Let $R(z)$ and $Q(z)$ be polynomial functions:
$$R(z) = r_0 + r_1 \cdot z + \cdots + r_n \cdot z^{n}$$ $$Q(z) = 1 + q_1 \cdot z + \cdots + q_m \cdot z^{m}$$
From fundamental theorem of algebra, $Q(z)$ has $m$ roots $z_{j} \in \mathbb{C}$, which are approximated by $w_j$
My objective is to compute the numerical integral $J$
$$J \approx I = \int_{0}^{1} \dfrac{R(x)}{Q(x)}\ dx$$
To compute $J$, I use Partial Fraction Decomposition
$$I = \sum_{j=1}^{m}\alpha_{j} \cdot \ln \left|\dfrac{1-z_j}{z_j}\right|\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ J = \sum_{j=1}^{m}\alpha_{j} \cdot \ln \left|\dfrac{1-w_j}{w_j}\right|$$
Question: How do I estimate the error $\left|J - I\right|$ from the distances $\left|w_j-z_j\right|$?
In fact, my objective is to know when I should stop refining my roots $w_j$ such error $\left|J - I\right|$ remains in a certain tolerance.
Notes:
- $0 \le n < m$
- All the coefficients of $R$ and $Q$ are real
- $Q(x)$ has no root in $\left[0, \ 1\right] \subset \mathbb{R}$, so $I$ is well defined.
Description:
As there are many methods to compute $J\approx I$, I want to compare them.
To be fair, given a tolerance $\varepsilon_0$, I will compare the computational time such.
$$\left|J - I\right| < \varepsilon_0$$
So far I have three methods. Two of them I have an estimative of the error, unfortunatelly for the first one (this question) I don't have any.
Partial fraction: (This question)
- Find the $m$ roots of $Q(x)$
- Decompose using Partial fraction decomposition
- Analitic integrate
The error is related to how distant the numeric roots are from their exact roots.
Composite rule: Inspired by composite simpson's rule
- Divide the domain in many subdomain $\left[x_{k}, \ x_{k+1}\right]$
- Interpolate $f(x)$ into $P(x)$ (usually polynomial) on $\left[x_k, \ x_{k+1}\right]$
- Analitic integrate $P(x)$ on $\left[x_k, \ x_{k+1}\right]$
- Sum all of them.
The error is related to the degree of $P(x)$ and the mesh. By using $P(x)$ quadratic and $p$ equally spaced intervals $$\left|J - I\right| \le \dfrac{1}{180p^4} \max_{\xi \in \left[0, \ 1\right]} f^{(4)}(\xi)$$ Given the tolerance $\varepsilon_0$, I only need to select $$p = \left\lceil\left(\dfrac{1}{180\varepsilon_0}\max_{\xi \in \left[0, \ 1\right]} \left|f^{(4)}(\xi)\right|\right)^{1/4}\right\rceil$$
Taylor expansion: I can approximate $f(x)$ by its Taylor expansion $P(x)$ around $(1/2)$.
- Compute the coefficients of Taylor expansion
- Analitic integrate
The error is related to the higher-order term of $P(x)$: $$P(x) = \sum_{i=0}^{p} \dfrac{f^{(i)}\left(\frac{1}{2}\right)}{i!} \cdot \left(x-\dfrac{1}{2}\right)^i$$ $$\left|P(x) - f(x)\right| \le \dfrac{1}{(p+1)!} \left(x-\dfrac{1}{2}\right)^{p+1}\cdot f^{(p+1)}(\xi)$$ $$\left|J - I\right| \le \dfrac{1}{2^{p+1} \cdot (p+2)!} \cdot \max_{\xi \in \left[0, \ 1\right]} f^{(p+1)}(\xi)$$ Given the tolerance $\varepsilon_0$, I only need to select $p$ such $$\dfrac{1}{2^{p+1} \cdot (p+2)!} \cdot \max_{\xi \in \left[0, \ 1\right]} f^{(p+1)}(\xi) \le \varepsilon_0$$