# Analytic Number Theory [lecture notes] by Jan-Hendrik Evertse

By Jan-Hendrik Evertse

Similar number theory books

The Atiyah-Patodi-Singer Index Theorem

According to the lecture notes of a graduate path given at MIT, this refined therapy results in a number of present learn issues and should certainly function a advisor to additional reviews.

Zero to Lazy Eight: The Romance Numbers

Did you ever ask yourself why a sew in time saves 9 and never, say, 4, or why the quantity seven is taken into account the luckiest, or how many the notice googol refers to? good, the Humez brothers, besides Joseph Maguire, have replied all of those questions and extra. In "Zero to Lazy Eight", they take us on a wacky and enlightening journey up the linguistic quantity scale from 0 to 13 and again when it comes to infinity, exhibiting us simply what numbers can let us know approximately our culture's prior, current, and destiny.

Extra resources for Analytic Number Theory [lecture notes]

Example text

27 and assume in addition that fn = −1 on U for every n 0. Then for the function F = ∞ n=0 (1 + fn ) we have ∞ F fn = . F 1 + f n n=0 Proof. Let Fm := m n=0 (1 + fn ). Then Fm → F uniformly on every compact subset of U . 28. 57 Chapter 3 Dirichlet series and arithmetic functions An arithmetic function is a function f : Z>0 → C. To such a function we associate its Dirichlet series ∞ f (n)n−s Lf (s) = n=1 where s is a complex variable. It is common practice (although this doesn’t make sense) to write s = σ + it, where σ = Re s and t = Im s.

Proof of π(x) 2x/ log x. Let again n = [x]. Since t/ log t is an increasing function of t, it suffices to prove that π(n) 2 · n/ log n for all integers n 3. We proceed by induction on n. It is straightforward to verify that π(n) 2 · n/ log n for 3 n 200. Let n > 200, and suppose that π(m) 2 · m/ log m for all integers m with 3 m < n. If n is even, then we can use π(n) = π(n − 1) and that t/ log t is increasing. Assume that n = 2k + 1 is odd. 6, we have 2k + 1 k p (k + 2)π(2k+1)−π(k+1) . 8, this leads to (k + 2)π(2k+1)−π(k+1) 22k .

Let ε > 0. Then there is N such that |fn (z) − fm (z)| < ε for all z ∈ K, m, n N . Choose m N . Then there is C > 0 such that |fm (z)| C for z ∈ K since fm is continuous. Hence |fn (z)| C + ε for z ∈ K, n N . 24. 26. let U ⊂ C be a non-empty open set, and {fn : U → C}∞ n=0 a sequence of analytic functions, converging to a function f pointwise on U and uniformly on every compact subset of U . Then fn (z) f (z) = n→∞ fn (z) f (z) lim for all z ∈ U with f (z) = 0, where the limit is taken over those n for which fn (z) = 0.