Properties of a Taylor Polynomial
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1 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial Constant: f (x) f (a) Linear: f (x) f (a) + f (a)(x a) Quadratic: f (x) f (a) + f (a)(x a) f (a)(x a) 2
2 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial Constant: f (x) f (a) Linear: f (x) f (a) + f (a)(x a) Quadratic: f (x) f (a) + f (a)(x a) f (a)(x a) 2 T n(a) = f (a) T n(a) = f (a) T n (a) = f (a). T n (n) (a) = f (n) (a) T n (n+1) (a) = 0
3 3.4.4: Still Better Approximations: Taylor Polynomials Taylor Polynomials Definition For a natural number n, n! = n.
4 3.4.4: Still Better Approximations: Taylor Polynomials Taylor Polynomials Definition For a natural number n, n! = n. By convention, 0! = 1.
5 3.4.4: Still Better Approximations: Taylor Polynomials Taylor Polynomials Definition For a natural number n, n! = n. By convention, 0! = 1. We write f (n) (x) to mean the nth derivative of f (x).
6 3.4.4: Still Better Approximations: Taylor Polynomials Taylor Polynomials Definition For a natural number n, n! = n. By convention, 0! = 1. We write f (n) (x) to mean the nth derivative of f (x). Given a function f (x) that is differentiable n times at a point a, the n-th degree Taylor polynomial for f (x) about a is T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n = n k=0 f (k) (a) (x a) k k! If a = 0, we call the function a Maclaurin polynomial.
7 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n
8 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a)
9 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a) T n(x) = f (a) ! f (a)(x a) ! f (a)(x a) 2 + n 1 n! f (n) (a)(x a) n 1 = f (a) + f (a)(x a) + 1 2! f (a)(x a) (n 1)! f (n) (a)(x a) n 1
10 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a) T n(x) = f (a) ! f (a)(x a) ! f (a)(x a) 2 + n 1 n! f (n) (a)(x a) n 1 = f (a) + f (a)(x a) + 1 2! f (a)(x a) T n(a) = f (a) 1 (n 1)! f (n) (a)(x a) n 1
11 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a) T n(x) = f (a) ! f (a)(x a) ! f (a)(x a) 2 + n 1 n! f (n) (a)(x a) n 1 T = f (a) + f (a)(x a) + 1 2! f (a)(x a) (n 1)! f (n) (a)(x a) n 1 T n(a) = f (a) n (x) = f (a) ! f 1 (a)(x a) + + (n 1) (n 1)! f (n) (a)(x a) n 2 = f (a) + f (a)(x a) (n 2)! f (n) (a)(x a) n 2
12 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a) T n(x) = f (a) ! f (a)(x a) ! f (a)(x a) 2 + n 1 n! f (n) (a)(x a) n 1 T = f (a) + f (a)(x a) + 1 2! f (a)(x a) (n 1)! f (n) (a)(x a) n 1 T n(a) = f (a) n (x) = f (a) ! f 1 (a)(x a) + + (n 1) (n 1)! f (n) (a)(x a) n 2 = f (a) + f (a)(x a) + + T n (a) = f (a) 1 (n 2)! f (n) (a)(x a) n 2
13 3.4.4: Still Better Approximations: Taylor Polynomials Properties of a Taylor Polynomial T n(x) = f (a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n T n(a) = f (a) T n(x) = f (a) ! f (a)(x a) ! f (a)(x a) 2 + n 1 n! f (n) (a)(x a) n 1 T = f (a) + f (a)(x a) + 1 2! f (a)(x a) (n 1)! f (n) (a)(x a) n 1 T n(a) = f (a) n (x) = f (a) ! f 1 (a)(x a) + + (n 1) (n 1)! f (n) (a)(x a) n 2 = f (a) + f (a)(x a) + + T n (a) = f (a) T n (n) (a) = f (n) (a) T (n+1) n (a) = 0 1 (n 2)! f (n) (a)(x a) n 2
14 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 1 Find the 7th degree Maclaurin 1 polynomial for e x. 1 Remember: a Maclaurin polynomial is just a Taylor polynomial centered about a = 0
