
MathTake quite a large number of terms and use trial and error (bifurcation, say): eg, solve `1 + (x/(1!))^2 + (x^2/(2!))^2 + (x^3/(3!))^2 + (x^4/(4!))^2 2.3 = 0` Crudely, from the first two terms,...

MathEuler's product formula states that `prod_(p quad "prime") 1/(1p^(s)) = sum_(k=1)^infty 1/k^s` So if we have that `sum_(k=1)^infty 1/k^2 = prod_(p quad "prime")1/(1p^(s))` for `s in...