It’s actually easier to do this first. 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. .9553364891 .9553375 Actual error = .9553364891 - .9553375 = .00000101087

It’s actually easier to do this first. 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) .9553364891 .9553375 Actual error = .9553364891 - .9553375 .00000101087 =

We’ll start with f(x) = cos x. 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 cos 0 1 (x-c)0/0 ! f’(x) = -sin x -sin 0 (x-c)1/1 ! f’’(x) = -cos x -cos 0 -1 (x-c)2/2 ! f’’’(x) = sin x -sin 0 (x-c)3/3 ! f4(x) = cos x cos 0 1 (x-c)4/4 ! f5(x) = -sin x -sin 0 (x-c)5/5 ! f6(x) = -cos x -cos 0 -1 (x-c)6/6 !

(I just did all that to verify the equation the book gave me.) 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: 1 (x-c)0/0 ! (I just did all that to verify the equation the book gave me.) -1 (x-c)2/2 ! 1 (x-c)4/4 ! Depending on who you listen to, one of these is the first neglected term. (x-c)5/5 ! -1 (x-c)6/6 ! The error will be less than or equal to the absolute value of this term.

.00002025 or .0000010125 -1 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: Since we’re dealing with trig functions, the maximum value the derivative can ever obtain is +1. f5(z)•(0.3-0)5 5 ! f6(z)•(0.3-0)6 6 ! (0.3)5 5 ! (0.3)6 6 ! Depending on who you listen to, one of these is the first neglected term. (x-c)5/5 ! Maximum possible error: -1 (x-c)6/6 ! .00002025 or .0000010125 The error will be less than or equal to the absolute value of this term. (Our textbook accepts both answers.)

- 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: .001615161 - 2.718281828 2.716666667

In this problem, it’s a Maclaurin series with c=0 and x=1. 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: .001615161 f(x) = ex e0 1 (x)0/0 ! f’(x) = ex e0 1 (x)1/1 ! f’’(x) = ex e0 1 (x)2/2 ! In this problem, it’s a Maclaurin series with c=0 and x=1. f’’’(x) = ex e0 1 (x)3/3 ! f4(x) = ex e0 1 (x)4/4 ! f5(x) = ex e0 1 (x)5/5 ! f6(x) = ex e0 1 (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. 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: .001615161 As we’ve seen, all the derivatives will be ez. Since e1 > e0, we’ll want to go with z=1 in order to maximize the error. f6(z)•(1)6 6 ! In this problem, it’s a Maclaurin series with c=0 and x=1. e1(1)6 6 ! Maximum possible error = .00378 1 (x)6/6 ! This time, this is definitely the first neglected term.

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:

Pn(x) = 0 + + + … + Rn(x) = f(x) = sin x sin 0 f’(x) = cos x cos 0 1 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 f(x) = sin x sin 0 f’(x) = cos x cos 0 1 f’’(x) = -sin x -sin 0 f’’’(x) = -cos x -cos 0 -1 Pn(x) = 0 + 1x1 1! + 0x2 2! -1x3 3! + … + fn(c) xn n! Rn(x) = fn+1(z) xn+1 (n+1) ! Since all derivatives are sin & cos, this maximum value will always be 1.

3 terms = Rn(x) = 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 Using the graphing calculator, we will start plugging in values for n until the error is less than 1/1000. = (1) (0.3)n+1 (n+1) ! x y (error) 1 .045 3 terms 2 .0045 3 .00034 Rn(x) = fn+1(z) xn+1 (n+1) ! Since all derivatives are sin & cos, this maximum value will always be 1. 4

In this case, e0.6 > e0. Therefore, we’ll use 0.6 for z. 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) e0.6 c = 0, x = 0.6 f(x) = ex e0 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.) f’(x) = ex e0 1 f’’(x) = ex e0 1 Pn(x) = 1 + 1x1 1! + 1x2 2! 1x3 3! + … + fn(c) xn n! In this case, e0.6 > e0. Therefore, we’ll use 0.6 for z. Rn(x) = fn+1(z) xn+1 (n+1) !

In this case, e0.6 > e0. Therefore, we’ll use 0.6 for z. 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) e0.6 c = 0, x = 0.6 = (e0.6) (0.6)n+1 (n+1) ! x y (error) 1 .32798 5 terms 2 .0656 3 .00984 4 .00118 5 .00012 In this case, e0.6 > e0. Therefore, we’ll use 0.6 for z. Rn(x) = fn+1(z) xn+1 (n+1) !

(Remember, x < 0 from the original instructions!) 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… R3(x) = f4(z) x4 4 ! . c = 0 n = 3 ≤ .001 z must be between c & x. Since c = 0 and x < 0, then z = 0 (in order to maximize z). All derivatives are ex. lx4l ≤ .024 e0 x4 4 ! . ≤ .001 1 l x l ≤ .3936 -.3936 ≤ x < 0 (Remember, x < 0 from the original instructions!) Solve using algebra:

-.3936 ≤ x ≤ .3936 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: -.3936 ≤ x ≤ .3936