Fundamentals of Neural Networks: Architectures, Algorithms, by Laurene V. Fausett

By Laurene V. Fausett

Supplying distinct examples of easy purposes, this new booklet introduces using neural networks. It covers basic neural nets for trend type; trend organization; neural networks in response to pageant; adaptive-resonance concept; and extra. For execs operating with neural networks.

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Additional resources for Fundamentals of Neural Networks: Architectures, Algorithms, and Applications

Example text

Assume that f is band-limited with [− 12 N ξ, 12 N ξ ], that is, max{|ξ | : f (ξ ) = 0} ≤ 12 N ξ and denote fk = f (k ξ ). We can recover f from f without any loss of information provided that it meets the Nyquist criterion: 1 1 = . ξ≤ N x FOV Hence, if ξ = 1/(N x), the DFT gives N/2−1 fn = fk e2π i(kn)/N , n=− k=−N/2 N N , . . , − 1. 2 2 Extending the sequence f = (f−N/2 , . . , f(N/2)−1 ) to the N-periodic sequence in such a way that fn+mN = fn , the discrete version of the Poisson summation formula is N/4−1 fn + fn+N/2 = f2k e2π i(2kn)/N , k=−N/4 n=− N N , .

2 2 If D 2 f (x) is a positive definite matrix, then, for a sufficiently small r, f (x) < f (x + h) + f (x − h) 2 for all |h| < r, Signal and System as Vectors 23 which leads to the sub-MVP f (x) < 1 |Br (x)| Br (x) f (y) dy. Similarly, the super-MVP can be derived for a negative definite matrix D 2 f (x). ⊂ Rn → R is a C 3 function and ∇f (x0 ) = 0. 2 Suppose f : 1. If f has a local maximum (minimum) at x0 , then the Hessian matrix D 2 f (x0 ) is negative (positive) semi-definite. 2. If D 2 f (x0 ) is negative (positive) definite, then f has a local maximum (minimum) at x0 .

2 Let A ∈ L(Rn , Rm ). Then • x∗ is called the least-squares solution of y = Ax if Ax∗ − y = infn Ax − y ; x∈R • x is called the minimum-norm solution of y = Ax if x† is a least-squares solution of y = Ax and † x† = inf{ x : x is the least-squares solution of y = Ax}. If x∗ is the least-squares solution of y = Ax, then Ax∗ is the projection of y on R(A), and the orthogonality principle yields 0 = Az, Ax∗ − y = zT (AT Ax∗ − AT y) If AT A is invertible, then for all z ∈ Rn . x∗ = (AT A)−1 AT y and the projection matrix on R(A) can be expressed as PA = A(AT A)−1 AT .

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