Use PYTHON

Pl could be approximated with the product of nested radicals:

Perhaps it might help to visualize the expression above as:

11 1 P Maybe in this other way:

2(0.5 sqrt(0.5) 2(0.5 sqrt(0.5+0.5 sqrt(0.5))) -2(0.5 sqrt(0.5+0.5 sqrt(0.5 +0.5 sqrt(0.5)))) #PYTHON CODE FILE: PI_nested_radicals.PY Created on Wed Feb 15 10:32:37 2023 @author: marco The operations have to be carried out starting with the inner radicals towards the outer ones. import math #variable to create the factors A=0.5 math.sqrt(0.5) P=1 # number of nested radicals N=10 for i in range(1,N.1): P=P*2*A #accumulator A=0.5*math.sqrt(0.5+A) # next factor PPI=2/P print("For",N,"nested radicals, PI=",PPI) Your job is to run the code, study it, understand it, and check if the number of radicals the code calls for is correct. In other words, if N=10 it effectively computes for 10 nested radicals.

Compute pi with nested radicals: https://mathworld.wolfram.com/PiFormulas.html Pl podra ser aproximado con el producto de radicales anidados: 2=2121+212121+2121+2121 Tal vez podra ayudar visualizar la expresin de arriba como: 2=2(2121)2(2121+2121)2(2121+2121+2121) Tal vez de esta otra manera: 2=2(0.5sqrt(0.5)2(0.5sqrt(0.5+0.5sqrt(0.5)))2(0.5sqrt(0.5+0.5sqrt(0.5+0.5sqrt(0.5)))) \# PYTHON CODE FILE:PI_nested_radicals.PY Created on Wed Feb 15 10:32:37 2023 @author: marco Las operaciones tienen que realizarse empezando con los radicales interiores hacia los exteriores. import math Tal vez de esta otra manera: 2=2(0.5sqrt(0.5)2(0.5sqrt(0.5+0.5sqrt(0.5)))2(0.5sqrt(0.5+0.5sqrt(0.5+0.5sqrt(0.5)))) \# PYTHON CODE FILE: PI_nested_radicals.PY Created on Wed Feb 15 10:32:37 2023 @author: marco Las operaciones tienen que realizarse empezando con los radicales interiores hacia los exteriores. import math A=0.5 math.sqrt(0.5) \# variable to create the factors P=1 N=10 \# number of nested radicals for i in range (1,N,1) : P=P2AA=0.5math.sqrt((0.5+A)#accumulator#nextfactor PPl=2/P print("For",N,"nested radicals, Pl=",PPI) Su trabajo es correr el codigo, estudiarlo, entenderlo y verificar si la cantidad de radicales que reclama el cdigo es la correcta. En otras palabras, si N=10 efectivamente calcula para 10 radicales anidados. Compute pi with nested radicals: https://mathworld.wolfram.com/PiFormulas.html PI podra ser aproximado con el producto de radicales anidados: 2=2121+212121+2121+2121 Tal vez podria ayudar visualizar la expresin de arriba como: Tal vez de esta otra manera: 2=2(0.5sqrt(0.5)2(0.5sqrt(0.5+0.5sqrt(0.5)))2(0.5sqrt(0.5+0.5sqrt(0.5+0.5sqrt(0.5)))) \# PYTHON CODE FILE: PI_nested_radicals.PY Created on Wed Feb 15 10:32:37 2023 @author:marco Las operaciones tienen que realizarse empezando con los radicales interiores hacia los exteriores. import math A=0.5 math.sart(0.5) H variable to create the factors P=1 Tal vez de esta otra manera: 2=2(0.5sqrt(0.5)2(0.5sqrt(0.5+0.5sqrt(0.5)))2(0.5sqrt(0.5+0.5sqrt(0.5+0.5sqrt(0.5)))) \# PYTHON CODE FILE: PI_nested_radicals.PY Created on Wed Feb 15 10:32:37 2023 @author: marco Las operaciones tienen que realizarse empezando con los radicales interiores hacia los exteriores. import math A=0.5 math. sqrt (0.5) P=1 N=10 for i in range (1,N,1) : P=P2A A=0.5 math sqrt (0.5+A) PPI=2/P print("For", N, "nested radicals, PPl=",PPI) \# variable to create the factors \# number of nested radicals \# accumulator \# next factor Su trabajo es correr el cdigo, estudiarlo, entenderlo y verificar si la cantidad de radicales que reclama el cdigo es la correcta. En otras palabras, si N=10 efectivamente calcula para 10 radicales anidados