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10 Nov 2019
come from? The Lagrange form for the remainder! This is R_n + 1 (x) = f^(n + 1) (c) x^n + 1/(n + 1)! for some c between 0 and x. Notice that f^(n+1) (x) = plusminus cos(x) or plusminus sin(x) so |f^(n+1) (c)| lessthanorequalto Using this we can bound the remainder term by |R_n + 1 (x)| lessthanorequalto |x|^n + 1/(n + 1)!. Hence p_n (x) - |x|^n + 1/(n + 1)! Lessthanorequalto cos(x) lessthanorequalto p_n (x) + |x|^n + 1(n + 1)!. Let's use this to calculate some values of cos(2). Rounded to 3 decimal places p_4(2) = -0.333. Our bound on Lagrange's form for the remainder gives us (rounded to 3 decimal places) 2^5/5! = 0.267. So, using equation (1): This does not tell us much about the value of cos(2). A better approximation to cos(2) will be given by the 8th degree Taylor polynomial (rounded to 5 decimal places) p_8(2) = -0.41587. And Lagrange's form for the remainder gives us the bound (rounded to 5 decimal places) 2^9/9! = 0.00141 which is quite small. So, using equation (1): These bounds agree in the first two decimal places! It is not possible to round this number since the third digit of cos(2) is not clear from this analysis. But we do know the value of cos(2) in the first two decimal places: cos(2) = (truncated to two decimal places).
come from? The Lagrange form for the remainder! This is R_n + 1 (x) = f^(n + 1) (c) x^n + 1/(n + 1)! for some c between 0 and x. Notice that f^(n+1) (x) = plusminus cos(x) or plusminus sin(x) so |f^(n+1) (c)| lessthanorequalto Using this we can bound the remainder term by |R_n + 1 (x)| lessthanorequalto |x|^n + 1/(n + 1)!. Hence p_n (x) - |x|^n + 1/(n + 1)! Lessthanorequalto cos(x) lessthanorequalto p_n (x) + |x|^n + 1(n + 1)!. Let's use this to calculate some values of cos(2). Rounded to 3 decimal places p_4(2) = -0.333. Our bound on Lagrange's form for the remainder gives us (rounded to 3 decimal places) 2^5/5! = 0.267. So, using equation (1): This does not tell us much about the value of cos(2). A better approximation to cos(2) will be given by the 8th degree Taylor polynomial (rounded to 5 decimal places) p_8(2) = -0.41587. And Lagrange's form for the remainder gives us the bound (rounded to 5 decimal places) 2^9/9! = 0.00141 which is quite small. So, using equation (1): These bounds agree in the first two decimal places! It is not possible to round this number since the third digit of cos(2) is not clear from this analysis. But we do know the value of cos(2) in the first two decimal places: cos(2) = (truncated to two decimal places).
Sixta KovacekLv2
15 Apr 2019