Pi greco – Rettangoli al centro

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Pi greco

Considera un cerchio di raggio unitario con centro nell’origine.
Sapendo che l’area di uno dei 4 settori circolari è \pi / 4 calcola un valore approssimato di pi greco utilizzando uno dei metodi di integrazione di tipo geometrico.

Metodo dei rettangoli con altezze nel punto medio

nhi*xi*yi*Area= \displaystyle h\cdot (y_0^*)
110*\displaystyle \frac{1}{2}\displaystyle \frac{\sqrt{3}}{2}= \displaystyle 1\cdot\left(\frac{\sqrt{3}}{2}\right)
h/2= \displaystyle \frac{\sqrt{3}}{2}
\displaystyle \frac{1}{2}Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{\sqrt{3}}{2}\right)
= \displaystyle 2\sqrt{3}
= 3,4641…
nhi*xi*yi*Area= \displaystyle h\cdot (y_0^*+y_1^*)
2\displaystyle \frac{1}{2}0*\displaystyle \frac{1}{4}\displaystyle \frac{\sqrt{15}}{4}= \displaystyle \frac{1}{2}\left(\frac{\sqrt{15}}{4}+\frac{\sqrt{7}}{4}\right)
h/21*\displaystyle \frac{3}{4}\displaystyle \frac{\sqrt{7}}{4}= \displaystyle \frac{\sqrt{7}}{8} + \frac{\sqrt{15}}{8}
\displaystyle \frac{1}{4}Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{\sqrt{7}}{8} + \frac{\sqrt{15}}{8}\right)
= \displaystyle \frac{\sqrt{7}}{2}+\frac{\sqrt{15}}{2}
= 3,259…
nhi*xi*yi*Area= \displaystyle h\cdot (y_0^*+y_1^*+y_2^*)
3\displaystyle \frac{1}{3}0*\displaystyle \frac{1}{6}\displaystyle \frac{\sqrt{35}}{6}= \displaystyle\frac{1}{3}\cdot\left(\frac{\sqrt{35}}{6}+\frac{\sqrt{3}}{2}+\frac{\sqrt{11}}{6}\right)
h/21*\displaystyle \frac{1}{2}\displaystyle \frac{\sqrt{3}}{2}= \displaystyle\frac{\sqrt{3}}{6}+\frac{\sqrt{11}}{18}+\frac{\sqrt{35}}{18}
\displaystyle \frac{1}{6}2*\displaystyle \frac{5}{6}\displaystyle \frac{\sqrt{11}}{6}Pi greco= 4\cdot Area
= \displaystyle 4\left(\frac{\sqrt{3}}{6}+\frac{\sqrt{11}}{18}+\frac{\sqrt{35}}{18}\right)
= \displaystyle \frac{2\sqrt{3}}{3}+\frac{2\sqrt{11}}{9}+\frac{2\sqrt{35}}{9}
= 3,2064…
nhi*xi*yi*Area= \displaystyle h\cdot (y_0^*+y_1^*+y_2^*+y_3^*)
4\displaystyle \frac{1}{4}0*\displaystyle \frac{1}{8}\displaystyle \frac{3\sqrt{7}}{8}= \displaystyle\frac{1}{4}\cdot\left(\frac{3\sqrt{7}}{8}+\frac{\sqrt{55}}{8}+\frac{\sqrt{39}}{8}+\frac{\sqrt{15}}{8}\right)
h/21*\displaystyle \frac{3}{8}\displaystyle \frac{\sqrt{55}}{8}= \displaystyle \frac{3\sqrt{7}}{32}+\frac{\sqrt{15}}{32}+\frac{\sqrt{39}}{32}+\frac{\sqrt{55}}{32}
\displaystyle \frac{1}{8}2*\displaystyle \frac{5}{8}\displaystyle \frac{\sqrt{39}}{8}Pi greco= 4\cdot Area
3*\displaystyle \frac{7}{8}\displaystyle \frac{\sqrt{15}}{8}= \displaystyle 4\left(\frac{3\sqrt{7}}{32}+\frac{\sqrt{15}}{32}+\frac{\sqrt{39}}{32}+\frac{\sqrt{55}}{32}\right)
= \displaystyle \frac{3\sqrt{7}}{8}+\frac{\sqrt{15}}{8}+\frac{\sqrt{39}}{8}+\frac{\sqrt{55}}{8}
= 3,1839…
5\displaystyle \frac{1}{5}
\displaystyle \frac{1}{10}Pi greco…
6\displaystyle \frac{1}{6}
\displaystyle \frac{1}{12}Pi greco…
100\displaystyle \frac{1}{100}
\displaystyle \frac{1}{200}Pi greco= 3,1419…