Download presentation

1
**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 = =

2
**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) Actual error = =

3
**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 !

4
**(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.

5
**.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 ! or The error will be less than or equal to the absolute value of this term. (Our textbook accepts both answers.)

6
**- 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: -

7
**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: 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.

8
**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: 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 = 1 (x)6/6 ! This time, this is definitely the first neglected term.

9
**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:

10
**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.

11
**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

12
**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) !

13
**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) !

14
**(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 ≤ ≤ x < 0 (Remember, x < 0 from the original instructions!) Solve using algebra:

15
**-.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: ≤ x ≤

Similar presentations

OK

Now that you’ve found a polynomial to approximate your function, how good is your polynomial? Find the 6 th degree Maclaurin polynomial for For what values.

Now that you’ve found a polynomial to approximate your function, how good is your polynomial? Find the 6 th degree Maclaurin polynomial for For what values.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on regular expression examples Ppt on solid dielectrics conference Ppt on isobars and isotopes Ppt on conservation of momentum experiment Ppt on non ferrous minerals Ppt on product lifecycle management Ppt on shell scripting interview Ppt on world diabetes day Free ppt on agriculture in india Ppt on local self government rural