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Lecture 18 Linear Approximation & Taylor's Formula

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Lecture 18 Linear Approximation & Taylor's Formula 1 Lecture 18 Linear approximation & Taylor’s formula Layali Al-Mashhadani E-post: [email protected] MAA149, Spring 2019 Mälardalen University 14 Mars, 2019 Contents of the lecture 1. Linear Approximations 1.1 Error Analysis 2. Taylor and Maclaurin Polynomials 2 3. Linear Approximations Linear Approximation (or Tangent Line Approximation): To estimate the value of a function near a certain known point, x =a, on its graph by substituting its tangent line in its place. Such a simplification is usually good only for a small interval around the given point, that is, when x is close to a. Suppose given a function 푓(푥), and suppose (푥표 , 푦표) = (푎, 푓 (푎)) is a known point on the graph of 푓(푥). Then the line tangent to the graph 푦 = 푓 (푥) at a has slope = 푓 ′(푎). Therefore, the point-slope form of the equation of tangent line is 푓(푥) = 푓 (푎) + 푓 ′(푎) (푥 − 푎) That is, when x is near a , 푓 (푥) ≈ 푓 (푎) + 푓 ′(푎) (푥 − 푎). The expression 퐿(푥) = 푓 (푎) + 푓 ′(푎) (푥 − 푎) is called the linearization of 푓 (푥) at a. In the picture, the tangent line to 푦 = 푥1/3 at 푥 = 8 is viewed as an approximation to the original curve. 푦 = 퐿(푥) is the equation of the tangent line. The error is the difference 퐿(푥) − 푥1/3 between the approximate and correct values 푬풓풓풐풓 (푬(풙)) = 풕풓풖풆 풗풂풍풖풆 – 풂풑풑풓풐풙풊풎풂풕풆 풗풂풍풖풆 Theorem Suppose that 푦 = 푓 (푥) is a differentiable curve at 푥 = 푎. Then the tangent line at 푥 = 푎 has equation 푦 = 푓(푥) = 푓 (푎) + 푓 ′(푎) (푥 − 푎) 3 We call the above equation the linear approximation or linearization of 푦 = 푓 (푥) at the point (푎, 푓 (푎)) and write 푓 (푥) ≈ 퐿(푥) = 푓 (푎) + 푓 ′(푎) (푥 − 푎). Example 1 Consider the cube root function above: 푦 = 푓 (푥) = 3√푥. We approximate near 푥 = 8. We have 1 1 푓(8) = 2, and 푓′(푥) = 푥−2/3 ⇒ 푓′(8) = 3 12 The linear approximation is 퐿(푥) = 푓 (8) + 푓 ′(8) (푥 − 8). This can be used, for example, to approximate cube roots without using a calculator: e.g. 3√8 = 2 + 0.00833 = 2.00833 Example 2 Find an approximation to √15 Solution Since 15 is much closer to 16 than to 9, we expect that the approximation 3.875 is the superior estimate. Using the calculator, √15 = 3.873 Errors The error in an approximation 푓 (푥) _ 퐿푎(푥) is the difference 퐸푎(푥) = 퐿푎(푥) − 푓 (푥). In the above example, the errors using the two approximations, are 4 퐸9(15) = 0.127, and 퐸16(15) = 0.002 1.1 