Anno 2008 Suppletiva – 10 – Trapezi

Pi greco

Tenuto conto che $\displaystyle \frac{\pi}{6}=\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\, dx$ si calcoli un’approssimazione di π, utilizzando uno dei metodi d’integrazione numerica studiati.

Metodo dei trapezi

\displaystyle \frac{h}{2}\cdot (y_0+ y_1)

n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{2}\cdot (y_0+ y_1)$
1	$\displaystyle \frac{1}{2}$	0	$0$	$1$		= $\displaystyle\frac{1}{4}\cdot\left(1+\frac{2\sqrt{3}}{3}}\right)$
		1	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$		= $\displaystyle\frac{1}{4}+\frac{\sqrt{3}}{6}}$
					Pi greco	= $6\cdot Area$
						= $\displaystyle 6\cdot \left(\frac{1}{4}+\frac{\sqrt{3}}{6}}\right)$
						= $\displaystyle\frac{3}{2}+\sqrt{3}$
						= 3,232051


n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+y_2)$
2	$\displaystyle \frac{1}{4}$	0	$0$	$1$		= $\displaystyle\frac{1}{8}\cdot\left(1+2\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)$
		1	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{4\sqrt{15}}{15}$		= $\displaystyle\frac{1}{8}+\frac{\sqrt{3}}{12}+\frac{\sqrt{15}}{15}$
		2	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$	Pi greco	= $6\cdot Area$
						= $\displaystyle 6\cdot \left(\frac{1}{8}+\frac{\sqrt{3}}{12}+\frac{\sqrt{15}}{15}\right)$
						= $\displaystyle\frac{3}{4}+\frac{\sqrt{3}}{2}+\frac{2\sqrt{15}}{5}$
						= 3,165219


n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+y_3)$
3	$\displaystyle \frac{1}{6}$	0	$0$	$1$		= $\displaystyle\frac{1}{12}\cdot\left(1+2\cdot\frac{6\sqrt{35}}{35}+2\cdot\frac{3\sqrt{2}}{4}+\frac{2\sqrt{3}}{3}\right)$
		1	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{6\sqrt{35}}{35}$		= $\displaystyle\frac{1}{12}+\frac{\sqrt{2}}{8}+\frac{\sqrt{3}}{18} + \frac{\sqrt{35}}{35}$
		2	$\displaystyle \frac{1}{3}$	$\displaystyle \frac{3\sqrt{2}}{4}$	Pi greco	= $6\cdot Area$
		3	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$		= $\displaystyle 6\cdot\left(\frac{1}{12}+\frac{\sqrt{2}}{8}+\frac{\sqrt{3}}{18} + \frac{\sqrt{35}}{35}\right)$
						= $\displaystyle\frac{1}{2}+\frac{3\sqrt{2}}{4}+\frac{\sqrt{3}}{3} + \frac{6\sqrt{35}}{35}$
						= 3,152196


n	h	i	x_i	y_i	Area	= $\displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+2\cdot y_3+ y_4)$
4	$\displaystyle \frac{1}{8}$	0	$0$	$1$		= $\displaystyle\frac{1}{16}\cdot\left( 1 + 2\cdot\frac{8\sqrt{7}}{21} + 2\cdot\frac{4\sqrt{15}}{15} +2\cdot \frac{8\sqrt{55}}{55} + \frac{2\sqrt{3}}{3}\right)$
		1	$\displaystyle \frac{1}{8}$	$\displaystyle \frac{8\sqrt{7}}{21}$		= $\displaystyle\frac{1}{16}+\frac{\sqrt{3}}{24}+\frac{\sqrt{7}}{21}+\frac{\sqrt{15}}{30}+ \frac{\sqrt{55}}{55}$
		2	$\displaystyle \frac{1}{4}$	$\displaystyle \frac{4\sqrt{15}}{15}$	Pi greco	= $6\cdot Area$
		3	$\displaystyle \frac{3}{8}$	$\displaystyle \frac{8\sqrt{55}}{55}$		= $\displaystyle 6\cdot\left(\frac{1}{16}+\frac{\sqrt{3}}{24}+\frac{\sqrt{7}}{21}+\frac{\sqrt{15}}{30}+ \frac{\sqrt{55}}{55}\right)$
		4	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{2\sqrt{3}}{3}$		= $\displaystyle\frac{3}{8}+\frac{\sqrt{3}}{4}+\frac{2\sqrt{7}}{7}+\frac{\sqrt{15}}{5} + \frac{6\sqrt{55}}{55}$
						= 3,147578