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III. Potensrækker – 3. Potensrækker 296 Divideres ledvis med x 4 fås cos(x 2 ) − 1 = (1 − x4 2! = − x4 2! + x8 4! + x8 4! − x12 6! − x12 6! + ... − 1 x4 x8 + − + ... 2! 4! 6! + ...) − 1 3.34. Opgave ☞ Matematik Alfa 1, August 2002 Opgave 4 - Løsning fortsat Dermed er ∞ n 1 f(x) = (−1) Det følger, at n=1 (2n)! x4(n−1) = − 1 1 + 2! 4! x4 − 1 6! x8 + 1 8! x12 − ... lim f(x) = −1 x→0 2 3.35. Potensrækkeløsning ☞ [S] 8.10 Using . . . diff. eq.
III. Potensrækker – 3. Potensrækker 297 Eksempel 1 Vi søger en løsning af form y ′′ + y = 0 y(x) = c0 + c1x + c2x 2 + c3x 3 + ... y ′ (x) = c1 + 2 · c2x + 3 · c3x 2 + 4 · c4x 3 + ... y ′′ (x) = 2 · c2 + 3 · 2 · c3x + 4 · 3 · c4x 2 + 5 · 4 · c5x 3 + ... 3.36. Potensrækkeløsning ☞ [S] 8.10 Using . . . diff. eq. Eksempel 1 - fortsat y(x) + y ′′ (x) = (c0 + 2c2) + (c1 + 3 · 2 · c3)x + (c2 + 4 · 3 · c4)x 2 + ... Fra y + y ′′ = 0 fås at alle koefficienterne må være 0, altså c0 + 2c2 = 0 c1 + 3 · 2c3 = 0 c2 + 4 · 3c4 = 0 3.37. Potensrækkeløsning ☞ [S] 8.10 Using . . . diff. eq.
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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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