Download Functional Equations and how to Solve Them by Christopher G. Small PDF

By Christopher G. Small

This ebook covers themes within the conception and perform of practical equations. targeted emphasis is given to tools for fixing sensible equations that seem in arithmetic contests, reminiscent of the Putnam pageant and the overseas Mathematical Olympiad. This e-book can be of specific curiosity to college scholars learning for the Putnam festival, and to school scholars operating to enhance their talents on arithmetic competitions on the nationwide and overseas point. arithmetic educators who educate scholars for those competitions will discover a wealth of fabric for education on useful equations difficulties. The ebook additionally offers a few short biographical sketches of a few of the mathematicians who pioneered the idea of useful equations. The paintings of Oresme, Cauchy, Babbage, and others, is defined in the context of the mathematical difficulties of curiosity on the time.

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3 has a number of variants. For example, Problem 1 at the end of the chapter asks the reader to solve Cauchy’s equation when the domain of f is restricted to be the nonnegative real numbers. A less immediate generalization is the following. 4. Let f : R → R satisfy Cauchy’s equation. Suppose in addition that there exists some interval [c, d] of real numbers, where c < d, such that f is bounded below on [c, d]. In other words, there exists a real number A such that f (x) ≥ A for all c ≤ x ≤ d. Then there exists a real number a such that f (x) = a x for all real numbers x.

So f (x) = f lim qi i→∞ = lim f (qi ) . i→∞ Similarly, g(x) = lim g(qi ) . i→∞ However, by assumption the functions f and g agree on all rational numbers. So f (qi ) = g(qi ) for all i = 1, 2, . .. It follows that f (x) = g(x) for all real x. 2 together gives us the result we need. 3. Let f : R → R be a continuous function satisfying Cauchy’s equation f (x + y) = f (x) + f (y) for all real values x and y. Then there exists a real number a such that f (x) = a x for all real numbers x. Proof. 1, we see that there exists a real number a such that f (q) = a q for all rational numbers q.

42) for all real values of x. 42) immediately implies that f (0) = 0. 41) together imply that f (n) = n for all integers n. We are part way there. All we need to do is fill in the gaps and prove this equation for noninteger values. As shown in future examples, extending such identities to all real values often involves using arguments involving limits or continuity. This problem is no exception. We next prove a statement that is weaker than the one we want. We show that if x ∈ [n, n + 1], then f (x) ∈ [n, n + 1].

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