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By William M. Boothby

Nice introductory differential geometry textual content! I used this publication to assist me move my qualifying examination. Yay Boothby!

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Read or Download An Introduction to Differentiable Manifolds and Riemannian Geometry, Revised, Volume 120, Second Edition (Pure and Applied Mathematics) PDF

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Extra resources for An Introduction to Differentiable Manifolds and Riemannian Geometry, Revised, Volume 120, Second Edition (Pure and Applied Mathematics)

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Von einer Funktion f sind zwei Funktionswerte f (1) = −1 und f (3) = 2 bekannt. Finden Sie eine Gerade g, die an den Stellen x = 1 und x = 3 mit f u ¨bereinstimmt. 6. Sie u ¨bernehmen ein Auto mit leerem Tank. Nach dem Tanken befinden sich 40 Liter im Tank. Nach 200 km steht die Tankuhr auf 34 “. Wo wird sie nach 300 ” km stehen? Nach wie vielen Kilometern m¨ ussen Sie sp¨ atestens wieder tanken? 2 7. Skizzieren Sie f (x) = (x − 3) grob ohne Wertetabelle, indem Sie u ¨berlegen: Welche Gestalt hat die Funktionskurve, wo sind die Nullstellen, was ist der Wert f (0)?

C) 3 = 1 + e−x , daraus 2 = e−x und damit x = − ln(2). 4. Wir logarithmieren beide Seiten der Ungleichung (da der Logarithmus eine streng monotone Funktion ist, bleibt dabei die Richtung des Ungleichungszeichens erhalten): 1 1 ). n · ln( ) < ln( 2 100 Nun dividieren wir beide Seite durch ln( 12 ). Dabei dreht sich die Richtung des Ungleichungszeichens um, da ln( 12 ) negativ ist: n> 1 ) ln( 100 1 . ln( 2 ) 1 ) = − ln(100) und ln( 12 ) = − ln(2) noch vereinfacht und Das kann wegen ln( 100 dann bequem mit dem Computer berechnet werden: n> Also ist ( 12 )n < 1 100 f¨ ur n ≥ 7.

Die beiden anderen L¨osungen sind damit π 1 w2 = 2ei 6 ei2π 3 = 2ei 5π 6 , π 2 w3 = 2ei 6 ei2π 3 = 2ei 9π 6 . 46 c)): π w1 = ei 4 , π 1 w2 = ei 4 ei2π 4 = ei 3π 4 , w3 = ei 5π 4 , w4 = ei 7π 4 . 5 Mit dem digitalen Rechenmeister Nullstellen Der Solve-Befehl in Mathematica versucht, eine Gleichung nach der angegebenen Variablen aufzul¨osen. Gleichungen werden mit einem doppelten Gleichheitszeichen eingegeben: In[1]:= Solve[x3 − 7x2 + 7x − 1 == 0, x] √ √ Out[1]= {{x → 1}, {x → 3 − 2 2}, {x → 3 + 2 2}} Mathematica hat also drei Nullstellen gefunden.

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