Analyse mathematique III: Fonctions analytiques, by Roger Godement

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see Ce vol. III disclose l. a. th?orie classique de Cauchy dans un esprit orient? bien davantage vers ses innombrables utilisations que vers une th?orie plus ou moins compl?te des fonctions analytiques. On montre ensuite remark les int?grales curvilignes ? l. a. Cauchy se g?n?ralisent ? un nombre quelconque de variables r?elles (formes diff?rentielles, formules de style Stokes). Les bases de los angeles th?orie des vari?t?s sont ensuite expos?es, principalement pour fournir au lecteur le langage "canonique" et quelques th?or?mes importants (changement de variables dans les int?grales, ?quations diff?rentielles). Un dernier chapitre montre touch upon peut utiliser ces th?ories pour construire los angeles floor de Riemann compacte d'une fonction alg?brique, sujet rarement trait? dans los angeles litt?rature non sp?cialis?e bien que n'?xigeant que des suggestions ?l?mentaires. Un quantity IV exposera, outre,l'int?grale de Lebesgue, un bloc de math?matiques sp?cialis?es vers lequel convergera tout le contenu des volumes pr?c?dents: s?ries et produits infinis de Jacobi, Riemann, Dedekind, fonctions elliptiques, th?orie classique des fonctions modulaires et l. a. model moderne utilisant los angeles constitution de groupe de Lie de SL(2,R).

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DX n will serve as a template for differentiation with respect to any set of variables. What we called its message is this: If F is immediately a function of Xl, ... ,Xn , then its (perhaps partial) derivatives relative to other variables are linear combinations of the derivatives of the X j relative to those others, using the coefficients ::. J As a final application of the chain rule, we give the product rule. Theorem 3. If f (x) and g (x) are differentiable at b, then: (a) h(x) == f(x)g(x) is differentiable at b.

Assume that g' (x) is 0 throughout a connected open set. Then g is constant there. Proof. Suppose 0 is the set and a and b are in O. By Theorem 2, there is a polygonal path POPI . Pk from a to b within O. By hypothesis, g is differentiable along each segment P jP j+ I. Applying Theorem 1 to each segment, we have g(Pj+l) - g(pj) = Vg(x/). (Pj+1 - because Vg is 0 everywhere. Hence g(pj) we conclude that g(a) = g(b). pj) = 0, = g(Pj+I), j = 0,1, ... ,k - 1, and D Exercises 1. Find a place c on the segment ab at which f(b) - f(a) = V f(c).

We stated that in tXk ' the two differentiations are done in reading order. In our work, however, it turns out that the order does not matter. 4. 3. Theorem 2. Assume that each mixed partial derivative of f is defined near b and continuous at b. Then the mixed partials are symmetric; that is, Proof. Assume that the mixed partials are defined in N(b, E). 3) with comers at b, c == b+se j, d == c+tek. a == b+tek. where s2 + t 2 < E2 (so all the points are in N(b, E». The idea of the proof is the following.

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