15 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 1 Find the 7th degree Maclaurin 1 polynomial for e x. Let f (x) = e x. Then every derivative of e x is just e x, and e 0 = 1. So: T 7(x) = f (0) + f (0)(x 0) f (0)(x 0) ! f (7) (0)(x 0) 7 = 1 + x + x 2 = 7 k=0 x k k! 2 + x 3 3! + x 4 4! + x 5 5! + x 6 6! + x 7 7! 1 Remember: a Maclaurin polynomial is just a Taylor polynomial centered about a = 0
16 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 1 Find the 7th degree Maclaurin 1 polynomial for e x. Let f (x) = e x. Then every derivative of e x is just e x, and e 0 = 1. So: T 7(x) = f (0) + f (0)(x 0) f (0)(x 0) ! f (7) (0)(x 0) 7 = 1 + x + x 2 = 7 k=0 x k k! 2 + x 3 3! + x 4 4! + x 5 5! + x 6 6! + x 7 7! e x approximations - link 1 Remember: a Maclaurin polynomial is just a Taylor polynomial centered about a = 0
17 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 2 Find the 8th degree Maclaurin 2 polynomial for f (x) = sin x. 2 Remember: a Maclaurin polynomial is just a Taylor polynomial centered about a = 0
18 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 2 Find the 8th degree Maclaurin 2 polynomial for f (x) = sin x. f (x) = sin x f (0) = 0 f (4) (0) = 0 f (8) (0) = 0 f (x) = cos x f (0) = 1 f (5) (0) = 1 f (x) = sin x f (0) = 0 f (6) (0) = 0 f (x) = cos x f (0) = 1 f (7) (0) = 1 T 8(x) = f (0) + f (0)(x 0) f (0)(x 0) ! f (8) (0)(x 0) 8 = x x 3 = 3 3! + x 5 5! x 7 7! x 2k+1 (2k + 1)! Link: sine approximations k=0 2 Remember: a Maclaurin polynomial is just a Taylor polynomial centered about a = 0
19 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 3 Find the 7th degree Taylor polynomial for f (x) = ln x, centered at a = 1.
20 3.4.5: Some Examples of Taylor Polynomials T n(a) = f (a) + f (a)(x a) + 1 2! f (a)(x a) n! f (n) (a)(x a) n Example: Taylor 3 Find the 7th degree Taylor polynomial for f (x) = ln x, centered at a = 1. f (x) = ln x f (1) = 0 f (4) (x) = 3!x 4 f (4) (1) = 3! f (x) = x 1 f (1) = 1 f (5) (x) = 4!x 5 f (5) (1) = 4! f (x) = x 2 f (1) 1 f (6) (x) = 5!x 6 f (6) (1) = 5! f (x) = 2x 3 f (1) = 2 f (7) (x) = 6!x 7 f (7) (1) = 6! T 8 (x) = f (1) + f (1)(x 1) f (1)(x 1) ! f (7) (1)(x 1) 7 = 0 + (1)(x 1) + ( 1) 1 2 (x 1)2 + (2) 1 3! (x 1)3 3! 1 (x 1)4 4! + 4! 1 5! (x 1)5 5! 1 6! (x 1)6 + 6! 1 (x 1)7 7! = (x 1) x 1 (x 1) k+1 (x 1)k = ( 1) k k=0 (x 1)3 4 + (x 1)4 5 (x 1)5 6 + (x 1)6 7!
21 Error: What Makes an Approximation Accurate? degree conditions formula 0 f (a) = T 0(a) T 0(x) = f (a) 1 f (a) = T 1(a) f (a) = T 1(a) T 1(x) = f (a) + f (a)(x a) 2 f (a) = T 2(a) f (a) = T 2(a) f (a) = T 2 (a) T 2(x) = f (a) + f (a)(x a) + 1 f (a)(x a) 2 2! f (a) = T n(a) n f (a) = T n(a) f (a) = T n (a) T n(x) = f (a) + f (a)(x a) + 1 f (a)(x a) 2 2! (a)(x a) n n! f (n) (a) = T n (n) (a)
22 Error Term: Taylor s Theorem Error The error in an estimation f (x) T n(x) is f (x) T n(x). We often use f (x) T n(x) if we don t care whether the approximation is too big or too little, but only that it is not too egregious. Taylor s Theorem For some c strictly between x and a, f (x) T n(x) = 1 (n + 1)! f (n+1) (c)(x a) n+1
23 Error Term: Taylor s Theorem Error The error in an estimation f (x) T n(x) is f (x) T n(x). We often use f (x) T n(x) if we don t care whether the approximation is too big or too little, but only that it is not too egregious. Taylor s Theorem For some c strictly between x and a, f (x) T n(x) = 1 (n + 1)! f (n+1) (c)(x a) n+1 The trick is bounding f (n+1) (c). It s usually OK to be sloppy here! Also, usually what we care about is the magnitude of the error: f (x) T n(x).
