By Robert Dalmasso

ISBN-10: 2100050176

ISBN-13: 9782100050178

Gasquet C., Witomski P. examine de Fourier et purposes (Dunod, )(fr)(ISBN 2100050176)(C)(366s)

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Extra info for Analyse de Fourier et applications: exercices corrigés

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Zur Probe wird in der Bestimmungsgleichung die Variable x durch das Lösungselement ersetzt und die Identität nachgewiesen. ¾ Die Lösungsmenge L ist in der Definitionsmenge D enthalten. 3. Keine Lösung Für diese Aussageformen gibt es in der angegebenen Grundmenge keine Lösungselemente. ¾ Die Lösungsmenge ist leer: L = { }. ): Gleichwertigkeit Andere Äquivalenzrelationen: Ähnlichkeit, Kongruenz und Parallelität Die Terme sind sich selbst gleich; sie unterscheiden sich höchstens in ihrer Beziehungsweise.

Für x2 – 2x + 2 = 0. 5 Für die Lösungen x1, x2 ∈ R der normierten quadratischen Gleichung x2 + px + q = 0 gilt x1 + x2 = – p und x1 · x2 = q . Beweis ⎛ ⎜ p x1 + x2 = ⎜ − + ⎜ 2 ⎝ ⎛ ⎜− p + x1 ⋅ x2 =⎜ 2 ⎝ ⎞ ⎛ ⎞ 2 2 ⎟ ⎜ p ⎟ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎜ ⎟ − q ⎟ + ⎜ − − ⎜ ⎟ − q ⎟ = 2⎜ − ⎟ = − p ; ⎝2⎠ ⎝2⎠ ⎟ ⎜ 2 ⎟ ⎝ 2⎠ ⎠ ⎝ ⎠ ⎞ ⎛ ⎤⎞ ⎛ p ⎞2 ⎛ p ⎞2 ⎛ p ⎞2 ⎡⎛ p ⎞2 p ⎟ ⎜ ⎟ ⎜ ⎟ − q ⎟⎜− − ⎜ ⎟ − q =⎜− ⎟ −⎢⎜ ⎟ − q ⎥⎟= q . ⎝2⎠ ⎝2⎠ ⎝ 2⎠ ⎣ ⎝2⎠ 2 ⎢ ⎥ ⎦ ⎠⎝ ⎠ Linearfaktorenzerlegung Der Satz von Viëta dient einmal dazu, die Lösungen einer quadratischen Gleichung einer Probe zu unterziehen.

B B ► Beispiel Graphisch darzustellen ist die Punktmenge g: 3x – 4y + 6 = 0. Lösung Eine Äquivalenzumformung ergibt die explizite Form y = 3 4 3 x+2 ; Festlegung des Ordinatenschnittpunktes Sy (0 | 3 ) und Verifikation 2 („Verwirklichung“) der Steigung liefern die verlangte Gerade. Hinweis: Das Steigungsdreieck einzuzeichnen ist eher unüblich . 21 y = g(x) = 3 x + 3 mit x ∈ R. 15 Zeichnen Sie die Graphen nachstehender linearer Funktionen (D = R) in ein gemeinsames Koordinatensystem 1 5 3 b) f2(x) = − x +1 ; c) f3(x) = − x − .

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Analyse de Fourier et applications: exercices corrigés by Robert Dalmasso


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