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By Hilborn R C

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H-ct R*f, where R*Rh operator. H, B Consider the equivalent equation E~. R(B) , then the iterative process E H, converges to a solution of the So the iterative process R*f, ho E H_ ct , converges in tion of the equation Bh = R*f H_ ct Bh n + h n = to the (unique) solu- and simultaneously to the solution of the equivalent equation Rh = f. 2. Investigation of the Scalar Equations 2. Investigation of the Scalar Equations 3S Here we prove Theorems 1-4 of Section 1. 1) is nonnegative definite, T > 0.

19) has no nontrivial solutions. z ( n-m ) . 9) hEiff,( -'2 n-m ). Hence 0 where = fD Rh a b h*dx = f1\ R(A)h(A) h*dp(A) • is the inner product in lR d . 10) By our assumption I. 10) it h(A) = 0 and hence o. 19) has a unique solution. 13). It remains to prove that the map R- l .. M ~(n-m) ... rtf-~(n-m) is continuous. proved that the map R: ft/ ... ft/. -~(n-m) surjective. As continuous. 1). Let some vector random process u = set) + net) of a filter with impulse response the useful signal, net) mean values zero.

X)V(j) , J x E are matrix coefficients of size J the function G(x) m, J, d x d. 19) are satisfied. It remains to prove that these conditions define + the vectors b:, c. J J uniquely. 19) has no nontrivial solutions. z ( n-m ) . 9) hEiff,( -'2 n-m ). Hence 0 where = fD Rh a b h*dx = f1\ R(A)h(A) h*dp(A) • is the inner product in lR d . 10) By our assumption I. 10) it h(A) = 0 and hence o. 19) has a unique solution. 13). It remains to prove that the map R- l .. M ~(n-m) ... rtf-~(n-m) is continuous.

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