Aproximando la primera derivada usando \( x=\frac{1}{2} \) y \( h=\frac{1}{10} \) $$ \begin{equation*} f'(x) \approx a_{1} f(x+h) + a_{2} f(x) \end{equation*} $$
Reemplazando $$ \begin{equation*} f'(x) \approx a_{1} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) \end{equation*} $$
Usando \( f(x) = 1 \) $$ \begin{align*} a_{1} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) &= 0 \\ a_{1} + a_{2} &= 0 \end{align*} $$
Usando \( f(x) = x \) $$ \begin{align*} a_{1} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) &= 1 \\ \frac{3}{5} a_{1} + \frac{1}{2} a_{2} &= 1 \end{align*} $$
Formando un sistema de ecuaciones $$ \begin{equation*} \begin{bmatrix} 1 & 1 \\ \frac{3}{5} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} a_{1} \\ a_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{equation*} $$
Resolviendo $$ \begin{align*} a_{1} &= 10 \\ a_{2} &= -10 \end{align*} $$
Reemplazando $$ \begin{equation*} f'(x) = 10 f(x+h) - 10 f(x) \end{equation*} $$
Reordenando $$ \begin{equation*} f'(x) = \frac{f(x+h) - f(x)}{\frac{1}{10}} = \frac{f(x+h) - f(x)}{h} \end{equation*} $$
Aproximando la segunda derivada usando \( x=\frac{1}{2} \) y \( h=\frac{1}{10} \) $$ \begin{equation*} f''(x) \approx a_{1} f(x+2h) + a_{2} f(x+h) + a_{3} f(x) \end{equation*} $$
Reemplazando $$ \begin{equation*} f''(x) \approx a_{1} f \biggr( \frac{7}{10} \biggl) + a_{2} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) \end{equation*} $$
Usando \( f(x) = 1 \) $$ \begin{align*} a_{1} f \biggr( \frac{7}{10} \biggl) + a_{2} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) &= 0 \\ a_{1} + a_{2} + a_{3} &= 0 \end{align*} $$
Usando \( f(x) = x \) $$ \begin{align*} a_{1} f \biggr( \frac{7}{10} \biggl) + a_{2} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) &= 0 \\ \frac{7}{10} a_{1} + \frac{3}{5} a_{2} + \frac{1}{2} a_{2} &= 0 \end{align*} $$
Usando \( f(x) = x^{2} \) $$ \begin{align*} a_{1} f \biggr( \frac{7}{10} \biggl) + a_{2} f \biggr( \frac{3}{5} \biggl) + a_{2} f \biggr( \frac{1}{2} \biggl) &= 2 \\ \frac{49}{100} a_{1} + \frac{9}{25} a_{2} + \frac{1}{4} a_{2} &= 2 \end{align*} $$
Formando un sistema de ecuaciones $$ \begin{equation*} \begin{bmatrix} 1 & 1 & 1 \\ \frac{7}{10} & \frac{3}{5} & \frac{1}{2} \\ \frac{49}{100} & \frac{9}{25} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} \end{equation*} $$
Resolviendo $$ \begin{align*} a_{1} &= 100 \\ a_{2} &= -200 \\ a_{3} &= 100 \end{align*} $$
Reemplazando $$ \begin{equation*} f''(x) = 100 f(x+2h) - 200 f(x+h) + 100 f(x) \end{equation*} $$
Reordenando $$ \begin{equation*} f''(x) = \frac{f(x+2h) - 2 f(x+h) + f(x)}{\frac{1}{100}} = \frac{f(x+2h) - 2 f(x+h) + f(x)}{h^{2}} \end{equation*} $$