24 Third degree Maclaurin polynomial for f (x) = e x :
25 Third degree Maclaurin polynomial for f (x) = e x : T 3(x) = f (0) + f (0)(x 0) + 1 2! f (0)(x 0) ! f (0)(x 0) 3 = e 0 + e 0 x + 1 2! e0 x ! e0 x 3 = 1 + x + x 2 2! + x 3 3!
26 Third degree Maclaurin polynomial for f (x) = e x : T 3(x) = f (0) + f (0)(x 0) + 1 2! f (0)(x 0) ! f (0)(x 0) 3 = e 0 + e 0 x + 1 2! e0 x ! e0 x 3 = 1 + x + x 2 2! + x 3 3! Example: Taylor 4 Bound the error associated with using T 3(x) to approximate e 1/10. Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a, and 2 < e < 3.
27 Third degree Maclaurin polynomial for f (x) = e x : T 3(x) = f (0) + f (0)(x 0) + 1 2! f (0)(x 0) ! f (0)(x 0) 3 = e 0 + e 0 x + 1 2! e0 x ! e0 x 3 = 1 + x + x 2 2! + x 3 3! Example: Taylor 4 Bound the error associated with using T 3(x) to approximate e 1/10. Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a, and 2 < e < 3. For some c strictly between 0 and.1: f (.1) T 3(.1) = 1 }{{} f (4) (c)(.1 0) 4 = 1 e c 4! 4!10 4 error For c in (0,.1), 1 e c < e 1 < 3, so 1 1 f (.1) T 4!10 3(.1) !10 }{{} 4 error f (.1) T 3(.1)
28 Example: Taylor 5 Suppose we use the 5th degree Taylor polynomial centered at a = π/2 to approximate f (x) = cos x. What could magnitude of the error be if we approximate cos(2)? Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a. We don t actually have to compute T 5(x), but if you want to as an exercise, click here to see it.
29 Example: Taylor 5 Suppose we use the 5th degree Taylor polynomial centered at a = π/2 to approximate f (x) = cos x. What could magnitude of the error be if we approximate cos(2)? Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a. We don t actually have to compute T 5(x), but if you want to as an exercise, click here to see it. For some c in (π/2, 2): f (2) T 5(2) = }{{} error 1 6! f (6) (c)(2 π/2) 6 Note f (6) (x) is going to be positive or negative sine or cosine, so f (6) (c) 1. Also, 2 π/2 < 1 2. Now: And 1 6! f (2) T 5(2) < 1 6! (1) ( 1 2 ) 6 = 1 6!2 6
30 Example: Taylor 6 Suppose we use a third degree Taylor polynomial centered at 4 to approximate f (x) = x. If we use this Taylor polynomial to approximate 4.1, give a bound for the magnitude of our error. Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a.
31 Example: Taylor 6 Suppose we use a third degree Taylor polynomial centered at 4 to approximate f (x) = x. If we use this Taylor polynomial to approximate 4.1, give a bound for the magnitude of our error. Recall f (x) T n(x) = 1 (n+1)! f (n+1) (c)(x a) n+1 for some c between x and a. For some c in (4, 4.1), f (4.1) T 3(4.1) = 1 4! f (4) (c)(4.1 4) 4 =.14 4! f (4) (c) So, let s investigate f (4) (c). First we find that the fourth derivative of f (x) = x 1/2 is f (4) (x) = 15 x 7/2. So, for c in (4, 4.1), we have 16 f (4) (c) = c = c 7 16 = So, the error is bounded by: f (4.1) T 3(4.1) !
32 Example: Taylor 7 Suppose you want to approximate the value of e, knowing only that it is somewhere between 2 and 3. You use a 4th degree Maclaurin polynomial for f (x) = e x to approximate f (1) = e 1 = e. Bound the magnitude of your error.
33 Example: Taylor 7 Suppose you want to approximate the value of e, knowing only that it is somewhere between 2 and 3. You use a 4th degree Maclaurin polynomial for f (x) = e x to approximate f (1) = e 1 = e. Bound the magnitude of your error. For some c in (0, 1): f (1) T 4(1) = 1 f (5) (c)(1 0) 5 5! = 1 5! ec 1 5! e1 < 3 = !
