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IX. Opgaver – 3. Januar 2004 578 Løsning. Problemstillingen er illustreret på figuren 1) Udregningen v v u ∈ U ⊥ u = projU(v) Ortogonal projektion på underrum u 1 · u 2 = 1 · 0 + 0 · 1 + 0 · 1 + 0 · 0 = 0 viser at u 1 og u 2 er ortogonale. Man får ([LA] Sætning 17) projektionen af v = (2,4,6,8) på u 1 = (1,0,0,0),u 2 = (0,1,1,0) projU(v) = proju 1 (v) + proju 2 (v) = v · u1 u1 + u1 · u1 v · u2 u2 u2 · u2 U
IX. Opgaver – 3. Januar 2004 579 2) Restvektoren har en længde, som er afstanden fra v til U = 2 10 (1,0,0,0) + 1 2 (0,1,1,0) = (2,5,5,0) . v − projU(v) = (2,4,6,8) − (2,5,5,0) = (0, −1,1,8) |(0, −1,1,8)| = √ 66 . Opgave 6. 1) Angiv egenværdierne for matricen 2 4 4 2 2) Undersøg om funktionen har et lokalt ekstremum i (0,0). . f(x,y) = x 2 + y 2 + 4xy + x 2 y Løsning. 1) Det karakteristiske polynomium er 2 − λ 4 4 2 − λ = λ2 − 4λ − 12 .
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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I. Differentiation - 2. Partielle a
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I. Differentiation - 3. Tangentplan
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I. Differentiation - 4. Kædereglen
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I. Differentiation - 6. Maksimum/mi
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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III. Potensrækker - 2. Talfølger
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III. Potensrækker - 4. Taylorpolyn
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1. Grafiske/numeriske metoder IV Di
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IV. Differentialligninger - 1. Graf
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1. Vektorer og matricer V Matricer
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V. Matricer - 1. Vektorer og matric
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V. Matricer - 2. Lineære afbildnin
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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VI. Egenvektorer og diagonalisering
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