Nội dung

SOME APPROXIMATION PROBLEMS BY FIRST-DEGREE DERIVATIVE Nguyen Van Dung and Giap Kim Anh Falculty of Basic Science - University of Transport and Communications Abstract. In this paper, I mention the approximate calculation of the function value at a point, based on the first derivative at its neighborhood points. The article shows which function classes can be applied to approximate. The paper also points out the importance of approximation as an estimate of the error of that calculation. Keywords. Function, Aapproximation, First derivative, Error. 1. Introduction In Mathematics, the approximation is very important because sometimes it is impossible to accurately calculate a specific value. The simplest problem is to approximate the function value at a point through the first derivative. This is a primary issue that the basic textbooks on calculus have all mentioned, but these textbooks often do not specify the details of this formula. In this article, we will go into solving those problems: Firstly, given the point x0 , we approximate the function value at the points "very close" to that point x0 The second, given the point x0 , we approximate the function value at the point x0 through the points "very close" x0 The third, given the point x0 , does choosing the point closer to x0 give us a better approximation? 2. Basic Knowledge Assuming f (x) is defined in the neighborhood of x0 , f (x) has a derivative at x0 , we have: f (x0 + ∆x) − f (x0 ) ∆x→0 ∆x f 0 (x0 ) = lim f (x0 + ∆x) − f (x0 ) − f 0 (x0 )∆x =0 ∆x→0 ∆x ⇒ ∀ > 0, ∃δ > 0 such that: ⇒ lim ∀|∆x| < δ we have: | f (x0 + ∆x) − f (x0 ) − f 0 (x0 )∆x |<δ ∆x 1 2 Nguyen Van Dung and Giap Kim Anh So, when ∆x → 0, |f (x0 + ∆x) − f (x0 ) − f 0 (x0 )∆x| → 0 or when ∆x is small enough: f (x0 + ∆x) ≈ f (x0 ) + f 0 (x0 )∆x (1) The Formula (1) is the formula for approximation by first-degree derivative; It allows us to approximate the value of the function at points "very close" to the given point x0 . √ Example 2.1. Approximation value A = 4, 01 √ 1 Let f (x) = x, we have f 0 (x) = √ 2 x Apply the Formula (1) with x0 = 4; ∆x = 0.01, we obtain: √ 1 A ≈ 4 + √ .0.01 = 2, 0025 2 4 3. Error When using approximate formulas, we are concerned with the error √ of the calculation. We use the Formula (1) to approximate two values: B = 1, 01365 and C = 10 1000 Calculating B, we apply the Formula (1) with f (x) = x365 , x0 = 1, ∆x = 0.01, we get: B ≈ 1365 + 365.1364 .0.01 = 4.65 Calculating C, we apply the Formula (1) with f (x) = C ≈2− √ 10 x, x0 = 1024, ∆x = −24 we get: 24 = 1, 99531.. 5120 Actual B ≈ 37.7834...; C ≈ 1, 99526.. Clearly B has a very large error while C has a very small error even though ∆x we choose at B is a lot smaller than when we choose at C. The above results show us that there are cases where we choose ∆x very small but in fact not "small enough" and there are cases where we choose ∆x quite large but in fact " small enough. " In other words, when we give the error , there are some functions we choose quite large ∆x and in the opposite there are functions we choose ∆x very small but yet unsatisfied. Returning to the derivative formula, f (x) has the derivative at x0 means: ∀ > 0, ∃δ > 0 such as ∀|∆x| < δ then : | or: f (x0 + ∆x) − f (x0 ) − f 0 (x0 )| <  ∆x |f (x0 + ∆x) − f (x0 ) − f 0 (x0 )∆x| < |∆x| So we have an approximate Formula (1): f (x0 + ∆x) ≈ f (x0 ) + f 0 (x0 )∆x with an error less than |∆x| Thus, there are cases where we choose ∆x quite large but in case  is very small, we can still give errors less than ∆x very small as we calculated the value of C above. Therefore, when we use the Formula (1) to approximate, the important thing is that the error is how much it is, not how small we choose ∆x. 3 Some approximation problems by first-degree derivative 4. Approximation In the approximation Formula (1), the condition is that the function has a derivative at the point x0 , it helps us to approximate the value at the "very close" point x0 ; Now we approach formula (1) according to another idea, given the total value of x0 + ∆x, we have to choose how x0 accordingly. For example, consider the function: ( 100 if x = 0 f (x) = 0 if x 6= 0 With this function, we approximate f (0) We see that f 0 (x) = 0 ∀x 6= 0 so when we choose any x0 6= 0, using the Formula (1) we get: f (0) ≈ f (x0 ) − f 0 (x0 )x0 = 0 while f (0) = 100, obviously in this case, we can not use the Formula (1) to calculate. The question arises: what happened in this case?? Where does this conflict occur, because obviously, for the Formula (1), the condition is that f (x) has a derivative at the point x0 , and the function f (x) above is complete all satisfy this condition. The answer here is that when we choose x0 fixed any other than 0 then ∆x = −x0 now is not "small enough" for us to apply the Formula (1); For example, if we choose x0 = 0.001 then ∆x = −0, 001 is still large for us to apply the Formula (1), ∆x must satisfy |∆x| < 0.001 then we can