By Gert K. Pedersen

ISBN-10: 0387967885

ISBN-13: 9780387967882

ISBN-10: 3540967885

ISBN-13: 9783540967880

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**Example text**

E. that N ˆ is the Assume that N ˆ ≥ 1 which implies that N ˆ2 ≥ N ˆ. largest positive integer. Thus, N 2 ˆ is a positive integer and N ˆ is the largest posiHowever, since N 2 ˆ ˆ ˆ2 ≤ N ˆ ≤N ˆ2 tive integer, it follows that N ≥ N . Consequently, N ˆ ≤1≤N ˆ so that N ˆ = 1. 14) has a solution, then one can (correctly) prove that the largest positive integer ˆ = 1. Of course the issue is a point of logic, where a false asis N sumption can be used to prove a false conclusion. If one assumes an optimizer exists and it does not, then necessary conditions can be used to produce incorrect answers.

In fact, this problem falls outside of the classical calculus of variations and to solve it, one must use the modern theory of optimal control. The fundamental new ingredient is that the control function u(t) satisfies the “hard constraint” |u(t)| ≤ 1. In particular, u(t) can take values on the boundary of the interval [−1, +1]. 5 Problem 5: Optimal Control in the Life Sciences Although many motivating problems in the classical calculus of variations and modern optimal control have their roots in the physical sciences and engineering, new applications to the life sciences is a very active area of current research.

Thus, we define the function spaces C k (I) = C k (I; R1 ) by C k (I) = C k (I; R1 ) = x : I ⊆ R1 −→ R1 : x(·) is C k on I . 3 Let I denote an interval and assume that x : I → Rn is a vector-valued function. We say that the function x(·) = T x1 (·) x2 (·) · · · xn (·) is continuous at ˆt if, for each > 0, there is a δ > 0 such that if t ∈ I and 0 <| t − tˆ |< δ, then x (t) − x tˆ < . The function x(·) is said to be a continuous function if it is continuous at every point in its domain I. ✐ ✐ ✐ ✐ ✐ ✐ “K16538” — 2013/7/25 — 10:41 — ✐ 34 ✐ Chapter 2.

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Categories: Calculus