Metodo di Esaustione (Quadrato)

01/05/202527/11/2021 di admin

Osserva

Il primo poligono è un quadrato
Il valore di $\Delta$ è necessario per il passo successivo
Il poligono successivo ha il doppio dei lati del precedente

	$l_n$ Lato	$p_n$ Perimetro	$\pi_n$ Pi greco	$a_n$ Apotema	$\Delta_n$ Delta
4	$\displaystyle l_4$	$\displaystyle 4\cdot l_4$	$\displaystyle \frac{p_4}{2 \cdot r}$	$\displaystyle \sqrt{r^2-\left(\frac{l_4}{2}\right)^2}$	$\displaystyle r-a_4$
8	$\displaystyle \sqrt{\left(\frac{l_4}{2}\right)^2+\Delta_4^2}$	$\displaystyle 8\cdot l_8$	$\displaystyle \frac{p_8}{2 \cdot r}$	$\displaystyle \sqrt{r^2-\left(\frac{l_8}{2}\right)^2}$	$\displaystyle r-a_8$
16	$\displaystyle \sqrt{\left(\frac{l_8}{2}\right)^2+\Delta_8^2}$	$\displaystyle 16\cdot l_{16}$	$\displaystyle \frac{p_{16}}{2 \cdot r}$	$\displaystyle \sqrt{r^2-\left(\frac{l_{16}}{2}\right)^2}$	$\displaystyle r-a_{16}$

Sostituendo il valore $\displaystyle l_4= \sqrt{2}\cdot r$

	$l_n$ Lato	$p_n$ Perimetro	$\pi_n$ Pi greco	$a_n$ Apotema	$\Delta_n$ Delta
4	$\displaystyle \sqrt{2}\cdot r$	$\displaystyle 4\cdot \sqrt{2}\cdot r$	$\displaystyle 2\cdot \sqrt{2}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2}\cdot r$	$\displaystyle \left(1-\frac{1}{2}\cdot\sqrt{2}\right)\cdot r$
8	$\displaystyle \sqrt{2-\sqrt{2}}\cdot r$	$\displaystyle 8 \cdot \sqrt{2-\sqrt{2}}\cdot r$	$\displaystyle 4 \cdot \sqrt{2-\sqrt{2}}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2+\sqrt{2}} \cdot r$	$\displaystyle \left(1-\frac{1}{2}\cdot\sqrt{2+\sqrt{2}}\right) \cdot r$
16	$\displaystyle \sqrt{2-\sqrt{2+\sqrt{2}}}\cdot r$	$\displaystyle 16 \cdot \sqrt{2-\sqrt{2+\sqrt{2}}}\cdot r$	$\displaystyle 8\cdot \sqrt{2-\sqrt{2+\sqrt{2}}}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot r$	$\displaystyle \left(1-\frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\right) \cdot r$

L’approssimazione di $\pi$ non dipende da $r$
$r=1$

	$l_n$ Lato	$p_n$ Perimetro	$\pi_n$ Pi greco	$a_n$ Apotema	$\Delta_n$ Delta
4	$\displaystyle \sqrt{2}$	$\displaystyle 4\cdot \sqrt{2}$	$\displaystyle 2 \cdot \sqrt{2}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2}$	$\displaystyle 1-\frac{1}{2}\cdot\sqrt{2}$
8	$\displaystyle \sqrt{2-\sqrt{2}}$	$\displaystyle 8 \cdot \sqrt{2-\sqrt{2}}$	$\displaystyle 4 \cdot \sqrt{2-\sqrt{2}}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2+\sqrt{2}}$	$\displaystyle 1-\frac{1}{2}\cdot\sqrt{2+\sqrt{2}}$
16	$\displaystyle \sqrt{2-\sqrt{2+\sqrt{2}}}$	$\displaystyle 16 \cdot \sqrt{2-\sqrt{2+\sqrt{2}}}$	$\displaystyle 8\cdot \sqrt{2-\sqrt{2+\sqrt{2}}}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}$	$\displaystyle 1-\frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}$
32	$\displaystyle \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$	$\displaystyle 32 \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$	$\displaystyle 16\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$	$\displaystyle \frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$	$\displaystyle 1-\frac{1}{2}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$

Approssimazioni

Osserva le approssimazioni successive di pi greco e una forma equivalente (produttoria di Viète)

Passo	#Lati	$\pi$	$\pi$	$\pi$
1	4	$\displaystyle 2 \cdot \sqrt{2}$	$\displaystyle 2 \cdot \frac{2}{\sqrt{2}}$	2,82842712474619
2	8	$\displaystyle 4 \cdot \sqrt{2-\sqrt{2}}$	$\displaystyle 2 \cdot \frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}$	3,06146745892072
3	16	$\displaystyle 8 \cdot \sqrt{2-\sqrt{2+\sqrt{2}}}$	$\displaystyle 2 \cdot \frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$	3,12144515225805
4	32	$\displaystyle 16\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$	…	3,13654849054594
5	64	$\displaystyle 32\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$	…	3,14033115695475
6	128	$\displaystyle 64\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}$	…	3,14127725093277
7	256	…	…	3,1415138011443
8	512	…	…	3,14157294036709
9	1024	…	…	3,14158772527716
10	2048	…	…	3,1415914215112
…	…	…	…	…

Approfondimenti

Wikipedia: Metodo di esaustione
Liu Hui, Zu Chongzhi