# Download Elliptic Functions and Rings of Integers by P Cassou-Nogues, Taylor PDF

By P Cassou-Nogues, Taylor

Книга Elliptic services and jewelry of Integers Elliptic services and jewelry of Integers Книги Математика Автор: Ph Cassou-Nogues Год издания: 1987 Формат: djvu Издат.:Birkhauser Страниц: 198 Размер: 3,1 ISBN: 0817633502 Язык: Английский0 (голосов: zero) Оценка:Elliptic services and jewelry of Integers

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Example text

Let R(x) be the ratio between P and Q: Q(x; y0 , y1 , . . , yn )= P x + 1; xy0 , xy1 + y0 , . . , xyn + ny(n−1) = R(x)P (x; y0 , y1 , . . , yn ). Suppose R(x) has a zero, say γ. Substitute γ into P (γ + 1; γy0 , γy1 + y0 , . . , γyn + ny(n−1) ) = 0 · P (x; y0 , y1 , . . , yn ) = 0. This last equality indicates that (z −(γ +1)) is a factor of P , contradicting the assumption that P was of minimal degree. ,hn ) (x) · (y0 )h0 · . . ,hn ) (x + 1) · (y0 )h0 · . . ,hn ) (x + 1). This equality cannot be satisfied for arbitrary x if R ≡ c, where c is constant.

6 holds for 1 < x ≤ 2 as well; repeatedly applying this procedure shows that it applies for all x > 0, as required. 1) is known as Euler’s second integral. 1. Beta Function. For Re(x), Re(y) > 0, define ˆ 1 B(x, y) := tx−1 (1 − t)y−1 dt 0 ˆ π/2 =2 sin(t)2x−1 cos(t)2y−1 dt 0 Γ(x)Γ(y) = = B(y, x). Γ(x + y) 47 48 This integral is commonly known as the Beta function. The definition above involves three equivalent identities – an integral over trigonometric functions, an integral over polynomials and a ratio of Gamma functions.

4. (H¨ older’s inequality). Let p and q be positive real numbers such that p1 + 1q = 1. Then for any integrable functions f, g : [a, b] → R, we have ˆ ˆ b 1/p b ˆ q |g| dx |f | dx f (x)g(x) dx ≤ . 4) Proof. 3 is satisfied. 3 as well as the Cauchy-Schwarz inequality, we have in the limit → 0+ , ˆ ˆ b f · g dx ≤ b− f · g dx + f (b)g(b) a a ≤ ´ b− a 1/p |f |p dx ´ b− a 1/q |g|q dx + |f (b)| · |g(b)| · 40 ˆ ˆ 1/p b p q |f | dx − |f (b)| · = 1/q b p q |g| dx − |g(b)| · a a + |f (b)| · |g(b)| · = ´b a 1/p p |f | dx ´b a 1/q 1/p q |g| dx 1− |f |p · ´b p a |f | dx 1/q 1− |g|q · ´b q a |g| dx + |f (b)| · |g(b)| · ≤ ´b a 1/p p |f | dx ´b a 1/q q |g| dx 1− + |f (b)| · |g(b)| · 1/p ˆ ˆ b b p = |f | dx |g|q dx a |f ||g| ´ ´ ( ab |f |p dx)1/p ( ab |g|q dx)1/q 1/q a and the proof is complete.