By Dan Rockmore

For one hundred fifty years the Riemann speculation has been the holy grail of arithmetic. Now, at a second while mathematicians are ultimately relocating in on an explanation, Dartmouth professor Dan Rockmore tells the riveting heritage of the quest for a solution.In 1859 German professor Bernhard Riemann postulated a legislation able to describing with an grand measure of accuracy the incidence of the major numbers. Rockmore takes us the entire means from Euclid to the mysteries of quantum chaos to teach how the Riemann speculation lies on the very middle of a few of the main state-of-the-art learn happening this present day in physics and arithmetic.

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**Extra resources for Stalking the Riemann Hypothesis: The Quest to Find the Hidden Law of Prime Numbers**

**Sample text**

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].