Error Analysis For any approximation error = true value − approximate value The error in approximating f(x) by its linearization L(x) about x = a is denoted E1(x) 퐸1(푥) = 푓(푥) − 퐿(푥) = 푓(푥) − 푓(푎) − 푓′(푎)(푥 − 푎): It is the vertical distance at x between the graph of f and the tangent line to that graph at x = a, as shown in the figure: 5 Theorem (An error formula for linearization) If 푓 ′′(푡) exists for all 푡 in an interval containing 푎 and 푥, then there exists some point 푋 between 푎 and 푥 such that the error 퐸1(푥) = | 푓(푥) − 퐿(푥)| in the linear approximation 푓(푥) ≈ 퐿(푥) = 푓(푎) + 푓 ′ (푎)(푥 − 푎) Satisfies 푓′′(푋) 퐸(푥) = (푥 − 푎)2 2 The following three corollaries are consequences of the last theorem: 6 Example 3 i. Use the linearization for √푥 about 푥 = 25 to find an approximate value for √26 ii. Determine the sign and estimate the size of the error in the approximation √26 solution i. ii. 7 2. Taylor’s formula The tangent line approximation of 푓 (푥) for 푥 near 푎 is called the first degree Taylor Polynomial of 푓 (푥) and is: 푃1(푥) = 퐿(푥) = 푓 (푎) + 푓 ′(푎) (푥 − 푎) We can obtain even better approximation to 푓(푥) by using quadratic or higher degree polynomials at 푥 = 푎, if 푓(푥) is twice differentiable near 푥 = 푎, then the polynomial is 푓′′(푎) 푃 (푥) = 푓 (푎) + 푓′(푎)(푥 − 푎) + (푥 − 푎)2 2 2 Taylor polynomial of the third degree is given by: 푓′′(푎) 푓′′′(푎) 푃 (푥) = 푓 (푎) + 푓′(푎)(푥 − 푎) + (푥 − 푎)2 + (푥 − 푎)3 3 2 3 In general if 푓(푥) be a function with derivatives of all orders throughout some interval containing 푎 as an interior point. Then the Taylor series at 푥 = 푎 is given by: 푓′(푎) 푓′′(푎) 푃 (푥) = 푓 (푎) + (푥 − 푎)1 + (푥 − 푎)2 푛 1! 2! 푓′′′(푎) + (푥 − 푎)3 + …. 3! 푓푛(푎) + (푥 − 푎)푛 푛! The error 퐸푛(푥) = 푓(푥) − 푃푛(푥) in the approximation 푓(푥) ≈ 푃푛(푥) is called the Error bound or Lagrange Remainder and given by: 푓(푛+1)(푋) 퐸 (푥) = (푥 − 푎)푛+1 푛 (푛 + 1)! Where X is some number between a and x Examples 4 Find the following Tylor polynomial: (a) 푃2(푥) for 푓(푥) = √푥 about 푥 = 25, then use it to approximate √26 and estimate the size of the error, and then specify the interval that you can be sure contains √26. (b) 푃3(푥) for 푓(푥) = ln 푥 about 푥 = 푒 8 Solution 푓′(25) 푓′′(25) (a) 푃 (푥) = 푓 (25) + (푥 − 25)1 + (푥 − 25)2 2 1! 2! 푓 (25) = √25 = 5 1 1 1 푓′ (푥) = 푓′ (25) = = 2 √푥 2 √25 10 3 1 − 1 1 1 1 1 1 1 푓′′ (푥) = − 푥 2 푓′′ (25) = − = − . = − . = − 4 4 3 4 3 4 125 500 푥2 (52)2 So, 1 1 − 푃 (푥) = 5 + (푥 − 25)1 + 500 (푥 − 25)2 2 10 2! 