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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: It’s actually easier to do this first Actual error = =

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: It’s actually easier to do this first Actual error = = (actual error)

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: We’ll start with f(x) = cos x. f(x) = cos x f’(x) = -sin x f’’(x) = -cos x f’’’(x) = sin x f 4 (x) = cos x f 5 (x) = -sin x f 6 (x) = -cos x cos 0 -cos 0 cos 0 -cos 0 -sin (x-c) 0 /0 ! (x-c) 1 /1 ! (x-c) 2 /2 ! (x-c) 3 /3 ! (x-c) 4 /4 ! (x-c) 5 /5 ! (x-c) 6 /6 !

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: 0 (x-c) 0 /0 ! (x-c) 2 /2 ! (x-c) 4 /4 ! (x-c) 5 /5 ! (x-c) 6 /6 ! 1 1 (I just did all that to verify the equation the book gave me.) Depending on who you listen to, one of these is the first neglected term. The error will be less than or equal to the absolute value of this term.

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: 0 (x-c) 5 /5 ! (x-c) 6 /6 ! Depending on who you listen to, one of these is the first neglected term. The error will be less than or equal to the absolute value of this term. f 5 (z)(0.3-0) 5 5 !5 ! f 6 (z)(0.3-0) 6 6 !6 ! Since we’re dealing with trig functions, the maximum value the derivative can ever obtain is +1. (0.3) 5 5 !5 ! (0.3) 6 6 !6 ! Maximum possible error: or (Our textbook accepts both answers.)

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: Exact error:

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: Exact error: f(x) = e x f’(x) = e x f’’(x) = e x f’’’(x) = e x f 4 (x) = e x f 5 (x) = e x f 6 (x) = e x e0e0 e0e0 e0e0 e0e0 e0e0 e0e0 e0e (x) 0 /0 ! (x) 1 /1 ! (x) 2 /2 ! (x) 3 /3 ! (x) 4 /4 ! (x) 5 /5 ! (x) 6 /6 ! This time, this is definitely the first neglected term. In this problem, it’s a Maclaurin series with c=0 and x=1.

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error: Exact error: (x) 6 /6 ! This time, this is definitely the first neglected term. In this problem, it’s a Maclaurin series with c=0 and x=1. f 6 (z)(1) 6 6 !6 ! As we’ve seen, all the derivatives will be e z. Since e 1 > e 0, we’ll want to go with z=1 in order to maximize the error. e 1 (1) 6 6 !6 ! Maximum possible error =.00378

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9.7--Taylor Polynomial Remainders (day #3) Use Taylor's Theorem to obtain an upper bound for the error of approximation, then calculate the exact value of the error:

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9.7--Taylor Polynomial Remainders (day #3) Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value to be less than.001 (Do not calculate the derivatives using technology!) 49) sin (0.3) f(x) = sin x f’(x) = cos x f’’(x) = -sin x f’’’(x) = -cos x sin 0 -sin 0 cos 0 -cos c = 0, x = 0.3 R n (x) = f n+1 (z) x n+1 (n+1) ! P n (x) = 0 + 1x11x1 1! + 0x20x2 2! + -1x 3 3! + … + f n (c) x n n! Since all derivatives are sin & cos, this maximum value will always be 1.

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9.7--Taylor Polynomial Remainders (day #3) Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value to be less than.001 (Do not calculate the derivatives using technology!) 49) sin (0.3) c = 0, x = 0.3 R n (x) = f n+1 (z) x n+1 (n+1) ! Since all derivatives are sin & cos, this maximum value will always be 1. = (1) (0.3) n+1 (n+1) ! Using the graphing calculator, we will start plugging in values for n until the error is less than 1/1000. x y (error) terms

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9.7--Taylor Polynomial Remainders (day #3) Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value to be less than.001 (Do not calculate the derivatives using technology!) 51) e 0.6 f(x) = e x f’(x) = e x f’’(x) = e x e0e0 e0e0 e0e c = 0, x = 0.6 P n (x) = 1 + 1x11x1 1! + 1x21x2 2! + 1x31x3 3! + … + f n (c) x n n! R n (x) = f n+1 (z) x n+1 (n+1) ! According to Taylor’s Theorem, there exists a z between 0 and 0.6 such that z maximizes the derivative (and thus, the error.) In this case, e 0.6 > e 0. Therefore, we’ll use 0.6 for z.

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9.7--Taylor Polynomial Remainders (day #3) Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value to be less than.001 (Do not calculate the derivatives using technology!) 51) e 0.6 c = 0, x = 0.6 R n (x) = f n+1 (z) x n+1 (n+1) ! In this case, e 0.6 > e 0. Therefore, we’ll use 0.6 for z. = (e 0.6 ) (0.6) n+1 (n+1) ! x y (error) terms

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9.7--Taylor Polynomial Remainders (day #3) Determine the values of x for which the function can be replaced by the Taylor polynomial if the error can't exceed.001: The error will be less than the absolute value of the next term… R 3 (x) = f 4 (z) x 4 4 !4 !. ≤.001 All derivatives are e x. c = 0 n = 3 z must be between c & x. Since c = 0 and x < 0, then z = 0 (in order to maximize z). e 0 x 4 4 !4 !. ≤ Solve using algebra: lx 4 l ≤.024 l x l ≤ ≤ x < 0 (Remember, x < 0 from the original instructions!)

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9.7--Taylor Polynomial Remainders (day #3) Determine the values of x for which the function can be replaced by the Taylor polynomial if the error can't exceed.001: ≤ x ≤.3936

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