e^(π.i) - 1 = 0. How does this most famous equation in Math =0? Thanks for help.
One easy way of getting -1, is by means of an infinite series:
e^pi.i = 1 + 3.141592i - pi^2/2! - pi^3i/3! + pi^4/4! + pi^5i/5!..........and so on to infinity.
So, using just 40 terms in Wolfram/Alpha, with this summation formula, gives this result:
sum (πi)^n / (n!), n=0 to 39 ≈-1.0000000000000000000000000000937273907911 - 7.1836938113×10^-30 i
As you can see, the "real" part is closing in on -1, and the "imaginary" part is disappearing towards zero. 100 terms gives =-1 -10^(-110i)
Actually, it is incorrect the way have it!!. It is: e^(pi.i) = - 1, or e^(pi.i) + 1 =0.
One easy way of getting -1, is by means of an infinite series:
e^pi.i = 1 + 3.141592i - pi^2/2! - pi^3i/3! + pi^4/4! + pi^5i/5!..........and so on to infinity.
So, using just 40 terms in Wolfram/Alpha, with this summation formula, gives this result:
sum (πi)^n / (n!), n=0 to 39 ≈-1.0000000000000000000000000000937273907911 - 7.1836938113×10^-30 i
As you can see, the "real" part is closing in on -1, and the "imaginary" part is disappearing towards zero. 100 terms gives =-1 -10^(-110i)
