#Please do not post homework
#Caroline Cote, Feb 15th, 2026, Assignment 7


#Problem 1:

HowManyDigits(JG1, 39);
#                              120
HowManyDigits(JG1, 35);
#                              108
HowManyDigits(JG1, 32);
#                              99
HowManyDigits(JG1, 33);
#                               102

#You would need to truncate the series after N = 33 to compute Pi to the 100 decimal digits.

#Problem 2

#JGcc1: from thesis page 6: = (9801 sqrt2)/(4Pi) 
JGcc1 := proc(N) local n,a:
a:= add(RF(1/2,n)*RF(1/4,n)*RF(3/4,n)/(n!^3)*(1/99)^(4*n)*(26390*n + 1103), n = 0..N):
(9801*sqrt(2))/(4*a):
end:

#JGcc2: from talk: = 32/Pi^2
JGcc2 := proc(N) local n, a:
a:= add(RF(1/2,n)^3*RF(1/4,n)*RF(3/4,n) / (n!^5) * (1/16)^n *(120*n^2 + 34*n + 3) , n = 0..N):
sqrt(32/a):
end:

#JGcc3: from thesis page 8: = 2sqrt2/Pi
JGcc3 := proc(N) local n, a:
a:= add((-1)^n / 2^(3*n) * RF(1/2,n)^3 / n!^3 *(6*n+1) , n = 0..N):
(2*sqrt(2))/a:
end:

#JGcc4: from thesis page 9: = 16sqrt3/(3Pi)
JGcc4:= proc(N) local n, a:
a := add((-1)^n / (2^(4*n)*3^n) * RF(1/2, n) *RF(1/4, n) *RF(3/4,n) / n!^3 * (28*n +3) , n = 0..N):
16*sqrt(3)/(3*a):
end:

#JGcc5: from thesis page 10: = 128 sqrt5/Pi^2
JGcc5:= proc(N) local n,a:
a:= add((-1)^n * RF(1/2,n)*RF(1/3,n)*RF(2/3,n)*RF(1/6,n)*RF(5/6,n) / (n!^5 *80^(3*n)) *(5418*n^2 + 693*n + 29) , n = 0..N):
sqrt(128*sqrt(5)/a):
end:

HowManyDigits(JGcc1, 100);
#                              806

HowManyDigits(JGcc2, 100);
#                              122

HowManyDigits(JGcc3, 100);
#                               91

HowManyDigits(JGcc4, 100);
#                              170

HowManyDigits(JGcc5, 100);
#                              576