34 Which Degree? Example: Taylor 8 Suppose you want to approximate sin 3 using a Taylor polynomial of f (x) = sin x centered at a = π. If the magnitude of your error must be less than 0.001, what degree Taylor polynomial should you use?
35 Which Degree? Example: Taylor 8 Suppose you want to approximate sin 3 using a Taylor polynomial of f (x) = sin x centered at a = π. If the magnitude of your error must be less than 0.001, what degree Taylor polynomial should you use? We want the magnitude of the error, so let s deal with absolute values. For some c in (3, π): 1 f (3) T n(3) = f (n+1) (c)(3 π) n+1 (n+1)! = 3 π n+1 (n+1)! f (n+1) (c) (.2)n+1 (n+1)! (1) =.2n+1 (n+1)! If we plug in n = 2, we get.2n+1 = , which is not SMALLER than If we (n+1)! plug in n = 3, we get.2n+1 = which IS smaller than So we have to (n+1)! use the degree 3 Taylor polynomial.
36 Example: Taylor 9 Suppose you want to approximate e 5 using a Maclaurin polynomial of f (x) = e x. If the magnitude of your error must be less than 0.001, what degree Maclaurin polynomial should you use?
37 Example: Taylor 9 Suppose you want to approximate e 5 using a Maclaurin polynomial of f (x) = e x. If the magnitude of your error must be less than 0.001, what degree Maclaurin polynomial should you use? The magnitude of the error means its absolute value. Our error is, for some c in (0, 5): f (5) T n(5) = 1 f (n+1) (c)(5 0) n+1 = 1 (n+1)! (n+1)! ec 5 n+1. We can bound e c for c in (0, 5) by 1 = e 0 < e c < e 5 < 3 5. So now: f (5) T n(5) 1 (n+1)! 35 5 n+1 1 We set (n+1)! 35 5 n+1 > 0.001, and by plugging in different values of n, we find the smallest n that makes the inequality true is n = 21. So we can use the 21st-degree Maclaurin polynomial and get our desired error.
38 Example: Taylor 10 Suppose you want to approximate ln 3 using a Taylor polynomial of f (x) = ln x centered at a = 1. If the magnitude of your error must be less than 0.001, what degree Taylor polynomial should you use?
39 Example: Taylor 10 Suppose you want to approximate ln 3 using a Taylor polynomial of f (x) = ln x centered at a = 1. If the magnitude of your error must be less than 0.001, what degree Taylor polynomial should you use? Your error will be: f (3) T n(3) = 1 f (n) (c)(3 1) n+1 for some c in (0, 3). So, we (n+1)! need to bound f (n) (c). by writing out a number of derivatives of natural log, we notice that for n 1, f (n) (c) = ( 1) n 1 (n 1)!c n. So, f (n+1) (c) = ( 1) n n!c n. For c in (0, 3): n! 1 = n! f (n+1) (c) n! n 3 n Now for the error: f (3) T n(3) 1 (n + 1)! n! 3 n 2n+1 = 2n+1 (n + 1)3 n Setting this < 0.001, we find by plugging in values of n that n = 5 is the smallest n that makes the inequality true. So, using T 5(x) will give us our desired error.
40 Taylor Error Example: Taylor 11 Let f (x) = 4 x. Suppose you use a second-degree Taylor polynomial of f (x) centered at a = 81 to approximate Bound your error, and tell whether T 2(10) is an overestimate or underestimate.
41 Taylor s formula tells us that, for some c in (81, 81.2) : f (10) T 2 (10) = 1 3! f (3) (c)( ) 3 = 1 6 ( 1 5 )3 f (3) (c) = f (3) (c) So, we should probably find out what f (3) (x) is. Since f (x) = x 1/4, it s not too hard to figure out f (x) = x 11/4. So, f (c) = Plugging in: c11/4 f (10) T 2 (10) = c 11/4 = c 11/ Now our job is to bound this, and we should use reasonable numbers. 81 c /4 c 11/ /4 ( 4 81) 11 c 11/ c 11/ c 11/ c 11/ So, f (x) T 2(x) And, since f (x) T 2(x) is positive, T 2 (x) is an underestimate.
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