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Unformatted text preview: Step 1: T(0.1) = 13 + 0.1*(0.5(13  (10 + 5sin(2π*0)))) = 12.85 °C Step 2: T(0.2) = 12.85 + 0.1*(0.5(12.85  (10 + 5sin(2π*0.1)))) = 12.8544 °C. (b) Because Euler's method is equivalent to using only the first two terms in the Taylor series, the error may be approximated by the next term in the Taylor series for T(t + Δt), which is (Δt) 2 T''(t)/2. In this case, T''(t) = dT ' dt = d dt − k T − 10 5sin 2πt = kT' +10πkcos(2πt) = k 2 (T  (10 + 5sin(2πt))) + 10πkcos(2πt) = 0.75 + 5π (for k = 0.5 / d, T = 13 °C, t = 0 d) ≈ 16 °C / d 2 E ≈ (Δt) 2 T''(t)/2 ≈ 0.08 °C (for Δt = 0.1 d)...
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 Spring '11
 Lybas
 Numerical Analysis, Derivative, Taylor Series, 0.08 °C, 12.85 °C, 12.8544 °C

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