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Bejing lectures in harmonic analysis : Proceedings of a by Stein E.M. (ed.)

By Stein E.M. (ed.)

The aim of this ebook is to explain a definite variety of effects regarding the learn of non-linear analytic dependence of a few functionals bobbing up clearly in P.D.E. or operator thought

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Extra resources for Bejing lectures in harmonic analysis : Proceedings of a summer school held in September 1984

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A. Abb. 13) ! 5 0 0 1 0 ! 1 1 ! 3 1 :: : 2 1 ! 2 2 ! 3 2 :: : :: : :: : ! 3 3 :: : Die Binomialkoeffizienten treten im binomischen Satz auf. a C b/ D k kD0 ! a C 0/ D k kD0 n ! a n k n 0 D 0 k ! a n 00 : Hier erweist sich nun die Festsetzung 00 D 1 als sinnvoll. Genauso: ! a C b/0 D 0 Wir beweisen den binomischen Satz durch vollständige Induktion. Für n D 0 ist die Behauptung offenbar richtig. Wir nehmen an, sie wäre für irgend ein n 0 richtig und schreiben: ! ! a C b/ 1 2 ! ! n n nC1 n 1 n 1 2 n a b C a b C :::Cab C D a 1 2 !

3 . 16 Man zeige für n 2 N: n X . 1/k kD0 ! a C b/ D k kD0 ! n an k bk : Wir setzen a D 1, b D 1 und bekommen für n 1: ! n n X X n 1n k . 1 1/n D 0 D . 1/k k kD0 kD0 ! 17 Man zeige für n 2 N: n X kD1 k ! n D n 2n k Wir formen um: ! n ! k 1/Š Nun lautet die Summe: n X kD1 k ! n X n n Dn k k kD1 ! n 1 X 1 n 1 Dn : 1 k kD0 Mit dem binomischen Satz bekommen wir: ! 1 C 1/n 1 D 2n 1 D k kD0 1 k 1k D ! 1 Begriff der Folge Eine Folge entsteht, wenn man jeder natürlichen Zahl eine reelle Zahl zuordnet.

1 n k n X n 1 D k kD0 k ! 6 Der Binomische Satz 35 Abb. 14 Summation einer Zeile im Pascalschen Dreieck 1 1– – 1 1 – 2 – 1 1 – 3 – 3 – 1 8 4 – 1 16 . . 1 – 4 – 6 . . 18 Wir zeigen für n 2 N und k 2 N0 (Abb. 15): n 1 X kCj k kD0 ! ! kCn kC1 k C :::C C D k k k 1 kCn D kC1 ! : Bei einem beliebigen k beweisen wir die Aussage durch vollständige Induktion über n. Für n D 1 gilt: ! ! nC1/ 1 kD0 kCj k ! 1 gilt die Behauptung. Dann bekommen wir mit ! ! kCn kCn kCn C D C k kC1 k ! n C 1/ kCnC1 : D D k k n 1 X kCj D k kD0 !

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