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Ergänzung zu der Abhandlung über die Entwicklung des Produktes ∏ n ( x ) = 1 ⋅ ( 1 + x ) ( 1 + 2 x ) ⋅ ⋅ ⋅ [ 1 + ( n − 1 ) x ] $$\mathop \prod \limits^n \left( x \right) = 1 \cdot \left( {1 + x} \right)\left( {1 + 2x} \right) \cdot \cdot \cdot \left[ {1 + \left( {n - 1} \right)x} \right]$$ nach den steigenden Potenzen von x

In: Gesammelte Mathematische Abhandlungen

Author

Listed:
  • H Hadwiger

Abstract

Zusammenfassung Im Beginn von IV dieser Abhandlung wird eine Form des Ausdrucks einer ganzen Funktion angenommen, wo der Koeffizient des Anfangsterms (Integrationskonstante bei der Integration einer Differenzengleichung) nur durch Induktion gefunden war. Am Ende von V, Gleichung (39), steht zwar die identische Gleichung, auf deren Beweis alles ankam; aber ich hatte wirklich vergessen, daß sie noch nicht bewiesen war, und sogar gesagt, der direkte Beweis möchte sehr schwer sein. Überhaupt führte damals der Versuch, durch Integration von Differenzengleichungen den Gegenstand zu erledigen, zu unnötiger Weitläufigkeit. Ich will nun den Fehler verbessern, indem ich zeige, daß die Sache weit einfacher ist.

Suggested Citation

  • H Hadwiger, 1953. "Ergänzung zu der Abhandlung über die Entwicklung des Produktes ∏ n ( x ) = 1 ⋅ ( 1 + x ) ( 1 + 2 x ) ⋅ ⋅ ⋅ [ 1 + ( n − 1 ) x ] $$\mathop \prod \limits^n \left( x \right) = 1 \cdot \left( {1 + x} \righ," Springer Books, in: Gesammelte Mathematische Abhandlungen, pages 136-141, Springer.
    • Handle: RePEc:spr:sprchp:978-3-0348-4117-7_3
    DOI: 10.1007/978-3-0348-4117-7_3
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