if we put x = sinφ we have

du= [1 − k²sin²φ]−1/2,

and if we call φ the amplitude of u, we may write φ = am(u), x = sin·am(u), which explains the origin of the notation sn(u). Similarly cn(u) is an abbreviation of cos·am(u), and dn(u) of Δam(u), where Δ(φ) meant (1 − k²sin²φ)1/2. The addition equation for each of the functions ƒ1(z), ƒ2(z), ƒ3(z) is very simple, being

ƒ(z + t) = ½ ( + ) log ƒ(z) + ƒ(t)= ƒ(z)ƒ′(t) − ƒ(t)ƒ′(z),
∂z ∂iƒ(z) − ƒ(t) ƒ²(z) − ƒ²(t)

where f1′(z) means dƒ1(z)/dz, which is equal to −ƒ2(z)·ƒ3(z), and ƒ²(z) means [ƒ(z)]2. This may be verified directly by showing, if R denote the right side of the equation, that ∂R/∂z = ∂R/∂t; this will require the use of the differential equation

[ƒ1′(z)]2 = [ƒ12(z) + e1 − e2] [ƒ12(z) + e1 − e3],

and in fact we find

( ∂2 ∂2) log [ƒ(z) + ƒ(t)] = ƒ2(z) − ƒ2(t) = ( ∂2 ∂2) log [ƒ(z) − ƒ(t)];
∂z2 dt2∂z2 dt2

hence it will follow that R is a function of z + t, and R is at once seen to reduce to ƒ(z) when t = 0. From this the addition equation for each of the functions sn(u), cn(u), dn(u) can be deduced at once; if s1, c1, d1, s2, c2, d2 denote respectively sn(u1), cn(u1), dn(u1), sn(u2), cn(u2), dn(u2), they can be put into the forms

sn(u1 + u2) = (s1c2d2 + s2c1d1) / D,cn(u1 + u2) = (c1c2 − s1s2d1d2) / D,dn(u1 + u2) = (d1d2 − k2s1s2c1c2) / D,