Suppose you have a table of the logarithm function $\ln x$ for positive integer values of $x$, and you compute $\ln 11.1$ by quadratic interpolation at $x_0 = 10, x_1 = 11, x_2 = 12$.
Estimate the relative error incurred.
Is this problem is about calculation of Error bound?
The goal is to interpolate three data points using quadratic interpolation for the natural log function using the three data points $x = 10, 11, 12$, yielding the data set:
$$(x, y) = (x, \ln x) = (10 , \ln 10), (11, \ln 11), (12 \ln 12)$$
We will use the following 10-digits of precision for these calculations:
$$(x, y) = (10 , 2.302585093), (11, 2.397895273), (12 , 2.484906650)$$
We want to interpolate these three data points with the quadratic:
$$y(x) = a_2 x^2 + a_1 x + a_0$$
In order to do this, we want to solve the linear system for $a_0, a_1, a_2$: $$a_2 x_1^2 + a_1x_1 + a_0 = y_1 \\ a_2 x_2^2 + a_1x_2 + a_0 = y_2 \\ a_2 x_3^2 + a_1x_3 + a_0 = y_3$$
In matrix form, we can show this as:
$$ \begin{bmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$$
We can now apply Gaussian Elimination (or whatever floats your boat) to solve for $a_0, a_1, a_2$, so we have:
$$ \begin{bmatrix} 1 & 10 & 100 \\ 1 & 11 & 121 \\ 1 & 12 & 144 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} 2.302585093 \\ 2.397895273 \\ 2.484906650 \end{bmatrix}$$
This yields:
$$a_0 = 0.893049 , a_1 = 0.182448 , a_2 = -0.0041494 $$
This gives us the quadratic interpolating polynomial of:
$$y(x) = -0.0041494 x^2 + 0.182448 x + 0.893049 $$
Using the data to compare, we arrive at:
$$\begin{array}{c|c|c|c} \text {x} & y(x) & \ln x & |y(x) - \ln x)| \\ \hline 10 & 2.30259 & 2.302585093 & 3.907005953873721 \times 10^{-6} \\ \hline 11 & 2.3979 & 2.397895273 & 4.327201629017452 \times 10^{-6} \\ \hline 11.1 & 2.40697 & 2.40695 & 0.0000291177 \\ \hline 12 & 2.48491 & 2.484906650 & 4.750211999748899 \times 10^{-6} \end{array}$$
To calculate the relative error at $x = 11.1$, we have:
$$ \mbox{Relative Error} = \left|\dfrac{\mbox{calculated - actual}}{\mbox{actual}}\right| = \left|\dfrac{2.40697 - 2.40695}{2.40695}\right| = 8.309271069065387 \times 10^{-6}$$
