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\title{Howework 4 -- Introduction to Computational Science}

\author{Claudio Maggioni}

\begin{document} \maketitle 
\section*{Question 1}
$$L_0(x) = \prod_{j = 0, j \neq 0}^n \frac{x - x_j}{x_i - x_j} = 
\frac{x - (-0.5)}{(-1) - (-0.5)} \cdot \frac{x - 0.5}{(-1) - 0.5} \cdot 
\frac{x - 1}{(-1) - 1} =  -\frac{2}{3}x^3 + \frac{2}{3}x^2 
+\frac{1}{6} x - \frac{1}{6}$$

$$L_1(x) = \prod_{j = 0, j \neq 1}^n \frac{x - x_j}{x_i - x_j} = 
\frac{x - (-1)}{(-0.5) - (-1)} \cdot \frac{x - 0.5}{(-0.5) - 0.5} \cdot 
\frac{x - 1}{(-0.5) - 1} = \frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}$$

$$L_2(x) = \prod_{j = 0, j \neq 2}^n \frac{x - x_j}{x_i - x_j} = 
\frac{x - (-1)}{0.5 - (-1)} \cdot \frac{x - (-0.5)}{0.5 - (-0.5)} \cdot 
\frac{x - 1}{0.5 - 1} = -\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}$$

$$L_3(x) = \prod_{j = 0, j \neq 3}^n \frac{x - x_j}{x_i - x_j} = 
\frac{x - (-1)}{1 - (-1)} \cdot \frac{x - (-0.5)}{1 - (-0.5)} \cdot 
\frac{x - 0.5}{1 - 0.5} = \frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}$$

$$p(x) = \sum_{i=0}^n y_i L_i(x) =
2 \cdot \left(-\frac{2}{3}x^3 + \frac{2}{3}x^2 
+\frac{1}{6} x - \frac{1}{6}\right) +
1 \cdot \left(\frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}\right) +
$$$$
0.5 \cdot \left(-\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}\right) + 
0.4 \cdot \left(\frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}\right) = -\frac{2}{5}x^3 + \frac{3}{5}x^2 - \frac{2}{5}x + \frac{3}{5}$$

$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} = 
\frac{\max_{x \in [-1,1]}|\frac{7680}{|2x+3|^6}|}{120} =
\max_{x \in [-1,1]}\frac{64}{|2x+3|^6} = 64$$

$$\max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
\left(x + \frac{1}{2}\right)(x+1)\right| = 
\max_{x \in [-1,1]} \left| x^4 - \frac{5}{4}x^2 + \frac{1}{4}\right| = \frac{1}{4}$$

$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 64 \cdot \frac{1}{4} = 8 \leq 8$$

The statement above is true so p satisfies the error estimate:

$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq 8$$
\end{document}
