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In mathematics, the Artin–Hasse exponential, introduced by , is the power series given by : ==Motivation== One motivation for considering this series to be analogous to the exponential function comes from infinite products. In the ring of formal power series Q :: where μ(n) is the Möbius function. This identity can be verified by showing the logarithmic derivative of the two sides are equal and that both sides have the same constant term. In a similar way, one can verify a product expansion for the Artin–Hasse exponential: :: So passing from a product over all ''n'' to a product over only ''n'' prime to ''p'', which is a typical operation in ''p''-adic analysis, leads from ''e''''x'' to ''E''''p''(''x''). The coefficients of ''E''''p''(''x'') are rational. We can use either formula for ''E''''p''(''x'') to prove that, unlike ''e''''x'', all of its coefficients are ''p''-integral; in other words, the denominators of the coefficients of ''E''''p''(''x'') are not divisible by ''p''. A first proof uses the definition of ''E''''p''(''x'') and Dwork's lemma, which says that a power series ''f''(''x'') = 1 + ... with rational coefficients has ''p''-integral coefficients if and only if ''f''(''x''''p'')/''f''(''x'')''p'' ≡ 1 mod ''p''Z''p'' A second proof comes from the infinite product for ''E''''p''(''x''): each exponent -μ(''n'')/''n'' for ''n'' not divisible by ''p'' is a ''p''-integral, and when a rational number ''a'' is ''p''-integral all coefficients in the binomial expansion of (1 - ''x''''n'')''a'' are ''p''-integral by ''p''-adic continuity of the binomial coefficient polynomials ''t''(''t''-1)...(''t''-''k''+1)/''k''! in ''t'' together with their obvious integrality when ''t'' is a nonnegative integer (''a'' is a ''p''-adic limit of nonnegative integers) . Thus each factor in the product of ''E''''p''(''x'') has ''p''-integral coefficients, so ''E''''p''(''x'') itself has ''p''-integral coefficients. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Artin–Hasse exponential」の詳細全文を読む スポンサード リンク
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