apply Formula (1) and it only helps us to approximate the values of f (x) at x 6= 0. This example leads to the question, for which functions satisfy which conditions we can approximate its value at a given point by selecting the point x0 "very close" to the given point. This is equivalent to when we have: lim (f (x) − f 0 (x)x) = f (0) x→0 (2) This means that we can approximate f (0) through values of f (x) and f 0 (x) when x near 0. Clearly the function: ( 100 if x = 0 f (x) = 0 if x 6= 0 is not satisfied (2), so it is not possible to use Formula (1) to approximate f (0). Theorem 4.1. If f (x) is continuous at 0, f (x) has a derivative bounded in the neighborhood of 0 (there is not necessarily a derivative at 0) then we have: lim (f (x) − f 0 (x)x) = f (0) x→0 According to Lagrange’s Theorem, ∃c ∈ (0; x) such that: f (x) − f (0) = f 0 (c)x Therefore, f (x) − f 0 (x)x = f (0) + (f 0 (c) − f 0 (x))x. Because f (x) has a bounded derivative, we have: lim (f (x) − f 0 (x)x) = f (0). x→0 The condition f (x) whose bounded derivative in the neighborhood of 0 is the sufficient condition of (2) but not the necessary condition of (2), consider: f (x) = x ln x; f (0) = 0 4 Nguyen Van Dung and Giap Kim Anh We see f 0 (x) = ln x + 1 for all x > 0, obviously limx→0 f 0 (x) = −∞ so f 0 (x) is not bounded in the neighborhood of 0. Other way: f (x) − f 0 (x)x = x ln x − (ln x + 1)x = −x obviously converges to f (0) = 0. Thus there exists a function f (x) whose derivative is not bounded in the neighborhood of 0 but (2) is satisfy. Besides, there exists a function f (x) whose derivative is not bounded in the neighborhood of 0 and (2) is not correct, for example:  x sin 1 if x 6= 0 f (x) = x 0 if x = 0 We have f 0 (x) = sin 1 1 1 − cos x x x 1 x 0 It’s easy to see that f (x) is not bounded in the neighborhood of 0 because f (x) − f 0 (x)x = cos lim f 0 ( k→∞ 1 ) = lim −2kπ = −∞ k→∞ 2kπ 1 x From (2), we can approximate the value of f (0) through the value of the function and its derivative at very close points 0 Note that the Formula (2) only helps us to approximate the value of f (0), meaning that when x is very close to 0 then: f (0) ≈ f (x) − f 0 (x)x and @ limx→0 f (x) − f 0 (x)x because @ limx→0 cos The Formula (2) means that for given , ∃δ such that ∀x : 0 < |x| < δ, the value f (x) − f 0 (x)x give us an approximate value of f (0) with error δ. This formula does not confirm between x1 and x2 very close to 0, which gives us approximately f (0) better; not really closer to zero gives us a more accurate approximation. Example, we consider function f (x) = x4 − 2x2 , its Graph is drawn below y 2 1 −1 x 2 O −2 1 3 −1 −2 −3 We approximate f (−1), we see that f (−1) = f (1) = −1 and f 0 (1) = 0. So when we choose any x0 (even though it’s very close to -1), it doesn’t give us an approximation better than we Some approximation problems by first-degree derivative 5 choose x0 = 1 We can choose function g(x) = f (103 x) if we want the distance δ "quite small". The above example shows that in a given δ , not the point closer gives us a better approximation, in this case, δ is fixed. But is that δ interval "small enough" yet to be sure the closer will give us a better approximation? The following example shows that for any δ interval small enough, not any closer will give a better approximation.  x2 sin 1 if x 6= 0 f (x) = x 0 if x = 0 we approximate f (0). We have: f 0 (x) = 2x sin So (2) is satisfy: 1 1 − cos ⇒ f 0 (x) is bounded in neighborhood of 0. x x lim (f (x) − f 0 (x)x) = f (0) x→0 1 1 , yn = . 2nπ −π/2 + 2nπ We have ∀ > 0, ∃n such that xn < yn <  1 1 f (xn ) − f 0 (xn )xn = and f (yn ) − f 0 (yn )yn = 2nπ (−π/2 + 2nπ)2 1 1 We see that > so yn give us a better approximation than xn though 2nπ (−π/2 + 2nπ)2 xn is closer to 0 than yn . Conclude. In this article, we have pointed out some important problems in approximation, namely selecting the appropriate x0 points. The paper shows that the function class can apply the approximate formula. Besides, it is pointed out that not just choosing the point near the point in question will give you a more accurate approximation. let xn = 6 Nguyen Van Dung and Giap Kim Anh REFERENCES [1] Nathan pflueger. Lecture 10: Linear Approximation. 2013 [2] Nguyen The Vinh, Nguyen Sy Anh Tuan, Le Hong Lan. Calculus 1. Transport publishing House. 2012 [3] Tran Binh. Calculus 1, Differential and integral calculations of a single function variable . Scientific and Technical publishing House 2009 [4] Bui Xuan Dieu. Department of Applied Mathematics, Ha Noi University of Science and Technology. Calculus 1. Ha Noi - 2009. Tóm tắt Trong bài báo này, tôi đề cập đến việc tính gần đúng giá trị hàm số tại một điểm, dựa vào đạo hàm cấp một tại những điểm trong lân cận của nó. Bài báo chỉ ra lớp hàm nào có thể áp dụng để tính gần đúng. Bài báo cũng chỉ ra tầm quan trọng của việc tính gần đúng là đánh giá sai số của phép tính đó. Từ khóa Hàm số, Tính gần đúng, Đạo hàm cấp một, Sai số