Pi greco – Trapezi

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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 trapezi

nhixiyiArea= \displaystyle \frac{h}{2}\cdot (y_0+y_1)
11001= \displaystyle \frac{1}{2}\cdot(1+0)
110= \displaystyle \frac{1}{2}
Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{1}{2}\right)
= 2,0
nhixiyiArea= \displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+y_2)
2\displaystyle \frac{1}{2}001= \displaystyle \frac{1}{4}\cdot \left(1+2\cdot \frac{\sqrt{3}}{2}+0\right)
1\displaystyle \frac{1}{2}\displaystyle \frac{\sqrt{3}}{2}= \displaystyle \frac{1}{4}+\frac{\sqrt{3}}{4}
210Pi greco= 4\cdot Area
= \displaystyle 4\cdot\left(\frac{1}{4}+\frac{\sqrt{3}}{4}\right)
= 1+\sqrt{3}
= 2,732…
nhixiyiArea= \displaystyle \frac{h}{2}\cdot (y_0+2\cdot y_1+2\cdot y_2+y_3)
3\displaystyle \frac{1}{3}001= \displaystyle \frac{1}{6}\cdot\left(1+ 2\cdot \frac{2\sqrt{2}}{3}+2\cdot \frac{\sqrt{5}}{3}+0 \right)
1\displaystyle \frac{1}{3}\displaystyle \frac{2\sqrt{2}}{3}= \displaystyle \frac{1}{6}+\frac{2\sqrt{2}}{9}+\frac{\sqrt{5}}{9}
2\displaystyle \frac{2}{3}\displaystyle \frac{\sqrt{5}}{3}Pi greco= 4\cdot Area
310= \displaystyle 4\cdot\left(\frac{1}{6}+\frac{2\sqrt{2}}{9}+\frac{\sqrt{5}}{9}\right)
= \displaystyle \frac{2}{3}+\frac{8\sqrt{2}}{9}+\frac{4\sqrt{5}}{9}
= 2,91755…
nhixiyiArea= \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}{4}001= \displaystyle \frac{1}{8}\cdot\left(1+2\cdot\frac{\sqrt{15}}{4}+2\cdot\frac{\sqrt{3}}{2}+2\cdot\frac{\sqrt{7}}{4}+0\right)
1\displaystyle \frac{1}{4}\displaystyle \frac{\sqrt{15}}{4}= \displaystyle \frac{1}{8}+\frac{\sqrt{3}}{8}+\frac{\sqrt{7}}{16}+\frac{\sqrt{15}}{16}\right)
2\displaystyle \frac{1}{2}\displaystyle \frac{\sqrt{3}}{2}Pi greco= 4\cdot Area
3\displaystyle \frac{3}{4}\displaystyle \frac{\sqrt{7}}{4}= \displaystyle 4\cdot\left(\frac{1}{8}+\frac{\sqrt{3}}{8}+\frac{\sqrt{7}}{16}+\frac{\sqrt{15}}{16}\right)
410= \displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{4}+\frac{\sqrt{15}}{4}\right)
= 2,9957…