By V. I. Smirnov and A. J. Lohwater (Auth.)

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Additional resources for A Course of Higher Mathematics. Volume IV

Example text

In other words, the resolvent proves to be a rational or meromorphic function of λ throughout the complex λ plane. We obtain D(X) by replacing the integral in equation (33) by a finite sum. Strictly speaking, this replacement is impermissible, but none of the working that follows is meant to have the force of a proof and merely serves as a guide so that we can guess the form of Ό(λ). We divide the interval [a, 6] into n equal parts, the length of each of which will be δ = (6 — a)/n. We introduce the following notation for the points of division and for the values of the functions appearing in (33) at these points: 8i = a+ i ~ a ; // = /(*/);

X = xm. We now form a new sequence by taking the first function from sequence (I), the second function from sequence (II), the third function from sequence (III), and so on: /(l) (X) = fti (X), /(*> (X) = tf> (*), /(3) {X) = /£» {x)t . . , /(Π)(^)=/(η")(^··· (*) We show t h a t this subsequence is now convergent at any point x = xk. In fact, let us take the point x = xk. e. all the functions /<*> (x) = /^) (*), /(*+« (x) = yg^D ( a . ) , . . e. the sequence of functions (**) is convergent at the point x = xk.

All the real rational numbers also form a denumerable set. For, suppose we take zero as the first number, then write down the above sequence with the number of opposite sign interposed after each number of the sequence, we obtain: 0> 1) 1> 2, — 2, —y- , —, 3, 3, ~Ö~ , 2~ > · · · If we strike out certain terms from any sequence uv u2, . . in such a way t h a t an infinity of terms remains, these remaining terms again form an infinite sequence unv un%, . . and they can be re-enumerated. I t follows from this t h a t any part of a denumerable set containing an infinite set of elements is itself a denumerable set.