Vol . 25, 1982 Two iterative functional equations for power series 241

So in the following we only have to consider those solutions Ф(д:, z) of (1) for which Ф(х,0) = 0.

THEOREM 2. Let 0(x,z)G Klx, zj be a solution with Ф(х,0) = 0 and Фу^ 0. Then

( a ) Ф(x,г) = exp(a.дc).z+2.^2ü.(x).z^ where aGK, ехр(д:) = S.^oJc"/«!, v,ix)GKlxl ü.(0) = 0, />2.

( b ) If a/0, there exists an invertible r(z)GK[zI such that

Фix , z ) =T ( exp { a . x ) . T - \z ) У

Proof Since Ф(х,0) = 0, we have Ф{x,z) = 2,^, v,{x)z\ v,(x)E Klxl From (1) it follows that

E t^.(^) (S v,(y)zn = 2 ü.(x +y)z'. (2)

This is only possible if ü,(jc)7^0 and

^ix ) 'V , { y ) =v , ix - \ - yl hence

üi ( x ) = exp(a.jc), a EK

Since ф(0,z) = z (see Lemma II.1) we have

i ; . ( 0 ) = 0, i^2.

Suppose now that a^O. Therefore i;,(jc)'- i;,(x)7^0 for />2. From (2) one deduces that

Hx + y ) = Vr{x),V2iy)-\-V2(x),Vr{y)\ V2(y-^x) = v,(y).V2{x)-^V2{y).v^(x)\ hence

Mx ) . ( v , { yy - v , ( y ) ) =V2 { y ) . { v , { xf - v , ( x ) % ^nd therefore

Hx ) = d.{v,(xy-v,(x)% dSK.