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By Bernadette Perrin-Riou

ISBN-10: 0821819461

ISBN-13: 9780821819463

Ordinarily, $p$-adic $L$-functions were created from advanced $L$-functions through precise values and Iwasawa concept. during this quantity, Perrin-Riou provides a concept of $p$-adic $L$-functions coming at once from $p$-adic Galois representations (or, extra mostly, from motives). This conception encompasses, specifically, a building of the module of $p$-adic $L$-functions through the mathematics conception and a conjectural definition of the $p$-adic $L$-function through its detailed values. because the unique ebook of this ebook in French (see ""Asterisque"" 229, 1995), the sphere has passed through major growth. those advances are famous during this English variation. additionally, a few minor advancements were made to the textual content.

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70. (a) First, prove that if k and l are two nonnegative integers, one has 1 lim n→∞ 0 n x k xl dx = 1 . (k + 1)(l + 1) Second, approximate f and g by polynomials. 24 1 Special Limits (b) Prove, by using the substitution n/x = y, that 1 n x f n 0 1 dx − 0 f (x)dx = n 1 0 f (t) ∞ n ∑ (k + t)2 − 1 dt k=n and calculate the limit of the last integral by using Lebesgue Convergence Theorem. 71. Follow the steps: Step 1. Prove that if n is an integer and k and m are two nonnegative integers, then n+1 n yk ({y})m dy = k ∑ p=0 k np .

Use that sin2 (nx) + cos2 (nx) = 1. 25. Observe that k a (n2 + kn + a)α n 2α 1 n 1 + n + n2 = 2β −1 · ∑ ∑ 2 β n k=1 n k=1 (n + kn + b) 1 + nk + nb2 α n β , and prove that if a, b, u ≥ 0 are real numbers and v > 0, one has k a 1 n 1 + n + n2 lim ∑ n→∞ n k=1 1 + k + b2 n n u v = 1 0 (1 + x)u−v dx. 26. Denote the value of the limit by Ln , use that fn (x) = exp( fn−1 (x) ln x), and apply Lagrange’s Mean Value Theorem to the exponential function to prove that Ln = −Ln−1 . 28. Use that 1/(i + j) = the order of summation.

However, n kα ∑ nβ n→∞ lim 1 n kα ∑ α n→∞ n k=1 n = · lim k=1 lim 1 n→∞ nβ −α −1 = 0. Case 3. α + 1 > β . )α , where the last inequality follows from √ the Arithmetic Mean–Geometric Mean Inequality. Using Stirling’s formula, n! ∼ 2π n(n/e)n , we get that limn→∞ xn = ∞, and the problem is solved. 16. The limit equals eα /p . Let (xn )n∈N be the sequence defined by n ∑ ln xn = 1+ k=1 α k p−1 . ,n ⎞ α k p−1 ≤ xn p p−1 k=1 n − k n ⎠∑ k ln 1 + n αp −k p−1 p−1 α k p−1 n p −k p−1 ⎞ α k p−1 . ,n ⎞ ⎛ ⎠=1 and p−1 lim ⎝ max n→∞ ⎞ k ln 1+ nαp −k p−1 ⎠ = 1.

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P-Adic L-Functions and P-Adic Representations by Bernadette Perrin-Riou

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