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φkDk.φk constant numbers an a power of D. Dn, meaning, is there a isomorphism X (from s onto s) such that Xφ(D)=DnX?. We prove that if φ(D) is equivalent to Dn, then φ(D) is of finite order, in fact a polynomial of degree n. The question of the equivalence of two differential operators of finite order in the space s is addressed too and solved completely when n=1.