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Numerical Mathematics - A Collection of Solved Problems

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imamoZa x ∈ (x 1 , x 2 ) = I imamoINTERPOLACIJA FUNKCIJA 185P 1 (1.93) = 0.69315 + 0.05130 (−0.07) = 0.65724.0.1M 2 = supξ∈I˛˛− 1ξ˛˛˛˛ 2 = 11.9 2 < 0.2771itako da je| log x − P 1 (x)| ≤ M 22! |(x − x 2)(x − x 1 )|,b) Zbogimamo|log x − P 1 (x)| < 0.2771 |(1.93 − 2)(1.93 − 1.9)| = 0.00029.2P 2 (x) = P 1 (x) + ∇2 log x 22!h 2 (x − x 2 )(x − x 1 ),P 2 (1.93) = 0.65724 + −0.00276 (−0.07) · 0.03 = 0.65753.2 · 0.12 Za x ∈ (x 0 , x 2 ) = I imamoM 3 = supξ∈I2˛ξ ˛˛˛˛ 3 = 21.8 3 < 0.343itako da je| log x − P 2 (x)| ≤ M 33! |(x − x 2)(x − x 1 )(x − x 0 )|,| log 1.93 − P 2 (1.93)| < 0.343 |(1.93 − 2)(1.93 − 1.9)(1.93 − 1.8)| < 0.00002.6Napomenimo da je tačna vrednost, na šest decimala, log 1.93 = 0.657520.6.1.25. Korišćenjem prve Gaussove, druge Gaussove i Stirlingove interpolacioneformule, izračunati vrednost f(0.95) na osnovu sledećih podatakax 0.5 0.7 0.9 1.1 1.3f(x) −0.6875 −0.8299 −0.9739 −0.9659 −0.6139

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