By Julii A. Dubinskii

One provider arithmetic has rendered the 'Et moi, ..., si j'avait su remark en revenir, je n'y serais aspect aIle: ' human race. It has positioned logic again Jules Verne the place it belongs, at the topmost shelf subsequent to the dusty canister labelled 'discarded non- The sequence is divergent; hence we will be sense'. in a position to do whatever with it. Eric T. Bell O. Heaviside arithmetic is a device for suggestion. A hugely priceless device in a global the place either suggestions and non linearities abound. equally, every kind of elements of arithmetic function instruments for different elements and for different sciences. employing an easy rewriting rule to the quote at the correct above one unearths such statements as: 'One provider topology has rendered mathematical physics ... '; 'One provider good judgment has rendered com puter technological know-how .. .'; 'One provider classification thought has rendered arithmetic ... '. All arguably precise. And all statements accessible this manner shape a part of the raison d'elre of this sequence.

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**Example text**

Then such a function defines a regular exponential functional acting according to the formula (u(z), 4>(z)) Since u(x ~f f Jri u(x + iy)4>(x + iy)dx. 1) n + iy) is rapidly decreasing along the real axes, this formula is well-defined. 1) do not depend on y E Rn, so that this regular functional may be identified PD-OPERATORS WITH CONSTANT ANALYTIC SYMBOLS 43 with its real "trace" u( x). The last functional is a bounded linear functional on the corresponding space EXPn(Rn), whose elements are entire functions of real variables x having an exponential continuation onto the whole complex space e~.

1) where A(D) is a PD-operator with analytic symbol. 1. Let A(() be an analytic function in some Runge domain Q C ee' where A(() i' 0, ( E Q. 1) has one and only one solution u(z) E EXPn(e~). Moreover, I u(z) = A(D) h(z). PD-OPERATORS WITH CONSTANT ANALYTIC SYMBOLS 29 Proof In fact, we have already noted that if A«() i= 0 in n, then the operator B(D) == I/A(D) is the inverse of the operator A(D). The theorem is proved. Example 1 (A method of selecting quasipolynomial solutions). \ is a complex number and P(z) is a polynomial.

Example 1. Let a E C 1 and lal = U(O, z) 1. Consider the Cauchy problem = ¢(z). Clearly, the solution of this problem can be written in the form where ¢(z) E Exp(C;), that is, is a function of arbitrary exponential type (for this example n= q). ;atO. 9) and the integration is performed over the straight line from PD-OPERATORS WITH CONSTANT ANALYTIC SYMBOLS 33 In fact, using the Taylor expansion of