\to \mathbb{R}$.
(For example, $f(x) = e^{-\ln^2 \left(3x-\frac{3}{2} \right)} \mathbb{I}_{\{\frac{1}{2}<x<2\}} $ over the support $[a,b] = \left[\frac{1}{2} ; 2 \right] $)

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I want to approximate the function $f(x)$, over its support $[a,b]$, by a sum of piecewise power function as follows
$$f(x) \approx \sum_{n=1}^N c_i x^{ r_i}\mathbb{I}_{\{a_i<x<b_i\}}$$
where $r_i, c_i,a_i,b_i \in \Bbb R$ and $a\leq a_i <b_i \leq b$ for $i=1,…,N$

The problem can be seen as:

Find the smallest $N$ such that there exists
$(r_i,c_i,a_i,b_i)_{i=1,..,N}$ satisfying
$$\underset{x\in[a,b]}{\text{max}} \left|f(x) – \sum_{n=1}^N c_i x^{r_i} \mathbb{I}_{\{a_i<x<b_i\}} \right| < \epsilon \tag{1} $$
with $\epsilon \in \Bbb R$ is given

Do you have any idea or reference for this problem?
Thank you in advance!

Best Answer

This is not exactly an answer but it is too long for a comment.

Function $$\large f(x)=e^{-\log ^2\left(3 x-\frac{3}{2}\right)}$$ shows a maximum at $x_*=\frac 56$.

Expanding as a Taylor series around $x_*$ is almost correct over a very limited range of $x$.

However, the $[2n,2n]$ Padé approximants $P_n$ do a pretty good job except very close to $\frac 12$. To give an idea, consider the norm $$\Phi_n=\int_{\frac{11}{20}}^2 \Big(f(x)-P_n\Big)^2\,dx$$

$$\left( \begin{array}{cc} n & \Phi_n\\ 1 & 7.4602\times 10^{-3} \\ 2 & 2.08463\times 10^{-5} \\ 3 & 1.72393\times 10^{-8} \\ 4 & 3.63413\times 10^{-12} \\ \end{array} \right)$$

For example, using $x=t+\frac 56$, $$\large P_2= \frac{1+6 t+\frac{1591 }{266}t^2-\frac{2409 }{266}t^3+\frac{19527 }{5320} t^4}{ 1+6 t+\frac{3985 }{266}t^2+\frac{4773}{266}t^3+\frac{7791}{760}t^4}$$

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