ANSWERS: 2
  • What again?
  • There's a problem there: Does the "degree 8" include the error term or not? By the wording I assume not. Here's the 8 degree polynomial expansion for you: p(x) = f(c) + (x-c) f'(c) + (x-c)^2/2 f''(c) + (x-c)^3/3! f'''(c) + (x-c)^4 /4! f^(4)(c) + (x-c)^5 /5! f^(5) (c) + (x-c)^6/6! f^(6)(c) + (x-c)^7/7! f^(7)(c) + (x-c)^8/8! f^(8)(c) Only a sadistic teacher will make you multiply it out, but you can simplify it because the derivatives of sin(x) are so easy: [sin(x) cos(x) -sin(x) -cos(x) etc] Result: p(x) = 1 - (x-c)^2/2 + (x-c)^4 /4! - (x-c)^6/6! + (x-c)^8/8! + E where E is the error term. With the assumption I stated above the error term is E= + (x-c)^9/9! cos(e) for some e in the range [c,x] Estimate the error using p(2) to aproximate sin(2) The answer is (2-pi/2)^9/9! cos(e) for some e beteen pi/2 and 2 (2-pi/2)^9/9! is 1.36 * 10^-9 Now what values does cos(e) take in the range pi/2 and 2? Answer: 0 to -0.416 Therefore the maximum error E is E = -5.66e^-10 p(x) will be between sin(2) and sin(2) + 5.66e^-10 However in this case why estimate? we can compute the error almost as easily: p(x) = 0.909297426884061867147020834536219 sin(2) = 0.909297426825681695396019865911745 So p(x) is too high by = 0.000000000058380171751000968624474 5.83x10^-11 which is inside the expected range.

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