By Christopher G. Small

Many books were written at the conception of useful equations, yet only a few support readers resolve sensible equations in arithmetic competitions and mathematical challenge fixing. This e-book fills that hole. each one bankruptcy encompasses a checklist of difficulties linked to the coated fabric. those range in hassle, with the simplest being obtainable to any highschool pupil who has learn the bankruptcy rigorously. the main tricky will problem scholars learning for the foreign Mathematical Olympiad or the Putnam festival. An appendix presents a springboard for extra research of the options of limits, countless sequence and continuity.

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**Extra resources for Functional Equations and How to Solve Them (Problem Books in Mathematics)**

**Example text**

4, above. Let f : R → R be a continuous function satisfying 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]. a) Show that f (n x) = n f (x) for all real x. b) Deﬁne p = d − c. Show that f is bounded below on the interval [0, p]. ) c) Deﬁne the function g(x) = f (x) − f (p) x. p Prove that g is also bounded below on the interval [0, p] and satisﬁes Cauchy’s equation. d) Show that g is periodic with period p in the sense that g(x + p) = g(x) for all real x.

12) to the equation g(x + y) = g(x) + g(y) which is Cauchy’s equation. Therefore, there exists a constant a such that g(x) = a x for all real x. Letting f (0) = b, we see that the general form for f must be f (x) = a x + b . 4 Linear functional equation A plausible extension of both Cauchy’s and Jensen’s equations is f (a x + b y + c) = p f (x) + q f (y) + r . 15), and what is the general form of the solution? The reader who wishes a challenge can solve this problem directly, ﬁnding thereby that the class of functions is not particularly rich.

24 1 An historical introduction The reader should not be under any misapprehensions about the function f (x). While the functional equation looks complex, it has a very simple solution. 38) we see that f (x) = x+1 solves the equation. However, we cannot expect this solution to be the only solution to our functional equation. 37). 36), we see that there is a family of functional equations here, satisfying [f (x)]2 = a x + (n + a)2 + x f (x + n) with solution f (x) = x + n + a . 38) in Chapter 3.