1 1 푃 (푥) = 5 + (푥 − 25)1 + (푥 − 25)2 2 10 1000 The required approximation is then: 1 1 √26 = 푓(26) ≈ 푃 (26) = 5 + (26 − 25)1 + (26 − 25)2 = 5.099 2 10 1000 The error in a Taylor approximation 풇(ퟐퟔ) ≈ 푷ퟐ(ퟐퟔ) is given by: 푓′′′(25) 퐸 (26) = (26 − 25)3 2 3! 3 3 3 3 푓′′′ (푥) = 푥−5/2 푓′′′ (25) = (25)−5/2 = = 8 8 8 × 55 8 × 3125 3 3 퐸 (26) = |푓′′′ (푋)| ≤ = = 0.00002 푓표푟 25 < 푋 < 26 2 8 × 3125 50,000 The interval that contains √ퟐퟔ is: (5.099 − 0.00002, 5.099 + 0.00002) = (5.09898, 5.09902) 푓′(푒) 푓′′(푒) 푓′′′(푒) (b) 푃 (푥) = 푓 (푒) + (푥 − 푒)1 + (푥 − 푒)2 + (푥 − 푒)3 3 1! 2! 3! 푓 (푒) = ln 푒 = 1 1 1 푓′ (푥) = 푓′ (푒) = 푥 푒 1 1 푓′′ (푥) = − 푓′′ (푒) = − 푥2 푒2 1 1 푓′′′ (푥) = 2 푓′′′ (푒) = 2 푥3 푒3 9 So, 푓′(푒) 푓′′(푒) 푓′′′(푒) 푃 (푥) = 푓 (푒) + (푒 − 푎)1 + (푥 − 푒)2 + (푥 − 푒)3 3 1! 2! 3! 1 1 1 1 푃 (푥) = + (푒 − 푎)1 − (푥 − 푒)2 + (푥 − 푒)3 3 푥 푒 2푒2 3푒3 2. Taylor and Maclaurin Polynomials Some Common Maclaurin Series with Errors in Big -O Form As 푥 → 0: 1 = 1 + 푥 + 푥2 + 푥3 + ⋯ + 푥푛 + 푂(푥푛+1) 1 − 푥 1 1 = = 1 − 푥 + 푥2 − 푥3 + ⋯ + (−1)푛+2푥푛 + 푂(푥푛+1) 1 + 푥 1 − (−푥) 푥2 푥3 푥푛 ln(1 + 푥) = 푥 − + − ⋯ + (−1)푛−1 + 푂(푥푛+1) 2 3 푛 푥2 푥3 푥푛 ln(1 − 푥) = ln(1 − (−푥)) = − 푥 − − − ⋯ + − + 푂(푥푛+1) 2 3 푛 푥2 푥3 푥푛 푒푥 = 1 + 푥 + + + ⋯ + + 푂(푥푛+1) 2! 3! 푛! 푒푎푥 = 푒푎(푥−1+1) = 푒푎. 푒푎(푥−1) = (푎(푥 − 1))2 (푎(푥 − 1))3 (푎(푥 − 1))푛 푒푎[1 + 푎(푥 − 1) + + + ⋯ + + 푂(푥푛+1) 2! 3! 푛! 푥3 푥5 푥(2푛+1) sin 푥 = 푥 − + − ⋯ + (−1)푛 + 푂(푥2푛+3) 3! 5! (2푛 + 1)! 푥2 푥4 푥2푛 cos 푥 = 1 − + − ⋯ + (−1)푛 + 푂(푥2푛+2) 2! 4! (2푛)! 푥3 푥5 푥(2푛+1) tan−1 푥 = 푥 − + − ⋯ + (−1)푛 + 푂(푥2푛+3) 3 5 2푛 + 1 10 Big-O Notation Suppose you have a function 푓(푥) with 푓(푎) = 0 and you want to consider how quickly the function goes to zero around 푎. Then ideally you would want to find a simple function 푔 (for example 푔(푥) = (푥 − 푎)푛 ) which also vanishes at a such that 푔 and 푓 are almost equal around a. The big-O notation tries to express something like this, but only states that 푓 goes to zero faster than 푔. Definition We write 푓(푥) = 푂( 푢(푥) ) as 푥 → 푎 (read is big -Oh of 푢(푥) as x approaches a) provided that |푓(푥)| ≤ 퐾|푢(푥)| holds for some constant K on some open interval containing 푥 = 푎. Similarly, 푓(푥) = 푔(푥) + 푂( 푢(푥) ) as 푥 → 푎 if 푓(푥) − 푔(푥) = 푂( 푢(푥) ) as 푥 → 푎, that is, if |푓(푥) − 푔(푥)| ≤ 퐾|푢(푥)| 푛푒푎푟 푥 = 푎 For example 푠푖푛 푥 = 푂(푥) as 푥 → 0 because |푠푖푛 푥| ≤ |푥| near 0. Another Example 푥3 푠푖푛 푥 − 푥 = 푂(푥3) 푎푠 푥 → 0 since 푃 (푥) for 푠푖푛 푥 is 푥 − , 3 3! Thus, we write 푠푖푛 푥 = 푥 + 푂(푥3) as 푥 → 0. 푥3 Using 푃 (푥) we get 푠푖푛 푥 = 푥 − + 푂(푥5) as 푥 → 0. 5 3! The following properties of the big-O notation 1.
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