b. Use the second-order Taylor series to approximate the function f(x) = 0.1 x* – 0.2 x3 – 0.5 x2 – 0.3x + 1.2 at x = 1. Assume that the we only know the value of the function and its derivatives at x = 0 (i.e., f(0) = 1.2). Hint: f(x;+1) = f(x;) + f'(x;)h+"G) h² + Lx) h³ + ...+ 2! 3! n! f'(x) = 0.4x3 – 0.6x2 – x – 0.3, thus f'(0) = -0.3 f"(x) = 1.2x2 - 1.2x - 1, thus f"(0) = -1 = 2.4x – 1.2, thus f"(0) = -1.2

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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b. Use the second-order Taylor series to approximate the function
f(x) = 0.1 x* – 0.2 x3 – 0.5 x² – 0.3x + 1.2 at x = 1. Assume that the we only know
the value of the function and its derivatives at x = 0 (i.e., f(0) = 1.2).
Hint: f(xị+1) = f(x;) + f'(x;)h +"&) h² + (x) h³ + ...+
2!
3!
n!
f'(x) = 0.4x3 – 0.6x2 – x – 0.3, thus f'(0) = -0.3
%3D
f"(x) = 1.2x2 - 1.2x – 1, thus f"(0) = -1
fmx) = 2.4x – 1.2, thus f"(0) = -1.2
Transcribed Image Text:b. Use the second-order Taylor series to approximate the function f(x) = 0.1 x* – 0.2 x3 – 0.5 x² – 0.3x + 1.2 at x = 1. Assume that the we only know the value of the function and its derivatives at x = 0 (i.e., f(0) = 1.2). Hint: f(xị+1) = f(x;) + f'(x;)h +"&) h² + (x) h³ + ...+ 2! 3! n! f'(x) = 0.4x3 – 0.6x2 – x – 0.3, thus f'(0) = -0.3 %3D f"(x) = 1.2x2 - 1.2x – 1, thus f"(0) = -1 fmx) = 2.4x – 1.2, thus f"(0) = -1.2
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